forming a function handles in matrix

10 次查看（过去 30 天）

显示更早的评论

Bathala Teja 2021-10-4

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1466044-forming-a-function-handles-in-matrix

评论： Bathala Teja 2021-10-6

在 MATLAB Online 中打开

I want to form a set of function handles in a row matrix.

i wrote script like below.

w = 2;

Nr = 20

nr = @(phi)zeros(1, Nr);

wr = @(phi)zeros(1, Nr);

for n = 1:Nr

for i = 1:w

Awr = 50*(cos(i*2)-cos(i*9));

nr(1, n) = @(phi)nr(1, n)+Awr*cos(i*1.7);

end

wr(1, n) = @(phi)nr(1, n)

end

Iam getting the below error

Nonscalar arrays of function handles are not allowed; use cell arrays instead.

how to rectify this??

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。

回答（1 个）

Steven Lord 2021-10-4

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1466044-forming-a-function-handles-in-matrix#answer_800809

MATLAB used to allow nonscalar arrays of function handles, but that functionality was removed probably 10 to 15 years ago. As the error message suggested, store your function handles in a cell array instead.

7 个评论
显示 5更早的评论隐藏 5更早的评论

Stephen23 2021-10-6

编辑：Stephen23 2021-10-6

在 MATLAB Online 中打开

@Bathala Teja: function handles cannot be multiplied (or have any numeric operation applied to them). But you can certainly evaluate the function handles (just as Steven Lord showed) and multiply their outputs:

f{1}(3)*f{2}(3)
%   ^^^     ^^^ Evaluate!

Bathala Teja 2021-10-6

在 MATLAB Online 中打开

See this

Here i performed multiflication of two function handles and after that i integrated total function.

What is the difference in between this one and above one?

m = 10;
nAi = @(phi)zeros(1, 1);
for i=1:2:m
    Ams = (2/(pi*i))*sin(pi*i/3)*(1+(2*cos(pi*i/9))); 
    nAi = @(phi)nAi(phi)+Ams*cos(2*i*phi);
end
nA = @(phi)50+nAi(phi) 
nA = function_handle with value:
    @(phi)50+nAi(phi)
wA = @(phi)nAi(phi)
wA = function_handle with value:
    @(phi)nAi(phi)
nBi = @(phi)zeros(1, 1);
for i=1:2:m
    Ams = (2/(pi*i))*sin(pi*i/3)*(1+2*cos(pi*i/9));
    nBi = @(phi)nBi(phi)+Ams*cos(2*i*(phi-((2*pi/3)/4)));
end
nB = @(phi)50+nBi(phi) 
nB = function_handle with value:
    @(phi)50+nBi(phi)
wB = @(phi)nBi(phi)
wB = function_handle with value:
    @(phi)nBi(phi)
Lab = integral(@(phi)(nA(phi).*wB(phi)), 0, 2*pi)
Lab = 3.9695