Frequency Response for Small $\zeta$

As can be seen from the pole-zero plot, when $\omega=\omega_d=\omega_n\sqrt{1-\zeta^2}$, the vector length $\vert v_1\vert=\vert j\omega-p_1\vert$ is minimized to $\zeta\omega_n$. If $\zeta$ is small ($\vert\zeta\vert<0.5$), $\vert v_1\vert \ll \vert v_2\vert$ and $v_1$ is dominant in $\vert H(j\omega)\vert$, and $\vert H(j\omega_d)\vert=\omega_n^2/(\vert v_1\vert\vert v_2\vert)$ is maximized when $\omega=\omega_d$. In this case, the system behaves like a band-pass filter which peaks at approximately $\omega_d$. Due to the second vector $\vert v_2\vert=\vert j\omega-p_2\vert$, the actual peak frequency $\omega_p$ is slightly lower than $\omega_d=\sqrt{1-\zeta^2} \omega_n$. The precise peak frequency $\omega_p$ can be found by solving

$$\frac{d}{d\omega} \vert H(j\omega)\vert=0$$

to be

$$\omega_p=\omega_n \sqrt{1-2\zeta^2}\approx \omega_d=\omega_n\sqrt{1-\zeta^2}$$

which is very close to $\omega_d$ for small $\zeta$. The bandwidth of the passing band is defined as the frequency range

$$\triangle \omega \stackrel{\triangle}{=}\omega_{c_2}-\omega_{c_1}$$

where $\omega_{c_1}$ and $\omega_{c_2}$ are the cutoff frequencies on either side of $\omega_p$ ( $\omega_{c_1} < \omega_p < \omega_{c_2}$) at which

$$\frac{\vert H(j\omega_{c_1})\vert}{\vert H(j\omega_p)\vert}=\... ...ega_{c_2})\vert}{\vert H(j\omega_p)\vert} =\frac{1}{\sqrt{2}}$$

or

$$\vert H(j\omega_c)\vert^2 = \vert H(j\omega_p)\vert^2 /2$$

i.e., when $\omega=\omega_c$, only half of the power contained in the signal can pass through the filter. To find $\triangle \omega$, consider the two frequencies in the pole-zero plot

$$\omega_{c_{1,2}}=\omega_d \pm \zeta \omega_n=\omega_n\sqrt{1-\zeta^2}\pm\omega_n\zeta =\omega_n (\sqrt{1-\zeta^2} \pm \zeta)$$

When $\omega$ is at either $\omega_{c_1}$ or $\omega_{c_2}$, the vector $j\omega_c-p_1$ in the denominator satisfies

$$\vert j\omega_c-p_1\vert=\sqrt{2} \vert j\omega_d-p_1\vert$$

and we have

$$\frac{\vert H(j\omega_c)\vert}{\vert H(j\omega_d)\vert}= \fr... ...omega_d-p_1\vert}{\vert j\omega_c-p_1\vert}=\frac{1}{\sqrt{2}}$$

Here we assumed $\vert j\omega_c-p_2\vert \approx \vert j\omega_d-p_2\vert$ as the second pole $p_2$ (in the 3rd quadrant) is farther away from both $j\omega_d$ and $j\omega_c$ than $p_1$ in the 2nd quadrant. We can now find approximately the bandwidth as

$$\triangle \omega =\omega_{c_1}-\omega_{c_2} =\omega_n (\sqrt... ...2}+\zeta)-\omega_n (\sqrt{1-\zeta^2}-\zeta) = 2\zeta \omega_n$$