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Metamorphic geode - The RuneScape Wiki
(Redirected from Metamorphic geodes)
A rare hollow rock with something inside.
Item JSON: {"edible":"no","members":"yes","stackable":"yes","stacksinbank":"yes","death":"reclaimable","name":"Metamorphic geode","bankable":"yes","gemw":false,"equipable":"no","disassembly":"yes","release_date":"7 January 2019","id":"44818","release_update_post":"Mining and Smithing Rework","lendable":"no","destroy":"Drop","highalch":1,"weight":0,"tradeable":"no","examine":"A rare hollow rock with something inside.","noteable":"no"}
Metamorphic geodes are occasionally received from members rocks: orichalcite, drakolith, necrite, phasmatite, concentrated coal, banite, light animica, and dark animica rocks. There is a 1% chance the player gets this type of geode instead of an igneous geode. The chance is increased to up to 3% if the player is wearing a luck enhancer.[1] The metamorphic geode always contains a random strange or golden rock the player has yet to earn, and another item from the possible rewards table below, with an equal chance of
{\displaystyle {\frac {1}{14}}}
each.[2]
Although a strange or golden rock is a guaranteed drop, there are some cases where a rock may not be rewarded, such as if the player already has all potential rocks in their bank or statue collection bag, or the player has toggled off the option to receive them.
Sealed elite clue scrolls can be received from metamorphic geodes, even if the player has the soft cap of 25 already.
When a metamorphic geode is opened, it yields one of the rewards below and is destroyed. The "open-all" option can be used by right-clicking the geode, and this automatically opens all types of geodes in the player's backpack, one every two ticks. Multiple geodes can be opened each tick by pressing the open option manually.
Strange rock (m) 1 Always[2] Not sold Not alchemisable
Golden rock (m) 1 Always[2] Not sold Not alchemisable
Help us improve the drop rates by submitting data (requires JavaScript)
The average monetary value of each metamorphic geode is estimated at 248,166 coins.
Starved ancient effigy 1 1/14 Not sold Not alchemisable
First Age coin 1 1/14 1,000,000 1,000,000
Anima crystal 1 1/14 Not sold Not alchemisable
Sealed clue scroll (elite) 1 99/1400[d 1] 1 1
Sealed clue scroll (master) 1 1/1400[d 1] 1 1
Banite stone spirit 25–75 1/14 8,000-24,000 Not alchemisable
Light animica stone spirit 10–30 1/14 5,990-17,970 Not alchemisable
Dark animica stone spirit 10–30 1/14 15,610-46,830 Not alchemisable
Corrupted ore 100–300 1/14 Not sold Not alchemisable
Concentrated alloy bar 1–10 (noted) 1/14 52,586-525,860 Not alchemisable
Enriched alloy bar 1 1/14 242,651 Not alchemisable
Hydrix bolt tips 25–50 1/14 203,725-407,450 150,000-300,000
Onyx dust 5–15 9/140[d 3] 49,635-148,905 54,000-162,000
Uncut onyx 1 1/140[d 3] 1,068,668 1,080,000
Dragon full helm 1 1/336[d 4] 2,380,620 90,000
Dragon helm 1 1/336[d 4] 58,384 60,000
Dragon chainbody 1 1/336[d 4] 155,711 150,000
Dragon platelegs 1 1/336[d 4] 161,737 162,000
Dragon plateskirt 1 1/336[d 4] 162,672 162,000
Dragon boots 1 1/336[d 4] 11,613 12,000
Dragon kiteshield 1 1/336[d 4] 326,260 330,000
Dragon sq shield 1 1/336[d 4] 841,404 300,000
Dragon 2h sword 1 1/336[d 4] 133,287 132,000
Dragon hasta 1 1/336[d 4] 28,050 30,000
Dragon halberd 1 1/336[d 4] 248,353 195,000
Dragon spear 1 1/336[d 4] 35,598 37,440
Dragon battleaxe 1 1/336[d 4] 116,485 120,000
Off-hand dragon battleaxe 1 1/336[d 4] 116,270 120,000
Dragon claw 1 1/336[d 4] 195,559 40,500
Off-hand dragon claw 1 1/336[d 4] 103,636 40,500
Dragon longsword 1 1/336[d 4] 58,046 60,000
Off-hand dragon longsword 1 1/336[d 4] 57,867 60,000
Dragon mace 1 1/336[d 4] 28,121 30,000
Off-hand dragon mace 1 1/336[d 4] 27,795 30,000
Dragon scimitar 1 1/336[d 4] 83,588 60,000
Off-hand dragon scimitar 1 1/336[d 4] 64,080 60,000
Dragon warhammer 1 1/336[d 4] 131,537 120,000
Off-hand dragon warhammer 1 1/336[d 4] 28,323 30,000
^ a b The chance of a clue scroll is 1/14. There is a 1% chance of receiving a master clue scroll instead of an elite clue scroll
^ a b The total chance of getting any type of onyx drop is 1/14. A second roll then decides whether you get onyx dust (9/10) or a whole uncut onyx (1/10).
^ a b c d e f g h i j k l m n o p q r s t u v w x The total chance of getting any piece of dragon equipment is 1/14. A second equal-chance roll then decides which specific piece of dragon equipment you get, at a 1/24 rate for any specific piece.[3]
Banite rock 80 1 Varies[dr 1]
Concentrated coal deposit 70 1 Varies[dr 1]
Dark animica rock 90 1 Varies[dr 1]
Drakolith rock 60 1 Varies[dr 1]
Light animica rock 90 1 Varies[dr 1]
Necrite rock 70 1 Varies[dr 1]
Orichalcite rock 60 1 Varies[dr 1]
Phasmatite rock 70 1 Varies[dr 1]
On release, a player who had obtained all rocks of one type (either strange or golden) but not all of the other type had the chance to not receive a rock from their geode. [4] This was fixed in the 21 January 2019 patch.
^ Mod Shauny. Lotd + Metamorphic Geode. Reddit. 10 January 2019. (Archived from the original on 13 January 2019.) "LOTD boosts your chance to receive a meta geode if a geode drops"
^ a b c Jagex. Mod Breezy's Twitter account. 11 November 2018. (Archived from the original on 11 November 2018.) Mod Breezy: "... you're also guaranteed one strange or golden rock that you don't yet have!"
^ Jagex. Mod Breezy's Twitter account. 24 January 2019. (Archived from the original on 24 January 2019.) Mod Breezy: "All equal chances" In response to the question: "For wiki citation, is this correct dragon equip from metamorphic geodes: Battleaxe, Claws, Hasta, Longsword, Mace, Scimitar, Warhammer, 2h, Halberd, Spear, Full/Med helm, Chainbody, Kite/Sq shield, Platelegs/skirt, Boots? Also, all equal chances or not?"
^ Jagex. Mod Jack's Twitter account. 16 January 2019. (Archived from the original on 16 January 2019.) Mod Jack: "Yes I'm going to bug this." In response to the question: "Can you please verify if metamorphic geodes are intended to be 100% guaranteed rocks as long as player needs at least 1 strange or golden? Seems like there is a bug where a player who already has all strange or all golden will not always receive a rock of needed type."
Retrieved from ‘https://runescape.wiki/w/Metamorphic_geode?oldid=35801674’
8m ago - AsahelFrost
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Data Smoothing - Maple Help
Home : Support : Online Help : Statistics and Data Analysis : Statistics Package : Example Worksheets : Data Smoothing
The Statistics package provides several functions for performing data smoothing - the process of extracting identifiable patterns from data and obscuring noise. The data smoothing functionality includes algorithms to produce smoothed data (MovingAverage, MovingStatistic, ExponentialFit) or to produce an estimation curve to approximate the distribution of the population (i.e. kernel density estimation).
The Statistics package includes several data filters for smoothing otherwise rough data including moving average, moving median, moving statistic, a general linear filter, exponential fit and weighted moving average.
This example demonstrates the use of data filters in analyzing stock prices.
\mathrm{restart}:
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{with}\left(\mathrm{Statistics}\right):
Consider the following function that generates a sample stock path over N time periods. The stock is considered to have initial cost S0, trend parameter r and fluctuation parameter sigma.
\mathrm{StockPath}:=\mathbf{proc}\left(N∷\mathrm{posint},\mathrm{S0}∷\mathrm{realcons},r∷\mathrm{realcons},\mathrm{σ}∷\mathrm{realcons}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{local} h,i,C,R,S;\phantom{\rule[-0.0ex]{0.0em}{0.0ex}} h≔\frac{1.}{N-1};\phantom{\rule[-0.0ex]{0.0em}{0.0ex}} C≔\mathrm{evalf}\left(\mathrm{exp}\left(r\cdot h-\frac{{\mathrm{σ}}^{2}\cdot h}{2}\right)\right);\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}R≔C\cdot \mathrm{exp}\left(\mathrm{σ}\cdot \sqrt{h}\cdot \mathrm{RandomVariable}\left(\mathrm{Normal}\left(0,1\right)\right)\right);\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}S:=\mathrm{Sample}\left(R,N+1\right);\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}S\left[1\right]:=\mathrm{S0};\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{return}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{CumulativeProduct}\left(S\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{end} \mathbf{proc}:\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}
Generate a sample stock path over 500 time periods and plot.
S:=\mathrm{StockPath}\left(1000,100.,0.15,0.2\right):
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{LineChart}\left(S,\mathrm{symbolsize}=4,\mathrm{thickness}=2\right)
The data smoothing functions provided in the Statistics library now give us a means to analyze the overall trend of the data while disregarding small fluctuations. Consider the moving average function, which calculates the average value of a window around each data point.
T:=\mathrm{MovingAverage}\left(S,20\right):\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}
\mathrm{LineChart}\left(T,\mathrm{symbolsize}=4,\mathrm{thickness}=2\right)
Exponential smoothing can also be applied. This method works by 'smoothing' out rough edges, generally caused by cyclic or irregular patterns in the data.
T:=\mathrm{ExponentialSmoothing}\left(S,0.9\right):
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{LineChart}\left(T,\mathrm{symbolsize}=4,\mathrm{thickness}=2\right)
1.2 Department Store Sales
This example demonstrates the use of data filters in analyzing sales at a department store.
\mathrm{restart};
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{with}\left(\mathrm{Statistics}\right):
Consider the following function that randomly generates the times of n sales at a department store. The rate of sales is represented by the parameter r and the deviation in this rate by the parameter theta.
\mathrm{SaleTimes}:=\mathbf{proc}\left(N∷\mathrm{realcons},r∷\mathrm{realcons},\mathrm{θ}∷\mathrm{realcons}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}} \mathbf{local} R,S,T,i;\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}} R≔r\cdot \mathrm{RandomVariable}\left(\mathrm{Exponential}\left(\mathrm{θ}\right)\right);\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}} S≔\mathrm{Sample}\left(R,N\right);\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}} \mathbf{return} \mathrm{CumulativeSum}\left(S\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}} \mathbf{end} \mathbf{proc}:
Consider the first 100 sales with rate parameter 0.5 and deviation parameter 0.2.
S:=\mathrm{SaleTimes}\left(100,0.5,0.2\right):
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{LineChart}\left(S,\mathrm{thickness}=2,\mathrm{symbolsize}=4\right)
The overall trend is readily apparent with the application of the moving average filter.
T:=\mathrm{MovingAverage}\left(S,20\right):
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{LineChart}\left(T,\mathrm{symbolsize}=4,\mathrm{thickness}=2\right)
The Statistics package provides algorithms for computing, plotting and sampling from kernel density estimates. A kernel density estimate is a continuous probability distribution used to approximate the population of a sample, constructed by considering a normalized sum of kernel functions for each data point.
The following is an example of Maple's kernel density estimation routines in action.
\mathrm{restart};
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{with}\left(\mathrm{Statistics}\right):
Consider the following bimodal data sample (hypothesized as bimodal since there appear to be two distinct clusterings of data - those in the range -1.2 to -0.8 and those in the range 0.7 to 0.9).
A:=\mathrm{Array}\left(\left[-1.18,-1.12,-1.06,-1.02,-0.84,0.72,0.78,0.89\right]\right):\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}
Z:=\mathrm{Array}\left(\left[0.\right]\right):
By applying kernel density estimation, we can create a function to interpolate the data. Since our data sample is relatively small, we can perform exact kernel density estimation. The exact method of kernel density estimation returns a probability density function which can then be evaluated at specific points.
F:=\mathrm{KernelDensity}\left(A,\mathrm{bandwidth}=0.4,\mathrm{kernel}=\mathrm{gaussian},\mathrm{method}=\mathrm{exact}\right):
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{evalf}\left(\left[F\left(-1.0\right),F\left(0.0\right),F\left(0.5\right),F\left(2.0\right)\right]\right)
\left[\textcolor[rgb]{0,0,1}{0.5947413597}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.08016057122}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.2829169446}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.004587682613}\right]
We can convert the kernel density estimate to a distribution using one of the standard RandomVariable constructors.
R:=\mathrm{RandomVariable}\left(\mathrm{Distribution}\left(\mathrm{PDF}=\left(x→F\left(x\right)\right)\right)\right):
\mathrm{evalf}\left(\left[\mathrm{PDF}\left(R,-1.0\right),\mathrm{PDF}\left(R,0.0\right),\mathrm{PDF}\left(R,0.5\right),\mathrm{PDF}\left(R,2.0\right)\right]\right)
\left[\textcolor[rgb]{0,0,1}{0.5947413597}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.08016057122}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.2829169446}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.004587682613}\right]
\mathrm{evalf}\left(\left[\mathrm{CDF}\left(R,-1.0\right),\mathrm{CDF}\left(R,0.0\right),\mathrm{CDF}\left(R,0.5\right),\mathrm{CDF}\left(R,2.0\right)\right]\right)
\left[\textcolor[rgb]{0,0,1}{0.3394631178}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.6303924803}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.7121675015}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.9994260712}\right]
This probability density function can also be plotted, in this case against the cumulative distribution function.
\mathrm{P1}:=\mathrm{plot}\left(\mathrm{PDF}\left(R,x\right),x=-2.5..2.5,\mathrm{thickness}=3\right):
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{P2}:=\mathrm{plot}\left(\mathrm{CDF}\left(R,x\right),x=-2.5..2.5,\mathrm{thickness}=3,\mathrm{color}=\mathrm{blue}\right):\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}
\mathrm{plots}\left[\mathrm{display}\right]\left(\mathrm{P2},\mathrm{P1}\right)
With the KernelDensitySample function, similar data can be quickly drawn from a data sample.
S:=\mathrm{KernelDensitySample}\left(A,100000,\mathrm{bandwidth}=0.4,\mathrm{kernel}=\mathrm{gaussian}\right):\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}
\mathrm{P1}:=\mathrm{Histogram}\left(S,\mathrm{averageshifted}=1,\mathrm{binwidth}=0.1,\mathrm{range}=-2.5..2.5\right):\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}
\mathrm{P2}:=\mathrm{plot}\left(\mathrm{PDF}\left(R,x\right),x=-2.5..2.5,\mathrm{thickness}=3,\mathrm{color}=\mathrm{red}\right):\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}
\mathrm{plots}\left[\mathrm{display}\right]\left(\mathrm{P1},\mathrm{P2}\right)
A kernel density estimate can be directly plotted using the KernelDensityPlot function. The following example demonstrates the difference between different choices of bandwidth.
\mathrm{P1}:=\mathrm{KernelDensityPlot}\left(A,\mathrm{bandwidth}=0.1,\mathrm{kernel}=\mathrm{biweight},\mathrm{method}=\mathrm{exact},\mathrm{color}=\mathrm{turquoise},\mathrm{thickness}=2,\mathrm{range}=-2..2\right):
\mathrm{P2}:=\mathrm{KernelDensityPlot}\left(A,\mathrm{bandwidth}=0.3,\mathrm{kernel}=\mathrm{biweight},\mathrm{method}=\mathrm{exact},\mathrm{color}=\mathrm{blue},\mathrm{thickness}=2,\mathrm{range}=-2..2\right):
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{P3}≔\mathrm{KernelDensityPlot}\left(A,\mathrm{bandwidth}=0.6,\mathrm{kernel}=\mathrm{biweight},\mathrm{method}=\mathrm{exact},\mathrm{color}=\mathrm{navy},\mathrm{thickness}=2,\mathrm{range}=-2..2\right):
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{plots}\left[\mathrm{display}\right]\left(\mathrm{P1},\mathrm{P2},\mathrm{P3}\right)
In most cases, only a few hundred samples are needed to roughly approximate the original probability distribution with a kernel density estimate.
B:=\mathrm{Sample}\left(\mathrm{StudentT}\left(2\right),600\right):
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{P1}≔\mathrm{Histogram}\left(B,\mathrm{range}=-5..5\right):
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{P2}≔\mathrm{DensityPlot}\left(\mathrm{StudentT}\left(2\right),\mathrm{color}=\mathrm{blue},\mathrm{thickness}=3,\mathrm{range}=-5..5\right):
\mathrm{P3}:=\mathrm{KernelDensityPlot}\left(B,\mathrm{kernel}=\mathrm{gaussian},\mathrm{method}=\mathrm{piecewise},\mathrm{color}=\mathrm{red},\mathrm{thickness}=3,\mathrm{range}=-5..5\right):
\phantom{\rule[-0.0ex]{0.0em}{0.0ex}} \mathrm{plots}\left[\mathrm{display}\right]\left(\mathrm{P1},\mathrm{P2},\mathrm{P3}\right)
Kernel density estimation requires the use of a kernel function - a normalized continuous function that is mapped to each data point. Five standard kernel functions are available with kernel density estimation.
2.1 Gaussian Kernel
The Gaussian kernel should be used with continuous data that is defined on the whole real line. It possesses the familiar bell shape and is based on the Gaussian probability density function.
\mathrm{KernelDensityPlot}\left(Z,\mathrm{kernel}=\mathrm{gaussian},\mathrm{method}=\mathrm{exact},\mathrm{thickness}=3\right);
\mathrm{KernelDensityPlot}\left(A,\mathrm{kernel}=\mathrm{gaussian},\mathrm{bandwidth}=0.4,\mathrm{method}=\mathrm{exact},\mathrm{thickness}=3\right)
2.2 Triangular Kernel
The triangular kernel is a piecewise function related to the triangular distribution. This kernel generally creates a kernel density estimate with sharp edges, although remaining relatively smooth.
\mathrm{KernelDensityPlot}\left(Z,\mathrm{kernel}=\mathrm{triangular},\mathrm{method}=\mathrm{exact},\mathrm{thickness}=3\right);
\mathrm{KernelDensityPlot}\left(A,\mathrm{kernel}=\mathrm{triangular},\mathrm{bandwidth}=0.4,\mathrm{method}=\mathrm{exact},\mathrm{thickness}=3\right)
2.3 Rectangular Kernel
The rectangular kernel is a piecewise function related to the uniform distribution. This kernel creates a kernel density estimate that resembles a staircase function.
\mathrm{KernelDensityPlot}\left(Z,\mathrm{kernel}=\mathrm{rectangular},\mathrm{method}=\mathrm{exact},\mathrm{thickness}=3\right);
\mathrm{KernelDensityPlot}\left(A,\mathrm{kernel}=\mathrm{rectangular},\mathrm{bandwidth}=0.4,\mathrm{method}=\mathrm{exact},\mathrm{thickness}=3\right)
2.4 Biweight Kernel
The biweight kernel is a smooth kernel that is defined on a finite interval, unlike the gaussian kernel. It should be used for bounded data that is smooth along the interval it is defined on.
\mathrm{KernelDensityPlot}\left(Z,\mathrm{kernel}=\mathrm{biweight},\mathrm{method}=\mathrm{exact},\mathrm{thickness}=3\right);
\mathrm{KernelDensityPlot}\left(A,\mathrm{kernel}=\mathrm{biweight},\mathrm{bandwidth}=0.4,\mathrm{method}=\mathrm{exact},\mathrm{thickness}=3\right)
2.5 Epanechnikov Kernel
The Epanechnikov kernel is the standard kernel for kernel density estimation. It generally provides the closest match to a probability density function under most circumstances. The kernel itself is a rounded function similar to the biweight, except it is not differentiable at its boundaries.
\mathrm{KernelDensityPlot}\left(Z,\mathrm{kernel}=\mathrm{epanechnikov},\mathrm{method}=\mathrm{exact},\mathrm{thickness}=3\right);
\mathrm{KernelDensityPlot}\left(A,\mathrm{kernel}=\mathrm{epanechnikov},\mathrm{bandwidth}=0.4,\mathrm{method}=\mathrm{exact},\mathrm{thickness}=3\right)
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Create Cox model object for lifetime probability of default - MATLAB - MathWorks France
LoanVars
ExtrapolationFactor
Create Cox Lifetime PD Model
Time Interval for Cox Models
Survival and Probability of Default for Cox Models
Create Cox model object for lifetime probability of default
Create and analyze a Cox model object to calculate lifetime probability of default (PD) using this workflow:
Use fitLifetimePDModel to create a Cox model object.
Use predict to predict the conditional PD and predictLifetime to predict the lifetime PD.
Use modelAccuracy to return the root mean square error (RMSE) of observed and predicted PD data. You can plot the results using modelAccuracyPlot.
CoxPDModel = fitLifetimePDModel(data,ModelType,AgeVar=agevar_value)
CoxPDModel = fitLifetimePDModel(___,Name=Value)
CoxPDModel = fitLifetimePDModel(data,ModelType,AgeVar=agevar_value) creates a Cox PD model object.
If you do not specify variable information for IDVar, LoanVars, MacroVars, and ResponseVar, then:
IDVar is set to the first column in the data input.
LoanVars is set to include all columns from the second to the second-to-last columns of the data input.
ResponseVar is set to the last column in the data input.
CoxPDModel = fitLifetimePDModel(___,Name=Value) sets optional properties using additional name-value arguments in addition to the required arguments in the previous syntax. For example, CoxPDModel = fitLifetimePDModel(data(TrainDataInd,:),"Cox",ModelID="Cox_A",Descripion="Cox_model",AgeVar="YOB",IDVar="ID",LoanVars="ScoreGroup",MacroVars={'GDP','Market'},ResponseVar="Default",TimeInterval=1) creates a CoxPDModel using a Cox model type. You can specify multiple name-value arguments.
Data, specified as a table, in panel data form. The data must contain an ID column and an Age column. The response variable must be a binary variable with the value 0 or 1, with 1 indicating default.
string with value "Cox" | character vector with value 'Cox'
Model type, specified as a string with the value "Cox" or a character vector with the value 'Cox'.
Example: CoxPDModel = fitLifetimePDModel(data(TrainDataInd,:),"Cox",ModelID="Cox_A",Descripion="Cox_model",AgeVar="YOB",IDVar="ID",LoanVars="ScoreGroup",MacroVars={'GDP','Market'},ResponseVar="Default",TimeInterval=1)
Required Cox Name-Value Argument
AgeVar — Age variable indicating which column in data contains loan age information
Age variable indicating which column in data contains the loan age information, specified as AgeVar and a string or character vector.
The required name-value argument AgeVar is not treated as a predictor in the Cox lifetime PD model. When using a Cox model, you must specify predictor variables using LoanVars or MacroVars. The AgeVar values are the event times for the underlying Cox proportional hazards model.
AgeVar values for each ID should be increasing. If there are nonpositive age increments, fitLifetimePDModel warns when you create a Cox model and removes the IDs with nonpositive age increments. By default, the TimeInterval value is set to the most common age increment in the training data.
Optional Cox Name-Value Arguments
Cox (default) | string | character vector
User-defined model ID, specified as ModelID and a string or character vector. The software uses the ModelID to format outputs and is expected to be short.
IDVar — ID variable indicating which column in data contains loan or borrower ID
1st column of data (default) | string | character vector
ID variable indicating which column in data contains the loan or borrower ID, specified as IDVar and a string or character vector.
LoanVars — Loan variables indicating which column in data contains loan-specific information
all columns of data that are not the first or last column (default) | string array | cell array of character vectors
Loan variables indicating which column in data contains the loan-specific information, such as origination score or loan-to-value ratio, specified as LoanVars and a string array or cell array of character vectors.
MacroVars — Macro variables indicating which column in data contains macroeconomic information
"" (default) | string array | cell array of character vectors
Macro variables indicating which column in data contains the macroeconomic information, such as gross domestic product (GDP) growth or unemployment rate, specified as MacroVars and a string array or cell array of character vectors.
ResponseVar — Variable indicating which column in data contains response variable
last column of data (default) | logical
Variable indicating which column in data contains the response variable, specified as ResponseVar and a logical value.
The response variable in the data must be a binary variable with 0 or 1 values, with 1 indicating default.
In Cox lifetime PD models, the ResponseVar values are define the censoring information for the underlying Cox proportional hazards model.
TimeInterval — Distance between age values in panel data input
set to most common AgeVar increment in the training data (default) | positive numeric
Distance between age values in training data in the panel data input, specified as TimeInterval and a positive numeric scalar.
Use the TimeInterval name-value argument to fit time-dependent models and also as the time interval for the PD computation when you use the predict function. For example, if the age data (AgeVar) is 1, 2, 3, ..., then the TimeInterval is 1; if the age data is 0.25, 0.5, 0.75,..., then the TimeInterval is 0.25. For more information, see Time Interval for Cox Models and Lifetime Prediction and Time Interval.
Unlike Logistic and Probit models, a Cox model requires an AgeVar variable. By default, if you do not specify a TimeInterval when creating a Cox model, the TimeInterval is inferred from the increments in the AgeVar values in the training data.
Probit (default) | string
Model — Underlying statistical model
Underlying statistical model, returned as a returned as a Cox proportional hazards model object. For more information, see fitcox and CoxModel.
Data Types: CoxModel
1st column of data (default) | string
ID variable indicating which column in data contains the loan or borrower ID, returned as a string.
Age variable indicating which column in data contains the loan age information, returned as a string.
all columns of data that are not the first or last column (default) | string array
Loan variables indicating which column in data contains the loan-specific information, returned as a string array.
Macro variables indicating which column in data contains the macroeconomic information, returned as a string array.
Variable indicating which column in data contains the response variable, returned as a string.
Distance between age values in panel data input, returned as a scalar positive numeric.
ExtrapolationFactor — Extrapolation factor
1 (default) | positive numeric between 0 and 1
Extrapolation factor, returned as a positive numeric scalar between 0 and 1.
By default, the ExtrapolationFactor is set to 1. For age values (AgeVar) greater than the maximum age observed in the training data, the conditional PD, computed with predict, uses the maximum age observed in the training data. In particular, the predicted PD value is constant if the predictor values do not change and only the age values change when the ExtrapolationFactor is 1. For more information, see Extrapolation for Cox Models, Extrapolation Factor for Cox Models, and Use Cox Lifetime PD Model to Predict Conditional PD.
This example shows how to use fitLifetimePDModel to create a Cox model using credit and macroeconomic data.
Load the credit portfolio data.
ID ScoreGroup YOB Default Year
__ __________ ___ _______ ____
1 Low Risk 1 0 1997
disp(head(dataMacro))
Year GDP Market
____ _____ ______
1997 2.72 7.61
1998 3.57 26.24
1999 2.86 18.1
2001 1.26 -10.51
2002 -0.59 -22.95
Join the two data components into a single data set.
Separate the data into training and test partitions.
Create a Cox Lifetime PD Model
Use fitLifetimePDModel to create a Cox model using the training data.
pdModel = fitLifetimePDModel(data(TrainDataInd,:),"Cox",...
AgeVar="YOB", ...
IDVar="ID", ...
LoanVars="ScoreGroup", ...
MacroVars={'GDP','Market'}, ...
ResponseVar="Default");
Display the underlying model.
Use modelDiscrimination to measure the ranking of customers by PD.
DiscMeasure = modelDiscrimination(pdModel,data(Ind,:),SegmentBy="ScoreGroup")
Cox, ScoreGroup=High Risk 0.64112
Cox, ScoreGroup=Medium Risk 0.61989
Cox, ScoreGroup=Low Risk 0.6314
modelDiscriminationPlot(pdModel,data(Ind,:),SegmentBy="ScoreGroup")
Use modelAccuracy to measure the accuracy (or calibration) of the predicted PD values. The modelAccuracy function requires a grouping variable and compares the accuracy of the observed default rate in the group with the average predicted PD for the group.
AccMeasure = modelAccuracy(pdModel,data(Ind,:),{'YOB','ScoreGroup'})
AccMeasure=table
Cox, grouped by YOB, ScoreGroup 0.0012471
Use modelAccuracyPlot to visualize the observed default rates compared to the predicted PD.
modelAccuracyPlot(pdModel,data(Ind,:),{'YOB','ScoreGroup'})
Predict Conditional and Lifetime PD
Use the predict function to predict conditional PD values. The prediction is a row-by-row prediction.
%dataCustomer1 = data(1:8,:);
CondPD = predict(pdModel,data(Ind,:));
Use predictLifetime to predict the lifetime cumulative PD values (computing marginal and survival PD values is also supported).
LifetimePD = predictLifetime(pdModel,data(Ind,:));
The Cox proportional hazards (PH) model is a survival model and it models the time until an event of interest occurs.
For probability of default (PD) models, the event of interest is the default on a credit obligation. Cox models need information on whether there was a default and when it happened. For other commonly used PD models, a binary variable indicating whether there was a default is enough. Cox PD models need that information, plus the age of the loan at the time of default.
The Cox proportional hazards (PH) model, also known as a Cox regression model, assumes the hazard rate is of the form
h\left(t;X\right)={h}_{0}\left(t\right)\mathrm{exp}\left(X\beta \right)
h0(t) is the baseline hazard rate.
X is the predictor data.
β is a vector of coefficients of the predictors.
exp(Xβ) is the hazard ratio.
The baseline hazard rate is a reference hazard level, common to all observations, and it does not depend on the predictor values. The hazard ratio is the factor that scales the baseline hazard value up or down, depending on the predictor values. For lower risk observations, the hazard ratio is less than 1 and this reduces the hazard rate. For higher risk observations, the hazard ratio increases the hazard rate.
In the hazard rate formula, the predictor values in X are fixed, or independent of time. This is the basic version of the Cox PH model. For PD models, the basic version of the Cox PH model includes predictors that have constant values, such as the origination score, or whether a property is for residential or commercial purposes.
The time-dependent Cox PH model allows predictor values to change over time. For example, the loan-to-value (LTV) ratio changes over the life of a loan, and the macroeconomic variables change from period to period. Therefore, the following hazard rate formula for time-dependent models includes predictor values that can be a function of time:
h\left(t;X\right)={h}_{0}\left(t\right)\mathrm{exp}\left(X\left(t\right)\beta \right)
The data input for fitLifetimePDModel must be in panel data form. For each ID (IDVar), there are multiple rows of data. The panel data input is required for both time-dependent and time -independent models.
For time-independent predictors, the predictor value is constant for each ID. For example, the score at origination for each customer is constant throughout the life of the loan, and this value is repeated for each row corresponding to the same ID in the panel data format.
For time-dependent predictors, the values may change from one row to the next for the same ID. The assumption is that the predictor values in each row are valid in the time interval defined by the age value (AgeVar) in the previous row and the age value in the current row.
Time is discretized into intervals, and predictor values in the training data (data input) are constant for each interval: X1 from t0 to t1; X2 from t1 to t2; and so forth.
The data input must be in panel data form, with multiple observations for each ID, with corresponding age information (the tk values, the AgeVar column) and the corresponding default indicator values (the ResponseVar column).
Assume that tk - tk - 1 = Δt for all k and this is the time interval. This time interval is the age increment for consecutive observations in the age data (AgeVar). The assumption is that these increments are regular and that the default indicator (ResponseVar) is defined consistently with this time interval, in the sense that a 1 means there was a default in a time interval of length Δt. The time interval Δt is also used for the computation of the probability of default. For more information, see Lifetime Prediction and Time Interval.
The survival function S(t) is a function of time, and gives the probability of surviving longer than a given time t.
S\left(t\right)=P\left(T>t\right)
T is the failure time, the random variable of interest, and in the Cox model case, the time to default.
t is the specific time of interest, for example, 1 year.
The main relationship between the survival function and the hazard rate is
S\left(t\right)=\mathrm{exp}\left(-{\int }_{0}^{t}h\left(u\right)du\right)
Higher values of the hazard rate cause the survival probability to drop faster. Conversely, lower values of the hazard rate cause the survival probability to rise faster.
The probability of default (PD) is the conditional probability of defaulting in a time interval, given that there has been no default prior to that interval. For example, the probability of default between time s and t, with s < t, is represented as:
\begin{array}{l}PD\left(s,t\right)=P\left(s<T\le t|T>s\right)\\ \text{ = }\frac{S\left(s\right)-S\left(t\right)}{S\left(s\right)}\\ \text{ = 1-}\frac{S\left(t\right)}{S\left(s\right)}\end{array}
In credit applications, the time interval of interest, Δt, is consistent with the training data and the definition of default in the response variable. The PD is a function of a single time variable t and the implicit time interval Δt:
PD\left(t\right)=1-\frac{S\left(t\right)}{S\left(t-\Delta t\right)}
[4] Roesch, Daniel and Harald Scheule. Deep Credit Risk: Machine Learning with Python. Independently published, 2020.
fitLifetimePDModel | Logistic | Probit
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Another pizza parlor had a super special. The parlor had a rack of small pizzas, each with three or four different toppings. The price was really low because an employee forgot to indicate on the boxes what toppings were on each pizza. Eight toppings were available, and one pizza was made for each of the possible combinations of three or four toppings.
How many pizzas were on the rack?
Find how many types of pizzas with three toppings can be made.
Find how many pizzas with four toppings can be made.
+
= 56 + 70 = 126
What was the probability of getting a pizza that had mushrooms on it?
If mushrooms are a known topping, then you choose one fewer topping from only
7
remaining toppings.
\frac{_{7}C_{3}+_{7}C_{2}}{126}=\frac{56}{126}=\frac{4}{9}
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\ge
Florina Caruntu1,2,†, Diana Aurora Bordejevic1,2,†, Mirela Cleopatra Tomescu1,2,*, Ioana Mihaela Citu1,2
1 Cardiology Clinic, Timisoara Municipal Clinical Emergency Hospital, 300024 Timișoara, Romania
2 Multidisciplinary Heart Research Center, Victor Babeș University of Medicine and Pharmacy, 300041 Timișoara, Romania
Submitted: 21 February 2021 | Revised: 10 May 2021 | Accepted: 21 June 2021 | Published: 24 September 2021
Older age is known as a negative prognostic parameter in patients with acute myocardial infarction (AMI). In this study, we aimed to investigate age-related differences in treatment protocols, in-hospital and 1-year mortality. This retrospective observational single-center study enrolled consecutive AMI patients with an urgent percutaneous coronary intervention (PCI) as the main method of myocardial revascularization. The patients divided were divided by age into group I (65 years) and group II (65 years). The primary endpoint was in-hospital mortality, the secondary endpoints were 1-year mortality and rehospitalization rates. Of the 522 admitted with AMI, 476 were enrolled in the study. The mean age was 67 13 years; 62% were men. Group I patients had a significantly lower rate of performed PCI (65% vs. 79%, P 0.001). 53 patients (12.3%) died during hospitalization, and this proportion was notably higher in the older population (20% vs. 6%, P 0.0001). The cardiac causes of death were more frequent in group I patients (12% vs. 5.6%, P = 0.016). The multivariate logistic regression selected two variables as independent predictors for the risk of in-hospital death: age 65 years (P = 0.0170), and Killip class at admission (P 0.0001). The 1-year mortality was 3.3%, slightly higher in group I patients (4.8% vs. 1.5%, P = 0.05). In conclusion, patients aged 65 years have three times higher in-hospital mortality, but similar 1-year mortality and readmission rates when compared with the younger patients. It is obvious that there is a large potential for improvement of the AMI care in this age group of patients.
Florina Caruntu, Diana Aurora Bordejevic, Mirela Cleopatra Tomescu, Ioana Mihaela Citu. Clinical characteristics and outcomes in acute myocardial infarction patients aged ≥65 years in Western Romania. Rev. Cardiovasc. Med. 2021, 22(3), 911–918. https://doi.org/10.31083/j.rcm2203098
Kaplan-Meier curves for in-hospital mortality.
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JEE Binomial Theorem | Brilliant Math & Science Wiki
This page will teach you how to master JEE binomial theorem. We highlight the main concepts, provide a list of examples with solutions, and include problems for you to try. Once you are confident, you can take the quiz to establish your mastery.
As per JEE syllabus, the main concepts under binomial theorem are binomial theorem expansion, numerically greatest term in the binomial expansion, binomial coefficients, and binomial series.
Binomial theorem expansion:
Binomial theorem expansion for positive integral index: If
n
(x+y)^n=^nC_0x^ny^0+^nC_1 x^{n-1}y^1+^nC_2x^{n-2}y^2+\cdots+^nC_n x^0y^n.
General term in the expansion: General term in the expansion of
(x+y)^n
(r+1)^\text{th}
term, i.e.
T_{r+1}=^nC_rx^{n-r}y^r.
Binomial theorem expansion for any index:
(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots.
Numerically greatest term in the binomial expansion:
Numerically greatest terms in the expansion of
(1+x)^n
T_p,T_{p+1}
(values of both these terms are equal), if
p
is an integer, where
p=\frac{(n+1)|x|}{|x|+1}.
Numerically greatest term in the expansion of
(1+x)^n
T_{c+1}
\frac{(n+1)|x|}{|x|+1}
is not an integer, where
c=\frac{(n+1)|x|}{|x|+1} .
Greatest binomial coefficient:
^nC_r
is maximum at
r= \begin{cases} \frac n2, &&\text{if } n \text{ is even} \\ \frac{n-1}{2}, \frac{n+1}{2}, &&\text{if } n \text{ is odd}. \end{cases}
Properties of binomial coefficients:
\begin{aligned} ^nC_0+^nC_1+^nC_2+\cdots+^nC_n &= 2^n \\ ^nC_0-^nC_1+^nC_2-\cdots+(-1)^n \ ^nC_n &=0\\ ^nC_1-2 \cdot ^nC_2+3 \cdot ^nC_3-\cdots+(-1)^{n-1} n \cdot ^nC_n&=0 ~~(\text{for } n>1). \end{aligned}
Use of differentiation and integration to find the sum of binomial coefficients
Bino-arithmetic series:
a \ ^nC_0+(a+d) \ ^nC_1+(a+2d) \ ^nC_2+\cdots+(a+nd) \ ^nC_n=(2a+nd)2^{n-1}
Bino-geometric series:
a \ ^nC_0+ab \ ^nC_1+ab^2 \ ^nC_2+\cdots+ab^n \ ^nC_n=a(1+b)^n
Bino-harmonic series:
\frac{^nC_0}{a}+\frac{^nC_1}{a+d}+\frac{^nC_2}{a+2d}+\cdots+\frac{^nC_n}{a+nd}=\displaystyle \int_{0}^1 x^{a-1} \left(1+x^d \right)^ndx
Bino-binomial series:
^nC_0 \ ^nC_n +^nC_1 \ ^nC_{n-1}+^nC_2 \ ^nC_{n-2}+\cdots+^nC_n \ ^nC_0=^{2n}C_n
\begin{array} { l l } A) \, & \quad \quad \quad \quad \quad & B) \, \\ C) \, & & D) \, \\ \end{array}
\begin{array} { l l } A) \, & \quad \quad \quad \quad \quad & B) \, \\ C) \, & & D) \, \\ \end{array}
\begin{array} { l l } A) \, & \quad \quad \quad \quad \quad & B) \, \\ C) \, & & D) \, \\ \end{array}
\begin{array} { l l } A) \, & \quad \quad \quad \quad \quad & B) \, \\ C) \, & & D) \, \\ \end{array}
Once you are confident of JEE Binomial Theorem, move on to JEE Multinomial Theorem.
Cite as: JEE Binomial Theorem . Brilliant.org. Retrieved from https://brilliant.org/wiki/jee-binomial-theorem/
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The limiting frictional force between two surface depends on
I. the normal reaction between the surfaces
II. the area of surface in contact
III. the relative velocity between the surfaces
IV. the nature of the surface
B. I & IV only
2. If a body moves with a constant speed and at the same time undergoes an acceleration, its motion is said to be
D. rectilinear
3. When blue and green colours of light are mixed, the resultant colour is
4. A metal rod has a length of 100cm at 200\(^oC\). At what temperature will its length be 99.4cm. If the linear expansivity of the material of the rod is 2 x 10\(^{-5}C^{-1}\)
A. 200\(^o\)C
B. 300\(^o\)C
C. 100\(^o\)C
D. -100\(^o\)C
5. According to kinetic molecular model, in gases
A. The molecule are very fast apart & occupy all the space made available
A. The particles vibrate about fixed positions and are held together by the strong intermolecular bond between them
D. The particles are closely packed together, they occupy minimum space & are usually arranged in a regular pattern
The value of T in the figure above is
7. A train has an initial velocity of 44m/s and an acceleration of -4m/s
{}^{2}
. Calculate its velocity after 10 seconds
8. Lamps in domestic lightings are usually in
9. During the transformation of matter from the solid to the liquid state, the heat supplied does not produce temperature increase because
A. all the heat is used to break the bonds holding the molecules of the solid together
B. the heat capacity has become very large as the substance melts
C. the heat energy is quickly conducted away
D. the heat gained is equal to the heat lost by the substance
10. In a slide wire bridge, the balance is obtained at a point 25cm from one end of wire 1m long. The resistance to be tested is connected to that end and a standard resistance of 3.6\(\Omega\) is connected to the other end of the wire. Determine the value of the unknown resistance
A. 3.2\(\Omega\)
B. 1.4\(\Omega\)
C. 3.21\(\Omega\)
D. 1.2\(\Omega\)
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Module_(mathematics) Knowpia
{\displaystyle r\cdot (x+y)=r\cdot x+r\cdot y}
{\displaystyle (r+s)\cdot x=r\cdot x+s\cdot x}
{\displaystyle (rs)\cdot x=r\cdot (s\cdot x)}
{\displaystyle 1\cdot x=x.}
The concept of a Z-module agrees with the notion of an abelian group. That is, every abelian group is a module over the ring of integers Z in a unique way. For n > 0, let n ⋅ x = x + x + ... + x (n summands), 0 ⋅ x = 0, and (−n) ⋅ x = −(n ⋅ x). Such a module need not have a basis—groups containing torsion elements do not. (For example, in the group of integers modulo 3, one cannot find even one element which satisfies the definition of a linearly independent set since when an integer such as 3 or 6 multiplies an element, the result is 0. However, if a finite field is considered as a module over the same finite field taken as a ring, it is a vector space and does have a basis.)
If R is any ring and n a natural number, then the cartesian product Rn is both a left and right R-module over R if we use the component-wise operations. Hence when n = 1, R is an R-module, where the scalar multiplication is just ring multiplication. The case n = 0 yields the trivial R-module {0} consisting only of its identity element. Modules of this type are called free and if R has invariant basis number (e.g. any commutative ring or field) the number n is then the rank of the free module.
If Mn(R) is the ring of n × n matrices over a ring R, M is an Mn(R)-module, and ei is the n × n matrix with 1 in the (i, i)-entry (and zeros elsewhere), then eiM is an R-module, since reim = eirm ∈ eiM. So M breaks up as the direct sum of R-modules, M = e1M ⊕ ... ⊕ enM. Conversely, given an R-module M0, then M0⊕n is an Mn(R)-module. In fact, the category of R-modules and the category of Mn(R)-modules are equivalent. The special case is that the module M is just R as a module over itself, then Rn is an Mn(R)-module.
If S is a nonempty set, M is a left R-module, and MS is the collection of all functions f : S → M, then with addition and scalar multiplication in MS defined pointwise by (f + g)(s) = f(s) + g(s) and (rf)(s) = rf(s), MS is a left R-module. The right R-module case is analogous. In particular, if R is commutative then the collection of R-module homomorphisms h : M → N (see below) is an R-module (and in fact a submodule of NM).
If R is a ring, we can define the opposite ring Rop which has the same underlying set and the same addition operation, but the opposite multiplication: if ab = c in R, then ba = c in Rop. Any left R-module M can then be seen to be a right module over Rop, and any right module over R can be considered a left module over Rop.
Submodules and homomorphismsEdit
If X is any subset of an R-module, then the submodule spanned by X is defined to be
{\textstyle \langle X\rangle =\,\bigcap _{N\supseteq X}N}
where N runs over the submodules of M which contain X, or explicitly
{\textstyle \left\{\sum _{i=1}^{k}r_{i}x_{i}\mid r_{i}\in R,x_{i}\in X\right\}}
, which is important in the definition of tensor products.[2]
{\displaystyle f(r\cdot m+s\cdot n)=r\cdot f(m)+s\cdot f(n)}
An R-module M is finitely generated if there exist finitely many elements x1, ..., xn in M such that every element of M is a linear combination of those elements with coefficients from the ring R.
A module is called a cyclic module if it is generated by one element.
A free R-module is a module that has a basis, or equivalently, one that is isomorphic to a direct sum of copies of the ring R. These are the modules that behave very much like vector spaces.
Projective modules are direct summands of free modules and share many of their desirable properties.
Injective modules are defined dually to projective modules.
A module is called flat if taking the tensor product of it with any exact sequence of R-modules preserves exactness.
A module is called torsionless if it embeds into its algebraic dual.
A simple module S is a module that is not {0} and whose only submodules are {0} and S. Simple modules are sometimes called irreducible.[4]
A semisimple module is a direct sum (finite or not) of simple modules. Historically these modules are also called completely reducible.
An indecomposable module is a non-zero module that cannot be written as a direct sum of two non-zero submodules. Every simple module is indecomposable, but there are indecomposable modules which are not simple (e.g. uniform modules).
A faithful module M is one where the action of each r ≠ 0 in R on M is nontrivial (i.e. r ⋅ x ≠ 0 for some x in M). Equivalently, the annihilator of M is the zero ideal.
A torsion-free module is a module over a ring such that 0 is the only element annihilated by a regular element (non zero-divisor) of the ring, equivalently rm = 0 implies r = 0 or m = 0.
A Noetherian module is a module which satisfies the ascending chain condition on submodules, that is, every increasing chain of submodules becomes stationary after finitely many steps. Equivalently, every submodule is finitely generated.
An Artinian module is a module which satisfies the descending chain condition on submodules, that is, every decreasing chain of submodules becomes stationary after finitely many steps.
A graded module is a module with a decomposition as a direct sum M = ⨁x Mx over a graded ring R = ⨁x Rx such that RxMy ⊂ Mx+y for all x and y.
Relation to representation theoryEdit
^ Dummit, David S. & Foote, Richard M. (2004). Abstract Algebra. Hoboken, NJ: John Wiley & Sons, Inc. ISBN 978-0-471-43334-7.
^ Mcgerty, Kevin (2016). "ALGEBRA II: RINGS AND MODULES" (PDF).
^ Ash, Robert. "Module Fundamentals" (PDF). Abstract Algebra: The Basic Graduate Year.
^ Jacobson (1964), p. 4, Def. 1; Irreducible Module at PlanetMath.
"Module", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
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Home : Support : Online Help : Mathematics : Linear Algebra : LinearAlgebra Package : Modular Subpackage : Rank
compute the rank of a mod m Matrix
compute the rank profile of a square mod m Matrix
Rank(m, A, meth)
RankProfile(m, A, meth)
The Rank function returns the rank of the input mod m Matrix, while the RankProfile function returns a list of 'rank' elements describing the rank profile of the input mod m Matrix.
The rank profile list is simply a list of the location of the first non-zero entry in each nontrivial row in the row reduced form of the Matrix.
Note that the two inplace methods available will destroy the data in the input Matrix, while the other two methods will generate a copy of the Matrix in which to perform the computation.
These commands are part of the LinearAlgebra[Modular] package, so they can be used in the form Rank(..) and RankProfile(..) only after executing the command with(LinearAlgebra[Modular]). However, they can always be used in the form LinearAlgebra[Modular][Rank](..) and LinearAlgebra[Modular][RankProfile](..).
\mathrm{with}\left(\mathrm{LinearAlgebra}[\mathrm{Modular}]\right):
p≔97
\textcolor[rgb]{0,0,1}{p}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{97}
M≔\mathrm{Mod}\left(p,[[2,11,1,80],[31,16,27,36],[8,32,32,31],[25,4,90,63]],\mathrm{integer}[]\right)
\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cccc}\textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{11}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{80}\\ \textcolor[rgb]{0,0,1}{31}& \textcolor[rgb]{0,0,1}{16}& \textcolor[rgb]{0,0,1}{27}& \textcolor[rgb]{0,0,1}{36}\\ \textcolor[rgb]{0,0,1}{8}& \textcolor[rgb]{0,0,1}{32}& \textcolor[rgb]{0,0,1}{32}& \textcolor[rgb]{0,0,1}{31}\\ \textcolor[rgb]{0,0,1}{25}& \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{90}& \textcolor[rgb]{0,0,1}{63}\end{array}]
\mathrm{Rank}\left(p,M\right),\mathrm{RankProfile}\left(p,M\right)
\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}]
With an inplace method the input Matrix is altered
\mathrm{Rank}\left(p,M,\mathrm{inplaceREF}\right),M
\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}[\begin{array}{cccc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{54}& \textcolor[rgb]{0,0,1}{49}& \textcolor[rgb]{0,0,1}{40}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{58}& \textcolor[rgb]{0,0,1}{26}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{35}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{1}\end{array}]
And a case that is not full rank
M≔\mathrm{Mod}\left(p,[[2,2,1,80],[31,31,27,36],[8,8,32,31],[25,25,90,63]],\mathrm{integer}[]\right)
\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cccc}\textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{80}\\ \textcolor[rgb]{0,0,1}{31}& \textcolor[rgb]{0,0,1}{31}& \textcolor[rgb]{0,0,1}{27}& \textcolor[rgb]{0,0,1}{36}\\ \textcolor[rgb]{0,0,1}{8}& \textcolor[rgb]{0,0,1}{8}& \textcolor[rgb]{0,0,1}{32}& \textcolor[rgb]{0,0,1}{31}\\ \textcolor[rgb]{0,0,1}{25}& \textcolor[rgb]{0,0,1}{25}& \textcolor[rgb]{0,0,1}{90}& \textcolor[rgb]{0,0,1}{63}\end{array}]
\mathrm{RankProfile}\left(p,M\right)
[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}]
\mathrm{Rank}\left(p,M\right)
\textcolor[rgb]{0,0,1}{3}
M≔\mathrm{Matrix}\left([[3,2],[2,3]],\mathrm{datatype}=\mathrm{integer}\right)
\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{3}\end{array}]
\mathrm{Rank}\left(6,M,\mathrm{REF}\right)
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Koopman Reduced Order Control for Three Body Problem ()
In this paper, we use a Circle Restricted Three-Body Problem (CRTBP) to simulate the motion of a satellite. Then we reformulate this problem with the controller into the description using Koopman eigenfunction. Although the original dynamical system is nonlinear, the Koopman eigenfunction behaves linearly. Choosing Jacobi integral as the Koopman eigenfunction and using the zero velocity curve as the reference for control, we are allowed to combine well-developed Linear Quadratic Regulator (LQR) controller to design a nonlinear controller. Using this approach, we design the low thrust orbit transfer strategy for the satellite flying from the earth to the moon or from the earth to the sun.
Circle Restricted Three-Body Problem, Koopman Eigenfunction, Zero Velocity Curve
Tang, H. (2019) Koopman Reduced Order Control for Three Body Problem. Modern Mechanical Engineering, 9, 20-29. doi: 10.4236/mme.2019.91003.
\frac{{\text{d}}^{2}x}{\text{d}{t}^{2}}-2\frac{\text{d}y}{\text{d}t}=x-\left(1-\mu \right)\frac{x-{x}_{1}}{{r}_{1}^{3}}-\mu \frac{x-{x}_{2}}{{r}_{2}^{3}}
\frac{{\text{d}}^{2}y}{\text{d}{t}^{2}}-2\frac{\text{d}x}{\text{d}t}=x-\left(1-\mu \right)\frac{y}{{r}_{1}^{3}}-\mu \frac{y}{{r}_{2}^{3}}
{r}_{1}=\sqrt{{\left({y}_{1}+\mu \right)}^{2}+{y}_{3}^{2}}
{r}_{2}=\sqrt{{\left({y}_{1}-1+\mu \right)}^{2}+{y}_{3}^{2}}
{y}_{1}=x,{y}_{2}=\stackrel{˙}{x},{y}_{3}=y,{y}_{4}=\stackrel{˙}{y}
\left(\begin{array}{l}{\stackrel{˙}{y}}_{1}\\ {\stackrel{˙}{y}}_{2}\\ {\stackrel{˙}{y}}_{3}\\ {\stackrel{˙}{y}}_{4}\end{array}\right)=\left(\begin{array}{c}{y}_{2}\\ 2{y}_{4}+{y}_{1}-\frac{\left(1-\mu \right)\left({y}_{1}+\mu \right)}{{r}_{1}^{3}}-\frac{\mu \left({y}_{1}-1+\mu \right)}{{r}_{2}^{3}}\\ {y}_{4}\\ -2{y}_{2}+{y}_{3}-\frac{\left(1-\mu \right){y}_{3}}{{r}_{1}^{3}}-\mu \end{array}\right):=f\left(y\right)
C=\left({x}^{2}+{y}^{2}\right)+2\left(\frac{1-\mu }{{r}_{1}}+\frac{\mu }{{r}_{2}}\right)-\left({\stackrel{˙}{x}}^{2}+{\stackrel{˙}{y}}^{2}\right)
\frac{\text{d}}{\text{d}t}y\left(t\right)=f\left(y\right)
\frac{\text{d}}{\text{d}t}\varphi \left(y\right)=\lambda \varphi \left(y\right)
\frac{\text{d}}{\text{d}t}\varphi \left(y\right)=\nabla \varphi \left(y\right)\cdot \stackrel{˙}{y}=\nabla \varphi \left(y\right)\cdot f\left(y\right)
\nabla \varphi \left(y\right)\cdot f\left(y\right)=\lambda \varphi \left(y\right)
\frac{\text{d}}{\text{d}t}y=f\left(y\right)+Bu
B=\left[\begin{array}{cc}0& 0\\ 1& 0\\ 0& 0\\ 0& 1\end{array}\right],u=\left[\begin{array}{c}{u}_{x}\\ {u}_{y}\end{array}\right]
\frac{\text{d}}{\text{d}t}\varphi \left(y\right)=\nabla \varphi \left(y\right)\cdot \stackrel{˙}{y}=\nabla \varphi \left(y\right)\cdot \left[f\left(y\right)+Bu\right]=\lambda \varphi \left(y\right)+\nabla \varphi \left(y\right)\cdot Bu
\varphi
\varphi
\frac{\text{d}C}{\text{d}t}=0\times C
\varphi
J\left(\varphi ,u\right)=\frac{1}{2}{\int }_{0}^{1}{\varphi }^{\text{T}}Q\varphi +{u}^{\text{T}}Ru
\varphi
u=-{K}_{\varphi }\varphi \left(y\right)
u=-{K}_{\varphi }\left(\varphi \left(x-{\varphi }_{ref}\right)\right)
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Based on a survey, assume that 27% of consumers are
Based on a survey, assume that 27% of consumers are comfortable having drones
Based on a survey, assume that 27% of consumers are comfortable having drones deliver their purchases. Suppose that we want to find the probability that when four consumers are randomly selected, exactly two of them are comfortable with delivery by drones. Identify the values of n, x, p, and q.
The value of n is?
A binomial experiment is a discrete probability experiment which is repeated for a fixed number of trials, and each of the trial is independent of the other trial. The possible outcomes for each trail are two and they are defined as success (S) and failure (F).
Assume that the random variable X follows a binomial distribution with parameters n and p. So, the binomial probability is,
P\left(X=x\right){=}^{n}{C}_{x}×{p}^{x}×{q}^{n-x}
Here, x is the number of successes that results from the binomial experiment, n is the number of trials and p is the probability of success on an individual trial.
\left(n\right)=4
Probability of success of each trial
\left(p\right)=0.27
Probability of failure of each trial
\left(q\right)=1-0.27=0.73
Number of successes
\left(x\right)=2
Consider, X be the random variable that represents the number of customers who are comfortable with delivery of their purchases by drones is binomially distributed.
The probability that exactly two of the customers are comfortable with delivery by drones can be computed as:
P\left(X=2\right){=}^{4}{C}_{2}×{0.27}^{2}×{0.73}^{42}
=6×0.0729×0.5329
=0.233
Assume that a procedure yields a binomial distribution. What is the probability of x successes if
n=64,\text{ }x=3
p=0.04
For a binomial distribution mean is 6 and variance is 2. Find the
P\left(X=4\right)
P\left(3\right)=
P\left(x\ge 4\right)=
P\left(x\le 2\right)=
P\left(4<X\le 8\right)
The selling price of a refrigerator is $584. If the markup is 25% of the dealers
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Vibration of string lattice | JVE Journals
Vladimir Astashev1 , Nikolay Andrianov2 , Vitaly Krupenin3
Vibroengineering PROCEDIA, Vol. 8, 2016, p. 97-101.
October 22, 2022 in Dubai, United Arab Emirates
A two-dimensional system oscillation with massive bodies located in lattice nodes is investigated in this paper. The results of theoretical analysis and of the performed experiments are given. Certain modes of the oscillation of lattices of different dimensions are described.
Keywords: string, lattice nodes, experimental stand, standing wave, natural frequencies, mode shapes.
String lattice is a two-dimensional analogue of the classical system known as “thread with beads”. It can be used as a model of 2D-objects such as membranes, thin plates, panels, lattice structures of various sorting and sieving machines and devices, as well as crystals and nanostructures [1].
Linear models of string lattices are families of intersected strings forming rectangular cells. Each cell node contains a point perfectly rigid body. Such models were first studied in papers [2, 3]. String lattice dynamics with obstacles was considered in [4, 5]. Here we give the first experimental results on oscillations of such systems.
2. The equations of motion and their analysis
Let us consider a rectangular lattice [2, 3], composed of two orthogonal families of identical linear elastic strings fixed at the ends. String lengths are
{l}_{1}
{l}_{2}
respectively (Fig. 1). Each string is numbered by indices
k=
0, 1, 2, …,
{N}_{1}
q=
{N}_{2}
. Point rigid bodies of equal masses m are placed into the lattice nodes.
Fig. 1. a) Model of lattice 4×4. Oscillations modes: b)
{\mathrm{\Theta }}_{11}
{\mathrm{\Theta }}_{22}
{\mathrm{\Theta }}_{12}
{\mathrm{\Theta }}_{21}
The rectangular lattice cells are equal; the string elements are inertialless. The strings are absolutely rigidly fixed in the nodes and tensions are so large that we can neglect possible changes of string lengths. A length of each cell “horizontal side” is
\mathrm{\Delta }{L}_{1}
; and of “vertical one” is
\mathrm{\Delta } {L}_{2}
. “Horizontal” sides have tension
{T}_{1}
, and “vertical” ones have tension
{T}_{2}
. Let us describe lattice deformation by functions of lattice nodes displacements
{u}_{kq}\left(t\right)
k=
{N}_{1}
q=
{N}_{2}
. Each of the functions
{u}_{kq}\left(t\right)
defines a displacement of the node along an axis perpendicular to the plane of lattice static equilibrium. Horizontal strings are numbered by the first index
k
, while vertical strings are numbered by the second index
q
External forces are denoted as
{g}_{kq}\left(p;t,{u}_{kq}\right)
p\equiv d/dt
. For all nodes we have
N=\left({N}_{1}–1\right)\left({N}_{2}–1\right)
m{\stackrel{¨}{u}}_{kq}+{c}_{1}\left(2{u}_{kq}-{u}_{\left(k-1,q\right)}-{u}_{\left(k+1,q\right)}\right)+{c}_{1}\left(2{u}_{kq}-{u}_{\left(k,q-1\right)}-{u}_{\left(k,q+1\right)}\right)={g}_{kq}\left(p;t,{u}_{kq}\right),
{c}_{1,2}=\frac{\mathrm{\Delta }{L}_{1,2}}{{T}_{1,2}}.
