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# Hello, how can I do a (loop) for this equation to find more than one value of the CF ? Specifically, I mean more than one value for (xm)

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hisham mohammed on 31 Dec 2022
Closed: Voss on 1 Jan 2023
function [CF]= Coherence_Factor(xm,N);
CF = ((abs (sum(xm))).^2) / (N* sum(abs(xm.^2)))
end

VBBV on 31 Dec 2022
Edited: VBBV on 31 Dec 2022
xm = rand(5,10);
N = 10;
[CF]= Coherence_Factor(xm,N)
CF = 1×10
0.4110 0.3906 0.3169 0.3752 0.2497 0.3349 0.3822 0.4066 0.4420 0.3975
function [CF]= Coherence_Factor(xm,N);
CF = ((abs (sum(xm))).^2) ./ (N* sum(abs(xm.^2)));
end
##### 2 CommentsShowHide 1 older comment
VBBV on 31 Dec 2022
for loop is not needed to find more than one value of CF or xm. However, you can still get such result with for loop also as
xm = randi([0 10],1,10);
N = 10;
for k = 1:length(xm)
CF(k)= Coherence_Factor(xm(k),N);
end
plot(CF) function [CF]= Coherence_Factor(xm,N);
CF = ((abs (sum(xm))).^2) / (N* sum(abs(xm.^2)));
end

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