tf and damp, all need the numerical coefficients of the transfer function

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But I want the symbolic operation, how can I do ?
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li hu
li hu on 15 May 2023
Thanks. If the denominator is a quadratic equation with one unknown, can you obtain the complex-conjugated poles ?
It is trivial for a numerical example.
But, how is it for a symbolic case ?
Walter Roberson
Walter Roberson on 15 May 2023
poles = solve(D, 'maxdegree', 4);
Typically this will create a quite long output. The exact symbolic solutions for the roots of degree 4 are complicated. When any of the coefficients are symbolic variables then there is little hope that you will get nice outputs. If all of the coefficients are numbers then from time to time you will get nice numeric solutions... but pretty much only for the case that someone deliberately created an easy form

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Answers (1)

Sam Chak
Sam Chak on 21 May 2023
Finding the poles (roots of the denominator) is like solving polynomial equations.
However, if it is 5th-degree and higher polynomial, then it cannot be solved algebraically in terms of a finite number of additions, subtractions, multiplications, divisions, and root extractions.

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