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Let's say we have discrete-time stationary random signals with Gaussian PDF of mean value 0 and variance 1, whose individual signal values are uncorrelated.

For such a signal, how can we determine ACF and the PSD?

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assuming ergodicity...

ACF:

$$\begin{align} R_x[k] &= \lim_{N \to \infty} \tfrac{1}{2N+1} \sum\limits_{n=-N}^{N} x[n] \, x[n+k] \\ &= \mathbb{E}\Big\{ x[n] \, x[n+k] \Big\} \\ &= \sigma_x^2 \ \delta[k] \\ &= 1\ \delta[k] \\ \end{align}$$

PSD:

$$ S_x(e^{j\omega}) = \sum\limits_{n=-\infty}^{\infty} R_x[n] \, e^{-j \omega n} $$

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  • $\begingroup$ Why assuming ergodicity? $\endgroup$ – Shady Nov 27 '18 at 20:11
  • $\begingroup$ i guess it's not an assumption, but an implication of a stationary process. it means that the averaging done in time (which is the top equation) can be replaced with a probabilistic averaging of the same quantity (which is the expectation). $\endgroup$ – robert bristow-johnson Nov 27 '18 at 20:14
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You have described a discrete-time Gaussian white noise to a T.

For a white stationary Gaussian random process $n[k]$, the autocorrelation $R_n[k]$ is: $$ R_n[k] = \mathbb E \left( n[m]n[m+k] \right) = \sigma^2 \delta[k], $$ where $\sigma^2$ is the variance.

The power spectral density $P(\omega)$ is just the DTFT of the autocorrelation, which is: $$ P(\omega) = \sum_{k=-\infty}^\infty R_n[k] e^{j\omega k} = \sigma^2 $$ i.e., it is the same for all frequencies ("white noise").

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    $\begingroup$ @robertbristow-johnson That's assuming it is interpreted as a continuous-time signal. In the DTFT domain it is not band-limited because the DTFT is $2\pi$-periodic $\endgroup$ – Robert L. Nov 27 '18 at 20:03
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    $\begingroup$ @robertbristow-johnson The assumption of integrating from $-\infty$ to $\infty$ to compute power is not contained or implied anywhere in this answer $\endgroup$ – Robert L. Nov 27 '18 at 20:08
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    $\begingroup$ @robertbristow-johnson I think that's a matter of taste. I prefer to think of the discrete-time word as being abstract, self-contained, and pure, independent of the ugly, dirty, cruel, analog world $\endgroup$ – Robert L. Nov 27 '18 at 20:12
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    $\begingroup$ okay. but unless this remains in the mind of the computer, eventually that white noise is coming out into the ugly, dirty, cruel, analog world. and it won't be white anymore, because white noise in the ugly, dirty, cruel, analog world has infinite power. $\endgroup$ – robert bristow-johnson Nov 27 '18 at 20:17
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    $\begingroup$ @robertbristow-johnson fair, but homework-esque problems like this usually stay inside the safety of computerland :) $\endgroup$ – Robert L. Nov 27 '18 at 20:19

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