Effacer les filtres
Effacer les filtres

How to obtain Real and Imaginary parts of symbolic polynomial

25 vues (au cours des 30 derniers jours)
Craig
Craig le 2 Jan 2021
I have the below code. I cannot figure out how to obtain the real and imaginary parts of the symbolic polynomial (I have already solved the complete problem in standard Matlab, trying to learn the Symbolic Toolbox). My code is:
polynomial = [1 10 45 105 105]; % This polynomial is the Bessel Filter
% polynomial computed above, and changes with the order of the filter.
p = poly2sym(polynomial, s)
pw = subs(p, 1*j*w)
% attempt from Walter Roberson's comment at:
% https://www.mathworks.com/matlabcentral/answers/
% 465739-real-and-imaginary-part-of-this-function-please
real_pw = simplify( rewrite(real(pw), 'exp') )
imag_pw = simplify( rewrite(imag(pw), 'exp') )
% NOT WORKING FOR ME
% Assuming I can get the above done, my next steps would be to convert the
% below (from standard Matlab) to symbolic. Any help with the below would
% also be appreciated.
sq_real = conv(real_pw, real_pw);
sq_imag = conv(imag_pw, imag_pw);
sq_denom = sq_real + sq_imag;
The above code gives me:
real_pw =
10*imag(w^3) - 45*real(w^2) + real(w^4) - 105*imag(w) + 105
imag_pw =
imag(w^4) - 45*imag(w^2) - 10*real(w^3) + 105*real(w)
Instead, I am expecting something like (obtained from my working plain Matlab code):
real_poly = 1 0 -45 0 105
imag_poly = 0 -10 0 105 0
Any help you can provide is appreciated!

Réponse acceptée

Paul
Paul le 2 Jan 2021
Modifié(e) : Paul le 2 Jan 2021
From the context of the code, it appear that w is a real number. Assuming it being so will help:
>> syms s
syms w real
polynomial = [1 10 45 105 105]; % This polynomial is the Bessel Filter
% polynomial computed above, and changes with the order of the filter.
p = poly2sym(polynomial, s);
pw = subs(p, 1j*w); % note use of 1j, not 1*j
real_pw = simplify( rewrite(real(pw), 'exp') )
imag_pw = simplify( rewrite(imag(pw), 'exp') )
% but this is simpler
real_pw = real(pw)
imag_pw = imag(pw)
real_pw =
w^4 - 45*w^2 + 105
imag_pw =
- 10*w^3 + 105*w
real_pw =
w^4 - 45*w^2 + 105
imag_pw =
- 10*w^3 + 105*w
I don't believe that conv is supported for symbolic polynomials (version 2019a)
>> conv(real_pw,real_pw)
Error using conv2
Invalid data type. First and second arguments must be numeric or logical.
Error in conv (line 43)
c = conv2(a(:),b(:),shape);
But conv is the same as polynomial multiplication, so maybe this is what you want:
>> sq_real = real_pw*real_pw
sq_real =
(w^4 - 45*w^2 + 105)^2
>> sq_imag = imag_pw*imag_pw
sq_imag =
(- 10*w^3 + 105*w)^2
>> sq_denom = sq_real + sq_imag
sq_denom =
(- 10*w^3 + 105*w)^2 + (w^4 - 45*w^2 + 105)^2
>> expand(sq_denom)
ans =
w^8 + 10*w^6 + 135*w^4 + 1575*w^2 + 11025
  7 commentaires
Paul
Paul le 3 Jan 2021
Walter may be able to work some symbolic magic; he's really good at that. The best I can do is get the numerical answer:
sq_denom(w)=simplify(pw*conj(pw));
gain = polynomial(end);
f(w) = sq_denom(w) - 2*gain^2; % this is the equation to solve for w
fplot(f(w)); % looks like there are two real roots, which makes sense from the form of f(w)
w0 = vpasolve(f,w);
w0(imag(w0)==0)
ans =
-2.1139176749042158430196891286339
2.1139176749042158430196891286339
Walter Roberson
Walter Roberson le 3 Jan 2021
solve(f, 'maxdegree', 4) I suspect
If so then the solutions are not likely to clarify anything, as they will be more than 1000 characters each.

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Symbolic Math Toolbox dans Help Center et File Exchange

Produits


Version

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by