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)

function [CF]= Coherence_Factor(xm,N);

CF = ((abs (sum(xm))).^2) / (N* sum(abs(xm.^2)))

end

Answer 1

[VBBV]

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

[hisham mohammed]

not this

[VBBV]

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