Fourier Transform of e^jkt, not giving an answer

38 vues (au cours des 30 derniers jours)
Aidan
Aidan le 7 Nov 2024 à 0:33
Modifié(e) : Paul le 7 Nov 2024 à 12:25
This is the code
syms t w K
S=exp(K*j*t)
F = fourier(S,t);
disp(F)
This is the output
F =fourier(exp(Y*t5*1i), t5, t)
It dose not seem to do anything at all for this one.
I was able to use it for sine and cosine, just not for exp()

Réponses (2)

Walter Roberson
Walter Roberson le 7 Nov 2024 à 1:05
Modifié(e) : Walter Roberson le 7 Nov 2024 à 1:06
The fourier transform of that function does not converge.
You need to take abs(t)
syms t w K
S=exp(K*j*abs(t))
F = fourier(S,t);
disp(char(F))
(K*2i)/(K^2 - t^2)
  1 commentaire
Paul
Paul le 7 Nov 2024 à 2:20
Modifié(e) : Paul le 7 Nov 2024 à 12:25
Hi Walter,
The transform variable should (probably) be w, not t.
Also, I'm surprised at the result. S only has a Fourier transform if Re(1j*K) < 0. I'm surprised that the SMT returned a solution absent that assumption (I'm not even sure that can be specified via assumptions.)
syms t w
S = exp(2*1j*abs(t))
F = fourier(S,w)
disp(char(F)) % no transform for K = 2
fourier(exp(abs(t)*2i), t, w)
Now that I think about it, I wonder if fourier has other cases where it makes an implicit assumption.

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Paul
Paul le 7 Nov 2024 à 1:00
Modifié(e) : Paul le 7 Nov 2024 à 1:02
Need to specify that K is real. Also, if only using two arguments to fourier the second argument is the transform variable, so should be w. I prefer using the three argument form to be clear what the function variable is and what the transform variable is.
syms t w K real
S=exp(K*j*t)
F = fourier(S,t,w)
disp(char(F))
2*pi*dirac(K - w)
Answers still not rendering symbolic output @Tushal Desai

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