why does the following code give error?

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u=[3 4 30 50];% Desired Vector
b=u;
[R,C]=size(b);
P=C/2;
M=2*C;
% calculate observed vector
xo=zeros(1,M);
for k=1:M 
    for i=1:P 
        xo(1,k)=xo(1,k)+1*exp(1i*((-pi/2)*sin(u(P+i))*(k-1)+(pi/(16*u(i)))*cos^2(u(P+i))*(k-1)^2)));
Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.
    end 
end
xe=zeros(1,M);
for k=1:M
     for i=1:P
            xe(1,k)=xe(1,k)+1*exp(1i*((k-1)*(-pi/2)*sin(b(P+i))+(pi/(16*b(i)))*cos^2(b(P+i))*(k-1)^2));
     end
end
abc=0.0;
for m1=1:M
  abc=abc+(abs(xo(1,m1)-xe(1,m1))).^2;  
end
abc=abc/M;
e=abc

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Answer 1

Torsten 2023-12-13

在 MATLAB Online 中打开

0 个投票

Use

xo(1,k)=xo(1,k)+1*exp(1i*((-pi/2)*sin(u(P+i))*(k-1)+(pi/(16*u(i)))*cos(u(P+i))^2*(k-1)^2));
xe(1,k)=xe(1,k)+1*exp(1i*((k-1)*(-pi/2)*sin(b(P+i))+(pi/(16*b(i)))*cos(b(P+i))^2*(k-1)^2));

instead of

xo(1,k)=xo(1,k)+1*exp(1i*((-pi/2)*sin(u(P+i))*(k-1)+(pi/(16*u(i)))*cos^2(u(P+i))*(k-1)^2)));
xe(1,k)=xe(1,k)+1*exp(1i*((k-1)*(-pi/2)*sin(b(P+i))+(pi/(16*b(i)))*cos^2(b(P+i))*(k-1)^2));

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Torsten 2023-12-13

编辑：Torsten 2023-12-13

Because even the order in which terms are multiplied can matter for the result.

Here, the terms that are summed are absolutly identical while in your previous code, the multiplicative order in both terms differs.

Sadiq 2023-12-13

Thanks a lot dear Torsten for your prompt response. Yes now I got it. Thank you once again.

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why does the following code give error?

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