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Description: The cart with an inverted pendulum, shown below, is \"bumped\" with an impulse
force, F. Determine the dynamic equations of motion for the system, and lin
earize about the pendulum s angle, theta = Pi (in other words, assume that p
endulum does not move more than a few degrees away from the vertical, chosen
to be at an angle of Pi). Find a controller to satisfy all of the design re
quirements given below.
-The cart with an inverted pendulum, shown below, is \"bumped\" with an impulse force, F. Determine the dynamic equations of motion for the system, and lin earize about the pendulum s angle, theta = Pi (in other words, assume that p endulum does not move more than a few degrees away from the vertical, chosen to be at an angle of Pi). Find a controller to satisfy all of the design re quirements given below.
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Size: 12186 |
Author: zhangfj_99 |
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Description: 最佳矩阵连乘 给定n个矩阵｛A1,A2,…An｝，其中Ai与A i+1是可乘的，i=1，2…，n-1。考察这n个矩阵的连乘积A1A2…An。矩阵A和B可乘的条件是矩阵A的列数等于矩阵B的行数。若A是一个p×q矩阵，B是一个q×r矩阵，则其乘积C=AB是一个p×r矩阵,需要pqr次数乘。
由于矩阵乘法满足结合律，故计算矩阵的连乘积可以有许多不同的计算次序。例如，设3个矩阵{A1,A2,A3}的维数分别为10×100，100×5，和5×50。若按加括号方式((A1A2)A3)计算，3个矩阵连乘积需要的数乘次数为10×100×5+10×5×50=7500。若按加括号方式(A1(A2A3))计算，3个矩阵连乘积总共需要10×5×50+10×100×50=75000次数乘。由此可见，在计算矩阵连乘积时，加括号方式，即计算次序对计算量有很大影响。
矩阵连乘积的最优计算次序问题，即对于给定的相继n个矩阵｛A1,A2,…An｝(其中矩阵Ai的维数为pi-1×p，i=1,2，…，n)，确定计算矩阵连乘积A1,A2,…An的计算次序，使得依此次序计算矩阵连乘积需要的数乘次数最少。
-best matrix continually multiply given n matrix (A1, A2, ... An), Ai and A is a mere i, i = 1, 2 ..., n-1. N explore the link matrix product ... An A1A2. Matrices A and B can either condition is out of the matrix A few matrix B is the number of rows. If A is a p q matrix B is a q-r-matrix, its product C = AB is a p r matrix, the number required by pqr. Because matrix multiplication meet the law of combination, it's even calculated matrix product can be calculated in many different priorities. For example, the matrix-based 3 (A1, A2, A3) dimension of 10 100, 100 5 5 and 50. If bracketed by the way ((A1A2) A3), even three product matrix multiplication in the number of 10 100 10 5 50 = 7,500. If bracketed by the way (A1 (A2A3)), three matrix product even need a total of 10 5 50
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Size: 6148 |
Author: 肿事右 |
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Description: C语言内箝汇编 圆周率，功能很强大，大家有兴趣的-C language within clamp compilation pi, functions very powerful, we are interested to s
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Size: 4621 |
Author: 张伟 |
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Description: Bi-dimensional Gabor filter with DC component compensation
This version of the 2D Gabor filter is basically a bi-dimensional Gaussian function centered at origin (0,0) with variance S modulated by a complex sinusoid with polar frequency (F,W) and phase P described by the following equation:
G(x,y,S,F,W,P)=k*Gaussian(x,y,S)*(Sinusoid(x,y,F,W,P)-DC(F,S,P)),
where:
Gaussian(x,y,S)=exp(-pi*S^2*(x^2+y^2))
Sinusoid(x,y,F,W,P)=exp(j*(2*pi*F*(x*cos(W)+y*sin(W))+P)))
DC(F,S,P)=exp(-pi*(F/S)^2+j*P)
File Id: 13776 Average rating: 0.0
Size: 1 KB # of reviews: 0
Submitted: 2007-01-26 Downloads: 274
Subscribers: 0
Keywords: gabor filter
Stiven Schwanz Dias
-Bi-dimensional Gabor filter with DC compo .. compensation This version of the 2D Gabor f ilter is basically a bi-dimensional Gaussian f unction centered at origin (0, 0) with variance S modulated by a complex sinuso id with polar frequency (F, W) and phase P described by the following equati on : G (x, y, S, F, W, P) = k * Gaussian (x, y, S) * (Sinusoid (x, y, F, W, P) - DC (F, S, P)), where : Gaussian (x, y, S) = exp (-pi * S * 2 ^ (x ^ 2 y ^ 2)) Sinusoid (x, y, F, W, P) = exp (j * (2 * pi * F * (x * cos (W) y * sin (W)) P))) D C (F, S, P) = exp (-pi * (F / S) ^ 2 * P j) File Id : 13776 Average rating : 0.0 Size : # 1 KB of reviews : 0 Submitted : 2007-01-26 Downloads : 274 Subscribers : 0 Keywords : gabor filter Stiven Schwanz Dias
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Size: 1298 |
Author: 石峰 |
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Description: The cart with an inverted pendulum, shown below, is "bumped" with an impulse
force, F. Determine the dynamic equations of motion for the system, and lin
earize about the pendulum s angle, theta = Pi (in other words, assume that p
endulum does not move more than a few degrees away from the vertical, chosen
to be at an angle of Pi). Find a controller to satisfy all of the design re
quirements given below.
