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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$ :

$\displaystyle 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$ :

$\displaystyle \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

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

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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition)
Copyright © 2024-05-20 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University