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I'm currently working on a problem that involves a Z-Transform. Basically, the essence of the problem is that if: \begin{equation} H\left(z\right) = \sum_{n=0}^{N-1}h\left(n\right)z^{-n} \end{equation} find: \begin{equation} Y\left(z\right)\:=\:H\left(z\right)\cdot H^*\left(\frac{1}{z^*}\right) \end{equation} The asterisk represents a complex conjugate. So, I know that

\begin{equation} H^*\left(\frac{1}{z^*}\right)\:=\:\sum _{n=0}^{N-1}h^*\left(n\right)z^n \end{equation} Am stuck at this part now: \begin{equation} Y\left(z\right)=\left(\sum_{n=0}^{N-1}h\left(n\right)z^{-n}\right)\left(\sum_{n=0}^{N-1}h^*\left(n\right)z^n\right) \end{equation}

I'm stuck at this part and don't know what to do from here. Thank you for your help in advance.

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  • $\begingroup$ You may try treating that product in Z as a convolution in time. $\endgroup$
    – Juancho
    Sep 2 '20 at 15:38
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This may help: the result is the cascade of an FIR filter with coefficients $h(n)$ with the reverse conjugate filter with coefficients $h^*(N-n)$. If we can limit $H(z)$ to be the unit circle (for computing the frequency response), then

$$H(z^*)|_{z=e^{j\omega}} = H(z^{-1})|_{z=e^{j\omega}}$$

since

for $z = e^{j\omega}$, $z^* = e^{-j\omega}$

And therefore

$$H^*(1/z)|_{z=e^{j\omega}}= H^*(z^*)|_{z=e^{j\omega}}$$

And

$$H^*(1/z^*)|_{z=e^{j\omega}}= H^*(z)|_{z=e^{j\omega}}$$

So

$$Y(z)|_{z=e^{j\omega}} = H(z)H^*(z)|_{z=e^{j\omega}} = |H(z)|^2|_{z=e^{j\omega}}$$


Also from the link the OP provided to me in the comments below, it is clear that the problem is dealing with the two sided Z-transform and the full problem has a solution given as

$$H(z)H^*(1/z^*)\Phi_{xx}(z) = \frac{\sigma_v^2}{(1-az^{-1})(1-a^*z)}$$

And we see from the earlier solution in the link that

$$H(z) = \frac{1}{1-az^{-1}}$$

Therefore

$$H^*(z) = \frac{1}{1-a^*(1/z^*)}$$

And

$$H^*(1/z) = \frac{1}{1-a^*(z^*)}$$

And

$$H^*(1/z^*) = \frac{1}{1-a^*z}$$

And

$$H(z)H^*(1/z^*)=\frac{1}{(1-az^{-1})(1-a^*z)}$$

Note confirming my earlier result, when we restrict z to be the unit circle (for determining the magnitude response):

$$H^*(z)|_{z=e^{j\omega}} = \frac{1}{1-a^*(1/z^*)}|_{z=e^{j\omega}}= \frac{1}{1-a^*(1/e^{-j\omega})} = \frac{1}{1-a^*(e^{j\omega})} = \frac{1}{1-a^*z}|_{z=e^{j\omega}}$$

Also to avoid any confusion, earlier in the linked solutions where it says "from 7... we arrive at (2.55) as follows" it gives a result that indicates

$$\Phi_{xx}(z) = \Phi^*_{xx}(1/z^*)$$

But this is not generally true for any z transform, (we cannot conclude that $H(z) = H^*(1/z^*)$ for all z) but it is in this case specific to the autocorrelation function $\phi_{xx}(k)$ since that function is Hermitian:

$$\phi^*_{xx}(k) = \phi_{xx}(-k)$$

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  • $\begingroup$ Thanks for the answer, but we can't assume an ROC. Also, this is not what the answer in the book is. $\endgroup$
    – JordenSH
    Sep 2 '20 at 22:43
  • $\begingroup$ @Raykh what is the answer in the book? $\endgroup$ Sep 2 '20 at 22:46
  • $\begingroup$ Well what I posted is a generic form of the problem, so the answer form would be different but was expecting the answer here to go inline with what the book has. $\endgroup$
    – JordenSH
    Sep 3 '20 at 2:17
  • $\begingroup$ @Raykh Yes could you provide that? $\endgroup$ Sep 3 '20 at 2:19
  • $\begingroup$ Sure thing. Please refer to below at problem 10 after the sentence that say "using 2.80" ebookyab.com/files/… $\endgroup$
    – JordenSH
    Sep 3 '20 at 3:16

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