Cannot extract real or imag part of a function
4 ビュー (過去 30 日間)
古いコメントを表示
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 件のコメント
採用された回答
Star Strider
2023 年 9 月 30 日
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 件のコメント
その他の回答 (1 件)
Walter Roberson
2023 年 9 月 30 日
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 件のコメント
Paul
2023 年 10 月 1 日
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.
参考
カテゴリ
Help Center および File Exchange で Polynomials についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!