# Phase Contribution of Zeros in Z-plane

I have been studying Proakis, and stumbled upon minimum/maximum/mixed phase concept.

He said that all zeros inside unit circle in Z-plane contributes zero phase change, and all zeros outside unit circle contribute $\pi$ phase change. That is if $\Omega(\omega)$ is phase of the Z-transform, then, $\Omega(\pi) - \Omega(0)=\pi$.

Geometrically this does not make sense to me. Take the following example (The circle in both the pictures is an unit circle). This example is based upon the following transform :

$$H(z) = z^{-1}(z+ 1/2)$$ In the above picture, it is a single Zero system. The Zero is denoted by $Z(=-1/2)$ above. Here the $\Omega$ is the phase angle of the transform. Now say we rotate by more than $\pi/2$. And we get to the following picture, Here $L_2$ is at say $\omega=3\pi/2$ angle. Then the new phase angle as $\omega$ approaches $\pi$ is going to be $\pi$. So from first picture at $\omega=0, \Omega=0$, and $\omega=\pi, \Omega=\pi$. So how come the phase difference $\Omega(\pi)-\Omega(0) = 0$ as claimed by the book for this transform. It also said that all zeros inside the unit circle would lead to zero phase change. But as described above, it does not make sense to me geometrically.

Can someone please guide me on where I am going wrong!

You're right that the angle $\Omega(\omega)$ in your sketch will equal $\pi$ for $\omega=\pi$. However, $\Omega(\omega)$ is not the phase angle of the system. In order to see this, let's write down the frequency response of a system with a single zero at $z_0=re^{j\theta}$:

$$H(e^{j\omega})=1-re^{j\theta}e^{-j\omega}=\frac{e^{j\omega}-re^{j\theta}}{e^{j\omega}}\tag{1}$$

From $(1)$, the phase angle is given by

$$\angle\{H(e^{j\omega})\}=\angle\{e^{j\omega}-re^{j\theta}\}-\angle\{e^{j\omega}\}=\Omega-\omega\tag{2}$$

So you indeed need the angle $\Omega$ as defined in your figures, but you also need to subtract the angular frequency $\omega$.

Let's assume $\theta=\pi$, as in your example. For $\omega=\pi$ (and $r<1$) you get $\Omega=\pi$, and, consequently, a phase of $\Omega-\omega=0$. Note that if $r>1$, you'll get $\Omega=0$ at $\omega=\pi$, and, consequently, a phase angle of $(-)\pi$.

• Excellent answer! I missed the denominator part! – user3001408 Nov 17 '16 at 6:54