{u}_{kq}=
k=
{N}_{1}
q=
{N}_{2}
. To write equations (1) in operator form we construct matrix dynamic compliance operators
\left[{L}_{kq,nj}\right]
. We use four-index numbering for the dynamic compliance matrix components
{L}_{kq,nj}\left(p\right)
because each node is numbered by a pair of indices. Expression
{L}_{kq,nj}\left(p\right)
denotes the dynamic compliance operator [3-6], which connects the force, applied to the point
\left(n,j\right)
, and node displacement
\left(k,q\right)
. For the considered system we write the principle of reciprocity as
{L}_{kq,nj}\left(p\right)={L}_{nj,kq}\left(p\right)
. The system of equations (1) in this case can be written in operator form:
{u}_{kq}\left(t\right)={\sum }_{n=1}^{{N}_{1}-1}{\sum }_{j=1}^{{N}_{2}-1}{L}_{kq,nj}\left(p\right){g}_{nj}\left(p;{u}_{nj},t\right).
If the forces
{g}_{nj}\left(p;{u}_{nj},t\right)
depend only on the time, Eq. (2) is linear and determines the solution of the problem. In general case we have a system of
N
nonlinear equations for the unknown displacements of nodes. The case of periodic oscillations [3-5] is most well-studied. In particular, if any obstacle is installed near the lattices the system is vibro-impact.
To determine Eq. (2) we should describe the operators
\left[{L}_{kq,nj}\right]
. The expressions for the dynamic compliance operators are completely determined by their natural frequencies
\left\{{\mathrm{\Omega }}_{kq}\right\}
and normalized coefficients
\left\{{\mathrm{\Theta }}_{kq}\right\}
of natural forms of a linear system [6]. Using the results from [2] we find:
{\mathrm{\Omega }}_{kq}^{2}=\frac{2{T}_{1}}{m\mathrm{\Delta }{L}_{1}}\left[1-\mathrm{cos}\left(k\pi {N}_{1}^{-1}\right)\right]+\frac{2{T}_{2}}{m\mathrm{\Delta }{L}_{2}}\left[1-\mathrm{cos}\left(q\pi {N}_{1}^{-1}\right)\right],
{\Theta }_{kq}=C\mathrm{s}\mathrm{i}\mathrm{n}\left(kn\pi {N}_{1}^{-1}\right)\mathrm{s}\mathrm{i}\mathrm{n}\left(qj\pi {N}_{2}^{-1}\right), C=4{\left[\left({N}_{1}-1\right)\left({N}_{2}-1\right)\right]}^{-1}.
Hence, [3]:
{L}_{kq,nj}\left(p\right)=\sum _{\alpha =1}^{{N}_{1}-1}\sum _{\beta =1}^{{N}_{2}-1}\mathrm{s}\mathrm{i}\mathrm{n}\left(k\alpha \pi {N}_{1}^{-1}\right)\mathrm{s}\mathrm{i}\mathrm{n}\left(q\beta \pi {N}_{2}^{-1}\right)\mathrm{s}\mathrm{i}\mathrm{n}\left(\alpha n\pi {N}_{1}^{-1}\right)\mathrm{s}\mathrm{i}\mathrm{n}\left(\beta j\pi {N}_{2}^{-1}\right){\left({\mathrm{\Omega }}_{\alpha \beta }^{2}+{p}^{2}\right)}^{-1}.
Thus, the parameters of oscillations in the linear case are fully defined and the necessary preliminary construction for the analysis in the case of non-linearities is carried out. Fig. 1(b-e) demonstrates the envelope surfaces of the four lower forms of free oscillations of 7×5 lattice, with nodal bodies not being shown.
Note that in case of isotropic lattices (
{N}_{1}={N}_{2}\equiv {N}_{0}
{T}_{1}={T}_{2}\equiv {T}_{0}
∆{L}_{1}=∆{L}_{2}\equiv ∆{L}_{0}
{c}_{1}={c}_{2}\equiv {c}_{0}
) the number of different natural frequencies
{\mathrm{\Omega }}_{kq}
{\mathrm{\Omega }}_{qk}
and modes
{\mathrm{\Theta }}_{kq}
{\mathrm{\Theta }}_{qk}
decreases because of the system symmetry:
{\mathrm{\Omega }}_{kq}^{2}=2{c}_{0}\frac{2{T}_{1}}{m\mathrm{\Delta }{L}_{1}}\left[2-\mathrm{c}\mathrm{o}\mathrm{s}\frac{k\pi }{{N}_{0}}-\mathrm{c}\mathrm{o}\mathrm{s}\frac{q\pi }{{N}_{0}}\right], {\mathrm{\Theta }}_{kq}=4{N}_{0}^{4}\mathrm{s}\mathrm{i}\mathrm{n}\frac{k\pi }{{N}_{0}}\mathrm{s}\mathrm{i}\mathrm{n}\frac{q\pi }{{N}_{0}}.
For experimental study of string lattices, we designed and constructed an experimental stand “Alligator Square” (Fig. 2), consisted of the working installation and of a control and registration system. Working installation includes a vibration exciter (V) and a system of interchangeable lattices (L) consisted of square aluminum frames (300×300 mm) with rubber bundles of 1mm in diameter. Bundles tensions and cells sizes are equal respectively. Lattice frame is fixed on the exciter platform. We manufactured lattices of 2×2, 3×3, 4×4. Lattices nodes are formed by 1.1 g washers placed at the points of bundles intersections. The washers were made of hardened steel. The installation enables the use of unilateral or bilateral obstacles made of glass sheets of 3mm thickness. The obstacles are installed on four supports (S) on a fixed base. The gaps values are set and regulated by micrometer screws.
The control and recording system consists of a signal generator (SG), a power amplifier (PA), a digital stroboscope (DS), a digital voltmeter (DV), measurer for the vibration amplitude of the exciter platform (VA). Registration of standing waves profiles is carried out by camera (C), operating in both a photographing mode and the mode of accelerated filming. Strobe lamp (SL) illuminates the lattice. A phase shifter built-in into the stroboscope control unit can "stop" the lattice and take its picture at any time. At a small mismatch of 0.3-5.7 Hz we obtain a picture of slow evolution of the profile of the two-dimensional standing wave. The general scheme of the stand “Alligator Square” is shown in Fig. 3.
Fig. 2. Working installation: a) scheme; b) draft; c) photo
Fig. 3. Scheme of the “Alligator Square” experimental stand
The aim of the experiments is to give a quality experimental analysis of oscillation modes of string lattices with periodic excitation of oscillations.
4. The experiments results
In our case, the excitation is of kinematic nature: the frame oscillates sinusoidally. By changing the frequencies of the excitation we can study resonance phenomena.
We studied the lattices of 2×2, 3×3, 4×4. The relationships between the natural frequencies were estimated by Eq. (4). The natural frequencies of all the lattices lie within the segments
\left[{\mathrm{\Omega }}_{11},{\mathrm{\Omega }}_{KK}\right]
K
= 2, 3, 4. Natural frequencies distribution within these segments is uneven.
Fig. 4. Lattice 2×2: mode
{\mathrm{\Theta }}_{11}
{f}_{11}=
15.1 Hz; the amplitude of excitation
\mu
In the case of a 2×2 lattice we receive three different frequencies
\left\{{\mathrm{\Omega }}_{11},{\mathrm{\Omega }}_{22},{\mathrm{\Omega }}_{12}\right\}
. Fig. 4 and Fig. 5 demonstrate oscillations with modes
{\mathrm{\Theta }}_{11}
{\mathrm{\Theta }}_{12}
{f}_{11}={\mathrm{\Omega }}_{11}/2\pi
{f}_{12}={\mathrm{\Omega }}_{12}/2\pi
where the final configurations of the systems over the half period are given.
In the case of a 3×3 lattice we receive six different frequencies
\left\{{\mathrm{\Omega }}_{11},{\mathrm{\Omega }}_{22},{\mathrm{\Omega }}_{33},{\mathrm{\Omega }}_{12},{\mathrm{\Omega }}_{13},{\mathrm{\Omega }}_{23}\right\}
. Fig. 6 shows a standing wave with oscillations of mode
{\mathrm{\Theta }}_{11}
. Two senior modes of oscillations are shown in Fig. 7.
{\mathrm{\Theta }}_{12}
{f}_{12}=
28.4 Hz; amplitude of excitation
\mu
= 5.2 mm
{\mathrm{\Theta }}_{11}
; frequency
{f}_{12}=
\mu
For a 4×4 lattice we receive a set of ten different natural frequencies
\left\{{\mathrm{\Omega }}_{11}
{\mathrm{\Omega }}_{22}
{\mathrm{\Omega }}_{33}
{\mathrm{\Omega }}_{44}
{\mathrm{\Omega }}_{12}
{\mathrm{\Omega }}_{13}
{\mathrm{\Omega }}_{14}
{\mathrm{\Omega }}_{23}
{\mathrm{\Omega }}_{24}
{\mathrm{\Omega }}_{34}
}, which belongs to the interval [
{\mathrm{\Omega }}_{11}
{\mathrm{\Omega }}_{44}
]. Fig. 8 demonstrates the evolution of the mode
{\mathrm{\Theta }}_{11}
When the number of nodes increases we face the problems with registration of the expected modes of oscillations, as we have distortions associated with the non-isotropy of the lattice and (or) non-linear factors. Fig. 9 shows the evolution of the standing wave at the implementation of oscillations on one of the highest forms. We can clearly see the transformation of the square lattice cells into the trapezoid ones. Such distortions can be seen also on Fig. 6 (a-b).
Fig. 7. Lattice 3×3: profiles at frequencies a)
f=
27.5 Hz, b) 36 Hz and c) amplitude of excitation
\mu
Fig. 8. Lattice 4×4: standing wave profiles; frequency and amplitude of excitation
{f}_{11}=
\mu
Fig. 9. Lattice 4×4: profiles of the standing wave at frequency
f=
29 Hz, amplitude of excitation
\mu
5. Remarks on vibro-impact processes study
We performed experiments to study vibro-impact regimes of 2×2 and 3×3 lattices. The researches were carried out at frequencies of excitation higher than
{\mathrm{\Omega }}_{11}/\text{2}\pi
for lattices of both types. The regimes of the synchronous claps [4, 5] were visually observed. We intend to accomplish a comprehensive interpretation in our subsequent works.
Experiments with two-dimensional lattice structures show that linear models in lattice constructions with a high number of nodes work well for analysis of the lower modes of oscillations. Analysis of the higher modes of oscillations demands taking into account non-linear factors. Therefore, it is required the creation of more general theories and the development of more advanced experimental stands.
Astashev V. K., Krupenin V. L., Perevezentsev V. N., Kolik L. V., Andrianov N. A. Properties of surface layers nanostructured by autoresonant ultrasonic turning. Journal of Machinery Manufacture and Reliability, Vol. 40, Issue 5, 2011, p. 68-72. [Search CrossRef]
Nagaev R. F., Khodzhaev K. Sh. Kolebaniya Mekhanicheskikh System s Periodicheskoi Strukturoi (Vibration of Mechanical Systems with Periodic Structure). FAN, Tashkent, 1973, p. 220, (in Russian). [Search CrossRef]
Krupenin V. L. On the calculation of vibratory processes in lattice two-dimensional constructions. Problemy Mashinostroeniya i Nadezhnosti Mashin, Vol. 2, 2006, p. 20-26, (in Russian). [Search CrossRef]
Krupenin V. L. Vibroimpact processes in two-dimensional lattice constructions. Problemy Mashinostroeniya i Nadezhnosti Mashin, Vol. 3, 2006, p. 16-22, (in Russian). [Search CrossRef]
Krupenin V. L. Analysis of singularized motion equations of latticed vibroimpact 2D systems in renouncing Newton’s hypothesis. Journal of Machinery Manufacture and Reliability, Vol. 45, Issue 2, 2016, p. 104-112. [Search CrossRef]
Babitsky V. I., Krupenin V. L. Vibration of Strongly Nonlinear Discontinuous Systems. Springer-Verlag, Berlin, Heidelberg, New York, 2001, p. 404. [Search CrossRef]
|
Programming Concepts: Recursive Techniques - Wikibooks, open books for an open world
Programming Concepts: Recursive Techniques
< A-level Computing | AQA | Paper 1 | Fundamentals of programming(Redirected from A-level Computing/AQA/Problem Solving, Programming, Operating Systems, Databases and Networking/Programming Concepts/Recursive Techniques)
← Stack Frames Recursion Paradigms Intro →
Recursion - Defining a sub routine in terms of itself.
Recursion is a key area in computer science that relies on you being able to solve a problem by the cumulation of solving increasingly smaller instances of the same problem.
A visual form of recursion known as the Droste effect. The woman in this image is holding an object which contains a smaller image of her holding the same object, which in turn contains a smaller image of herself holding the same object, and so forth.
An example of recursion in a name is the GNU Operating System, you might well ask what GNU stands for:
Wait a minute, they have defined the name by restating the name!:
GNU is Not Unix is Not Unix
GNU is Not Unix is Not Unix is Not Unix
GNU is Not Unix is Not Unix is Not Unix is Not Unix
Luckily for us, with computer code our recursion should tend towards an end as we are solving increasingly smaller instances of the same problem, we should be tending towards an end. With the GNU example the name always remains the same size, but with the following example there is an end:
Example: A recursive story
function revise(essay) {
read(essay);
get_feedback_on(essay);
apply_changes_to(essay);
revise(essay) unless essay.complete?;
Which is pseudocode to describe the process of revising an essay, the revision process involves reading the essay, getting feedback, applying the changes, then revising the slightly better essay. You keep doing this until the essay is complete.
Let's take a look at some computer science examples
2 Fibonacci number example
3 Recursion summary
Factorial example[edit | edit source]
A classic computer example of a recursive procedure is the function used to calculate the factorial of a natural number:
{\displaystyle n!=n\times (n-1)\times (n-2)\times ...\times 1}
{\displaystyle 3!=3\times 2\times 1=6\ }
{\displaystyle 5!=5\times 4\times 3\times 2\times 1=120\ }
{\displaystyle 10!=10\times 9\times 8\times 7\times 6\times 5!=3628800\ }
Did you notice what I did with the final solution? Instead of writing 5 * 4 * 3 * 2 * 1, I used 5! which elicits the same result. Looking back our definition of why recursion is used, we seem to solve big problems by solving smaller instances of the same problem, so factorials are ripe for recursion:
10! = 10 * 9!
As we can see, each n! is the product of n * (n-1)!. In summary:
{\displaystyle \operatorname {fact} (n)={\begin{cases}1&{\mbox{if }}n=1\\n\cdot \operatorname {fact} (n-1)&{\mbox{if }}n>1\\\end{cases}}}
1. if n is >= 1, return [ n × factorial(n-1) ]
2. otherwise, return 1
Now let's have a look at how we would write code to solve this:
function factorial(ByVal n as integer)
return n * factorial(n-1) 'recursive call
console.writeline(factorial(10))
It looks very simple and elegant. But how does it work?
Let's build a trace table and see what happens. This trace table will be different from the ones that you have built before as we are going to have to use a stack. If you haven't read up on stacks you must do so before continuing:
All is going well so far until we get to line 3. Now what do we do? We'll soon have two values of n, one for Function call 1 and one for Function call 2. Using the trace table as a stack (with the bottom of the stack at the top and the top of the stack at the bottom) we'll save all the data about the function call including its value of n and make note of what line we need to return to when we have finished with factorial(3).
We now have a similar situation to before, let's store the return address and go to factorial(2)
Now we have another problem, we have found an end to the factorial(1). What line do we go to next? As we are treating our trace table as a stack we'll just pop the previous value off the top and look at the last function call we stored away, that is function call 3, factorial(2), and we even know what line to return to, line 3:
We know that factorial(1) = 1 from the previous returned value. Therefore factorial(2) returns 2 * 1 = 2
Again we'll pop the last function call from the stack leaving us with function call 2, factorial(3) and line 3.
We know that factorial(3) = 6 from the previous returned value. Therefore factorial(4) returns 4 * 6 = 24
We reach the end of function call 1. But where do we go now? There is nothing left on the stack and we have finished the code. Therefore the result is 24.
Fibonacci number example[edit | edit source]
Fibonacci numbers are found in nature and have fascinated mathematicians for centuries
Another good example is a method to calculate the Fibonacci numbers. By definition, the first two Fibonacci numbers are 0 and 1, and each subsequent number is the sum of the previous two.:
{\displaystyle 0,\;1,\;1,\;2,\;3,\;5,\;8,\;13,\;21,\;34,\;55,\;89,\;144,\;\ldots \;}
For example, take the 6th number, 5. This is the sum of the 5th number, 3, and the 4th number, 2.
In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recursive statement
{\displaystyle F_{n}=F_{n-1}+F_{n-2},\!\,}
with the first two values being
{\displaystyle F_{0}=0\quad {\text{and}}\quad F_{1}=1.}
Let's try and create a code version of this:
function fib(n as integer)
Recursion summary[edit | edit source]
Recursion does have some issues though, consider how much data we had to store on the stack for just 4 function calls. If we were to perform 1000, the memory used would be incredibly large.
Recursion can produce simpler, more natural solutions to a problem
Recursion takes up large amounts of computer resources storing return addresses and states
Defining a sub routine in terms of itself
Give the output of the following recursive function call recur(6):
function recur(ByVal n as integer)
Draw a trace table for the following recursive function
Name one benefit and one drawback of recursive solutions:
↑ thesecretmaster on StackExchange https://cseducators.stackexchange.com/questions/17/analogy-for-teaching-recursion
Retrieved from "https://en.wikibooks.org/w/index.php?title=A-level_Computing/AQA/Paper_1/Fundamentals_of_programming/Recursion&oldid=3676432"
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Turingery - Wikipedia
Manual codebreaking method
Turingery[1] or Turing's method[2] (playfully dubbed Turingismus by Peter Ericsson, Peter Hilton and Donald Michie[3]) was a manual codebreaking method devised in July 1942[4] by the mathematician and cryptanalyst Alan Turing at the British Government Code and Cypher School at Bletchley Park during World War II.[5][6] It was for use in cryptanalysis of the Lorenz cipher produced by the SZ40 and SZ42 teleprinter rotor stream cipher machines, one of the Germans' Geheimschreiber (secret writer) machines. The British codenamed non-Morse traffic "Fish", and that from this machine "Tunny" (another word for the tuna fish).
Reading a Tunny message required firstly that the logical structure of the system was known, secondly that the periodically changed pattern of active cams on the wheels was derived, and thirdly that the starting positions of the scrambler wheels for this message—the message key—was established.[7] The logical structure of Tunny had been worked out by William Tutte and colleagues[8] over several months ending in January 1942.[9] Deriving the message key was called "setting" at Bletchley Park, but it was the derivation of the cam patterns—which was known as "wheel breaking"—that was the target of Turingery.
German operator errors in transmitting more than one message with the same key, producing a "depth", allowed the derivation of that key. Turingery was applied to such a key stream to derive the cam settings.[10]
1 The SZ40 and SZ42
2 Differencing
2.1 Bit-level differencing
3 Turing's method
The SZ40 and SZ42[edit]
Main article: Lorenz cipher
The logical functioning of the Tunny system was worked out well before the Bletchley Park cryptanalysts saw one of the machines—which only happened in 1945, shortly before the allied victory in Europe.[11]
BP wheel name[12]
{\displaystyle \psi }
{\displaystyle \psi }
{\displaystyle \psi }
{\displaystyle \psi }
{\displaystyle \psi }
{\displaystyle \mu }
{\displaystyle \mu }
{\displaystyle \chi }
{\displaystyle \chi }
{\displaystyle \chi }
{\displaystyle \chi }
{\displaystyle \chi }
Number of cams (pins)
The SZ machines were 12-wheel rotor cipher machines which implemented a Vernam stream cipher. They were attached in-line to standard Lorenz teleprinters. The message characters were encoded in the 5-bit International Telegraphy Alphabet No. 2 (ITA2). The output ciphertext characters were generated by combining a pseudorandom character-by-character key stream with the input characters using the "exclusive or" (XOR) function, symbolised as "
{\displaystyle \oplus }
" in mathematical notation. The relationship between the plaintext, ciphertext and cryptographic key is then:
{\displaystyle \mathrm {ciphertext} =\mathrm {plaintext} \oplus \mathrm {key} }
Similarly, for deciphering, the ciphertext was combined with the same key to give the plaintext:
{\displaystyle \mathrm {plaintext} =\mathrm {ciphertext} \oplus \mathrm {key} }
This produces the essential reciprocity to allow the same machine with the same settings to be used for both enciphering and deciphering.
Each of the five bits of the key for each character was generated by the relevant wheels in two parts of the machine. These were termed the chi (
{\displaystyle \chi }
) wheels, and the psi (
{\displaystyle \psi }
) wheels. The chi wheels all moved on one position for each character. The psi wheels also all moved together, but not after each character. Their movement was controlled by the two mu (
{\displaystyle \mu }
) or "motor" wheels.[13]
The key stream generated by the SZ machines thus had a chi component and a psi component that were combined with the XOR function. So, the key that was combined with the plaintext for enciphering—or with the ciphertext for deciphering—can be represented as follows.[13]
{\displaystyle \mathrm {key} ={\textit {chi}}\mathrm {{\mbox{-}}key} \oplus {\textit {psi}}\mathrm {{\mbox{-}}key} }
{\displaystyle K=\chi \oplus \psi }
The twelve wheels each had a series of cams (or "pins") around them. These cams could be set in a raised or lowered position. In the raised position they generated a "mark" "×" ("1" in binary), in the lowered position they generated a "space" "·" ("0" in binary). The number of cams on each wheel equalled the number of impulses needed to cause them to complete a full rotation. These numbers are all co-prime with each other, giving the longest possible time before the pattern repeated. With a total of 501 cams this equals 2501 which is approximately 10151, an astronomically large number.[14] However, if the five impulses are considered independently, the numbers are much more manageable. The product of the rotation period of any pair of chi wheels gives numbers between 41×31=1271 and 26×23=598.
Differencing[edit]
Cryptanalysis often involves finding patterns of some sort that provide a way into eliminating a range of key possibilities. At Bletchley Park the XOR combination of the values of two adjacent letters in the key or the ciphertext was called the difference (symbolised by the Greek letter delta
{\displaystyle \Delta }
) because XOR is the same as modulo 2 subtraction (without "borrow")—and, incidentally, modulo 2 addition (without "carry"). So, for the characters in the key (K), the difference
{\displaystyle \Delta K}
was obtained as follows, where underline indicates the succeeding character:
{\displaystyle \Delta K=K\oplus {\underline {K}}}
(Similarly with the plaintext, the ciphertext, and the two components of the key).
The relationship amongst them applies when they are differenced. For example, as well as:
{\displaystyle K=\chi \oplus \psi }
{\displaystyle \Delta K=\Delta \chi \oplus \Delta \psi }
If the plaintext is represented by P and the cipertext by Z, the following also hold true:
{\displaystyle \Delta Z=\Delta P\oplus \Delta \chi \oplus \Delta \psi }
{\displaystyle \Delta P=\Delta Z\oplus \Delta \chi \oplus \Delta \psi }
The reason that differencing provided a way into Tunny was that, although the frequency distribution of characters in the ciphertext could not be distinguished from a random stream, the same was not true for a version of the ciphertext from which the chi element of the key had been removed. This is because, where the plaintext contained a repeated character and the psi wheels did not move on, the differenced psi character (
{\displaystyle \Delta \psi }
) would be the null character ("·····" or 00000), or, in Bletchley Park terminology, "/". When XOR-ed with any character, this null character has no effect, so in these circumstances,
{\displaystyle \Delta \chi =\Delta K}
. Repeated characters in the plaintext were more frequent, both because of the characteristics of German (EE, TT, LL and SS are relatively common),[15] and because telegraphists frequently repeated the figures-shift and letters-shift characters[16] as their loss in an ordinary telegraph message could lead to gibberish.[17]
Turingery introduced the principle that the key differenced at one, now called
{\displaystyle \Delta K}
, could yield information unobtainable from ordinary key. This
{\displaystyle \Delta }
principle was to be the fundamental basis of nearly all statistical methods of wheel-breaking and setting.[1]
Bit-level differencing[edit]
As well as applying differencing to the full 5-bit characters of the ITA2 code, it was also applied to the individual impulses (bits). So, for the first impulse, that was enciphered by wheels
{\displaystyle \chi _{1}}
{\displaystyle \psi _{1}}
, differenced at one:
{\displaystyle \Delta K_{1}=K_{1}\oplus {\underline {K_{1}}}}
And for the second impulse:
{\displaystyle \Delta K_{2}=K_{2}\oplus {\underline {K_{2}}}}
It is also worth noting that the periodicity of the chi and psi wheels for each impulse (41 and 43 respectively for the first one) is reflected in its pattern of
{\displaystyle \Delta K}
. However, given that the psi wheels did not advance for every input character, as did the chi wheels, it was not simply a repetition of the pattern every 41 × 43 = 1763 characters for
{\displaystyle \Delta K_{1}}
, but a more complex sequence.
Turing's method[edit]
In July 1942 Turing spent a few weeks in the Research Section.[18] He had become interested in the problem of breaking Tunny from the keys that had been obtained from depths.[3] In July, he developed the method of deriving the cam settings from a length of key.[1] It involved an iterative, almost trial-and-error, process. It relied on the fact that when the differenced psi character is the null character ("·····" or 00000), /, then XOR-ing this with any other character does not change it. Thus the delta key character gives the character of the five chi wheels (i.e.
{\displaystyle \Delta \chi =\Delta K}
Given that the delta psi character was the null character half of the time on average, an assumption that
{\displaystyle \Delta K=\Delta \chi }
had a 50% chance of being correct. The process started by treating a particular
{\displaystyle \Delta K}
character as being the Δ
{\displaystyle \chi }
for that position. The resulting putative bit pattern of × and · for each chi wheel, was recorded on a sheet of paper that contained as many columns as there were characters in the key, and five rows representing the five impulses of the
{\displaystyle \Delta \chi }
. Given the knowledge from Tutte's work, of the periodicity of each of the wheels, this allowed the propagation of these values at the appropriate positions in the rest of the key.
A set of five sheets, one for each of the chi wheels, was also prepared. These contained a set of columns corresponding in number to the cams for the appropriate chi wheel, and were referred to as a 'cage'. So the
{\displaystyle \chi _{3}}
cage had 29 such columns.[19] Successive 'guesses' of
{\displaystyle \Delta \chi }
values then produced further putative cam state values. These might either agree or disagree with previous assumptions, and a count of agreements and disagreements was made on these sheets. Where disagreements substantially outweighed agreements, the assumption was made that the
{\displaystyle \Delta \psi }
character was not the null character "/", so the relevant assumption was discounted. Progressively, all the cam settings of the chi wheels were deduced, and from them the psi and motor wheel cam settings.
As experience of the method developed, improvements were made that allowed it to be used with much shorter lengths of key than the original 500 or so characters.[1]
^ a b c d Good, Michie & Timms 1945, p. 313 in Testery Methods 1942–1944
^ Government Code and Cypher School 1944, p. 89
^ a b Copeland 2006, p. 380
^ Good, Michie & Timms 1945, p. 309 in Early Hand Methods
^ Churchhouse 2002, p. 4
^ Good 1993, p. 161
^ Good, Michie & Timms 1945, p. 6 in German Tunny
^ a b Good, Michie & Timms 1945, p. 7 in German Tunny
^ Churchhouse 2002, p. 158
^ Carter, p. 3
^ Copeland 2006, p. 385 which reproduces a
{\displaystyle \chi _{3}}
cage from the General Report on Tunny
Carter, Frank, Bletchley Park Technical Papers: Colossus and the Breaking of the Lorenz Cipher (PDF), retrieved 28 January 2011
Copeland, Jack (2006), "Turingery", in Copeland, B. Jack (ed.), Colossus: The Secrets of Bletchley Park's Codebreaking Computers, Oxford: Oxford University Press, ISBN 978-0-19-284055-4
Good, Jack (1993), "Enigma and Fish", in Hinsley, F.H.; Stripp, Alan (eds.), Codebreakers: The inside story of Bletchley Park, Oxford: Oxford University Press, ISBN 978-0-19-280132-6
Government Code and Cypher School (1944), The Bletchley Park 1944 Cryptographic Dictionary formatted by Tony Sale (PDF), retrieved 7 October 2010
Hodges, Andrew (1992), Alan Turing: The Enigma, London: Vintage, ISBN 978-0-09-911641-7
Newman, Max (c. 1944), "Appendix 7: Δ
{\displaystyle \chi }
-Method", in Copeland, B. Jack (ed.), Colossus: The Secrets of Bletchley Park's Codebreaking Computers, Oxford: Oxford University Press, ISBN 978-0-19-284055-4
Tutte, William T. (2006), "My Work at Bletchley Park", in Copeland, B Jack (ed.), Colossus: The Secrets of Bletchley Park's Codebreaking Computers, Oxford: Oxford University Press, ISBN 978-0-19-284055-4
Retrieved from "https://en.wikipedia.org/w/index.php?title=Turingery&oldid=1063877878"
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Robust Performance Measure for Mu Synthesis - MATLAB & Simulink - MathWorks í•œêµ
Upper Bound of μ
For a system T(s), the robust H∞ performance μ is the smallest value γ such that the peak gain of T remains below γ for uncertainty up to 1/γ, in normalized units. For example:
μ = 0.5 means that ||T(s)||∞ remains below 0.5 for uncertainty up to twice the uncertainty specified in T. The worst-case gain for the specified uncertainty is typically smaller.
μ = 2 means that ||T(s)||∞ remains below 2 for uncertainty up to half the uncertainty specified in T. For this value, the worst-case gain for the full specified uncertainty can be much larger. It can even be infinite, meaning that the system does not remain stable over the full range of the specified uncertainty.
The quantity μ is the peak value over frequency of the structured singular value μ(ω) for the uncertainty specified in T. This quantity is a generalization of the singular value for uncertain systems. It depends on the structure of the uncertainty in the system. In practice, μ is difficult to compute exactly, so the software instead computes lower and upper bounds,
\underset{¯}{\mathrm{μ}}
\stackrel{¯}{\mathrm{μ}}
\stackrel{¯}{\mathrm{μ}}
\stackrel{¯}{\mathrm{μ}}
\stackrel{¯}{\mathrm{μ}}
Use musynperf evaluate the robust performance of an uncertain system. This function returns lower and upper bounds on μ, the uncertainty values that yield the peak μ, and other information about the closed-loop robust performance.
To understand the computation of robust H∞ performance, consider an uncertain system T(s), modeled as a fixed portion T0 and an uncertain portion Δunc/γ, such that
T\left(s\right)=\text{LFT}\left({\mathrm{Î}}_{unc}/\mathrm{γ},{T}_{0}\right)
Δunc collects the uncertain elements {Δ1,…,ΔN}.
{\mathrm{Î}}_{unc}=\left(\begin{array}{ccc}{\mathrm{Î}}_{1}& & \\ & â±& \\ & & {\mathrm{Î}}_{N}\end{array}\right).
Each Δj is an arbitrary real, complex, or dynamic uncertainty that is normalized such that ||Δj||∞ ≤ 1. The factor γ adjusts the level of uncertainty.
||T||∞ ≤ γ for all ||Δunc||∞ ≤ 1.
By the small-gain theorem (see [1]), this robust performance condition is equivalent to stating that the system of diagram (b), LFT(Δperf/γ,T), is stable for all for all ||Δperf||∞ ≤ 1.
Δperf is called the performance block. Expand T as in diagram (a), and group Δperf with the uncertain blocks Δunc to define a new block Δ,
\mathrm{Î}â\left(\begin{array}{cc}{\mathrm{Î}}_{perf}& 0\\ 0& {\mathrm{Î}}_{unc}\end{array}\right).
{âTâ}_{\infty }â¤\mathrm{γ}\text{â}\text{for all}\text{â}{â{\mathrm{Î}}_{unc}â}_{\infty }â¤1\text{â}â\text{â}{âLFT\left(\mathrm{Î}/\mathrm{γ}\right),{T}_{0}â}_{\infty }\text{â}\text{stable for all}\text{â}{â\mathrm{Î}â}_{\infty }â¤\text{â}1.
The robust performance μ is the smallest γ for which this stability condition holds. Equivalently, 1/μ is the largest uncertainty level 1/γ for which the system of diagram (c) is robustly stable. In other words, 1/μ is the robust stability margin of the feedback loop of diagram (c) for the augmented uncertainty Δ. (For more information on robust stability margins, see Robustness and Worst-Case Analysis.)
To obtain an estimate on the upper bound of μ, the software introduces scalings. If the system in diagram (c) is stable for all ||Δ||∞ ≤ 1, then the system of the following diagram is also stable, for any invertible D.
If D commutes with Δ, then the system of diagram (d) is the same as the system in the following diagram.
The matrices D that structurally commute with Δ are called D scalings. They can be frequency dependent, which is denoted by D(ω).
\stackrel{¯}{\mathrm{μ}}
\stackrel{¯}{\mathrm{μ}}â\underset{D\left(\mathrm{Ï}\right)}{\mathrm{inf}}{âD\left(\mathrm{Ï}\right){T}_{0}\left(j\mathrm{Ï}\right)D{\left(\mathrm{Ï}\right)}^{â1}â}_{\infty }.
For the optimal D*(ω), and any γ ≥
\stackrel{¯}{\mathrm{μ}}
{â{D}^{*}\left(\mathrm{Ï}\right){T}_{0}\left(j\mathrm{Ï}\right){D}^{*}{\left(\mathrm{Ï}\right)}^{â1}â}_{\infty }â¤\text{â}\text{â}\mathrm{γ}.
Therefore, by the small-gain theorem, the system of diagram (e) is stable for all ||Δ||∞ ≤ 1. It follows that 1/γ ≤ 1/μ, or γ ≤ μ, because 1/μ is the robust stability margin. Consequently, μ ≤
\stackrel{¯}{\mathrm{μ}}
\stackrel{¯}{\mathrm{μ}}
is an upper bound for the robust performance μ. This upper bound
\stackrel{¯}{\mathrm{μ}}
When all the uncertain elements Δj are complex or LTI dynamics, the software approximates
\stackrel{¯}{\mathrm{μ}}
by picking a frequency grid {ω1,…,ωN}. At each frequency point, the software solves the optimal scaling problem
{\stackrel{¯}{\mathrm{μ}}}_{i}=\underset{{D}_{i}}{\mathrm{inf}}â{D}_{i}{T}_{0}\left(j{\mathrm{Ï}}_{i}\right){D}_{i}{}^{â1}â.
\stackrel{¯}{\mathrm{μ}}
\stackrel{¯}{\mathrm{μ}}=\underset{i}{\mathrm{max}}{\stackrel{¯}{\mathrm{μ}}}_{i}.
When some Δj are real, it is possible to obtain a less conservative upper bound by using additional scalings called G scalings. In this case,
\stackrel{¯}{\mathrm{μ}}
{\stackrel{¯}{\mathrm{μ}}}_{i}
{\left(\begin{array}{c}{T}_{0}\left(j{\mathrm{Ï}}_{i}\right)\\ I\end{array}\right)}^{H}\left(\begin{array}{cc}{D}_{r}\left({\mathrm{Ï}}_{i}\right)& âj{G}_{cr}^{H}\left({\mathrm{Ï}}_{i}\right)\\ j{G}_{cr}\left({\mathrm{Ï}}_{i}\right)& â{\stackrel{¯}{\mathrm{μ}}}_{i}^{2}{D}_{c}\left({\mathrm{Ï}}_{i}\right)\end{array}\right)\left(\begin{array}{c}{T}_{0}\left(j{\mathrm{Ï}}_{i}\right)\\ I\end{array}\right)â¤0
for some Dr(ωi), Dc(ωi), and Gcr(ωi). These frequency-dependent matrices are the D and G scalings.
\stackrel{¯}{\mathrm{μ}}
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We consider the motion by curvature of a network of smooth curves with multiple junctions in the plane, that is, the geometric gradient flow associated to the length functional. Such a flow represents the evolution of a two-dimensional multiphase system where the energy is simply the sum of the lengths of the interfaces, in particular it is a possible model for the growth of grain boundaries. Moreover, the motion of these networks of curves is the simplest example of curvature flow for sets which are “essentially” non regular. As a first step, in this paper we study in detail the case of three curves in the plane meeting at a single triple junction and with the other ends fixed. We show some results about the existence, uniqueness and, in particular, the global regularity of the flow, following the line of analysis carried on in the last years for the evolution by mean curvature of smooth curves and hypersurfaces.
Classification : 53C44, 53A04, 35K55
author = {Mantegazza, Carlo and Novaga, Matteo and Tortorelli, Vincenzo Maria},
title = {Motion by curvature of planar networks},
AU - Novaga, Matteo
AU - Tortorelli, Vincenzo Maria
TI - Motion by curvature of planar networks
Mantegazza, Carlo; Novaga, Matteo; Tortorelli, Vincenzo Maria. Motion by curvature of planar networks. Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, Série 5, Tome 3 (2004) no. 2, pp. 235-324. http://www.numdam.org/item/ASNSP_2004_5_3_2_235_0/
[1] D. Kinderlehrer - C. Liu, Evolution of grain boundaries, Math. Models Methods Appl. Sci. 11 (2001), 713-729. | MR 1833000 | Zbl 1036.74041
[2] J. Langer, A Compactness Theorem for Surfaces with
{L}_{p}
-Bounded Second Fundamental Form, Math. Ann. 270 (1985), 223-234. | MR 771980 | Zbl 0564.58010
[3] A. Lunardi, “Analytic semigroups and optimal regularity in parabolic problems,” Birkhäuser Verlag, Basel, 1995. | MR 1329547 | Zbl 0816.35001
[4] A. Lunardi - E. Sinestrari - W. Von Wahl, A semigroup approach to the time dependent parabolic initial-boundary value problem, Differential Integral Equations 5 (1992), 1275-1306. | MR 1184027 | Zbl 0758.34048
[5] L. Simon, “Lectures on Geometric Measure Theory”, Australian National University, Proc. Center Math. Anal., Vol. 3, Canberra, 1983. | MR 756417 | Zbl 0546.49019
[6] V. A. Solonnikov, “Boundary value problems of mathematical physics”. VIII, American Mathematical Society, Providence, R.I., 1975. | MR 369867
[7] H. M. Soner, Motion of a set by the curvature of its boundary, J. Differential Equations 101 (1993), 313-372. | MR 1204331 | Zbl 0769.35070
[8] A. Stone, A density function and the structure of singularities of the mean curvature flow, Calc. Var. Partial Differential Equations 2 (1994), 443-480. | MR 1383918 | Zbl 0833.35062
[9] A. Stone, A boundary regularity theorem for mean curvature flow, J. Differential Geom. 44 (1996), 371-434. | MR 1425580 | Zbl 0873.58066
|
Invariant Principle | Brilliant Math & Science Wiki
Alexander Katz, Pi Han Goh, Geoff Pilling, and
Generally speaking, an invariant is a quantity that remains constant during the execution of a given algorithm. In other words, none of the allowed operations changes the value of the invariant. The invariant principle is extremely useful in analyzing the end result (or possible end results) of an algorithm, because we can discard any potential result that has a different value for the invariant as impossible to reach.
Consider a set of states
S = (s_1, s_2, \dots, s_n)
, and a set of transitions
T \subseteq S \times S
(s_i, s_j) \in T
if and only if we can transition from state
s_i
s_j
. An invariant with respect to
T
f: S \rightarrow \mathbb{R}
(s_i,s_j) \in T \implies f(s_i)=f(s_j)
In particular, given a starting state
s_1
and a rule for transitions, invariants allow us to determine which states we can reach from
s_1
Many problems give the starting state and the transition rule, then ask whether a certain result is achievable (or equivalently, ask which results are achievable). For example,
Alice writes the numbers 1, 2, 3, 4, 5, and 6 on a blackboard. Bob selects two of these numbers, erases both of them, and writes down their sum on the blackboard. For example, if Bob chose the numbers 3 and 4, the blackboard would contain the numbers 1, 2, 5, 6, and 7. Bob continues until there is only one number left on the board. What are the possible values of that number?
In this problem, the invariant is the sum of the numbers on the blackboard,
n
. If Bob chooses to erase the numbers
and
b
, he will write
a+b
on the blackboard, making the new sum
n-a-b+(a+b)=n
n
is indeed an invariant. This means that at any time during the process, the sum of the numbers on the blackboard will be
n=1+2+3+4+5+6=21
, which means that the final number must be 21.
In a more formal sense, the states of this problem are possible sets of numbers on the blackboard, and the starting state is
s_1=\{1, 2, 3, 4, 5, 6\}.
The transition in this problem is erasing two numbers and writing their sum. The invariant function,
f(S)
, is the sum of the numbers in
S,
and the invariant rule is verified as above. Therefore, since
f(s_1)=21,
the end state
S_{\text{final}}
must also satisfy
f(S_{\text{final}})=21,
S_{\text{final}}
has only one number, it must be 21.
_\square
Although the invariant was able to determine precisely what would happen in the previous problem, they are usually only able to determine what cannot happen. For example, if the problem was slightly changed as follows:
Alice writes the numbers 1, 2, 3, 4, 5, and 6 on a blackboard. Bob selects two of these numbers, erases both of them, and writes down their positive difference on the blackboard. For example, if Bob chose the numbers 3 and 4, the blackboard would contain the numbers 1, 1, 2, 5, and 6. Bob continues until there is only one number left on the board. What are the possible values of that number?
Then many different results are possible. However, invariants are still useful in excluding possibilities:
If Bob chooses the numbers
and
b
a \geq b,
the sum changes from
n
n-a-b+(a-b)=n-2b
. Therefore, the sum always changes by an even number, meaning
n \pmod{2}
is an invariant. Originally, the sum of the numbers on the board is
1+2+3+4+5+6=21
, so at any point of the process, the sum of the numbers must be odd. Therefore, the final number cannot be even.
_\square
Note that the invariant says nothing about whether all odd final results are possible; it merely says that no even results are possible.
A monovariant is very similar to an invariant, but instead of remaining unchanged under transitions, monovariants either increase or decrease under transitions. They are very useful in showing algorithms terminate; for example, if a particular monovariant is a positive integer that decreases at every step of an algorithm, the algorithm must eventually terminate since decreasing integers cannot stay positive forever.
In more formal language, a monovariant with respect to
T
f: S \rightarrow \mathbb{R}
(s_i,s_j) \in T \implies f(s_i)>f(s_j)
, or a function
f: S \rightarrow \mathbb{R}
(s_i,s_j) \in T \implies f(s_i)<f(s_j)
Note that the definition uses
>
<
\geq
\leq
. This is because if an algorithm can leave the monovariant equal to its previous value, it becomes much less useful at demonstrating the algorithm terminates. However, if it can be shown that the algorithm must change the value of the monovariant in a finite number of steps, then the non-strict inequality would be sufficient.
As a general rule, invariants are useful whenever several different actions are possible, and especially when a problem asks whether a specific result is possible. In particular, invariants are especially helpful in the analysis of combinatorial games, where the potential transitions are given by legal moves, and the result asked about is the winner of the game.
Alice and Bob have a large chocolate bar, in the shape of a
10 \times 10
grid. Each turn, a player may either eat an entire bar of chocolate, or break any chocolate bar into two smaller rectangular chocolate bars along a grid line. The player who moves last loses. Who wins this game? (adapted from NIMO)
Every turn, the number of chocolate bars either increases by one (if the player breaks a chocolate bar into two chocolate bars), or decreases by one (if the player eats a chocolate bar). Therefore, the number of chocolate bars Alice will have to choose from is invariant modulo 2. At the beginning of the game, Alice has only one chocolate bar to choose from. Since the players cannot break the chocolate bar forever (since they must break the chocolate along grid lines), eventually Alice will have to eat the final piece of chocolate, so Bob wins regardless of how the players choose to play the game.
_\square
In more advanced problems, the use of invariants will not be set up in such an obvious manner. In those cases, it is usually necessary to transform the problem into a transitional one in some way. Many coloring problems fall under this category. For example, a problem asking whether is it possible to tile some shape with dominoes can be considered a transitional problem, where the starting state is the whole shape, the set of transitions is removing two adjacent squares, and the end state is an empty board.
Two opposite corners are removed from a standard chessboard. Is it possible to tile the resulting board with dominoes?
Suppose that the removed corners were both white squares. In every transition, we remove one white square and one black square, so
S=(\text{black squares})-(\text{white squares})
is invariant. Originally, there are 32 black squares and 30 white squares, so
S=2
at any point. The empty board has
S=0
, so it is impossible to reach it from the original board; in other words, the board cannot be tiled by dominoes.
_\square
Many problems will also use monovariants in a non-obvious way, usually by asking to prove that some arrangement is possible. These problems are often solved by beginning with a random arrangement, describing an algorithm that improves the situation at each step, and showing the algorithm terminates with the help of a monovariant.
In the parliament of Sikinia, each member has at most three enemies. Prove that the house can be separated into two houses, so that each member has at most one enemy in his own house. (Engel, Problem Solving Strategies, p.2)
Initially, separate the members into two houses in any manner. Let
H
be the total sum of all enemies each member has in his own house. Now suppose there is a person
A
who has at least two enemies in his own house. Then he has at most one enemy in the other house, so if
A
switches houses,
H
will decrease. Since
H
is necessarily a non-negative integer, it cannot decrease forever, so at some point this process ends. This means we cannot any longer find a person who has at least two enemies in their own house, which is precisely the assignment requested.
_\square
Another big clue that a problem involves invariants is being asked to prove something for all possible inputs, such as points in a plane. A common strategy to try in these problems is picking a random point, and examining what changes when the point is moved around.
As the first two problems showed, the sum of all the numbers (possibly modulo some constant) is a common invariant to try. More generally, a weighted sum is often useful, such as
a+2b+4c+8d+16e
a-b+c-d+e-f
. When dealing with transforming lists of numbers, another common invariant is the number of inversions, or pairs
(i,j)
i<j
i^\text{th}
element is bigger than the
j^\text{th}
element. For example, the list
(5, 1, 2, 3, 4)
has four inversions.
A common monovariant to look out for, when dealing with tuples of numbers, is distance from the origin. In many cases, the set of transitions will either cause the distance from the origin to always decrease, often demonstrating that the end result must be
(0,0,\dots, 0)
; in some cases, this is even an invariant.
(a) and (b) (a) but not (b) (b) but not (a) Neither (a) nor (b)
Start with the set
\{2, 3, 4\}
. In each step, choose two numbers
and
b
, and replace them with
0.6a-0.8b
0.8a+0.6b
(a) Is it possible to reach the set
\{1, 3, 5\}
(b) Is it possible to reach the set
\{0, 2, 5\}
(adapted from Engel, Problem Solving Strategies, p.9)
Both (a) and (b) (b) but not (a) Neither (a) nor (b) (a) but not (b)
On the island of Camelot live 13 gray, 15 brown and 17 crimson chameleons. If two chameleons of different colors meet, they both simultaneously change color to the third color (e.g. if a gray and a brown chameleon meet each other, they both change to crimson).
(a) Is it possible that they will eventually all be the same color?
(b) Is it possible that there will eventually be the same numbers of gray, brown, and crimson chameleons?
(1984 Tournament of Towns, Problem 1)
2(12!)
The game never ends
12!
12^2
There are 12 boys seated around a big round table. They play a game with 12 cards. Initially a boy
A_1
has all the 12 cards with him.
Every minute, if any boy has 2 or more cards with him, he passes a card to the boy on the left, and a card to the boy on the right. The game ends when each and every boy has 1 and only 1 card with him.
How many minutes does it take for this game to end?
In a chess tournament there are 12 players participating. Every player has to play against every other player in the first round. Winners get 3 points , losers get -1 point and the players who tie get 1 point each.
After first round what is the sum of points of all players?
Never Always Insufficient information May or may not die, as per the knight's actions
A dragon has 100 heads. A knight can cut off 15, 17, 20, or 5 heads, respectively, with one blow of his sword. In each of these cases 24, 2, 14, or 17 new heads grow on its shoulders, respectively. If all heads are blown off, the dragon dies. Can the dragon ever die?
There is a mob of 3000 people, 1 of whom is a zombie while the other 2999 are not.
Every minute, all 3000 people form 1000 groups of three.
If there are any zombies in a group of three, all three end up zombies.
What is the probability that after 5 minutes there will be exactly 100 zombies in the mob?
Please provide up to 5 decimal places as necessary.
Cite as: Invariant Principle. Brilliant.org. Retrieved from https://brilliant.org/wiki/invariant-principle-definition/
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The Laplace transform function for the output voltage of a network is expressed in the following form:V_0(s) = (12(s+2))/(s(s+1)(s+3)(s+4))
{V}_{0}\left(s\right)=\frac{12\left(s+2\right)}{s\left(s+1\right)\left(s+3\right)\left(s+4\right)}
Determine the final value of this voltage. that is,
{\upsilon }_{0}\left(t\right)
t\to \mathrm{\infty }
f\left(\mathrm{\infty }\right)=\underset{s\to o}{lim}sF\left(s\right)
{\upsilon }_{0}\left(\mathrm{\infty }\right)=\underset{s\to o}{lim}s\left(\frac{12\left(s+2\right)}{s\left(s+1\right)\left(s+3\right)\left(s+4\right)}\right)
=\underset{s\to o}{lim}\left(\frac{12\left(s+2\right)}{s\left(s+1\right)\left(s+3\right)\left(s+4\right)}\right)
=\frac{\left(12\right)\left(2\right)}{\left(1\right)\left(3\right)\left(4\right)}
=2
Solve no.4 inverse laplace
{L}^{-1}\left\{s\mathrm{ln}\left(\frac{s}{\sqrt{{s}^{2}+1}}\right)+{\mathrm{cot}}^{-1s}\right\}
F\left(s\right)=\frac{2s}{{s}^{2}+9}
Applications of First Order Differential Equations: A dead body was found within a closed room of a house where the temperature was a constant
{24}^{\circ }C
. At the time of discovery, the core temperature of the body was determined to be
{28}^{\circ }C
. One hour later, a second measurement showed that the core temperature of the body was
{26}^{\circ }C
. The core temperature of the body at the time of death is
{37}^{\circ }C
. A. Determine how many hours elapsed before the body was found. B. What is the temperature of the body 5 hours from the death of the victim?
{x}^{\left(3\right)}+x\text{-6x'=0, x(0)=0 , x'(0)=x}\left(0\right)=7
Determine the Laplace transform of the given function f.
f\left(t\right)=\left(t-1{\right)}^{2}{u}_{2}\left(t\right)
L\left\{t-{e}^{-3t}\right\}
which of the laplace transform is
1.\right)\text{ }L\left\{t-{e}^{-3t}\right\}=\frac{1}{{s}^{2}}+\frac{1}{s-3}
2.\right)\text{ }L\left\{t-{e}^{-3t}\right\}=\frac{1}{{s}^{2}}-\frac{1}{s-3}
3.\right)\text{ }L\left\{t-{e}^{-3t}\right\}=\frac{1}{{s}^{2}}+\frac{1}{s+3}
4.\right)\text{ }L\left\{t-{e}^{-3t}\right\}=\frac{1}{{s}^{2}}-\frac{1}{s+3}
{y}^{″}+5{y}^{\prime }+4y=0
y\left(0\right)=1
{y}^{\prime }\left(0\right)=0
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Express the definite integral as an infinite series and find
Express the definite integral as an infinite series and find its value to within
{10}^{-4}
{\int }_{0}^{1}\mathrm{cos}\left({x}^{2}\right)dx
Term-by-Term Differentiation and Integration
F\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}{\left(x-c\right)}^{n}
R>0
. Then F is differentiable on
\left(c-R,c+R\right)
. Furtehermore, we can integrate and differentiate term by term. For
x\in \left(c-R,c+R\right)
{F}^{\prime }\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}n{a}_{n}{\left(x-c\right)}^{n-1}
\int F\left(x\right)dx=A+\sum _{n=0}^{\mathrm{\infty }}\frac{{a}_{n}}{n+1}{\left(x-c\right)}^{n+1}
(A any constant)
These series have the same radius of convergence R
Here we need to find the value of
F\left(x\right)={\int }_{0}^{1}\mathrm{cos}\left({x}^{2}\right)dx
we will use the expansion of the
\mathrm{cos}x
From table 2, We have Maclaurin Series
f\left(x\right)=\mathrm{cos}x=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}\frac{{x}^{2n}}{\left(2n\right)!}=1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+\dots
converges for for all x here x is replaced with
{x}^{2}
{\mathrm{cos}x}^{2}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}\frac{{\left({x}^{2}\right)}^{2n}}{\left(2n\right)!}
{\mathrm{cos}x}^{2}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}\frac{{x}^{4n}}{\left(2n\right)!}dx
{\int }_{0}^{1}{\mathrm{cos}x}^{2}dx=\sum _{n=0}^{\in }ft{\left(-1\right)}^{n}\frac{1}{\left(4n+1\right)\left(2n\right)!}
Now we need to find out F(1) error less than 0.0001
{\int }_{0}^{1}{\mathrm{cos}x}^{2}dx=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}\frac{1}{\left(4n+1\right)\left(2n\right)!}
Above is alternating series with
Use ana appropriate test to determine whether the series converges.
\sum _{k=1}^{\mathrm{\infty }}\left(\frac{k!}{{20}^{k}{k}^{k}}\right)
Evaluating series. Evaluate the following infinite series or state that the series diverges.
\sum _{k=1}^{\mathrm{\infty }}\frac{9}{\left(3k-2\right)\left(3k+1\right)}
9+\frac{117}{4}+\frac{1521}{16}+\frac{19773}{64}+...
does this series converge or diverge? If the series converges, find the sum of the series.
How to get the sum of the series
\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(4{n}^{2}-1\right)}^{2}}
\mathrm{csc}\left(x\right)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}}{x+k\pi }
\sum _{n=1}^{\mathrm{\infty }}\frac{5}{{4}^{\mathrm{ln}n}}
\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n+1}}{\sqrt{n+4}}
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{\displaystyle e^{i\omega n\Delta t}}
{\displaystyle b_{0}}
{\displaystyle {\begin{aligned}b_{0}e^{i\omega n\Delta {\rm {t}}}=b_{0}\left({\rm {\ cos\ }}\omega n\Delta t+i{\rm {\ sin\ }}\omega n\Delta t\right)=b_{0}{\rm {\ cos\ }}\omega n\Delta t+ib_{0}{\rm {\ sin\ }}\omega n\Delta t.\end{aligned}}}
{\displaystyle b_{0}}
{\displaystyle b_{0}=0.5}
{\displaystyle b_{0}=0.5}
{\displaystyle b_{0}}
{\displaystyle b_{0}}
{\displaystyle b_{0}=-{0.5}}
{\displaystyle b_{0}=-0.5}
{\displaystyle -{0.5}e^{-i\omega n\Delta t}}
{\displaystyle -{1=}e^{i\pi }}
{\displaystyle {\begin{aligned}{0.5}\left(-{1}\right)e^{i\omega n\Delta t}=0.5e^{i\pi }e^{i\omega n\Delta t}=0.5e^{i\left(\omega n\Delta t+\pi \right)}\end{aligned}}}
{\displaystyle e^{i\left(\omega n\Delta t+\pi \right)}}
{\displaystyle {\begin{aligned}e^{i\left(\omega n\Delta t+\pi \right)}={\rm {\ cos\ }}\left(\omega n\Delta t+\pi \right)+i{\rm {\ sin\ }}\left(\omega n\Delta t+\pi \right)\end{aligned}}}
{\displaystyle {\begin{aligned}e^{i\pi }={\rm {\ cos\ }}\pi +i{\rm {\ sin\ }}\pi =-{1},\end{aligned}}}
{\displaystyle \pi }
{\displaystyle \pi }
{\displaystyle e^{i\omega n\Delta t}}
{\displaystyle {18}0^{\rm {o}}}
{\displaystyle e^{i\left(\omega n\Delta t+\pi \right)}}
{\displaystyle b_{0}=-{0.5}}
{\displaystyle {\begin{aligned}-{0.5}e^{i\omega n\Delta t}=0.5e^{i\left(\omega n\Delta t+\pi \right)}=0.5{\rm {\ cos\ }}\left(\omega n\Delta t+\pi \right)+i0.5{\rm {\ sin\ }}\left(\omega n\Delta t+\pi \right)\end{aligned}}}
{\displaystyle \pi }
{\displaystyle {0.5}e^{-i\left(\omega n\Delta t+\pi \right)}}
{\displaystyle e^{i\omega n\Delta t}}
{\displaystyle {\begin{aligned}{\frac {\rm {Output}}{\rm {Input}}}={\frac {{0.5}e^{i\left(\omega n\Delta t+\pi \right)}}{e^{l\omega n\Delta {\rm {t}}}}}=0.5e^{i\pi }.\end{aligned}}}
{\displaystyle \omega }
{\displaystyle \omega =2\pi f}
{\displaystyle b_{0}=0.5}
{\displaystyle {\begin{aligned}{\frac {\rm {Output}}{\rm {Input}}}=B\left(f\right)={\frac {{0.5}e^{i\left({2}\pi fn\Delta f+\pi \right)}}{e^{i{2}\pi fn\Delta t}}}=0.5e^{i\pi }.\end{aligned}}}
{\displaystyle B\left(f\right)={|}B\left(f\right){|}e^{i\psi (f)}}
{\displaystyle {|}B\left(f\right){|}}
{\displaystyle \psi \left(f\right)}
{\displaystyle b_{0}=-{0.5}}
{\displaystyle \pi }
{\displaystyle \pi }
{\displaystyle \pi }
{\displaystyle {18}0^{\rm {o}}}
{\displaystyle b_{0}=-{0.5}}
{\displaystyle b_{0}}
{\displaystyle {\begin{aligned}B\left(f\right)={\frac {\rm {Output}}{\rm {Input}}}={\frac {b_{0}e^{i{2}\pi fn\Delta t}}{e^{j{2}\pi f\Delta t}}}=b_{0},\end{aligned}}}
{\displaystyle b_{0}}
{\displaystyle {\begin{aligned}B\left(f\right)={\frac {\rm {Output}}{\rm {Input}}}={\frac {e^{i{2}\pi f\left(n-{1}\right)\Delta t}}{e^{i{2}\pi fn\Delta t}}}=e^{-i{2}\pi f\Delta t}.\end{aligned}}}
{\displaystyle e^{-i{2}\pi f\Delta t}}
{\displaystyle e^{-i2{\pi }f\Delta t}}
{\displaystyle e^{-i{2\pi }f\Delta t}}
{\displaystyle {-2}\pi f\Delta t}
{\displaystyle {-2}\pi f\Delta t}
{\displaystyle 0\leq 2\pi f\Delta t\leq \pi }
{\displaystyle 0\leq f\leq 1/(2\Delta t)}
{\displaystyle \Delta t=0.004}
{\displaystyle 0\leq f\leq 125}
{\displaystyle b_{0}}
{\displaystyle b_{0}}
{\displaystyle e^{-i{2}\pi f\Delta t}}
{\displaystyle b_{1}Z}
{\displaystyle {\begin{aligned}B\left(f\right)={\frac {\rm {Output}}{{\rm {In}}{\rm {put}}}}={\frac {b_{1}e^{j{2}\pi f\left(n-{1}\right)\Delta f}}{e^{i{2}\pi fn\Delta {\rm {t}}}}}=b_{1}e^{-i{2}\pi f\Delta t}.\end{aligned}}}
{\displaystyle b_{0}+b_{1}Z}
{\displaystyle {\begin{aligned}{\frac {\rm {Output}}{\rm {Input}}}={\frac {b_{0}e^{i{2}\pi fn\Delta t}+b_{l}e^{i{2}\pi f\left(n-{1}\right)\Delta t}}{e^{i{2}\pi fn\Delta {\rm {t}}}}}=b_{0}+b_{1}e^{-i{2}\pi f\Delta t}.\end{aligned}}}
Retrieved from "https://wiki.seg.org/index.php?title=Magnitude_spectrum_and_phase_spectrum/en&oldid=167187"
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What are the Di parameters in the power grid model? - Murray Wiki
{\displaystyle D_{i}}
parameters in the power grid model are the damping coefficients for the different rotors; there is a typo in the equations, as each rotor has its own damping coefficient (the first equation notation is
{\displaystyle D_{1}}
, the second has to be corrected to
{\displaystyle D_{2}}
Retrieved from "https://murray.cds.caltech.edu/index.php?title=What_are_the_Di_parameters_in_the_power_grid_model%3F&oldid=6628"
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Differential Costs - Course Hero
Managerial Accounting/Differential Analysis/Differential Costs
Differential Approach to Decision-Making
Differential analysis focuses only on relevant costs.
