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Cannot extract real or imag part of a function

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Fine
Fine el 30 de Sept. de 2023
Comentada: Paul el 1 de Oct. de 2023
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);

Respuesta aceptada

Star Strider
Star Strider el 30 de Sept. de 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 = 
f_ft = matlabFunction(f_FT)
f_ft = function_handle with value:
@(omega)pi.*dirac(omega).*(3.076923076923077e-2-2.461538461538462e-1i)+pi.*exp(omega).*(sign(omega)-1.0).*8.0
omegav = linspace(0, pi, 25);
ft = f_ft(omegav);
R = real(f_ft(omegav))
R = 1×25
Inf 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
I = imag(f_ft(omegav))
I = 1×25
-Inf 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
The presence of the term makes a plot essentially impossible.
.

Más respuestas (1)

Walter Roberson
Walter Roberson el 30 de Sept. de 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 comentario
Paul
Paul el 1 de Oct. de 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)
f_FT = 
[real(f_FT) imag(f_FT)].'
ans = 
syms w real
[real(f_FT) imag(f_FT)].'
ans = 
But taking the matlabFunction at this point might not be useful because of the diracs.

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