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

But I want the symbolic operation, how can I do ?

[Sam Chak]

Hi @li hu

You need to at least provide the desired characteristics of the system behavior. With that, you design something about the coefficient of the transfer function.

[li hu]

I provide the coefficients of the transfer function, but they are symbols instead of numerics.

tf and damp of MATLAB seem to accept numerics only.

Any solution?

[Walter Roberson]

There is nothing in the Control System Toolbox that accepts symbolic expressions, unfortunately. And the Symbolic Toolbox provides little in the way of tools for symbolic process control.

Two key symbolic toolbox functions are laplace and ilaplace . The "transfer function" form with ratio of two s polynomials, is the laplace transform of the control function.

I am not familiar with the calculations involved for damp .. broadly speaking you would be looking to solve() the denominator of the laplace expression of the function in order to find the poles. numden to split numerator and denominator. I do not know at the moment what you would do to find the time constant.

[Sam Chak]

Hi @li hu

The transfer function is merely a fraction if you want to deal with symbols. What exactly do you want to solve in MATLAB?

syms s b a1 a2
G = b/(a1*s + a2)

G = 

[li hu]

Then, how to get the complex-conjucated poles and the damping ratio ?

[Walter Roberson]

syms s b a1 a2
G = b/(a1*s + a2)
[N, D] = numden(G)
poles = solve(D)
rate_constant = -1./poles

I would need to research the calculation of the damping ratio. Also, the time constant shown here might possibly only be valid for linear denominators.

[li hu]

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]

 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

Answer 1

[Sam Chak]

Hi @li hu

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.

See Abel's Impossibility Theorem.