-The cart with an inverted pendulum, shown below, is "bumped" with an impulse force, F. Determine the dynamic equations of motion for the system, and lin earize about the pendulum s angle, theta = Pi (in other words, assume that p endulum does not move more than a few degrees away from the vertical, chosen to be at an angle of Pi). Find a controller to satisfy all of the design re quirements given below.
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Size: 12288 |
Author: |
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Description: 这是我写的一个51内核对APR6008语音储存播放ic的底层驱动程式!spi通讯方式.在89s52上面调试通过!-This is the one I wrote to the 51 core players APR6008 voice storage drive the bottom ic formula! S pi forms of communication. in 89s52 above debugging through!
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Size: 41984 |
Author: king |
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Description: Bi-dimensional Gabor filter with DC component compensation
This version of the 2D Gabor filter is basically a bi-dimensional Gaussian function centered at origin (0,0) with variance S modulated by a complex sinusoid with polar frequency (F,W) and phase P described by the following equation:
G(x,y,S,F,W,P)=k*Gaussian(x,y,S)*(Sinusoid(x,y,F,W,P)-DC(F,S,P)),
where:
Gaussian(x,y,S)=exp(-pi*S^2*(x^2+y^2))
Sinusoid(x,y,F,W,P)=exp(j*(2*pi*F*(x*cos(W)+y*sin(W))+P)))
DC(F,S,P)=exp(-pi*(F/S)^2+j*P)
File Id: 13776 Average rating: 0.0
Size: 1 KB # of reviews: 0
Submitted: 2007-01-26 Downloads: 274
Subscribers: 0
Keywords: gabor filter
Stiven Schwanz Dias
-Bi-dimensional Gabor filter with DC compo .. compensation This version of the 2D Gabor f ilter is basically a bi-dimensional Gaussian f unction centered at origin (0, 0) with variance S modulated by a complex sinuso id with polar frequency (F, W) and phase P described by the following equati on : G (x, y, S, F, W, P) = k* Gaussian (x, y, S)* (Sinusoid (x, y, F, W, P)- DC (F, S, P)), where : Gaussian (x, y, S) = exp (-pi* S* 2 ^ (x ^ 2 y ^ 2)) Sinusoid (x, y, F, W, P) = exp (j* (2* pi* F* (x* cos (W) y* sin (W)) P))) D C (F, S, P) = exp (-pi* (F/S) ^ 2* P j) File Id : 13776 Average rating : 0.0 Size :# 1 KB of reviews : 0 Submitted : 2007-01-26 Downloads : 274 Subscribers : 0 Keywords : gabor filter Stiven Schwanz Dias
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Size: 1024 |
Author: 石峰 |
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Description: 中国人民保险公司Informix培训教材。值得收藏。-China People s Insurance Company of Informix training materials. Worthy of collection.
