Cannot extract real or imag part of a function
4 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
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 Commenti
Risposta accettata
Star Strider
il 30 Set 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 Commenti
Più risposte (1)
Walter Roberson
il 30 Set 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 Commento
Paul
il 1 Ott 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.
Vedere anche
Categorie
Scopri di più su Polynomials in Help Center e File Exchange
Prodotti
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!