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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 =
f_FT_condition =

Accepted Answer

Paul
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))
ans = 
If that's the expected result check out
doc heaviside
to understand why f(x) is defined that way.
  3 Comments
Emmanuel Rodriguez
Emmanuel Rodriguez on 7 Aug 2021
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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