# How to define partial fraction expansion?

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mir 2023 年 9 月 5 日
コメント済み: mir 2023 年 9 月 9 日
Hi everyone,
I would like to expand the development of the partial fractions of a function F starting from [r,p,k] = residue(F).
Is there a function that allows you to do this?
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Dyuman Joshi 2023 年 9 月 5 日
"I would like to explain the development of the partial fractions of a function F starting from [r,p,k] = residue(F)."
What exactly do you mean by "explain the developement"?
mir 2023 年 9 月 5 日

s =tf('s')
s = s Continuous-time transfer function.
F = (s-1)/(s+2)
F = s - 1 ----- s + 2 Continuous-time transfer function.
N_f = F.Numerator{1}
N_f = 1×2
1 -1
D_f = F.Denominator{1}
D_f = 1×2
1 2
[r,p,k] = residue(N_f,D_f)
r = -3
p = -2
k = 1
I would see the function:
F_residue = 1 - 3/(s+2)
so as a sum of terms

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### 回答 (4 件)

Dyuman Joshi 2023 年 9 月 5 日
"I would see the function:
F_residue = 1 - 3/(s+2)
so as a sum of terms"
One method would be to use symbolic variables (Note - requires symbolic toolbox) -
s =tf('s');
F1 = (s-1)/(s+2);
F_res1 = expr(F1)
F_res1 = F2 = (-4*s+8)/(s^2+6*s+8);
F_res2 = expr(F2)
F_res2 = function F_res = expr(F)
N_f = F.Numerator{1};
D_f = F.Denominator{1};
[r,p,k] = residue(N_f,D_f);
syms s
if isempty(k)
k=0;
end
F_res = k + sum(r./(s-p));
end
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Dyuman Joshi 2023 年 9 月 9 日
You are right, I did not incorporate poles with multiplicity more than 1.
It would be better to use partfrac, rather than trying to modify the code I wrote.
%Directly use partfrac
syms s
F = (s-4)/(s*(s+2)^2);
out = partfrac(F)
out = %or via transfer function
s = tf('s');
F = (s-4)/(s*(s+2)^2);
N_f = F.Numerator{1};
D_f = F.Denominator{1};
n = numel(N_f);
syms s
F0 = (s.^(n-1:-1:0)*N_f')/(s.^(n-1:-1:0)*D_f')
F0 = out0 = partfrac(F0)
out0 = mir 2023 年 9 月 9 日
@Dyuman Joshi don't worry, I solved it
Thank you so much!

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Ayush Anand 2023 年 9 月 5 日
Hi Mir,
I understand you are looking for a function that can compute the partial fraction decomposition of a function. MATLAB provides a function called "residue" that can be used for this purpose. The "residue" function takes the coefficients of the numerator and denominator polynomials of the rational function as inputs and returns the residues, poles, and constant term of the partial fraction expansion.
You can read more about the syntax and use cases of "residue" here:
I hope this helps!
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Shubham 2023 年 9 月 5 日
Hi,
MATLAB provides a function called residue that can be used to perform partial fraction decomposition of a rational function. The residue function calculates the residues and poles of a given function.
Hope this helps. Thanks
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John D'Errico 2023 年 9 月 5 日

Therefore your question is truly to "explain the development" of that partial fraction expansion. For that, you can read the docs surrounding residue, thus by reading here: residue.
You might look at the code itself in the function residue. If you do that, then PLEASE, PLEASE, DO NOT EDIT THE FUNCTION. We see so many times where some one has edited a MATLAB supplied function, and not reaizing they did so, make some silly edit to the code by accident, and then save the function. If you want to look at the code, then do this:
type residue
However, my guess is you won't get far there, as the code has a lot in it, and is probably not illuminating. A new user to MATLAB will just find themselves confused.
Or, you might want to do some reading, to understand how a partial fraction expansion is performed. Maybe start here in the reference provided in residue itself:
Reference: A.V. Oppenheim and R.W. Schafer, Digital
Signal Processing, Prentice-Hall, 1975, p. 56-58.
Or you might look here:
But I'm afraid, if by "explain" you want something that will take you step by step through the specific mathematical computations that were performed, there is no tool in MATLAB that will do that, thus teaching you how to perform a specific partial fraction expansion on some specific problem. (I think Mathematica/Alpha will do things like that, for at least some problems, but I don't use that tool.)

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