When companies use differential analysis to help them make decisions, they focus only on the relevant costs of the various alternatives. Relevant costs are those that differ among the alternatives. Limiting analysis this way helps keep irrelevant information from becoming part of the decision-making process and inadvertently leading to a poor decision.
For example, a company that produces widgets on an assembly line may consider renting an enhanced machine. This widget-making machine would allow managers to eliminate 10 workers from the assembly line. These workers have been making widgets by hand. The wages and benefits of the 10 assembly-line workers are relevant costs to the decision-making process, as those would be eliminated with the rental of the machine. However, the overhead the company pays for the building where the assembly line is located is an irrelevant cost in this case. That cost will be the same whether the business rents the machine or not.
Renting the widget-making machine also increases relevant costs associated with the rental price of the machine plus taxes, delivery, setup, and so on. There may also be expenses associated with operating, insuring, and maintaining the machine. The company may need to hire employees who know how to operate and maintain the machine. Those expenses would reduce the comparative wage savings of eliminating 10 assembly-line jobs.
Finally, the widget-making machine will run for only a certain number of years. After that, the company will need to decide whether to rent a new one or again hire assembly-line workers.
The differential analysis approach to decision-making focuses on variables related only to relevant costs. It excludes all irrelevant cost data from the decision-making process.
Total Costs versus Differential Costs
Some costs are not relevant to the decision being made.
The differential approach of decision analysis focuses solely on analyzing the relevant costs and benefits of the various alternatives. This is in contrast to the total cost approach to decision-making, which involves consideration of all of the costs and benefits, relevant or not, to the decision-making process. When performed correctly, the two methods should provide the same answer.
In the example of a company that is deciding whether to lease a new widget-making machine, many inputs will remain the same: the number of units sold and produced, the selling price of the units, the costs of materials to produce the widgets, and fixed expenses such as the overhead of the factory where the widgets are produced. The primary differences between leasing a widget-making machine and continuing to have people make the widgets by hand will involve differences in labor costs and differences in the rental expense and upkeep associated with the machine.
A manager using the differential cost approach would compare the cost of renting the new machine (which would be $5,000 monthly) to labor savings (which would be $15,000 monthly). The manager would conclude that the company should rent the machine, as there would be a net improvement to the bottom line of
\$\rm{15{,}000}-\$\rm{5}{,}{000}=\$\rm{10}{,}{000}
Sample Analysis Showing Differential and Total Costs
This example shows the process many companies go through when looking at differential and total costs. Renting a machine would cost $5,000 per month in fixed expenses but would save $15,000 per month in variable expenses.
A total cost approach would account for figures that do not change either way, such as the direct materials expense, fixed overhead for the factory, and selling price of the widgets produced. Therefore, the same decision is reached regardless of which method of analysis the manager uses. However, using the differential approach of only considering relevant costs involves fewer calculations.
The total cost approach includes factors that do not change the costs associated with each possible decision. Although this approach is more thorough, it takes more time and increases the chance of error.
Using relevant costs when possible makes the most sense for a business that is applying differential analysis. The managers are more likely to have on hand, or be able to easily obtain, the information needed to compare relevant costs across the alternative choices. Making considerations based on the total costs would require a full income statement's worth of information for every alternative. Also, focusing on just the relevant data helps to hone the decision-making process and makes sure managers are not considering irrelevant information that may lead to poor decision-making.
<Decision-Making Using Differential Analysis>Adding or Dropping Product Lines
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Automatic Differentiation Background - MATLAB & Simulink - MathWorks Deutschland
Automatic Differentiation in Optimization Toolbox
Automatic differentiation (also known as autodiff, AD, or algorithmic differentiation) is a widely used tool in optimization. The solve function uses automatic differentiation by default in problem-based optimization for general nonlinear objective functions and constraints; see Automatic Differentiation in Optimization Toolbox.
Forward mode automatic differentiation evaluates a numerical derivative by performing elementary derivative operations concurrently with the operations of evaluating the function itself. As detailed in the next section, the software performs these computations on a computational graph.
As many researchers have noted (for example, Baydin, Pearlmutter, Radul, and Siskind [1]), for a scalar function of many variables, reverse mode calculates the gradient more efficiently than forward mode. Because an objective function is scalar, solve automatic differentiation uses reverse mode for scalar optimization. However, for vector-valued functions such as nonlinear least squares and equation solving, solve uses forward mode for some calculations. See Automatic Differentiation in Optimization Toolbox.
f\left(x\right)={x}_{1}\mathrm{exp}\left(-\frac{1}{2}\left({x}_{1}^{2}+{x}_{2}^{2}\right)\right).
\begin{array}{c}\frac{df}{d{x}_{1}}=\frac{d{u}_{6}}{d{x}_{1}}\\ =\frac{\partial {u}_{6}}{\partial {u}_{-1}}+\frac{\partial {u}_{6}}{\partial {u}_{5}}\frac{\partial {u}_{5}}{\partial {x}_{1}}\\ =\frac{\partial {u}_{6}}{\partial {u}_{-1}}+\frac{\partial {u}_{6}}{\partial {u}_{5}}\frac{\partial {u}_{5}}{\partial {u}_{4}}\frac{\partial {u}_{4}}{\partial {x}_{1}}\\ =\frac{\partial {u}_{6}}{\partial {u}_{-1}}+\frac{\partial {u}_{6}}{\partial {u}_{5}}\frac{\partial {u}_{5}}{\partial {u}_{4}}\frac{\partial {u}_{4}}{\partial {u}_{3}}\frac{\partial {u}_{3}}{\partial {x}_{1}}\\ =\frac{\partial {u}_{6}}{\partial {u}_{-1}}+\frac{\partial {u}_{6}}{\partial {u}_{5}}\frac{\partial {u}_{5}}{\partial {u}_{4}}\frac{\partial {u}_{4}}{\partial {u}_{3}}\frac{\partial {u}_{3}}{\partial {u}_{1}}\frac{\partial {u}_{1}}{\partial {x}_{1}}.\end{array}
{\stackrel{˙}{u}}_{i}
{\stackrel{˙}{u}}_{i}
To compute the partial derivative with respect to x2, you traverse a similar computational graph. Therefore, when you compute the gradient of the function, the number of graph traversals is the same as the number of variables. This process can be slow for many applications, when the objective function or nonlinear constraints depend on many variables.
Reverse mode uses one forward traversal of a computational graph to set up the trace. Then it computes the entire gradient of the function in one traversal of the graph in the opposite direction. For problems with many variables, this mode is far more efficient.
{\overline{u}}_{i}=\frac{\partial f}{\partial {u}_{i}}.
\begin{array}{c}\frac{\partial f}{\partial {u}_{-1}}=\frac{\partial f}{\partial {u}_{1}}\frac{\partial {u}_{1}}{\partial {u}_{-1}}+\frac{\partial f}{\partial {u}_{6}}\frac{\partial {u}_{6}}{\partial {u}_{-1}}\\ ={\overline{u}}_{1}\frac{\partial {u}_{1}}{\partial {u}_{-1}}+{\overline{u}}_{6}\frac{\partial {u}_{6}}{\partial {u}_{-1}}.\end{array}
{u}_{1}={u}_{-1}^{2}
{\overline{u}}_{-1}={\overline{u}}_{1}2{u}_{-1}+{\overline{u}}_{6}{u}_{5}.
{\overline{u}}_{6}=\frac{\partial f}{\partial f}=1
, the reverse mode computation obtains the adjoint values for all variables. Therefore, reverse mode computes the gradient in just one computation, saving a great deal of time compared to forward mode.
f\left(x\right)={x}_{1}\mathrm{exp}\left(-\frac{1}{2}\left({x}_{1}^{2}+{x}_{2}^{2}\right)\right).
{\overline{u}}_{0}=\frac{\partial f}{\partial {u}_{0}}=\frac{\partial f}{\partial {x}_{2}}
{\overline{u}}_{-1}=\frac{\partial f}{\partial {u}_{-1}}=\frac{\partial f}{\partial {x}_{1}}
[1] Baydin, Atilim Gunes, Barak A. Pearlmutter, Alexey Andreyevich Radul, and Jeffrey Mark Siskind. "Automatic Differentiation in Machine Learning: A Survey." The Journal of Machine Learning Research, 18(153), 2018, pp. 1–43. Available at https://arxiv.org/abs/1502.05767.
solve | prob2struct
Books on Automatic Differentiation
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To calculate: The single radical of the expression \sqrt{5}\cdot\sqrt[3]
\sqrt{5}\cdot \sqrt[3]{4}
\sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}
\sqrt[n]{a}={a}^{1/n}
\sqrt{5}\cdot \sqrt[3]{4}=\left(5{\right)}^{\frac{1}{2}×\frac{3}{3}}\cdot \left(4{\right)}^{\frac{1}{3}×\frac{2}{2}}
={\left(5\right)}^{\frac{3}{6}}\cdot {\left(4\right)}^{\frac{2}{6}}
=\sqrt[6]{{5}^{3}}\cdot \sqrt[6]{{4}^{2}}
\sqrt{5}\cdot \sqrt[3]{4}=\sqrt[6]{\left(5{\right)}^{3}\left(4{\right)}^{2}}
=\sqrt[6]{125\cdot 16}
=\sqrt[6]{2000}
Domains Find the domain of the following vector-valued function.
r\left(t\right)=\sqrt{4-{t}^{2}}i+\sqrt{t}j-\frac{2}{\sqrt{1+t}}k
Which of the following functions f has a removable discontinuity at a? If the discontinuity is removable, find a function g that agrees with f for
x\ne a
and is continuous at a.
f\left(x\right)=\frac{{x}^{4}-1}{x-1}\text{ },a=1
Pets Plus and Pet Planet are having a sale on the same aquarium. At Pets Plus the aquarium is on sale for 30% off the original price and at Pet Planet it is discounted by 25%.
in a chart there is the numbers: 118 and 110
If the sales tax rate is 8%, which store has the lower sale price?
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Algorithms - CS Notes Algorithms | CS Notes
An algorithm is a procedure, designed to accomplish a specific task, that takes input and produces an output.
An algorithm should solve a general, well-specified problem. A specification for an algorithm should include the complete set of instances it will operate on, as well as the algorithm’s output after running on each of these instances [1, P. 3].
Algorithms are commonly expressed either in English, as pseudocode, or in a programming language [1, P. 12].
A good algorithm is:
It’s not always possible to meet these goals [1, P. 4].
The most important property of an algorithm is that it’s correct.
One way to demonstrate the correctness of an algorithm is a mathematical proof.
A mathematical proof consists of four parts:
A statement of what you’re trying to achieve.
A set of assumptions you take to be true.
A chain of reasoning that takes you from the assumptions to the statement you are attempting to prove.
A little square (∎) or QED to denote the end of the proof.
In order to demonstrate correctness, your problem must be well-specified.
The set of allowed input instances.
The required properties of the algorithm output.
You should avoid asking ill-defined questions. Asking “what is the best path?” is ill-defined: what does best mean? You should be more specific, for example: “which is the fastest path to take?” [1, P. 13].
You can prove an algorithm incorrect by providing an input that produces the incorrect output. These are called counter-examples [1, P. 13].
Good counter-examples are verifiable and simple [1, P. 13].
Failure to prove an algorithm as incorrect does not make it correct. Mathematical induction is a common method to prove the correctness of an algorithm.
The way to prove a predicate
P
through induction is to:
Prove the case
P(0)
P(k)
P(k+1)
Note: see this video teaching proof by induction for further explanation.
Summations are common in algorithm analysis.
Sigma notation is a way of expressing summation formulas. For example, the sum of 1 to
n
in sigma notation is:
Note: see this video explaining sigma notation for further explanation.
Modelling is the process of formulating your application in terms of well-defined, well-understood problems. Modelling can eliminate the need to create your own algorithm, since you can rephrase your problem to use a pre-written algorithm [1, P. 19].
Most problems are real-world problems. For example, you might need to create a system to route traffic. Algorithms don’t work on real-world objects, they work abstractions, like a graph. In order to write effective algorithms you must learn how to describe your problems in terms of abstract structures [1, P. 19].
In order to model a problem, you should have a solid understanding of the data structures available to you.
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@uttertechnology
Are you Salman Khan fan and want to know about Salman Khan Net Worth 2021: His earnings, property, lifestyle and cars ? In this post, you'll learn all about Salman Khan's net worth, height, age, weight, biography, and family. You may find Salman Khan's net worth here. He is now known as the Bhaijaan of Bollywood (a nickname meaning big brother) for his extreme levels of success.
Over the course of his career, Salman Khan has earned more than 100 crore at the box office as an actor. In the span of his lifetime, he was honored with the National Film Award, the Filmfare Award, and numerous other accolades. Forbes named Khan, an Indian celebrity, as the top-earning Indian entertainer in 2018, as he earned $37.7 million that year.
One of the biggest Bollywood stars in the world, Salman Khan, has a net worth of
360 million in 2021. That is around
34 billion USD. He gets a massive Rs 200 crore from endorsements and movies. According to several sources, this is his annual take.
Salman Khan earned over
3.3 million in 2019, an amount that increased to over
5 million in 2020. Salman Khan had a movie called Biwi Ho To Aisi that grossed $22 million in 1988. He's done a lot of work on shows, as well. One of Bollywood's best flicks is his film. The Bajrangi Bhaijaan movie made by Salman Khan is the third highest-grossing film in India.
Profession Acting, Modeling,Producer
Salman Khan childhood & family
View this post on InstagramA post shared by Arpita Khan Sharma (@arpitakhansharma)
He is the eldest son of screenwriter Salim Khan and his first wife Sushila Charak, and his father is known for his collaborations with superstar actor, scriptwriter, and musician, Sanjay Dutt. His grandpa was Muslim and had moved to India, settling in Indore, Madhya Pradesh. He was originally from Afghanistan and went to live in India.
Helen, an actress in the past, who co-starred with him in some movies, is his stepmother. He is the oldest of four children, with two brothers and two sisters. The actress Alvira is married to director Atul Agnihotri, who also happens to be an actor. Though his younger brothers attended the same school, Salman and Arbaaz graduated from St. Stanislaus High School in Bandra, Mumbai. The youngest of his two brothers, Arbaaz, attended The Scindia School alongside him.
Family All Members Father – Salim Khan Mother – Sushila Charak Brother – Arbaaz Khan
Sohail Khan Sister – Arpita Khan
Salman Khan's Professional Achievements
His second film, Maine Pyar Kiya, earned him a Filmfare Award for Best Male Debut for the role he played in it, but his first big break came when he was cast in the lead in Biwi Ho To Aisi. While his movie career started to fade in the 2000s, Shah Rukh was still a star that time, acting in very popular movies like the box office hits Chak De! India (2007), Om Shanti Om (2007), and Don 2 (2011).
He is one of the highest-paid actors in India, placing in the top five. He first made his mark in Bollywood in the 1990s. When Bollywood was overwhelmed with fierce rivalry. When he started, he had the opportunity to work on projects like ‘Hum Saath Saath He'. He was beginning to be known in Bollywood because of his rising popularity. Salman Khan is expected to have a net worth of $360 million in 2021. This figure includes his movie and brand earnings.
Over the past few years, every movie Salman has starred in has earned tremendous box-office numbers. The Bollywood film “Sultan” performed especially well in 2016, after which it became the second-highest grossing movie in India. It's no surprise that Salman Khan earned a spot on the top-earning celebrities list because he's at the top of the highest paid celebs rankings.
Besides his film career, Salman hosts the Hindi TV series Big Boss and performs in theatrical shows. Salman Khan's net worth has increased significantly as a result of his success as a Bigg Boss host. From Bigg Boss season 4 to 6, his average earnings every episode were over 2.5 crores.
As Bigg Boss's fame and ratings soared, Salman Khan boosted his fee to 7-8 crores per episode. For the 11th season, this figure was upped to 11 crores each episode. A television host in India, he was the highest-paid.
Salman Khan personal favorites
Sylvester Stallone, Dilip Kumar, Govinda & Dharmender
Favourite restaurant Cafe Noorani at Haji Ali (Mumbai)
Favorite Outfit T-shirt and Jeans
Favorite Perfume Obsession by Calvin Klein
Favorite Brands Gianni Versace and Giorgio Armani
Favorite Song Tu Hi Tu (Kick)
Favorite Cricketer Harbhajan Singh, Yuvraj Singh and Ashish Nehra
Hobbies Painting, Writing, Travelling, Acting, Life Enjoying, Riding
Assets & Properties of Salman Khan in India
The Galaxy Apartment in Bandra is 16 crore property where Salman Khan's family lives.
Owns a triplex apartment worth around Rs 80 crores in another portion of Bandra.
When he reached 50, he bought a yacht that is now worth Rs 3 Crores, and he got it in 2016.
Credits Filmfare- He owns a farm house in Panvel spread accross 150 acres and has 3 bungalows approximately 100 million rupees are worth of the property.
- Not only that, the actor also has a 5 BHK Gorai Beach farmhouse sprawled over 100 acres. The farmhouse is surrounded by a gym, pool and a theater hall.
Credits India Today- Khan owns a production firm and a distribution company, the latter of which handles the distribution of the films he produces. Among SKF's productions are Dr. Cabbie, Dabanng, Bajrangi Bhaijaan, and Race 3.
Dr. Cabbie, the first film produced by SKF, was shot in Canada. The picture earned $350,452 on opening day, and was the second highest-earning movie in Canada.
Love Yatri, a film in which Ayush Sharma and Warina Hussain play main roles, was made by the actor in his sister's husband's capacity as producer.
The Salman Khan Being Human Productions production firm was started in 2011.
The awards earned by Salman Khan
Salman debuted as an actor in Maine Pyar Kiya in 1990 and got the Best Male Debut award for his performance.
He got the Best Supporting Actor award for his role in Kuch Kuch Hota Hai in 1999.
Bajrangi Bhaijaan, in 2016, received the award for Best Wholesome Entertainment, and it also won the award for Best Children's Film in 2012.
He was additionally recognized in 2008 with the Rajiv Gandhi Award for exceptional accomplishment in entertainment.
Salman Khan's Motorcycle and Car Collection
Salman Khan has a massive collection of automobiles and motorcycles. He's far more passionate about cars than motorcycles. Millions of rupees worth of fancy cars belong to him. It includes most well-known companies in the industry: the Lexus LX 470, Mercedes Benz GL-Class, BMW X5, Range Rover Vogue, BMW X6, Audi R8, Audi Q7, W221 Mercedes Benz S-Class and many others.
On the day the RS7 was launched, he made his purchase. He really likes Audis. He personally announced this during an interview.
Additionally, he owns a stellar bike collection. He has a variety of bikes, from more expensive manufacturers like Suzuki Intruder M1800 RZ limited edition motorbike and Suzuki Hayabusa and others
The bicycle he has is valued approximately Rs 4.32 lakhs and is named the Giant Propel 2014 XTC.
Salman Khan had Aishwarya Rai, Sangeeta Bijlani, Somi Ali, Katrina Kaif, Faria Alam, and Lulia Vantur.
## Some facts about Salman Khan
Despite having over 50 years of age and 25 years in Bollywood, Salman is single, and the national news dubs him the nation's National Bachelor.
It is reported that Aamir Khan's residence includes various paintings done by Salman Khan. He also had his artwork on the movie Jai Ho promotional posters.
When he showers, Salman Khan makes use of multiple kinds of soap.
The fact that Salman and his father wear matching tortoise stone bracelets, and view them as highly beneficial to them, reveals a lot about their personal history.
In the film Baazigar, Salman Khan was approached to portray a villain, but he declined, and his character was eventually offered to Shahrukh.
Salman's father hoped that he would grow up to be a cricket player, but Salman's ambitions were different; he was striving to become a writer. He wrote Veer and Chandramukhi, the movie.
Salman Khan's Donations
Khan proposed to free 400 convicts in January 2012 in exchange for *4 million via his non-governmental organization. Their charges had previously been resolved, and they were unable to pay their legal fines due to financial constraints. It is a not-for-profit organization dedicated to education and healthcare. Salman Khan established the foundation in the form of a registered trust.
Khan contributed the most money of any Bollywood actor, contributing Rs 12 crore to flood victims in Kerala.
Read More | Sanjay Dutt Net Worth 2021: Car, Salary, Income, Assets, Bio
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Category inductance +> CalculatePlus
Category inductance
Inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it. The flow of electric current creates a magnetic field around the conductor. The field strength depends on the magnitude of the current, and follows any changes in current. From Faraday's law of induction, any change in magnetic field through a circuit induces an electromotive force (EMF) (voltage) in the conductors, a process known as electromagnetic induction. This induced voltage created by the changing current has the effect of opposing the change in current. This is stated by Lenz's law, and the voltage is called back EMF.
Inductance is defined as the ratio of the induced voltage to the rate of change of current causing it. It is a proportionality factor that depends on the geometry of circuit conductors and the magnetic permeability of nearby materials. An electronic component designed to add inductance to a circuit is called an inductor. It typically consists of a coil or helix of wire.
The term inductance was coined by Oliver Heaviside in 1886. It is customary to use the symbol
{\displaystyle L}
for inductance, in honour of the physicist Heinrich Lenz. In the SI system, the unit of inductance is the henry (H), which is the amount of inductance that causes a voltage of one volt, when the current is changing at a rate of one ampere per second. It is named for Joseph Henry, who discovered inductance independently of Faraday.
Latest from category inductance
4 aH to ZH (attohenry to zettahenry)
500 uH to yoctohenry (microhenry to yH)
50 mH to aH (millihenry to attohenry)
4 uH to centihenry (microhenry to cH)
5 nanohenry to henry (nH to H)
5 henry to millihenry (H to mH)
10 dH to henry (decihenry to H)
100 mH to henry (millihenry to H)
20 uH to myriahenry (microhenry to myH)
microhenry to henry (μH to H)
4 henry to microhenry (H to μH)
5 H to decahenry (henry to daH)
millihenry to decahenry (mH to daH)
100 H to mH (henry to millihenry)
200 MH to H (megahenry to henry)
2 petahenry to kH (PH to kilohenry)
YH to hH (yottahenry to hectohenry)
8 EH to attohenry (exahenry to aH)
0.21 MH to millihenry (megahenry to mH)
microhenry to millihenry (μH to mH)
1,200 uH to H (microhenry to henry)
9 microhenry to nH (μH to nanohenry)
1,000 uH to mH (microhenry to millihenry)
2 uH to mH (microhenry to millihenry)
6 nanohenry to pH (nH to picohenry)
5 mH to gigahenry (millihenry to GH)
30 MH to millihenry (megahenry to mH)
6 dH to exahenry (decihenry to EH)
10 GH to H (gigahenry to henry)
9 nanohenry to microhenry (nH to μH)
5 microhenry to millihenry (μH to mH)
10 mH to uH (millihenry to microhenry)
2 microhenry to dH (μH to decihenry)
330 uH to MH (microhenry to megahenry)
330 nH to ZH (nanohenry to zettahenry)
0.02 MH to H (megahenry to henry)
daH to zeptohenry (decahenry to zH)
uH to pH (microhenry to picohenry)
fH to dH (femtohenry to decihenry)
4 microhenry to henry (μH to H)
daH to decihenry (decahenry to dH)
yoctohenry to henry (yH to H)
4 fH to H (femtohenry to henry)
10 GH to aH (gigahenry to attohenry)
95 TH to MH (terahenry to megahenry)
9 henry to uH (H to microhenry)
68 nH to uH (nanohenry to microhenry)
10 uH to cH (microhenry to centihenry)
MH to pH (megahenry to picohenry)
680 nH to mH (nanohenry to millihenry)
1,000 microhenry to henry (μH to H)
9 dH to fH (decihenry to femtohenry)
330 nH to uH (nanohenry to microhenry)
1 pH to petahenry (picohenry to PH)
7 yottahenry to microhenry (YH to μH)
MH to GH (megahenry to gigahenry)
8 henry to petahenry (H to PH)
nH to H (nanohenry to henry)
4 aH to MH (attohenry to megahenry)
8 millihenry to henry (mH to H)
1 uH to H (microhenry to henry)
6 nanohenry to TH (nH to terahenry)
9 daH to decihenry (decahenry to dH)
3 cH to henry (centihenry to H)
4 zeptohenry to mH (zH to millihenry)
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The dynamics of a nonautonomous oscillator with friction memory | JVE Journals
L. A. Igumnov1 , V. S. Metrikin2 , M. V. Zaytzev3
3National Research Lobachevsky State University of Nizhni Novgorod, Nizhny Novgorod, Russia
In this article, an oscillator is considered in which one of its component moves periodically and for which static friction is taken to be time-dependent. The dynamics of the oscillator are analyzed using Poincare map to find shifts between periodic and chaotic motion with the change of the model parameters (amplitude and frequency of velocity of the periodic motion and the coefficient of static friction as a function of the time of stationary contact).
Keywords: mathematical model, Poincare map, bifurcation diagram, time-dependent static friction, chaos.
A. Y. Ishlinskiy and I. V. Kregelskiy [1] presented a hypothesis that the coefficient of static friction is not a constant, but a monotonously increasing function of the time of the contact between two bodies. After a substantial delay, it attracted the attention of scientists working on systems with friction [2-4]. It has been shown, that systems with time-dependent static friction can display different types of dynamic behavior, including complex periodic motion and chaos. Similar systems to the one studied here have also been studied in other works (e.g., [5-7]), but they did not use time-dependent static friction and used the classic model of static friction instead.
The model considered in this article is depicted in Fig. 1(a).
Fig. 1. Physical model of a vibration system
The model consists of a mass
m
, that rides on a moving belt with dry friction. The belt moves with periodically changing velocity
V\left(t\right)
. The mass is attached to inertial space by a spring with stiffness
k
. The coefficient of static friction, in accordance with the hypothesis of Ishlinskiyi A. Yi., Kragelskiyi I. V. [1], is considered to be a continuous monotonically increasing function of the time
{t}_{k}
of stationary contact (stick motion of the mass and the belt) as depicted in Fig. 1(b). The Coulomb model is used for non-static dry friction.
The motion of the mass in the model is described by the Eq. (1), (2):
m\stackrel{˙}{x}=-kx-{f}_{\mathrm{*}}P\mathrm{s}\mathrm{i}\mathrm{g}\mathrm{n}\left(\stackrel{˙}{x}-V\left(t\right)\right), \stackrel{˙}{x}\ne V\left(t\right)\mathrm{ },
\left|kx+m\stackrel{˙}{V}\left(t\right)\right|\le f\left({t}_{k}\right)P, \stackrel{˙}{x}=V\left(t\right).
Eq. (1) describes the slip motion of the mass with non-static dry friction with coefficient
{f}_{*}
. Eq. (2) describes the condition for the stick motion.
{f}_{\mathrm{п}}\left({t}_{k}\right)
is the coefficient of static friction as a function of the stick motion time
{t}_{k}
V\left(t\right)=const
, the Eq. (1) takes the form of the corresponding equation in [4].
With dimensionless time
\tau =t{\omega }_{0}
\xi =xk/{f}_{\mathrm{*}}P
{\omega }_{0}=\sqrt{k/m}
\theta \left(\tau \right)=V\sqrt{km}/{f}_{\mathrm{*}}P
, the Eq. (1), (2) will take the form:
\stackrel{¨}{\xi }+\xi =-\mathrm{s}\mathrm{i}\mathrm{g}\mathrm{n}\left(\stackrel{˙}{\xi }-\theta \left(\tau \right)\right), \stackrel{˙}{\xi }\ne \theta \left(\tau \right),
\left|\xi +\stackrel{˙}{\theta }\left(\tau \right)\right|\le 1\mathrm{ }+\epsilon \left({\tau }_{k}\right), \stackrel{˙}{\xi }=\theta \left(\tau \right),
\epsilon \left({\tau }_{k}\right)=\left({f}_{\mathrm{п}}\left({\tau }_{k}\right)-{f}_{\mathrm{*}}\right){{f}_{\mathrm{*}}}^{-1}
– dimensionless coefficient of static friction).
1.2. Structure of the phase space
Since the system is non-autonomous and can be described by second-order differential equation, its state is a triplet
\left\{\xi ,\stackrel{˙}{\xi },\tau \right\}
(Fig. 2), and its phase space is accordingly 3-dimensional. The phase space is split into 2 half-spaces
{\mathrm{\Phi }}_{1}\left(\xi ,\stackrel{˙}{\xi }>\theta ,\tau \right)
{\mathrm{\Phi }}_{2}\left(\xi ,\stackrel{˙}{\xi }<\theta ,\tau \right)
by the surface
\mathrm{\Pi }\left(\stackrel{˙}{\xi }=\theta \left(\tau \right)\right)
The shape of phase trajectories in each half-space is described by the Eq. (5), (6):
\stackrel{¨}{\xi }+\xi =-1,\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\stackrel{˙}{\xi }>\theta \left(\tau \right),
\stackrel{¨}{\xi }+\xi =+1,\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\stackrel{˙}{\xi }<\theta \left(\tau \right).
It can be shown, that on the surface
П
there is a strip of stick motions
{\mathrm{П}}_{C}
bounded by curves
{\mathrm{Г}}_{1}
{\mathrm{Г}}_{2}
{\mathrm{\Gamma }}_{1}:\left\{\begin{array}{l}\xi =1-\stackrel{˙}{\theta }\left(\tau \right),\\ \stackrel{˙}{\xi }=\theta \left(\tau \right),\end{array}\right\ {\mathrm{\Gamma }}_{2}:\left\{\begin{array}{l}\xi =-1-\stackrel{˙}{\theta }\left(\tau \right),\\ \stackrel{˙}{\xi }=\theta \left(\tau \right).\end{array}\right\
On the strip
{\mathrm{П}}_{C}
a phase trajectory is described by Eq. (8):
\xi \left(\tau \right)=\stackrel{\tau }{\underset{{\tau }_{\mathrm{п}}}{\int }}\theta \left(\eta \right)d\eta +{\xi }_{\mathrm{п}},\left\{{\xi }_{\mathrm{п}},{\tau }_{\mathrm{п}}\right\}\in {\mathrm{П}}_{C}.
An example of a phase space trajectory with stick motion intervals is shown in Fig. 3.
Fig. 2. Phase space
S
Fig. 3. The qualitative form of the phase trajectories
1.3. Dynamics of the system
For the purpose of modeling the dynamics of system Eqs. (1-2), the function for the dimensionless belt speed was taken as follows:
\theta \left(\tau \right)=A\mathrm{c}\mathrm{o}\mathrm{s}\left(\mathrm{\Omega }\tau \right)+B
\epsilon \left({\tau }_{k}\right)
for the dimensionless coefficient of static friction was taken as follows:
\epsilon \left({\tau }_{k}\right)=\left\{\begin{array}{l}{\tau }_{k},\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\tau }_{k}\le {\epsilon }_{\mathrm{*}},\\ {\epsilon }_{\mathrm{*}},\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\tau }_{k}\ge {\epsilon }_{\mathrm{*}}.\end{array}\right\
{M}_{i}\left({\tau }_{i},{\xi }_{i}\right)
i=0,1,..,n
– a sequence of point on the surface
П
, which does not belong to the strip
{\mathrm{П}}_{C}
and which is determined by the Eq. (5) for
i=2k<n
k=
1, 2,… and Eq. (6) for
i=2m+1<n
m=\mathrm{ }
0, 1,…. Suppose as well, that the coordinates of
{M}_{0}
\tau ={\tau }_{0}
\xi =1+\epsilon \left({\tau }_{k,c}\right)
\stackrel{˙}{\xi }=\theta \left({\tau }_{0}\right)
. For such sequence, a number n can be found, for which
{M}_{n+1}\left({\tau }_{p},{\xi }_{p}\right)
will belong to the strip
{\mathrm{П}}_{C}
and the interval of phase trajectory is described by Eq. (9) until the point determined from condition Eq. (4). Let’s take
{Τ}_{+}
to denote the mapping
{M}_{2k+1}\to {M}_{2k+2}
k=\text{0,}\text{ }\text{1,}\text{ }\text{2,}\text{…}<n
{Τ}_{-}
– the mapping
{M}_{2m}\to {M}_{2m+1}
m=\text{1, }\text{2,}\text{…}<n
{M}_{0}\to {M}_{n+1}
can thus be described by
{Τ}_{1}\left(j,l,n\right)={\left({\left({Τ}_{-}\right)}^{j}{\left({Τ}_{+}\right)}^{l}\right)}^{\left[\frac{n}{2}\right]}
l
j=\mathrm{ }
n
. The connection between two adjacent stick motion intervals
{\tau }_{k}
{\tau }_{k+1}
can be described as follows
\psi \left({\tau }_{k+1}\right)=\phi \left({\tau }_{k}\right)
\psi \left(\tau \right)=-\epsilon \left(\tau \right)+\frac{A}{\mathrm{\Omega }}\left(\mathrm{s}\mathrm{i}\mathrm{n}\left(\mathrm{\Omega }\tau \right)-\mathrm{s}\mathrm{i}\mathrm{n}\left(\mathrm{\Omega }{\tau }_{p}\right)\right)-A\mathrm{\Omega }\mathrm{s}\mathrm{i}\mathrm{n}\left(\mathrm{\Omega }\tau \right)+B\left(\tau -{\tau }_{p}\right),
\phi \left(\tau \right)=1+\left(-1{\right)}^{j+1}{\xi }_{p}\left({\tau }_{0},{\tau }_{1},...,{\tau }_{n},\tau \right),\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }2\left(j-1\right)\le \epsilon \left(\tau \right)\le 2,\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }j=1,2,\dots .
Since the system state trajectory will always end on the sliding surface, they dynamics of the system can be examined using Poincare map of the boundary
{\mathrm{Г}}_{1}\left({\mathrm{Г}}_{2}\right)
into itself, or using the sequence of durations of stick motions
{\tau }_{k}
k=\mathrm{ }
1, 2, 3… To study the dynamics of the system, we developed an application allowing to perform calculations of trajectories of the system, the Poincare maps, and bifurcation diagrams, using different values of the system parameters. All calculations are done using double precision floating point arithmetic. When calculating a single trajectory, different points on the trajectory are calculated with progressively increasing time distance from the initial point, until the trajectory reaches a boundary between areas where the system behavior is described by different equations, e.g. when the trajectory reaches the surface
\mathrm{\Pi }\left(\stackrel{˙}{\xi }=\theta \left(\tau \right)\right)
. Then the accuracy of the calculated point of the transition to a different area is further improved using bisection method, until the desired accuracy is reached. After that, a new trajectory piece is calculated the same way starting from the calculated transition point as the initial point. Trajectory pieces are calculated until the time or the number of stick motion interval reaches the defined limit.
To build a bifurcation diagram, one of the parameters is selected to be varied and its range is selected. Then for each value in the range its own trajectory is calculated, with all the other parameters having the same values for each trajectory. To build the graph, a certain number (usually 30) of last intervals of stick motion are taken and displayed for each trajectory.
1.4. The results of numerical calculations
Fig. 4 displays a bifurcation diagram with
\mathrm{Ω}
as the variable parameter. The other parameters’ values are:
\theta \left(\tau \right)=0.1\mathrm{c}\mathrm{o}\mathrm{s}\left(\mathrm{\Omega }\tau \right)+1.41
{\epsilon }_{*}= \text{3}.
Fig. 5(a) displays a diagram with parameter
Ω
range from 1 to 5, the other subfigures show different subintervals of that.
Fig. 4 shows that depending on the frequency of the belt movement, the system can display different types of dynamic behavior, including chaos. It should be noted, that with the increase of the frequency chaos become less frequent.
Fig. 5 displays diagrams with
A
as the variable parameter and
\theta \left(\tau \right)=A\mathrm{c}\mathrm{o}\mathrm{s}\left(\mathrm{\Omega }\tau \right)+1.41
{\epsilon }_{\mathrm{*}}=\mathrm{ }
\mathrm{\Omega }=\mathrm{ }
1.466 on Fig. 5(a) and
\mathrm{\Omega }=\mathrm{ }
3.23 on Fig. 5(b).
Fig. 5 shows the influence of the increase of the amplitude of the belt oscillations on the dynamics of the system. Two conclusions can be made based on it. One, the increase of the amplitude from zero and within a small vicinity of zero changes the dynamics of the system gradually, without sudden qualitative changes.
Fig. 4. Bifurcation diagrams in frequency
Ω
tape oscillation
Fig. 5. The bifurcation diagram of the amplitude
A
fluctuations tape
This paper studied the dynamics of a vibration system with hereditary friction. Under the influence of an external periodic force. The obtained bifurcation diagrams of the system revealed the main regimes of periodic and stochastic motions as functions of the governing parameters of the system:
1) As the amplitude of the external force increases, the bifurcation diagrams get “blurred” in relation to the case of no external force. This indicates the appearance of chaos, limited by the amplitude of the external force. As the amplitude of the external force grows further, periodic motions are born with one long stop. In the last case the system behaves as if there was no external force.
2) Increase of the frequency of the external force leads to chaotization of the motions. High frequencies lead to the birth of periodic motions with a finite number of long stops
Ishlinskiyi A. Yi., Kragelskiyi I. V. About racing in friction. Journal of Technical Physics, Vol. 4/5, Issue 14, 1944, p. 276-282, (in Russian). [Search CrossRef]
Kahenevskiyi L. Ia. Stochastic auto-oscillations with dry friction. Inzh-fiz Journal, Vol. 47, Issue 1, 1984, p. 143-147, (in Russian). [Search CrossRef]
Vetukov M. M., Dobroslavskiyi S. V., Nagaev R. F. Self-oscillations in a system with dry friction characteristic of hereditary type. Proceedings of the USSR Academy of Solid Mechanics, Vol. 1, 1990, p. 23-28, (in Russian). [Search CrossRef]
Metrikin V. S., Nagaev R. F., Stepanova V. V. Periodic and stochastic self-oscillations in a system with dry friction hereditary type. Journal Applied Mathematics and Mechanics, Vol. 5, Issue 60, 1996, p. 859-864, (in Russian). [Search CrossRef]
Leine R. I., van Campen D. H., Kraker A. D. E. Stick-slip vibrations induced by alternate friction models. Journal Nonlinear Dynamics, Vol. 16, 1998, p. 41-54. [Search CrossRef]
Leine R. I., van Campen D. H., Kraker A. D. E. An approximate analysis of dryfriction-induced stick-slip vibrations by a smoothing procedure. Journal Nonlinear Dynamics, Vol. 19, 1999, p. 157-169. [Search CrossRef]
Leine R. I., van Campen D. H. Discontinuous fold bifurcations in mechanical systems. Archive of Applied Mechanics, Vol. 72, 2002, p. 138-146. [Search CrossRef]
Feygin M. I. Forced Oscillations of Systems with Discontinuous Nonlinearities. Science, 1994, p. 285, (in Russian). [Search CrossRef]
Neymark Iy. I. The Method of Point Mappings in the Theory of Nonlinear Oscillations. Science, 1972, p. 471, (in Russian). [Search CrossRef]
Shuster G. Deterministic Chaos. Peace, 1988, p. 237, (in Russian). [Search CrossRef]
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Water flows through a water hose at a rate of Q_{1}=680cm^{3}/s, the diameter of the hose is d_{1}=2.2cm. A nozzle is attached to the water hose.
{Q}_{1}=680c\frac{{m}^{3}}{s}
, the diameter of the hose is
{d}_{1}=2.2cm
. A nozzle is attached to the water hose. The water leaves the nozzle at a velocity of
{v}_{2}=9.2\frac{m}{s}
a) Enter an expression for the cross-sectional area of the hose,
{A}_{1}
, in terms of its diameter,
{d}_{1}
b) Calculate the numerical value of
{A}_{1},
in square centimeters.
c) Enter an expression for the speed of the water in the hose,
{v}_{1}
, in terms of the volume floe rate
{Q}_{1}
{A}_{1}
d) Calculate the speed of the water in the hose,
{v}_{1}
in meters per second.
e) Enter an expression for the cross-sectional area of the nozzle,
{A}_{2}
{v}_{1},{v}_{2}
{A}_{1}
f) Calculate the cross-sectional area of the nozzle,
{A}_{2}
{A}_{1}=\pi ×{d}_{1}^{\frac{2}{4}}
{A}_{1}=\pi ×{2.2}^{\frac{2}{4}}
=3.80c{m}^{2}
{Q}_{1}={A}_{1}×{v}_{1}⇒{v}_{1}=\frac{{Q}_{1}}{{A}_{1}}
{v}_{1}=\frac{680}{3.80}
=179c\frac{m}{s}
=1.79\frac{m}{s}
e) Using continuty equation,
{A}_{2}×{v}_{2}={A}_{1}×{v}_{1}
{A}_{2}={A}_{1}×\frac{{v}_{1}}{{v}_{2}}
{A}_{2}\right\}=3.80×\frac{1.79}{9.2}
=0.739c{m}^{2}
Makes inferences about populations using data drawn from the population. Instead of using the entire population to gather the data, the statistician will collect a sample or samples from the millions of residents and make inferences about the entire population using the sample. histogram population mean descriptive statistics inferential statistics
List the assumptions necessary for each of the following inferential techniques: a. Large-sample inferences about the difference
\left({\mu }_{1}-{\mu }_{2}\right)
between population means using a two-sample z-statistic b. Small-sample inferences about
\left({\mu }_{1}-{\mu }_{2}\right)
using an independent samples design and a two-sample t-statistic c. Small-sample inferences about
\left({\mu }_{1}-{\mu }_{2}\right)
using a paired difference design and a single-sample t-statistic to analyze the differences d. Large-sample inferences about the differences
\left({\mu }_{1}-{\mu }_{2}\right)
between binomial proportions using a two sample z-statistic e. Inferences about the ratio
\frac{{\sigma }_{1}^{2}}{{\sigma }_{2}^{2}}
of two population variances using an F-test.
Given Vallias' value for
{\int }_{0}^{\sqrt{a}}{x}^{2}dx
calculate his value for
{\int }_{0}^{a}\sqrt{x}dx,
using graph.
Test whether the claim that the two sample groups come from populations with the same mean or not.
In a one-way ANOVA, what does it mean to reject the statement in the null hypothesis if three treatment groups are being compared?
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Difference between revisions of "CFD Simulations AC2-10" - KBwiki
Difference between revisions of "CFD Simulations AC2-10"
m (Dave.Ellacott moved page Lib:CFD Simulations AC2-10 to CFD Simulations AC2-10 over redirect)
For this compressible case, the filtered governing equations for mass, momentum and total enthalpy
{\displaystyle {(H=h(T,p)+0.5u_{j}^{2})}}
are solved. They are defined as:
{\displaystyle {{\dfrac {\partial {\overline {\rho }}}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}=0}}
{\displaystyle {\text{(4.1)}}}
{\displaystyle {{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{i}\right)}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{i}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}=-{\dfrac {\partial {\overline {p}}}{\partial x_{i}}}+{\dfrac {\partial }{\partial x_{j}}}\left(\left(\mu +\mu _{t}\right){\dfrac {\partial {\widetilde {u}}_{i}}{\partial x_{j}}}\right)}}
{\displaystyle {\text{(4.2)}}}
{\displaystyle {{\dfrac {\partial \left({\overline {\rho }}{\widetilde {H}}\right)}{\partial t}}-{\dfrac {\partial {\overline {p}}}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{j}{\widetilde {H}}\right)}{\partial x_{j}}}={\dfrac {\partial }{\partial x_{j}}}\left(\lambda {\dfrac {\partial {\widetilde {T}}}{\partial x_{j}}}+{\dfrac {\mu _{t}}{Pr_{t}}}{\dfrac {\partial {\widetilde {h}}}{\partial x_{j}}}\right)}}
{\displaystyle {\text{(4.3)}}}
with the thermal conductivity
{\displaystyle {\lambda }}
and the turbulent Prandtl number
{\displaystyle {Pr_{t}}}
(set to 0.9 in the following). A non-density-weighted filtered variable is denoted by
{\displaystyle {\overline {\Phi }}}
{\displaystyle {\widetilde {\Phi }}}
is the density-weighted filtered variable. This system is complemented by the filtered equation of state for perfect gases, defined as:
{\displaystyle {{\overline {p}}={\overline {\rho }}{\dfrac {R}{W}}{\widetilde {T}}}}
{\displaystyle {\text{(4.4)}}}
with the molecular weight
{\displaystyle {W}}
and the universal gas constant
{\displaystyle {R}}
. The eddy viscosity (
{\displaystyle {\mu _{t}}}
) is calculated based on the scale-adaptive simulation (SAS-SST) turbulence model [27]. In case of insufficient spatial or temporal resolution (for a SRS), it reverts to the SST model [26] and maintains a valid base of modelling. Its key element is the von Kármàn length scale [38]. (
{\displaystyle {L_{vK}}}
), introduced into the scale-determining
{\displaystyle {\omega }}
equation. In unstable flows,
{\displaystyle {L_{vK}}}
adjusts the eddy viscosity to a level which allows the formation of a turbulent spectrum [27, 45, 39, 25].
{\displaystyle {L_{vK}}}
{\displaystyle {L_{vK}=\kappa {\dfrac {\sqrt {2{\widetilde {S}}_{ij}{\widetilde {S}}_{ij}}}{{\widetilde {u}}''}}}}
{\displaystyle {\text{(4.5)}}}
{\displaystyle {{\widetilde {S}}_{ij}={\dfrac {1}{2}}\left({\dfrac {\partial {\widetilde {u}}_{i}}{\partial x_{j}}}+{\dfrac {\partial {\widetilde {u}}_{j}}{\partial x_{i}}}\right){\text{,}}\quad {\widetilde {u}}''={\sqrt {{\dfrac {\partial ^{2}{\widetilde {u}}_{i}}{\partial x_{k}^{2}}}{\dfrac {\partial ^{2}{\widetilde {u}}_{i}}{\partial x_{j}^{2}}}}}}}
{\displaystyle {\text{(4.6)}}}
and the von Kármán constant
{\displaystyle {\kappa }}
{\displaystyle {{\dfrac {\partial {\overline {\rho }}}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}=0}}
{\displaystyle {\text{(4.7)}}}
{\displaystyle {{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{i}\right)}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{i}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}={\dfrac {\partial {\overline {\tau }}_{ij}}{\partial x_{j}}}+{\dfrac {\partial \tau _{ij}^{sgs}}{\partial x_{j}}}-{\dfrac {\partial {\overline {p}}}{\partial x_{i}}}}}
{\displaystyle {\text{(4.8)}}}
{\displaystyle {{\dfrac {\partial \left({\overline {\rho }}{\widetilde {e}}\right)}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{j}{\widetilde {e}}\right)}{\partial x_{j}}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {K}}\right)}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {K}}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}={\dfrac {\partial }{\partial x_{j}}}\left(\alpha _{eff}{\dfrac {\partial {\widetilde {e}}}{\partial x_{j}}}\right)-{\dfrac {\partial }{\partial x_{j}}}\left({\overline {p}}{\widetilde {u}}_{j}\right)}}
{\displaystyle {\text{(4.9)}}}
Here, the overline denotes LES filtered and the tilde Favre filtered quantities. Further,
{\displaystyle {{\overline {\tau }}_{ij}}}
denotes the viscous stress tensor. Detailed information regarding equations 4.7–4.9 and the utilized variables can be found in [21, 22].
{\displaystyle {{\dfrac {\partial {\overline {\rho }}}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}=0}}
{\displaystyle {\text{(4.10)}}}
{\displaystyle {{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{i}\right)}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{i}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}}=-{\dfrac {\partial {\overline {P}}}{\partial x_{i}}}+{\dfrac {\partial }{\partial x_{j}}}\left(\left({\overline {\mu }}+\mu _{t}\right)\left({\dfrac {\partial {\widetilde {u}}_{i}}{\partial x_{j}}}+{\dfrac {\partial {\widetilde {u}}_{j}}{\partial x_{i}}}-{\dfrac {2}{3}}{\dfrac {\partial {\widetilde {u}}_{k}}{\partial x_{k}}}\delta _{ij}\right)\right)+{\overline {\rho }}g_{i}}
{\displaystyle {\text{(4.11)}}}
{\displaystyle {{\text{with}}\quad {\overline {P}}={\overline {p}}+{\dfrac {1}{3}}{\overline {\rho }}\tau _{kk}^{sgs}}}
{\displaystyle {{\dfrac {\partial \left({\overline {\rho }}{\widetilde {e}}\right)}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{j}{\widetilde {e}}\right)}{\partial x_{j}}}=-{\overline {p}}{\dfrac {\partial {\widetilde {u}}_{j}}{\partial x_{j}}}+{\widetilde {\tau }}_{ij}{\dfrac {\partial {\widetilde {u}}_{i}}{\partial x_{j}}}+{\dfrac {\partial }{\partial x_{j}}}\left({\dfrac {\left({\overline {\mu }}+{\dfrac {c_{v}}{c_{p}}}\mu _{t}\right)c_{p}}{Pr}}{\dfrac {\partial {\widetilde {T}}}{\partial x_{j}}}\right)}}
{\displaystyle {\text{(4.12)}}}
The internal energy
{\displaystyle {e}}
{\displaystyle {e=h-{\dfrac {p}{\rho }}=e_{0}+\int _{T_{0}}^{T}c_{v}dT-{\dfrac {p}{\rho }}\quad {\text{, with}}\quad h=h_{0}+\int _{T_{0}}^{T}c_{p}dT\quad {\text{and}}\quad c_{v}=\left.{\dfrac {\partial e}{\partial T}}\right|_{v}}}
{\displaystyle {\text{(4.13)}}}
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Forecast univariate autoregressive integrated moving average (ARIMA) model responses or conditional variances - MATLAB forecast - MathWorks India
{y}_{t}=1+0.3{y}_{t-1}+2{x}_{t}+{\epsilon }_{t},
{\epsilon }_{\mathit{t}}
{\mathit{x}}_{\mathit{t}}
{\mathit{y}}_{0}
E\left({y}_{t}\right)=\frac{1+2\left(1\right)}{1-0.3}.
\left(1,0,0\right){\left(1,1,0\right)}_{4}
\left(1,0,0\right){\left(1,1,0\right)}_{4}
\left(1-0.5L\right)\left(1-0.2{L}^{4}\right)\left(1-{L}^{4}\right){y}_{t}=1+{\epsilon }_{t},
{\epsilon }_{\mathit{t}}
\mathit{t}
\begin{array}{l}{y}_{t}=0.073+0.138{y}_{t-1}+{\epsilon }_{t}\\ {\sigma }_{t}^{2}=0.022+0.873{\sigma }_{t-1}^{2}+0.119{\epsilon }_{t-1},\end{array}
{\epsilon }_{\mathit{t}}
{\left[\begin{array}{cc}{y}_{T-K-1}& {y}_{T-K}\end{array}\right]}^{\prime }
\left[\begin{array}{ccc}{x}_{1,\left(T-K+1\right):T}& {x}_{2,\left(T-K+1\right):T}& {x}_{3,\left(T-K+1\right):T}\end{array}\right]
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Correct symbol timing clock skew - Simulink - MathWorks América Latina
Normalized timing error output port
The Symbol Synchronizer block corrects symbol timing clock skew for PAM, PSK, QAM, or OQPSK modulation schemes between a single-carrier transmitter and receiver. For more information, see Symbol Synchronization Overview.
The input signal operates on a sample rate basis, while the output signal operates on a symbol rate basis.
Input samples, specified as a scalar or column vector of a PAM, PSK, QAM, or OQPSK modulated single-carrier signal. This port in unnamed on the block.
Sym — Output signal symbols
Output signal symbols, returned as a variable-size scalar or column vector that has the same data type as the input. For an input with dimensions of Nsamp-by-1, the output at Sym has dimensions of Nsym-by-1. Nsym is approximately equal to Nsamp divided by the Nsps. Nsps is equal to the Samples per symbol parameter value. The output length is truncated if it exceeds the maximum output size of
⌈\frac{{N}_{\text{samp}}}{{N}_{\text{sps}}}×1.1⌉
This port is unnamed when Normalized timing error output port is not selected.
Err — Estimated timing error
Estimated timing error for each input sample, returned as a scalar or column vector with values in the range [0, 1]. The estimated timing error is normalized by the input sample time. Err has the same data type and size as the input signal.
To enable this port, select Normalized timing error output port.
Modulation type — Modulation type
PAM/PSK/QAM (default) | OQPSK
Modulation type, specified as PAM/PSK/QAM, or OQPSK.
Timing error detector — Type of timing error detector
Type of timing error detector, specified as Zero-Crossing (decision-directed), Gardner (non-data-aided), Early-Late (non-data-aided), or Mueller-Muller (decision-directed). This parameter assigns the timing error detection scheme used in the synchronizer.
For more information, see Timing Error Detection (TED).
Samples per symbol, specified as a positive integer greater than 1. For more information, see Nsps in Loop Filter.
Damping factor — Damping factor of the loop filter
Normalized loop bandwidth — Normalized bandwidth of loop filter
0.01 (default) | positive scalar less than 1
Normalized bandwidth of the loop filter, specified as a positive scalar less than 1. The loop bandwidth is normalized by the sample rate of the input signal. For more information, see BnTs in Loop Filter.
To ensure that the symbol synchronizer locks, set the Normalized loop bandwidth parameter to a value less than 0.1.
Detector gain — Phase detector gain
Normalized timing error output port — Enable normalized timing error output port
Select this parameter to output normalized timing error data at the output port Err.
QPSK Signal Timing Offset Correction
Correct a fixed symbol timing offset on a noisy QPSK signal by using the Symbol Synchronizer block. The number of symbols output by the Symbol Synchronizer block is variable size. If a fixed size signal is required for downstream processing, you can use a Selector (Simulink) block to convert the Symbol Synchronizer output to a fixed size signal.
Recover frame synchronization from a QPSK system impaired by a variable timing error.
x\left(k{T}_{\text{s}}+\stackrel{^}{\tau }\right)
e\left(k\right)=x\left(\left(k-1/2\right){T}_{s}+\stackrel{^}{\tau }\right)\left[{\stackrel{^}{a}}_{0}\left(k-1\right)-{\stackrel{^}{a}}_{0}\left(k\right)\right]+y\left(\left(k-1/2\right){T}_{s}+\stackrel{^}{\tau }\right)\left[{\stackrel{^}{a}}_{1}\left(k-1\right)-{\stackrel{^}{a}}_{1}\left(k\right)\right]
e\left(k\right)=x\left(\left(k-1/2\right){T}_{s}+\stackrel{^}{\tau }\right)\left[x\left(\left(k-1\right){T}_{s}+\stackrel{^}{\tau }\right)-x\left(k{T}_{s}+\stackrel{^}{\tau }\right)\right]+y\left(\left(k-1/2\right){T}_{s}+\stackrel{^}{\tau }\right)\left[y\left(\left(k-1\right){T}_{s}+\stackrel{^}{\tau }\right)-y\left(k{T}_{s}+\stackrel{^}{\tau }\right)\right]
e\left(k\right)=x\left(k{T}_{s}+\stackrel{^}{\tau }\right)\left[x\left(\left(k+1/2\right){T}_{s}+\stackrel{^}{\tau }\right)-x\left(\left(k-1/2\right){T}_{s}+\stackrel{^}{\tau }\right)\right]+y\left(k{T}_{s}+\stackrel{^}{\tau }\right)\left[y\left(\left(k+1/2\right){T}_{s}+\stackrel{^}{\tau }\right)-y\left(\left(k-1/2\right){T}_{s}+\stackrel{^}{\tau }\right)\right]
e\left(k\right)={\stackrel{^}{a}}_{0}\left(k-1\right)x\left(k{T}_{s}+\stackrel{^}{\tau }\right)-{\stackrel{^}{a}}_{0}\left(k\right)x\left(\left(k-1\right){T}_{s}+\stackrel{^}{\tau }\right)+{\stackrel{^}{a}}_{1}\left(k-1\right)y\left(k{T}_{s}+\stackrel{^}{\tau }\right)-{\stackrel{^}{a}}_{1}\left(k\right)y\left(\left(k-1\right){T}_{s}+\stackrel{^}{\tau }\right)
x\left(k{T}_{\text{s}}+\stackrel{^}{\tau }\right)
y\left(k{T}_{\text{s}}+\stackrel{^}{\tau }\right)
\stackrel{^}{\tau }
{\stackrel{^}{a}}_{0}\left(k\right)
{\stackrel{^}{a}}_{1}\left(k\right)
x\left(k{T}_{\text{s}}+\stackrel{^}{\tau }\right)
y\left(k{T}_{\text{s}}+\stackrel{^}{\tau }\right)
{K}_{1}=\frac{-4\zeta \theta }{\left(1+2\zeta \theta +{\theta }^{2}\right){K}_{p}}
{K}_{2}=\frac{-4{\theta }^{2}}{\left(1+2\zeta \theta +{\theta }^{2}\right){K}_{p}}\text{\hspace{0.17em}}.
\theta =\frac{\frac{{B}_{\text{n}}{T}_{\text{s}}}{{N}_{\text{sps}}}}{\zeta +\frac{1}{4\zeta }}\text{\hspace{0.17em}},
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{\displaystyle {\ce {Al_2(SO_4)_3}}}
{\displaystyle {\ce {H-{\overset {\displaystyle H \atop |}{\underset {| \atop \displaystyle H}{C}}}-{\overset {\displaystyle H \atop |}{\underset {| \atop \displaystyle H}{C}}}-{\overset {\displaystyle H \atop |}{\underset {| \atop \displaystyle H}{C}}}-{\overset {\displaystyle H \atop |}{\underset {| \atop \displaystyle H}{C}}}-H}}}
{\displaystyle {\ce {H-{\overset {\displaystyle H \atop |}{\underset {| \atop \displaystyle H}{C}}}-{\overset {\displaystyle H \atop |}{\underset {| \atop \displaystyle H}{C}}}-{\overset {\displaystyle H \atop |}{\underset {| \atop \displaystyle H}{C}}}-{\overset {\displaystyle H \atop |}{\underset {| \atop \displaystyle H}{C}}}-H}}}
^ "Law of Constant Composition". Everything Math and Science. SIYAVULA. Retrieved 31 March 2016.
