Do we analyze real poles the same way you analyze complex conjugate poles?

Question

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?

Answer 1

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:

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

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.

Answer 2

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:

$$H(s)=\frac{(s-z_1)(s-z_2)}{(s-p_1)(s-p_2)}$$

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.