Fourier Transform of e^jkt, not giving an answer

4 visualizaciones (últimos 30 días)
Aidan
Aidan el 7 de Nov. de 2024
Editada: Paul el 7 de Nov. de 2024
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()

Respuestas (2)

Walter Roberson
Walter Roberson el 7 de Nov. de 2024
Editada: Walter Roberson el 7 de Nov. de 2024
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 comentario
Paul
Paul el 7 de Nov. de 2024
Editada: Paul el 7 de Nov. de 2024
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.

Iniciar sesión para comentar.


Paul
Paul el 7 de Nov. de 2024
Editada: Paul el 7 de Nov. de 2024
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

Categorías

Más información sobre Programming en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by