How to solve the poles of (z^2 + 4*z + 3) / ((z^5 + 4*z^4 + 3*z^3 + 2*z^2 + 5*z + 2) *exp(-5*z))?

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Abalone
Abalone 2024 年 11 月 5 日 14:43
編集済み: Shashi Kiran 2024 年 11 月 5 日 15:20
I am solving the poles of (z^2 + 4*z + 3) / ((z^5 + 4*z^4 + 3*z^3 + 2*z^2 + 5*z + 2) *exp(-5*z)),and the code is below:
syms z ;
f = (z^2 + 4*z + 3) / (z^5 + 4*z^4 + 3*z^3 + 2*z^2 + 5*z + 2) *exp(-5*z) ;
[Poles, Orders, Residues] = poles(f);
Warning: Unable to determine poles.
disp(Poles);
disp(Orders);
disp(Residues);
As the waring says,why can't i get the poles?
Is this because the denominator is a fifth-degree polynomial and we can't solve its zeros in radical form?
Any help would be appreciated.

採用された回答

Shashi Kiran
Shashi Kiran 2024 年 11 月 5 日 15:15
編集済み: Shashi Kiran 2024 年 11 月 5 日 15:20
The warning appears because MATLAB's "poles" function finds it hard to get exact poles when the expression involves non-trivial components like .
The exponential term does not add any poles or zeros; it is smooth everywhere. It expands to
This series has no poles or zeros.
So, to find the poles, we can ignore the exponential term and focus only on the polynomial in the denominator.
syms z ;
f = (z^2 + 4*z + 3) / (z^5 + 4*z^4 + 3*z^3 + 2*z^2 + 5*z + 2);
[Poles, Orders, Residues] = poles(f);
disp(Poles);
Hope this helps.

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