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Let $G(z)$ be a real coefficient stable all-pass transfer-function with degree greater than zero. Then it can be shown that $|G(z)| < 1$ for $|z| > 1$: Given that the poles occur in complex conjugate pairs, one way of establishing this property is as a natural generalization of the case when

$$G_1(z) = \frac{1 − \alpha^*z}{z − \alpha}$$

where the superscript $∗$ denotes the complex conjugate operation. By considering $1 − |G_1(z)|^2$ or otherwise show that $G_1(z)$ has this property, i.e., $|G_1(z)| < 1$ for $|z| > 1$ .

I try to demonstrate $|G(z)|^2 = G(z)G(z^{-1}) > 1$, but i don't know how to deal with the complex conjugate $\alpha$ and $\alpha^*$. or my solution is totally wrong?

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  • $\begingroup$ would you like me to show you how to present the question that has a lotta specific math in it, using $\LaTeX$ markup for the math? $\endgroup$ – robert bristow-johnson Jan 10 '17 at 3:22
  • $\begingroup$ Actually, $|G(z)|^2 = G(z)G^*(z) \neq G(z)G(z^{-1})$. $\endgroup$ – Tendero Jan 10 '17 at 3:48
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First, observe that it's sufficient to show the property for a first-order all-pass filter, because any higher order all-pass filter can be written as a cascade of first-order all-pass sections. And if

$$|G_i(z)|<1\quad\text{for}\quad |z|>1$$

then also

$$\left|\prod_iG_i(z)\right|<1\quad\text{for}\quad |z|>1$$

must be satisfied.

Now derive an expression for the squared magnitude of the first-order all-pass transfer function $G_1(z)$:

$$|G_1(z)|^2=\frac{1-2\text{Re}\{\alpha^*z\}+|\alpha|^2|z|^2}{|z|^2-2\text{Re}\{\alpha^*z\}+|\alpha|^2}\tag{1}$$

Using $(1)$, the inequality $|G_1(z)|^2<1$ can be written as

$$\begin{align}1-2\text{Re}\{\alpha^*z\}+|\alpha|^2|z|^2&<|z|^2-2\text{Re}\{\alpha^*z\}+|\alpha|^2\\ 1+|\alpha|^2|z|^2&<|z|^2+|\alpha|^2\\0&<(|z|^2-1)(1-|\alpha|^2)\tag{2}\end{align}$$

Note that if the all-pass filter is stable, we must have $|\alpha|<1$, because $\alpha$ is the pole location, and a causal and stable filter must have all its poles inside the unit circle of the complex plane. Consequently, the inequality $(2)$ is satisfied for $|z|>1$ because both factors on the right-hand side are positive.

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  • $\begingroup$ Thanks a lot, after your help i understand that how to express the $(α*z)^2$. it is really an important step for this Q. I just need the Real value when unfold the squared equation with complex value. With appreciation. $\endgroup$ – Haoming Li Jan 11 '17 at 10:41
  • $\begingroup$ We just need the Real part of the $(α^*z)$ cause it is the magnitude value of the TF? $\endgroup$ – Haoming Li Jan 11 '17 at 10:48
  • $\begingroup$ @HaomingLi: Yes, the general formula for $|u+v|^2$ with complex $u$ and $v$ is: $|u+v|^2=|u|^2+2\text{Re}\{uv^*\}+|v|^2$ $\endgroup$ – Matt L. Jan 11 '17 at 11:43

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