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The state runs a lottery once every week in which six numbers are randomly selected from 16 without replacement. A playe
The state runs a lottery once every week in which six number
The state runs a lottery once every week in which six numbers are randomly selected from 16 without replacement. A player chooses six numbers before the state’s sample is selected. The player wins if all 6 numbers match. If a player enters one lottery each week, what is the probability that he will win at least once in the next 200 weeks? Report answer to 3 decimal places.
In this question, there is a lottery in which a person needs to select the 6 numbers if those 6 numbers are in state’s sample then he/she win the lottery. We need to find the probability that he will win at least once in next 200 weeks.
Let’s divide the numbers in two group,
Group 1, Those numbers which are in state’s sample
Group 2, Those numbers which are not in state’s sample
The player selects the 6 numbers he wins when all 6 numbers are in state’s sample,
Therefore, Probability of winning a lottery;
Calculations have shown below;
=\frac{6{C}_{6}×10{C}_{0}}{16{C}_{6}}
=\frac{1}{8008}
Now, probability of in next 200 lotteries a person will wins at least once;
We will use the binomial distribution to get the required probability,
P\left(x\ge 1\right)=1-P\left(x=0\right)
=1-200{C}_{0}×{\left(\frac{1}{8008}\right)}^{0}×{\left(1-\frac{1}{8008}\right)}^{200}
=0.025
Hyper geometric distribution:
A hyper geometric distribution is a discrete probability distribution that determines the probability of getting k successes in n draws (without replacement) from a finite population of size N that contains exactly K success states.
Denote the total number of successes as k,
Denote the total number of objects that are drawn without replacement as n,
Denote the population size as N,
Denote the total number of success states in the population as K.
The probability distribution of k is a hyper geometric distribution with parameters (N, K, n) and the probability mass function (pmf) of k is given as:
P\left(k\right)=\frac{\left(\begin{array}{c}K\\ k\end{array}\right)×\left(\begin{array}{c}N-K\\ n-k\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}
Parameters of hyper geometric distribution for the given situation:
In a state’s lottery, 6 numbers are randomly selected from a total of 17 numbers without replacement.
N=17
The number of draws is
n=6
Furthermore, it is given that, a player chooses 6 numbers before the state’s sample is selected and the player wins, if all the 6 numbers match with the state’s sample.
Thus, the number of success states in the population is
K=6
The probability distribution of k successes is a hyper geometric distribution with parameters (
N=17,K=6,n=6
) and the probability mass function (pmf) of k is given as:
P\left(k\right)=\frac{\left(\begin{array}{c}6\\ k\end{array}\right)×\left(\begin{array}{c}17-6\\ 6-k\end{array}\right)}{\left(\begin{array}{c}17\\ 6\end{array}\right)}
Probability that exactly 3 of the 6 numbers chosen by a player appear in the state’s sample:
It is given that, 3 of the 6 chosen numbers have to appear in the state’s lottery.
That is, the number of successes is
k=3
The probability that exactly 3 of the 6 numbers chosen by a player appear in the state’s sample is obtained as 0.267 from the calculation given below:
P\left(k=3\right)=\frac{\left(\begin{array}{c}6\\ 3\end{array}\right)×\left(\begin{array}{c}17-6\\ 6-3\end{array}\right)}{\left(\begin{array}{c}17\\ 6\end{array}\right)}
=\frac{20×165}{12,376}
=\frac{3,300}{12,376}
=0.2666
=0.267
(Rounded to 3 decimal places)
The probability that exactly 3 of the 6 numbers chosen by a player appear in the state’s sample is 0.267.
Use the normal approximation to the binomial distribution to determine (to four decimals) the probability of getting 7 heads and 7 tails in 14 flips of a balanced coin. Also refer to the binomial probabilities table of “Statistical Tables” to find the error of this approximation.
An oil company conducts a geologival survey that indicates an exploratory oil well should have a 23% change of striking oil.
A. What is the probability that the 2nd strike will occur on the 5th attempt?
B. What is the probability that the 7th strike will occur on the 12th attempt?
C. What is the probability that the 11th strike will occur between the 15th and 18th attempt?
If you know that it rains during ten days out of every thirty days in a particular city. Using the binomial distribution, what is the probability that no rain will fall during a given week?
A component may come from any one of three manufacturers with probabilities
p1=0.25,p2=0.50,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }p3=0.25
. The probabilities that the components will function properly are a function of the manufacturer and are 0.1, 0.2, and 0.4 for the first, second, and third manufacturer, respectively.
a. Compute the probability that a randomly chosen component will function properly.
b. Compute the probability that three components in series randomly selected will function properly
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Rayleigh wave equations with couple stress: Modeling and dispersion characteristicCouple-stress Rayleigh waves | Geophysics | GeoScienceWorld
Rayleigh wave equations with couple stress: Modeling and dispersion characteristic
, Department of Mathematics and Center of Geophysics, Harbin 150001, China. E-mail: 18b912018@stu.hit.edu.cn.
, School of Earth and Space Science, Beijing 100871, China and
, Department of Mathematics and Center of Geophysics, Harbin 150001, China. E-mail: jwm@pku.edu.cn (corresponding author).
Yanqi Wu, Jianwei Ma; Rayleigh wave equations with couple stress: Modeling and dispersion characteristic. Geophysics 2021;; 87 (1): T1–T13. doi: https://doi.org/10.1190/geo2020-0890.1
In elastostatics, the scale effect is a phenomenon in which the elastic parameters of a medium vary with the specimen size when the specimen is sufficiently small. Linear elasticity cannot explain the scale effect because it assumes that the medium is a continuum and does not consider microscopic rotational interactions within the medium. In elastodynamics, wave-propagation equations are usually based on linear elasticity. Thus, nonlinear elasticity must be introduced to study the scale effect on wave propagation. We have developed one of the generalized continuum theories, a so-called couple-stress theory, into solid earth geophysics to build a more practical model of the underground medium. The first-order velocity-stress wave equation is derived to simulate the propagation of Rayleigh waves. Body and Rayleigh waves are compared using elastic theory and couple-stress theory in a homogeneous half-space and a layered space. The results indicate that couple stress causes the dispersion of surface waves and S-waves even in a homogeneous half-space. The effect is enhanced by increasing the source frequency and characteristic length, despite its insufficiently clear physical meaning. Rayleigh waves are more sensitive to the couple-stress effect than are body waves. Based on the phase-shifting method, it is determined that Rayleigh waves exhibit different dispersion characteristics in the couple-stress theory than in the conventional elastic theory. For the fundamental mode, dispersion curves tend to move to a lower frequency with an increase in characteristic length
l
. For the higher modes, the dispersion curve energy is stronger with a greater characteristic length
l
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Find the cross product a X b and verify that
Find the cross product a X b and verify that it is orthogonal to both a and b.
Find the cross product
a×b
and verify that it is orthogonal to both a and b.
a=\left(2,3,0\right),b=\left(1,0,5\right)
To find the cross product of two vectors a and b, where the components of both vectors are as follows
a=<{a}_{1},{a}_{2},{a}_{3}>
b=<{b}_{1},{b}_{2}{,}_{3}>
We use the following formula,
c=a×b
=|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ {a}_{1}& {a}_{2}& {a}_{3}\\ {b}_{1}& {b}_{2}& {b}_{3}\end{array}|
=\stackrel{^}{i}|\begin{array}{cc}{a}_{2}& {a}_{3}\\ {b}_{2}& {b}_{3}\end{array}|-\stackrel{^}{j}|\begin{array}{cc}{a}_{1}& {a}_{3}\\ {b}_{1}& {b}_{3}\end{array}|+\stackrel{^}{k}|\begin{array}{cc}{a}_{1}& {a}_{2}\\ {b}_{1}& {b}_{2}\end{array}|
=\stackrel{^}{i}\left({a}_{2}{b}_{3}-{a}_{3}{b}_{2}\right)-\stackrel{^}{j}\left({a}_{1}{b}_{3}-{a}_{3}{b}_{1}\right)+\stackrel{^}{k}\left({a}_{1}{b}_{2}-{a}_{2}{b}_{1}\right)
In order to prove that this vector is orthogonal to both vectors a and b, we need to show that the dot product of vector c to each vector is zero, i.e we need to show that
c\cdot a=0
c\text{ }\cdot b=0
We use equation, in order to find vector c which is the cross product of the two vectors a and b as follows
c=a×b
=|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 2& 3& 0\\ 1& 0& 5\end{array}|
=\stackrel{^}{i}|\begin{array}{cc}3& 0\\ 0& 5\end{array}|-\stackrel{^}{j}|\begin{array}{cc}2& 0\\ 1& 5\end{array}|+\stackrel{^}{k}|\begin{array}{cc}2& 3\\ 1& 0\end{array}|
=\stackrel{^}{i}\left(\left(3\right)\left(5\right)-\left(0\right)\left(0\right)\right)-\stackrel{^}{j}\left(\left(2\right)\left(5\right)-\left(0\right)\left(1\right)\right)+\stackrel{^}{k}\left(\left(2\right)\left(0\right)-\left(3\right)\left(1\right)\right)
=15\stackrel{^}{i}-10\stackrel{^}{j}-3\stackrel{^}{k}
Thus, the cross product of the two vector is the following vector
c=<15,-10,-3>
We find the dot product of vector c with vector a, and then b in order to find out whether if its orthogonal to both or now
c\cdot a=<15,-10,-3>\cdot <2,3,0>
=30-30+0
=0
And, the dot product of the vector c and b
r\left(t\right)=<{t}^{2},\frac{2}{3}{t}^{3},t>
<4,-\frac{16}{3},-2>
\left(1,3,0\right),\left(-2,0,2\right),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(-1,3,-1\right)
Find a unit vector that is orthogonal to both u = (1, 0, 1) and v = (0, 1, 1).
Vectors u and v are orthogonal. If
u=⟨3,1+b⟩
v=⟨5,1-b⟩
find all possible values for b.
The image of the point (2,1) under a translation is (5,-3). Find the coordinates of the image of the point (6,6) under the same translation.
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Loss of naive Bayes incremental learning classification model on batch of data - MATLAB loss - MathWorks í•œêµ
Loss of naive Bayes incremental learning classification model on batch of data
loss returns the classification loss of a configured naive Bayes classification model for incremental learning (incrementalClassificationNaiveBayes object).
L = loss(Mdl,X,Y) returns the minimal cost classification loss for the naive Bayes classification model for incremental learning Mdl using the batch of predictor data X and corresponding responses Y.
L = loss(Mdl,X,Y,Name,Value) uses additional options specified by one or more name-value arguments. For example, you can specify the classification loss function.
Three different ways to measure performance of an incremental model on streaming data exist:
Create a naive Bayes classification model for incremental learning. Specify the class names and a metrics window size of 1000 observations. Configure the model for loss by fitting it to the first 10 observations.
Mdl = incrementalClassificationNaiveBayes('ClassNames',unique(Y),'MetricsWindowSize',1000);
canComputeLoss = (size(Mdl.DistributionParameters,2) == Mdl.NumPredictors) + ...
(size(Mdl.DistributionParameters,1) > 1) > 1
canComputeLoss = logical
Mdl is an incrementalClassificationNaiveBayes model. All its properties are read-only.
Simulate a data stream, and perform the following actions on each incoming chunk of 500 observations:
mc = array2table(zeros(nchunk,3),'VariableNames',["Cumulative" "Window" "Chunk"]);
mc{j,["Cumulative" "Window"]} = Mdl.Metrics{"MinimalCost",:};
mc{j,"Chunk"} = loss(Mdl,X(idx,:),Y(idx));
Now, Mdl is an incrementalClassificationNaiveBayes model object trained on all the data in the stream. During incremental learning and after the model is warmed up, updateMetrics checks the performance of the model on the incoming observations, and then the fit function fits the model to those observations. loss is agnostic of the metrics warm-up period, so it measures the minimal cost for every chunk.
xline(Mdl.MetricsWarmupPeriod/numObsPerChunk + 1,'r-.')
The yellow line represents the minimal cost on each incoming chunk of data. After the metrics warm-up period, Mdl tracks the cumulative and window metrics.
Fit a naive Bayes classification model for incremental learning to streaming data, and compute the multiclass cross entropy loss on the incoming chunks of data.
Create a naive Bayes classification model for incremental learning. Configure the model as follows:
Specify a metrics window size of 2000 observations.
Track the multiclass cross entropy loss to measure the performance of the model. Create an anonymous function that measures the multiclass cross entropy loss of each new observation, and include a tolerance for numerical stability. Create a structure array containing the name CrossEntropy and its corresponding function handle.
Compute the classification loss by fitting the model to the first 10 observations.
crossentropy = @(z,zfit,w,cost)-log(max(zfit(z),tolerance));
ce = struct("CrossEntropy",crossentropy);
Mdl = incrementalClassificationNaiveBayes('ClassNames',unique(Y),'MetricsWarmupPeriod',1000, ...
'MetricsWindowSize',2000,'Metrics',ce);
Call loss to compute the cross entropy on the incoming chunk of data. Whereas the cumulative and window metrics require that custom losses return the loss for each observation, loss requires the loss for the entire chunk. Compute the mean of the losses within a chunk.
tanloss = array2table(zeros(nchunk,3),'VariableNames',["Cumulative" "Window" "Chunk"]);
tanloss{j,1:2} = Mdl.Metrics{"CrossEntropy",:};
tanloss{j,3} = loss(Mdl,X(idx,:),Y(idx),'LossFun',@(z,zfit,w,cost)mean(crossentropy(z,zfit,w,cost)));
Mdl is an incrementalClassificationNaiveBayes model object trained on all the data in the stream. During incremental learning and after the model is warmed up, updateMetrics checks the performance of the model on the incoming observations, and then the fit function fits the model to those observations.
Plot the performance metrics to see how they evolve during incremental learning.
h = plot(tanloss.Variables);
ylabel('Cross Entropy')
legend(h,tanloss.Properties.VariableNames)
updateMetrics computes the window metrics after processing 2000 observations (20 iterations).
Because Mdl is configured to predict observations from the beginning of incremental learning, loss can compute the cross entropy on each incoming chunk of data.
Otherwise, you must fit the input model Mdl to data that contains all expected classes. That is, Mdl.DistributionParameters must be a cell matrix with Mdl.NumPredictors columns and at least one row, where each row corresponds to each class name in Mdl.ClassNames.
Batch of predictor data with which to compute the loss, specified as an n-by-Mdl.NumPredictors floating-point matrix.
Y — Batch of labels
Batch of labels with which to compute the loss, specified as a categorical, character, or string array; logical or floating-point vector; or cell array of character vectors.
If Y contains a label that is not a member of Mdl.ClassNames, loss issues an error. The data type of Y and Mdl.ClassNames must be the same.
Example: 'LossFun','classiferror','Weights',W specifies returning the misclassification error rate, and the observation weights W.
'mincost' (default) | string vector | function handle | cell vector | structure array | ...
Loss function, specified as a built-in loss function name or function handle.
Minimal expected misclassification cost
lossval = lossfcn(C,S,W,Cost)
The output argument lossval is an n-by-1 floating-point vector, where n is the number of observations in X. The value in lossval(j) is the classification loss of observation j.
C is an n-by-K logical matrix with rows indicating the class to which the corresponding observation belongs. K is the number of distinct classes (numel(Mdl.ClassNames), and the column order corresponds to the class order in the ClassNames property. Create C by setting C(p,q) = 1, if observation p is in class q, for each observation in the specified data. Set the other element in row p to 0.
Cost is a K-by-K numeric matrix of misclassification costs.
Example: 'LossFun',"classiferror"
Prior class probabilities, specified as a value in this numeric vector. Prior has the same length as the number of classes in Mdl.ClassNames, and the order of the elements corresponds to the class order in Mdl.ClassNames. loss normalizes the vector so that the sum of the result is 1.
Chunk of observation weights, specified as a floating-point vector of positive values. loss weighs the observations in X with the corresponding values in Weights. The size of Weights must equal n, the number of observations in X.
Classification loss, returned as a numeric scalar. L is a measure of model quality. Its interpretation depends on the loss function and weighting scheme.
\underset{j=1}{\overset{n}{â}}{w}_{j}=1.
L=\underset{j=1}{\overset{n}{â}}{w}_{j}\mathrm{log}\left\{1+\mathrm{exp}\left[â2{m}_{j}\right]\right\}.
L=\underset{j=1}{\overset{n}{â}}{w}_{j}{c}_{{y}_{j}{\stackrel{^}{y}}_{j}},
{\stackrel{^}{y}}_{j}
{c}_{{y}_{j}{\stackrel{^}{y}}_{j}}
{\stackrel{^}{y}}_{j}
L=\underset{j=1}{\overset{n}{â}}{w}_{j}I\left\{{\stackrel{^}{y}}_{j}â {y}_{j}\right\},
L=â\underset{j=1}{\overset{n}{â}}\frac{{\stackrel{Ë}{w}}_{j}\mathrm{log}\left({m}_{j}\right)}{Kn},
{\stackrel{Ë}{w}}_{j}
L=\underset{j=1}{\overset{n}{â}}{w}_{j}\mathrm{exp}\left(â{m}_{j}\right).
L=\underset{j=1}{\overset{n}{â}}{w}_{j}\mathrm{max}\left\{0,1â{m}_{j}\right\}.
L=\underset{j=1}{\overset{n}{â}}{w}_{j}\mathrm{log}\left(1+\mathrm{exp}\left(â{m}_{j}\right)\right).
{\mathrm{γ}}_{jk}={\left(f{\left({X}_{j}\right)}^{â²}C\right)}_{k}.
{\stackrel{^}{y}}_{j}=\underset{k=1,...,K}{\text{argmin}}{\mathrm{γ}}_{jk}.
L=\underset{j=1}{\overset{n}{â}}{w}_{j}{c}_{j}.
L=\underset{j=1}{\overset{n}{â}}{w}_{j}{\left(1â{m}_{j}\right)}^{2}.
For each conditional predictor distribution, loss computes the weighted average and standard deviation.
If the prior class probability distribution is known (in other words, the prior distribution is not empirical), loss normalizes observation weights to sum to the prior class probabilities in the respective classes. This action implies that the default observation weights are the respective prior class probabilities.
If the prior class probability distribution is empirical, the software normalizes the specified observation weights to sum to 1 each time you call loss.
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About formation of the stable modes of the movement of multilink mechanical systems | JVE Journals
A. S. Gorobtzov1 , E. N. Ryzhov2 , A. S. Polyanina3
3Kamyshin Technological Institute, State Educational Institution of Higher Education Volgograd State Technical University, Kamyshin, Russia
The problem of synthesis of the controlled movement of the stepping robot, the movers of that move on trajectories with the sites, which are close to rectilinear, is considered. Stabilization of movement of points on the movers of stepping robot on a rectilinear trajectory will allow synthesizing algorithm of stepping without jumps of the accelerations in points of change of phases of a support of movers. In this case, under the control object, the point of mover is understood, and the controlled coordinate is a corresponding coordinate of such point.
Keywords: self-oscillations, control, asymptotic stability.
In this work, the scheme of synthesis of the self-oscillatory modes of the movement of multivariate mechanical system, which allows receiving the periodic movements of links, for example, in problems of the movement of multivariate robotic devices of any structure, is considered. It is offered a methodology of synthesis of nonlinear generators of asymptotically stable trajectories, which provide the stable movement of the set points of the control object on such trajectories.
The scheme of synthesis can be used in the trajectory problems of robotics, at the construction of generators of oscillations and frequency converters.
Unlike the methods of control that exercise the control on a deviation, the offered generators of self-oscillations give stable movement on the set trajectories and require measurement of smaller number of parameters of the controlled movement.
So in robotic problems at a control by movement on the closed trajectories, it is possible to use general characteristics of coordinates – amplitude, period etc., for finding by method of inverse problem of the controlling functions, for example, of force in drives. Necessary condition to which has to satisfy the controlling functions is providing of stability that for the nonlinear system is a non-trivial problem.
The scheme of synthesis of the self-oscillatory modes of the movement supposes that the system is supplemented by the blocks, which are carrying out the solution of the nonlinear differential equations on which entrance some general parameters of the movement of object arrive, and at the exit, the controlling functions for executive drives are formed. The controlling functions of such nonlinear generator can be internal variables, for example, at the use in the converters of tension, or some regenerate sizes, which are received on the basis of internal variables, for example, by the method of inverse problem in robotics.
2. Problem of receiving the stable movement of control object on the set trajectories
For a task of control object in the form of spatial mechanical system dynamics equations in the form of the differentially algebraic equations are usually used [2, 3].
For example, movement of stepping robot described by differentially-algebraic equations of kind:
\left\{\begin{array}{l}\mathbf{M}\stackrel{¨}{\mathbf{x}}-{\mathbf{D}}^{T}\mathbf{p}=\mathbf{f}\left(\stackrel{˙}{\mathbf{x}},\mathbf{x},t\right)+\mathbf{u}\left(t\right),\\ \mathbf{D}\stackrel{¨}{\mathbf{x}}=\mathbf{h}\left(\stackrel{˙}{\mathbf{x}},\mathbf{x}\right).\end{array}\right\
\mathbf{x}=\left({x}_{1},{x}_{2},...,{x}_{n}{\right)}^{T}
– is the vector of general coordinates of the controlled system (object of control),
\mathbf{M}
is the matrix of inertia of the system of bodies,
{\mathbf{f}}_{0}\left(\mathbf{x},\stackrel{˙}{\mathbf{x}},\mathbf{t}\right)
is the vector of external and internal forces of control object,
\mathbf{u}\left(t\right)
is the vector of controlling forces, which are forces in the drives.
\mathbf{D}
is the matrix of variable coefficients of equations of kinematics connections by a dimension
k×n
\mathbf{h}\left(\stackrel{˙}{\mathbf{x}},\mathbf{x}\right)
is the vector of right parts of equations of connections,
\mathbf{p}
is the vector of Lagrange multipliers. The settlement scheme of stepping robot (Fig. 1) is contain 25 rigid bodies with 6 DOF (matrix
\mathbf{M}
), 120 kinematics constraints (matrix
\mathbf{D}
\mathbf{h}\left(\stackrel{˙}{\mathbf{x}},\mathbf{x}\right)
), weight forces (vector
{\mathbf{f}}_{0}\left(\mathbf{x},\stackrel{˙}{\mathbf{x}},\mathbf{t}\right)
) and 24 control forces in drives (vector
\mathbf{u}\left(t\right)
Fig. 1. The settlement scheme of the stepping robot
One of universal methods of determination of forces in drives
\mathbf{u}\left(t\right)
is a method of reverse problem. For the system Eq. (1) the method of reverse problem is taken to the task of additional kinematics connections, providing the movement on the set program trajectories. Control of system is reduced to movement of her points on the trajectories
\mathbf{w}\left(t\right)
, found, for example, by the methods of optimal control.
Action of controlling forces
\mathbf{u}\left(t\right)
in Eq. (1) of movement can be replaced with the equations of connections. Then Eq. (1) becomes:
\left\{\begin{array}{l}\mathbf{M}\stackrel{¨}{\mathbf{x}}-{\mathbf{D}}^{T}\mathbf{p}-{\mathbf{D}}_{w}^{T}{\mathbf{p}}_{w}=\mathbf{f}\left(\stackrel{˙}{\mathbf{x}},\mathbf{x},t\right),\\ \begin{array}{l}\mathbf{D}\stackrel{¨}{\mathbf{x}}=\mathbf{h}\left(\stackrel{˙}{\mathbf{x}},\mathbf{x}\right),\\ {\mathbf{D}}_{w}\stackrel{¨}{\mathbf{x}}=\stackrel{¨}{\mathbf{w}}\left(t\right).\end{array}\end{array}\right\
{\mathbf{D}}_{w}
is the matrix of variable coefficients of equations of connections for points movement of that is set,
\stackrel{¨}{\mathbf{w}}\left(t\right)
is the vector of accelerations of the specified points,
{\mathbf{p}}_{w}
is the vector of Lagrange multipliers, corresponding to connections with the set program trajectories. Components are included in the vector
\mathbf{w}\left(t\right)
, which describe trajectories of the movement of corps of robot and end-point of the movers.
We will set the rectilinear movement of corps with a constant speed and the periodic movement of end-point of feet. Thereby we get the movement of the robot. The program movement of point is shown on Fig. 2 and consists of the sites, which are close to rectilinear. The trajectory in Fig. 2 is piece-analytical. It is made of various functions, which are analytical on each site. In points of joining of sites, function has a break on derivatives.
Fig. 2. The trajectory of point of foot
From the decision of system Eq. (2) it is possible to find controlling functions for every drive. An empiric piece-nonlinear task is thus assumed of functions, which describe trajectories of end-point of stepping movers. It will result in the saltatory changes of forces in drives.
In this connection, an important problem is providing of stable movement of all system Eq. (1) at the set closed trajectories of the movement of separate points. For the case of stepping machine such closed trajectories have a form near to rectangular, i.e. trajectories consist of almost rectilinear areas.
3. Analytical constructing of generators of self-oscillations
To provide the stable movement of the set points of object of control it is offered to add system Eq. (1) by means of the equations of connections with the nonlinear differential equations. These equations are generators of program trajectories and their solutions are closed, asymptotically stable curves of the set form [4-6]. The system Eq. (1) is supplemented with the Eq. (3):
\left\{\begin{array}{l}{\stackrel{¨}{x}}^{*}={\mathbf{f}}^{\mathbf{*}}\left({\stackrel{˙}{\mathbf{x}}}^{*},{\mathbf{x}}^{\mathbf{*}}\right),\\ \mathrm{\Phi }\left(\stackrel{˙}{\mathbf{x}},\mathbf{x},{\stackrel{˙}{\mathbf{x}}}^{*},{\mathbf{x}}^{\mathbf{*}}\right)=0,\end{array}\right\
{\mathbf{x}}^{*}={\left({x}_{1}^{*},{x}_{2}^{*},\dots ,{x}_{k}^{*}\right)}^{T}
is the vector of general coordinates of the generator of periodic movements,
{\mathbf{f}}^{\mathrm{*}}\left({\mathbf{x}}^{\mathrm{*}},{\stackrel{˙}{\mathbf{x}}}^{\mathrm{*}}\right)
is the vector of the right parts of the generator,
\mathrm{\Phi }\left(\mathbf{x},\stackrel{˙}{\mathbf{x}},{\mathbf{x}}^{\mathrm{*}},{\stackrel{˙}{\mathbf{x}}}^{\mathrm{*}}\right)
and is the vector of equations of connections between the generator and the object of control. Generally, these connections are nonholonomic, nonintegrable.
For definition of structure a vector – functions
{\mathbf{f}}^{\mathrm{*}}\left({\mathbf{x}}^{\mathrm{*}},{\stackrel{˙}{\mathbf{x}}}^{\mathrm{*}}\right)
of system Eq. (3) the problem of synthesis of generators of self-oscillations is solved with the movement trajectories with the sites, which are close to rectilinear:
\left\{\begin{array}{l}\left\{\begin{array}{l}{\stackrel{˙}{x}}_{2i-1}={\alpha }_{2i}{x}_{2i}^{2m-1}+{\beta }_{2i}{x}_{2i}^{2k-1}+{\gamma }_{2i}{x}_{2i}^{2l-1},\\ {\stackrel{˙}{x}}_{2i}={\alpha }_{2i-1}{x}_{2i-1}^{2m-1}+{\beta }_{2i-1}{x}_{2i-1}^{2k-1}+{\gamma }_{2i-1}{x}_{2i-1}^{2l-1}+{U}_{i}\left({x}_{2i-1},{x}_{2i}\right)\\ \mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }+\sum _{\begin{array}{l}j=1,\\ j\ne i\end{array}}^{n}{U}_{i,j}\left({x}_{2j-1},{x}_{2j},{x}_{2i}\right),\end{array}\right\\\ \underset{t\to +\mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\sum _{i=1}^{2n}\left(\frac{{x}_{i}^{2m}\left(t\right)}{{a}_{i}^{2m}}+\frac{{x}_{i}^{2k}\left(t\right)}{{b}_{i}^{2k}}+\frac{{x}_{i}^{2l}\left(t\right)}{{c}_{i}^{2l}}\right)=1,\end{array}\right\
i=\text{1,}\text{ }\text{2,...,}\text{ }n
m
k
l\in \mathrm{Ν}
The solution of the problem is based on construction of Lyapunov function [1]. Approach to the solution of a problem of synthesis consists in finding of the stabilizing intrasystem controls
{U}_{i}\left({x}_{2i-1},{x}_{2i}\right)
in phase space of each of
n
the subsystems and intersystem controls
{\mathrm{U}}_{\mathrm{i},\mathrm{j}}\left({x}_{2j-1},{x}_{2j},{x}_{2i}\right)
which to connect these subsystems. The construction of the intrasystem control solves a problem of synthesis of the self-oscillatory mode in the corresponding subspace. Geometrically it corresponds to the birth in a phase subspace of a stable limit cycle with the movement sites, which are close to rectilinear (Fig. 3(a)). Introduction of intersystem controls provides the required dynamic descriptions of all system overall. We will search controlling functions in a next kind:
{U}_{i}\left({x}_{2i-1},{x}_{2i}\right)={\sigma }_{i}{x}_{2i}+{\beta }_{2i-1,2i}^{m}{x}_{2i-1}^{2m}{x}_{2i}+{\beta }_{2i,2i}^{m}{x}_{2i}^{2m+1}+{\beta }_{2i-1,2i}^{k}{x}_{2i-1}^{2k}{x}_{2i}
\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }+{\beta }_{2i,2i}^{k}{x}_{2i}^{2k+1}+{\beta }_{2i-1,2i}^{l}{x}_{2i-1}^{2l}{x}_{2i}+{\beta }_{2i,2i}^{l}{x}_{2i}^{2l+1},
{\mathrm{U}}_{i,j}\left({x}_{2j-1},{x}_{2j},{x}_{2i}\right)={\beta }_{2j-1,2i}^{m}{x}_{2j-1}^{2m}{x}_{2i}+{\beta }_{2j,2i}^{m}{x}_{2j}^{2m}{x}_{2i}+{\beta }_{2j-1,2i}^{k}{x}_{2j-1}^{2k}{x}_{2i}
\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }+{\beta }_{2j,2i}^{k}{x}_{2j}^{2k}{x}_{2i}+{\beta }_{2j-1,2i}^{l}{x}_{2j-1}^{2l}{x}_{2i}+{\beta }_{2j,2i}^{l}{x}_{2j}^{2l}{x}_{2i},
i
j=\text{1, 2, …,}\text{ }n
j\ne i
Such nonlinearity allows increasing significantly flexibility of controls concerning change of geometry of limit cycles (Fig. 3).
Fig. 3. Stable limit cycles
Using the condition of invariance of a surface [4, 5]
\partial {\mathbf{D}}^{2n}
\sum _{\mathbit{i}=1}^{2\mathbit{n}}\left(\frac{{\mathbit{x}}_{\mathbit{i}}^{2\mathbit{m}}}{{\mathbit{a}}_{\mathbit{i}}^{2\mathbit{m}}}+\frac{{\mathbit{x}}_{\mathbit{i}}^{2\mathbit{k}}}{{\mathbit{b}}_{\mathbit{i}}^{2\mathbit{k}}}+\frac{{\mathbit{x}}_{\mathbit{i}}^{2\mathbit{l}}}{{\mathbit{c}}_{\mathbit{i}}^{2\mathbit{l}}}\right)=1,
correlations on the coefficients of the stabilizing controls are found:
\left\{\begin{array}{l}{\mathbit{\beta }}_{2\mathbit{j}-1,\mathbf{ }2\mathbit{i}}^{\mathbit{m}}=-{\mathbit{\sigma }}_{\mathbit{i}}{\mathbit{a}}_{2\mathbit{j}-1}^{-2\mathbit{m}},\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbit{\beta }}_{2\mathbit{j}-1,\mathbf{ }2\mathbit{i}}^{\mathbit{k}}=-{\mathbit{\sigma }}_{\mathbit{i}}{\mathbit{b}}_{2\mathbit{j}-1}^{-2\mathbit{k}},\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbit{\beta }}_{2\mathbit{j}-1,\mathbf{ }2\mathbit{i}}^{\mathbit{l}}=-{\mathbit{\sigma }}_{\mathbit{i}}{\mathbit{c}}_{2\mathbit{j}-1}^{-2\mathbit{l}},\\ {\mathbit{\beta }}_{2\mathbit{j},\mathbf{ }2\mathbit{i}}^{\mathbit{m}}=-{\mathbit{\sigma }}_{\mathbit{i}}{\mathbit{a}}_{2\mathbit{j}}^{-2\mathbit{m}},\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbit{\beta }}_{2\mathbit{j},\mathbf{ }2\mathbit{i}}^{\mathbit{k}}=-{\mathbit{\sigma }}_{\mathbit{i}}{\mathbit{b}}_{2\mathbit{j}}^{-2\mathbit{k}},\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbit{\beta }}_{2\mathbit{j},\mathbf{ }2\mathbit{i}}^{\mathbit{l}}=-{\mathbit{\sigma }}_{\mathbit{i}}{\mathbit{c}}_{2\mathbit{j}}^{-2\mathbit{l}},\end{array}\right\
i
j=\text{1, 2, …,}\text{ }n
j\ne i
When performing conditions Eq. (5) on coefficients of functions of control the border
\partial {\mathbf{D}}^{2n}
is asymptotically attracting for trajectories, with the entry conditions defined in variety
{\mathbf{D}}^{2n}
and in some
\delta
- layer adjoining a surface from the outside i.e. the set
{B}_{\delta }^{-}\left(\partial {\mathbf{D}}^{2n}\right)\cup {B}_{\delta }^{+}\left(\partial {\mathbf{D}}^{2n}\right)
is area of asymptotic stability of invariant variety of system Eq. (5).
In the work, use of the offered differential equations in systems of generation of program movements of links of robots is considered. At the same time, the attracting invariant varieties provide the stable movement of object on the set trajectories.
The movement of the stepping machine with four movers is given in Fig. 1. The controlled points coincide with end-point of fingers of movers. The feature of this application is a presence of a few controlled links, which make the movements displaced in time.
In Fig. 4 the schedules of vertical movements of the controlled point of foot of the stepping drive are shown. Schedules are received by the method of analytical approximation and usual piecewise linear approximation.
The trajectory, which is received by the offered method, contains sites of constant level that it is necessary for work of stepping drive.
Fig. 4. Vertical movement of foot
In this paper, the procedure of embedding of generators of asymptotically stable trajectories with the sites, which are close to rectilinear in multivariate objects of control, is received. Thus, stability of movement of all system as a whole is provided and nonlinear properties of multivariate object of control of any structure are considered. The function, which sets the program movement, is analytical in space of conditions of a control system, is asymptotically stable and is close to a program trajectory.
Lyapunov A. The General Problem about Stability of the Movement. Gostekhizdat, Moscow, 1950, (in Russian). [Search CrossRef]
Shabana A. Dynamics of Multibody Systems. Cambridge University Press, New York, NY, 2005. [Search CrossRef]
Fumagalli Alessandro, Gaias Gabriella, Masarati Pierangelo A simple approach to kinematic inversion of redundant mechanisms. ASME International Design Engineering Technical Conferences and Computers and Information in Engineering Conference, 2007, p. 1931-1939. [Search CrossRef]
Gorobtsov A., Ryzhov E., Churzina A. Lame – manifolds in problems of synthesis of nonlinear oscillatory modes. Journal of Vibroengineering, Vol. 10, Issue 4, 2008, p. 456-459. [Search CrossRef]
Gorobtsov A., Ryzhov E., Churzina A. Multipurpose generators of self-oscillations. Izvestia VSTU, Vol. 11, Issue 9, 2009, p. 19-22, (in Russian). [Search CrossRef]
Gorobtsov A., Grigoryeva O., Ryzhov E. Attracting ellipsoids and synthesis of oscillatory regimes. Automation and Remote Control, Vol. 70, Issue 8, 2009, p. 1301-1308. [Search CrossRef]
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Comparing bond yields can be daunting, mainly because they can have varying frequencies of coupon payments. And, because fixed-income investments use a variety of yield conventions, you have to convert the yield to a common basis when comparing different bonds.
Taken separately, these conversions are straightforward. But when a problem contains both compounding period and day-count conversions, the correct solution is harder to reach.
Factors to Consider when Comparing Bond Yields
U.S. Treasury bills (T-bills) and corporate commercial paper investments are quoted and traded in the market on a discount basis. The investor does not receive any coupon interest payments. The profit is in the difference between its current purchase price and its face value at maturity. That is the implicit interest payment.
The amount of the discount is stated as a percentage of the face value, which is then annualized over a 360-day year.
Investors in T-bills don't get interest payments. The return is the difference between the purchase price and face value at maturity.
To complicate matters, that rate is based on a hypothetical year of 360 days.
In CDs, the annual percentage rate (APR) understates return. The better figure is annual percentage yield (APY), which takes compounding into account.
There are baked-in problems with rates quoted on a discount basis. For one thing, discount rates understate the true rate of return over the term to maturity. This is because the discount is stated as a percentage of face value.
It is more reasonable to think of a rate of return as the interest earned divided by the current price, not the face value. Since the T-bill is purchased at less than its face value, the denominator is overly high and the discount rate is understated.
The second problem is that the rate is based on a hypothetical year that has only 360 days.
The Yields on Bank CDs
The returns of bank certificates of deposit historically were quoted on a 360-day year also, and some are to this day. However, since the rate is modestly higher using a 365-day year, most retail CDs are now quoted using a 365-day year.
The returns are posted with their annual percentage yield (APY). This is not to be confused with the annual percentage rate (APR), which is the rate which most banks quote with their mortgages.
In APR calculations, the interest rates received during the period are simply multiplied by the number of periods in a year. But the effect of compounding is not included with APR calculations—unlike APY, which takes the effects of compounding into account.
A six-month CD that pays 3% interest has an APR of 6%. However, the APY is 6.09%, calculated as follows:
APY = (1 + 0.03)^2 - 1 = 6.09\%
APY=(1+0.03)2−1=6.09%
Yields on Treasury notes and bonds, corporate bonds, and municipal bonds are quoted on a semi-annual bond basis (SABB) because their coupon payments are made semi-annually. Compounding occurs twice per year, using a 365-day year.
Bond Yield Conversions
In order to properly compare the yields on different fixed-income investments, it’s essential to use the same yield calculation. The first and easiest conversion changes a 360-day yield to a 365-day yield. To change the rate, simply "gross up" the 360-day yield by the factor 365/360. A 360-day yield of 8% is equal to a 365-day yield of 8.11%. That is:
8\% \times \frac{365}{360} = 8.11\%
8%×360365=8.11%
Discount rates, commonly used on T-bills, are generally converted to a bond-equivalent yield (BEY), sometimes called a coupon-equivalent or an investment yield. The conversion formula for "short-dated" bills with a maturity of 182 or fewer days is the following:
\begin{aligned} &BEY = \frac{365 \times DR}{360 - (N \times DR)}\\ &\textbf{where:}\\ &BEY=\text{the bond-equivalent yield}\\ &DR=\text{the discount rate (expressed as a decimal)}\\ &N=\text{\# of days between settlement and maturity}\\ \end{aligned}
BEY=360−(N×DR)365×DRwhere:BEY=the bond-equivalent yieldDR=the discount rate (expressed as a decimal)N=# of days between settlement and maturity
So-called "long-dated" T-bills have a maturity of more than 182 days. In this case, the usual conversion formula is a little more complicated because of compounding. The formula is:
BEY = \frac{-2N}{365} + 2[(\frac{N}{365})^2 + (\frac{2N}{365} - 1)(\frac{N \times DR}{360 - (N \times DR)})]^{1/2} \div 2N - 1
BEY=365−2N+2[(365N)2+(3652N−1)(360−(N×DR)N×DR)]1/2÷2N−1
For short-dated T-bills, the implicit compounding period for the BEY is the number of days between settlement and maturity. But the BEY for a long-dated T-bill does not have any well-defined compounding assumption, which makes its interpretation difficult.
BEYs are systematically less than the annualized yields for semi-annual compounding. In general, for the same current and future cash flows, more frequent compounding at a lower rate corresponds to less frequent compounding at a higher rate.
A yield for more frequent than semiannual compounding (such as is implicitly assumed with both short-dated and long-dated BEY conversions) must be lower than the corresponding yield for actual semiannual compounding.
BEYs and the Treasury
BEYs reported by the Federal Reserve and financial market institutions should not be used as a comparison to the yields on longer-maturity bonds. The problem isn’t that the widely used BEYs are inaccurate. They serve a different purpose—namely, to facilitate comparison of yields on T-bills, T-notes, and T-bonds maturing on the same date.
To make an accurate comparison, discount rates should be converted to a semiannual bond basis (SABB), because that is the basis commonly used for longer maturity bonds.
To calculate SABB, the same formula to calculate APY is used. The only difference is that compounding happens twice a year. Therefore, APYs using a 365-day year can be directly compared to yields based on SABB.
A discount rate (DR) on an N-day T-bill can be converted directly to a SABB with the following formula:
SABB = \frac{360}{360-\left ( N \times DR \right )} \times \frac{182.5}{N-1} \times 2
SABB=360−(N×DR)360×N−1182.5×2
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For each exercise, indicate which area under the appropriate normal
For each exercise, indicate which area under the appropriate normal curve would be determined to approximate the specified binomial probability.
P\left(7<X\le 10\right)
{X}_{i}
{X}_{i}
{X}_{i}
{p}_{1},{p}_{2},\dots ,{p}_{n}
{N}_{t}
According to a 2020 article, 69% of men do not wash their hands after using a public restroom. Suppose a random sample of 40 men are selected. What is the probability that exactly 30 will not wash their hands after using a public restroom?
n=33
p=0.10
The probability of rain on any given day in a region is found to be 0.1 based on the Binomial probability distribution.
a) What is the probability of 4 days with rain in one week?
b) What is the probability that it will rain in the next 4 days?
c) What is the probability that it will rain 4 consecutive days after 3 days with no rain in one week?
Can a binomial random variable X have a mean
=7
=11
. Why or why not?
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Thermodynamic temperature is a base dimension in the International System of Units. The SI unit of thermodynamic temperature is the kelvin, defined as the fraction
\frac{1}{273.16}
of the thermodynamic temperature of the triple point of water (13th CGPM, 1967).
A degree Celsius is defined as
1
A degree Rankine is defined as
\frac{5}{9}
A degree Fahrenheit is defined as
\frac{5}{9}
A degree centigrade is defined as
\frac{1}{100}
of the thermodynamic temperature interval between the freezing and boiling points of water at standard pressure. It is approximately equal to
0.99975
A degree Reaumur is defined as
\frac{1}{80}
0.7998
A planck temperature is defined as the square root of: the planck constant times the speed of light to the fifth power, divided by twice
\mathrm{\pi }
times the Newtonian gravitational constant times the Boltzmann constant.
\mathrm{convert}\left('\mathrm{kelvin}','\mathrm{dimensions}','\mathrm{base}'=\mathrm{true}\right)
\textcolor[rgb]{0,0,1}{\mathrm{thermodynamic_temperature}}
\mathrm{convert}\left(10,'\mathrm{units}','\mathrm{degF}','\mathrm{degC}'\right)
\frac{\textcolor[rgb]{0,0,1}{50}}{\textcolor[rgb]{0,0,1}{9}}
\mathrm{convert}\left(10,'\mathrm{temperature}','\mathrm{degF}','\mathrm{degC}'\right)
\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{110}}{\textcolor[rgb]{0,0,1}{9}}
\mathrm{convert}\left(10,'\mathrm{units}','\mathrm{degR}','\mathrm{kelvin}'\right)
\frac{\textcolor[rgb]{0,0,1}{50}}{\textcolor[rgb]{0,0,1}{9}}
\mathrm{convert}\left(10,'\mathrm{temperature}','\mathrm{degR}','\mathrm{kelvin}'\right)
\frac{\textcolor[rgb]{0,0,1}{50}}{\textcolor[rgb]{0,0,1}{9}}
\mathrm{convert}\left(23.325,'\mathrm{units}','\mathrm{degF}','\mathrm{kelvin}'\right)
\textcolor[rgb]{0,0,1}{12.95833333}
\mathrm{convert}\left(23.325,'\mathrm{temperature}','\mathrm{degF}','\mathrm{kelvin}'\right)
\textcolor[rgb]{0,0,1}{268.3305555}
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Generalizing The Circular Functions | Brilliant Math & Science Wiki
Generalizing The Circular Functions
Kishlaya Jaiswal and A Former Brilliant Member contributed
The basic idea behind this wiki is to demonstrate that how we can evaluate inverse trigonometric functions outside their domain using Complex Analysis And Euler's Formula.
Everyone of us knows that the range of circular functions :
\sin x
\cos x
[-1,1]
So did you ever tried solving the equation
\sin x = 2
or in particular, finding the solutions for :
\sin x = n\ , \forall\ n \in \mathbb{R}
Yeah, you guessed it right, the solution is bit too complex!!!
Since, we are solving an equation, for such a point which is outside the range of a function, we know that there won't exist a real solution.
So, how should we start...?
In fact, because we're dealing with both complex numbers and trigonometric functions, that gives us a clue of starting with the Euler's Identity :
e^{ix} = \cos x + i\sin x
Today, I am going to introduce you to a method with which you can easily evaluate
\arccos (x)
\arcsin (x)
for all real value of
x
First of all, by Euler Formula, we have :
e^{ix} = \cos x + i\sin x
e^{-ix} = \cos x - i\sin x
Subtracting them up, gives
\sin x = \frac{e^{ix}-e^{-ix}}{2i}
Now, we wish to find the solutions for
\sin x = n
n = \frac{e^{ix}-\frac{1}{e^{ix}}}{2i}
So, now can you look up the quadratic coming around...?
No! ok, I'll just a use a simple substitution here, which makes the work tidier and easier to see the quadratic.
e^{ix} = t
n = \frac{t-\frac{1}{t}}{2i}
\Rightarrow t^2 - 2int - 1 = 0
Now, that's a quadratic in
t
whose solutions are -
t = e^{ix} = in\pm\sqrt{1-n^2}
Taking natural logarithm and multiplying by
-i
on both
L.H.S
R.H.S
yields -
x = -i\ln{\left(in\pm\sqrt{1-n^2}\right)}
\boxed{\arcsin(n) = -i\ln{\left(in\pm\sqrt{1-n^2}\right)}}
\frac{-\pi}{2} \leq \arcsin(x) \leq \frac{\pi}{2}
\arcsin(x)
lies in the first and fourth quadrants, and hence we do not need to make any changes in our formula.
1. Evaluate :
\arcsin(2015)
x
\cos x = n\ ,\ \forall\ n \in \mathbb{R}
Cite as: Generalizing The Circular Functions. Brilliant.org. Retrieved from https://brilliant.org/wiki/generalizing-the-circular-functions/
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Proper time - Wikipedia
1.2 In general relativity
2 Examples in special relativity
2.1 Example 1: The twin "paradox"
2.2 Example 2: The rotating disk
3 Examples in general relativity
3.1 Example 3: The rotating disk (again)
3.2 Example 4: The Schwarzschild solution – time on the Earth
In special relativity[edit]
{\displaystyle \eta _{\mu \nu }={\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}},}
{\displaystyle (x^{0},x^{1},x^{2},x^{3})=(ct,x,y,z)}
{\displaystyle ds^{2}=c^{2}dt^{2}-dx^{2}-dy^{2}-dz^{2}=\eta _{\mu \nu }dx^{\mu }dx^{\nu },}
and separates points on a trajectory of a particle (think clock). The same interval can be expressed in coordinates such that at each moment, the particle is at rest. Such a frame is called an instantaneous rest frame, denoted here by the coordinates
{\displaystyle (c\tau ,x_{\tau },y_{\tau },z_{\tau })}
for each instants. Due to the invariance of the interval (instantaneous rest frames taken at different times are related by Lorentz transformations) one may write
{\displaystyle ds^{2}=c^{2}d\tau ^{2}-dx_{\tau }^{2}-dy_{\tau }^{2}-dz_{\tau }^{2}=c^{2}d\tau ^{2},}
since in the instantaneous rest frame, the particle or the frame itself is at rest, i.e.,
{\displaystyle dx_{\tau }=dy_{\tau }=dz_{\tau }=0}
. Since the interval is assumed timelike (ie.
{\displaystyle ds^{2}>0}
), taking the square root of the above yields[10]
{\displaystyle ds=cd\tau ,}
{\displaystyle d\tau ={\frac {ds}{c}}.}
{\displaystyle \Delta \tau =\int _{P}d\tau =\int {\frac {ds}{c}}.}
{\displaystyle {\begin{aligned}\Delta \tau &=\int _{P}{\frac {1}{c}}{\sqrt {\eta _{\mu \nu }dx^{\mu }dx^{\nu }}}\\&=\int _{P}{\sqrt {dt^{2}-{dx^{2} \over c^{2}}-{dy^{2} \over c^{2}}-{dz^{2} \over c^{2}}}}\\&=\int {\sqrt {1-{\frac {1}{c^{2}}}\left[\left({\frac {dx}{dt}}\right)^{2}+\left({\frac {dy}{dt}}\right)^{2}+\left({\frac {dz}{dt}}\right)^{2}\right]}}dt\\&=\int {\sqrt {1-{\frac {v(t)^{2}}{c^{2}}}}}dt=\int {\frac {dt}{\gamma (t)}},\end{aligned}}}
{\displaystyle \Delta \tau =\int {\sqrt {\left({\frac {dt}{d\lambda }}\right)^{2}-{\frac {1}{c^{2}}}\left[\left({\frac {dx}{d\lambda }}\right)^{2}+\left({\frac {dy}{d\lambda }}\right)^{2}+\left({\frac {dz}{d\lambda }}\right)^{2}\right]}}\,d\lambda .}
{\displaystyle \Delta \tau ={\sqrt {\left(\Delta t\right)^{2}-{\frac {\left(\Delta x\right)^{2}}{c^{2}}}-{\frac {\left(\Delta y\right)^{2}}{c^{2}}}-{\frac {\left(\Delta z\right)^{2}}{c^{2}}}}},}
In general relativity[edit]
{\displaystyle \Delta \tau =\int _{P}\,d\tau =\int _{P}{\frac {1}{c}}{\sqrt {g_{\mu \nu }\;dx^{\mu }\;dx^{\nu }}}.}
{\displaystyle \Delta \tau =\int _{P}d\tau =\int _{P}{\frac {1}{c}}{\sqrt {g_{00}}}dx^{0}.}
Examples in special relativity[edit]
Example 1: The twin "paradox"[edit]
For a twin paradox scenario, let there be an observer A who moves between the A-coordinates (0,0,0,0) and (10 years, 0, 0, 0) inertially. This means that A stays at
{\displaystyle x=y=z=0}
for 10 years of A-coordinate time. The proper time interval for A between the two events is then
{\displaystyle \Delta \tau _{A}={\sqrt {(10{\text{ years}})^{2}}}=10{\text{ years}}.}
{\displaystyle \Delta \tau _{leg}={\sqrt {({\text{5 years}})^{2}-({\text{4.33 years}})^{2}}}={\sqrt {6.25\;\mathrm {years} ^{2}}}={\text{2.5 years}}.}
{\displaystyle \Delta \tau _{B}=2\Delta \tau _{leg}={\text{5 years}}.}
Thus it is shown that the proper time equation incorporates the time dilation effect. In fact, for an object in a SR (special relativity) spacetime traveling with a velocity of v for a time
{\displaystyle \Delta T}
, the proper time interval experienced is
{\displaystyle \Delta \tau ={\sqrt {\Delta T^{2}-\left({\frac {v_{x}\Delta T}{c}}\right)^{2}-\left({\frac {v_{y}\Delta T}{c}}\right)^{2}-\left({\frac {v_{z}\Delta T}{c}}\right)^{2}}}=\Delta T{\sqrt {1-{\frac {v^{2}}{c^{2}}}}},}
Example 2: The rotating disk[edit]
An observer rotating around another inertial observer is in an accelerated frame of reference. For such an observer, the incremental (
{\displaystyle d\tau }
) form of the proper time equation is needed, along with a parameterized description of the path being taken, as shown below.
Let there be an observer C on a disk rotating in the xy plane at a coordinate angular rate of
{\displaystyle \omega }
and who is at a distance of r from the center of the disk with the center of the disk at x = y = z = 0. The path of observer C is given by
{\displaystyle (T,\,r\cos(\omega T),\,r\sin(\omega T),\,0)}
{\displaystyle T}
is the current coordinate time. When r and
{\displaystyle \omega }
are constant,
{\displaystyle dx=-r\omega \sin(\omega T)\,dT}
{\displaystyle dy=r\omega \cos(\omega T)\,dT}
. The incremental proper time formula then becomes
{\displaystyle d\tau ={\sqrt {dT^{2}-\left({\frac {r\omega }{c}}\right)^{2}\sin ^{2}(\omega T)\;dT^{2}-\left({\frac {r\omega }{c}}\right)^{2}\cos ^{2}(\omega T)\;dT^{2}}}=dT{\sqrt {1-\left({\frac {r\omega }{c}}\right)^{2}}}.}
So for an observer rotating at a constant distance of r from a given point in spacetime at a constant angular rate of ω between coordinate times
{\displaystyle T_{1}}
{\displaystyle T_{2}}
, the proper time experienced will be
{\displaystyle \int _{T_{1}}^{T_{2}}d\tau =(T_{2}-T_{1}){\sqrt {1-\left({\frac {r\omega }{c}}\right)^{2}}}=\Delta T{\sqrt {1-v^{2}/c^{2}}},}
Examples in general relativity[edit]
Example 3: The rotating disk (again)[edit]
{\displaystyle r={\sqrt {x^{2}+y^{2}}}}
{\displaystyle \theta =\arctan \left({\frac {y}{x}}\right)-\omega t.}
{\displaystyle d\tau ={\sqrt {\left[1-\left({\frac {r\omega }{c}}\right)^{2}\right]dt^{2}-{\frac {dr^{2}}{c^{2}}}-{\frac {r^{2}\,d\theta ^{2}}{c^{2}}}-{\frac {dz^{2}}{c^{2}}}-2{\frac {r^{2}\omega \,dt\,d\theta }{c^{2}}}}}.}
{\displaystyle d\tau =dt{\sqrt {1-\left({\frac {r\omega }{c}}\right)^{2}}},}
{\displaystyle d\tau ={\sqrt {\left[1-\left({\frac {R\omega }{c}}\right)^{2}\right]dt^{2}-\left({\frac {R\omega }{c}}\right)^{2}\,dt^{2}+2\left({\frac {R\omega }{c}}\right)^{2}\,dt^{2}}}=dt.}
Example 4: The Schwarzschild solution – time on the Earth[edit]
{\displaystyle d\tau ={\sqrt {\left(1-{\frac {2m}{r}}\right)dt^{2}-{\frac {1}{c^{2}}}\left(1-{\frac {2m}{r}}\right)^{-1}dr^{2}-{\frac {r^{2}}{c^{2}}}d\phi ^{2}-{\frac {r^{2}}{c^{2}}}\sin ^{2}(\phi )\,d\theta ^{2}}},}
For the Earth, M = 5.9742×1024 kg, meaning that m = 4.4354×10−3 m. When standing on the north pole, we can assume
{\displaystyle dr=d\theta =d\phi =0}
(meaning that we are neither moving up or down or along the surface of the Earth). In this case, the Schwarzschild solution proper time equation becomes
{\textstyle d\tau =dt\,{\sqrt {1-2m/r}}}
. Then using the polar radius of the Earth as the radial coordinate (or
{\displaystyle r={\text{6,356,752 metres}}}
{\displaystyle d\tau ={\sqrt {\left(1-1.3908\times 10^{-9}\right)\;dt^{2}}}=\left(1-6.9540\times 10^{-10}\right)\,dt.}
At the equator, the radius of the Earth is r = 6378137 m. In addition, the rotation of the Earth needs to be taken into account. This imparts on an observer an angular velocity of
{\displaystyle d\theta /dt}
of 2π divided by the sidereal period of the Earth's rotation, 86162.4 seconds. So
{\displaystyle d\theta =7.2923\times 10^{-5}\,dt}
. The proper time equation then produces
{\displaystyle d\tau ={\sqrt {\left(1-1.3908\times 10^{-9}\right)dt^{2}-2.4069\times 10^{-12}\,dt^{2}}}=\left(1-6.9660\times 10^{-10}\right)\,dt.}
Retrieved from "https://en.wikipedia.org/w/index.php?title=Proper_time&oldid=1064274383"
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Dynamic Walking Challenge: Go the Distance! - OpenSim Documentation - Global Site
Note: This example is compatible with OpenSim version 4.1.
In this exercise you will use the OpenSim software to design and simulate a dynamic walker. As a starting point, you will be given a Passive Dynamic Walker Model and an arena with obstacles. The goal of the exercise is to maximize the distance the walker can travel on increasingly challenging terrain by adjusting the model's parameters and adding new model components. You will use the OpenSim graphical user interface (GUI) and Matlab scripting commands to add model components, adjust component properties, visualize dynamic simulations, and make plots of your simulation results.
All scripts and model files for this exercise are included in the OpenSim 4.1 distribution and are found in your OpenSim resources directory; <Resources-Dir>/Code/Matlab/Dynamic_Walker_Challenge/.
II. Explore the OpenSim GUI and Model
In this section, you will familiarize yourself with the OpenSim GUI and explore the editable properties of the walker and environment. You will use the GUI to examine the bodies, joints, contact geometry, and forces that make up the model. In the following sections, you will learn how to add bodies and joints and adjust the model properties to create your own robust walker.
A. Launch the OpenSim Program
Launch the OpenSim application. You will see multiple information panels and an empty main Visualizer Window. You can learn more about the Graphical User Interface (GUI) in the OpenSim User's Guide.
B. Explore the model components
In the OpenSim GUI, select File>Open Model... to open the WalkerModelTerrain.osim.
The View panel is used to control the model's visual display. You can move the view camera around the model using a combination of mouse and trackpad gestures.
In the Navigator panel on the left side of the screen, use the + icon to expand the list of model components. Explore the individual components in the model by clicking on them in the Navigator Panel.
As you click on each component, scroll through the component's properties in the Property Editor panel underneath the Navigator panel.
Observe each of the following:
Bodies: A platform, pelvis, thigh, and shank bodies with mass, inertia, and visual geometry.
Joints: A Planar joint connecting the model with the Platform and Pin joints between all the segments' pin joints.
Contact Geometry: Contact spheres and a contact half space (plane), which are used to generate contact forces.
Contact forces: Forces that are computed and applied due to the interaction between contact geometries.
Other Forces: Coordinate limit forces, which enforce the desired range of motion of the joints. Springs and other forces.
The Coordinates panel lists the current values of the model's generalized coordinates.
The values of the coordinates (in meters or degrees) can be changed with the sliders or by entering a value in the left text box.
The initial coordinate speeds (in meters/second or radians/second) are used as the initial conditions forward simulations. These can be set in the speed text box on the right hand side of each row.
The value of a coordinate can be locked by toggling the lock icon.
To reset the model to the default pose, select Poses>Default. You can also uses the Poses menu to change the default pose and add custom poses.
III. Simulate and Visualize the Walker
In this section, you will run a dynamic simulation of the model, visualize the resulting motion, plot the output, and save a movie of the motion using the OpenSim GUI.
