Cannot extract real or imag part of a function
4 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I Fourier-transformed a bymbolic expression and turned it into a function, but cannot use real or imag functions for it. The error is: Incorrect number or types of inputs or outputs for function real.
syms x
f = 1/(1+28*1i)+28*1i/(x-1i);
f_FT = fourier(f);
f_ft = matlabFunction(f_FT);
R = real(f_ft);
I = imag(f_ft);
0 commentaires
Réponse acceptée
Star Strider
le 30 Sep 2023
You are taking the real and imag parts of a function handle. It is necessary to evaluate the function handle first.
Try this —
syms x omega
f = 1/(1+8*1i)+8*1i/(x-1i);
f_FT = fourier(f, omega)
f_ft = matlabFunction(f_FT)
omegav = linspace(0, pi, 25);
ft = f_ft(omegav);
R = real(f_ft(omegav))
I = imag(f_ft(omegav))
The presence of the term makes a plot essentially impossible.
.
0 commentaires
Plus de réponses (1)
Walter Roberson
le 30 Sep 2023
f_ft is a function handle. The only operations supported for function handles are copying, assignment, invocation, display, functions() which returns information.
You could take the real() of the symbolic expression and matlabFunction that, or you could invoke the handle on specific values and real() the result.
1 commentaire
Paul
le 1 Oct 2023
Before taking real() and imag() of the symbolic expression, the transform variable should be declared as real
syms x
f = 1/(1+28*1i)+28*1i/(x-1i);
f_FT = fourier(f)
[real(f_FT) imag(f_FT)].'
syms w real
[real(f_FT) imag(f_FT)].'
But taking the matlabFunction at this point might not be useful because of the diracs.
Voir également
Catégories
En savoir plus sur Polynomials dans Help Center et File Exchange
Produits
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!