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zeros

  • 计算全息close all clc clear A=zeros(64) A(15:20,20:40)=1 A(15:50,20:25)=1 A(45:50,20:40)=1 A(30:34,

    计算全息close all clc clear A=zeros(64) A(15:20,20:40)=1 A(15:50,20:25)=1 A(45:50,20:40)=1 A(30:34,20:35)=1 % ppp=exp(rand(64)*pi*2*i) A=A.*ppp % Author s email: zjliu2001@163.com figure imshow(abs(A),[]) Fa=fft2(fftshift(A)) Fs=fftshift(Fa) Am=abs(Fs) % amplitude Ph=angle(Fs) % phase s=11 % 这表示边长吗? cgh=zeros(64*s) th=max(max(abs(Fs)))

    标签: 20 close clear zeros

    上传时间: 2014-10-13

    上传用户:wweqas

  • 卡尔曼滤波器matlab程序

    load initial_track  s; % y:initial data,s:data with noiseT=0.1; % yp denotes the sample value of position% yv denotes the sample value of velocity% Y=[yp(n);yv(n)];% error deviation caused by the random acceleration % known dataY=zeros(2,200);Y0=[0;1];Y(:,1)=Y0;A=[1 T    0 1];          B=[1/2*(T)^2 T]';H=[1 0]; C0=[0 0    0 1];C=[C0 zeros(2,2*199)];Q=(0.25)^2; R=(0.25)^2;

    标签: matlab 卡尔曼滤波器 程序

    上传时间: 2014-12-28

    上传用户:asaqq

  • The Reed-Somolon code is specified by the finite field, the length (length <= 2^m-1), the numbe

    The Reed-Somolon code is specified by the finite field, the length (length <= 2^m-1), the number of redundant symbols (length-k), and the initial zero of the code, init_zero, such that the zeros are: init_zero, init_zero+1, ..., init_zero+length-k-1

    标签: length the Reed-Somolon specified

    上传时间: 2014-07-31

    上传用户:skfreeman

  • 平均因子分解法

    平均因子分解法,适用于正定矩阵First, let s recall the definition of the Cholesky decomposition: Given a symmetric positive definite square matrix X, the Cholesky decomposition of X is the factorization X=U U, where U is the square root matrix of X, and satisfies: (1) U U = X (2) U is upper triangular (that is, it has all zeros below the diagonal). It seems that the assumption of positive definiteness is necessary. Actually, it is "positive definite" which guarantees the existence of such kind of decomposition.

    标签: 分解

    上传时间: 2013-12-24

    上传用户:啊飒飒大师的

  • 标准的遗传算法代码

    标准的遗传算法代码,下面是程序:function y=fitness(chrom,p,aim) global P_cross P_mutation [Popsize len]=size(chrom) fitness_gene=zeros(Popsize,1) in_he=zeros(4,1) out_he=zeros(4,1) in_out=0 out_out=0

    标签: 标准 代码 算法

    上传时间: 2013-12-08

    上传用户:pkkkkp

  • function [U,center,result,w,obj_fcn]= fenlei(data) [data_n,in_n] = size(data) m= 2 % Exponent fo

    function [U,center,result,w,obj_fcn]= fenlei(data) [data_n,in_n] = size(data) m= 2 % Exponent for U max_iter = 100 % Max. iteration min_impro =1e-5 % Min. improvement c=3 [center, U, obj_fcn] = fcm(data, c) for i=1:max_iter if F(U)>0.98 break else w_new=eye(in_n,in_n) center1=sum(center)/c a=center1(1)./center1 deta=center-center1(ones(c,1),:) w=sqrt(sum(deta.^2)).*a for j=1:in_n w_new(j,j)=w(j) end data1=data*w_new [center, U, obj_fcn] = fcm(data1, c) center=center./w(ones(c,1),:) obj_fcn=obj_fcn/sum(w.^2) end end display(i) result=zeros(1,data_n) U_=max(U) for i=1:data_n for j=1:c if U(j,i)==U_(i) result(i)=j continue end end end

    标签: data function Exponent obj_fcn

    上传时间: 2013-12-18

    上传用户:ynzfm

  • 44b0公版的测试程序

    44b0公版的测试程序, ******************************************************* * NAME : 44BINIT.S * * Version : 10.JAn.2003 * * Description: * * C start up codes * * Configure memory, Initialize ISR ,stacks * * Initialize C-variables * * Fill zeros into zero-initialized C-variables *

    标签: 44b0 测试程序

    上传时间: 2013-12-22

    上传用户:teddysha

  • FFTGUI Demonstration of Finite Fourier Transform. FFTGUI(y) plots real(y), imag(y), real(fft(y)

    FFTGUI Demonstration of Finite Fourier Transform. FFTGUI(y) plots real(y), imag(y), real(fft(y)) and imag(fft(y)). FFTGUI, without any arguments, uses y = zeros(1,32). When any point is moved with the mouse, the other plots respond. Inspired by Java applet by Dave Hale, Stanford Exploration Project, http://sepwww.stanford.edu/oldsep/hale/FftLab.html

    标签: FFTGUI real Demonstration Transform

    上传时间: 2017-06-05

    上传用户:anng

  • 批处理感知器算法

    批处理感知器算法的代码matlab w1=[1,0.1,1.1;1,6.8,7.1;1,-3.5,-4.1;1,2.0,2.7;1,4.1,2.8;1,3.1,5.0;1,-0.8,-1.3;     1,0.9,1.2;1,5.0,6.4;1,3.9,4.0]; w2=[1,7.1,4.2;1,-1.4,-4.3;1,4.5,0.0;1,6.3,1.6;1,4.2,1.9;1,1.4,-3.2;1,2.4,-4.0;     1,2.5,-6.1;1,8.4,3.7;1,4.1,-2.2]; w3=[1,-3.0,-2.9;1,0.5,8.7;1,2.9,2.1;1,-0.1,5.2;1,-4.0,2.2;1,-1.3,3.7;1,-3.4,6.2;     1,-4.1,3.4;1,-5.1,1.6;1,1.9,5.1]; figure; plot(w3(:,2),w3(:,3),'ro'); hold on; plot(w2(:,2),w2(:,3),'b+'); W=[w2;-w3];%增广样本规范化 a=[0,0,0]; k=0;%记录步数 n=1; y=zeros(size(W,2),1);%记录错分的样本 while any(y<=0)     k=k+1;     y=a*transpose(W);%记录错分的样本     a=a+sum(W(find(y<=0),:));%更新a     if k >= 250         break     end end if k<250     disp(['a为:',num2str(a)])      disp(['k为:',num2str(k)]) else      disp(['在250步以内没有收敛,终止']) end %判决面:x2=-a2*x1/a3-a1/a3 xmin=min(min(w1(:,2)),min(w2(:,2))); xmax=max(max(w1(:,2)),max(w2(:,2))); x=xmin-1:xmax+1;%(xmax-xmin): y=-a(2)*x/a(3)-a(1)/a(3); plot(x,y)

    标签: 批处理 算法matlab

    上传时间: 2016-11-07

    上传用户:a1241314660

  • DTFT的计算

    已知系统函数为H(z)=1/[(1-0.2z^-1)(1-0.3z^-1)(1+0.4z^-1)]。试用长除法求h(n)的6点输出。 答案:clc;clear all;b=1;a=poly([0.2,0.3,-0.4]);x=deconv([1,zeros(1,6+4-1-1)],a)

    标签: DTFT 计算

    上传时间: 2017-10-21

    上传用户:zhouhua