Caucer fraction expansion fraction expansion for filter synthesis
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2s^2+2s+1|2s^3 +2s^2 +2s+1 | s
2s^3+2s^2+s
-------------------------------------------------------
s+1|2s^2+2s+1| 2s
So want to know how to pereform Caucer fraction expansion fraction expansion for filter synthesis. For a N/M polynomial. so i get a constant*s mainly not s+1,or s-1 which residue gives sometimes
The example below is the one i am trying to figure out
N = 18889465931478580854784*s^13 + 47283973869184172490752*s^12 + 380144960515421138583552*s^11 + 706526257926373225921184*s^10 + 2291017465069233730748416*s^9 + 2885191461803855148054208*s^8 + 4836216961066520318836736*s^7 + 3887431410705412021850560*s^6 + 3598519003466394276724736*s^5 + 1708028481397562746282496*s^4 + 850144437757157128011776*s^3 + 199942013009514701125133*s^2 + 45117763022779280523264*s + 2734820905916453355520; % Example numerator
M = 47283965291789216579584*s^12 + 117964927714854248120320*s^11 + 706526147746757277255008*s^10 + 1167883982950116513284096*s^9 + 2885191063569152216303936*s^8 + 3033824177148917979283456*s^7 + 3887430981188531521547840*s^6 + 2475385520141079517593600*s^5 + 1708028355427532732161536*s^4 + 587964405289678977105920*s^3 + 199942002958859758602739*s^2 + 26228297618450611699712*s + 2734820905916453355520; % Example denominator want to do it for this
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Answers (2)
Star Strider
on 3 Oct 2024
Edited: Star Strider
on 3 Oct 2024
The Cauer decomposition is useful. Your post is only the second time I’ve encountered a question about it here. The residue function appears to be the only opttion. I experimented with the Symbolic Math Toolbox, and while is ‘sort of’ works, it only does so with textbook examples.
Try this —
syms s
H = (2*s^5+40*s^3+128*s)/(s^4+10*s^2+9)
[Hn1,Hd1] = numden(H)
T1 = quorem(Hn1,Hd1)
H2 = Hn1/Hd1 - T1
[Hn2,Hd2] = numden(H2)
T2 = quorem(Hd2,Hn2)
H3 = Hd2/Hn2 - T2
[Hn3,Hd3] = numden(H3)
T3 = quorem(Hd3,Hn3)
H4 = Hd3/Hn3 - T3
[Hn4,Hd4] = numden(H4)
T4 = quorem(Hd4,Hn4)
H5 = Hd4/Hn4 - T4
[Hn5,Hd5] = numden(H5)
T5 = quorem(Hd5,Hn5)
This gives the same result as in W-K Chen, Passive and Active Filters Wiley 1986 (ISBN 0-471-82352-X), P. 113. (Yes, it’s been a while since I’ve done either Cauer or Foster decompositions.)
I stepped through all this manually, to be sure that it gave the correct result. You can probably automate it with a loop, either using this approach or deconv and residue. The ‘T’ values are the component values.
With a ‘real world’ transfer function, it may not produce such a neat result.
EDIT — There still seems to be something wrong with Answers, in that it does not produce the typeset LaTeX results from the Symbolic Math Toolbox in submitted posts, although they are visible while the code is being run initially. (I have reported this to the powers-that-be in MathWorks.) It doesn’t even produce ASCII results, so you’ll have to run this again here or on your own computer to see the results. This seems to be isolated to Answers, since I get the same result in Firefox and DuckDuckGo, so it seems to be browser-independent.
.
8 Comments
Star Strider
on 6 Nov 2024 at 19:33
My pleasure!
I’m not certain what magic to use to perhaps transform or map your transfer function iinto a version that could work. One of the experiments I tried was creating polynomials from the imaginary parts of the poles and zeros, since the Cauer factorisation requires that (no real parts).
