why does the following code give error?

1 次查看（过去 30 天）

Sadiq 2023-12-13

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2060109-why-does-the-following-code-give-error

评论： Sadiq 2023-12-13

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

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Torsten 2023-12-13

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2060109-why-does-the-following-code-give-error#answer_1370699

在 MATLAB Online 中打开

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));