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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: 基于求函数f(x,y,z)=xyz*sin(xyz)最大值问题的演示程序-Based on the demand function f (x, y, z) = xyz* sin (xyz) max demo issues
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Size: 7441408 |
Author: lty |
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Description: 三： 针对带表头结点的单链表，试编写下列函数。
(1) 定位函数Locate：在单链表中寻找第i个结点。若找到，则函数返回第i个结点的地址；若找不到，则函数返回NULL。
(2) 求最大值函数max：通过一趟遍历在单链表中确定值最大的结点。
(3) 统计函数number：统计单链表中具有给定值x的所有元素。
(4) 建立函数create：根据一维数组a[n]建立一个单链表，使单链表中各元素的次序与a[n]中各元素的次序相同，要求该程序的时间复杂性为O(n)。
(5) 整理函数tidyup：在非递减有序的单链表中删除值相同的多余结点。-III: Table for the first node with a single list, try to prepare the following function. (1) positioning function Locate: find a single list in the first i nodes. If found, the function returns the first i nodes of the address if not, the function returns NULL. (2) for the maximum function max: traverse through the trip in a single list to determine the value of the largest node. (3) statistical function number: single-chain statistics with a given value of x of all elements. (4) the establishment of function create: the basis of one-dimensional array a [n] the creation of a single list, so that a single list in the order of each element with a [n] in the order of the elements the same time require the complexity of the procedure for O (n). (5) collate function tidyup: in non-descending order to remove a single list of surplus value of the same node.
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Size: 16384 |
Author: 卡米力江 |
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Description: 台式机上网时可动态传输视频，包括动态保存，用户控制更新频率-a desktop internet camera application for Linux, Max OS X, and Windows
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Size: 1219584 |
Author: 邵隆 |
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Description: Description
Scientific calculator. Allows to perform caclulation with high precicion and implements most populatr mathematical functions: sin, cos, tan, asin, acon, atan, exp, log, sqr, floor and ceil. Also it make it possible to define your own function, store results in variables and use variable sin expressions. Calculator store al formuls you have entered. Plot function can be used to draw graph of function with single argument.
More detailed description of calculator is here.
-
Notes
* Write formula using normal infix notation, for example (x^2+ y^2)/1.5. To execute statement use Evaluate command.
* Use variable to write generic formulas, by changing values assigned to the variables you wil be able to repeat calculation for all input data with minimal typing. Define your own functions which also can speedup your work and reduce probability of making an error.
* It is possible to write small functional programs using conditional ?: operator. For example sum of arithmetic progression can be calculated by the following function
sum(x)=x>0?x+sum(x-1):0
Calculator has only < and > comparison operators (because comparing floating point number for equality is any case bad idea). But it is also to check that value belongs to specified range using min<x>max operator.
* All statement you have entered and stored in database. You can select proper statement using List command. Statements are marked with three differen
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Size: 245760 |
Author: tang |
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Description: CRC Code:
Input: polynomial p(x) space bar and caption (max 10000 chars).
Output: CRC Code
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Size: 1024 |
Author: kaguarek |
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Description: 在图像利用M*N的窗 运行2维最大滤波器
在图像利用M*N的窗 运行2维最小滤波器
-performs two-dimensional maximum filtering on the image X using an M-by-N window. performs the 1D max/min filtering of the row vector X using a N-length filter. performs two-dimensional minimum filtering on the image X using an M-by-N window
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Size: 4096 |
Author: Alex |
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Description: Function direct I Filter
Call:
[Y,Zf]=FDI(B,A,X,Zi)
Parameters A(N), B(M), X(L)and Y(L) as in FILTER.
Input Parameters:
Zi(max(N,M)-1,2): Conditions iniciales.
Zi(:,1): from zeros.
Zi(:,2): from polos.
Output Parameters:
Zf(max(N,M)-1,2): final conditions.
Zf(:,1): from zeros.
Zf(:,2): from polos.-Function direct I Filter
Call:
[Y,Zf]=FDI(B,A,X,Zi)
Parameters A(N), B(M), X(L)and Y(L) as in FILTER.
Input Parameters:
Zi(max(N,M)-1,2): Conditions iniciales.
Zi(:,1): from zeros.
Zi(:,2): from polos.
Output Parameters:
Zf(max(N,M)-1,2): final conditions.
Zf(:,1): from zeros.
Zf(:,2): from polos.
