Why am I not able to obtain the Fourier Transform of exponent expression using Symbolic math?
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Emmanuel J Rodriguez
on 6 Aug 2021
Commented: Emmanuel J Rodriguez
on 7 Aug 2021
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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Accepted Answer
Paul
on 6 Aug 2021
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 Comments
Paul
on 6 Aug 2021
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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