How to do this unusual Fourier transform?
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I am trying to compute a sine transform:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1464487/image.png)
I'm not sure about the non-standard bound of the integral (why it's not infinity???).
Anyway, let's say f_t = rand(40,1). Then,
N = 1000;
omega = logspace(-2,2,N);
for ii = 1:N
f_omega{ii} = omega(ii).*integral(@(t) f_t.*sin(omega(ii).*t),0,2*pi./omega(ii),'ArrayValued',true);
end
gives a vector of length 40 for each omega, which doesn't seem right to me. I should have just a vector of length N at the end. How do I do this integral? And should it matter what range in t I choose for f(t)?
3 个评论
Star Strider
2023-8-25
I could not get it to work numerically, even using the non-random sine function for ‘f_t’ so that I could be certain that it had a non-random result. I did the symbolic calculation and plot to see what the correct result would be (sort of like looking in the back of the book to see what the correct answer is).
回答(1 个)
Walter Roberson
2023-8-25
Anyway, let's say f_t = rand(40,1).
Let's not say that.
f is a function of t. By saying that f_t = rand(40,1) you are saying that f is constant in t, and is a 40-dimensional (constant) point. When you integrate that, of course you are going to end up with a 40-dimensional result.
5 个评论
Walter Roberson
2023-8-25
Using that notation, but with the same functionality as Star Strider already posted:
syms f(t) G(omega) varepsilon_0 e_1
Pi = sym(pi);
sympref('AbbreviateOutput',false);
f(t) = cos(5*Pi*t)
G(omega) = omega/(varepsilon_0 * Pi) * int((f(t)*e_1) * sin(omega*t), t, 0, 2*Pi/omega);
G = simplify(G, 500)
G_representative = subs(G, {e_1, varepsilon_0}, {1,1})
figure
fplot(G_representative,[0.01 1000])
grid
Ax = gca;
Ax.XScale = 'log';
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