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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
hisham mohammed 2022-12-31
关闭： Voss 2023-1-1
function [CF]= Coherence_Factor(xm,N);
CF = ((abs (sum(xm))).^2) / (N* sum(abs(xm.^2)))
end

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VBBV
VBBV 2022-12-31
编辑：VBBV 2022-12-31
Ran in:
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 个评论
VBBV
VBBV 2022-12-31
Ran in:
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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