forming a function handles in matrix

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Bathala Teja
Bathala Teja 2021-10-4
评论： Bathala Teja 2021-10-6
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??

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回答（1 个）

Steven Lord
Steven Lord 2021-10-4
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.
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Stephen23
Stephen23 2021-10-6
编辑：Stephen23 2021-10-6
@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
Bathala Teja 2021-10-6
Ran in:
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

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