I am not certain what you want witth respect to an equivalent circuit. I believe you intend circuit synthesis or circuit realisation. If so, this may do what you want
First, using your data, identify the system using the System Identification Toolbox, specifically using the iddata and tfest functions to return a transdfer function.
When you have the transfer funciton that gives an appropriate fit to your data (check this using the using the compare function) you can then use circuit synthesis to identify the component values.
Illustrating that with a Cauer expansion (you will probably need to extract the transfer function numerator and denominator polynomials from the identified transfer function and then use the poly2sym function on each of them first) —
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)
That can be coded as a loop —
denpoly = [1 1]; % Define First (Used To Terminate The ‘while’ Loop)
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);
if k > 50 % Fail-Safe To Prevent Infinite Recursion
disp("Exiting loop (k = "+k+" )")
break
end
end
fprintf(['Component Values: \n',repmat('\t\t\t%s\n',1, numel(Comp)), '\n'], Comp)
These would be configured as:
>---- 2 H ----- 40/9 H ---- 70/9 H ---->
| |
1/20 F 9/40 F
| |
>-------------------------------------->
This is a textbook example, and gives the same result as in W-K Chen, Passive and Active Filters Wiley 1986 (ISBN 0-471-82352-X), P. 113. There are other approaches that may also be appropriate, such as the Foster expansion.
This may not always work as desired in ‘real world’ transfer functions, in that the component (‘Comp’ vector) values may not be simple functions of s as they are here, and may instead be vectors. Those results would not be valid.
(There is some sort of bug in Answers such that the results of symbolic calculations do not appear in the posted results, although they appear in the results while being run or edited. This has been reported, although apparently not yet fixed.)
.
