Cannot extract real or imag part of a function

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Fine 2023-9-30

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2027709-cannot-extract-real-or-imag-part-of-a-function

评论： Paul 2023-10-1

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);

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Answer 1

Star Strider 2023-9-30

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在 MATLAB Online 中打开

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.

.

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Answer 2

Walter Roberson 2023-9-30

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https://ww2.mathworks.cn/matlabcentral/answers/2027709-cannot-extract-real-or-imag-part-of-a-function#answer_1322844

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.