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Size: 1024 |
Author: carlos12 |
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Description: man sin pso-pso max sin-man sin pso-pso ma x sin-
-man sin pso-pso max sin-man sin pso-pso max sin-man sin pso-pso max sin-
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Size: 1024 |
Author: ebrahim |
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Description: matlab用于求解规划问题的工具包
推荐把这个工具整合到matlab中去，这个工具是私人的，不过大家都可以免费下载使用。下载后，只要在matlab中添加路径就可以使用这工具箱。
正在吸引我的是，这个工具箱建立了一种新的数据类型，使所有规划问题都整合在一起。
举例如下：
已知非线性整数规划为：
Max z=x1^2+x2^2+3*x3^2+4*x4^2+2*x5^2-8*x1-2*x2-3*x3-x4-2*x5
s.t.
0<=xi<=99(i=1,2,...,5)
x1+x2+x3+x4+x5<=400
x1+2*x2+2*x3+x4+6*x5<=800
2*x1+x2+6*x3<=800
x3+x4+5*x5<=200
在matlab中输入 x=intvar（1，5）；
f=[1 1 3 4 2]*(x .^2)-[8 2 3 1 2]*x F=set(0<=x<=99)
F=F+set([1 1 1 1 1]*x <=400)+set([1 2 2 1 6]*x <=800)+set(2*x(1)+x(2)+6*x(3)<=800)
F=F+set(x(3)+x(4)+5*x(5)<=200) solvesdp(F,-f)
double(f) 80199
double(x) 53 99 99 99 0
intvar（m，n）:生成整数型变量；
sdpvar（m，n）：生产变量；
solvesdp（F，f）：求解最优解（最小值），其中F为约束条件（用set连接），f为目标函数
double：显示求解的答案
intvar，sdpvar，生成的变量可以像矩阵一样使用，如例题显示。 -matlab用于求解规划问题的工具包
推荐把这个工具整合到matlab中去，这个工具是私人的，不过大家都可以免费下载使用。下载后，只要在matlab中添加路径就可以使用这工具箱。
正在吸引我的是，这个工具箱建立了一种新的数据类型，使所有规划问题都整合在一起。
举例如下：
已知非线性整数规划为：
Max z=x1^2+x2^2+3*x3^2+4*x4^2+2*x5^2-8*x1-2*x2-3*x3-x4-2*x5
s.t.
0<=xi<=99(i=1,2,...,5)
x1+x2+x3+x4+x5<=400
x1+2*x2+2*x3+x4+6*x5<=800
2*x1+x2+6*x3<=800
x3+x4+5*x5<=200
在matlab中输入 x=intvar（1，5）；
f=[1 1 3 4 2]*(x .^2)-[8 2 3 1 2]*x F=set(0<=x<=99)
F=F+set([1 1 1 1 1]*x <=400)+set([1 2 2 1 6]*x <=800)+set(2*x(1)+x(2)+6*x(3)<=800)
F=F+set(x(3)+x(4)+5*x(5)<=200) solvesdp(F,-f)
double(f) 80199
double(x) 53 99 99 99 0
intvar（m，n）:生成整数型变量；
sdpvar（m，n）：生产变量；
solvesdp（F，f）：求解最优解（最小值），其中F为约束条件（用set连接），f为目标函数
double：显示求解的答案
intvar，sdpvar，生成的变量可以像矩阵一样使用，如例题显示。
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Size: 807936 |
Author: wrui |
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Description: clear
num=[0,0,10]
den=[1,2,10]
p=roots(den)
[u,w]=solve( w^2=10 , 2*w*u=2 , u,w )
[y,x,t]=step(num,den)
plot(t,y)
[yss,n]=max(y)
finalvalue=dcgain(num,den)
percentovershoot=100*(yss-finalvalue)/finalvalue
timetopeak=t(n)
k=length(t)
while (y(k)>0.98*finalvalue)&(y(k)<1.02*finalvalue)
k=k-1
end
settlingtime=t(k)
p
w=vpa(w,4)
u=vpa(u,4)
yss
timetopeak
finalvalue
settlingtime
-clear num = [0,0,10] den = [1,2,10] p = roots (den) [u, w] = solve ( ' w ^ 2 = 10' , ' 2* w* u = 2 ' ,' u, w ' ) [y, x, t] = step (num, den) plot (t, y) [yss, n] = max (y) finalvalue = dcgain (num , den) percentovershoot = 100* (yss-finalvalue)/finalvalue timetopeak = t (n) k = length (t) while (y (k)> 0.98* finalvalue) & (y (k) < 1.02* finalvalue) k = k-1 end settlingtime = t (k) p w = vpa (w, 4) u = vpa (u, 4) yss timetopeak finalvalue settlingtime
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Size: 2919424 |
Author: 董森 |
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Description: 针对3ds max8 的插件，用于.x文件的导出，可以设计游戏的3d模型，放在3ds max安装目录的/plugins下。-Plug-ins for 3ds max8 for. X file export, you can design the game 3d model, on the 3ds max installation directory/plugins next.
