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?


2 Answers 2


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 $\sigma = 0$.


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).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .