So I am analyzing a filter and I came up with the following expression for the transfer function in its canonical form:

$$\frac{987654 + 4666.67 s + 1. s^2}{2.48519 \times10^8 + 52316.2 s + s^2}$$

which of course is of the type

$$\frac{\omega_z^2 + \frac{\omega_z}{Q_z} s + s^2}{\omega_p^2 + \frac{\omega_p}{Q_p} s + s^2}$$

Now I went on to calculate the poles and the zeros and came up with real poles and real zeros.

$$z_1=-222.222$$ $$z_2=-4444.44$$ $$p_1=-5284$$ $$p_2=-47032.2$$

Now if these were complex conjugate poles and/or zero I would proceed to check for the natural oscillation frequency (\$\omega\$) and quality factor (Q).

Now I feel that talking about the quality factor actually as, since it is below 0.5 it will match the fact that we have real poles and or zeros. But does it make sense to calculate the frequency. Because since the poles/zeros have different real parts, they oscillate in different frequencies.

I actually checked for the Bode plots on this site http://www.onmyphd.com/?p=bode.plot.online.generator and it indeed shows, in the asymptotic one, that we have 4 different frequencies affecting the slopes. So what should be my correct interpretation? I feel there are some underlying concepts here that I'm missing? Can someone help me organize my thoughts on this?

  • \$\begingroup\$ "Because since the poles/zeros have different real parts, they oscillate in different frequencies". Can you describe what you mean by this? Since they are all on the negative real axis, there isn't really an oscillation associated with them. \$\endgroup\$
    – AJN
    Sep 27, 2020 at 5:44
  • \$\begingroup\$ Both zeroes are at a lower frequency than the poles. The high frequency gain is near unity. DC gain is much below unity. I think this is either a high pass filter meant for attenuating low frequencies, or a lead filter meant for providing phase lead. Can you add the context of where you found this filter? \$\endgroup\$
    – AJN
    Sep 27, 2020 at 5:46
  • \$\begingroup\$ "So what should be my correct interpretation?" - what are you ultimately trying to achieve that causes you to feel you are so far unsuccessful in your goal? \$\endgroup\$
    – Andy aka
    Sep 27, 2020 at 8:49
  • \$\begingroup\$ @AJN you are right, the term "oscillating" is not well used here, since oscillators happen only when poles are on the imaginary axis. Yes, it is a high pass filter with two levels: -48 dB (aprox.) at low frequency and 0 dB at high frequency. \$\endgroup\$ Sep 27, 2020 at 9:43
  • \$\begingroup\$ @Andyaka I don't think I'm correctly interpreting what it means to have real negative poles instead of complex conjugated ones. Mainly I don't know if it makes sense to talk about a "common" frequency for the poles (as we did with complex conjugated) since they are indeed at different frequencies as we understand in the asymptotic bode diagram \$\endgroup\$ Sep 27, 2020 at 9:44

2 Answers 2


I would recommend to factor the original formula in a low-entropy form and match the normalized polynomial of a second-order system which is: \$H(s)=\frac{a_0}{b_0}\frac{1+a_1s+a_2s^2}{1+b_1s+b_2s^2}\$ which after proper rearrangement leads to the correct form where a leading term defines the dc gain in your case: \$H(s)=H_0\frac{1+\frac{s}{\omega_{0N}Q_{0N}}+(\frac{s}{\omega_{0N}})^2}{1+\frac{s}{\omega_{0D}Q_{0D}}+(\frac{s}{\omega_{0D}})^2}\$ where the \$N\$ and \$D\$ subscripts respectively refer to the numerator and denominator.

If the quality factor is below 0.5, it implies that roots are real and not coincident: there are no imaginary parts and the system is well damped. For a low quality factor, the poles (or zeroes) are well spread and one of them dominates the low-frequency spectrum while the other one the upper portion. In this case, you can apply the so-called low-\$Q\$ approximation:

\$1+\frac{s}{\omega_{0}Q_{0}}+(\frac{s}{\omega_{0}})^2\ \approx (1+\frac{s}{\omega_{p1}})(1+\frac{s}{\omega_{p2}})\$ where \$\omega_{p1}=\omega_0Q\$ and \$\omega_{p2}=\frac{\omega_0}{Q}\$.

In your case, you can rework the expression by factoring 987654 in the numerator and \$2.48\;10^8\$ in the denominator. This shows a dc attenuation of 48 dB followed by the two zeroes over the two poles. A Mathcad sheet shows the typical response of the factored form and compares it with the original expression:

enter image description here

With the given values, you see a boost in phase around 1kHz, perhaps to compensate a control system?

enter image description here

It is important to properly format transfer functions following a design-oriented analysis or D-OA: the transfer function should describe a system and unveil if it has gains, poles or zeroes. Respecting the format associating a leading term (if any) with a fraction starting by 1+... type of format naturally fulfills this requirement in my opinion.


The reason for breaking down a long transfer function (t.f.) into 2nd order sections is to avoid numerical instabilities, or tame the quality factor of the roots. The general formula for a 2nd order t.f. is only valid if it cannot be reduced any further, i.e. the roots of the polynomials forming the t.f. are complex (conjugate, for Hurwitz). But since your t.f. has only real poles and zeroes, then it can be split further into two 1st order sections:


This means that, since you no longer have complex roots, you can no longer talk about having a quality factor: you have two basic 1st order cells in series. For a short explanation about why, see this answer.

  • 2
    \$\begingroup\$ "since you no longer have complex roots, you can no longer talk about having a quality factor". Well, you can still talk about quality factor, if you define it as a suitable ratio of the equation's coefficients. Basically Q = 1/(2 zeta) where zeta is the damping ratio. And damping ratio can be expressed as a function of pole separation p2/p1 even for real poles. As a matter of fact zeta > 1 for real distinct poles, and this gives Q < 1/2 as @VerbalKint states in his answer. \$\endgroup\$ Sep 27, 2020 at 23:52
  • \$\begingroup\$ @SredniVashtar Strictly speaking, yes, you can talk about a quality factor, but it would quite useless since the moment you have a real pole you know precisely that you can only have one possible value. This is the reason I split it up into single, real poles, to show that such a 2nd order can be split into 1st order sections, showing that there are, in fact, two poles, instead of a "complex" pole. Does this fit into the over damped category? Of course, but it also shows that overdamped really means two distinct, simple, real poles, nothing more. \$\endgroup\$ Sep 28, 2020 at 7:02
  • \$\begingroup\$ I am not sure what you are referring to with "only one possible value". I want to be clear I am not antagonizing you. When I first learned about this 'generalization' of Q I was quite surprised it could be still used in non-oscillating systems, but in the end it makes sense because, just like the damping ratio, it's a measure of damping even if the system is not oscillating. Q just loses the meaning of ratio between energy stored and energy dissipated per cycle (which I believe is the agreeable sense of your answer) and becomes a measure of how far apart the real distinct poles are. \$\endgroup\$ Sep 28, 2020 at 15:06
  • \$\begingroup\$ @SredniVashtar It's all very true, and, as I agreed, it's at the basis of the overdamped case for 2nd order systems. The difference I tried to make is that the overdamped system can be further reduced into simple 1st order systems, where there is no damping due to there being real-valued roots, only. It's similar to RC, or RL cells, they just end up cramming roots on the real axis, as opposed to making use of the whole complex plane. I suppose it's just another way of looking at it. The way I see it, this is a discussion like any other. As long as some sort of truth comes out of it, why not? \$\endgroup\$ Sep 28, 2020 at 15:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.