the question is attached I have to answer, but I'm having problems.
I tried writing like this
fun = @(t,w)exp((-1i).*w.*t);
q=integral(@(t)fun(t,w),-1,1);
int() is recommended, but can it be written like above?

 採用された回答

Star Strider
Star Strider 2018 年 5 月 20 日

1 投票

You are close to solving it. You need to express ‘q’ as an anonymous function of ‘w’, then evaluate that function over the interval (-10,10):
fun = @(t,w)exp(-1i.*w.*t);
q = @(w) integral(@(t)fun(t,w),-1,1, 'ArrayValued',true);
wv = linspace(-10, 10);
figure(1)
plot(wv, real(q(wv)), '-b', wv, imag(q(wv)), '-r')
grid
So yes, you can use integral to evaluate the Fourier transform of your function. (It never occurred to me to do this, so I learned something.)

2 件のコメント

doyoun Kim
doyoun Kim 2018 年 5 月 21 日
編集済み: doyoun Kim 2018 年 5 月 21 日
Thx a lot
Star Strider
Star Strider 2018 年 5 月 21 日
As always, my pleasure.

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