# forming a function handles in matrix

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Bathala Teja 2021년 10월 4일
댓글: Bathala Teja 2021년 10월 6일
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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### 답변(1개)

Steven Lord 2021년 10월 4일
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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Bathala Teja 2021년 10월 6일
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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R2021b

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