A. Run the Forward Tool
Before you run the forward tool, go to the Coordinates tab in the left panel and choose a set of initial coordinate values and coordinate velocities for your simulation.
Select Tools>Forward Dynamics from the top menu in the GUI
In the Time Pane, set the Time Range to Process from 0 to 2 seconds
In the Output Pane, append a directory called FWD to the currently listed directory. You can leave all the other settings at their default value.
To quickly set up future runs of the forward tool, save your settings to a file (e.g., Setup_Forward.xml) by clicking the Save... button
Click the Run button to begin the simulation.
B. Visualize the motion
After running the forward tool, three storage files (.sto) are written to the output directory specified with the time history of the control signals and the model states. The model state histories, which for this model contain the coordinate values and velocities, are automatically loaded into the program (Results) and can be accessed in the Navigator panel under Motions.
If you wish to run several motions with different starting conditions, make sure to Rename the motion from the default name Results by right clicking on the item in the Navigator (e.g. Default_Results).
At the top middle section of the GUI, the blue movie controls can be used to playback and visualize the results. On the left hand side, the playback speed can be adjusted with the arrow icons or by entering a value in the textbox. In the center, you can use the control icons to play a movie forward or backwards, either continuously, or frame-by-frame.
Hit the blue play button to visualize the results from your forward simulation.
C. Create a movie in the GUI
In the Navigator panel under Motions, double click a motion to make it the current motion (bold name).
On the left side of the Visualizer Window, click the video camera icon.
Set the time and speed of your motion using the view controls.
Play the resulting motion (Note: you can pause the movie before completion if you want a specific interval). You can pause and rotate the model and play again to get a movie with two views of the walker.
Click the video camera icon again. This will end the capture and turn the camera icon blue again.
A dialog will appear. Type in your desired name for your movie file and hit the Save button.
D. Plot in GUI
The GUI has a Plotter tool for plotting results. We'll plot the coordinate velocities for the right leg of the model.
Select Tools>Plot from the top menu in the GUI
In the bottom panel, select Y-Quantity... and select the Results(deg.) file which was created by the forward tool.
In the Select Motion Quantity window, select the generalized coordinate velocities for the right hip (RHip_rz_u) and the right knee (RKnee_rz_u) and hit OK.
In the bottom panel, select X-Quantity... and select time.
In the bottom right hit Add to plot the data.
More information regarding the plotting tool is available by hitting the help button in the GUI or by going to the help page here.
E. Iteratively modify the model to maximize walking distance
The following is a partial list of the common parameters you can change. Try a few different changes to see if you can improve the walker. But don't worry too much of your walker doesn't go very far. In the sections that follow, we will discuss how to generate a more stable walker.
Modify the mass and inertia parameters of the bodies.
The mass and inertia properties of the body can be accessed through the Navigator panel. Open the bodies group by clicking on the + icon and selecting a body from the list. In the properties window you can select the mass, the mass_center location, and the six unique elements of the inertia matrix for the body.
The mass_center is the location of the mass center from the body origin measured in the body reference frame and the inertia values are the inertia of the body about the body center of mass measured in the body reference frame.
Modify the segment length in the joints.
Access the joint properties in the Joints set in the Navigator panel. The length of the segment is set by the adjacent joints. To change the length of the right thigh segment, edit the translation property of the appropriate RightThigh_offset frame (located under Bodies > RightThigh in the Navigator). Read more about OpenSim joints on the OpenSim Models page or in the OpenSim Doxygen.
Modify the initial starting conditions by entering the initial coordinate values and the initial coordinate speeds in the Coordinate panel.
In the above configuration of the Forward Tool, the initial conditions of the system will be taken from the listed values in the Coordinates panel.
Changing the joint locations to length the segments will not change the visual object in the GUI. To change the visual object, open the Body set, click on the body, and click on ... for the Displayer element in the Properties panel.
Use the keyboard shortcuts (Ctrl+Z for undo, Ctrl+Y for redo) or the orange and blue arced arrows in the Top Left section of the GUI to manage your model changes.
IV. Extending the Model with the Matlab Scripting interface
The base model has trouble maintaining knee extension during stance. In this section, you will use the Matlab scripting interface to extend the base model in two ways. First, you will add a force component to the model that will always act to extend the knee. Second, you will add a more substantial foot to the model. Before getting started, make sure your working directory in Matlab is the UserFunctions directory
A. Add a Magnet Force around the Knee Joint in Matlab
With the base dynamic walker model, the knee often flexes during the stance phase. One way to maintain knee extension during the stance phase is to add a magnet force around the knee which will contribute to extension of the knee.
OpenSim provides a library of actuators, from simple springs up to efficient muscle models. Here, we add a magnet force with a general purpose force called an ExpressionBasedPointToPointForce. The ExpressionBasedPointToPointForce calculates the relative distance (d) between two points and its time derivative (ddot) and allows the user to specify the mathematical expression for the force using the variables d and ddot. The function for the ExpressionBasedPointToPointForce can be written as a string such as +,-,*,/ and common functions such as exp, pow, sqrt, sin, cos, and tan. Importantly, the expression may not contain whitespace.
The magnet forces will be calculated as f = \frac{c}{d^2}
//<![CDATA[ \begin{array}{l}f = \frac{c}{d^2}\end{array} //]]>
where the constant, c, is 0.01 Nm^{2}
//<![CDATA[ \begin{array}{l}Nm^{2}\end{array} //]]>
AddExpressionPointToPointForceMagnets.m is a partially completed script that adds these magnet like forces to the right and left knee. The script adds a ExpressionBasedPointToPointForce for the right knee, open the script in Matlab and complete the TODO sections using the already completed code and Doxygen documentation as a guide to add an ExpressionBasedPointToPointForce to the left knee. Once you have run the completed script, open the new model in the GUI and observe the affect of the ExpressionBasedPointToPointForce on the right and left knee motion.
B. Add a custom foot mesh
Another way to add an extension moment during stance is to add a more substantial foot with a large radius of curvature. During stance, the contact forces create a moment on the lower leg which acts to extend the knee. By moving the contact point further ahead on the foot, the moment on the foot created by the ground contact forces can be used to inhibit knee flexion in stance. This section demonstrates how to add a custom foot object using the Matlab.
AddCustomFeet.m adds ContactMesh components for the feet from an included mesh file and creates an ElasticFoundationForce component to represent the contact between the foot and the ground. The mesh file (.obj) is of a simple cylindrical foot design, created in the open source program Blender.
OpenSim comes with a large library of geometry files for visualizing elements of the human skeletal system, as well as simple generic shapes such as, spheres, cylinders, and boxes. If the base library is not sufficient for your model, it is easy to add additional geometry files. OpenSim can use Geometry types with the following file extensions: .vtp, .stl and .obj. By default, OpenSim looks for Geometry files in the Model's local directory and in the OpenSim Geometry folder. You can add additional locations to look for Geometry..
V. Design and Simulate your own Walker Model
The UserFunctions directory includes example scripts to help you iteratively build your own walker model and perform forward simulations. These scripts are designed for convenience and to demonstrate key parts of the functionality of the Matlab scripting interface. You can access the documentation for each function by using Matlab's internal help interface (e.g., help RunForwardTool). You can use tab completion or methodsview (e.g. methodsview(Millard2012EquilibriumMuscle) to access available methods for OpenSim Classes.
The below scripts can be used to augment your model by adding different types of force components. Running each script will generate a new model with the corresponding component attached. Each script edits the default, unedited, model by default. You can progressively add additional components to your model by changing the input name of the model in the script. Ideally, you should mix and match any number of components to generate a model of your liking.
Related Functions/Classes
AddClutchedPathSpring.m Adds a clutched path spring to the model. The clutch is set based on length. ClutchedPathSpring
AddCustomFoot.m Adds a custom mesh object to the base model. WeldJoint, ContactMesh, ElasticFoundationForce
AddExpressionPointToPointForceMagnets.m Adds a magnet force between the thigh and shank to add a knee extension torque. ExpressionBasedPointToPointForce
AddPathSpring.m Adds a path spring to the model. PathSpring
AddSpringGeneralizedForce.m Adds a spring generalized force to the model. SpringGeneralizedForce
CreateWalkingModelAndEnvironment.m Takes a basic walking model and adds obstacles. ContactSphere, ContactHalfSpace, HuntCrossleyForce
As part of the design process, you will want to iteratively perform forward simulations with your new model to see how additional components, as well altering initial model coordinate values and speeds, improves the perform of the walking model. Use the DesignMainStarter.m script to quickly and iteratively perform multiple simulations of your model. This script allows you to change the initial coordinate values and speeds of your model, decide if you want to view the simulation using the SimTK visualizer, and plot results from the simulation. The SimTK visualizer is a handy tool for being able to observe simulation results without having to leave your development environment (Matlab or Python).
DesignMainStarter.m
The starting point for design iteration. The script loads a model, allows you to change the initial coordinate values and speeds, and perform a forward dynamic simulation using Matlab.
Change the path to appropriate model you wish to simulate. Initially, the model file is set to the default osimModel = Model('../Model/WalkerModelTerrain.osim').
To use the SimTK Visualizer to view the simulation, set visualize = true.
Once you have the model walking, change endTime to set the simulation time length.
The coordinate values and speeds are all initially set to the default value of the model, change these values to alter the initial pose of the model.
Results of the simulation are written to ResultsFWD/simulation_states.sto
PlotOpenSimData.m Generates plots from the simulations results of DesignMainStarter. Reads state values from simulation_states.sto, and plots some variables of interest.
Plots are only performed for Pelvis X-translation, right Hip rotation, and right knee rotation. Edit the file to plot other states of interest. You can view what is plotable by opening the contents ResultsFWD/simulation_states.sto in a text editor (or Excel).
The original exercise was created by Daniel A. Jacobs. Ajay Seth, Chris Dembia, Jen Hicks, James Dunne, Tom Uchida contributed to the scripts library used in this example.
Millard, M., Uchida, T., Seth, A., Delp, S.L. (2013) Flexing computational muscle: modeling and simulation of musculotendon dynamics. ASME Journal of Biomechanical Engineering, 135(2):021005.
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Data structures - CS Notes Data structures | CS Notes
Data structures affect how long it takes to access and update data. Selecting appropriate data structures for a problem is in an important skill for software engineers to develop.
Contiguous vs Linked data structures
A data structure is a representation of data and a collection of associated operations [1, P. 5].
Data structures are implementations of abstract data types [1, P. 4].
ADTs (Abstract Data Types) define the set of operations supported by a data structure [1, P. 4].
An ADT is the interface of a data structure, it doesn’t define how a data structure implements the operations [1, P. 4].
There can be many implementations of an ADT [1, P. 4].
Data structures can be either contiguous in memory, or linked with pointers.
Contiguously allocated structures are built from a single slab of memory. They include arrays, matrices, and heaps [2, P. 66].
Linked data structures are built from chunks of memory linked together via pointers. They include lists, trees, and graph adjacency lists [2, P. 66].
Arrays are the fundamental contiguously-allocated data structure. Arrays are fixed-size, and their items can be accessed using an index.
The advantages of contiguously allocated arrays are:
Constant time access. As long as you have the index, you can access an array item in constant time.
Space efficiency. Arrays only consist of data. No space is spent on pointers or other formatting information.
Memory locality. It’s common to iterate over all items of a data structure. Arrays are very good for this because they exhibit good memory locality: each item exists directly after the previous item in memory. This works well with the cache system used in modern computer architectures.
Static arrays can’t have their size modified during execution.
Dynamic arrays can increase and decrease in size during execution. A simple implementation is to initialize an array with a size of 1. Whenever the array runs out of space, its size is doubled by allocating double the previous size in memory and copying over the old array items to the lower half of the new array [2, P. 67].
A pointer “represents the address of a location in memory”. Pointers connect linked pieces of a data structure together [2, P. 67].
Linked lists are the simplest linked data structure. A linked list is made up of nodes that contain data. As well as data, each node contains a pointer to the next next node in the list.
Figure: Singly linked list [3, P. 86]
A generic linked list definition includes a next pointer, and a data value that contains some data:
Generally, linked data structures share common properties:
Each node contains one or more data fields.
Each node contains at least one pointer to another node (although the pointer can be empty). This can mean that much of the data space of linked structures is taken up with pointers, not data.
The advantages of arrays include:
Random (constant) access.
Better space efficieny (since they don’t need to store pointers).
The advantages of linked structures over static arrays include:
No overflows unless memory is full.
Insertions and deletions are simpler than for contiguous arrays.
[2, Pp. 71-2]
Both of these data types can be thought of as recursive objects. Removing the first element from a linked list leaves a smaller linked list, and splitting elements from an array creates two smaller arrays. Divide-and-conquer algorithms work well on these recursive data structures [2, P. 71].
Stacks and queues are both container data structures that allow you to add and retrieve data.
Stacks support retrieval based on LIFO (last-in-first-out). That is, the last item added to a stack is the first item to be removed. LIFO is often compared to a pile of cafeteria trays. A worker adds more trays to a pile by placing them on top of the existing trays and a customer takes their tray from the top of the pile.
The put and get operations for stacks are usually called push() and pop() [2, P. 71].
Queues support retrieval based on FIFO (first-in-first-out). They work the same way as real-world queues.
The put and get operations for queues are usually called enqueue() and dequeue() [2, P. 71].
Figure: A queue [3, P. 86]
Stacks and queues are commonly implemented with either arrays or linked lists [2, P. 71].
A dictionary data type enables access to data items by their content, or by a key value.
Dictionaries generally support the following operations:
Some dictionary data structures include other useful operations:
max() or min().
predecessor(k) or successor(k).
Hash tables are an efficient way of maintaining a dictionary.
A hash table works by using a hashing function to map a key to an integer. The integer is used as an index to store and retrieve an item from an array (or from a list stored in an array) [2, P. 89].
Normally the integer produced by the hashing function (
H
) is larger than the number of slots available in the hash table (
m
). The large integer can be converted to an integer in the range of the hash table slots by calculating the remainder of
H(K)/m
using the modulo (%) operator [2, P. 89].
Hash tables often suffer from collisions, where multiple distinct keys hash to the same value. There are different strategies that can be used in the case of collisions [2, P. 89].
One approach is chaining. In chaining the hash table is represented as an array of linked lists. Each time an item is added to the hash table, it’s inserted as a list node [2, P. 89].
Figure: Hash table using chaining [3, P. 86]
Binary search trees are a data structure that enable fast search.
Binary search trees are built on rooted binary trees. A rooted binary tree is recursively defined as either empty, or a node (the root) with two child rooted binary trees, known as the left subtree and right subtree [2, P. 77].
Figure: Binary tree [3, P. 86]
A binary search tree is a rooted binary tree where all nodes in the left subtree have a value < the root, and all nodes in the right subtree have trees > the root [2, P. 77].
Figure: Binary search tree [3, P. 86]
Binary search trees offer
O(h)
search, insertion, and deletion, where
h
is the height of the tree. If the search tree is balanced (the difference between the depth of the bottom subtrees is at most 1), then this is
\lg n
n
is the number of items. The problem is that binary search trees will not always be naturally balanced [2, Pp. 81-2].
Balanced binary search trees are data structures that maintain a balanced tree by doing extra work during insertion and deletion. For example, red-black trees and splay trees [2, P. 82].
[1] P. Morin, Open Data Structures, 1st ed. AU Press, 2013.
[2] S. Skiena, The Algorithm Design Manual, 2nd ed. Springer, 2008.
[3] L. R., Linux Kernel Development (Developer’s Library), 3rd ed. Addison-Wesley Professional, 2010.
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If \sum a_{n} and \sum b_{n} are both divergent, is
If \sum a_{n} and \sum b_{n} are both divergent, is \sum(a_{n}+b_
yogi55hr 2021-11-10 Answered
\sum {a}_{n}
\sum {b}_{n}
\sum \left({a}_{n}+{b}_{n}\right)
\sum {a}_{n}=\sum n\text{ }And\text{ }\sum {b}_{n}=\sum -n
are divergent series
\sum {a}_{n}+{b}_{n}=\sum n-n=\sum 0=0
, which is convergent
A tow truck pulls a car that is stuck in the mud, with a force of 2500 N as in Fig P5.27 The tow cable is under tension and therefor pulls downward and to the left on the pin at its upper end. The light pin is held in equilibrium by forces exerted by the two bars A and B. Each bar is a strut that is, each is a bar whose weight is small compared to the forces it exerts, and which exerts forces only through hinge pins at its ends. Each strut exerts a force directed parallel to its length. Determine the force of tension or compression in each strut. Proceed as follows Make a guess as to which way (pushing or pulling ) each force acts on the top pin Draw a free-body diagram of the pin. Use the condition for equilibrium of the pin to translate the free-body diagram into equations. From the equations calculate the forces exerted by struts A and B. If you obtain a positive answer, you correctly guessed the direction of the force. A negative answer means the direction should be reversed, but the absolute value correction should be reversed, but the absolute value correctly gives the magnitude of the force. If a strut pulls of a pin, it is in tension. If it pushes, the strut is in compression. Identify whether each strut is in tension or in compression.
Assign a binary code in some orderly manner to the 52 playingcards. Use the minimum number of bits.
2 triangles are proportional to each other.
\underset{1}{b}=6m
\underset{2}{b}=34m
\underset{1}{h}=5.5m
\underset{2}{h}
The energy flow to the earth from sunlight is about
1.4{\left\{k\frac{W}{m}\right\}}^{2}
(a) Find the maximum values of the electric and magnetic fields for a sinusoidal wave of this intensity.
(b) The distance from the earth to the sun is about
1.5×{10}^{11}m
Find the total power radiated by the sun.
Jane charges $5 per hour for babysitting. Last week, she earned a total for $55. Which of following equations could be used to find the number of hours h she babysat last week?
\frac{h}{55}=5
5h=55
5+h=55
55-h=5
|
There are 10 balls in a bag: 4 red, 3
There are 10 balls in a bag: 4 red, 3 blue, 1 green, and 2 yellow. What is the probability that you will draw yellow first and blue second?
There are 10 balls in a bag: 4 red, 3 blue, 1 green, and 2 yellow. Answer each question by reporting probability as a simplified fraction, a decimal, or a percent. What is the probability that you will draw yellow first and blue second?
How do you determine the probability distribution of
P\left(x>5\right)
\begin{array}{|cccccc|}\hline x& 4& 5& 6& 7& 8\\ P\left(X=x\right)& 0.1& 0.1& 0.2& 0.1& 0.5\\ \hline\end{array}
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Tritium Production and Permeation in High-Temperature Reactor Systems | HT | ASME Digital Collection
Hans Schmutz,
Carl Stoots,
Carl Stoots
Sabharwall, P, Schmutz, H, Stoots, C, & Griffith, G. "Tritium Production and Permeation in High-Temperature Reactor Systems." Proceedings of the ASME 2013 Heat Transfer Summer Conference collocated with the ASME 2013 7th International Conference on Energy Sustainability and the ASME 2013 11th International Conference on Fuel Cell Science, Engineering and Technology. Volume 4: Heat and Mass Transfer Under Extreme Conditions; Environmental Heat Transfer; Computational Heat Transfer; Visualization of Heat Transfer; Heat Transfer Education and Future Directions in Heat Transfer; Nuclear Energy. Minneapolis, Minnesota, USA. July 14–19, 2013. V004T19A001. ASME. https://doi.org/10.1115/HT2013-17036
Tritium (
H13
) is a radioactive isotope of hydrogen formed by ternary fission events (rare emissions of three nuclides rather than two during a fission) and neutron absorption (and subsequent decay) of predecessor radionuclides, particularly 6Li and 7Li. Also in fusion, the concept of breeding tritium during the fusion reaction is of significance for the future needs of a large-scale fusion power plant. Tritium is of special interest among the fission products created in next-generation nuclear reactors such as gas cooled reactors and molten salt reactors, because of the large quantities produced when compared with conventional light-water reactors (LWR) and the higher temperatures of operation for these systems enhances permeation. To prevent the tritium contamination of proposed reactor buildings and surrounding sites, this paper examines the root causes and potential solutions for mitigation of permeation of this radionuclide, including materials selection and inert gas sparging. A model is presented that can be used to predict permeation rates of hydrogen through metallic alloys at temperatures from 450–750°C. Results of the diffusion model are presented along with mitigation strategies for tritium permeation.
High temperature, Nuclear reactors, Nuclear fission, Radioisotopes, Hydrogen, Light water reactors, Temperature, Absorption, Alloys, Contamination, Diffusion (Physics), Emissions, Gas cooled reactors, Molten salt reactors, Neutrons, Nuclear fusion, Nuclides, Power stations, Structures
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Influence of the resonance characteristics of free-yaw small wind turbines on the performance | JVE Journals
Nadezda A. Afanasyeva1 , Vitali V. Dudnik2 , Vladimir L. Gaponov3
1, 2, 3Don State Technical University, Rostov-on-Don, Russian Federation
Variability of behavior within very short time periods is typical of a wind flow. Thus, the upwind horizontal axis wind turbine with passive yaw system represents a torsional oscillation system. The article aims to determine how the yaw oscillation impacts the wind turbine efficiency. Results of experimental study indicated that there are significant yaw angle fluctuations caused by a resonance phenomenon. Appearing of a resonant excitation leads to disproportional fluctuations of yaw angle about the mean value of 11.6° achieving the angle of 40°. Mathematical simulation of experimental wind turbine for conditions of observed phenomena showed a decrease of the efficiency at about 7 % achieving 47 % respectively.
Keywords: horizontal axis wind turbine, passive yawing, torsional oscillation, wind turbine performance.
Wind energy is one of the most perspective renewable energy sources. Small wind turbine is an attractive alternative for off-grid electrification, both as stand-alone utility and in combination with other energy technologies. One obstacle for using a wind power is the instability of its speed and direction, and, consequently, of output electric power. At the same time, the wind has not only a long-term and seasonal variability, but also changes its behavior within very short periods of time (instantaneous velocity pulsations, wind gusts, etc.).
Designers of a horizontal axis wind turbine (HAWT) yaw mechanisms are faced with a difficult decision. Use of a yaw-controlled rotor leads to increase in initial cost and decreases the reliability. On the other hand, a free-yaw rotor represents a high-risk design with unknown and random yaw characteristics [1].
The small wind turbines are often equipped by the passive mechanisms of yawing (free-yaw), such as lifting forces acting on a tail vane in the case of an upwind turbine or axial forces acting on the rotor in the case of a downwind turbine. The upwind free-yaw HAWT represents a torsional oscillation system. In fact, HAWT is a pendulum oscillator. The elastic element's stiffness of such system depends on wind speed and yaw angle.
The history of wind energy development is rife with accounts of yaw related problems. The most serious problems are structural failures due to yaw loans. Yaw-driven systems have had many instances of excessive yaw loads demerging the yaw drive mechanism. Many free-yaw rotors consistently operate at small (5°-15°) yaw errors and occasionally operate at lager yaw angles [1, 2].
Thus, a resonance excitation can appear during the HAWT operation that has a great influence on the performance. That excitation can induce amplified oscillations in the system that will lead to great output power losses and can cause damage of yaw drive mechanism. These processes and the effect of periodic unsteady fluctuations on the energy efficiency of wind turbines were devoted to the study, the results of which are presented below.
The object of the research was considered experimental small upwind HAWT equipped by a free-yaw mechanism PE-250 (Fig. 1(а)). In wind turbines with free-yaw mechanism the aligning of rotor axis and wind direction carried out without any active controllers. This approach is simple and provides operational reliability, but, in the same time, may lead to unacceptable levels of mechanical loads [1].
The results of dynamic simulation of the HAWT PE-250 operation, by the method described in [3], showed a significant influence of yaw angle on the wind turbine rotor operational efficiency. (Fig. 1(b)). Thus, the yaw angle of 10° brings to a loss in generated power amount of about 4 %, then at 20° – already 14 %.
Fig. 1. a) HAWT PE-250; b) theoretical output power generated by the wind turbine rotor (for the wind speed equals 6 m/s) depending on the yaw angle
3. Parameters of oscillation process in the HAWT yaw mechanism
The HAWT PE-250 in yaw operating conditions represents a torsional oscillator with elastic element of a lifting force acting on tail vane. Introducing such a system in general, the basic kinematic scheme of yawing HAWT can be obtained.
Like a torsional oscillator the HAWT performs torsional oscillation about the fixed yaw axis. The turning of the wind turbine nacelle around yaw axis is described by the general equation of dynamics [4, 5]:
I\frac{d\omega }{dt}={M}^{T},
{M}^{T}
– the torsional torque about the rotation axis,
I
– the moment of inertia of the wind turbine relative to the same axis,
d\omega /dt
– the angular acceleration.
If wind direction changes relative to the nacelle axis the aerodynamic forces acting on the tail vane cause a torque. This torque tends to return the nacelle in position coaxially with new wind direction.
Factually, there is a pendulum oscillating system schematically shown on the Fig. 2. Stiffness of such system depends on the magnitude of wind speed and yaw angle.
Fig. 2. Schematic of the pendulum oscillating system of HAWT in yaw operating conditions (top view)
The natural frequency of this system can be determined by the classical formula of torsional oscillations [6, 7]:
f=\frac{1}{2\pi }\sqrt{\frac{C}{I}},
C
– the torsional spring constant of HAWT yaw mechanism.
In case of the wind turbine yaw mechanism the value of torsional spring constant corresponds to the ratio of total torsional torque and yaw angle. The torque relative to the yaw angle, at the certain magnitude of wind speed, can be determined by the complex dynamic model presented in [3]. An example of that ratio for the wind speed of 6 m/s is presented on the Fig. 3.
Based on the obtained ratio the natural torsional oscillation frequencies relative to the yaw angle can be determined. For a wind speed of 6 m/s, the graph of that ratio is presented on Fig. 4.
Fig. 3. Total torsional torque
{M}^{T}
depending on the yaw angle
\psi
(wind speed – 6 m/s)
Fig. 4. The natural oscillation frequencies of HAWT PE-250 depending on the yaw angle at the wind speed of 6 m/s
In order to determine the actually occurring yaw angles the field tests of the experimental free-yaw HAWT was performed. The considered experimental wind turbine represents a small-size machine. However, it has all the properties of the standard HAWTs with the passive mechanism of yawing. Thus, the test results can be applied to somewhat greater wind turbines. The technical characteristics of experimental HAWT are given in Table 1.
Comparing with the yaw mechanisms of small HAWTs from known manufacturers, the experimental wind turbine has a higher integral index of relative efficiency (Table 2).
Table 1. Technical characteristics of experimental HAWT
Tip chord length
Root chord length
Geometric twist law (°)
\text{170.47}\cdot {r}^{4}-\text{527.4}\cdot {r}^{3}+\text{605.49}\cdot {r}^{2}-\text{307.27}\cdot r+\text{58.742}
Radius to the blade root
Table 2. The passive yaw mechanism characteristics of experimental HAWT compared to the small HAWTs from known manufacturers
Name of the HAWT from known manufacturer
Eoltec Scirocco E5.6-6
Experimental HAWT
The distance between the yaw axis and the tail vane center relative to the distance between the yaw axis and the rotor center
Relative area of the tail vane
Integral index of relative efficiency
A specialize measuring complex was developed to detect the operating parameters of experimental HAWT. The automatic braking system was installed in the spinner of nacelle. The system stops the rotor when the rotational speed exceeds the permissible value.
5. Results of the yaw angle measurements on an experimental HAWT PE-250
The field test results show that the conditions of wind flow misalignment (yaw error) are constantly present during the operation of the free-yaw HAWT. In some moments of time the value of misalignment achieves a significant amount. An example of recorded azimuthal parameters of the wind turbine nacelle and the wind flow are shown on Fig. 2. It could be seen that during the first period of time from 66 to 110 s the yaw error is relative small and on average equals 5.4° with frequency of wind turbine yawing equals 0.26 Hz. The frequency of wind flow oscillation is 0.46 Hz. From the other side the average yaw error during the second period of time from 110 to 165 s equals on average 25.6° with frequency of yawing equals 0.15 Hz. While the frequency of wind flow oscillation decreases to 0.32 Hz. It can be seen also that the wind speed is relatively constant. Thus it should be obvious that there are resonance excitations caused by the decreased frequency.
In some moments the coincidence of frequencies and phases leads to resonance oscillations of significant amplitudes. The coincidence of phases presented on Fig. 5 leads to a yaw error of about 40-50°. These behaviors of the HAWT will lead to great output power losses.
It can be concluded that, the upwind HAWT with a passive mechanism of yawing is constantly in conditions of azimuthal misalignment with wind flow direction. The experimental results show that the mean square deviation in the azimuthal directions is about 68.08 %. The arithmetic value of misalignment is equal to 11.6°. Consequently, the amount of decrease in efficiency for the angle 11.6° is equal to 7.1 %. Additionally, the misalignment of 40° caused by the resonance excitation, results in the efficiency decrease of 47.4 %. Thus the discovered values of misalignment angle caused by the resonance excitations lead to a significant decrease in efficiency and cause increase in operational loads. It reduces the life of equipment and makes it necessary to take in to account the resonance effects at the stage of designing.
Fig. 5. An example of recorded azimuthal directions of wind flow and wind turbine with a high amplitude misalignment angle
Hansen A. C. Yaw Dynamics of Horizontal Axis Wind Turbines-Final Report WE21.8202. National Renewable Energy Laboratory 1617 Cole Blvd. Golden, CO 80401University of Utuh Salt Lake City, Utah, 1992. [Search CrossRef]
Wind Turbines, Part 2: Design Requirements for Small Wind Turbines. British Standard BS EN 61400-2:2006, 2006. [Search CrossRef]
Afanaseva N. A., Dudnik V. V., Gaponov V. L. Energy supply and energy efficiency in agriculture. Proceedings of the 10th International Scientific and Technical Conference, Ecology, Moscow, 2016, p. 371-376, (in Russian). [Search CrossRef]
Zhukov V. M., Kostin A. A., Fedyushin V. B., Chernih L. M. Physics. Oscillations and Waves: Laboratory Practical. SPbSUT, 2014, (in Russian). [Search CrossRef]
Probst O., Martínez J., Elizondo J., Monroy O. Small Wind Turbine Technology, Wind Turbines. InTech, 2011. [Search CrossRef]
Singh M., Muljadi E., Jonkman J., Gevorgian V., Girsang I., Dhupia J. Simulation for Wind Turbine Generators – With FAST and MATLAB-Simulink. Modules Technical Report NREL/TP-5D00-59195, Denver West Parkway Golden, 2014. [Search CrossRef]
Burton T., Sharpe D., Jenkins N., Bossanyi E. Wind Energy Handbook. John Wiley and Sons, New York, 2001. [Search CrossRef]
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A rubber ball of mass m is dropped from a cliff. As theball falls. it is subject
A rubber ball of mass m is dropped from a cliff. As theball falls. it is subject to air drag (a resistive force caused bythe air). The drag force on t
A rubber ball of mass m is dropped from a cliff. As theball falls. it is subject to air drag (a resistive force caused bythe air). The drag force on the ball has magnitude
b{v}^{2}
, where b is a canstant drag coefficient andv is the instantaneous speed of the ball. The dragcoefficient b is directly proportional to the cross-sectional areaof the ball and the density of the air and does not depend on themass of the ball. As the ball falls, its speedapproaches a constant value called the terminal speed.
a. Write, but do Not solve, a differentialequation for the instantaneous speed v of the ball in terms of timet, the given quantities quantities, and fundamentalconstants.
b. Determine the terminal speed vt interms of the given quantities and fundamental constants.
c. Detemine the energy dissipated by the dragforce during the fall if the ball is released at height h andreaches its reminal speed before hitting the ground, in terms ofthe given quantities and fundamental constants.
\sigma F=ma
w-{F}_{\eta }=ma
mg-b{v}^{2}=ma
mg-kA\rho {v}^{2}=ma
where k is the proportionalityconstant
a=\frac{dv}{dt}=g-\left(\frac{kA\rho }{m}\right){v}^{2}
\frac{dv}{dt}+\frac{kA\rho }{m}{v}^{2}=g
I don't know if this is correct. The derivativeis with respect to time t, but is it considered "in terms oftime t?"
(b) When the terminal velocity is reached,there is no acceleration
mg-kA\rho {v}^{2}=ma=0
v=\sqrt{\frac{mg}{kA\rho }}
{30}^{\circ }
The Drawing Shows A Baggage Carousel At An Airport
Construct all random samples consisting two observations from the given data. You are asked to guess the average weight of the six watermelons by taking a random sample without replacement from the population.
\begin{array}{|ccccccc|}\hline Watermelon& A& B& C& D& E& F\\ \text{Weight (in pounds)}& 19& 14& 15& 9& 10& 17\\ \hline\end{array}
A particle travels along a straight line with a velocity of
v=\left(4t-3{t}^{2}\right)\frac{m}{s}
, where t is in seconds. Determine the position of the particle when
t=4s.\text{ }s=0
t=0
|
Copy each expression below and circle the terms. Then calculate the value of each expression.
6+4(2+3)
It may help to review some of the knowledge you have about the Order of Operations.
This is what your circles should look like. Be sure to add up the sum and show your work!
(6+4)(2+3)
Because the addition signs are inside parentheses, we call this just one term. The parentheses tell us to simplify the sums before multiplying.
50
6+4·2+3
First rewrite the expression. Now, it's time to draw the circles around any terms you see. Remember, these terms are separated by addition (
+
6+4·2+3
Now, can you simplify the terms and find the sum?
|
An urn contains 3 red and 7 black balls. Players A and B withdraw balls from the urn consecutively until a red ball is selected.
An urn contains 3 red and 7 black balls.
An urn contains 3 red and 7 black balls. Players A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B, and so on. There is no replacement of the balls drawn.)
A wins if the first red ball is drawn 1st, 3rd, 5th, or 7th.
We will calculate the number of events for each first apparition of a red ball.(i.e. a red ball is drawn first, therefore there are (9C2) places in which the other 2 red balls can be placed. In other words, there are (9C2) events in which A wins on the first draw)
E(1)=(9C2)
When the sum up the number of favorable events and divide by the number of total events.
S=(10C3)
E(x): The number of favorable events (position of the first red ball)
S: Total number of events (all possible combinations of the balls)
P(A wins)=
\frac{\left(9C2\right)+\left(7C2\right)+\left(5C2\right)+\left(3C2\right)}{10C3}
P(A wins)=.05833
\left\{x\in {R}^{\prime }\mid 0<x<1\right\}
\left\{x\in R\mid x\le 0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x⇒1\right\}
\left\{n\in Z\mid n\text{ }is\text{ }a\text{ }factor\text{ }of\text{ }6\right\}
\left\{n\in Z\cdot \mid n\text{ }is\text{ }a\text{ }factor\text{ }of\text{ }6\right\}
If a student jumps straight up to a height of 0.440m, what is his initial speed as it leaves the ground?How long is he in the air?
{\int }_{-2}^{2}f\left(x\right)dx=4,{\int }_{2}^{5}f\left(x\right)dx=3,{\int }_{-2}^{5}g\left(x\right)dx=2
f\left(x\right)\le g\left(x\right)
-2\le x\le 5
The two-way table summarizes data on the gender and eye color of students in a college statistics class. Imagine choosing a student from the class at random. Define event
A: student is male and event
B: student has blue eyes.
\text{Gender}\text{ }\text{Eye color}\begin{array}{lccc}& \text{ Male }& \text{ Female }& \text{ Total }\\ \text{ Blue }& & & 10\\ \text{ Brown }& & & 40\\ \text{ Total }& 20& 30& 50\end{array}
Copy and complete the two-way table so that events A and B are independent.
|
An egg distributer determines that the probability that any individual
An egg distributer determines that the probability that any individual egg has a
Some characteristics of binomial distribution:
Each trial has exactly two possible outcomes.
There are n number of trials, in which p is the probability of success and
q\left(=1-p\right)
is the probability of failure.
The probability of success in each trial remains constant.
Formula for binomial probability:
The formula for binomial probability is as follows.
P\left(X=x\right){=}^{n}{c}_{x}{p}^{x}{q}^{\left(n-x\right)}
q is the probability of failure,
n is the number of questions
a) Binomial probability formula to determine the probability that exactly x eggs of n eggs are cracked:
Here, the event of success is “eggs that have cracks”.
The probability of success (an individual egg has a crack) is,
p=0.15
The probability of failure is,
q=1-0.15=0.85
The binomial probability formula to determine the probability that exactly x eggs of n eggs are cracked is as follows.
P\left(X=x\right){=}^{n}{c}_{x}{\left(0.15\right)}^{x}{\left(0.85\right)}^{\left(n-x\right)}
b) Binomial probability formula to determine the probability that exactly 2 eggs in a one-dozen egg carton are cracked:
In general, number of eggs in one-dozen egg carton is,
n=12
The binomial probability formula to determine the probability that exactly 2 eggs in a one-dozen egg carton are cracked is as follows.
P\left(X=2\right){=}^{12}{C}_{2}{\left(0.15\right)}^{2}{\left(0.85\right)}^{\left(12-2\right)}
{=}^{12}{C}_{2}{\left(0.15\right)}^{2}{\left(0.85\right)}^{10}
What value of p will give the following probability distribution if x is Binomially distributed with
n=5
and probability of success p?
\text{Misplaced &}
Group of answer choices 30% 32% 34% 36% 38%
Assume the n trails is independent and that p is the probability of success on a given trail. Use the binomial probability formula to find
P\left(x\right)n=5,x=3,p=0.3
What is the probability of 24 succeses? p=.20 and n=100
P\left(x\right)
n=4,\text{ }x=1,\text{ }p=0.6
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Dynamics of gas turbine engines rotors taking into account non-linear effects | JVE Journals
O. Repetckii1 , I. Ryzhikov2 , Tien Quyet Nguyen3
1Irkutsk State Agrarian University named after A.A. Ezhevsky, Irkutsk, Russia
2, 3Irkutsk National Research Technical University, Irkutsk, Russia
Different types of non-linearity must be considered in solving the problem of determining the fatigue life of gas turbine engines rotors on the base of numerical methods. When turbomachinery rotors are simulated for assessment their fatigue life it is necessary take into account geometrical and physical nonlinearity of the model. As a result of partial steam or gas admission and edge tracks from stator blades it is important to consider the non-linear nature of the loads impacting on the rotor elements during the transient conditions of the work. While determining real stresses at the last stage of fatigue life assessment it is also necessary to take into account their non-linear nature.
Keywords: gas turbine engines rotors, stress, fatigue life, natural frequencies, mode shapes.
The assessment of fatigue life of gas turbine engines or steam-gas rotors requires follow-up of main stages [1, 2]:
1) The calculation of static stress. Calculation of natural frequencies and mode shapes, taking into account the rotation, uneven heating and gas forces. Their adjustment takes into account a mistuning [3, 4, 6-9, 15].
2) Numerical and experimental determination of the exciting load on the stationary and transient operating conditions. The main excitation sources for compressor and turbine rotor elements are perturbations due to partiality for supplying steam or gas and edge traces from stator blades.
3) An assessment of damping in material, structural damping, aerodynamic damping and damping from shock effects. A major role in turbines plays material and structural damping. The leading place in compressors belongs to the aerodynamic damping.
4) Calculation of a response on steady and unstable modes. Adjusted two or three-dimensional stress analysis in possible areas of stress concentration. Summation of static and dynamic stresses takes into account loading history.
5) Assessment of a fatigue life of rotor systems. Prediction of time of formation of a crack and destruction of constructions.
Solution of the problem of blade structures fatigue life assessment can be represented by scheme on Fig. 1.
2. Considering of geometrical and physical nonlinearity
The statics equation in finite element method for constant speed of rotation and temperature is:
\left(\left[K\right]+\left[{K}_{G}\right]+\left[{K}_{R}\right]\right)\left\{\delta \right\}={f}_{\mathrm{\Omega }}+{f}_{t},
\left[K\right]
– a stiffness matrix,
\left[{K}_{R}\right]
– the supplementary stiffness matrix arising from rotation,
\left[{K}_{G}\right]
– a matrix of a geometrical stiffness,
\left\{\delta \right\}
– node displacements,
{f}_{\mathrm{\Omega }}+{f}_{t}
– vectors of temperature and centrifugal loadings.
In case of the free vibration without damping we have [5]:
\left[M\right]\left\{\stackrel{¨}{\delta }\right\}+\left[{M}_{C}\right]\left\{\stackrel{˙}{\delta }\right\}+\left(\left[K+{K}_{G}+{K}_{R}\right]\right)\left\{\delta \right\}=0,
\left[{M}_{C}\right]
– Coriolis matrix and [
M
] – mass matrix.
For taking into account physical nonlinearity of the stress-strained state of structures there are two main theories: the straining theory of plasticity and the theory of a current. The equations of the straining theory of plasticity establish connection between stresses and deformations, and the equations of the theory of a current - between infinitely small increments of these values. In case of a simple loading both theories give identical result, however, for a number of tasks, for example, for a problem of a thermo plasticity, the theory of a current reflects loading history more fully. We find a solution of the Eq. (1), (2) according to iteration algorithm [1, 5, 10].
Fig. 1. Assessment of fatigue life scheme
3. Considering of nonlinearity in determination of real stresses
For calculation of fatigue life, it is necessary to find real, but not elastic stresses. It is possible to execute it by means of numerical methods, for example, by the technique described in [12]. But this technique requests considerable computing resources (time, memory). Far less resources are requested by the so-called Neuber rule [12, 13] describing connection between elastic and real stresses and deformations (Fig. 2):
\sigma \epsilon =\frac{{\sigma }_{e}^{2}}{E}.
Fig. 2. Recalculation of elastic stresses in elastic-plastic (Neuber rule)
Local elastic stresses from equations for deformation and stresses can be recalculated into real elastic – plastic deformations (Fig. 2) where
{\epsilon }_{\alpha }
– amplitude of variable deformations:
{\epsilon }_{\alpha }=\frac{\sigma }{E}+{\left(-\frac{\sigma }{{K}^{\mathrm{\text{'}}}}\right)}^{\frac{1}{n{\mathrm{\text{'}}}^{ }}}.
K\mathrm{\text{'}}
n\mathrm{\text{'}}
turn out from the experimental coefficients of vibration strength, and
{\sigma }_{f}\mathrm{\text{'}}
{\epsilon }_{f}\mathrm{\text{'}}
as well as an exponent
b
c
n\mathrm{\text{'}}=\frac{b}{c},
K\mathrm{\text{'}}=\frac{{\sigma }_{f}^{\mathrm{\text{'}}}}{\left({\epsilon }_{f}^{\mathrm{\text{'}}}{\right)}^{n\mathrm{\text{'}}}}.
It is possible to apply damageability parameters to the considering of a relaxation and creep, for example, on Morrow [13]:
{\epsilon }_{\alpha }=\frac{{\sigma }_{f}^{\mathrm{\text{'}}}-{\sigma }_{m}}{E}\cdot \left(2{N}_{f}{\right)}^{b}+{\epsilon }_{f}^{\mathrm{\text{'}}}\cdot \left(2{N}_{f}{\right)}^{c}.
Further from work [14]:
{\epsilon }_{\alpha }=\frac{{\sigma }_{f}^{\mathrm{\text{'}}}}{E\cdot {\sigma }_{\mathrm{m}\mathrm{a}\mathrm{x}}}\cdot \left(2{N}_{f}{\right)}^{2b}+\frac{{\sigma }_{f}^{\mathrm{\text{'}}}\cdot {\epsilon }_{f}}{{\sigma }_{\mathrm{m}\mathrm{a}\mathrm{x}}}\cdot \left(2{N}_{f}{\right)}^{b+c},
{\sigma }_{\mathrm{m}\mathrm{a}\mathrm{x}}={\sigma }_{\alpha }+{\sigma }_{m}
N=2{N}_{f}
To find from the Eq. (8) real stresses, it is necessary to know real deformation. For this purpose it is possible to use a technique of Rieger [14], which possesses an essential shortcoming, namely – need of the experimental determination of cycle parameters for each material. In this regard it is obviously necessary to use other dependence – the empirical formula of Manson connecting a range of the complete deformation and number of cycles before destruction [10]. Using the assumption of the linear relation of fatigue strength from average stress (
{\sigma }_{\mathrm{m}\mathrm{a}\mathrm{x}}\ge 0
) and in view of asymmetry of a cycle, we will receive [10]:
\mathrm{\Delta }\epsilon =\frac{3,5\cdot \left({\sigma }_{B}\left(t\right)-{\sigma }_{m}\right)}{E\left(t\right)}\cdot {N}_{f}^{-0,12}+{\left(\mathrm{l}\mathrm{n}\frac{100}{100-\psi \left(t\right)}\right)}^{0,6}\cdot {N}_{f}^{-0,6},
\psi
– contraction ratio in %, and
t
– temperature of a cyclic deformation.
Birger I. A. suggested to lead Eq. (9) to a look:
\mathrm{\Delta }\epsilon =\frac{2{\sigma }_{-1}}{E\left(t\right)}\cdot {\left(\frac{{N}_{D}}{{N}_{f}}\right)}^{\frac{1}{k}}+{\left(\mathrm{l}\mathrm{n}\frac{100}{100-\psi \left(t\right)}\right)}^{0,6}\cdot {N}_{f}^{-0,6},
k
– an exponent of fatigue curve.
{\sigma }_{m}\le 0
expression is used:
\mathrm{\Delta }\epsilon =3,5\cdot \frac{{\sigma }_{B}\left(t\right)}{E}\cdot {N}_{f}^{-0,12}+{\left(\frac{{N}_{f}}{\mathrm{l}\mathrm{n}\frac{100}{100-\psi \left(t\right)}}\right)}^{0,6}.
The total deformation
\mathrm{\Delta }\epsilon
(Fig. 3) is the sum of elastic and plastic components which are schematically represented in Fig. 4.
Change of stresses and deformations ranges from cycle to cycle comes to an end through rather small number of cycles, and the main decrease in fatigue life happens at the stabilized state (constant ranges of stresses). Thus, ranges of stresses or deformations of the stabilized state are used in calculations. After calculation
\mathrm{\Delta }\epsilon =2{\epsilon }_{a}
{\epsilon }_{a}
in the Eq. (3) it is possible to receive the real stresses (Fig. 2).
Fig. 3. Diagram “stress-deformation”
Fig. 4. Summing of elastic and plastic deformation
For the complex spatial structures like compressor fan blade with curvilinear fixing it is necessary to consider a non-linear geometrical stiffness in conditions of centrifugal stresses in the median plane. One of such blades model is presented in Fig. 5. Input data: an elastic modulus 1,1×105 MPa; density 4,54×103 kg/m3; a Poisson’s ratio – 0,3; rotation speed 1121 s-1. On a transition curve of a blade profile part to a disk rim rigid fixing is simulated. The results of calculations in the linear and geometrically non-linear statements are given in Fig. 6. Decrease in the maximal stresses for pressure side of blade approximately for 10 % is noted. For geometrically non-linear decision the maximal stresses in this area are 122×107 Pa, lengthening and the angles of elastic promotion of a blade at the same time decreased more, than twice.
Fig. 5. Finite element model of blade
Fig. 6. Stresses in wide blade (- linear, - - - non-linear solution)
a) Pressure side
The second example – a cooled blade of the helicopter gas-turbine engine impeller. Mechanical characteristics of a blade: density 8,4×103 kg/m3; a Poisson’s ratio – 0,3; the elastic modulus 2,135×105 MPa. The effect of geometrical nonlinearity consideration under maximal rotation speed was analyzed. Decrease of natural frequencies under speed increasing is noted. That it is connected with existence of the compression sections on a blade back (Table 1).
Table 1. Research results
n=
n=
717 s-1
With geometrical non-linearity
{f}_{1}
{f}_{2}
{f}_{3}
{f}_{4}
{f}_{5}
The efficient technique is developed for an assessment of fatigue life of gas turbine engines rotors being under heavy static and cyclic loads causing a low-cyclic and multi-cycle material fatigue during the work. This technique is developed taking into account geometrical nonlinearity of finite element model, physical nonlinearity of material and non-linear effects during the transient conditions of the work. The technique was successfully tested on test models and real constructions [1-5, 10-12] that means its operability.
Repetckii O. V. Automation of Strength Calculations of Turbomachines. Irkutsk Publishing House, Irkutsk, 1990, p. 100, (in Russian). [Search CrossRef]
Repetckii O. V., Buy Man Kyong A question of the choice of a numerical method of the analysis of stresses at an assessment of a multi-cycle fatigue of blades of transport turbomachines. News of IGEA, Vol. 6, 2010, p. 153-158, (in Russian). [Search CrossRef]
Repetckii O. V., Do Man Tung Mathematical model operation and numerical analysis of vibrations of ideal cyclic and symmetric systems finite element method. IGEA News, Vol. 3, 2012, p. 149-153, (in Russian). [Search CrossRef]
Repetckii O. V., Do ManTung Investigation of the characteristics of mistuned turbomachinery bladed discs vibration based on the reduced-order modeling by finite element method. Bulletin SibSAU, Vol. 1, Issue 53, 2014, p. 60-66, (in Russian). [Search CrossRef]
Repetckii O. V. Computer Analysis of the Dynamics and Strength of Machines. Irkutsk State Technical University Publishing House, Irkutsk, 1999, p. 301, (in Russian). [Search CrossRef]
Beirow B. Grundlegende Untersuchungen zum Schwingungsverhalten von Verdichterlaufrädern in Integralbauweise. Shaker Verlag, Aachen, 2009. [Search CrossRef]
Klauke T. Schaufelschwingungen Realer Integraler Verdichterräder im Hinblick auf Verstimmung und Loaklisierung. Der Andere Verlag, Cottbus, 2008. [Search CrossRef]
Maywald T., Beirow B., Kühhorn A. Mistuning und Dämpfung II. FVV Abschlussbericht, 2014. [Search CrossRef]
Maywald T., Beirow B., Kühhorn A. Mistuning und Dämpfung II. FVV Informationstagung Turbomaschinen, 2014. [Search CrossRef]
Heiman B., Gerdt V., Popp K., Repetckii O. V. Mechatronics: Components, Methods, Examples. Publishing House of the SB RAS, Novosibirsk, 2010, p. 602. [Search CrossRef]
Zainchkovsky K. S., Repetckii O. V. Sensitivity Analysis of Gas Turbine Engine Blades by Finite Element Method. Intercollege Scientific Collection IrSTU “The Asymptotic Methods in Problems of Aerodynamics and Design of Aircraft”, Irkutsk State Technical University, Irkutsk, 1996, p. 83-88, (in Russian). [Search CrossRef]
Irretier H., Repetski O. Vibration and Life Estimation of Rotor Structure. IFToMM Conference on Rotor Dynamics, Darmstadt, 1998. [Search CrossRef]
Kayser A. Entwicklung eines Programmes zur Lebensdauer berechnung von Turbinenschaufeln. B.Sc. Thesis, Institute of Mechanics, University of Kassel, Kassel, 1990, p. 110. [Search CrossRef]
Rieger N. F., Steele J. M., Lara T. C. T. Turbine blade life prediction computer program. Proceedings of EPRI Workshop on Steam Turbine Blade Reliability, 1982. [Search CrossRef]
Wei S. T., Pierre C. Localization phenomena in mistuned assemblies with cyclic symmetry. Part 1: Free vibrations. Journal of Vibration, Acoustics, Stress, and Reliability in Design, Vol. 110, Issue 4, 1987, p. 429-438. [Search CrossRef]
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Sometime, a problem may contain information which is either not needed at all to solve the problem, or needed but only if the problem is to be solved in the slowest of ways. Regardless of the case, irrelevant information can be misleading and confusing. If we are able to identify it when choosing a solution strategy, we are one step closer to getting the problem right.
\ \
\ \
\ \
\ \
\ \
All of (A), (B) and (C)
In order to compute the total amount Danny spent in the store, we need to know how many bags of chips he bought, and how much one bag of chips costs. So, we can eliminate choices (A), (C), (D), and (E).
Leonard drives a total of 500 miles, stopping at 20 gas stations along the way. If he eats at 25 different restaurants and his trip lasts 10 days, what is the average distance (in miles) Leonard travels each day?
Leonard drives 500 miles in 10 days, hence on average he travels
\frac{500} { 10} = 50
miles per day.
This is the total distance that Leonard drives, and not the average distance per day.
This is the number of restaurants Leonard visits.
This is the number of gas stations along the way.
This is the number of days that he drove, not the average distance per day.
When bus 1729 leaves the depot, there are 14 people on board. At the first stop, 10 people get on. At the second stop, 8 people get on and 4 people get off. At the third stop, 9 people get off. How many people are on the bus just after the first stop?
\ \ 10
\ \ 14
\ \ 17
\ \ 24
\ \ 28
Cite as: Irrelevant Information. Brilliant.org. Retrieved from https://brilliant.org/wiki/sat-irrelevant-information/
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Experience - OSRS Wiki
The RuneScape Wiki also has an article on: rsw:ExperienceThe RuneScape Classic Wiki also has an article on: classicrsw:Experience
Experience, commonly abbreviated as EXP or XP, is a measure of progress in a certain skill. It is obtained by performing tasks related to that skill. Experience can also be gained through other means, such as quests, the book of knowledge from the Surprise Exam random event, a lamp from the genie random event, certain mini-games, and lamps for completing parts of the Achievement Diary and Combat Achievements.
After gaining a certain amount of experience, players will advance to the next level in that skill, which can result in new abilities, and the chance to try more quests. The amount of experience needed for the next level is approximately 10% more than the last level. For example, 83 experience is required for advancement to level 2, while 91 experience is required for advancement to level 3. Reaching level 99 in a skill requires a total of 13,034,431 experience. By around level 30, the exponential factor predominates, so that the amount of experience required doubles for each 7th level. Accordingly, level 92 is nearly the exact halfway mark to level 99, requiring 6,517,253 experience, and level 85 requires very nearly one quarter of the experience needed for level 99.
Experience is stored as a 32-bit integer with one decimal point,[1] although the game does not display decimal values; for example, if a player receives two experience drops of 2.5, the first is shown as 2 and the second as 3 (or vice versa depending on their existing experience points). Experience values that would have multiple decimal points, such as through multiplication with experience-boosting sets, are rounded down to one decimal. The maximum experience that can be obtained in one skill is 200,000,000. The skill can still be trained afterwards, but no experience will be received.
1.2 Virtual levels
The experience difference between level
{\textstyle L-1}
and level
{\textstyle L}
{\textstyle {\frac {1}{4}}\left\lfloor L-1+300\cdot 2^{\frac {L-1}{7}}\right\rfloor }
. The tables below show this experience difference for each level and also the cumulative experience from level 1 to level
{\textstyle L}
1 0 N/A 0.00 26 8,740 898 0.07 51 111,945 10,612 0.86 76 1,336,443 126,022 10.25
2 83 83 0.00 27 9,730 990 0.07 52 123,660 11,715 0.95 77 1,475,581 139,138 11.32
3 174 91 0.00 28 10,824 1,094 0.08 53 136,594 12,934 1.05 78 1,629,200 153,619 12.50
4 276 102 0.00 29 12,031 1,207 0.09 54 150,872 14,278 1.16 79 1,798,808 169,608 13.80
10 1,154 185 0.01 35 22,406 2,182 0.17 60 273,742 25,856 2.10 85 3,258,594 307,221 25.00
22 5,624 606 0.04 47 75,127 7,144 0.58 72 899,257 84,812 6.90 97 10,692,629 1,008,052 82.03
24 7,028 737 0.05 49 91,721 8,707 0.70 74 1,096,278 103,383 8.41 99 13,034,431 1,228,825 100.00
25 7,842 814 0.06 50 101,333 9,612 0.78 75 1,210,421 114,143 9.29
Virtual levels[edit | edit source]
103 19,368,992 1,826,016 112 47,221,641 4,451,840 121 115,126,838 10,853,671
The formula needed to calculate the amount of experience needed to reach level
{\textstyle L}
{\displaystyle {\text{Experience}}=\left\lfloor {{\frac {1}{4}}\sum _{\ell =1}^{L-1}}\left\lfloor {\ell +300\cdot 2^{\ell /7}}\right\rfloor \right\rfloor }
{\displaystyle {\text{Experience}}\approx {\frac {1}{8}}\left({L}^{2}-L+600\,{\frac {{2}^{L/7}-2^{1/7}}{{2}^{1/7}-1}}\right)}
The experience required for each level can be graphed. Graphing the experience required on a linear scale shows that the experience required is essentially exponential.
Graphing the same data on a logarithmic scale shows that the function deviates from being exponential before around level 10.
Quest experience rewards (skill experience earned for the completion of quests)
Experience rate (maximum possible hourly experience rates vary for each skill)
^ Jagex. Mod Ash's Twitter account. 29 January 2018. (Archived from the original on 24 May 2020.) Mod Ash: "It's stored as a 32-bit INT that's treated as a fixed-point number with one decimal place. So it'll take numbers up to 2 billion, which it treats as numbers up to 200,000,000.0, hence the 200 million XP cap you see."
Retrieved from ‘https://oldschool.runescape.wiki/w/Experience?oldid=14282491’
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Brainfuck - Simple English Wikipedia, the free encyclopedia
The brainfuck programming language is an esoteric (weird and unusual) programming language. It was created by Urban Müller in 1993.[1][2] It has eight instructions (commands) which operate (do things on) a tape.[1] Instructions are done one by one, in order.[1] The tape has multiple sections.[1] Each section is a number.[1][2] Each section is, in the beginning, zero.[1] Brainfuck is like a Turing machine. [1]
1 Instruction table
Instruction table[change | change source]
+ Add one to the current tape section.
- Subtract one from the current tape section.
< Move to the tape section to the left of the current one.
> Move to the tape section to right left of the current one.
. Print the value of the current tape section as an ASCII symbol.
, Read an ASCII symbol into the current tape section as a number.
[ If the current tape section is zero, go to the matching ], skipping the instructions in between.
] If the current tape section is not zero, go back to the matching [, and do the code after it again.
If the current section is not zero, these three commands subtract one until the current section is zero. Otherwise, they leave it at zero.
These five commands first add three to the current section. Then, they subtract two from the current section. Since
{\displaystyle n+3-2=n+1}
, these five commands are the same as "+" alone.
Derivatives[change | change source]
As a result of brainfuck's fame, many derivatives (versions) of brainfuck have been created. These include Brain-Flak[3], pbrain[4], and tinyBF[5]. Most are also Turing complete, just like brainfuck.
↑ 1.0 1.1 1.2 1.3 1.4 1.5 1.6 "brainfuck - Esolang". esolangs.org. Retrieved 2022-04-23.
↑ 2.0 2.1 262588213843476. "Basics of BrainFuck". Gist. Retrieved 2022-04-23. {{cite web}}: CS1 maint: numeric names: authors list (link)
↑ "Brain-Flak - Esolang". esolangs.org. Retrieved 2022-04-24.
↑ "pbrain - Esolang". esolangs.org. Retrieved 2022-04-24.
↑ "tinyBF - Esolang". esolangs.org. Retrieved 2022-04-24.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Brainfuck&oldid=8190776"
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The current in a wire varies with time according to the relation I=55A?(0.65A/s2)t2. How
The current in a wire varies with time according to the relation I=55A?(0.65A/s2)t2. How many coulombs of charge pass a cross section of the wire in t
I=55A?\left(0.65A/{s}^{2}\right){t}^{2}
t=0
t=8.5s
q={\int }_{{t}_{1}}^{{t}_{2}}Itd
\left(55A-\left(0.65A/{s}^{2}\right){t}^{2}\right)
for I, 0s for
{t}_{1}
, and 8.5s for
{t}_{2}
to find the charge.
q={\int }_{0s}^{8.5s}\left(55A-\left(0.65A/{s}^{2}\right){t}^{2}\right)dt
=\left[55tA{\right]}_{0s}^{8.5s}-\left[\frac{0.65{t}^{3}}{2}{\right]}_{0s}^{8.5s}
=467.5C-133.1C
=334.1C
I=\frac{Q}{\mathrm{△}t}
=\frac{Q}{{t}_{2}-{t}_{1}}
Substitute 334.1 C for Q and 8.5s for
{t}_{2}
and 0s for
{t}_{1}
to find I.
