Hello, how can I do a (for 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
评论： Walter Roberson 2023-1-1
function [CF]= Coherence_Factor(xm,N);
CF = ((abs (sum(xm))).^2) / (N* sum(abs(xm.^2)))
end
%Function call window
xm= [-1:4],N=length(xm);
xm=[-5:-2],N=length(xm);
xm=[-2:1],N=length(xm)
[CF]= Coherence_Factor(xm,N)

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回答（2 个）

VBBV
VBBV 2022-12-31
Ran in:
xm= {[-1:4], [-5:-2],[-2:1]};
xm = 1×3 cell array
{[-1 0 1 2 3 4]} {[-5 -4 -3 -2]} {[-2 -1 0 1]}
for k = 1:numel(xm)
N=length(xm(k));
CF(k)= Coherence_Factor(xm(k),N);
end
CF
CF = 1×3
2.6129 3.6296 0.6667
function [CF]= Coherence_Factor(xm,N);
CF = ((abs (sum(cell2mat(xm)))).^2) ./ (N* sum(abs(cell2mat(xm).^2)));
end
Voss
Voss 2023-1-1
Ran in:
xm = {-1:4, -5:-2, -2:1};
Nx = numel(xm);
CF = zeros(1,Nx);
for k = 1:Nx
N = numel(xm{k});
CF(k) = Coherence_Factor(xm{k},N);
end
CF
CF = 1×3
0.4355 0.9074 0.1667
function CF = Coherence_Factor(xm,N)
CF = abs(sum(xm)).^2 / (N*sum(abs(xm.^2)));
end
Note that if the xm are always guaranteed to be real-valued (as opposed to complex-valued), then the abs() calls are unnecessary:
xm = {-1:4, -5:-2, -2:1};
Nx = numel(xm);
CF = zeros(1,Nx);
for k = 1:Nx
N = numel(xm{k});
CF(k) = Coherence_Factor(xm{k},N);
end
CF
CF = 1×3
0.4355 0.9074 0.1667
function CF = Coherence_Factor(xm,N)
CF = sum(xm).^2 / (N*sum(xm.^2));
end
And if N is always supposed to be the number of elements in xm, you can define N inside the function Coherence_Factor and avoid having the second input argument:
xm = {-1:4, -5:-2, -2:1};
Nx = numel(xm);
CF = zeros(1,Nx);
for k = 1:Nx
CF(k) = Coherence_Factor(xm{k});
end
CF
CF = 1×3
0.4355 0.9074 0.1667
function CF = Coherence_Factor(xm)
N = numel(xm);
CF = sum(xm)^2 / (N*sum(xm.^2));
end
  1 个评论
Walter Roberson
Walter Roberson 2023-1-1
I wonder if it is supposed to be sum(abs(xm).^2) or abs(sum(xm.^2)) ?
The current sum(xm) having priority has different results if values could be a mix of negative and positive

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