Classic Analog Phase Shifters

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

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

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

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

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., $\omega_b=g_1$ . 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 $\omega_b=g_1=2\pi100$ , 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.