I=\frac{334.1C}{8.5s-0s}
=39.3A
Find the relation between the given quadric surfaces and conic sections.
{x}^{2}\sqrt{x}
Find the equation of a plane passing through Ox and intersecting a hyperbolic paraboloid
\frac{{x}^{2}}{p}-\frac{{y}^{2}}{q}=2z\left(p>0,q>0\right)
along a hyperbola with equal semi-axes.
My attempt: The equation of a plane passing through Ox is
By+Cz=0
z=\frac{-By}{C}
z=\frac{-By}{C}
into the equation of hyperbolic paraboloid
\frac{{x}^{2}}{p}-\frac{{y}^{2}}{q}=\frac{-2By}{C}
and transform to
\frac{{x}^{2}}{p}-\frac{{\left(y-\frac{Bq}{C}\right)}^{2}}{q}=-\frac{{B}^{2}q}{{C}^{2}}
. What to do next? I don't understand how to find B and C if we assume
\frac{{x}^{2}}{p}-\frac{{\left(y-\frac{Bq}{C}\right)}^{2}}{q}=-\frac{{B}^{2}q}{{C}^{2}}
is a hyperbola with equal semi-axes.
r=\frac{8}{4+\mathrm{cos}\theta }
To find the vertices and foci of the conic section:
\frac{{\left(x\text{ }-\text{ }4\right)}^{2}}{{5}^{2}}\text{ }-\text{ }\frac{{\left(y\text{ }+\text{ }3\right)}^{2}}{{6}^{2}}=1
{\left(x+2\right)}^{2}+{\left(y-1\right)}^{2}=4
and also describe the translation of the conic from the standard position.
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Protocol Design - Work - Nano Documentation
Protocol Design - Work
Work algorithm details
Protocol Design - Work¶
Spam resistance¶
A spam transaction is loosely defined as a block broadcasted with the intention of saturating the network, reducing its availability for other network participants, or increasing the size of the ledger. In order to make spam attempts more costly, each valid block in Nano requires a proof-of-work solution to be attached to it - similar to the original proposition of Hashcash1. Participants can compute the required work in the order of seconds. The cost of spamming the network then increases linearly with the number of spam transactions, thus reducing the impact of spam transactions from theoretically infinite to a manageable amount.
With this design, there is an added step of verifying a block's work. As one could spam invalid blocks (in this context, blocks with invalid work), one key requirement is that the cost of verifying work is negligible.
Work algorithm details¶
Every block includes a work field that must be correctly populated. Valid work is obtained by randomly guessing a nonce such that:
H(\text{nonce} || \text{x}) \ge \text{threshold}
where H
is an algorithm, usually in the form of a hash function, ||
is the concatenation operator, threshold
is a parameter of the network that relates to the resources spent to obtain a valid work, and x
The account's public key, in the case of the first block on the account, or
The previous block's hash
The following image illustrates the process by which valid work is obtained for Block 2.
The work field is not used when signing a block. This design has two consequences:
A block can be securely signed locally, while the work is requested from a remote server, with larger resources. This is especially important for devices with low resources.
Since all inputs are known before generating a block, a user can precompute the work for the next block, eliminating any time between creating and broadcasting a block. After a block is created, the next block's work can be computed immediately, using the last block's hash as input.
Choosing an algorithm¶
While the specific algorithm used is an implementation decision, there is a minimal set of requirements that must be met for compatibility with the Nano protocol.
Asymmetry. Verifying work should take the least amount of resources (including time) as possible.
Small proof size. Work should take a minimal amount of a block's size compared to the resources required to generate it, in order to reduce overhead and maximize throughput.
Amortization-free. The cost of obtaining work for multiple blocks should scale linearly with the number of blocks. This ensures fairness for all participants.
Progress-free. Any attempt at obtaining work should follow a stochastic process, with no dependence on previous attempts.
Additional requirements of parameter flexibility, constrained parallelism, and being optimization-free, are desired but not required 2.
Basics - PoW
Nano How 4: Proof of Work
For more details on these requirements, refer to A. Biryukov, "Equihash: Asymmetric Proof-of-Work Based on the Generalized Birthday Problem" 2017. [Online]. Available: https://doi.org/10.5195/ledger.2017.48 ↩
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Time-series databases - CS Notes Time-series databases | CS Notes
Coordination and time
Memory managmement
A time-series database stores time-series data.
Time series data is a sequence of data points collected over time intervals. Time series data is often alerted on and dashboarded for monitoring purposes.
An example time series for the number of server requests over 10s intervals:
// [server_req_count, unix_timestamp_ms]
[41, 1646680930000],
[15, 1646681050000]
Popular open-source time series DBs include Whisper (Graphite), Prometheus, InfluxDB, and Prometheus.
To improve retention, older time series data is often rolled up into increasingly low-resolution data points.
Gorilla is an in-memory time-series database created by Meta.
Gorilla was designed to support the following requirements:
Low latency queries (10s of ms)
Fine-grained aggregations over short time windows
Meta observed that the workload for their time series system was write-heavy. They also observed that the majority of reads were for recent data (~85% of requests could be satisfied with data <26 hours old). Based on these observations, Gorilla was built as a write-through cache storing recent data, with the rest of the data stored in HBase.
The data is a triple: <key: string, timestamp: uint64_t, value: double>. Data is sharded based on the key.
[1, Pp. 1816-9]
Gorilla uses two novel encodings: delta-of-delta encoding for integer timestamps and XOR encoding for floating point values.
In delta-of-delta encoding the delta between previous values is calculated, and then the delta between the previous delta is calculated and stored.
Timestamp 1645503469 1645503569 1645503669 1645503679
Delta 1645503469 100 100 10
Delta of delta 1645503469 100 0 10
Gorilla compresses timestamps using delta-of-delta encoding, based on the observation that timestamp data is often at fixed intervals (leading to small delta-of-deltas).
~96% of Meta’s timestamps could be encoded in a single bit using the following variable-bit scheme:
0 - store 0b0
[-63, 64] store 0b10 followed by value (7 bits)
[-255, 256] store 0b110 followed by value (9 bits)
[-2047, 2048] store 0b1110 followed by value (12 bits)
Else store 0b1111 followed by value (32 bits)
A block header stores the timestamp
t-1
, which is aligned to a two hour window. The first timestamp is then stored as the delta from
t-1
in 14 bits.
Data point values (doubles) are stored using XOR encoding, where data is based on an XOR of previous values.
The first value is stored without compression.
For the rest of the data, the following scheme is used:
If XOR with the previous value is 0, then store 0b0
If XOR is nonzero, calculate the number of leading and trailing zeros in the XOR. Store 0b1 followed by:
Control bit 0b0—if the block of meaningful bits is within the same block as the previous meaningful bits (there are at least as many leading zeros and trailing zeros as the previous one), then use that information for the block position and just store the meaningful XOR value.
Control bit 0b1—store the length of the number of leading 0s in the next 5 bits, then store the length of the meaningful XORed value in the next 6 bits, then store the meaningful bits of the XORed value.
Meta’s found ~51% of values are stored within a single bit, ~30% are stored with control bits 0b10, and the remainder are stored with control bits 0b11.
Tha main in-memory data structure for Gorilla is a TimeSeriesMap consisting of a vector of stdlib shared pointers to TimeSeries objects and a case-insensitive case-preserving map from time series names to a TimeSeries entry:
struct TimeSeriesMap {
ReadWriteLock *lock;
vector<sharded_ptr<TimeSeries>> *ts_vec;
unorderd_map<string, shared_ptr<TS>> *ts_map;
SpinLock *lock;
string open_block;
vector<string> closed_blocks;
struct ShardMap {
vector<unique_ptr<TSmap>>;
A ShardMap maps shard IDs to TSMaps. Null pointers are stored in the shard map if the shard isn’t held by the node [1, P. 1821].
A TimeSeries contains a sequence of closed blocks for data older than two hours and a single open block which is an append-only string where new values are added (it’s often reallocated as its size changes). When a block is closed, it’s moved to slab-allocated memory where it is left untouched until it’s deleted from memory [1, P. 1822].
Data is read by copying all data blocks that could contain data for a query’s key and time range directly into the output RPC structure. Decompression is done outside of Gorilla [1, P. 1822].
Gorilla achieves persistence by storing data in GlusterFS with 3X replication.
A Gorilla host owns multiple shards of data. It maintains a single directory per shard. A directory contains four types of files:
Key lists (map of key string to integer identifying index in in-memory vector)
Append-only logs
Complete block files
Each shard represents about 16GB on-disk storage [1, P. 1823].
New keys are appended to the key list. Gorilla scans all keys for each shard in order to re-write the file.
When data is streamed to Gorilla, it is stored in a log file in compressed format. Keys are interleaved, and so a timestamp-value pair is stored along with its 32-bit integer ID [1, P. 1822].
Gorilla doesn’t offer ACID guarantees (it’s not a WAL). Gorilla buffers ~64KB of data before writing it to the log file. The buffer is flushed on a clean shutdown but a crash can cause a small amount of data loss [1, P. 1822].
Every two hours, Gorilla copies the compressed block data to disk. The block has two sections: a set of consecutive 64KB slabs of compressed data blocks, and a list of <time_series_ID, data_block_pointer> pairs. When a block file is complete, Gorilla creates a checkpoint file (marking when a complete block file is flushed to disk) and deletes the corresponding logs [1, P. 1822].
If a block file isn’t flushed to disk on a crash then the new Gorilla process will find that the checkpoint file doesn’t exist. In this case it will read from the log file only [1, Pp. 1822-3].
Region failures are handled by having two Gorilla instances in separate DCs. Data is streamed to both instances and there is no attempt to guarantee consistency. In the case one instance fails, traffic is routed to the redundant instance.
Single node failure is handled using ShardManager—a Paxos-based system. When a node fails, ShardManager distributes its shards among the remaining nodes in the cluster. During shard movement, write clients buffer their incoming data (the buffer holds 1 minute of data and older data is dropped), this works for routine shard reassignment. If a Gorilla host crashes in a region, writes are buffered by the client and the Gorilla cluster attempts to resurrect the host. If the shard movement takes too long, reads can be pointed to the corresponding Gorilla host in the other region.
When a shard is added to a host, the host reads all the data from GlusterFS. A host can read all the data it needs to be fully functional in about 5 minutes. While the host is reading data, it accepts incoming data points and puts them in a queue to be processed. When shards are reassigned, clients drain their buffers by writing to the new node. In the case of a crash, as soon as a new host is assigned a shard it begins accepting streaming writes, so no in-flight data is lost. If a host shuts down gracefully then it flushes data to disk before exiting meaning that no data is lost (software upgrades can be handled via rolling upgrades using this mechanism).
If a host crashes before flushing the data to disk, the data is lost. In practice this is rare and only a few seconds of data will be lost, and so the increased write throughput is considered worth the tradeoff.
After a node failure, queries return partial data. When a client library receives a partial result, it will try the redundant region. In the case that both results are partial, the client returns the partial data with some flags so that users can be alerted to the status of the data.
[1] T. Pelkonen et al., “Gorilla: A fast, scalable, in-memory time series database,” Proceedings of the VLDB Endowment, vol. 8, no. 12, pp. 1816–1827, 2015.
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In writings on relativity time, the various relations are only changed by the transverse shift. This paper proves that the axial Doppler shift does that as well and gives some impacts of that on common differential relations in physics. When a modulated signal lasting a time = T is subjected to an optical Doppler shift K (either axial or transverse or both), where K is shifted frequency/original frequency, the Doppler shifted signal will last T/K. This because all shifted harmonics of its Fourier series (with a fundamental period of T) will last 1/K times the period of the original harmonic. The reader can graph any Fourier series and then graph its shifted series. The reader will see the shifted period is T/K. The Fourier series of the original repeats when time is greater than T and the shifted one when time is greater than T/K, which means the original series only represents the signal from time = 0 to T and the shifted series represents the shifted signal from time = 0 to T/K. Hence, the shifted one has all of the information in T/K as the original has in T. Therefore everything in the series including information is T/K long in the shifted series. Therefore, both the axial and the transverse Doppler shift change time periods in a vacuum, independent of material involved. That has not been obvious for over 100 years the axial shift changes time from the definition of frequency = 1/time.
Space Time, Relativity, Doppler, Time, Information Rates, Information Transfer, Missing Dimensions
Reich, S. (2019) A Fourier Series Proof That the Axial like the Transverse Optical Doppler Shift Impacts Time and Information Rates. Journal of High Energy Physics, Gravitation and Cosmology, 5, 992-994. doi: 10.4236/jhepgc.2019.54054.
i\hslash \text{d}\psi /\text{d}t=H\psi
ih\int \left({\partial }^{j}\left({}^{w}\psi \right)/\partial {t}^{j}\right)\text{d}j={}^{w}H{}^{w}\psi
i\hslash \underset{}{\overset{j}{\sum }}{\partial }^{j}{}^{w}\psi /\partial {t}^{j}={}^{w}H{}^{w}\psi
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The current entering the positive terminal of a device is i(t) = 6e
Josh Sizemore 2021-12-22 Answered
The current entering the positive terminal of a device is
i\left(t\right)=6{e}^{-2t}
mA and the voltage across the device is v(t) = 10di/dt V. (a) Find the charge delivered to the device between t = 0 and t = 2 s. (b) Calculate the power absorbed. (c) Determine the energy absorbed in 3 s.
{\int }_{0}^{2}i\left(t\right)dt={\int }_{0}^{2}6\cdot {e}^{-2t}dt=6\cdot {\int }_{0}^{2}{e}^{-2t}dt=6\cdot \left(-\frac{1}{2}{e}^{-2t}\right){\mid }_{0}^{2}=
=6\cdot \left(-\frac{1}{2}\cdot \left({e}^{-2\cdot 2}-{e}^{-2\cdot 0}\right)\right)=-3\cdot \left({e}^{-4}-1\right)=2,945
v\left(t\right)=10\cdot \frac{di}{dt}=10\cdot \frac{d}{dt}\cdot \left(6{e}^{-2t}\right)=10\cdot \left(-12{e}^{-2t}\right)=-120\cdot {e}^{-2t}
p\left(t\right)=v\left(t\right)\cdot i\left(t\right)=-120\cdot {e}^{-2t}\cdot 6{e}^{-2t}=-720\cdot {e}^{-4t}\text{ }\mu W
W={\int }_{0}^{3}p\left(t\right)dt=-720\cdot {\int }_{0}^{3}{e}^{-4t}dt=-720\cdot \left(-\frac{1}{4}{e}^{-4t}\right){\mid }_{0}^{3}=
=180{e}^{-4t}{\mid }_{0}^{3}=180\left({e}^{-12}-{e}^{0}\right)=180\left({e}^{-12}-1\right)=-180\text{ }\mu J
{F}_{A}
{F}_{B}
{F}_{A}=4200
{F}_{B}
{F}_{A}+{F}_{B}
{12.0}^{\circ }
In the following, simplify using absolute value signs as needed.
\sqrt[3]{80{x}^{7}{y}^{6}}
Match the parametric equations with the graphs labeledI-VI. Give reasons for your choices. (Do not use a graphingdevice).
x={t}^{4}-t+1,\text{ }y={t}^{2}
x={t}^{2}-2t,\text{ }y=\sqrt{t}
x=\mathrm{sin}2t,\text{ }y=\mathrm{sin}\left(t+\mathrm{sin}2t\right)
x=\mathrm{cos}5t,\text{ }y=\mathrm{sin}2t
x=t+\mathrm{sin}4t,\text{ }y={t}^{2}+\mathrm{cos}3t
x=\frac{\mathrm{sin}2t}{4+{t}^{2}},\text{ }y=\frac{\mathrm{cos}2t}{4+{t}^{2}}
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Option price and sensitivities by Bates model using FFT and FRFT - MATLAB optSensByBatesFFT - MathWorks Australia
\mathrm{max}\left(St-K,0\right)
\mathrm{max}\left(K-St,0\right)
\begin{array}{l}d{S}_{t}=\left(r-q-{\lambda }_{p}{\mu }_{J}\right){S}_{t}dt+\sqrt{{v}_{t}}{S}_{t}d{W}_{t}+J{S}_{t}d{P}_{t}\\ d{v}_{t}=\kappa \left(\theta -{v}_{t}\right)dt+{\sigma }_{v}\sqrt{{v}_{t}}d{W}_{t}\\ \text{E}\left[d{W}_{t}d{W}_{t}^{v}\right]=pdt\\ \text{prob(}d{P}_{t}=1\right)={\lambda }_{p}dt\end{array}
\mathrm{ln}\left(1+{\mu }_{J}\right)-\frac{{\delta }^{2}}{2}
\frac{1}{\left(1+J\right)\delta \sqrt{2\pi }}\mathrm{exp}\left\{{\frac{-\left[\mathrm{ln}\left(1+J\right)-\left(\mathrm{ln}\left(1+{\mu }_{J}\right)-\frac{{\delta }^{2}}{2}\right]}{2{\delta }^{2}}}^{2}\right\}
{W}_{t}^{v}
{\lambda }_{p}
{\lambda }_{p}
{f}_{Bate{s}_{j}\left(\varphi \right)}
\begin{array}{l}{f}_{Bates\left(\varphi \right)}=\mathrm{exp}\left({C}_{j}+{D}_{j}{v}_{0}+i\varphi \mathrm{ln}{S}_{t}\right)\mathrm{exp}{\left({\lambda }_{p}\tau \left(1+{\mu }_{J}\right)}^{{m}_{j}+\frac{1}{2}}\left[{\left(1+{\mu }_{j}\right)}^{i\varphi }{e}^{{\delta }^{2}\left({m}_{j}i\varphi +\frac{{\left(i\varphi \right)}^{2}}{2}\right)}-1\right]-{\lambda }_{p}\tau {\mu }_{J}i\varphi \right)\\ {m}_{j}=\left\{\begin{array}{l}{m}_{1}=\frac{1}{2}\\ {m}_{2}=-\frac{1}{2}\end{array}\right\}\\ {C}_{j}=\left(r-q\right)i\varphi \tau +\frac{\kappa \theta }{{\sigma }_{v}{}^{2}}\left[\left({b}_{j}-p{\sigma }_{v}i\varphi +{d}_{j}\right)\tau -2\mathrm{ln}\left(\frac{1-{g}_{j}{e}^{{d}_{j}\tau }}{1-{g}_{j}}\right)\right]\\ Dj=\frac{{b}_{j}-p{\sigma }_{v}i\varphi +{d}_{j}}{{\sigma }_{v}^{2}}\left(\frac{1-{e}^{{d}_{j}\tau }}{1-{g}_{j}{e}^{{d}_{j}\tau }}\right)\\ {g}_{j}=\frac{{b}_{j}-p{\sigma }_{v}i\varphi +{d}_{j}}{{b}_{j}-p{\sigma }_{v}i\varphi -{d}_{j}}\\ {d}_{j}=\sqrt{{\left({b}_{j}-p{\sigma }_{v}i\varphi \right)}^{2}-{\sigma }_{v}^{2}\left(2{u}_{j}i\varphi -{\varphi }^{2}\right)}\\ \text{where for }j=1,2:\\ {u}_{1}=\frac{1}{2},{u}_{2}=-\frac{1}{2},{b}_{1}=\kappa +{\lambda }_{VolRisk}-p{\sigma }_{v},{b}_{2}=\kappa +{\lambda }_{VolRisk}\end{array}
\begin{array}{l}{C}_{j}=\left(r-q\right)i\varphi \tau +\frac{\kappa \theta }{{\sigma }_{v}{}^{2}}\left[\left({b}_{j}-p{\sigma }_{v}i\varphi -{d}_{j}\right)\tau -2\mathrm{ln}\left(\frac{1-{\epsilon }_{j}{e}^{-{d}_{j}\tau }}{1-{\epsilon }_{j}}\right)\right]\\ Dj=\frac{{b}_{j}-p{\sigma }_{v}i\varphi -{d}_{j}}{{\sigma }_{v}^{2}}\left(\frac{1-{e}^{-{d}_{j}\tau }}{1-{\epsilon }_{j}{e}^{-{d}_{j}\tau }}\right)\\ {\epsilon }_{j}=\frac{{b}_{j}-p{\sigma }_{v}i\varphi -{d}_{j}}{{b}_{j}-p{\sigma }_{v}i\varphi +{d}_{j}}\end{array}
\begin{array}{l}Call\left(k\right)=\frac{{e}^{-\alpha k}}{\pi }{\int }_{0}^{\infty }\mathrm{Re}\left[{e}^{-iuk}\psi \left(u\right)\right]du\\ \psi \left(u\right)=\frac{{e}^{-r\tau }{f}_{2}\left(\varphi =\left(u-\left(\alpha +1\right)i\right)\right)}{{\alpha }^{2}+\alpha -{u}^{2}+iu\left(2\alpha +1\right)}\\ Put\left(K\right)=Call\left(K\right)+K{e}^{-r\tau }-{S}_{t}{e}^{-q\tau }\end{array}
\mathrm{ln}\left({S}_{t}\right)-\frac{N}{2}\Delta k
\mathrm{ln}\left({S}_{t}\right)+\left(\frac{N}{2}-1\right)\Delta k
{S}_{t}\mathrm{exp}\left(-\frac{N}{2}\Delta k\right)
{S}_{t}\mathrm{exp}\left[\left(\frac{N}{2}-1\right)\Delta k\right]
Call\left({k}_{n}\right)=\Delta u\frac{{e}^{-\alpha {k}_{n}}}{\pi }\sum _{j=1}^{N}\mathrm{Re}\left[{e}^{-i\Delta k\Delta u\left(j-1\right)\left(n-1\right){e}^{i{u}_{j}}\left[\frac{N\Delta k}{2}-\mathrm{ln}\left({S}_{t}\right)\right]}\psi \left({u}_{j}\right)\right]{w}_{j}
\Delta k\Delta u=\left(\frac{2\pi }{N}\right)
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Orientation from accelerometer, gyroscope, and magnetometer readings - Simulink - MathWorks Australia
Initial process noise
Accelerometer noise ((m/s2)2)
Gyroscope noise ((rad/s)2)
Magnetometer noise (μT2)
Gyroscope drift noise (rad/s)
Linear acceleration noise ((m/s2)2)
Magnetic disturbance noise (μT2)
Linear acceleration decay factor
Magnetic disturbance decay factor
Magnetic field strength (μT)
Navigation Toolbox / Multisensor Positioning / Navigation Filters
Sensor Fusion and Tracking Toolbox / Multisensor Positioning / Navigation Filters
The AHRS Simulink® block fuses accelerometer, magnetometer, and gyroscope sensor data to estimate device orientation.
Accel — Accelerometer readings in sensor body coordinate system (m/s2)
Accelerometer readings in the sensor body coordinate system in m/s2, specified as an N-by-3 matrix of real scalars. N is the number of samples, and the three columns of Accel represent the [x y z] measurements, respectively.
Gyro — Gyroscope readings in sensor body coordinate system (rad/s)
Gyroscope readings in the sensor body coordinate system in rad/s, specified as an N-by-3 matrix of real scalars. N is the number of samples, and the three columns of Gyro represent the [x y z] measurements, respectively.
Mag — Magnetometer readings in sensor body coordinate system (µT)
Magnetometer readings in the sensor body coordinate system in µT, specified as an N-by-3 matrix of real scalars. N is the number of samples, and the three columns of magReadings represent the [x y z] measurements, respectively.
Orientation — Orientation of sensor body frame relative to navigation frame
M-by-4 array of scalar | 3-by-3-by-M-element rotation matrix
Orientation of the sensor body frame relative to the navigation frame, return as an M-by-4 array of scalars or a 3-by-3-by-M array of rotation matrices. Each row the of the N-by-4 array is assumed to be the four elements of a quaternion. The number of input samples, N, and the Decimation Factor parameter determine the output size M.
Angular Velocity — Angular velocity in sensor body coordinate system (rad/s)
M-by-3 array of real scalar (default)
Angular velocity with gyroscope bias removed in the sensor body coordinate system in rad/s, returned as an M-by-3 array of real scalars. The number of input samples, N, and the Decimation Factor parameter determine the output size M.
Decimation factor by which to reduce the input sensor data rate, specified as a positive integer.
The number of rows of the inputs –– Accel, Gyro , and Mag –– must be a multiple of the decimation factor.
Initial process noise — Initial process noise
ahrsfilter.defaultProcessNoise (default) | 12-by-12 matrix of real scalar
Initial process noise, specified as a 12-by-12 matrix of real scalars. The default value, ahrsfilter.defaultProcessNoise, is a 12-by-12 diagonal matrix as:
Orientation format — Orientation output format
Output orientation format, specified as 'quaternion' or 'Rotation matrix':
'quaternion' –– Output is an M-by-4 array of real scalars. Each row of the array represents the four components of a quaternion.
'Rotation matrix' –– Output is a 3-by-3-by-M rotation matrix.
The output size M depends on the input dimension N and the Decimation Factor parameter.
Accelerometer noise ((m/s2)2) — Variance of accelerometer signal noise ((m/s2)2)
Gyroscope noise ((rad/s)2) — Variance of gyroscope signal noise ((rad/s)2)
Magnetometer noise (μT2) — Variance of magnetometer signal noise (μT2)
Gyroscope drift noise (rad/s) — Variance of gyroscope offset drift ((rad/s)2)
Linear acceleration noise ((m/s2)2) — Variance of linear acceleration noise (m/s2)2
Magnetic disturbance noise (μT2) — Variance of magnetic disturbance noise (μT2)
Linear acceleration decay factor — Decay factor for linear acceleration drift
Decay factor for linear acceleration drift, specified as a scalar in the range [0,1). If linear acceleration changes quickly, set this parameter to a lower value. If linear acceleration changes slowly, set this parameter to a higher value. Linear acceleration drift is modeled as a lowpass-filtered white noise process.
Magnetic disturbance decay factor — Decay factor for magnetic disturbance
Magnetic field strength (μT) — Magnetic field strength (μT)
Magnetic field strength in μT, specified as a real positive scalar. The magnetic field strength is an estimate of the magnetic field strength of the Earth at the current location.
The AHRS block uses the nine-axis Kalman filter structure described in [1]. The algorithm attempts to track the errors in orientation, gyroscope offset, linear acceleration, and magnetic disturbance to output the final orientation and angular velocity. Instead of tracking the orientation directly, the indirect Kalman filter models the error process, x, with a recursive update:
{x}_{k}=\left[\begin{array}{c}{\theta }_{k}\\ {b}_{k}\\ {a}_{k}\\ {d}_{k}\end{array}\right]={F}_{k}\left[\begin{array}{c}{\theta }_{k-1}\\ {b}_{k-1}\\ {a}_{k-1}\\ {d}_{k-1}\end{array}\right]+{w}_{k}
\begin{array}{l}{x}_{k}^{-}={F}_{k}{x}_{k-1}^{+}\\ {P}_{k}^{-}={F}_{k}{P}_{k-1}^{+}{F}_{k}^{T}+{Q}_{k}\\ \\ {y}_{k}={z}_{k}-{H}_{k}{x}_{k}^{-}\\ {S}_{k}\text{\hspace{0.17em}}={R}_{k}+{H}_{k}{P}_{k}^{-}{H}_{k}{}^{T}\\ {K}_{k}={P}_{k}^{-}{H}_{k}^{T}{\left({S}_{k}\right)}^{-1}\\ {x}_{k}^{+}={x}_{k}^{-}+{K}_{k}{y}_{k}\\ {P}_{k}^{+}={P}_{k}{}^{-}-{K}_{k}{H}_{k}{P}_{k}^{-}\end{array}
\begin{array}{l}{x}_{k}^{-}=0\\ {P}_{k}^{-}={Q}_{k}\\ \\ {y}_{k}={z}_{k}\\ {S}_{k}\text{\hspace{0.17em}}={R}_{k}+{H}_{k}{P}_{k}^{-}{H}_{k}{}^{T}\\ {K}_{k}={P}_{k}^{-}{H}_{k}^{T}{\left({S}_{k}\right)}^{-1}\\ {x}_{k}^{+}={K}_{k}{y}_{k}\\ {P}_{k}^{+}={P}_{k}{}^{-}-{K}_{k}{H}_{k}{P}_{k}^{-}\end{array}
\Delta {\phi }_{N×3}=\frac{\left(gyroReading{s}_{N×3}-gyroOffse{t}_{1×3}\right)}{fs}
where N is the decimation factor specified by the Decimation factor and fs is the sample rate.
\Delta {Q}_{N×1}=\mathrm{quaternion}\left(\Delta {\phi }_{N×3},\text{'}rotvec\text{'}\right)
{q}_{1×1}^{-}=\left({q}_{1×1}^{+}\right)\left(\prod _{n=1}^{N}\Delta {Q}_{n}\right)
{g}_{1×3}={\left(rPrior\left(:,3\right)\right)}^{T}
gAcce{l}_{1×3}=accelReading{s}_{1×3}-linAccelprio{r}_{1×3}
mGyr{o}_{1×3}={\left(\left(rPrior\right)\left({m}^{T}\right)\right)}^{T}
{z}_{g}=g-gAccel
{z}_{m}=mGyro-magReadings
mErro{r}_{3×1}={\left(\left(K{\left(\text{10:12},:\right)}_{3×6}\right){\left({z}_{1×6}\right)}^{T}\right)}^{T}
tf=\left\{\begin{array}{c}\text{true}\\ \text{false}\end{array}\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{else}\end{array}\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sum |mError|}^{2}>\left(4\right){\left(\text{ExpectedMagneticFieldStrength}\right)}^{2}\\ \text{\hspace{0.17em}}\end{array}
{H}_{3×9}=\left[\begin{array}{cccccccccccc}0& {g}_{z}& -{g}_{y}& 0& -\kappa {g}_{z}& \kappa {g}_{y}& 1& 0& 0& 0& 0& 0\\ -{g}_{z}& 0& {g}_{x}& \kappa {g}_{z}& 0& -\kappa {g}_{x}& 0& 1& 0& 0& 0& 0\\ {g}_{y}& -{g}_{x}& 0& -\kappa {g}_{y}& \kappa {g}_{x}& 0& 0& 0& 1& 0& 0& 0\\ 0& {m}_{z}& -{m}_{y}& 0& -\kappa {m}_{z}& -\kappa {m}_{y}& 0& 0& 0& -1& 0& 0\\ -{m}_{z}& 0& {m}_{x}& \kappa {m}_{z}& 0& -\kappa {m}_{x}& 0& 0& 0& 0& -1& 0\\ {m}_{y}& -{m}_{x}& 0& -\kappa {m}_{y}& \kappa {m}_{x}& 0& 0& 0& 0& 0& 0& -1\end{array}\right]
where gx, gy, and gz are the x-, y-, and z-elements of the gravity vector estimated from the a priori orientation, respectively. mx, my, and mz are the x-, y-, and z-elements of the magnetic vector estimated from the a priori orientation, respectively. κ is a constant determined by the Sample rate and Decimation factor properties: κ = Decimation factor/Sample rate.
{S}_{6x6}={R}_{6x6}+\left({H}_{6x12}\right)\left({P}_{{}_{12x12}}^{-}\right){\left({H}_{6x12}\right)}^{T}
{R}_{6×6}=\left[\begin{array}{cccccc}acce{l}_{\text{noise}}& 0& 0& 0& 0& 0\\ 0& acce{l}_{\text{noise}}& 0& 0& 0& 0\\ 0& 0& acce{l}_{\text{noise}}& 0& 0& 0\\ 0& 0& 0& ma{g}_{\text{noise}}& 0& 0\\ 0& 0& 0& 0& ma{g}_{\text{noise}}& 0\\ 0& 0& 0& 0& 0& ma{g}_{\text{noise}}\end{array}\right]
acce{l}_{\text{noise}}=\text{AccelerometerNoise}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\text{LinearAccelerationNoise}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}{\kappa }^{2}\left(\text{GyroscopeDriftNoise}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\text{GyroscopeNoise}\right)
ma{g}_{\text{noise}}=\text{MagnetometerNoise}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\text{MagneticDisturbanceNoise}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}{\kappa }^{2}\left(\text{GyroscopeDriftNoise}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\text{GyroscopeNoise}\right)
{P}_{{}_{12×12}}^{+}={P}_{{}_{12×12}}^{-}-\left({K}_{12×6}\right)\left({H}_{6×12}\right)\left({P}_{{}_{12×12}}^{-}\right)
Q=\left[\begin{array}{cccccccccccc}{P}^{+}\left(1\right)+{\kappa }^{2}{P}^{+}\left(40\right)+\beta +\eta & 0& 0& -\kappa \left({P}^{+}\left(40\right)+\beta \right)& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& {P}^{+}\left(14\right)+{\kappa }^{2}{P}^{+}\left(53\right)+\beta +\eta & 0& 0& -\kappa \left({P}^{+}\left(53\right)+\beta \right)& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& {P}^{+}\left(27\right)+{\kappa }^{2}{P}^{+}\left(66\right)+\beta +\eta & 0& 0& -\kappa \left({P}^{+}\left(66\right)+\beta \right)& 0& 0& 0& 0& 0& 0\\ -\kappa \left({P}^{+}\left(40\right)+\beta \right)& 0& 0& {P}^{+}\left(40\right)+\beta & 0& 0& 0& 0& 0& 0& 0& 0\\ 0& -\kappa \left({P}^{+}\left(53\right)+\beta \right)& 0& 0& {P}^{+}\left(53\right)+\beta & 0& 0& 0& 0& 0& 0& 0\\ 0& 0& -\kappa \left({P}^{+}\left(66\right)+\beta \right)& 0& 0& {P}^{+}\left(66\right)+\beta & 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& {\nu }^{2}{P}^{+}\left(79\right)+\xi & 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& {\nu }^{2}{P}^{+}\left(92\right)+\xi & 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& {\nu }^{2}{P}^{+}\left(105\right)+\xi & 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& {\sigma }^{2}{P}^{+}\left(118\right)+\gamma & 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& {\sigma }^{2}{P}^{+}\left(131\right)+\gamma & 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& {\sigma }^{2}{P}^{+}\left(144\right)+\gamma \end{array}\right]
κ –– Decimation factor divided by sample rate.
β –– Gyroscope drift noise.
η –– Gyroscope noise.
ν –– Linear acceleration decay factor.
ξ –– Linear acceleration noise.
σ –– Magnetic disturbance decay factor.
γ –– Magnetic disturbance noise.
{K}_{12×6}=\left({P}_{{}_{12×12}}^{-}\right){\left({H}_{6×12}\right)}^{T}{\left({\left({S}_{6×6}\right)}^{T}\right)}^{-1}
{x}_{12×1}=\left({K}_{12×6}\right){\left({z}_{1×6}\right)}^{T}
{x}_{9×1}=\left(K\left(\text{1:9},\text{1:3}\right){\left({z}_{g}\right)}^{T}
{q}^{+}=\left({q}^{-}\right)\left({\theta }^{+}\right)
linAccelPrior=\left(linAccelPrio{r}_{k-1}\right)\nu -{b}^{+}
ν –– Linear acceleration decay factor
gyroOffset=gyroOffse{t}_{k-1}-{a}^{+}
angularVelocit{y}_{1×3}=\frac{\sum gyroReading{s}_{N×3}}{N}-gyroOffse{t}_{1×3}
mErrorNE{D}_{1×3}={\left({\left(rPos{t}_{3×3}\right)}^{T}{\left(mErro{r}_{1×3}\right)}^{T}\right)}^{T}
Μ=m-mErrorNED
inclination=\text{atan2}\left(Μ\left(3\right),Μ\left(1\right)\right)
\begin{array}{l}m\left(1\right)=\left(\text{ExpectedMagneticFieldStrength}\right)\left(\mathrm{cos}\left(inclination\right)\right)\\ m\left(2\right)=0\\ m\left(3\right)=\left(\text{ExpectedMagneticFieldStrength}\right)\left(\mathrm{sin}\left(inclination\right)\right)\end{array}
ahrsfilter | ecompass | imufilter | imuSensor | gpsSensor
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Nonlinear damping in vibration of CFRP plates | JVE Journals
Olga Kazakova1 , Igor Smolin2 , Iosif Bezmozgiy3
1, 2, 3Tomsk State University, Tomsk, 634050, Russia
1, 3S. P. Korolev Rocket and Space Public Corporation Energia, Korolev, Moscow Area, 141070, Russia
2Institute of Strength Physics and Materials Science SB RAS, Tomsk, 634055, Russia
The article describes research results of damping properties of carbon fiber reinforced plastics (CFRP). The effect of stress/strain levels on the damping value is studied. Research is conducted on flat samples (plates) with different lay-up schemes from 1-layered to 12-layered. The paper contains information about the modal and harmonic tests on the samples and their numerical modeling.
Keywords: modal analysis, harmonic analysis, finite element model, non-linear damping properties, carbon fiber reinforced plastics.
Modern trends in dynamic strength in rocket space technology dictate avoiding expensive model tests in favor of protoflight tests. These tests are performed on protoflight hardware under qualification and acceptance testing conditions. One of the requirements necessary for conducting such tests is to provide a reliable verified dynamic finite element model (FEM) of the hardware. Creating of models of structures from composite materials raises the question of the choice of the damping coefficient for the calculation. Using of a common approach with linear damping determination without stress dependency cannot create reliable models that will give accurate results if hardware composition or testing conditions are changed. In this regard it is required to investigate the dependence between damping ratio and stress value. The dependence of the damping ration on the vibration amplitude was mentioned also in [1].
In order to study the dissipative properties of CFRP, 12 types of flat samples were made with gradual increasing of structure complexity from 1 to 12 layers (3 samples of each type). Corresponding FEM for each sample type was created using ANSYS software. An example of the sample and its FEM are shown in Fig. 1.
Fig. 1. Layered composite sample: a) the test sample, b) finite element model of the sample
The samples were exposed to different types of dynamic tests. This article contains the results of modal and harmonic testing along with computer modeling.
The purpose of the tests is to determine the vibration characteristics (natural frequencies and mode shapes) of a structure. As a result of setting a sample FEM stiffness property of material and boundary conditions are specified according to modal test. Tests were conducted using Polytec scanning laser vibrometer in the frequency range from 0 to 5000 Hz. The experimental and calculated natural frequencies and mode shapes of the sample are shown on Fig. 2. Frequency error in calculations is less than 5 percent. Third mode is the exception; its frequency error is around 11 percent. This verified modal tests model of a sample was also used for harmonic analysis.
Fig. 2. Experimental (left) and calculated (right) natural frequencies and mode shapes: first (a, b), second (c, d), third (e, f) and forth (g, h) modes
Fr
= 20.7 Hz
Fr
Fr
= 215.2 Hz
Fr
Fr
Fr
Fr
Fr
The purpose of the harmonic test is to determine structure response to the input load. Testing scheme is shown on Fig. 3.
Fig. 3. Testing scheme for a sample with three-axis vibration transducer (1) and strain gauge transducer (2)
The samples were exposed to sinusoidal input action in a frequency range from 0 to 100 Hz and in a wide amplitude range from 0.2 g to 1.5 g. Sensor data were recorded by three-axis vibration transducer and strain gauge transducer.
Figure 4 shows an example of obtained amplitude-frequency acceleration response; similar graphs of amplitude-frequency strain response were also obtained.
Fig. 4. Experimental amplitude response registered on two samples: a) input 1.0 g, b) input 1.5 g
The values of the first natural frequency of the sample, damping ratio, acceleration amplitude in the direction of input force (Z-axis) and the strain amplitude obtained from the experiments are presented in Table 1.
Table 1. Numerical characteristics of the response registered on samples
Impact, g
Natural frequency, Hz
Acceleration amplification factor
Strain, μE
Harmonic analysis was also accompanied by computer modelling. Setting damping parameters is required for calculation. The ANSYS software has multiple damping parameters, but all they do not depend on stress-stain state of construction [2]. It is possible to use only the frequency-dependent damping.
Harmonic analysis was conducted by superposition method which makes it possible to set the frequency-independent damping DMPR. Calculations enable the response of the structure in the place of three-axis vibration transducer installation to be determined. Damping ratio is determined using the half-power method from previously received amplitude frequency response plot [3]. When the same damping parameter DMPR for all materials of the computational model is assigned, attenuation values will be equal to the specified DMPR value. Graphs of acceleration amplification factor versus frequency for DMPR = 0.04 and in relation to DMPR on resonance frequency are shown on Fig. 5.
Fig. 5. Calculated amplitude response (a); the dependence of the amplitude response on damping ratio (b)
Experimental acceleration amplification factors registered on two samples were equal to 14.1 and 15.0, whereas the numerical modeling gives the value of 13.6.
In calculation a single impact equal to 1.0 g is applied to model. In experiments the impact varies. To compare experimental and calculation outputs, the main deformation obtained from test data was divided by the value of the input action. Thus for input 1.5 g the experimental main deformation corresponds to 685 μE, whereas the calculated one is 990 μE. The proximity of the experimental and calculated acceleration amplification factors and the difference between the values of the principal strain on the root of the sample may be indicative of the different modes which are connected with the nonuniformity of the damping field due to dependence of the damping ratio upon the stress-strain state of material.
In this regard samples’ FEM were modified. Various materials depending on the stress/strain level have been assigned to elements of computational samples models. Fig. 6 shows the distribution of principal strain in the sample, obtained by calculation with the general damping DMPR 0.04 and sample FEM with different kinds of materials (M1–M8).
Fig. 6. Calculated principal strain of the sample (a), sample model with different materials (b)
This model allows specifying different damping values for the model elements depending on the stress levels in the element. This approach allows determining the effect of damping in each zone on the stress (strain) distribution over the sample surface, and, after verification of the model from the test results, to identify the relationship between the stress levels and damping ratio.
For these purposes, parameter DMPR was alternately varied for each material at a constant zero damping in other materials of the sample and test calculations were carried out. As a result of test calculations the responses at the location of vibration transducer were defined, using which the effective damping ratio by the half-power method and the natural frequency were identified. The results are presented in the form of dependency graphs between the effective damping ratio and DMPR parameter of different materials and as impact of damping ration on the first resonant frequency value shown on Fig. 7.
Fig. 7. Relationships between damping ratio and DMPR (a) and first natural frequency (b)
The next step of model verification was to compare differences in the strain (stress) distribution on the sample with the same value of the effective damping ratio (equal to 0.04), which was accomplished by alternately setting the damping parameter DMPR to materials M5–M8 with zero damping in other areas.
Fig. 8 shows the distribution of the principal strain (stress) at the sample central nodes along X-axis for calculation cases 1–4 of Table 2 and using the general damping parameter DMPR (0.04) for all materials. The figure shows the strain difference of 110 μE in the root of the sample obtained for the calculation with damping at the root (M8) by comparison to calculations with damping in other materials (M5-M7).
Table 2. Numerical characteristics of the calculated response for different materials
Fig. 8. The change in principal strain on sample central nodes
1) Damping ratio has a significant effect on the response of the structure.
2) Damping value in various zones of the sample affects the shape of the sample modes and insignificantly affects first natural frequency.
3) If the deformation field of the sample during vibration is determined, relationships between strain (stress) and the damping coefficient can be identified.
4) To apply the obtained relationships in calculations an iterative-based algorithm for considering damping non-linearity is needed.
The research was supported by “The Tomsk State University Academic D. I. Mendeleev Fund Program”, Grant No. 8.2.19.2015.
Khan S. U., Li C. Y., Siddiqui N. A., Kim J.-K. Vibration damping characteristics of carbon fiber-reinforced composites containing multi-walled carbon nanotubes. Composites Science and Technology, Vol. 71, Issue 12, 2011, p. 1486-1494. [Search CrossRef]
Mechanical APDL. Release 16.1, Help System, Structural Analysis Guide, ANSYS, Inc. [Search CrossRef]
Heylen W., Lammens S., Sas P. Modal Analysis Theory and Testing. Division of Production Engineering, Machine Design and Automation, Leuven (Heverlee), Belgium, 2008. [Search CrossRef]
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Lyman-alpha_emitter Knowpia
A Lyman-alpha emitter (LAE) is a type of distant galaxy that emits Lyman-alpha radiation from neutral hydrogen.
A Lyman alpha emitter (left) and an artist's impression of what one might look like if viewed at a relatively close distance (right).
Most known LAEs are extremely distant, and because of the finite travel time of light they provide glimpses into the history of the universe. They are thought to be the progenitors of most modern Milky Way type galaxies. These galaxies can be found nowadays rather easily in narrow-band searches by an excess of their narrow-band flux at a wavelength which may be interpreted from their redshift:
{\displaystyle 1+z={\frac {\lambda }{1215.67\mathrm {\AA} }}}
where z is the redshift,
{\displaystyle \lambda }
is the observed wavelength, and 1215.67 Å is the wavelength of Lyman-alpha emission. The Lyman-alpha line in most LAEs is thought to be caused by recombination of interstellar hydrogen that is ionized by an ongoing burst of star-formation. Such Lyman alpha emission was first suggested as a signature of young galaxies by Bruce Partridge and P. J. E. Peebles in 1967.[1] Experimental observations of the redshift of LAEs are important in cosmology[2] because they trace dark matter halos and subsequently the evolution of matter distribution in the universe.
Lyman-alpha emitters are typically low mass galaxies of 108 to 1010 solar masses. They are typically young galaxies that are 200 to 600 million years old, and they have the highest specific star formation rate of any galaxies known. All of these properties indicate that Lyman-alpha emitters are important clues as to the progenitors of modern Milky Way type galaxies.
Lyman-alpha emitters have many unknown properties. The Lyman-alpha photon escape fraction varies greatly in these galaxies. This is what portion of the light emitted at the Lyman-alpha line wavelength inside the galaxy actually escapes and will be visible to distant observers. There is much evidence that the dust content of these galaxies could be significant and therefore is obscuring the brightness of these galaxies. It is also possible that anisotropic distribution of hydrogen density and velocity play a significant role in the varying escape fraction due to the photons' continued interaction with the hydrogen gas (radiative transfer).[3] Evidence now shows strong evolution in the Lyman Alpha escape fraction with redshift, most likely associated with the buildup of dust in the ISM. Dust is shown to be the main parameter setting the escape of Lyman Alpha photons.[4] Additionally the metallicity, outflows, and detailed evolution with redshift is unknown.
Importance in cosmologyEdit
LAEs are important probes of reionization,[5] cosmology (BAO), and they allow probing of the faint end of the luminosity function at high redshift.
The baryonic acoustic oscillation signal should be evident in the power spectrum of Lyman-alpha emitters at high redshift.[6] Baryonic acoustic oscillations are imprints of sound waves on scales where radiation pressure stabilized the density perturbations against gravitational collapse in the early universe. The three-dimensional distribution of the characteristically homogeneous Lyman-alpha galaxy population will allow a robust probe of cosmology. They are a good tool because the Lyman-alpha bias, the propensity for galaxies to form in the highest overdensity of the underlying dark matter distribution, can be modeled and accounted for. Lyman-alpha emitters are over dense in clusters.
Lyman-break galaxy
^ Partridge, R. B.; Peebles, P. J. E. (1967). "Are Young Galaxies Visible?". The Astrophysical Journal. 147: 868. Bibcode:1967ApJ...147..868P. doi:10.1086/149079. ISSN 0004-637X.
^ Nilsson (2007). "The Lyman-alpha Emission Line as a Cosmological Tool". arXiv:0711.2199. Bibcode:2007PhDT.......106N. {{cite journal}}: Cite journal requires |journal= (help)
^ Zheng, Zheng; Wallace, Joshua (2014). "Anisotropic Lyman-Alpha Emission". The Astrophysical Journal. 794 (2): 116. arXiv:1308.1405. Bibcode:2014ApJ...794..116Z. doi:10.1088/0004-637X/794/2/116. S2CID 119308774.
^ Blanc, Guillermo A.; Gebhardt, K.; Hill, G. J.; Gronwall, C.; Ciardullo, R.; Finkelstein, S.; Gawiser, E.; HETDEX Collaboration (2012). "HETDEX: Evolution of Lyman Alpha Emitters". American Astronomical Society Meeting Abstracts #219. 219: 424.13. Bibcode:2012AAS...21942413B.
^ Clément, B.; Cuby, J.-G.; Courbin, F.; Fontana, A.; Freudling, W.; Fynbo, J.; Gallego, J.; Hibon, P.; Kneib, J.-P.; Le Fèvre, O.; Lidman, C.; McMahon, R.; Milvang-Jensen, B.; Moller, P.; Moorwood, A.; Nilsson, K. K.; Pentericci, L.; Venemans, B.; Villar, V.; Willis, J. (2012). "Evolution of the observed Lyαluminosity function from z = 6.5 to z = 7.7: Evidence for the epoch of reionization?". Astronomy & Astrophysics. 538: A66. arXiv:1105.4235. Bibcode:2012A&A...538A..66C. doi:10.1051/0004-6361/201117312. S2CID 56301110.
^ [1] Constraining Cosmology with Lyman-alpha Emitters a Study Using HETDEX Parameters
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Revision as of 09:46, 1 November 2018 by Wibke (talk | contribs) (→Introduction)
{\displaystyle {(H=h(T,p)+0.5u_{j}^{2})}}
{\displaystyle {{\dfrac {\partial {\overline {\rho }}}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}=0}}
{\displaystyle {\text{(4.1)}}}
{\displaystyle {{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{i}\right)}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{i}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}=-{\dfrac {\partial {\overline {p}}}{\partial x_{i}}}+{\dfrac {\partial }{\partial x_{j}}}\left(\left(\mu +\mu _{t}\right){\dfrac {\partial {\widetilde {u}}_{i}}{\partial x_{j}}}\right)}}
{\displaystyle {\text{(4.2)}}}
{\displaystyle {{\dfrac {\partial \left({\overline {\rho }}{\widetilde {H}}\right)}{\partial t}}-{\dfrac {\partial {\overline {p}}}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{j}{\widetilde {H}}\right)}{\partial x_{j}}}={\dfrac {\partial }{\partial x_{j}}}\left(\lambda {\dfrac {\partial {\widetilde {T}}}{\partial x_{j}}}+{\dfrac {\mu _{t}}{Pr_{t}}}{\dfrac {\partial {\widetilde {h}}}{\partial x_{j}}}\right)}}
{\displaystyle {\text{(4.3)}}}
{\displaystyle {\lambda }}
{\displaystyle {Pr_{t}}}
{\displaystyle {\overline {\Phi }}}
{\displaystyle {\widetilde {\Phi }}}
{\displaystyle {{\overline {p}}={\overline {\rho }}{\dfrac {R}{W}}{\widetilde {T}}}}
{\displaystyle {\text{(4.4)}}}
{\displaystyle {W}}
{\displaystyle {R}}
{\displaystyle {\mu _{t}}}
{\displaystyle {L_{vK}}}
{\displaystyle {\omega }}
{\displaystyle {L_{vK}}}
{\displaystyle {L_{vK}}}
{\displaystyle {L_{vK}=\kappa {\dfrac {\sqrt {2{\widetilde {S}}_{ij}{\widetilde {S}}_{ij}}}{{\widetilde {u}}''}}}}
{\displaystyle {\text{(4.5)}}}
{\displaystyle {{\widetilde {S}}_{ij}={\dfrac {1}{2}}\left({\dfrac {\partial {\widetilde {u}}_{i}}{\partial x_{j}}}+{\dfrac {\partial {\widetilde {u}}_{j}}{\partial x_{i}}}\right){\text{,}}\quad {\widetilde {u}}''={\sqrt {{\dfrac {\partial ^{2}{\widetilde {u}}_{i}}{\partial x_{k}^{2}}}{\dfrac {\partial ^{2}{\widetilde {u}}_{i}}{\partial x_{j}^{2}}}}}}}
{\displaystyle {\text{(4.6)}}}
{\displaystyle {\kappa }}
{\displaystyle {{\dfrac {\partial {\overline {\rho }}}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}=0}}
{\displaystyle {\text{(4.7)}}}
{\displaystyle {{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{i}\right)}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{i}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}={\dfrac {\partial {\overline {\tau }}_{ij}}{\partial x_{j}}}+{\dfrac {\partial \tau _{ij}^{sgs}}{\partial x_{j}}}-{\dfrac {\partial {\overline {p}}}{\partial x_{i}}}}}
{\displaystyle {\text{(4.8)}}}
{\displaystyle {{\dfrac {\partial \left({\overline {\rho }}{\widetilde {e}}\right)}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{j}{\widetilde {e}}\right)}{\partial x_{j}}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {K}}\right)}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {K}}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}={\dfrac {\partial }{\partial x_{j}}}\left(\alpha _{eff}{\dfrac {\partial {\widetilde {e}}}{\partial x_{j}}}\right)-{\dfrac {\partial }{\partial x_{j}}}\left({\overline {p}}{\widetilde {u}}_{j}\right)}}
{\displaystyle {\text{(4.9)}}}
{\displaystyle {{\overline {\tau }}_{ij}}}
{\displaystyle {{\dfrac {\partial {\overline {\rho }}}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}=0}}
{\displaystyle {\text{(4.10)}}}
{\displaystyle {{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{i}\right)}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{i}{\widetilde {u}}_{j}\right)}{\partial x_{j}}}}=-{\dfrac {\partial {\overline {P}}}{\partial x_{i}}}+{\dfrac {\partial }{\partial x_{j}}}\left(\left({\overline {\mu }}+\mu _{t}\right)\left({\dfrac {\partial {\widetilde {u}}_{i}}{\partial x_{j}}}+{\dfrac {\partial {\widetilde {u}}_{j}}{\partial x_{i}}}-{\dfrac {2}{3}}{\dfrac {\partial {\widetilde {u}}_{k}}{\partial x_{k}}}\delta _{ij}\right)\right)+{\overline {\rho }}g_{i}}
{\displaystyle {\text{(4.11)}}}
{\displaystyle {{\text{with}}\quad {\overline {P}}={\overline {p}}+{\dfrac {1}{3}}{\overline {\rho }}\tau _{kk}^{sgs}}}
{\displaystyle {{\dfrac {\partial \left({\overline {\rho }}{\widetilde {e}}\right)}{\partial t}}+{\dfrac {\partial \left({\overline {\rho }}{\widetilde {u}}_{j}{\widetilde {e}}\right)}{\partial x_{j}}}=-{\overline {p}}{\dfrac {\partial {\widetilde {u}}_{j}}{\partial x_{j}}}+{\widetilde {\tau }}_{ij}{\dfrac {\partial {\widetilde {u}}_{i}}{\partial x_{j}}}+{\dfrac {\partial }{\partial x_{j}}}\left({\dfrac {\left({\overline {\mu }}+{\dfrac {c_{v}}{c_{p}}}\mu _{t}\right)c_{p}}{Pr}}{\dfrac {\partial {\widetilde {T}}}{\partial x_{j}}}\right)}}
{\displaystyle {\text{(4.12)}}}
{\displaystyle {e}}
{\displaystyle {e=h-{\dfrac {p}{\rho }}=e_{0}+\int _{T_{0}}^{T}c_{v}dT-{\dfrac {p}{\rho }}\quad {\text{, with}}\quad h=h_{0}+\int _{T_{0}}^{T}c_{p}dT\quad {\text{and}}\quad c_{v}=\left.{\dfrac {\partial e}{\partial T}}\right|_{v}}}
{\displaystyle {\text{(4.13)}}}
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Component: H1
ddtime H1 =- betaIP H1 + A
ddtime H2 = betaIP H1 - betaCP H2 + betaIS H2
ddtime H3 = betaIS H2 - betaCS H3
ddtime A = Amax 1.0 -ⅇ- lamda time - t_on 1.0 -ⅇ- lamda t_off - t_on if time < t_off ∨ time ≥ t_on Amax ⅇ- alpha time - t_off if time ≥ t_off ddtime A_ = A 1.0 - m B_
Component: Process_L
alph = alph_0 I I0 p I I + 100.0 B_ = G 1.0 - n alph ddtime n = 60.0 alph 1.0 - n - beta n ddtime I = 0.0 if time ≥ 0.0 ∧ time < 4.75 9500.0 if time ≥ 4.75 ∧ time < 11.2 0.0 otherwise
Component: Process_P
ddtime x =π 12.0 xc + mu 1.0 3.0 x + 4.0 3.0 x 3.0 - 256.0 105.0 x 7.0 + B ddtime xc =π 12.0 q B xc - x 24.0 0.99729 tau_x 2.0 + k B B = B_ 1.0 - 0.4 x 1.0 - 0.4 xc
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Swept-frequency cosine - MATLAB chirp - MathWorks Nordic
Quadratic Chirp
Convex Quadratic Chirp
Symmetric Concave Quadratic Chirp
Logarithmic Chirp
Complex Chirp
y = chirp(t,f0,t1,f1)
y = chirp(t,f0,t1,f1,method)
y = chirp(t,f0,t1,f1,method,phi)
y = chirp(t,f0,t1,f1,'quadratic',phi,shape)
y = chirp(___,cplx)
y = chirp(t,f0,t1,f1) generates samples of a linear swept-frequency cosine signal at the time instances defined in array t. The instantaneous frequency at time 0 is f0 and the instantaneous frequency at time t1 is f1.
y = chirp(t,f0,t1,f1,method) specifies an alternative sweep method option.
y = chirp(t,f0,t1,f1,method,phi) specifies the initial phase.
y = chirp(t,f0,t1,f1,'quadratic',phi,shape) specifies the shape of the spectrogram of a quadratic swept-frequency signal.
y = chirp(___,cplx) returns a real chirp if cplx is specified as 'real' and returns a complex chirp if cplx is specified as 'complex'.
Generate a chirp with linear instantaneous frequency deviation. The chirp is sampled at 1 kHz for 2 seconds. The instantaneous frequency is 0 at t = 0 and crosses 250 Hz at t = 1 second.
y = chirp(t,0,1,250);
Compute and plot the spectrogram of the chirp. Divide the signal into segments such that the time resolution is 0.1 second. Specify 99% of overlap between adjoining segments and a spectral leakage of 0.85.
pspectrum(y,1e3,'spectrogram','TimeResolution',0.1, ...
'OverlapPercent',99,'Leakage',0.85)
Generate a chirp with quadratic instantaneous frequency deviation. The chirp is sampled at 1 kHz for 2 seconds. The instantaneous frequency is 100 Hz at t = 0 and crosses 200 Hz at t = 1 second.
Generate a convex quadratic chirp sampled at 1 kHz for 2 seconds. The instantaneous frequency is 400 Hz at t = 0 and crosses 300 Hz at t = 1 second.
y = chirp(t,fo,1,f1,'quadratic',[],'convex');
Generate a concave quadratic chirp sampled at 1 kHz for 4 seconds. Specify the time vector so that the instantaneous frequency is symmetric about the halfway point of the sampling interval, with a minimum frequency of 100 Hz and a maximum frequency of 500 Hz.
t = -2:1/1e3:2;
y = chirp(t,fo,1,f1,'quadratic',[],'concave');
pspectrum(y,t,'spectrogram','TimeResolution',0.1, ...
Generate a logarithmic chirp sampled at 1 kHz for 10 seconds. The instantaneous frequency is 10 Hz initially and 400 Hz at the end.
