Free Books

Classic Analog Phase Shifters

Setting $ s=j\omega$ in Eq.$ \,$(8.19) gives the frequency response of the analog-phaser transfer function to be


$\displaystyle H_a(j\omega) \eqsp \frac{j\omega-\omega_b}{j\omega+\omega_b},
$

where the `$ a$' subscript denotes ``analog'' as opposed to ``digital''. The phase response is readily found to be

$\displaystyle \Theta_a(\omega) \eqsp \pi - 2\tan^{-1}\left(\frac{\omega}{\omega_b}\right).
$

Note that the phase is always $ \pi$ at dc ($ \omega=0$), meaning each allpass section inverts at dc. Also, at $ \omega=\infty$ (remember we're talking about analog here), we get a phase of zero. In between, the phase falls from $ \pi$ to 0 as frequency goes from 0 to $ \infty$. In particular, at $ \omega=\omega_b$, the phase has fallen exactly half way, to $ \pi /2$. We will call $ \omega=\omega_b$ the break frequency of the allpass section.9.23 Figure 8.24a shows the phase responses of four first-order analog allpass filters with $ \omega_b$ set to $ 2\pi[100,200,400,800]$. Figure 8.24b shows the resulting normalized amplitude response for the phaser, for $ g=1$ (unity feedfoward gain). The amplitude response has also been normalized by dividing by 2 so that the maximum gain is 1. Since there is an even number (four) of allpass sections, the gain at dc is $ [1+(-1)(-1)(-1)(-1)]/2 = 1$. Put another way, the initial phase of each allpass section at dc is $ \pi$, so that the total allpass-chain phase at dc is $ 4\pi$. As frequency increases, the phase of the allpass chain decreases. When it comes down to $ 3\pi$, the net effect is a sign inversion by the allpass chain, and the phaser has a notch. There will be another notch when the phase falls down to $ \pi$. Thus, four first-order allpass sections give two notches. For each notch in the desired response we must add two new first-order allpass sections.
Figure 8.24: (a) Phase responses of first-order analog allpass sections with break frequencies at 100, 200, 400, and 800 Hz. (b) Corresponding phaser amplitude response.
\includegraphics[width=\twidth]{eps/phaser1a}
From Fig.8.24b, we observe that the first notch is near $ f=100$ Hz. This happens to be the frequency at which the first allpass pole ``breaks,'' i.e., $ \omega=g_1$. Since the phase of a first-order allpass section at its break frequency is $ \pi /2$, the sum of the other three sections must be approximately $ 2\pi + \pi/2$. Equivalently, since the first section has ``given up'' $ \pi /2$ radians of phase at $ \omega=g_1=2\pi100$, the other three allpass sections combined have given up $ \pi /2$ radians as well (with the second section having given up more than the other two). In practical operation, the break frequencies must change dynamically, usually periodically at some rate.
Next Section:
Classic Virtual Analog Phase Shifters
Previous Section:
Matlab Code for Inverse Filtering