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Size: 123904 |
Author: 林宇斌 |
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Description: c代码，用遗传算法求函数最大值f(x)=x^2 x 从0到30，希望有用-c code, using genetic algorithms function max f (x) = x ^ 2 x from 0 to 30, hope helpful
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Size: 22528 |
Author: lijuan3623 |
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Description: C代码：遗传算法求函数最大值f(x)=x^2 x 从0到30，希望有用-C code: genetic algorithms function max f (x) = x ^ 2 x from 0 to 30 seek to help
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Size: 12288 |
Author: lijuan3623 |
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Description: 可以把3dmax模型文件导入到虚幻引擎3-3dmax file to unreal engine3
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Size: 2046976 |
Author: saiatsai |
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Description: 由3D Max建模，导出的.X文件的读取，渲染-By the 3D Max modeling, exported. X file read, rendering
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Size: 907264 |
Author: 风轻云淡 |
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Description: 用于MIMO检测的sphere decoder程序- This function implements a soft output sphere decoder.
Based on the paper: C. Studer, M. Wenk, A. Burg, and H. Blocskei: "Soft-Output Sphere
Decoding: Performance and Implementation Aspects", Asilomar 2006
R ... is an upper triangular matrix obtained from the QR decomposition of the MIMO channel H
s ... received symbol (column) vector, s=Q^H*y
dist_ZF ... Distance for the zero forcing solution
symbols_ZF ... ZF solution
M ... number of bits encoded in every layer [1 x nTA]
symbol_alphabet ... for the demapping [nTA x 2^max(M)], filled with zeros for smaller symbol alphabets
bittable ... matrix containing the bits according to the symbol_alphabet [sum(M) x 2^max(M)]
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Size: 3072 |
Author: bennytang |
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Description: Avetis Ioannisyan
avetis@60ateight.com
Last Updated: 11/30/05
LMS Channel Adaptation
reset randomizers
randn( state ,sum(100*clock))
rand( state ,sum(100*clock))
numPoints = 5000
numTaps = 10 channel order
Mu = 0.001:0.001:0.01 iteration step size
input is guassian
x = randn(numPoints,1) + j*randn(numPoints,1)
choose channel to be random uniform
h = rand(numTaps, 1) + i*rand(numTaps, 1)
h = [1 0 0 0 1] testing only
h = h/max(h) normalize channel
convolve channel with the input
d = filter(h, 1, x)
initialize variables
w = []
y = []
in = []
e = [] error, f- Avetis Ioannisyan
avetis@60ateight.com
Last Updated: 11/30/05
LMS Channel Adaptation
reset randomizers
randn( state ,sum(100*clock))
rand( state ,sum(100*clock))
numPoints = 5000
numTaps = 10 channel order
Mu = 0.001:0.001:0.01 iteration step size
input is guassian
x = randn(numPoints,1) + j*randn(numPoints,1)
choose channel to be random uniform
h = rand(numTaps, 1) + i*rand(numTaps, 1)
h = [1 0 0 0 1] testing only
h = h/max(h) normalize channel
convolve channel with the input
d = filter(h, 1, x)
initialize variables
w = []
y = []
in = []
e = [] error, f
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Size: 1024 |
Author: josh |
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Description: Generate a 3D point plot of L=(x,y,z) using the values in the vector c to
determine the color of each point. If c is empty, then z (column 3 of L)
is used to color the plot. The data points are sorted so that plot3 is
only called once for each group of points that map to the same color. The upper and lower limits of the color range (and the z axis) can be defined with crange. This is useful for creating a series of plots with the same
coloring. The colormap (but not the colorbar) is flipped upside down if
crange is given as [max min] instead of [min max]. The figure handle is
returned if an output argument is given.-Generate a 3D point plot of L=(x,y,z) using the values in the vector c to
determine the color of each point. If c is empty, then z (column 3 of L)
is used to color the plot. The data points are sorted so that plot3 is
only called once for each group of points that map to the same color. The upper and lower limits of the color range (and the z axis) can be defined with crange. This is useful for creating a series of plots with the same
coloring. The colormap (but not the colorbar) is flipped upside down if
crange is given as [max min] instead of [min max]. The figure handle is
returned if an output argument is given.
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Size: 4096 |
Author: Jeff |
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Description: 可以直接汇出X档使用,利用3D MAX来制作好的模组可以直接汇出X档使用-X files can be exported directly used to make good use of 3D MAX modules can be directly used to export X Files
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Size: 38912 |
Author: kic530 |
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