This is that experiment, however it takes too long to run here —
syms s
N = 18889465931478580854784*s^13 + 47283973869184172490752*s^12 + 380144960515421138583552*s^11 + 706526257926373225921184*s^10 + 2291017465069233730748416*s^9 + 2885191461803855148054208*s^8 + 4836216961066520318836736*s^7 + 3887431410705412021850560*s^6 + 3598519003466394276724736*s^5 + 1708028481397562746282496*s^4 + 850144437757157128011776*s^3 + 199942013009514701125133*s^2 + 45117763022779280523264*s + 2734820905916453355520; % Example numerator
M = 47283965291789216579584*s^12 + 117964927714854248120320*s^11 + 706526147746757277255008*s^10 + 1167883982950116513284096*s^9 + 2885191063569152216303936*s^8 + 3033824177148917979283456*s^7 + 3887430981188531521547840*s^6 + 2475385520141079517593600*s^5 + 1708028355427532732161536*s^4 + 587964405289678977105920*s^3 + 199942002958859758602739*s^2 + 26228297618450611699712*s + 2734820905916453355520; % Example denominator want to do it for this example
H = N/M
poles = vpa(solve(M == 0))
zeros = vpa(solve(N == 0))
format long
impoles = poly2sym(double(imag(poles)),s);
imzeros = poly2sym(double(imag(zeros)),s);
H = expand((imzeros * s)) / impoles;
H = vpa(H, 5)
denpoly = [1 1];
% syms s
% H = (2*s^5+40*s^3+128*s)/(s^4+10*s^2+9)
k = 0;
H = 1/H;
while numel(denpoly) > 1 % Loop Version (Can Be A Function)
k = k + 1;
% disp("k = "+k)
[Hn,Hd] = numden(1/H);
denpoly = sym2poly(Hd);
Comp(k) = quorem(Hn,Hd);
H = Hn/Hd - Comp(k);
end
fprintf(['Component Values: \n',repmat('\t\t\t%s\n',1, numel(Comp)), '\n'], Comp)
Multiplying the numerator by s is necessary so that the numeerator has a power that is one greater than the denominator polynomial.
.
Hitesh
on 5 Oct 2024
Hi Liam Jacobs,
You can use MATLAB's built-in functions, "quorem" or "residue," to obtain the constant value. Ensure these functions are used within the loop where you continuously update the numerator and denominator of the polynomials. MATLAB's symbolic toolbox can handle high-order polynomials effectively.
Please refer to below code for your reference:
% Define the symbolic variable
syms s;
% Define the numerator and denominator polynomials
% Example: 14th-order numerator with large coefficients
N = 1e5*s^14 + 2e4*s^13 + 3e3*s^12 + 4e2*s^11 + 5e1*s^10 + 6e0*s^9 + ...
7e5*s^8 + 8e4*s^7 + 9e3*s^6 + 10e2*s^5 + 11e1*s^4 + 12e0*s^3 + ...
13e5*s^2 + 14e4*s + 15e3; % Example numerator
M = 2*s^2 + 2*s + 1; % Example denominator
% Initialize the continued fraction expansion
cauerExpansion = sym([]);
% Perform polynomial division iteratively
while true
% Perform polynomial division
[Q, R] = quorem(N, M, s);
% Append the quotient to the continued fraction expansion
cauerExpansion = [cauerExpansion, Q];
% Check if the remainder is zero
if R == 0
break;
end
% Check if the remainder is a constant
coeffs = sym2poly(R);
if length(coeffs) == 1
fprintf('Remainder is a constant: %s\n', char(R));
break;
end
% Update the numerator and denominator for the next iteration
N = M;
M = R;
end
% Display the continued fraction expansion
disp('Cauer Continued Fraction Expansion:');
disp(cauerExpansion);
% Format the continued fraction for display
cfString = sprintf('%s', char(cauerExpansion(1)));
for i = 2:length(cauerExpansion)
cfString = sprintf('%s + 1/(%s)', cfString, char(cauerExpansion(i)));
end
disp('Formatted Continued Fraction:');
disp(cfString);
For more information on "quorem" and "residue" function, kindly refer the following documentation link:
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