I am trying to solve the following question

Somehow I am getting the variance{u(n)} equal to '0' !! This is the case when I take the coefficient 'a' as real. As it is not mentioned in the question I need to find the solution to this question for a complex 'a' also. I would be glad I anybody could help me solve it. I waold also like to clear my concepts about the complex coefficients, as in how does the Z and inverse Z-transform takes place when 'a' is complex. This is a back problem of the second chapter in the book - Adaptive Filters: Theory and Application by Farhang-Boroujeny


Consider the following LTI system with impulse response $h[n]$

$$ \{v[n]\} \longrightarrow \boxed{H(z)} \longrightarrow \{u[n]\} $$

From the analysis of LTI systems with WSS random inputs, the following results deduced. Given the input $v[n]$ which is zero mean white WSS random process, its ACS (Auto-Correlation Sequence) sequence is: $$\phi_{vv}[m] = \sigma_v^2 \delta[m]$$ where $\sigma_v^2$ variance of the input.

Then the output ACS is given by: $$\phi_{uu}[m] = h[m] \star h[-m]^{*} \star \phi_{vv}[m]$$

Similary the input and output PSD (Power Spectral Densities) are related by: $$S_{uu}(w) = H(w) H(w)^{*} S_{vv}(w) = |H(w)|^2 S_{vv}(w) $$ which follows from the relation that PSD of a WSS RP is the DTFT of its ACS; i.e., $$ S_{uu}(w) =\sum_{m=-\infty}^{\infty} \phi_{uu}[m] e^{-j w m} \longleftrightarrow \phi_{uu}[m] = \frac{1}{2\pi} \int_{-\pi}^{\pi} S_{uu}(w) e^{j w m} dw$$

Then the variance of the output is given by the following: $$ \sigma_{uu}^2 = \phi_{uu}[0] = \frac{1}{2\pi} \int_{-\pi}^{\pi} S_{uu}(w)dw$$

Now using the given information $H(z) = \frac{1}{1-a z^{-1}}$ and $\phi_{vv}[m] = 1 \delta[m]$ you can compute the variance in either way:

$$ \sigma_u^2 = \left( h[m] \star h[-m]^{*} \star \delta[m] \right) |_{m=0} $$ or $$ \sigma_u^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |\frac{1}{1-a e^{-jw}}|^2 dw $$

For example chosing the first approach with $h[n] = a^n u[n]$ yields: $$ \begin{align} \sigma_u^2 &= \left( h[m] \star h[-m]^{*} \star \delta[m] \right) |_{m=0} \\ \sigma_u^2 &= \left( h[m] \star h[-m]^{*} \right) |_{m=0} \\ \sigma_u^2 &= \sum_{k=-\infty}^{\infty} h[m-k]h[-k]^{*} |_{m=0} \\ \sigma_u^2 &= \sum_{k=-\infty}^{\infty} h[-k]h[-k]^{*} \\ \sigma_u^2 &= \sum_{k=-\infty}^{\infty} a^{-k} u[-k] (a^{-k})^{*} u[-k]\\ \end{align} $$

$$ \sigma_u^2 = \sum_{k=-\infty}^{0} (a a^{*})^{-k} = \sum_{k=-\infty}^{0} |a|^{-2k} = \sum_{k=0}^{\infty} (|a|^{2})^k = \frac{1}{1-|a|^2}$$

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    $\begingroup$ Thank you very much @Fat32. I really appreciate your efforts. I am trying to figure out every little detail of this. Will let you know for sure if I have trouble understanding anything. $\endgroup$ – Copernicus Sep 15 '17 at 3:56
  • $\begingroup$ @Rohan have you finished your analysis? $\endgroup$ – Fat32 Sep 17 '17 at 21:47
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    $\begingroup$ Yes, and it is so apparent to me now. Thank you very much @Fat32. $\endgroup$ – Copernicus Sep 17 '17 at 22:26
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    $\begingroup$ @Rohan I'm glad I could help. Please note that for the last infinite sum to converge we should have $|a| < 1$, as gven in the question. $\endgroup$ – Fat32 Sep 17 '17 at 22:44
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    $\begingroup$ I was not considering the time reversal property, I took only the delay ($x[n − k] <=> z^−kX(z)$) into account. I dont know why it didnt work!! $\endgroup$ – Copernicus Sep 17 '17 at 22:53

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