Effect of a simple pole vs complex conjugate poles

Question

If $H(s)$ is a transfer function and it has just one pole in $s = p$, $p \in \mathbf{R}$,

$$H(s) = \displaystyle \frac{H_0}{(s - p)}$$

the frequency response is $20 \log_{10} |H(j\omega)|$. With $\tau = -1/p$, the frequency response is a constant for $\omega \ll \tau$ and decreases as 20 dB/dec for $\omega \gg \tau$.

But if $H(s)$ has complex conjugate poles,

$$H(s) = \displaystyle \frac{H_0}{(s - p)(s - p^*)}$$

with $p = \sigma_p + j\omega_p$, and $\omega_n = \sqrt{\sigma_p^2 + \omega_p^2}$, the frequency response is constant for $\omega \ll \omega_n$ and it decreases as 40 dB/dec for $\omega \gg \omega_n$, so double the case of a single-pole.

This can analitically be proved. But how could this be figured out?

I was expecting that there is no difference in the surface $|H(s)|$ between the effect of a single pole (located along the real axis) and a pole located in $s = p$ (belonging to a couple of complex-conjugate poles).

What is the "geometrical" reason behind this result?

Answer 1

The effect that the slope doubles from a single real-valued pole to a complex-conjugate pair can be illustrated by noting that the multiplication of the two transfer functions squares the frequency and by doing so doubles the slope:

$$H_1(j\omega)H_2(j\omega) =|H_1(\omega)|\text{e}^{j\phi_1(\omega)}|H_2(\omega)|\text{e}^{j\phi_2(\omega)} =\underbrace{|H_1(\omega)||H_2(\omega)|}_{\text{magnitude}} \overbrace{\text{e}^{j(\phi_1(\omega)+\phi_2(\omega))}}^{\text{phase}}. $$

Plotting the magnitude in a log-scale yields

$$ 20 \lg |H_1(\omega)||H_2(\omega)| = 20 \lg \frac{1}{\omega^2} = -40 \lg \omega.$$

for $H_1 = \frac{1}{j\omega}$ and $H_2 = \frac{1}{-j\omega}$. Resulting in a slope proportional to $\lg\omega$ ($-40\text{dB}$ every decade of $\omega$).

Generally speaking, two transfer functions multiplied in the frequency domain are added in the magnitude plot:

$$ 20 \lg |H_1(\omega)||H_2(\omega)| = 20 \lg |H_1(\omega)| + 20 \lg |H_2(\omega)|. $$ Note:
If you're wondering why I skip the real part of the poles, there's a theorem stating that the complete system information is comprised in the bode plot for $\sigma = 0$.

Answer 2

Yes, there contribution of each complex pole to the magnitude plot is similar (for high frequencies) to that of a real pole. But your 2nd H(s) has two poles, so the slope is double that of the 1st H(s).