Given this problem enter image description here

I know that the autcorrelation and power spectrum are fourier pairs, so when taking taking fourier transform of Rxx, one should end up with Sxx

However, when I take the transform of $R_{xx}(m)$, I end up with $S_{xx}(\omega)=\frac{2\alpha }{\alpha ^{2} + \omega ^{2}}$, which also matches up with fourier transform pairs in the book. Here's by hand:

$S_{xx}(\omega )= \int_{-\infty}^{\infty} R_{xx}(m)e^{-i\omega m} dm=\widehat{R}_{xx}(\omega)$

$=\int_{-\infty}^{\infty} R_{xx}(m)e^{-i\omega m} dm=\int_{-\infty}^{\infty} e^{-\alpha |m|}e^{-i\omega m} dm=$ $=\int_{-\infty}^{0} e^{\alpha m}e^{-i\omega m} dm + \int_{0}^{\infty} e^{-\alpha m}e^{-i\omega m} dm=$

$ =\int_{-\infty}^{0} e^{m(\alpha -i\omega)} dm + \int_{0}^{\infty} e^{-m(\alpha +i\omega )} dm=$

$=\frac{1}{(\alpha -i\omega)} - \frac{0}{(\alpha -i\omega)}+\frac{0}{(\alpha +i\omega)}-\frac{1}{(\alpha +i\omega)}$

$=\frac{1}{(\alpha -i\omega)}+\frac{1}{(\alpha +i\omega)}=\frac{\alpha +i\omega+\alpha -i\omega}{(\alpha -i\omega)(\alpha +i\omega)}=\frac{2\alpha}{\alpha^{2} +\omega^{2}} $

Which doesn't look like equation given in the problem, unless I'm missing some identities or something here.

Also, as for problem h), that should be pretty straight forward. Avg. power of $X=R_{xx}(0)$ right?

  • 1
    $\begingroup$ You can show that the given $S_{XX}(w)$ is equal to your answer by Taylor's series expanding the numerator and the denominator terms separately and ignoring second order and higher terms. Also yes, average power is indeed equal to $R_{xx}(0)$. $\endgroup$
    – Sudarsan
    Sep 17, 2013 at 17:46

1 Answer 1


That question was posted quite some time ago! Have you tried the discrete time Fourier transform of $R_{XX}[m]$, instead of the Fourier transform? I believe this is what was meant, there (the text talks about a "sequence").

An "easy" way to solve this, according to me, is to compute the Z-transform, then identify $S_{XX}(\omega) = Z[R_{XX}](z=e^{j\omega})$. The Z-transform is given as: \begin{align} Z[R_{XX}](z) &= \sum_{m=-\infty}^{\infty} R_{XX}[m] z^{-m} \\ &= \sum_m e^{-\alpha |m|} z^{-m} \\ &= \frac{1 - e^{-2\alpha}}{(1-e^{-\alpha}z^{-1})(1 - e^{-\alpha}z)} \end{align}

Hope that still helps!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.