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I have been staring at this problem for a week now... Suppose $H(e^{j \omega})$ is the frequency response of a stable and causal minimum-phase discrete-time system with $h[0]=1$ ($h[n]$ is the impulse response of the system). Show that:

$$ \int\limits_{-\pi}^{\pi} \ln \big| H(e^{j \omega})\big|^2 \, \mathrm{d}\omega = 0 $$

So:

  • Minimum phase means poles and zeros are inside the unit circle
  • Stable and causal is redundant since the system is minimum phase
  • What is conceptually meant by $h[0] = 1$?
  • What is the significance of $\ln$(magnitude of freq response)?
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    $\begingroup$ hay, want me to show you how to do $\LaTeX$ math in stack exchange? $\endgroup$ – robert bristow-johnson May 11 '18 at 7:44
  • $\begingroup$ Please! How would you represent that integral? $\endgroup$ – user35668 May 11 '18 at 8:01
  • $\begingroup$ is the question saying in advance that it's minimum phase? or is that something you have to derive somehow? $\endgroup$ – robert bristow-johnson May 11 '18 at 8:03
  • $\begingroup$ It is known to be minimum phase. I just need to prove the relationship shown given that the system is minimum phase and h[0] = 1. $\endgroup$ – user35668 May 11 '18 at 8:06
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The transfer function of a discrete-time minimum phase system described by a linear difference equation with constant coefficients can be written as

$$H(z)=h[0]\cdot\frac{\prod_{k=1}^M(1-c_kz^{-1})}{\prod_{k=1}^N(1-d_kz^{-1})},\qquad |c_k|\le 1,\quad |d_k|<1\tag{1}$$

where $h[0]$ is the first sample of the corresponding impulse response, and $c_k$ and $d_k$ are the locations of the zeros and poles of $H(z)$, respectively. Note that factoring out $h[0]$ makes sure that the numerator and denominator polynomials are monic (i.e., have a leading coefficient equal to $1$).

From $(1)$, the squared magnitude of $H(z)$ is

$$|H(z)|^2=h^2[0]\cdot\frac{\prod_{k=1}^M|1-c_kz^{-1}|^2}{\prod_{k=1}^N|1-d_kz^{-1}|^2}\tag{2}$$

and its logarithm is

$$\log |H(z)|^2=\log\left(h^2[0]\right) +\sum_{k=1}^M\log|1-c_kz^{-1}|^2 - \sum_{k=1}^N\log|1-d_kz^{-1}|^2\tag{3}$$

Evaluating on the unit circle $z=e^{j\omega}$ and integrating over the interval $[-\pi,\pi]$ gives

$$\int_{-\pi}^{\pi}\log |H(e^{j\omega})|^2d\omega=2\pi\log\left(h^2[0]\right) +\\+\sum_{k=1}^M\int_{-\pi}^{\pi}\log|1-c_ke^{-j\omega}|^2d\omega - \sum_{k=1}^N\int_{-\pi}^{\pi}\log|1-d_ke^{-j\omega}|^2d\omega\tag{4}$$

The integrals on the right-hand side of $(4)$ are of the form

$$\int_{-\pi}^{\pi}\log|1-ae^{j\theta}e^{-j\omega}|^2d\omega\tag{6}$$

with real-valued $a$ and $0\le a\le 1$. With

$$|1-ae^{j\theta}e^{-j\omega}|^2=1+a^2-2a\cos(\omega-\theta)\tag{7}$$

the integral $(6)$ can be written as

$$\int_{-\pi}^{\pi}\log\left(1+a^2-2a\cos(\omega-\theta)\right)d\omega\tag{8}$$

Since the integrand is $2\pi$-periodic and we're integrating over one period, the integral is independent of the angle $\theta$, which finally gives

$$\int_{-\pi}^{\pi}\log\left(1+a^2-2a\cos(\omega)\right)d\omega\tag{9}$$

As shown in the answers to this question on math.SE, the integral $(9)$ equals zero if $0\le a\le 1$, i.e., if $H(z)$ is minimum phase. This means that if $H(z)$ is a minimum phase system, we get from $(4)$

$$\int_{-\pi}^{\pi}\log |H(e^{j\omega})|^2d\omega=2\pi\log\left(h^2[0]\right) \tag{10}$$

and with $h[0]=1$ (which is just an appropriate scaling) we finally arrive at

$$\int_{-\pi}^{\pi}\log |H(e^{j\omega})|^2d\omega=0 \tag{11}$$

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  • $\begingroup$ Thank you, I can stop NOW writing a more tedious answer. Only one detail: can a transfer function of a discrete-time minimum phase system always be written as a rational function? I thought the equivalence was not set yet $\endgroup$ – Laurent Duval May 11 '18 at 10:50
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    $\begingroup$ @LaurentDuval: You're right that in theory we could also have non-rational transfer functions, but the result remains valid also in the general case. It would be valuable if you could show that in an answer :) $\endgroup$ – Matt L. May 11 '18 at 11:09
  • $\begingroup$ Ok, that's what I remembered. There are a few subtleties on the integrability, if my memories are accurate, thank you for the challenge. $\endgroup$ – Laurent Duval May 11 '18 at 11:15

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