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Size: 72704 |
Author: unix3646 |
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Description: ％radon transform
clear all
％
N=800
n=1:N
fs=200
t=n/fs
x1=exp(j*2*pi*(5*t+0.5*5*t.^2))
x2=exp(j*2*pi*(5*t+0.5*15*t.^2))
x=x1+x2
％N=length(x)
％ ambifunb(x )
％*****************************************RAT
naf=ambifunb(x)
htl(abs(naf))
％ [wh,rho,theta]=htl(abs(naf))
colormap([0,0,0])
％ xlabel( 极半径 )
％ ylabel( 角度 )
％**************************************％找出峰值点的坐标,计算初始频率和调频斜率(正确)
％找出峰值点的坐标
b=max(max(wh))
[u,a]=find(wh>=0.8*b)
- Radon transformclear all N = 800 n = 1: N fs = 200 t = n/fs x1 = exp (j* 2* pi* (5* t+ 0.5* 5* t. ^ 2)) x2 = exp ( j* 2* pi* (5* t+ 0.5* 15* t. ^ 2)) x = x1+ x2 N = length (x) ambifunb (x)***************************************** RATnaf = ambifunb (x) htl (abs (naf)) [wh, rho, theta ] = htl (abs (naf)) colormap ([0,0,0]) xlabel (polar radius) ylabel (angle)************************************** to find the coordinates of the peak point, calculating the initial slope of the frequency and FM (right) find the peak point of the coordinates b = max (max ( wh)) [u, a] = find (wh
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Size: 1024 |
Author: abcde |
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Description: PIC16F639的SPI操作源代码
PIC16F639的SPI操作源代码PI C16F639的SPI操作源代码-PIC16F639 source code of the SPI to operate the SPI operation PIC16F639 source PI C16F639 the SPI to operate the source code
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Size: 1024 |
Author: 苏海涛 |
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Description: 使用C#调用PI-SDK进行基于PI的开发 - 使用PI-SDK建立与PI数据库的连接-The use of C# Called PI-SDK-based PI development- the use of PI-SDK and PI to establish the database connection
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Size: 268288 |
Author: hwj |
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Description: 利用布丰投针的思想，用数值随机算法估计计算PI的值。-Buffon' s needle to use for thought, random algorithm is estimated by numerical calculation of the value of PI.
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Size: 8192 |
Author: 吴凡 |
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Description: 用Visual C++开发的图形化圆周率Pi计算程序，共有“反正切
级数展开”、“反正弦级数展开”、“Marchin公式”、“Shank
s公式”和“Gauss公式”五种方法可供选择。-Using Visual C++ development of graphical Pi pi calculation program, a total of " arctangent series expansion," " arcsine Series" , " Marchin formula" , " Shank s formula" and " Gauss formula," the five options .
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Size: 67584 |
Author: 张晓彬 |
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Description: 国内最大的PI实时数据库代理商中电飞华的技术白皮书，很好的PI实时数据库学习资料。-PI the nation' s largest real-time database agents FibrLINK technical white papers, a very good learning materials PI real-time database.
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Size: 3215360 |
Author: James |
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Description: PI的api相关sdk包、lib、includes等还有就是各个的help文件-PI' s api related sdk package, lib, includes all of the help and so there is a document
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Size: 6735872 |
Author: lcy |
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Description: optimal PI controller design and simulation of static var compensator using MATLAB s simulink
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Size: 431104 |
Author: anif |
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Description: fit_ML_normal - Maximum Likelihood fit of the log-normal distribution of i.i.d. samples!.
Given the samples of a log-normal distribution, the PDF parameter is found
fits data to the probability of the form:
p(x) = sqrt(1/(2*pi))/(s*x)*exp(- (log(x-m)^2)/(2*s^2))
with parameters: m,s
format: result = fit_ML_log_normal( x,hAx )
input: x - vector, samples with log-normal distribution to be parameterized
hAx - handle of an axis, on which the fitted distribution is plotted
if h is given empty, a figure is created.
output: result - structure with the fields
m,s - fitted parameters
CRB_m,CRB_s - Cram?r-Rao Bound for the estimator value
RMS - RMS error of the estimation
type - ML - fit_ML_normal - Maximum Likelihood fit of the log-normal distribution of i.i.d. samples!.
Given the samples of a log-normal distribution, the PDF parameter is found
fits data to the probability of the form:
p(x) = sqrt(1/(2*pi))/(s*x)*exp(- (log(x-m)^2)/(2*s^2))
with parameters: m,s
format: result = fit_ML_log_normal( x,hAx )
input: x - vector, samples with log-normal distribution to be parameterized
hAx - handle of an axis, on which the fitted distribution is plotted
if h is given empty, a figure is created.
output: result - structure with the fields
m,s - fitted parameters
CRB_m,CRB_s - Cram?r-Rao Bound for the estimator value
RMS - RMS error of the estimation
type - ML
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Size: 1024 |
Author: resident e |
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Description: 多核技术求PI,在多核技术支持的系统下求出PI，体现单核与多核的计算差异。此程序仅供参考，头文件为未给出。-This is a small program about how to count PI in multi core system. It s just
a test program within some problem.
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Size: 10377216 |
Author: 杰 |
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Description: The number π is a mathematical constant that is the ratio of a circle s circumference to its diameter and is approximately equal to 3.14159. It has been represented by the Greek letter "π" since the mid-18th century, though it is also sometimes romanized as "pi
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Size: 1024 |
Author: ramin |
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Description: 计算三种方法计算圆周率的效率、循环计算次数，测试计算机性能。-The program calculates the three methods of calculating the efficiency ratio of the circumference, calculation of the number of cycles to test your computer s performance.
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Size: 185344 |
Author: ZKL |
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