Why am I not able to obtain the Fourier Transform of exponent expression using Symbolic math?
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The answer should be a closed-form solution.
% Practice, Problem 7 from Kreyszig sec 10.10, p. 575
syms f(x)
f(x) = x*exp(-x);
f_FT = fourier(f(x))
% Doesn't find transform
assume(x>0)
f_FT_condition = fourier(f(x))
assume(x,'clear')
ans:
f_FT =
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/704962/image.png)
f_FT_condition =
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/704962/image.png)
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采纳的回答
Paul
2021-8-6
Based on the assumption, I'm going to assume that f(x) = x*exp(-x) for x>=0 and f(x) = 0 for x < 0. In which case
syms f(x)
f(x) = x*exp(-x)*heaviside(x);
fourier(f(x))
If that's the expected result check out
doc heaviside
to understand why f(x) is defined that way.
3 个评论
Paul
2021-8-6
The scaling on the Fourier transform is arbitrary, but must be consistent with the scaling on the inverse transform. This scaling is controlled via sympref() (look at its doc page before you use it). The default is a scaling of 1 on the Fourier transform. But you can change that:
syms f(x)
f(x) = x*exp(-x)*heaviside(x);
sympref('FourierParameters',[1/(sqrt(2*sym(pi))) -1]);
fourier(f(x))
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