Relating Pole Radius to Bandwidth

Consider the continuous-time complex one-pole resonator with $s$ -plane transfer function

$$H(s) = \frac{-\sigma_p}{s-p}.$$

where $s=\sigma + j\omega$ is the Laplace-transform variable, and $p\isdef \sigma_p+j\omega_p$ is the single complex pole. The numerator scaling has been set to $-\sigma_p$ so that the frequency response is normalized to unity gain at resonance:

$$H(j\omega_p) = \frac{-\sigma_p}{j\omega_p-\sigma_p-j\omega_p} = \frac{-\sigma_p}{-\sigma_p} = 1.$$

The amplitude response at all frequencies is given by

$$G(\omega) \isdef \left\vert H(j\omega)\right\vert = \frac{\left\vert\sigma_p\right\vert}{\left\vert j\omega-p\right\vert} = \frac{\left\vert\sigma_p\right\vert}{\sqrt{(\omega-\omega_p)^2 + \sigma_p^2}}.$$

Without loss of generality, we may set $\omega_p=0$ , since changing $\omega_p$ merely translates the amplitude response with respect to $\omega$ . (We could alternatively define the translated frequency variable $\nu\isdef \omega-\omega_p$ to get the same simplification.) The squared amplitude response is now

$$G^2(\omega) = \frac{\sigma_p^2}{\omega^2+\sigma_p^2}.$$

Note that

$$\begin{eqnarray*} G^2(0) &=& 1 = 0 \hbox{ dB},\\ G^2(\pm\sigma_p) &=& \frac{1}{2} = - 3 \hbox{ dB}. \end{eqnarray*}$$

This shows that the 3-dB bandwidth of the resonator in radians per second is $2\left\vert\sigma_p\right\vert$ , or twice the absolute value of the real part of the pole. Denoting the 3-dB bandwidth in Hz by $B$ , we have derived the relation $2\pi B = 2\left\vert\sigma_p\right\vert$ , or

$$\zbox {B=\frac{\left\vert\sigma_p\right\vert}{\pi}=\frac{\left\vert\mbox{re}\left\{p\right\}\right\vert}{\pi}.}$$

Since a $-3$ dB attenuation is the same thing as a power scaling by $1/2$ , the 3-dB bandwidth is also called the half-power bandwidth.

It now remains to ``digitize'' the continuous-time resonator and show that relation Eq.(8.7) follows. The most natural mapping of the $s$ plane to the $z$ plane is

$$z = e^{sT},$$

where $T$ is the sampling period. This mapping follows directly from sampling the time axis in the Laplace transform to obtain the z transform. It is also called the impulse invariant transformation [68, pp. 216-219], and for digital poles it is the same as the matched z transformation [68, pp. 224-226]. Applying the matched z transformation to the pole $p$ in the $s$ plane gives the digital pole

$$p_d = R_d e^{j\theta_d} \isdef e^{p T} = e^{(\sigma_p+j\omega_p)T} = e^{\sigma_p T} e^{j\omega_p T}$$

from which we identify

$$R_d = e^{\sigma_p T} = e^{-\pi B T}$$

and the relation between pole radius $R_d$ and analog 3-dB bandwidth $B$ (in Hz) is now shown. Since the mapping $z=e^{sT}$ becomes exact as $T\to 0$ , we have that $B$ is also the 3-dB bandwidth of the digital resonator in the limit as the sampling rate approaches infinity. In practice, it is a good approximate relation whenever the digital pole is close to the unit circle ( $R_d \approx 1$ ).