Why am I not able to obtain the Fourier Transform of exponent expression using Symbolic math?

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Emmanuel J Rodriguez
评论： Emmanuel J Rodriguez 2021-8-7
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 =
f_FT_condition =

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Paul
Paul 2021-8-6
Ran in:
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))
ans = 
If that's the expected result check out
doc heaviside
to understand why f(x) is defined that way.
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Paul
Paul 2021-8-6
Ran in:
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))
ans = 
Emmanuel J Rodriguez
Emmanuel J Rodriguez 2021-8-7
Oh, right! I remember now that the scaling term can be applied to either one of the Fourier pairs. HUGE help! Thank you so much!

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