Use a logarithmic scale for the frequency axis. The spectrogram becomes a line, with high uncertainty at low frequencies.
Generate a complex linear chirp sampled at 1 kHz for 10 seconds. The instantaneous frequency is –200 Hz initially and 300 Hz at the end. The initial phase is zero.
fo = -200;
y = chirp(t,fo,t(end),f1,'linear',0,'complex');
Verify that a complex chirp has real and imaginary parts that are equal but with
{90}^{\circ }
phase difference.
x = chirp(t,fo,t(end),f1,'linear',0) + 1j*chirp(t,fo,t(end),f1,'linear',-90);
pspectrum(x,t,'spectrogram','TimeResolution',0.2, ...
Time array, specified as a vector.
f0 — Instantaneous frequency at time 0
0 (default) | real scalar in Hz
Initial instantaneous frequency at time 0, specified as a real scalar expressed in Hz.
t1 — Reference time
1 (default) | positive scalar in seconds
Reference time, specified as a positive scalar expressed in seconds.
f1 — Instantaneous frequency at time t1
100 (default) | real scalar in Hz
Instantaneous frequency at time t1, specified as a real scalar expressed in Hz.
method — Sweep method
'linear' (default) | 'quadratic' | 'logarithmic'
Sweep method, specified as 'linear', 'quadratic', or 'logarithmic'.
'linear' — Specifies an instantaneous frequency sweep fi(t) given by
{f}_{i}\left(t\right)={f}_{0}+\beta t,
\beta =\left({f}_{1}-{f}_{0}\right)/{t}_{1}
and the default value for f0 is 0. The coefficient β ensures that the desired frequency breakpoint f1 at time t1 is maintained.
'quadratic' — Specifies an instantaneous frequency sweep fi(t) given by
{f}_{i}\left(t\right)={f}_{0}+\beta {t}^{2},
\beta =\left({f}_{1}-{f}_{0}\right)/{t}_{1}{}^{2}
and the default value for f0 is 0. If f0 > f1 (downsweep), the default shape is convex. If f0 < f1 (upsweep), the default shape is concave.
'logarithmic' — Specifies an instantaneous frequency sweep fi(t) given by
{f}_{i}\left(t\right)={f}_{0}×{\beta }^{t},
\beta ={\left(\frac{{f}_{1}}{{f}_{0}}\right)}^{\frac{1}{{t}_{1}}}
and the default value for f0 is 10–6.
phi — Initial phase
0 (default) | positive scalar in degrees
Initial phase, specified as a positive scalar expressed in degrees.
shape — Spectrogram shape of quadratic chirp
'convex' | 'concave'
Spectrogram shape of quadratic chirp, specified as 'convex' or 'concave'. shape describes the shape of the parabola with respect to the positive frequency axis. If not specified, shape is 'convex' for the downsweep case with f0 > f1, and 'concave' for the upsweep case with f0 < f1.
cplx — Output complexity
Output complexity, specified as 'real' or 'complex'.
y — Swept-frequency cosine signal
Swept-frequency cosine signal, returned as a vector.
cos | diric | gauspuls | pulstran | rectpuls | sawtooth | sin | sinc | square | tripuls
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When a person stands on tiptoe (a strenuous position), the position of the foot
When a person stands on tiptoe (a strenuous position), the position of the foot is as shown in Figure (a). The total gravitational force on the body,
Explantion is given below
\left(1-{\frac{t}{40}}^{2}\right)5000=V0\le t\le 40
(a) In an RLC circuit, can the amplitude of the voltage across an inductor be greater than the amplitude of the generator emf? (b) Consider an RLC circuit with emf amplitude
{\xi }_{m}=10V
R=10\mathrm{\Omega }
L=1.0H
, and capacitance
C=1.0\mu F
. Find the amplitude of the voltage across the inductor at resonance.
The electric field strength between two parallel conducting plates separated by 4.00 cm is
7.50×{10}^{4}V
. (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be zero volts. What is the potential 1.00 cm from that plate and 3.00 cm from the other?
Enter the solubility-product expression for
Al\left(OH\right)3\left(s\right)
\left[A{l}^{3+}\right]{\left[O{H}^{-}\right]}^{3}
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Difference between revisions of "Problem 4 - How do we decompose H(w)?" - Murray Wiki
Difference between revisions of "Problem 4 - How do we decompose H(w)?"
'''Q:''' How is it possible to decompose <math>S(\omega)</math>?
'''A:''' There is a general observation to make. Given a transfer function <math>H(s)=\frac{1}{s+a}</math>, its power spectral density will be <math>S(\omega)=\frac{1}{-\omega^2+a^2}</math>. If we define <math>\lambda:=-\omega^2</math>, then we see that we have <math>S(\lambda)=\frac{1}{\lambda+a^2}</math>. Qualitatively, we can argue that poles of <math>H(s)</math> in <math>-a</math> are mapped to poles of <math>S(\lambda)</math> in <math>-a^2</math>.
'''A:''' There is a general observation to make. Given a transfer function <math>H(s)=\frac{1}{s+a}</math>, its power spectral density will be <math>S(\omega)=\frac{1}{\omega^2+a^2}</math>. If we define <math>\lambda:=\omega^2</math>, then we see that we have <math>S(\lambda)=\frac{1}{\lambda+a^2}</math>. Qualitatively, we can argue that poles of <math>H(s)</math> in <math>-a</math> are mapped to poles of <math>S(\lambda)</math> in <math>-a^2</math>.
Same for transfer functions having more than one pole and zeros.
In the exercise you should therefore substitute <math>-\omega^2</math> with <math>\lambda</math>, find its poles and zeros, and then map back to a guess for <math>H(s)</math>. Such guess will not be unique in general, but it is if one assumes certain properties regarding the phase!
In the exercise you should therefore substitute <math>\omega^2</math> with <math>\lambda</math>, find its poles and zeros, and then map back to a guess for <math>H(s)</math>. Such guess will not be unique in general, but it is if one assumes certain properties regarding the phase!
Q: How is it possible to decompose
{\displaystyle S(\omega )}
A: There is a general observation to make. Given a transfer function
{\displaystyle H(s)={\frac {1}{s+a}}}
, its power spectral density will be
{\displaystyle S(\omega )={\frac {1}{\omega ^{2}+a^{2}}}}
{\displaystyle \lambda :=\omega ^{2}}
, then we see that we have
{\displaystyle S(\lambda )={\frac {1}{\lambda +a^{2}}}}
. Qualitatively, we can argue that poles of
{\displaystyle H(s)}
{\displaystyle -a}
are mapped to poles of
{\displaystyle S(\lambda )}
{\displaystyle -a^{2}}
. Same for transfer functions having more than one pole and zeros.
In the exercise you should therefore substitute
{\displaystyle \omega ^{2}}
{\displaystyle \lambda }
, find its poles and zeros, and then map back to a guess for
{\displaystyle H(s)}
. Such guess will not be unique in general, but it is if one assumes certain properties regarding the phase!
Retrieved from "https://murray.cds.caltech.edu/index.php?title=Problem_4_-_How_do_we_decompose_H(w)%3F&oldid=5533"
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Laws Of Motion, Popular Questions: CBSE Class 11-science PHYSICS, Physics Part I - Meritnation
The speed limit of a car over a roadways bridge in the form of a vertical arc is 9.8 ms-1. Calculate the diameter of the arc.
Effective C between X and Y?
A\right) \frac{10}{3}m/{s}^{2} B\right) \frac{7}{3}m/{s}^{2} C\right) 0.5 m/{s}^{2} D\right) 1.0 m/{s}^{2}
Please give very detailed solution for the following question
Onaiza asked a question
A mass m1 hanging at the end string, draws a mass m2 along the surface of a smooth table if the mass on the table be doubled the tension in string becomes 1.5 times then m1/m2 is.
A circular race track of radius 400 m is banked at an angle of 10°. If the coefficient of friction between the wheels of a race car and the road is 0.2 , what is the (i) optimum speed of the car to avoid wear and tear on its tyres (ii) maximum permissible speed to avoid slipping, g = 9.8 m/s2.
Anjani Bahl asked a question
Minimum value of F so that 'm' falls freely is given by? And also find acceleration with which Wedge M moves and the contact force b/w m and M
a shot fired from cannon explodes in air . what will be the change in momentum and the kinetic energy ?
In the figure given below ,the force F to be applied on triangular block of mass M so that the block of mass m placed on it appears stationary w.r.t wedge is ( coeff of friction is 0 between mass and wedge And between wedge and ground also)
a ) mg tan 'theta' b ) ( M + m ) g tan ' theta'
c ) ( M + m ) g cos 'theta' d ) ( M + m ) g sin 'theta'
figure shows two block system 4 kg block rests on a smooth horizontal surface upper surface of 4 kg is rough a block of mass 2 kg is placed on a surface surface the acceleration of upper Block with respect to earth when 4 kg mass is pulled by a force of 13 newton is
The motion of a particle of mass m is described by ut + 1/2 gt2. Find the force acting on the particle.............????? Plzzz......experts or ne1 else........!!!!!!
Abhiram Sankar asked a question
Find the value of T1 and T2 for the system shown in figure.
state the laws of friction. te coefficient of static frictin between block A mass=2. and the table is 0.2. what would be the maximum mass of block B so that the two blocks do not move. (the string and the pulley are assumed to be smooth and massless, g=10m/s2)
A curve road has the banking angle calculated for 80 km/hr however road is coverd with ice and you plan to creep around the highest lane at 20 km/hr what may happenn to car ?why?
a ball is released from the top of a tower. the ratio of work done by force of gravity in first,second and third second of the motion of the ball is,
a) 1:2:3 b) 1:4:9 c) 1:3:5 d) 1:5:3
What is the minimum value of F needed so that block begins to move upward on frictionless incline plane as shown:-
a )Mg tan (θ/2)
b) Mg cot (θ/2)
c) Mg sin θ/(1 + sin θ)
d) Mg sin (θ/2)
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A piecewise function has the following features:
Represent the line segment from P to Q by a vector-valued function and by a set of parametric equations P(−3, −6, −1), Q(−1, −9, −8).
{X}_{xx}=\left(1+\delta \left\{\left(x\right)\right\}\right)X
{u}_{xx}=\left(1+\delta \left(x\right)\right){u}_{tt}
\left(1+\delta \left(x\right)\right)
{X}_{xx}=\left(1+\delta \left(x\right)\right)X
\to \mathrm{\infty }
\left\{\begin{array}{l}{y}^{\prime }\left(t\right)=\frac{{t}^{2}}{2+\mathrm{sin}\left({t}^{2}\right)}cos\left(y\left(t{\right)}^{2}\right)\\ y\left(0\right)=0\end{array}
Find the indefinite integral and check the results by differentiation.
\int \frac{dx}{{x}^{7}}
\int \frac{{x}^{4}+3{x}^{2}-5}{{x}^{3}}dx
{\int }_{2}^{5}\left(4{x}^{3}-8x+7\right)dx
{\int }_{o}^{-\mathrm{\pi }}\left(5\mathrm{cos} \theta d\theta \right)
Continuing Solutions of
\stackrel{˙}{x}=\frac{{t}^{2}{x}^{5}}{1+{x}^{2}+{x}^{4}}
to entire number line
can be continued to the whole real line.
I know that this ODE is seperable as follows
\frac{1+{x}^{2}+{x}^{4}}{{x}^{5}}dx={t}^{2}dt
Thus, giving the solution
-\frac{1}{4{x}^{4}}-\frac{1}{2{x}^{2}}+\mathrm{ln}\left(|x|\right)+C={\frac{13}{t}}^{3}
However, forom here, it is not clear to me how any solution x(t) can be continued to the entire real number line.
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Kuhn poker - Wikipedia
Kuhn poker is an extremely simplified form of poker developed by Harold W. Kuhn as a simple model zero-sum two-player imperfect-information game, amenable to a complete game-theoretic analysis. In Kuhn poker, the deck includes only three playing cards, for example a King, Queen, and Jack. One card is dealt to each player, which may place bets similarly to a standard poker. If both players bet or both players pass, the player with the higher card wins, otherwise, the betting player wins.
3 Generalized versions
3.1 3-player Kuhn Poker
Game description[edit]
In conventional poker terms, a game of Kuhn poker proceeds as follows:
Each player antes 1.
Each player is dealt one of the three cards, and the third is put aside unseen.
Player one can check or bet 1.
If player one checks then player two can check or bet 1.
If player two checks there is a showdown for the pot of 2 (i.e. the higher card wins 1 from the other player).
If player two bets then player one can fold or call.
If player one folds then player two takes the pot of 3 (i.e. winning 1 from player 1).
If player one calls there is a showdown for the pot of 4 (i.e. the higher card wins 2 from the other player).
If player one bets then player two can fold or call.
If player two folds then player one takes the pot of 3 (i.e. winning 1 from player 2).
If player two calls there is a showdown for the pot of 4 (i.e. the higher card wins 2 from the other player).
The game has a mixed-strategy Nash equilibrium; when both players play equilibrium strategies, the first player should expect to lose at a rate of −1/18 per hand (as the game is zero-sum, the second player should expect to win at a rate of +1/18). There is no pure-strategy equilibrium.
Kuhn demonstrated there are infinitely many equilibrium strategies for the first player, forming a continuum governed by a single parameter. In one possible formulation, player one freely chooses the probability
{\displaystyle \alpha \in [0,1/3]}
with which he will bet when having a Jack (otherwise he checks; if the other player bets, he should always fold). When having a King, he should bet with the probability of
{\displaystyle 3\alpha }
(otherwise he checks; if the other player bets, he should always call). He should always check when having a Queen, and if the other player bets after this check, he should call with the probability of
{\displaystyle \alpha +1/3}
The second player has a single equilibrium strategy: Always betting or calling when having a King; when having a Queen, checking if possible, otherwise calling with the probability of 1/3; when having a Jack, never calling and betting with the probability of 1/3.
Complete tree of Kuhn poker including probabilities for mixed-strategy Nash equilibrium. Dotted lines mark subtrees for dominated strategies.
Generalized versions[edit]
In addition to the basic version invented by Kuhn, other versions appeared adding bigger deck, more players, betting rounds, etc., increasing the complexity of the game.
3-player Kuhn Poker[edit]
A variant for three players was introduced in 2010 by Nick Abou Risk and Duane Szafron. In this version, the deck includes four cards (adding a ten card), from which three are dealt to the players; otherwise, the basic structure is the same: while there is no outstanding bet, a player can check or bet, with an outstanding bet, a player can call or fold. If all players checked or at least one player called, the game proceeds to showdown, otherwise, the betting player wins.
A family of Nash equilibria for 3-player Kuhn poker is known analytically, which makes it the largest game with more than two players with analytic solution.[1] The family is parameterized using 4–6 parameters (depending on the chosen equilibrium). In all equilibria, player 1 has a fixed strategy, and he always checks as the first action; player 2's utility is constant, equal to –1/48 per hand. The discovered equilibrium profiles show an interesting feature: by adjusting a strategy parameter
{\displaystyle \beta }
(between 0 and 1), player 2 can freely shift utility between the other two players while still remaining in equilibrium; player 1's utility is equal to
{\displaystyle -{\frac {1+2\beta }{48}}}
(which is always worse than player 2's utility), player 3's utility is
{\displaystyle {\frac {1+\beta }{24}}}
It is not known if this equilibrium family covers all Nash equilibria for the game.
Kuhn, H. W. (1950). "Simplified Two-Person Poker". In Kuhn, H. W.; Tucker, A. W. (eds.). Contributions to the Theory of Games. Vol. 1. Princeton University Press. pp. 97–103.
James Peck. "Perfect Bayesian Equilibrium" (PDF). Ohio State University. Retrieved 2 September 2016. : 19–29
^ Szafron, Duane; Gibson, Richard; Sturtevant, Nathan (May 2013). "A Parameterized Family of Equilibrium Profiles forThree-Player Kuhn Poker" (PDF). In Ito; Jonker; Gini; Shehory (eds.). Proceedings of the 12th International Conference on Autonomous Agents and Multiagent Systems (AAMAS 2013). Saint Paul, Minnesota, USA.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Kuhn_poker&oldid=1052287080"
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A bag contains three red balls numbered 1, 2, 3
A bag contains three red balls numbered 1, 2, 3 and six white balls numbered 4
A bag contains three red balls numbered 1, 2, 3 and six white balls numbered 4, 5, 6, 7, 8, 9. A ball is drawn. What is the probability that the ball is the following? (Enter your probabilities as fractions.)
(a) red and even-numbered
(b) red or even-numbered
(c) white or odd-numbered
A bag contains three red balls numbered 1,2,3 and six white balls numbered 4,5,6,7,8, and 9.
(a) The total number of balls is 9.
Find the probability of red and even numbered balls.
Set of red and even numbered balls
=2
P=\frac{\text{Number of outcomes}}{\text{Total outcomes}}
P=\frac{1}{9}
(b) The total number of balls is 9.
Find the probability of red or even numbered balls.
Set of red or even numbered balls=1,2,3,4,6,8.
P=\frac{\text{Number of outcomes}}{\text{Total outcomes}}
P=\frac{6}{9}
P=\frac{2}{3}
(c) The total number of balls is 9.
Find the probability of white or odd numbered balls.
Set of white or odd numbered balls=1,2,4,5,6,7,8,9.
P=\frac{\text{Number of outcomes}}{\text{Total outcomes}}
P=\frac{8}{9}
\int \frac{1}{u\left({u}^{2}+1\right)}du
1=A\left({u}^{2}+1\right)+B\left(u\right)
1=A\left({u}^{2}+1\right)+B\left(u\right)
A=1
u=0
B
{u}^{2}+1
I'm reading a textbook in which an equation is rearranged and I'm failing to see how they've done it. I've tried writing it down step by step in my notebook but can't come up with the right answer. It's frustrating because I'm trying to learn calculus and differential equations - something left out of my education. After reading, I understand the idea behind calculus and what is being done and what it means, but I'm bad and working individual examples out.
It's a differential equation with the output set to 0 to find the initial value of
{x}_{0}
{r}_{0}\left(1-\frac{{x}_{0}}{{k}_{0}}\right)-{d}_{0}=0
{x}_{0}=\frac{{k}_{0}}{{r}_{0}}\left({r}_{0}-{d}_{0}\right)
It's the part with the fraction in the brackets that's throwing me, I can sort of see where they got
\left({r}_{0}-{d}_{0}\right)
from.
I think to start with I need to get rid of the brackets so as:
{r}_{0}-\frac{{r}_{0}{x}_{0}}{{r}_{0}{k}_{0}}-{d}_{0}=0
Then I think the
{r}_{0}
on the top and bottom cancel?
{r}_{0}-\frac{{x}_{0}}{{k}_{0}}-{d}_{0}=0
To multiply two rational expressions, we multiply their ? together and multiply their ? together.
\frac{2}{x+1}\cdot \frac{x}{x+3}
is the same as ?.
Is it possible to rationalize a denominator containing two cube roots?
The fraction in question is
-\frac{12}{\sqrt[3]{12\sqrt{849}+108}-\sqrt[3]{12\sqrt{849}-108}}
And was reached in calculating the solution to
{x}^{4}-x-1=0
. I've tried all the standard methods, including
\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}
, but that doesn't work for cube roots, because once you have the square of one the two middle terms will not cancel each other out.
\frac{5}{6}+\frac{1}{6}=
Can someone elaborate the rule of common denominators?
a) What is
\frac{3}{8}+\frac{1}{4}?
b) What is
\frac{3}{12}-\frac{2}{15}?
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Loan Life Coverage Ratio – LLCR Definition
LLCR is similar to the debt service coverage ratio (DSCR), but it is more commonly used in project financing because of its long-term nature. The DSCR captures a single point in time, whereas the LLCR addresses the entire span of the loan.
The Formula for the Loan Life Coverage Ratio (LLCR) Is
\begin{aligned} &\frac{\sum_{t=s}^{s+n}\frac{CF_t}{\left(1 + i\right)^t} + DR}{O_t}\\ &\textbf{where:}\\ &CF_t = \text{Cash-flows available for debt service at year t}\\ &t = \text{The time period}\left(\text{year}\right)\\ &s = \text{The number of years expected to pay the debt back}\\ &i = \text{The weighted average cost of capital}\left(\text{WACC}\right)\\ &\text{expressed as an interest rate}\\ &DR = \text{Cash reserve available to repay the debt}\\ &\left(\text{the debt reserve}\right)\\ &O_t = \text{The debt balance outstanding at the time of}\\ &\text{evaluation}\\ \end{aligned}
Ot∑t=ss+n(1+i)tCFt+DRwhere:CFt=Cash-flows available for debt service at year tt=The time period(year)s=The number of years expected to pay the debt backi=The weighted average cost of capital(WACC)expressed as an interest rateDR=Cash reserve available to repay the debt(the debt reserve)Ot=The debt balance outstanding at the time ofevaluation
How to Calculate the Loan Life Coverage Ratio
The LLCR can be calculated using the above formula, or by using a shortcut: dividing the NPV of project free cash flows by the present value of the debt outstanding.
In this calculation, the weighted average cost of debt is the discount rate for the NPV calculation and the project "cash flows" are more specifically the cash flows available for debt service (CFADS).
What Does the Loan Life Coverage Ratio Tell You?
LLCR is a solvency ratio. The loan life coverage ratio is a measure of the number of times over the cash flows of a project can repay an outstanding debt over the life of a loan. A ratio of 1.0x means that LLCR is at a break-even level. The higher the ratio, the less potential risk there is for the lender.
Depending on the risk profile of the project, sometimes a debt service reserve account is required by the lender. In such a case, the numerator of LLCR would include the reserve account balance. Project financing agreements invariably contain covenants that stipulate LLCR levels.
The loan life coverage ratio (LLCR) is a financial ratio used to estimate the solvency of a firm, or the ability of a borrowing company to repay an outstanding loan.
The loan life coverage ratio is a measure of the number of times over the cash flows of a project can repay an outstanding debt over the life of a loan.
The higher the ratio, the less potential risk there is for the lender.
The Difference Between LLCR and DSCR
In corporate finance, the Debt-Service Coverage Ratio (DSCR) is a measure of the cash flow available to pay current debt obligations. The ratio states net operating income as a multiple of debt obligations due within one year, including interest, principal, sinking-fund and lease payments. However, DSCR captures just a single point in time, while LLCR allows for several time periods, which is more suitable for understanding liquidity available for loans of medium to long time horizons.
LLCR is used by analysts to assess the viability of a given amount of debt and consequently to evaluate the risk profile and the related costs. It has a less immediate explanation compared to DSCR, but when LLCR has a value greater than one, this is usually a strong reassurance for investors.
Limitations of LLCR
One limitation of the LLCR is that it does not pick up weak periods because it basically represents a discounted average that can smooth out rough patches. For this reason, if a project has a steady cash flows with a history of loan repayment, a good rule of thumb is that the LLCR should be roughly equal to the average debt service coverage ratio.
Cash available for debt service (CADS) is a ratio that measures the amount of cash a company has on hand to pay obligations due within a year.
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Hydraulic head - 3D Tool - 3D Manufacturing
Hydraulic head (7833 views - Tooling)
Hydraulic head or piezometric head is a specific measurement of liquid pressure above a geodetic datum. It is usually measured as a liquid surface elevation, expressed in units of length, at the entrance (or bottom) of a piezometer. In an aquifer, it can be calculated from the depth to water in a piezometric well (a specialized water well), and given information of the piezometer's elevation and screen depth. Hydraulic head can similarly be measured in a column of water using a standpipe piezometer by measuring the height of the water surface in the tube relative to a common datum. The hydraulic head can be used to determine a hydraulic gradient between two or more points.
1 "Head" in fluid dynamics
2 Components of hydraulic head
2.1 Fresh water head
3 Hydraulic gradient
4 Hydraulic head in groundwater
5 Head loss
6 Analogs to other fields
A mass free falling from an elevation
{\displaystyle z\,>\,0\,}
(in a vacuum) will reach a speed
{\displaystyle v={\sqrt {{2g}{z}}},}
when arriving at elevation z=0, or when we rearrange it as a head:
{\displaystyle h={\frac {v^{2}}{2g}}}
{\displaystyle g}
is the acceleration due to gravity
{\displaystyle {\frac {v^{2}}{2g}}}
is called the velocity head, expressed as a length measurement. In a flowing fluid, it represents the energy of the fluid due to its bulk motion.
{\displaystyle h=\psi +z\,}
{\displaystyle h}
is the hydraulic head (Length in m or ft), also known as the piezometric head.
{\displaystyle \psi }
is the pressure head, in terms of the elevation difference of the water column relative to the piezometer bottom (Length in m or ft), and
{\displaystyle z}
is the elevation at the piezometer bottom (Length in m or ft)
{\displaystyle \psi ={\frac {P}{\gamma }}={\frac {P}{\rho g}}}
{\displaystyle P}
is the gauge pressure (Force per unit area, often Pa or psi),
{\displaystyle \gamma }
is the unit weight of the liquid (Force per unit volume, typically N·m−3 or lbf/ft³),
{\displaystyle \rho }
is the density of the liquid (Mass per unit volume, frequently kg·m−3), and
{\displaystyle g}
is the gravitational acceleration (velocity change per unit time, often m·s−2)
{\displaystyle h_{\mathrm {fw} }=\psi {\frac {\rho }{\rho _{\mathrm {fw} }}}+z}
{\displaystyle h_{\mathrm {fw} }\,}
is the fresh water head (Length, measured in m or ft), and
{\displaystyle \rho _{\mathrm {fw} }\,}
is the density of fresh water (Mass per unit volume, typically in kg·m−3)
{\displaystyle i={\frac {dh}{dl}}={\frac {h_{2}-h_{1}}{\mathrm {length} }}}
{\displaystyle i}s the hydraulic gradient (dimensionless),
{\displaystyle dh}
is the difference between two hydraulic heads (Length, usually in m or ft), and
{\displaystyle dl}
is the flow path length between the two piezometers (Length, usually in m or ft)
{\displaystyle \nabla h=\left({\frac {\partial h}{\partial x}},{\frac {\partial h}{\partial y}},{\frac {\partial h}{\partial z}}\right)={\frac {\partial h}{\partial x}}\mathbf {i} +{\frac {\partial h}{\partial y}}\mathbf {j} +{\frac {\partial h}{\partial z}}\mathbf {k} }
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This article uses material from the Wikipedia article "Hydraulic head", which is released under the Creative Commons Attribution-Share-Alike License 3.0. There is a list of all authors in Wikipedia
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Nodal analysis - Wikipedia
Kirchhoff's current law is the basis of nodal analysis.
In analyzing a circuit using Kirchhoff's circuit laws, one can either do nodal analysis using Kirchhoff's current law (KCL) or mesh analysis using Kirchhoff's voltage law (KVL). Nodal analysis writes an equation at each electrical node, requiring that the branch currents incident at a node must sum to zero. The branch currents are written in terms of the circuit node voltages. As a consequence, each branch constitutive relation must give current as a function of voltage; an admittance representation. For instance, for a resistor, Ibranch = Vbranch * G, where G (=1/R) is the admittance (conductance) of the resistor.
Nodal analysis is possible when all the circuit elements' branch constitutive relations have an admittance representation. Nodal analysis produces a compact set of equations for the network, which can be solved by hand if small, or can be quickly solved using linear algebra by computer. Because of the compact system of equations, many circuit simulation programs (e.g., SPICE) use nodal analysis as a basis. When elements do not have admittance representations, a more general extension of nodal analysis, modified nodal analysis, can be used.
2.2 Supernodes
3 Matrix form for the node-voltage equation
Note all connected wire segments in the circuit. These are the nodes of nodal analysis.
Select one node as the ground reference. The choice does not affect the element voltages (but it does affect the nodal voltages) and is just a matter of convention. Choosing the node with the most connections can simplify the analysis. For a circuit of N nodes the number of nodal equations is N−1.
Assign a variable for each node whose voltage is unknown. If the voltage is already known, it is not necessary to assign a variable.
For each unknown voltage, form an equation based on Kirchhoff's Current Law (i.e. add together all currents leaving from the node and mark the sum equal to zero). The current between two nodes is equal to the voltage of the node where the current exits minus the voltage of the node where the current enters the node, both divided by the resistance between the two nodes.
If there are voltage sources between two unknown voltages, join the two nodes as a supernode. The currents of the two nodes are combined in a single equation, and a new equation for the voltages is formed.
Solve the system of simultaneous equations for each unknown voltage.
Basic case[edit]
Basic example circuit with one unknown voltage, V1.
The only unknown voltage in this circuit is
{\displaystyle V_{1}}
. There are three connections to this node and consequently three currents to consider. The direction of the currents in calculations is chosen to be away from the node.
Current through resistor
{\displaystyle R_{1}}
{\displaystyle (V_{1}-V_{S})/R_{1}}
{\displaystyle R_{2}}
{\displaystyle V_{1}/R_{2}}
Current through current source
{\displaystyle I_{S}}
{\displaystyle -I_{S}}
With Kirchhoff's current law, we get:
{\displaystyle {\frac {V_{1}-V_{S}}{R_{1}}}+{\frac {V_{1}}{R_{2}}}-I_{S}=0}
This equation can be solved with respect to V1:
{\displaystyle V_{1}={\frac {\left({\frac {V_{S}}{R_{1}}}+I_{S}\right)}{\left({\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}\right)}}}
Finally, the unknown voltage can be solved by substituting numerical values for the symbols. Any unknown currents are easy to calculate after all the voltages in the circuit are known.
{\displaystyle V_{1}={\frac {\left({\frac {5{\text{ V}}}{100\,\Omega }}+20{\text{ mA}}\right)}{\left({\frac {1}{100\,\Omega }}+{\frac {1}{200\,\Omega }}\right)}}={\frac {14}{3}}{\text{ V}}}
Supernodes[edit]
In this circuit, VA is between two unknown voltages, and is therefore a supernode.
In this circuit, we initially have two unknown voltages, V1 and V2. The voltage at V3 is already known to be VB because the other terminal of the voltage source is at ground potential.
The current going through voltage source VA cannot be directly calculated. Therefore, we cannot write the current equations for either V1 or V2. However, we know that the same current leaving node V2 must enter node V1. Even though the nodes cannot be individually solved, we know that the combined current of these two nodes is zero. This combining of the two nodes is called the supernode technique, and it requires one additional equation: V1 = V2 + VA.
The complete set of equations for this circuit is:
{\displaystyle {\begin{cases}{\frac {V_{1}-V_{\text{B}}}{R_{1}}}+{\frac {V_{2}-V_{\text{B}}}{R_{2}}}+{\frac {V_{2}}{R_{3}}}=0\\V_{1}=V_{2}+V_{\text{A}}\\\end{cases}}}
{\displaystyle V_{2}={\frac {(R_{1}+R_{2})R_{3}V_{\text{B}}-R_{2}R_{3}V_{\text{A}}}{(R_{1}+R_{2})R_{3}+R_{1}R_{2}}}}
Matrix form for the node-voltage equation[edit]
In general, for a circuit with
{\displaystyle N}
nodes, the node-voltage equations obtained by nodal analysis can be written in a matrix form as derived in the following. For any node
{\displaystyle k}
, KCL states
{\textstyle \sum _{j\neq k}G_{jk}(v_{k}-v_{j})=0}
{\displaystyle G_{kj}=G_{jk}}
is the negative of the sum of the conductances between nodes
{\displaystyle k}
{\displaystyle j}
{\displaystyle v_{k}}
is the voltage of node
{\displaystyle k}
{\textstyle 0=\sum _{j\neq k}G_{jk}(v_{k}-v_{j})=\sum _{j\neq k}G_{jk}v_{k}-\sum _{j\neq k}G_{jk}v_{j}=G_{kk}v_{k}-\sum _{j\neq k}G_{jk}v_{j}}
{\displaystyle G_{kk}}
is the sum of conductances connected to node
{\displaystyle k}
. We note that the first term contributes linearly to the node
{\displaystyle k}
{\displaystyle G_{kk}}
, while the second term contributes linearly to each node
{\displaystyle j}
connected to the node
{\displaystyle k}
{\displaystyle G_{jk}}
with a minus sign. If an independent current source/input
{\displaystyle i_{k}}
is also attached to node
{\displaystyle k}
, the above expression is generalized to
{\textstyle i_{k}=G_{kk}v_{k}-\sum _{j\neq k}G_{jk}v_{j}}
. It is readily shown that one can combine the above node-voltage equations for all
{\displaystyle N}
nodes, and write them down in the following matrix form
{\displaystyle {\begin{pmatrix}G_{11}&G_{12}&\cdots &G_{1N}\\G_{21}&G_{22}&\cdots &G_{2N}\\\vdots &\vdots &\ddots &\vdots \\G_{N1}&G_{N2}&\cdots &G_{NN}\end{pmatrix}}{\begin{pmatrix}v_{1}\\v_{2}\\\vdots \\v_{N}\end{pmatrix}}={\begin{pmatrix}i_{1}\\i_{2}\\\vdots \\i_{N}\end{pmatrix}}}
{\textstyle \mathbf {Gv} =\mathbf {i} .}
{\displaystyle \mathbf {G} }
on the left hand side of the equation is singular since it satisfies
{\displaystyle \mathbf {G1} =0}
{\displaystyle \mathbf {1} }
{\displaystyle 1\times N}
column matrix. This corresponds to the fact of current conservation, namely,
{\textstyle \sum _{k}i_{k}=0}
, and the freedom to choose a reference node (ground). In practice, the voltage is the reference node is taken to be 0. Consider it is the last node,
{\displaystyle v_{N}=0}
. In this case, it is straightforward to verify that the resulting equations for the other
{\displaystyle N-1}
nodes remain the same, and therefore one can simply discard the last column as well as the last line of the matrix equation. This procedure results in a
{\displaystyle (N-1)\times (N-1)}
dimensional non-singular matrix equation with the definitions of all the elements stay unchanged.
Ybus matrix
P. Dimo Nodal Analysis of Power Systems Abacus Press Kent 1975
Wikiversity has learning resources about Nodal analysis
Online four-node problem solver
Simple Nodal Analysis Example
Retrieved from "https://en.wikipedia.org/w/index.php?title=Nodal_analysis&oldid=1065428401"
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x{e}^{yzds}
\left(0,0,0\right)
\left(1,2,3\right)
Let C the line segment from
\left(0,0,0\right)
\left(1,2,3\right)
. The parametric equation for this line in vector form is
r\left(t\right)={r}_{0}+td
{r}_{0}=⟨0,0,0⟩r
and the direction vector d is
d=⟨1,2,3⟩-⟨0,0,0⟩=⟨1,2,3⟩
r\left(t\right)=⟨t,2t,3t⟩
and we have in scalar form the parametric equations
for C :
x=t
y=2t
z=3t
0\le t\le 1
To detemine the line integral:
I={\int }_{C}x{e}^{yz}ds
we first determine ds:
ds=\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dz}{dt}\right)}^{2}}dt
=\sqrt{{\left(1\right)}^{2}+{\left(2\right)}^{2}={\left(3\right)}^{2}}dt
=\sqrt{1+4+9}dt
=\sqrt{14}dt
{\int }_{C}x{e}^{yz}ds={\int }_{0}^{1}t{e}^{6{t}^{2}}\sqrt{14}dt=\sqrt{14}{\int }_{0}^{1}{e}^{6{t}^{2}}tdt
Now changing to the variable
u=6{t}^{2}
du=12tdt=12tdt
I=\sqrt{14}{\int }_{0}^{6}\frac{1}{12}{e}^{u}du=\frac{\sqrt{14}}{12}{\int }_{0}^{6}{e}^{u}du
=\frac{\sqrt{14}}{12}{e}^{u}\overline{)\begin{array}{c}{}^{6}\\ {}_{0}\end{array}}=\frac{\sqrt{14}}{12}\left({e}^{6}-{e}^{0}\right)
=\frac{\sqrt{14}}{12}\left({e}^{6}-1\right)
\int \frac{\mathrm{arctan}c}{1+{x}^{2}}dx
How would one integrate the following?
\int \frac{{x}^{n-2}}{{\left(1+x\right)}^{n}}dx
{\int }_{0}^{1}\sqrt[3]{1+7x}dx
\int y3\text{ }ds,\text{ }C÷x=t3,\text{ }y=t,\text{ }0?\text{ }t?\text{ }3
{\int }_{0}^{2}\frac{dx}{\sqrt{x}\left(x-1\right)}
{\int }_{0}^{1}\frac{\mathrm{ln}\left(x\right)}{{x}^{2}-1}dx=\frac{{\pi }^{2}}{8}
\int \frac{1+{x}^{2}}{\left(1-{x}^{2}\right)\sqrt{1+{x}^{4}}}dx
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Math nerd, Physics boffin, Programming geek.
Will solve problems for food.
Algebra Equality And Elixir Match Operator
As the main driver of Moore’s Law today seems to be the increase of processing cores instead of a processor’s clock speed, modern software has to truly embrace parallel computing in order to fully use the available resources. In this scenario, functional programming started becoming increasingly relevant. More than just the newest buzzword, some are heralding it as the next step in programming evolution.
Is Homer Simpson even our final form?
Curiously, due to my background in Mathematics and the computer science professor I had in college, my first computer science lessons were in functional programing more than 10 years ago. As much as I enjoyed writing code in a functional style, I never envisioned it gaining traction, as I felt it had a more academic flavor to it. Personally moving from functional to object-oriented programming was a revolution and enabled me to speak a more widespread dialect. It is funny how certain things play out.
Recently I have been investing some time looking more closely into Elixir. After having read some blog posts, gone through a couple of tutorials, and played the obligatory CodeSchool challenges, I decided to get myself a copy of Dave Thomas’ Programming Elixir 1.3. Most resources start by explaining the match operator. As most languages use the = symbol to represent assignment, some relearning has to be done to understand what = does in Elixir.
In case you haven’t heard of it before, here is quick run through of the how the equals sign work in Elixir.
Looks pretty standard right? Than what about this?
What kind of black magic is 1 = a? Well, turns out the = symbol in Elixir kind of acts first as an assertion (==), then if the assertion in not downright false, it tries to act as an assignment (:=). I like to think of it this way:
Is the assertion true?
If not, are there variables on the left side whose value could be changed to make the assertion pass? If so, bind that value to that variable.
If all the above fail, blow up.
With this in mind let’s run though the previous example.
a = 1 is false as a is undefined. Can we assign a value to a such that the assertion would be true? In this case, yes, so let’s bind the value 1 to the variable a.
What value is bound to a? 1.
Is 1 = a? Yes it is.
So what would happen, if we then asked 2 = a?
We get a error as no match between the left and the right hand side were possible. (Did I tell you how much I like the error messages in Elixir?).
Why was Erlang (the language Elixir builds upon) designed like this? Because this opens the door to pattern matching, which can allow you to do some non-trivial assignments in a way that can be quite elegant. Here is a random example (no elegance there).
iex(1)> [1,b,c]=[1,2,[3,4]]
Algebra Equality
As this behavior may differ from what some people are accustomed to, a common metaphor used to explain it is the equality in Algebra.
Joe Armstrong, Erlang’s creator, compares the equals sign in Erlang to that used in algebra. When you write the equation x = a + 1, you are not assigning the value of a + 1 to x. Instead you’re simply asserting that the expressions x and a + 1 have the same value. If you know the value of x, you can work out the value of a, and vice versa.
While I think this is a good enough analogy and that the equals signs in Elixir indeed allows to do some powerful things, it still falls short if compared to the equality in Algebra, and \(x = a + 1\) is not really a good example to showcase their similarities.
In Algebra, the equality is a special binary relation called an equivalence relation. For a binary relation to be an equivalence relation, it must be:
Reflexive. a must be equal to a.
a = a
Symmetrical. If a equals b, then b equals a.
a = b \Longrightarrow b = a
Transitive. If a equals b, and b equals c, then a equals c.
\begin{cases} a = b\\ b = c \end{cases} \Longrightarrow a = c
Of those three, only the transitivity is assured by the match operator. This means that with the quoted example \(x = a + 1\) Elixir will not be able to determine the value of a if it knows the value of x.
iex(2)> x = a + 1
warning: variable "a" does not exist and is being expanded to "a()", please use parentheses to remove the ambiguity or change the variable name
** (CompileError) iex:2: undefined function a/0
As the undefined variable is on the right hand side of the match, Elixir does not try to bind a value to it. Instead here it assumes a is a function, which is obviously undefined, hence the CompileError.
But trying to determine the value of x when the value of a is known works.
Integrating A Rails Blog With Facebook Instant Articles
© 2018 Igor de Alcântara Barroso
Rivolicons by Hadrien.
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Surrogate optimization for global minimization of time-consuming objective functions - MATLAB surrogateopt - MathWorks France
\underset{x}{\mathrm{min}}f\left(x\right)\text{ such that }\left\{\begin{array}{l}\text{lb}\le x\le \text{ub}\\ A·x\le b\\ \text{Aeq}·x=\text{beq}\\ c\left(x\right)\le 0\\ {x}_{i}\text{ integer, }i\in \text{intcon}\text{.}\end{array}
100\left(x\left(2\right)-x\left(1{\right)}^{2}{\right)}^{2}+\left(1-x\left(1\right){\right)}^{2}
\left(x\left(1\right)-1/3{\right)}^{2}+\left(x\left(2\right)-1/3{\right)}^{2}\le \left(1/3{\right)}^{2}
c\left(x\right)\le 0
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Solve nonlinear curve-fitting (data-fitting) problems in least-squares sense - MATLAB lsqcurvefit - MathWorks Nordic
\underset{x}{\mathrm{min}}{‖F\left(x,xdata\right)-ydata‖}_{2}^{2}=\underset{x}{\mathrm{min}}\sum _{i}{\left(F\left(x,xdat{a}_{i}\right)-ydat{a}_{i}\right)}^{2},
F\left(x,xdata\right)=\left[\begin{array}{c}F\left(x,xdata\left(1\right)\right)\\ F\left(x,xdata\left(2\right)\right)\\ ⋮\\ F\left(x,xdata\left(k\right)\right)\end{array}\right].
x\left(1\right)
x\left(2\right)
\text{ydata}=x\left(1\right)\mathrm{exp}\left(x\left(2\right)\text{xdata}\right).
y=\mathrm{exp}\left(-1.3t\right)+\epsilon ,
t
\epsilon
y=x\left(1\right)\mathrm{exp}\left(x\left(2\right)\text{xdata}\right)
0\le x\left(1\right)\le 3/4
-2\le x\left(2\right)\le -1.
x\left(1\right)
x\left(2\right)
\text{ydata}=x\left(1\right)\mathrm{exp}\left(x\left(2\right)\text{xdata}\right).
x\left(1\right)
x\left(2\right)
\text{ydata}=x\left(1\right)\mathrm{exp}\left(x\left(2\right)\text{xdata}\right).
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===Influence===
In {{game|HeartGold and SoulSilver|s}}, all Pokémon found in {{jo|Safari Zone}} and during the [[Bug Catching Contest]] will have their IVs rerolled up to 4 times if none of the generated IVs are 31.
In {{g|X and Y}}, all Pokémon in the {{egg3|Undiscovered}} are guaranteed to have a perfect 31 in at least three of their individual values when caught in the wild or obtained as a gift (except via [[Mystery Gift]]). In {{g|Omega Ruby and Alpha Sapphire}}, this only applies to {{pkmn2|Legendary}} and [[Mythical Pokémon]]. In [[Generation VII]], this applies to Legendary Pokémon, Mythical Pokémon, and [[Ultra Beast]]s.
In {{g|X and Y}}, all Pokémon in the {{egg3|No Eggs Discovered}} are guaranteed to have a perfect 31 in at least three of their individual values when caught in the wild or obtained as a gift (except via [[Mystery Gift]]). In {{g|Omega Ruby and Alpha Sapphire}}, this only applies to {{pkmn2|Legendary}} and [[Mythical Pokémon]]. In [[Generation VII]], this applies to Legendary Pokémon, Mythical Pokémon, and [[Ultra Beast]]s.
In {{g|Sun and Moon}}, [[Hyper Training]] allows the player to make a Pokémon's stats act as if the Pokémon had maximum IVs. However, this does not actually change the IVs, so its true IVs are still used for the purposes of {{m|Hidden Power}} and {{pkmn|breeding}}.
! Minimum IV
! style="{{roundytr|5px}}" | Perfect IV probability
! 1 in 3375 (0.0296%)
! 1 in 1000 (0.1%)
! 1 in 216 (0.463%)
! style="{{roundybr|5px}} | 1 in 64 (1.56%)
<math>Stat = (base + IV) \times cpMult</math>
If a Pokémon is transferred from Pokémon GO to [[Pokémon: Let's Go, Pikachu! and Let's Go, Eevee!]] or [[Pokémon HOME]], the IVs will be recalculated directly based off the IVs it had in Pokémon GO.
If a Pokémon is transferred from Pokémon GO to [[Pokémon: Let's Go, Pikachu! and Let's Go, Eevee!]] or [[Pokémon HOME]], the IVs will be recalculated directly based on the IVs it had in Pokémon GO.
* IV for HP will equal <math>2 \times IV_{HP}+1</math>
* IVs for Attack and Sp. Attack will both equal <math>2 \times IV_{Attack}+1</math>
* IVs for Attack and Sp. Atk will both equal <math>2 \times IV_{Attack}+1</math>
* IVs for Defense and Sp. Defense will both equal <math>2 \times IV_{Defense}+1</math>
* IVs for Defense and Sp. Def will both equal <math>2 \times IV_{Defense}+1</math>
* IV for Speed will be randomly determined
|fr=Stats individuelles
|it=Punti individuali
|pl=Indywidualne możliwości
|ru=Индивидуальные характеристики ''Individual'nyye kharakteristiki''
|es=Fortalezas individuales
[[de:Determinant Values]]
[[de:Individuelle Stärken]]
[[es:Genética Pokémon]]
[[fr:IV]]
{\displaystyle HP={\Biggl \lfloor }{{\Biggl (}(Base+DV)\times 2+{\biggl \lfloor }{\tfrac {{\bigl \lceil }{\sqrt {STATEXP}}{\bigr \rceil }}{4}}{\biggr \rfloor }{\Biggr )}\times Level \over 100}{\Biggr \rfloor }+Level+10}
{\displaystyle OtherStat={\Biggl \lfloor }{{\Biggl (}(Base+DV)\times 2+{\biggl \lfloor }{\tfrac {{\bigl \lceil }{\sqrt {STATEXP}}{\bigr \rceil }}{4}}{\biggr \rfloor }{\Biggr )}\times Level \over 100}{\Biggr \rfloor }+5}
{\displaystyle HP={\Bigl \lfloor }{(2\times Base+IV+\lfloor {\tfrac {EV}{4}}\rfloor )\times Level \over 100}{\Bigr \rfloor }+Level+10}
{\displaystyle OtherStat={\Biggl \lfloor }{\biggl (}{\Bigl \lfloor }{(2\times Base+IV+\lfloor {\tfrac {EV}{4}}\rfloor )\times Level \over 100}{\Bigr \rfloor }+5{\biggr )}\times Nature{\Biggr \rfloor }}
{\displaystyle Stat=(base+IV)\times cpMult}
{\displaystyle 2\times IV_{HP}+1}
{\displaystyle 2\times IV_{Attack}+1}
{\displaystyle 2\times IV_{Defense}+1}
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Interpreting Rates of Change | Brilliant Math & Science Wiki
Janae Pritchett and Anton Kriksunov contributed
A rate of change is the ratio between the change in one quantity to the change in another quantity. Linear relationships have a constant rate of change.
Interpreting Rates of Change from Situations and Tables
Interpreting Rates of Change from Equations
The tile pattern below is growing by three tiles per figure. Therefore, the tile pattern has a growth rate of 3.
A runner travels
\SI{80}{feet}
\SI{8}{seconds}
. What is the runner's rate of change?
The runner is traveling
\dfrac{\SI{80}{feet}}{\SI{8}{seconds}}
\SI{10}{feet.per.second}
Is the relationship between the inputs and outputs below linear?
If the rate of change between pairs of inputs and outputs in constant, then the relationship is linear.
The rate of change between the pairs
(0,11)
(3,5)
\dfrac{11-5}{0-} = \dfrac{6}{-3}=-2.
(3,5)
(5,1)
\dfrac{5-1}{3-5} = \dfrac{4}{-2}=-2.
(5,1)
(8,-5)
\dfrac{1-(-5)}{5-8} = \dfrac{6}{-3}=-2.
The relationship is linear with a rate of change of -2.
Plant A Plant B Plants A and B are growing at the same rate
Plants A and B are both growing at a constant rate. Which plant is growing faster?
The slope of a line shows the rate of change in a linear relationship.
For example, the graph below shows a rate of change of 10 liters per second. The slope of the line is
\frac{10}{1}=10.
Which car has the greatest gas mileage?
Line A is the steepest so it has the greatest rate of change so car A has the greatest gas mileage.
We can see that car A travels about 50 miles on one gallon of gas, car B travels about 30 miles on one gallon of gas, and car C travels about 10 miles on one gallon of gas.
0.75 liters per second 2 liters per second 2.25 liters per second 3 liters per second
A cylindrical container is being filled with water. The volume of water over time is shown below. What is the rate of change in the water volume?
Equations of lines in the form
y=mx+b
represent linear functions with constant rates of change. The rate of change in the relationship is represented by
m.
y=5,000x+12,0000
represents the total number of miles on Zen's car,
y,
each year that she owned it,
x.
How many miles does Zen drive per year?
The growth rate in this equation is 5,000. Therefore, Zen drives 5000 miles per year. Zen's car had 12,000 miles on it when she purchased it.
In pattern A, the number of dots,
y,
x
y=6x+7.
In pattern B the number of dots,
y,
x
y=7x+6.
Which dot pattern is growing more quickly?
In pattern A, the growth rate is 6. In pattern B, the growth rate is 7. Therefore, pattern B is growing more quickly.
Cite as: Interpreting Rates of Change. Brilliant.org. Retrieved from https://brilliant.org/wiki/interpret-rates-of-change/
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The value of respiratory quotient is infinite during anaerobic respiration Give reason Explain with proper equations and RQ - Biology - Respiration in Plants - 8717851 | Meritnation.com
The value of respiratory quotient is infinite during anaerobic respiration .Give reason. Explain with proper equations and RQ values . Plzz experts help !!!
Respiratory quotient is defined as the ratio of the volume of carbon dioxide released to the volume of oxygen consumed in respiration over a period of time.
\mathrm{RQ}\quad =\quad \frac{\mathrm{Volume}\quad \mathrm{of}\quad \mathrm{carbon}\quad \mathrm{dioxide}\quad \mathrm{released}}{\mathrm{Volume}\quad \mathrm{of}\quad \mathrm{oxygen}\quad \mathrm{absorbed}}
During anaerobic respiration, there is no consumption of oxygen. Carbon dioxide is produced in most of the cases. Hence, respiratory quotient is infinity.
\mathrm{RQ}\quad =\quad \frac{\mathrm{Volume}\quad \mathrm{of}\quad \mathrm{carbon}\quad \mathrm{dioxide}\quad \mathrm{released}}{\mathrm{Volume}\quad \mathrm{of}\quad \mathrm{oxygen}\quad \mathrm{absorbed}}\quad =\quad \frac{\mathrm{Volume}\quad \mathrm{of}\quad \mathrm{carbon}\quad \mathrm{dioxide}}{0}\quad =\quad \mathrm{Infinity}\quad
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(Created page with "{{Publication | published = true | date = 2020-02-07 | authors = Zhou Zhao, Nicolas Boutry, Élodie Puybareau, Thierry Géraud | title = A Two-Stage Temporal-Like Fully Convol...")
| abstract = Automatic segmentation of the left ventricle (LV) of a living human heart in a magnetic resonance (MR) image (2D+t) allows to measure some clinical significant indices like the regional wall thicknesses (RWT), cavity dimensions, cavity and myocardium areas, and cardiac phase. Here, we propose a novel framework made of a sequence of two fully convolutional networks (FCN). The first is a modified temporal-like VGG16 (the "localization network") and is used to localize roughly the LV (filled-in) epicardium position in each MR volume. The second FCN is a modified temporal-like VGG16 too, but devoted to segment the LV myocardium and cavity (the "segmentation network"). We evaluate the proposed method with 5-fold-cross-validation on the MICCAI 2019 LV Full Quantification Challenge dataset. For the network used to localize the epicardium, we obtain an average dice index of 0.8953 on validation set. For the segmentation network, we obtain an average dice index of 0.8664 on validation set (there, data augmentation is used). The mean absolute error (MAE) of average cavity and myocardium areas, dimensions, RWT are 114.77~mm<math>^2</math>; 0.9220~mm; 0.9185~mm respectively. The computation time of the pipeline is less than 2~s for an entire 3D volume. The error rate of phase classification is 7.6364%, which indicates that the proposed approach has a promising performance to estimate all these parameters.
| abstract = Automatic segmentation of the left ventricle (LV) of a living human heart in a magnetic resonance (MR) image (2D+t) allows to measure some clinical significant indices like the regional wall thicknesses (RWT), cavity dimensions, cavity and myocardium areas, and cardiac phase. Here, we propose a novel framework made of a sequence of two fully convolutional networks (FCN). The first is a modified temporal-like VGG16 (the "localization network") and is used to localize roughly the LV (filled-in) epicardium position in each MR volume. The second FCN is a modified temporal-like VGG16 too, but devoted to segment the LV myocardium and cavity (the "segmentation network"). We evaluate the proposed method with 5-fold-cross-validation on the MICCAI 2019 LV Full Quantification Challenge dataset. For the network used to localize the epicardium, we obtain an average dice index of 0.8953 on validation set. For the segmentation network, we obtain an average dice index of 0.8664 on validation set (there, data augmentation is used). The mean absolute error (MAE) of average cavity and myocardium areas, dimensions, RWT are <math>114.77~\text{mm}^2</math>; 0.9220~mm; 0.9185~mm respectively. The computation time of the pipeline is less than 2~s for an entire 3D volume. The error rate of phase classification is 7.6364%, which indicates that the proposed approach has a promising performance to estimate all these parameters.
RWT are 114.77~mm$^2$; 0.9220~mm; 0.9185~mm respectively.
RWT are $114.77~\text<nowiki>{</nowiki>mm<nowiki>}</nowiki>^2$; 0.9220~mm; 0.9185~mm
The computation time of the pipeline is less than 2~s for
an entire 3D volume. The error rate of phase classification
is 7.6364\%, which indicates that the proposed approach has
a promising performance to estimate all these parameters.<nowiki>}</nowiki>
all these parameters.<nowiki>}</nowiki>
{\displaystyle 114.77~{\text{mm}}^{2}}
|
Isaac Newton | Brilliant Math & Science Wiki
Adam Strandberg, Sravanth C., God Ly, and
Sir Isaac Newton (1642-1727) was one of the world's most famous and influential thinkers. He founded the fields of classical mechanics, optics and calculus, among other contributions to algebra and thermodynamics. His concept of a universal law--one that applies everywhere and to all things--set the bar of ambition for physicists since.
Newton held the position of Lucasian Professor of Mathematics at Cambridge University in England, a prestigious professorship later shared by Charles Babbage, George Gabriel Stokes, and Stephen Hawking, among others.
Work outside Science and Mathematics
The binomial theorem is a formula used to expand out expressions of the form
(x + y)^{r}
While Blaise Pascal had already developed the binomial theorem for the case where
r
is a nonnegative integer, Newton derived the general case for which
r
could be any rational number in 1655, while spending time away from Cambridge avoiding an outbreak of the plague [1]:
(x+y)^r = \sum\limits_{k=0}^{r} {r \choose k} x^{r-k} y^{k},
{r \choose k} = \frac{r (r-1) (r-2) \ldots (r-k+1)}{k!}
r
{r \choose k} = 0
k > r
. Then the infinite sum in the formula becomes a finite sum, and the expression reduces to the ordinary binomial theorem.
The expressions generated by these expansions were especially useful for calculating approximations of functions. Newton used the formula to calculate the value of
\pi
out to 16 decimal places [2].
f(x) = \sqrt{1 + x}
is expanded out in terms of powers of
x
f(x) = \sum\limits_{k=0}^{\infty} a_{k} x^{k},
what's the coefficient
a_{3}
x^{3}
Newton discovered three laws that combined would in principle determine the motion of any object. He published his laws in 1687 in the first volume of the Principia Mathematica (Latin for "Mathematical Principles"). These laws explain how any objects will move given the forces acting between them, and the initial position and velocity of the objects.
First Law: An object moving at some velocity will stay at that velocity unless acted upon by some force.
Second Law: The acceleration
\vec{a}
of an object is given by
\vec{F} = m\vec{a}
m
is its mass and
\vec{F}
is the net force on the object.
Third Law: Every action has an equal and opposite reaction.
The first law was in contrast to Aristotelian mechanics, which held that every object had a natural place, and that all objects would tend to go towards their natural place. Newton replaced this goal-centered view of the world with a mechanical, local one. The laws describe a perfectly deterministic universe, one in which the motion and behavior of all objects are theoretically exactly specified given a set of initial conditions and rules for determining the force between objects.
Because of Newton's contribution to the idea of force, the metric unit for force
\text{N} \equiv (\text{kg}\times \text{m})/\text{s}^2
is called the Newton.
Newton's laws of motion describe how objects accelerate given specific forces. In order to determine how the position of an object changes from a description of its acceleration, Newton needed to develop a new field of mathematics known as calculus.
Solving for the position, velocity, and acceleration of a moving particle
If a particle has a constant velocity
v
t
, its position at a slightly later time
t + \Delta t
x(t) + v \Delta t
. But if the particle is accelerating, this is not quite true. The velocity changes during the period
\Delta t
. In order to account for this, the time
\Delta t
could be split into two intervals, and the velocity at each of those points is used to calculate the position. But this doesn't help, as again the velocity changes between
t
t + \frac{1}{2} \Delta t
. Thinking in this way leads to an infinite regress.
Newton introduced calculus as a way of formalizing this reasoning and allowing calculations of position and velocity by considering how these functions behaved as
\Delta t
became very small, otherwise known as taking the limit of the function. The process of finding velocity from acceleration or position from velocity is called integration. The process of finding velocity from position or acceleration from velocity is called differentiation.
The mathematician and philosopher Gottfried Leibniz also invented calculus around the same time. There was a large fight in the scientific community over whether Leibniz or Newton had invented it first, or indeed whether one had stolen the ideas from the other. However, the consensus now is that they genuinely did develop the idea independently. Because of this, they used different notation styles for expressing calculus, both of which are in use today. In Newton's notation, the derivative of
x
with respect to time is given by
\dot{x}
, whereas in Leibniz notation, the derivative is
\frac{dx}{dt}
In the third volume of the Principia, Newton described his theory of universal gravitation. His two main insights were that masses attract each other along the line between them, and that every mass attracts every other mass, no matter how large or small. For instance, the force that you exert on the Earth is the same magnitude as the force the Earth exerts on you, just in the opposite direction. Mathematically, this is expressed as
\vec{F}_{g} = - \frac{Gm_{1}m_{2}}{r^{2}} \hat{r},
F_{g}
is the force of gravity,
G
m_1
m_2
are the masses of the two objects, and
\hat{r}
is the direction of the line between them. The net gravitational force on an object is the vector sum of the forces exerted on it by all other objects in the universe.
One popular story holds that Newton came up with the idea for gravitation when he was sitting under a tree and got hit on the head by an apple. This is not well supported by historical documents, but Newton did at least use falling apples as an analogy for explaining radial forces: "Therefore does this apple fall perpendicularly or towards the centre? If matter thus draws matter, it must be proportion of its quantity. Therefore the apple draws the Earth, as well as the Earth draws the apple." [3].
