Factorize generic transfer function

This is not homework. I'm just trying to understand concepts and methods, and for that I'm using a generic Transfer Function for an Active Filter like $$\\mathcal{H}(s) = \frac{1}{1+BAs + A^2s^2}\$$.

1. How can I factor de denominator to the form $$\(s-z_1)(s-z_2)\$$ in order to find the poles of this "generic" transfer function? I can factor out pretty easy quadratic forms, but is there a way to be systematic technique over more complicated quadratic expressions like this generic one?

2. When factoring I'll find the complex-conjugate poles for this transfer function: but what are their actual meaning for the frequency response? As far as I understand it, the transfer function "explodes" - does that mean there is instability in the filter?

3. If the function was something like $$\\mathcal{H}'(s) = \frac{(As)^2}{1+BAs + A^2s^2}\$$ would the zero be at 0 frequency? Does it mean the filter is stable at zero frequency? (that would't make much sense to me)

4. What would be the Quality Factor in a generic-type transfer function like this one?

• You can factor any quadratic using the quadratic formula to find its roots. The same goes for cubics and quartics using the cubic and quartic formulas (yes, those exist). It's a famous result in Galois theory that no formula for quintics and beyond exists, so you'd have to use other methods to factor those. Commented Oct 2, 2022 at 18:16
• ludicrous, See some of the development here, where I wrote to you earlier. In general the roots in the numerator (zeros) have been replaced with gain fractions for each of the standard form terms. In short, there are no zeros in the standard form.
– jonk
Commented Oct 2, 2022 at 20:11

How can I factor de denominator to the form (s−z1)(s−z2) in order to find the poles of this "generic" transfer function?

(1), (2) and (4) answered together: -

$$\dfrac{1}{1 +sB +s^2A^2}\longrightarrow \dfrac{\frac{1}{A^2}}{s^2 +s\frac{B}{A} +\frac{1}{A^2}}$$

And is of the form: $$\\dfrac{\omega_n^2}{s^2 +s2\zeta\omega_n +\omega_n^2}\$$

• where $$\\omega_n\$$ is the natural resonant frequency (aka $$\\omega_0\$$ or "pole frequency")
• and, $$\\zeta\$$ is the damping ratio or $$\\frac{1}{2Q}\$$.

So, s can be calculated like this: -

$$s = \dfrac{-2\zeta\omega_n \pm \sqrt{4\zeta^2\omega_n^2-4\omega_n^2}}{2}$$

$$s = -\zeta\omega_n \pm \sqrt{\zeta^2\omega_n^2-\omega_n^2}$$

$$s = -\zeta\omega_n \pm \omega_n\cdot\sqrt{\zeta^2-1}$$

Then, depending on whether $$\\zeta\$$ is greater or less than 1 you would proceed in one of two ways. The "less than 1" (more interesting) scenario: -

$$s = -\zeta\omega_n \pm j\omega_n\cdot\sqrt{1-\zeta^2}$$

Did you notice that I multiplied the terms inside the square root by -1 and brought-out the square root of -1 in the form of "j" the complex operator?

And this result is clearly a low-pass filter with conjugate poles: -

Image taken from my basic website. 3-D view of positive frequency pole: -

(3) Yes, there will be a pole at the origin on the pole-zero diagram

Does it mean the filter is stable at zero frequency?

It means that it is a band-pass or high-pass filter

• Equation at the top to the right of the arrow. The denominator constant should be 1/A^2 Commented Oct 2, 2022 at 20:02
• Andy, thank you for your comprehensive response! But how could you immediately tell it was a low passa filtres from the poles? Commented Oct 2, 2022 at 20:17
• @RussellH oops yes... fixing. Cheers. Commented Oct 2, 2022 at 20:47
• @ludicrous a low pass filter has generally not got an "s" in the numerator. If there's an "s" then it's a BP filter; if it's $s^2$ then it's a HP filter. Commented Oct 2, 2022 at 20:49
• @Andyaka would be a way to "prove" it by doing the limits as $s \rightarrow 0$ and $s \rightarrow \infty$ for each one of the cases? Commented Oct 2, 2022 at 20:59