How to find damping and the natural response in a system with a zero and complex pole pair

I want to know how to get damping and the natural response in a system with a zero and complex pole pair. I calculated the following transfer function:
(from an active filter composed of 2 capacitors and 3 resistors)

$$\frac{As}{Bs^2+Cs+D}$$

Where the poles of s are complex ($$\s\rightarrow \alpha\pm j\beta\$$) and $$\A \neq D\$$, I've seen solutions where $$\A =D\$$ and the system takes the form of:

$$\frac{\omega_n^2 s}{s^2+2\zeta\omega_n s+\omega_n^2}$$ But that's not the case in my transfer function.

• Did you mean to say $A\neq D$?
– AJN
Sep 17 '21 at 17:00
• Clearly A doesn’t equal C. Sep 17 '21 at 17:10
• Yes, I ment $A \neq D$. I've updated the question. Sep 17 '21 at 19:16
• For damping you already have what you need: make the denominator monic and solve for $\zeta$. As for the response just take the inverse Laplace. You might not even need to perform any calculations since the 2nd order transfer function is well known. Sep 17 '21 at 19:30
• After some research and thinking, I've come to the following answer of how to transform the function into the standard form: $\dfrac{s/B}{s^2+C/B s+ D/B}$, but I'm not sure if I can just ignore the numerator in the transfer function Sep 17 '21 at 19:51

A second-order polynomial can be put under the following normalized form: $$\D(s)=1+b_1s+b_2s^2\$$. Then, you can equate this expression with the normalized form where the quality factor $$\Q\$$ appears and find the correspondence between the terms: $$\D(s)=1+\frac{s}{\omega_oQ}+(\frac{s}{\omega_o})^2\$$. From there, if you do the maths ok, the you find $$\\omega_o=\frac{1}{\sqrt{b_2}}\$$ and $$\Q=\frac{\sqrt{b_2}}{b_1}\$$.
So, in your case, you have $$\H(s)=\frac{\omega_n^2 s}{s^2+2\zeta\omega_n s+\omega_n^2}\$$. Factor $$\\omega_n^2\$$ in the numerator and the denominator and you will have the above form for the identification of your coefficients.