6.371 \times 10^6 \text{ m}
8.8 \times 10^3 \text{ m}
Newton showed that his law could reproduce Kepler's laws of planetary motion, which described how planets move in fixed ellipses around the sun. He then generalized these laws by showing that the paths of objects acting under the gravity of the sun could be any conic sections, including ellipses, but also parabolas, hyperbolas, and lines.
Because the law of gravitation describes a force between two objects no matter how far away they are, Leibniz accused Newton of invoking "spooky action at a distance." [4] This was against a popular philosophy of science at the time, which held that all effects needed to result from local interactions. Today, Einstein's theory of general relativity has replaced Newton's theory, in some sense proving Leibniz right. General relativity agrees with many of the predictions of Newton's theory, but doesn't have action at a distance. Gravitational effects (in the form of warped spacetime) propagate at the speed of light.
Light refracting in a prism: not just an album cover
As with his laws of motion, Newton also overturned Aristotelian beliefs about light. It was thought that white light was completely pure, but Newton showed that it was composed of every other wavelength of light by refracting white light through a prism. The white light splits into distinct beams of light corresponding to all the colors of the rainbow. Newton also invented the color wheel, which arranged all those colors in order, but with violet next to red to show the way humans perceive color:
Newton's depiction of the color wheel in Opticks
Newton's law of cooling holds that the rate at which an object will change temperature is directly proportional to the temperature difference between it
(T_{obj})
and its environment
(T_{env}):
\frac{dT_{obj}}{dt} = k (T_{env} - T_{obj}).
If the environment remains at constant temperature, this implies that
T_{obj}
will asymptotically approach
T_{env}:
T_{obj} = T_{env} + \big(T_{obj}(0) - T_{env}\big) e^{-kt},
which can be shown using differential equations.
80^\circ F
60^\circ F
50^\circ F
(
^\circ F)?
In addition to his lasting scientific discoveries, Newton also investigated alchemy, the study of turning one element into another. While the techniques that Newton investigated led nowhere, alchemy was in a sense rediscovered in the form of nuclear physics. It is now strictly possible to turn lead into gold using a particle accelerator. However, at an estimated quadrillion dollars per ounce, it would be a poor financial choice [5].
Newton was devoutly religious and would frequently study the Bible, attempting to make predictions based on its contents. He once wrote that the world would end no sooner than the year 2060 based on the Book of John [6].
[1] Westfall, Richard. Never at Rest: A Biography of Isaac Newton. p. 143. 1983.
[2] Newton's Generalization of the Binomial Theorem. Retrieved from http://www.wwu.edu/teachingmathhistory/docs/psfile/newton1-student.pdf on February 22, 2016.
[3] Connor, Steve. The Core of Truth Behind Sir Newton's Apple. The Independent. January 17, 2010. Retrieved from http://www.independent.co.uk/news/science/the-core-of-truth-behind-sir-isaac-newtons-apple-1870915.html on February 22, 2016.
[4] Leibniz's Philosophy of Physics. Stanford Encyclopedia of Philosophy. Published December 17. 2007. Retrieved from http://plato.stanford.edu/entries/leibniz-physics/ on February 22, 2016.
[5] Matson, John. Fact Or Fiction?: Lead Can Be Turned Into Gold. Scientific American. January 31, 2014. Retrieved from http://www.scientificamerican.com/article/fact-or-fiction-lead-can-be-turned-into-gold/ on February 22, 2016.
[6] Newton, Sir Isaac. Sir Isaac Newton's Daniel and the Apocalypse. 1733. Retrieved from http://publicdomainreview.org/collections/sir-isaac-newtons-daniel-and-the-apocalypse-1733/ on February 22, 2016.
Cite as: Isaac Newton. Brilliant.org. Retrieved from https://brilliant.org/wiki/isaac-newton/
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ENVI Financial Ratios - FinancialModelingPrep
\dfrac{Current Assets}{Current Liabilities}
\dfrac{Cash and Cash Equivalents + Short Term Investments + Account Receivables}{Current Liabilities}
\dfrac{Cash and Cash Equivalents}{Current Liabilities}
\dfrac{(Account Receivable (start) + Account Receivable (end))/2}{Revenue/365}
\dfrac{(Inventories (start) + Inventories (end))/2}{COGS/365}
\dfrac{DSO + DIO}{}
\dfrac{(Accounts Payable (start) + Accounts Payable (end))/2}{COGS/365}
\dfrac{DSO + DIO − DPO}{}
\dfrac{Gross Profit}{Revenue}
\dfrac{Operating Income}{Revenue}
\dfrac{Income Before Tax}{Revenue}
\dfrac{Net Income}{Revenue}
\dfrac{Provision For Income Taxes}{Income Before Tax}
\dfrac{Net Income}{Average Total Assets}
\dfrac{Net Income}{Average Total Equity}
\dfrac{EBIT}{Average Total Asset − Average Current Liabilities}
\dfrac{Net Income}{EBT}
\dfrac{EBT}{EBIT}
\dfrac{EBIT}{Revenue}
\dfrac{Total Liabilities}{Total Assets}
\dfrac{Total Debt}{Total Equity}
\dfrac{Long−Term Debt}{Long−Term Debt + Shareholders Equity}
\dfrac{Total Debt}{Total Debt + Shareholders Equity}
\dfrac{EBIT}{Interest Expense}
\dfrac{Operating Cash Flows}{Total Debt}
\dfrac{Total Assets}{Total Equity}
\dfrac{Revenue}{NetPPE}
\dfrac{Revenue}{Total Average Assets}
\dfrac{Operating Cash Flow}{Revenue}
\dfrac{Free Cash Flow}{Operating Cash Flow}
\dfrac{Operating Cash Flow}{Total Debt}
\dfrac{Operating Cash Flow}{Short-Term Debt}
\dfrac{Operating Cash Flow}{Capital Expenditure}
\dfrac{Operating Cash Flow}{Dividend Paid + Capital Expenditure}
\dfrac{DPS (Dividend per Share)}{EPS (Net Income per Share Number}
\dfrac{Stock Price per Share}{Equity per Share}
\dfrac{Stock Price per Share}{Operating Cash Flow per Share}
-423.31 The price/cash flow ratio is used by investors to evaluate the investment attractiveness, from a value standpoint, of a company's stock.
\dfrac{Stock Price per Share}{EPS}
\dfrac{Price Earnings Ratio}{Expected Revenue Growth}
428.94 The PEG ratio is a refinement of the P/E ratio and factors in a stock's estimated earnings growth into its current valuation.The general consensus is that if the PEG ratio indicates a value of 1, this means that the market is correctly valuing (the current P/E ratio) a stock in accordance with the stock's current estimated earnings per share growth. If the PEG ratio is less than 1, this means that EPS growth is potentially able to surpass the market's current valuation. In other words, the stock's price is being undervalued. On the other hand, stocks with high PEG ratios can indicate just the opposite - that the stock is currently overvalued.
\dfrac{Stock Price per Share}{Revenue per Share}
\dfrac{Dividend per Share}{Stock Price per Share}
\dfrac{Entreprise Value}{EBITDA}
-185.49 Overall, this measurement allows investors to assess a company on the same basis as that of an acquirer. As a rough calculation, enterprise value multiple serves as a proxy for how long it would take for an acquisition to earn enough to pay off its costs in years(assuming no change in EBITDA).
\dfrac{Stock Price per Share}{Intrinsic Value}
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Bias and Variability in the Periodogram - MATLAB & Simulink - MathWorks France
y\left(n\right)-0.75y\left(n-1\right)+0.5y\left(n-2\right)=\epsilon \left(n\right),
\epsilon \left(n\right)
is a zero mean white noise sequence with some specified variance. In this example, assume the variance and the sampling period to be 1. To simulate the preceding AR(2) process, create an all-pole (IIR) filter. View the filter's magnitude response.
y\left(n\right)-2.7607y\left(n-1\right)+3.8106y\left(n-2\right)-2.6535y\left(n-3\right)+0.9238y\left(n-4\right)=\epsilon \left(n\right).
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Consider a normal population distribution with the value of \sigma known? c.What value of
Consider a normal population distribution with the value of \sigma known?c.What value of Z\frac{\aplha}{2} in the CI formula (7.5)results in a confidence level of 99.7%?d. Answer the question posed in part (c) for a confidence level of75%?
Consider a normal population distribution with the value of
\sigma
known?
c.What value of
\frac{\alpha }{2}
in the CI formula (7.5)results in a confidence level of 99.7%?
d. Answer the question posed in part (c) for a confidence level of 75%?
You randomly survey students about participating in the science fair. The two-way table shows the results. How many female students do not participate in the science fair?
\begin{array}{|ccc|}\hline \text{Gender}& \text{No}& \text{Yes}\\ \text{Female}& 15& 22\\ \text{Male}& 12& 32\\ \hline\end{array}
A titanium bicycle frame displaces
0.314L
of water and has a mass of
1.41kg
. What is the density of the titanium in
\frac{g}{c}{m}^{3}
(a) The molar solubility of
PbB{r}_{2}\text{ }at\text{ }{25}^{\circ }C\text{ }is\text{ }1.0×{10}^{-2}\frac{mol}{L}
. Calculate Kp- (b) If 0.0490 g of
AgI{O}_{3}
dis- sp• solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate Ksp value from Appen- dix D, calculate the pH of a saturated solution of
Ca{\left(OH\right)}^{2}
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2-dehydro-3-deoxygluconokinase - Wikipedia
2-Keto-3-deoxygluconate kinase homohexamer, Thermus thermophilus
In enzymology, a 2-dehydro-3-deoxygluconokinase (EC 2.7.1.45) is an enzyme that catalyzes the chemical reaction
ATP + 2-dehydro-3-deoxy-D-gluconate
{\displaystyle \rightleftharpoons }
ADP + 6-phospho-2-dehydro-3-deoxy-D-gluconate
Thus, the two substrates of this enzyme are ATP and 2-dehydro-3-deoxy-D-gluconate, whereas its two products are ADP and 6-phospho-2-dehydro-3-deoxy-D-gluconate.[1]
This enzyme belongs to the family of transferases, specifically those transferring phosphorus-containing groups (phosphotransferases) with an alcohol group as acceptor. The systematic name of this enzyme class is ATP:2-dehydro-3-deoxy-D-gluconate 6-phosphotransferase. Other names in common use include 2-keto-3-deoxygluconokinase, 2-keto-3-deoxy-D-gluconic acid kinase, 2-keto-3-deoxygluconokinase (phosphorylating), 2-keto-3-deoxygluconate kinase, and ketodeoxygluconokinase. This enzyme participates in pentose phosphate pathway and pentose and glucuronate interconversions.
As of late 2007, only one structure has been solved for this class of enzymes, with the PDB accession code 1WYE.
^ Cynkin MA, Ashwell G (June 1960). "Uronic acid metabolism in bacteria. IV. Purification and properties of 2-keto-3-deoxy-D-gluconokinase in Escherichia coli". The Journal of Biological Chemistry. 235: 1576–9. PMID 13813474.
Retrieved from "https://en.wikipedia.org/w/index.php?title=2-dehydro-3-deoxygluconokinase&oldid=1039180449"
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(→CFD results and Comparison against Experiments)
m (Dave.Ellacott moved page Gold:CFD Simulations AC2-01 to CFD Simulations AC2-01 over redirect)
{\displaystyle {{\partial Q} \over {\partial t}}+{{\partial f_{1}} \over {\partial x_{1}}}={{\partial g_{1}} \over {\partial x_{1}}}+S}
{\displaystyle f_{1}}
{\displaystyle g_{1}}
{\displaystyle S}
{\displaystyle f_{1}}
{\displaystyle g_{1}}
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Hemoglobin | Brilliant Math & Science Wiki
Jordan Calmes, Satyabrata Dash, Christopher Williams, and
Hemoglobin (Hb or Hgb) is a molecule found in red blood cells. It carries oxygen from the lungs to the body's tissues and carbon dioxide from the tissues back to the lungs. Hemoglobin is important to both the structure and function of red blood cells.
Hemoglobin. [1]
Binding Affinity and Carbon Monoxide Poisoning
Chemical structure of a heme group
Hemoglobin is made up of four polypeptide chains of two types (two alpha and two beta). Each chain contributes one iron-containing ring structure, or heme group, with one ferrous ion (
\ce{Fe^{2+}}
) attached in the center, so each molecule of iron can transport four molecules of
\ce{O_2}
In fetuses and infants, the hemoglobin molecule is made up of two alpha chains and two gamma chains. Fetal hemoglobin has a higher binding affinity for oxygen than adult hemoglobin does, allowing the fetus to transport oxygen from the mother's blood supply and utilize it in utero. After birth, the growth of gamma chains is inhibited. As the infant grows, the gamma chains are gradually replaced by beta chains, forming the adult hemoglobin structure.
Carbon Iron Oxygen Hydrogen
When red blood cells are oxygenated, they have a vibrant red color. Which element do the oxygen atoms bind to in hemoglobin to create that color?
Beyond supplying oxygen to the tissues, hemoglobin serves several important physiological functions.
Hemoglobin is involved in buffering the blood by carrying excess
\ce{CO_2}
to the lungs.
Hemoglobin also plays an important role in maintaining the shape of the red blood cells. Normal red blood cells are round with narrow centers; they resemble a donut without a hole in the middle. Sickle-cell-anemia is a hereditary condition caused by a defect in hemoglobin structure. Affected individuals have a single base pair that is different in the gene coding for hemoglobin shape. This single amino acid substitution causes the red blood cells to taper and flatten into a C or sickle shape. The sickled cells polymerize (stick together) more easily, impeding blood flow and causing the cells to rupture.[2]
Generally people with sickle-cell diseases, like sickle-cell anemia, are more prone to infection, so treatments include preventative measures like vaccination, antibiotics, and dietary supplements. More serious treatments include blood transfusions, certain medications, and, in select cases, bone marrow transplants.
In the 120-day lifespan of red blood cells, sugar molecules attach to the hemoglobin (a process known as glycosylation). People with persistent high blood sugar (those with diabetes) will have higher-than-average levels of the glycosylated hemoglobin called Hemoglobin A1c (
\ce{HgbA_{1c}}
\ce{HgbA_{1c}}
levels are a useful monitoring parameter. While a fasting blood glucose level gives information about a patient's blood sugar at a single point in time,
\ce{HgbA_{1c}}
tells the average blood sugar level over the past 3-4 months, giving a more complete picture of a patient's nutrition and overall health and a better idea of whether or not their disease state is controlled.
Carbon monoxide (
\ce{CO}
) has a binding affinity for hemoglobin that is over 200 times higher than oxygen's. Once carbon monoxide is bound, the hemoglobin is unable to release it and pick up an oxygen molecule. Additionally, hemoglobin that's bound to (
\ce{CO}
) (called carboxyhemoglobin) makes it harder for oxygen to dissociate from oxyhemoglobin, meaning that oxygen already in the bloodstream is less accessible to tissues. Carbon monoxide is a colorless, odorless gas. It has many sources, including incomplete combustion of gasoline, defective household furnaces, and cigarette smoking.
CO levels will be higher in the fetus than in the mother No CO will reach the fetus CO levels will be the same in the fetus and the mother CO levels will be lower in the fetus than in the mother
Pregnant women are often warned to avoid smoking cigarettes. If a woman does smoke, how high will the levels of carbon monoxide (CO) be in the fetus compared to the levels in the mother?
Lowlanders visiting a mountainous area often become more tired from the same amount of activity than they would at home. Some even develop altitude sickness.
At sea level, the partial pressure of oxygen is higher than it is at higher elevation. As a result, when gas exchange takes place in the lungs at altitude, the alveolar air, arterial blood, and venous blood all have a lower oxygen content than they would in the same person at sea level if the difference is severe enough, the person experiences hypoxia, a state where the body is no longer meeting its tissues' oxygen needs.
Individuals who move from low elevation to high elevation can acclimatize in several ways, including sustained hyperventilation, increased cardiac output, and increased red blood cell production. Athletes will often train at altitude to improve their oxygen utilization when they compete, as these physiological effects can last for several days after returning to lower elevations.
Acclimatization is short-term, but communities who live at high altitude for multiple generations sometimes develop adaptations to hypoxia. The Aymara of the Andes and the Tibetans are two ethnic groups that live at high altitude, and their genetic adaptations illustrate multiple ways of surviving in a lower-oxygen environment.[3]
The Aymara have much higher erythropoietin levels at altitude, which stimulates red blood cell production, resulting in higher hemoglobin concentrations. This physiological difference leads directly to a higher oxygen saturation compared to populations born at lower elevations. However, this adaptation also has a down side: the Aymara also have high rates of pulmonary hypertension, a type of high blood pressure that can eventually lead to blocked arteries around the lungs and heard.
Tibetans have chronically increased ventilation rates compared to either the Aymara or lowlanders, and their oxygen saturation levels tend to be lower than average. However, they do not suffer from hypoxia. Their body compensates by reducing the pressure in their lungs and increasing capillary density, which allows the tissues to better bind oxygen without increasing hemoglobin.
BerserkerBen, . Hemoglobin_t-r_state_ani. Retrieved August 24, 2016, from https://en.wikipedia.org/wiki/Hemoglobin#/media/File:Hemoglobin_t-r_state_ani.gif
National Heart, Lung, and Blood Institute, . What is Sickle Cell Disease?. Retrieved from http://www.nhlbi.nih.gov/health/health-topics/topics/sca/
Beall, C. Tibetan and Andean Patterns of Adaptation to High-Altitude Hypoxia. Human Biology, 72(1), 201-228.
Cite as: Hemoglobin. Brilliant.org. Retrieved from https://brilliant.org/wiki/hemoglobin/
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An Empirical Analysis of the Risk Taking Channel of Monetary Policy in China—Base on Evidence from Chinese Listed Bank
An Empirical Analysis of the Risk Taking Channel of Monetary Policy in China―Base on Evidence from Chinese Listed Bank
{\text{Risk}}_{i,t}={a}_{0}{\text{Risk}}_{i,t-1}+{a}_{1}{\text{MP}}_{t}+{a}_{2}{\text{GDP}}_{t}+{a}_{3}{\text{Bank}}_{it}+{a}_{4}{\text{Mar}}_{it}+{v}_{i}+{u}_{it}
{a}_{1}
{a}_{1}
\begin{array}{l}{\text{Risk}}_{i,t}={a}_{0}{\text{Risk}}_{i,t-1}+{a}_{1}{\text{MP}}_{t}+{a}_{2}{\text{GDP}}_{t}+{a}_{3}{\text{Size}}_{i,t-1}+{a}_{4}{\text{Roa}}_{i,t-1}+{a}_{5}{\text{Cap}}_{i,t-1}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{v}_{i}+{u}_{it}\end{array}
Chen, H.T. (2019) An Empirical Analysis of the Risk Taking Channel of Monetary Policy in China―Base on Evidence from Chinese Listed Bank. American Journal of Industrial and Business Management, 9, 1033-1044. https://doi.org/10.4236/ajibm.2019.94071
1. Angeloni, I., Faia, E. and Duca, M.L. (2015) Monetary Policy and Risk Taking. Journal of Economic Dynamics & Control, 52, 285-307. https://doi.org/10.1016/j.jedc.2014.12.001
2. The General Office of the State Council (2013) Guidelines on Financial Support for Economic Restructuring and Transformation and Upgrading. Chinese Financiers, No. 8, 17-19.
3. Zhao, S. (2017) Research on the Existence of Risk Bearing Channels of Monetary Policy—From the Perspective of New Inter-Bank Business. Research on Technology Economy and Management, No. 7, 70-76.
4. Borio, C. and Zhu, H. (2009) Capital Regulation, Risk-Taking and Monetary Policy: A Missing Link in the Transmission Mechanism? Journal of Heart & Lung Transplantation the Official Publication of the International Society for Heart Transplantation, 8, 73-84.
5. Dubecq, S., Mojon, B. and Ragot, X. (2009) Fuzzy Capital Requirements, Risk-Shifting and the Risk Taking Channel of Monetary Policy. https://doi.org/10.2139/ssrn.1630317
6. Laeven, L., Dell’Ariccia, G. and Marquez, R. (2011) Monetary Policy, Leverage and Bank Risk Taking. Cepr Discussion Papers, 975-1009. https://doi.org/10.2139/ssrn.1740763
7. Valencia, F. (2014) Monetary Policy, Bank Leverage, and Financial Stability. Journal of Economic Dynamics and Control, 47, 20-38. https://doi.org/10.1016/j.jedc.2014.07.010
8. De Nicolò, G., Dell’Ariccia, G., Laeven, L., et al. (2010) Monetary Policy and Bank Risk Taking. https://doi.org/10.2139/ssrn.1654582
9. Adrian, T. and Shin, H.S. (2009) Money, Liquidity, and Monetary Policy. American Economic Review, 99, 600-605. https://doi.org/10.1257/aer.99.2.600
10. Adrian, T. and Shin, H.S. (2010) Liquidity and Leverage. Journal of Financial Intermediation, 19, 418-437. https://doi.org/10.1016/j.jfi.2008.12.002
11. Rajan, R.G. (2005) Has Financial Development Made the World Riskier? Social Science Electronic Publishing, NBER Working Paper No. 11728, 371-379. https://doi.org/10.3386/w11728
12. Dell’Ariccia, G. and Marquez, R. (2006) Lending Booms and Lending Standards. The Journal of Finance, 61, 2511-2546. https://doi.org/10.1111/j.1540-6261.2006.01065.x
13. Campbell, J.Y. and Cochrane, J.H. (1999) By Force of Habit: A Consumption-Based Explanation of Aggregate Stock Market Behavior. Journal of Political Economy, 107, 205-251. https://doi.org/10.1086/250059
14. Angeloni, I. and Faia, E. (2013) Capital Regulation and Monetary Policy with Fragile Banks. Journal of Monetary Economics, 60, 311-324.
15. Altunbas, Y., Gambacorta, L. and Marques-Ibanez, D. (2010) Does Monetary Policy Affect Bank Risk-Taking. https://doi.org/10.2139/ssrn.1577075
16. López, M., Tenjo, F. and Zárate, H. (2011) The Risk-Taking Channel and Monetary Transmission Mechanism in Colombia. Ensayossobre Política Económica, 29, 212-234.
17. López, M., Tenjo, F. and Zárate, H. (2012) The Risk-Taking Channelin Colombia Revisited. Ensayos Sobre Política Económica, 30, 274-295.
18. Zhang, H. (2012) Monetary Policy Position and Bank Risk Taking: An Empirical Study Based on China’s Banking Industry (2000-2010). Economic Research, No. 5, 31-44.
19. Zhang, H. (2012) Risk-Taking Channels of Monetary Policy: Conduction Path, Asymmetry and Intrinsic Mechanism. Financial Review, No. 1, 71-81.
20. Delis, M.D. and Kouretas, G.P. (2011) Interest Rates and Bank Risk-Taking. Journal of Banking & Finance, 35, 840-855. https://doi.org/10.1016/j.jbankfin.2010.09.032
21. Xu, C. (2012) Currency Environment, Capital Adequacy Ratio and Risk Acceptance of Commercial Banks. Financial Research, No. 7, 48-62.
22. Aggarwal, R. and Jacques, K.T. (1998) Assessing the Impact of Prompt Corrective Action on Bank Capital and Risk. Economic Policy Review, 4, 23-32. https://doi.org/10.2139/ssrn.1024839
24. Yin (2013) Research on Risk Bearing Channels of Monetary Policy. Shandong University, Jinan.
25. Jin, P. (2014) Bank Risk-Taking Channels, Monetary Policy and Macro-Prudential Supervision: Research Review and Outlook. Southern Finance, No. 8, 13-20.
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Developing a New Reformulation of Single Level Capacitated Lot Sizing Problem (SLCLSP) with Set up, Shortage and Inventory Costs
Developing a New Reformulation of Single Level Capacitated Lot Sizing Problem (SLCLSP) with Set up, Shortage and Inventory Costs ()
Department of Industrial & Management Engineering, Indian Institute of Technology, Kanpur, India.
Formulation of SLCLSP given by Pochet and Wolsey [1] had set up, variables, inventory and shortage cost. We give a new reformulation where SLCLSP is reduced to set up and inventory variables. We find that this reformulation has less number of real variables than the reformulation of Pochet and Wolsey [1]. It is argued that this leads to computations advantages, and this is supported by the empirical investigation that we carried out.
Reformulation of SLCLSP
Sharma, R. , Kumar, V. and Khan, N. (2017) Developing a New Reformulation of Single Level Capacitated Lot Sizing Problem (SLCLSP) with Set up, Shortage and Inventory Costs. American Journal of Operations Research, 7, 272-281. doi: 10.4236/ajor.2017.75019.
Capacitated lot sizing problem (CLSP) is well studied in literature, see Verma [2] , and Verma and Sharma [3] [4] [5] for a summary of recent works on CLSP. For literature on reformulation of CLSP, see Pochet and Wolsey [1] and Miller and Nemhauser et al. [6] for a detailed exposition on reformulations of CLSP. In this paper we give a new approach which leads to a better reformulation of CLSP.
2. Formulation by Pochet and Wolsey [1]
t: Set of the Time period from 1, ・・・, n, for which we are taking decisions;
{f}_{t}
: Fixed setup cost in time period “t”;
{p}_{t}
: Per unit production cost in time period “t”;
{d}_{t}
: Demand in time period “t”, here demand is independent;
{h}_{t}
: Per unit inventory carrying cost in time period “t”;
s{h}_{t}
: Per unit shortage cost in time period “t”;
{c}_{t}
: Production capacity in the time period “t”;
{x}_{t}
: Number of product produced in time period “t”;
{y}_{t}
: Binary variable takes value “1” if machine setup to produce in time period “t”, “0”, otherwise;
{I}_{t}
: In stock inventory at the end of time period “t”;
{s}_{t}
: Backlog in the end of period “t”;
\text{Minimize}\text{\hspace{0.17em}}Z=\underset{t=1}{\overset{n}{\sum }}{f}_{t}\ast {y}_{t}+\underset{t=1}{\overset{n}{\sum }}{p}_{t}\ast {x}_{t}+\underset{t=0}{\overset{n-1}{\sum }}{h}_{t}\ast {I}_{t}+\underset{t=1}{\overset{n}{\sum }}s{h}_{t}\ast {s}_{t}
Production balance constraints
{x}_{t}+\left({s}_{t}-{s}_{t-1}\right)={d}_{t}+\left({I}_{t}-{I}_{t-1}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\le t\le n
{x}_{t}\le {c}_{t}\ast {y}_{t},\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\le t\le n
{I}_{o}={s}_{n}=0
Pochet and Wolsey [1] gave the following constraint that lead to reformulation:
{x}_{t}={y}_{t}\ast {c}_{t}
{x}_{t},{r}_{t},{s}_{t}\ge 0
SLCLSP as given by Pochet and Wolsey [1] is Model A1: Min (1); s.t. (2), (3), (4) and (6). By using (5) in place of (3) lead to reformulation (called Model A2: min (1); s.t. (2), (4), (5) and (6). Model A1 has less number of variables as variable “x” is eliminated.
We add a new constraint given below (see, [7] ) that can be used in place of (2):
{I}_{0}+\underset{t=1}{\overset{{t}_{1}}{\sum }}{x}_{t}+{s}_{{t}_{1}}=\underset{t=1}{\overset{{t}_{1}}{\sum }}{D}_{t}+{I}_{{t}_{1}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\forall {t}_{1}=1,\cdots ,T
Using (5) we get the following (called Model A3): Min Z1 (or (8)); s.t. (4), (5), (6), (7).
\text{Min}Z1=\underset{t=1}{\overset{T}{\sum }}{f}_{t}\ast {y}_{t}+\underset{t=1}{\overset{T}{\sum }}{p}_{t}\ast {c}_{t}\ast {y}_{t}+\underset{t=1}{\overset{T}{\sum }}{h}_{t}\ast {I}_{t}+\underset{t=1}{\overset{T}{\sum }}s{h}_{t}\ast {s}_{t}
We use (7) to eliminate st from the problem A2 to get: Min Z2 (or (9)); s.t. (4), (5), (6).
\text{Min}Z2=\underset{t=1}{\overset{T}{\sum }}{f}_{t}\ast {y}_{t}+\underset{t=1}{\overset{T}{\sum }}{p}_{t}\ast {c}_{t}\ast {y}_{t}+\underset{t=1}{\overset{T}{\sum }}{h}_{t}\ast {I}_{t}
+\underset{t=1}{\overset{T}{\sum }}s{h}_{t}\ast \left(\underset{t=1}{\overset{{t}_{1}}{\sum }}{D}_{t}+\text{}{I}_{{t}_{1}}-{I}_{0}-\underset{t=1}{\overset{{t}_{1}}{\sum }}{c}_{t}\ast {y}_{t}\right)
It can be seen that model A3 has least number of variables; it is followed by A2 that has less number of variables compared to model A1 which is well known reformulation (Pochet and Wolsey [1] . We solve model A1, model A2, and
Table 1. Problem for 50 time period (Z value and No. of node processed in brand and bound procedure).
Table 2. Problem for 50 time period (Iteration and execution time in GAMS).
Table 3. Problem for 60 time period (Z value and No. of nodes).
Table 4. Problem for 60 time period (Iteration and execution time).
Table 5. Problem for 100 time period (Z value and No. of nodes).
Table 6. Problem for 100 time period (Iteration and execution time).
model A3 by using student version of GAMS available at IIT Kanpur; and find that our reformulations (model A3 and A2) have superior computational advantages than model has A1.
3. Preparing Test Problems and Results
We created problem instances with set up, inventory carrying, shortage and production cost are normally distributed with mean and variance given below:
Fixed cost: mean 100000 and variance 10000
Shortage cost: mean 5000 and variance 500
Inventory carrying cost: mean 600 and variance 60
Variable Production cost: mean1500 and variance 100
Demand and capacity were chosen from uniform distribution in the range of 10,000 - 15,000. In the case of infeasible solution, the capacity values are increased or demand values are decreased keeping other costs same. We created 50 problem instances each for periods 50, 60 and 100. Models A1, A2 and A3 were coded in GAMS and were solved in GAMS; and they were run in branch and bound mode. The GAMS solver returns a satisfactory solution obtainable in reasonable time. It is to be noted that these problems are NP-HARD and will take few billion centuries to come to optimal solution. Detailed data are given in appendix see Tables 1-6; and consolidated results of “t” test are given in Tables 7-9 below. Models A1, A2 and A3 are compared on the criteria of execution time,
Table 7. 50 Time period: t values.
Table 9. 100 time period: t values.
*Significant at 0.05 level; **significant at 0.01 level; ***significant at 0.001 level; In Table 7: 2.742**means that model A3 takes less time than model A1 and is significant at 0.01 level.
number of iterations and number of nodes evaluated in the search tree). On an average, A3 is superior to A1 and A2 and A2 is superior to A1 on most criteria; and large positive “t” values in the Tables 7-9 give adequate support in favor of A3.
Thus it can be seen that model A3 has superior results in general (except for the case of execution time in 60 period problems) (here A3 is better than A1, but not statistically significant). This shows that the new formulation given by us is superior to models available in literature. This is the useful contribution we make. The three reformulations presented in this paper use Equation (5) and this leads
\underset{t=1}{\overset{T}{\sum }}{x}_{t}\ge \underset{t=1}{\overset{T}{\sum }}{D}_{t}
\underset{t=1}{\overset{T}{\sum }}{x}_{t}=\underset{t=1}{\overset{T}{\sum }}{D}_{t}
, we need to develop good heuristics, and
show that the duality gap is as less as possible. We have already started work on this, and will come back with results as soon as possible.
[1] Pochet, Y. and Wolsey, L.A. (1991) Solving Multi-Item Lot-Sizing Problems Using Strong Cutting Planes. Management Science, 37, 53-67.
[2] Mayank, V. (2012) Capacitated Lot Sizing with Back Orders in Multilevel Situations. Ph.D. Thesis, Indian Institute of Technology, Kanpur.
[3] Mayank, E. and Sharma, R.R.K. (2009) Relaxations and Equivalence of Two Formulations of the Capacitated Lot Sizing Problem with Back-Orders and Setup Times. Proceedings of the Global Conference on Business and Finance, 4, 42-53.
[4] Mayank, V. and Sharma, R.R.K. (2010) A New Lagrangian Relaxation Based Approach to solve Capacitated Lot-Sizing Problem with Backlogging. Global Business and Management Research, Universal-Publishers, Boca Raton, Vol. 2, 285-295.
[5] Mayank, V. and Sharma, R.R.K. (2015) Lagrangian Based Approach to Solve a Two Level Capacitated Lot Sizing Problem. Cogent Engineering, 2, 108861.
[6] Miller, A.J., Nemhauser, G.L. and Savelsbergh, M.W.P (2000) On the Capacitated Lot Sizing and Continuous 0-1 Knapsack Polyhedral. European Journal of operational Research, 125, 298-315.
[7] Kumar, V. (2012) Equal Distribution of Shortages in Supply Chain of Food Corporation of India: Using Lagrangian Relaxation Methodology. M. Tech Dissertation, Indian Institute of Technology, Kanpur. (Unpublished)
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Solving Equations: Level 2 Challenges Practice Problems Online | Brilliant
Five brothers stayed in the house with their mother. One day, their mother brought home some mangos.
Alex woke up first. As he was hungry, he ate
\frac{1}{6}
of the mangoes, and headed out.
Brian woke up next. As he was hungry, he ate
\frac{1}{5}
of what was left, and headed out.
Charles woke up next. As he was hungry, he ate
\frac{1}{4}
Danny woke up next. As he was hungry, he ate
\frac{1}{3}
Euler woke up next. As he was hungry, he ate
\frac{1}{2}
Their mother came home and saw 3 mangoes in the basket. How many mangoes were there initially?
A teacher writes 6 consecutive integers on the blackboard.
He erases one of the integers and the sum of the remaining five integers is 2016.
What integer was erased?
There are 100 people in a room, exactly 99% are physicist. How many physicist must leave the room to bring down the percentage of physicist to exactly 98%?
\text{If } ac = bd \text{ and } c = d, \text{ then it is certain that } a = b.
If we take a certain 2-digit integer and reverse its digits to form another 2-digit integer, the absolute difference between these two numbers is always divisible by which of the following numbers?
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Health Risk Assessment from Exposure to Heavy Metals in Surface and Groundwater Resources within Barkin Ladi, North Central Nigeria
1Department of Minerals & Petroleum Resources Engineering, Plateau State Polytechnic, Barkin Ladi, Nigeria
\text{ADD}=\frac{\text{C}\times \text{IR}\times \text{ED}\times \text{EF}}{\text{BW}\times \text{AT}\times 365}
\left(\text{HQ}\right)=\frac{\text{ADD}}{\text{RfD}}
\left(\text{HI}\right)=\sum {\text{HQ}}_{i}
Ramadan, J. A., & Haruna, A. I. (2019). Health Risk Assessment from Exposure to Heavy Metals in Surface and Groundwater Resources within Barkin Ladi, North Central Nigeria. Journal of Geoscience and Environment Protection, 7, 1-21. https://doi.org/10.4236/gep.2019.72001
1. Adamu, C. I., Nganje, T. N., & Edet, A. (2014). Heavy Metal Contamination and Health Risk Assessment Associated with Abandoned Barite Mines in Cross River State, Southeastern Nigeria. Journal of Environmental Nanotechnology, Monitoring & Management, 3, 10-21. [Paper reference 1]
2. Ajayi, P. O. (2003). Comprehensive Geography for Senior Secondary Schools. Lagos: Johnson Publishers Ltd. [Paper reference 1]
3. Apambire, W. B., Boyle, D. R., & Michel, F. A. (1997). Geochemistry, Genesis and Health Implications of Flouriferous Groundwaters in the Upper Regions of Ghana. Environmental Geology, 33, 13-24. https://doi.org/10.1007/s002540050221 [Paper reference 1]
4. Ayantobo, O. O., Awomeso, J. A., Oluwasanya, G. O., Bada, B. S., & Taiwo, A. M. (2014). Non-Cancer Human Health Risk Assessment from Exposure to Heavy Metals in Surface and Groundwater in Igun-Ijesha, Southwest, Nigeria. American Journal of Environmental Sciences, 10, 301-311. https://doi.org/10.3844/ajessp.2014.301.311 [Paper reference 2]
5. Elumalai, V., Brindha, K., & Lakshmanan, E. (2017). Human Exposure Risk Assessment Due to Heavy Metals in Groundwater by Pollution Index and Multivariate Statistical Methods. A Case Study from South Africa. Water Journal, 9, 234. [Paper reference 1]
6. ENHIS (European Environment and Health Information System) (2007). Exposure of Children to Chemical Hazards in Food. Fact Sheet No. 4.4.CODE. RPG4_Food_EXI, World Health Organization. [Paper reference 1]
7. Fordyce, J., Smith, B., Appleton, D., Johnson, C., Smedley, P., & Williams, L. (1999). Environmental Geochemistry and Health-Global Perspectives. British Geological Survey. [Paper reference 1]
8. Iloeje, N. P. (1981). A New Geography of Nigeria. Lagos: Longman Nig. Ltd. [Paper reference 1]
9. Lim, H. S., Lee, J. S., Chon, H. T., & Sager, H. (2008). Heavy Metal Contamination and Health Risk Assessment in the Vicinity of Abandoned Songcheon Au-Ag Mine in Korea. Journal of Geochemical Exploration, 96, 223-230. https://doi.org/10.1016/j.gexplo.2007.04.008 [Paper reference 1]
10. MacLeod, W. N., & Berridge, N. G. (1971): Geology of the Jos Plateau. Geological Survey of Nigeria. [Paper reference 2]
11. Montuori, P., Lama, P., Aurino, S., Naviglio, D., & Triassi, M. (2013). Metals Loads into the Mediterranean Sea: Estimate of Sarno River Inputs and Ecological Risk. Ecotoxicology, 22, 295-307. https://doi.org/10.1007/s10646-012-1026-9 [Paper reference 2]
12. Ramadan, J. A., & Haruna, A. I. (2018). Assessment of Heavy Metal Contamination in Surface and Groundwater Resources Using Pollution Indices in Parts of Barkin Ladi, North Central Nigeria. IOSR Journal of Applied Geology and Geophysics, 6, 25-40. [Paper reference 1]
13. Roychowdhury, T., Tokunaga, H., & Ando, M. (2003). Survey of Arsenic and Heavy Metals in Food Composites and Drinking Water and Estimation of Dietary Intake by the Villagers from an Arsenic Infected Area of West Bengal, India. Science of the Total Environment, 308, 15-35. https://doi.org/10.1016/S0048-9697(02)00612-5 [Paper reference 2]
14. USEPA (US Environmental Protection Agency) (1999). A Risk Assessment Multiway Exposure Spreadsheet Calculation Tool. Washington DC: United States Environmental Protection Agency. [Paper reference 17]
15. USEPA (US Environmental Protection Agency) (2001). Baseline Human Health Risk Assessment, Vasquez Boulevard and 1-70 Superfund Site, Denver. http://www.epa.gov/region8/superfund/sites/VB-170-Risk.pdf. [Paper reference 1]
16. USEPA IRIS (US Environmental Protection Agency’s Integrated Risk Information System) (2011). Environmental Protection Agency Region I. Washington DC. 20460. [Paper reference 2]
17. Wongsasuluk, P., Chotpantarat, S., Siriwong, W., & Robson, M. (2014). Heavy Metal Contamination and Health Risk Assessment in Drinking Water from Shallow Groundwater Wells in an Agricultural Area in Ubon Ratchathani Province, Thailand. Environmental Geochemistry and Health, 36, 169-182. https://doi.org/10.1007/s10653-013-9537-8 [Paper reference 1]
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Convert from sone to phon - MATLAB sone2phon - MathWorks í•œêµ
sone2phon
Convert Sone to Phon
ISO 532-1: Zwicker Method
ISO 532-2: Moore-Glasberg Method
Convert from sone to phon
phon = sone2phon(sone)
phon = sone2phon(sone,standard)
phon = sone2phon(sone) converts sone to phon, according to ISO 532-1:2017(E).
phon = sone2phon(sone,standard) specifies the standard used to convert sone to phon.
Plot the relationship between loudness (sone) and loudness levels (phon), as specified in ISO 532-1.
s = (0.51:0.01:1.8).^10;
p1 = sone2phon(s);
semilogx(s,p1)
xlabel('Loudness (sone)')
ylabel('Loudness Level (phon)')
title('Relation Between Sone and Phon (ISO 532-1)')
axis([0 s(end) 0 130])
p2 = sone2phon(s,'ISO 532-2');
sone — Input loudness in sone
nonnegative scalar | vector of nonnegative values | matrix of nonnegative values | multidimensional array of nonnegative values
Input loudness in sone, specified as a scalar, vector, matrix, or multidimensional array of nonnegative values.
standard — Reference standard for unit conversion
Reference standard for unit conversion, specified as 'ISO 532-1' or 'ISO 532-2'.
phon — Output loudness level in phon
Output loudness level in phon, returned as a scalar, vector, matrix, or multidimensional array the same size as sone.
The Zwicker method of conversion from sone to phon is given by this equation in [1]:
phon=\left\{\begin{array}{c}40{\left(sone\right)}^{0.35}\\ 40+10{\mathrm{log}}_{2}\left(sone\right)\end{array}\begin{array}{c}\text{â}\\ \text{â}\end{array}\begin{array}{c}\text{if}\text{â}sone<1\\ \text{otherwise}\text{â}\end{array}
In the Moore-Glasberg method, conversion from sone to phon is prescribed according to this table (table 5 in [2]).
Loudness Level (phon)
Calculated Loudness (sone)
The sone2phon function uses interpolation for values not specified in the table.
[2] ISO 532-2:2017(E). "Acoustics – Methods for calculating loudness – Part 2: Moore-Glasberg method." International Organization for Standardization.
phon2sone | acousticLoudness
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Gain_(laser) Knowpia
The gain can be defined as the derivative of logarithm of power
{\displaystyle ~P~}
as it passes through the medium. The factor by which an input beam is amplified by a medium is called the gain and is represented by G.
{\displaystyle G={\frac {\rm {d}}{{\rm {d}}z}}\ln(P)={\frac {{\rm {d}}P/{\rm {d}}z}{P}}}
{\displaystyle ~z~}
is the coordinate in the direction of propagation. This equation neglects the effects of the transversal profile of beam.
{\displaystyle 2ik{\frac {\partial E}{\partial z}}=\Delta _{\perp }E+2\nu E+iGE}
{\displaystyle ~\nu ~}
is variation of index of refraction (Which is supposed to be small),
{\displaystyle ~E~}
is complex field, related to the physical electric field
{\displaystyle ~E_{\rm {phys}}~}
with relation
{\displaystyle ~E_{\rm {phys}}={\rm {Re}}\left({\vec {e}}E\exp(ikz-i\omega t)\right)~}
{\displaystyle ~{\vec {e}}~}
is vector of polarization,
{\displaystyle ~k~}
is wavenumber,
{\displaystyle ~\omega ~}
is frequency,
{\displaystyle ~\Delta _{\rm {\perp }}=\left({\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}\right)~}
is transversal Laplacian;
{\displaystyle ~{\rm {Re~}}}
means real part.
Gain in quasi two-level systemEdit
In the simple quasi two-level system, the gain can be expressed in terms of populations
{\displaystyle ~N_{1}~}
{\displaystyle ~N_{2}~}
of lower and excited states:
{\displaystyle ~G=\sigma _{\rm {e}}N_{2}-\sigma _{\rm {a}}N_{1}~}
{\displaystyle ~\sigma _{\rm {e}}~}
{\displaystyle ~\sigma _{\rm {a}}~}
are effective emission and absorption cross-sections. In the case of non-pumped medium, the gain is negative.
Round-trip gain means gain multiplied by the length of propagation of the laser emission during a single round-trip. In the case of gain varying along the length, the round-trip gain can be expressed with integral
{\displaystyle g=\int G{\rm {d}}z}
. This definition assumes either flat-top profile of the laser beam inside the laser, or some effective gain, averaged across the beam cross-section.
The amplification coefficient
{\displaystyle ~K~}
can be defined as ratio of the output power
{\displaystyle ~P_{\rm {out}}}
to the input power
{\displaystyle ~P_{\rm {in}}}
{\displaystyle ~K=P_{\rm {out}}/P_{\rm {in}}}
It is related with gain;
{\displaystyle ~K=\exp \left(\int G{\rm {d}}z\right)~}
The gain and the amplification coefficient should not be confused with the magnification coefficient. The magnification characterizes the scale of enlarging of an image; such enlargement can be realized with passive elements, without gain medium. [1]
Alternative terminology and notationsEdit
There is no established terminology about gain and absorption. Everyone is free to use own notations, and it is not possible to cover all the systems of notations in this article.
In radiophysics, gain may mean logarithm of the amplification coefficient.
In many articles on laser physics, which do not use the amplification coefficient
{\displaystyle ~K~}
defined above, the gain is called Amplification coefficient, in analogy with Absorption coefficient, which is actually not a coefficient at all; one has to multiply it to the length of propagation (thickness), change the signum, take inverse of the exponential, and only then get the coefficient of attenuation of the sample.
Some publications use term increment instead of gain and decrement instead of absorption coefficient to avoid the ambiguity,[2] exploiting the analogy between paraxial propagation of quasi-monochromatic waves and time evolution of a dynamic system.
Round-trip gain, gain multiplied by the length of propagation of the laser emission during a single round-trip
Effective cross-sections
^ A.E.Siegman (1986). Lasers. University Science Books. ISBN 0-935702-11-3. Archived from the original on 2016-12-06. Retrieved 2007-04-09.
^ D.Yu.Kuznetsov (1995). The transformation of the transverse structure of monochromatic light in the non-linear media. In book: -- Optics and Lasers. ed: G.G.Petrash.
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Priya Karna1, Roma Karna2, Sunil Karna1
1Department of Natural Science, Union College, Barbourville, KY, USA.
2Barbourville High School, Barbourville, KY, USA.
The experiment was conducted to demonstrate the behavior of fire propagation in wildlands using a matchsticks forest model. A model forest was designed on a flame resistant clay, on top of which matchsticks were inserted and kept vertical to the ground by keeping space between them constant with the help of aluminum grid. The data for distance travelled by fire with time were taken at wide range of slopes from downhill of ﹣25° to uphill of 45° on a model forest of packing ratios 0.08 and 0.04. The minimum rate of fire spread was observed around 15° downhill. The data collected from this experiment follow tan2θ and agree with the Rothermel’s mathematical model of fire propagation except at elevation above 35° for low packing ratio.
Wildfire, Packing Ratio, Slope, Fire Propagation
Karna, P. , Karna, R. and Karna, S. (2017) Effect of Slope and Packing Ratio on the Behavior of Matchsticks Burnings. Open Access Library Journal, 4, 1-6. doi: 10.4236/oalib.1103737.
Wildfire is a natural periodic event that cleans up the wild vegetation and is necessary for soil fertilization [1] . However, wildfire often becomes devastating and causes tremendous losses every year. A faster deforestation, wildlife chaos, uncompensated environmental loss, and economical imbalances are some of the impacts of wildlife on society. Fire behavior is a complex phenomenon of aerodynamics, thermodynamics, and combustion physics [2] . Understanding the behavior of fire spread in bush may help control fire spreading, modern forestation, and wildlife husbandry. Many studies have been conducted in past to predict wildfire propagation and to protect wildland from fire damage [3] [4] [5] [6] . Some of these reports state that slopes have relatively low effect in the absence of wind on fire spread [7] . However, other reports indicate that slope can significantly affect the rate of fire spread [4] , the average size of flames increases with the slope [4] , and the rate of fire spread follows tan2(slope angle) if there is no wind [6] . Although these studies have reported many important aspects of fire propagation, our studies will further aid understanding of the behavior of fire propagation and provide verification of Rothermel’s theoretical model. The rate of fire spread in Rothermel’s model is given by the following equation
R=A\left(1+{\phi }_{\omega }+{\phi }_{s}\right)
, where A is a constant that depends upon the heat source, fuel density, and a reaction intensity,
{\phi }_{\omega }
is a wind coefficient, and
{\phi }_{s}=5.275{\beta }^{-0.3}{\mathrm{tan}}^{2}\theta
, a slope factor, where β indicates packing ratio. Since our experiment was performed in no wind condition the rate of fire spread will be taken as
R=A\left(1+{\phi }_{s}\right)=A\left(1+B{\mathrm{tan}}^{2}\theta \right)
, Where A and B are constants that can be determined from the line of best fit.
In this study, we aim to further investigate and understand fire propagation in forests by designing sets of experiments with a matchsticks forest model at different slopes and packing ratios. Two sets of experiment were designed to measure the rate of fire spread on a matchsticks fuel bed with slopes ranging from −25˚ to 45˚ and with packing ratios of 0.08 and 0.04. All tests were conducted in an open space with temporary side walls to block any wind, and were maintained at ~58˚F ± 3 ambient temperature, ~76% ± 4 relative humidity, and at still wind velocity. Arrays of protruding matchsticks in clay bed were used to investigate the behavior of fire spread based on fuel sticks spacing and the fuel bed elevations. Flame length, headfire, and backfire speeds were also considered in predicting fire propagation behavior. Some of the outliers in the data may be assumed to be more influential and were significantly affecting the behavior of fire spread. After analyzing the data by Cook’s distance plot with the help of CRAN R software the outliers were recognized and removed from the data table to predict alternate curve fitting. The data with outliers are shown in results & discussion section in this article. Since the experiments were conducted only in a laboratory based setup at still wind, these data may or may not predict real forest fire behavior.
Two sets of experiments were conducted on a heat resistant clay as a fuel bed using kitchen matchsticks as a fuel. The uniformity of the matchsticks’ fuel size was not verified in this experiment. Matchsticks were arranged in a regular array of 10 cm by 5 cm fuel bed. The gap between sticks was 0.50 cm and 0.82 cm in set I and set II experiments, respectively. Matchsticks were inserted on the clay bed in such a way that they were vertical to the ground irrespective of the slope. Aluminum grids were used to make equidistant holes on the grid. Grid I consisted of holes of 0.24 cm at 0.50 cm apart, and grid II consisted of holes of 0.24 cm at 0.82 cm apart. An inclined plane was used to elevate fuel bed to different slopes to replicate hills. Precautions were taken to maintain the gap between the sticks when they were arranged on the fuel bed. Each set of experiments was performed on a same fuel bed. The fuel bed has a thickness of 0.50 cm. The wind velocity and ambient temperature readings were taken by an HP866B anemometer. However, the experiments were conducted only in absence of wind. Packing ratio, the ratio of volume of fuel to the volume of fuel bed including sticks was maintained at 0.08 and 0.04 in set I and set II experiments, respectively. One experiment was also performed on 0.02 ratio but almost no fire spread was recorded on a level bed. One extra matchstick was inserted on the fuel bed in the middle of the first row to ignite the matchsticks’ fuel. Flame height, headfire, and backfire speed (not presented here) were also measured to understand a behavior of fire spread.
Videos of burning matchsticks on the fuel bed were taken at slow motion mode of 120 frame per second with an iPhone 6s camera. For easy understanding of fire propagation, we have expressed the fire speed in an arbitrary unit (au). Rate of fire spread was recorded using Tracker, a video analysis and modeling tool software in a manual tracking mode. Figures 1(a)-(d) below show the experimental setup of the fuel bed before, during, and after fuel burnings. In Figure 1(d) a vertical line represents the height of the flame. All experiments were conducted in an open atmosphere where air flow was blocked with the help of temporary side walls, and were maintained at ~58˚F ± 3 ambient temperature, ~76% ± 4 relative humidity, and at no wind velocity.
Figure 2(a) shows the data generated by Tracker tool as a distance travelled by the fire progressing with time along the x-axis on a fuel bed, and Figure 2(b) is the best linear fit to determine the rate of fire spread. We used a R software to analyze the data once the data were generated from Tracker. The equation of line of best fit in Figure 2(b) is
y=a+bx
, as determined by R software, where
b=14.71\pm 0.29
a=0.31\pm 1.2
for the 40˚ slope, and
b=7.65\pm 0.08
a=-22.69\pm 0.94
for the −10˚ slope with the coefficient of determination 99% for both slopes. Hence, the rate of fire spread was 14.71 ± 0.29 for slope at 40˚ and 7.65 ± 0.08 for the slope −10˚. Similarly, the rate of fire spread was determined for all the other slopes. The slope of a line of best fit was taken as a rate of fire spread or fire propagation velocity. Slope of lines measured from R was
Figure 1. Experimental Setup. (a)-(d) represent experimental setup for matchsticks inserted on a clay bed, curled up burnt sticks, fuel bed at 40˚ slope, and burning sticks, respectively.
Figure 2. (a) Data generation by Tracker tool as fire spread with time; (b) Line of best fit for distance travelled by fire with time. Lines I and II represent uphill 40˚, and downhill −10˚, respectively.
nearly the same as a slope given by a Tracker tool.
Figure 3(a) and Figure 4(a) are curve fittings on raw data for rate of fire spread with elevation. The equation of curve fittings is given as
y=a+bx+c{x}^{2}+d{x}^{3}
, where the coefficients of polynomial are
a=12.7\pm 1.3,b=0.01,c=-0.002,
d=0.00006
a=9.33\pm 0.7,b=0.1,c=0.001,
d=-0.00001
in Figure 4(a). The nature of curves indicates that the rate of fire spread remains almost constant from slope zero to 15˚ uphill elevation, increases slowly as elevation increases thereafter, and decreases for downhill as steepness increases as shown in Figure 3(a). The curve in Figure 4(a) shows that the rate of fire spread increases almost linearly with the uphill slope and decreases linearly as downhill slope increases. Such results do not agree with the Rothermel’s mathematical model [6] . However, we observed that the fire propagation is slow initially as downslope increases, but becomes faster after burning of a couple of rows. Thus, we decided to analyze our data to locate and remove outliers using Cook’s distance plot as shown in Figure 3(b) and Figure 4(b) and found that there were two high influential points in data set I and one point in data set II. Figure 3(c) and Figure 4(c) represent the curve fittings after removal of these outliers from our data set I and II respectively. These curves agree with Rothermel’s model of
a\left(1+b\cdot {\mathrm{tan}}^{2}\theta \right)
a=10.47\pm 0.3,
b=0.03\pm 0.01
chosen from the curve fitting of Figure 3(c), and
a=8.3\pm 0.3,
b=0.1\pm 0.02
chosen from the curve fitting of Figure 4(c). For comparison purpose, we have shown our experimental data as blue and red curves together with the mathematical model
a\left(1+b\cdot \text{tan}\theta \right)
a\left(1+b\cdot {\mathrm{tan}}^{2}\theta \right)
curves as green and black lines in Figure 3(d) and Figure 4(d). The shaded region around the curves are standard error in Figure 3(d) and Figure 4(d). We did not find any portion of our data that matches with
a\left(1+b\cdot \text{tan}\theta \right)
. Figure 4(d) indicates that the fire spread grows rapidly as the slope increases, but slows down at higher elevation above 35˚ for a low packing ratio. Such slowing down of spread may be due to the increment in height difference between fuel materials as slope increases. Such an effect was not observed for high packing ratio of 0.08, which may be due to
Figure 3. (a) Curve fitting on raw data of set I experiment (packing ratio = 0.08). The equation of curve fitting is given as
y=a+bx+c{x}^{2}+d{x}^{3}
a=12.7,b=0.01,c=-0.002,\text{and}\text{\hspace{0.17em}}d=0.00006.
(b) Cook’s distance plot for set I data to find high influential points that affect the curve shape. (c) Curve fitting after removal of high influential points from data set I. The equation of curve fitting is given as
y=a+bx+c{x}^{2}+d{x}^{3}
a=10.47,b=0.03,c=0.002,\text{and}\text{\hspace{0.17em}}d=-0.000008.
(d) Comparing the nature of set I data with theoretical models. Lines I, II, and III represent experimental data of set I, theoretical models tanθ, and tan2θ, respectively.
Figure 4. (a) Curve fitting on raw data of set II experiment (packing ratio = 0.04). The equation of curve fitting is given as
y=a+bx+c{x}^{2}+d{x}^{3}
a=9.33,b=0.1,c=0.001,\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}d=-0.00001
. (b) Cook’s distance plot for set II data to find high influential points that affect the curve shape. (c) Curve fitting after removal of high influential point from set II data. The equation of curve is given as
y=a+bx+c{x}^{2}+d{x}^{3}
a=8.29,b=0.1,c=0.003,\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}d=-0.00005
. (d) Comparing the nature of set II data with theoretical models. Lines I, II, and III represent experimental data set II, theoretical models tanθ, and tan2θ, respectively.
backfire and headfire that help fire to propagate rapidly in a densely packed fuel bed. The spewing of backfires, headfires, and the whirling of flames may be the causes of outliers in data sets. The factors such as backfire, headfire, non-uniform fuel size, local humidity, and the moisture in the fuel beds may develop a slightly different environment for heat transfer which ultimately deviate this experimental data from a theoretical model.
In this experiment, we studied the behavior of fire propagation on uphill and downhill forest slopes by designing a clay fuel bed imbedded with matchsticks. Two different packing ratios 0.08 and 0.04 of matchsticks fuel beds were included in this experiment. Our data follow the pattern of tan2θ within the experimental error as predicted by Rothermel [6] except at high elevation for low packing ratio. At low packing ratio, our data indicates the rate of fire propagation slows down with the slopes above 35˚. However, at high packing ratio, the rate of fire spread increases with the increase of upslope and is proportional to tan2θ. The minimum rate of fire spread was observed around 15˚ downslope.
I would like to thank Faculty Research Committee at Union College, Barbourville, KY for providing fund for this project. I would also like to thank Dr. Rice Melinda, Dr. Dan Covington, Department of safety, and Physical Plant at Union College for their assistance with this project.
[1] Punckt, C., Bodega, P., Kaira, P. and Rotermund, H. (2015) Wildfires in the Lab: Simple Experiment and Models for the Exploration of Excitable Dynamics. Journal of Chemistry Education, 92, 1330-1337.
http://doi.org/10.1021/ed500714f
[2] Countryman, C. (1972) The Fire Environment Concept. National Wildfire Coordinating Group.
http://caminosfire.com/wordpress/wp-content/uploads/2010/10/
pms433_fire_environment_countryman.pdf
[3] Weise, D. and Biging, G. (1997) A Qualitative Comparison of Fire Spread Models Incorporating Wind and Slope Effects. Forest Science, 43, 170-180.
https://www.fs.fed.us/psw/publications/weise/psw_1997_weise000.pdf
[4] Butler, B., Anderson, W. and Catchpole, E. (2007) Influence of Slope on Fire Spread Rate. USDA Forest Service, Rocky Mountain Research Station, Proceedings RMRS-P-46CD, 75-82.
https://www.fs.fed.us/rm/pubs/rmrs_p046/rmrs_p046_075_082.pdf
[5] Wagner, C. (1988) Effect of Slope on Fires Spreading Downhill. Canadian Journal of Forest Research, 18, 818-820.
http://www.cfs.nrcan.gc.ca/bookstore_pdfs/23539.pdf
[6] Rothermel, R. (1972) A Mathematical Model for Predicting Fire Spread in Wildland Fuels. USDA Forest Service, Research Paper INT-115, Ogden, Utah, USA, 40.
[7] Curry, J. and Fons, W. (1940) Forest-Fire Behavior Studies. Mechanical Engineering, 62, 219-225.
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