Q of a Complex Resonator

The simplest case is a complex one-pole resonator having its single pole at $s=p_0=\sigma_0+j\omega_0$ :

$$H(s) = \frac{1}{s-p_0}, \quad p_0=\sigma_0+j\omega_0$$

The maximum gain is readily found to be $G(\omega_0)=\vert H(j\omega_0)\vert=1/\vert\sigma_0\vert$ , i.e., it is obtained at the radian frequency $\omega=\omega_0$ . We therefore define the imaginary part of the pole $\omega_0$ as the resonance frequency.

Next, we need to find the 3dB-bandwidth. Without loss of generality, we can set $\omega_0=0$ , since bandwidth is not changed for a complex one-pole resonator by a vertical translation along the $j\omega$ axis. (Consider the graphical method for calculating amplitude response from poles and zeros.) It is now easy to see that the -3dB points of a ``dc resonance $H(s) = 1/(s-\sigma_0)$ are at $\omega=\pm\sigma_0$ :

$$\frac{\vert H[j(0\pm\sigma_0)]\vert^2}{\vert H(j0)\vert^2} \eqsp \left.\left\vert\frac{1}{\pm j\sigma_0 - \sigma_0}\right\vert^2 \right/ \left\vert\frac{1}{ - \sigma_0}\right\vert^2 \eqsp \frac{\sigma_0^2}{\sigma_0^2+\sigma_0^2} = \frac{1}{2}.$$

We have thus found that the $Q$ of a complex one-pole resonator with pole at $p=\sigma_0+j\omega_0$ is given by

$$\zbox {Q = \frac{\omega_0}{2\vert\sigma_0\vert} = \frac{\mbox{Resonance\_Frequency}}{\mbox{3dB\_Bandwidth}.}}$$