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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.
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.
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Matlab Code for Inverse Filtering