Welcome I have a CF equation that performs a three-dimensional summation process and gives a two-dimensional result I did this process using (For Loops) in MATLAB. I want to c

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hisham mohammed
hisham mohammed 2023-3-30
编辑： Dyuman Joshi 2023-3-30
clc
xm(:,:,1)= [1 2 3 4;2 3 4 5;3 4 5 6;4 5 6 7 ]
xm(:,:,2)= [2 3 4 5;3 4 5 6;4 5 6 7;5 6 7 8]
tic
[cf] =coherence_factor(xm)
toc
function [cf] = coherence_factor(xm)
for i=1:size(xm,1);
for j=1:size(xm,2);
n=size(xm,3);
cf(i,j)=(abs (sum(xm(i,j,:))).^2) / (n* sum(abs(xm(i,j,:).^2)));
end
end
end

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采纳的回答

Dyuman Joshi
Dyuman Joshi 2023-3-30
编辑：Dyuman Joshi 2023-3-30
Ran in:
%Data
xm(:,:,1)= [1 2 3 4;2 3 4 5;3 4 5 6;4 5 6 7 ];
xm(:,:,2)= [2 3 4 5;3 4 5 6;4 5 6 7;5 6 7 8];
n=size(xm,3);
%Loop Method
for i=1:size(xm,1);
for j=1:size(xm,2);
cf1(i,j)=(abs (sum(xm(i,j,:))).^2) / (n* sum(abs(xm(i,j,:).^2)));
end
end
cf1
cf1 = 4×4
0.9000 0.9615 0.9800 0.9878 0.9615 0.9800 0.9878 0.9918 0.9800 0.9878 0.9918 0.9941 0.9878 0.9918 0.9941 0.9956
%Vectorized method
cf2 = abs(sum(xm,3).^2)./(n*sum(abs(xm.^2),3))
cf2 = 4×4
0.9000 0.9615 0.9800 0.9878 0.9615 0.9800 0.9878 0.9918 0.9800 0.9878 0.9918 0.9941 0.9878 0.9918 0.9941 0.9956

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