フィルターのクリア

How can I multiply 2D matrix with it hermitian?

2 ビュー (過去 30 日間)
zayed
zayed 2011 年 12 月 15 日
Hi,
How can I do this in matlab?I want to multiply a matrix with it's hermitian,as the following equation:
M=(1/K)*sum(zkd*zkd'),
zkd is (1000*5).zkd is from a signal(size 5000) and i divide it to blocks with block size equal to 1000,and number of blocks 5.
clear all;
clc
load lab2_data;
x = radar_noise; % input signal
K = 1000; % K CAPITAL is Block Size
L = length(x) - mod(length(x),K); % only full blocks
zkd = reshape(x(1:L), K, []);
Md=zeros(L,L); % M ZEROS covariance matrix L*L
for i=1:size(zkd,1); % LOOP covariance matrix calculation
Mz=zkd(i,:)*zkd(i,:)';
Md=Md+Mz;
end
Md=Md/K;
  2 件のコメント
Jan
Jan 2011 年 12 月 15 日
And what is wrong with the posted command?
zayed
zayed 2011 年 12 月 15 日
The problem is when I check the results, M is(1000*1000)with all elements in have the same value.

サインインしてコメントする。

採用された回答

Walter Roberson
Walter Roberson 2011 年 12 月 15 日
>> zkd = rand(30,3) + i*rand(30,3);
>> R = sum(zkd * zkd');
>> size(R)
ans =
1 30
>> R
R =
Columns 1 through 2
43.7252707339352 - 14.3845494385565i 37.9012903137849 - 1.21833325457709i
Columns 3 through 4
44.0278780546839 - 5.91428978829146i 40.0850287858898 - 10.9783386248396i
The exact values are not important, but it is clear that all of the values are not the same.
If your zkd is 1000 by 5 then zkd*zkd' is going to be 1000 by 1000, and sum(zkd*zkd') is going to be 1 by 1000, in contradiction to your indication that M is 1000 by 1000.
  5 件のコメント
Walter Roberson
Walter Roberson 2011 年 12 月 15 日
Md=zeros(L,L);
zayed
zayed 2011 年 12 月 15 日
But L=5000.and this give the following error :
??? Error using ==> unknown
Matrix dimensions must agree.
Error in ==> cov3 at 11
Md=Md+Mz;

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeLogical についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by