I am new to the study of time series. Recently I have asked a question about the covariance of real and imaginary part of a real(in time domain) stochastic time series and I have received an answer for it. The problem is that for continuous time series the variance of each point in frequency domain is infinite. I've been told there that its the reason that they use power spectrum. Now the confusion comes from the fact that I don't know whether power spectrum is time averaged or not? More precisely as wiki says: $$S_{xx}(\omega)=\lim\limits_{T\to \infty}\mathbf{E} \left[ | \hat{x}_T(\omega) |^2 \right] = \lim\limits_{T\to \infty}\mathbf{E} \left[ \frac{1}{T} \int\limits_0^T x^*(t) e^{i\omega t}\, dt \int\limits_0^T x(t') e^{-i\omega t'}\, dt' \right] = \lim\limits_{T\to \infty}\frac{1}{T} \int\limits_0^T \int\limits_0^T \mathbf{E}\left[x^*(t) x(t')\right] e^{i\omega (t-t')}\, dt\, dt$$ OR $$S_{xx}(\omega)=\mathbf{E} \left[ | \hat{x}_T(\omega) |^2 \right] = \mathbf{E} \left[ \int\limits_0^\infty x^*(t) e^{i\omega t}\, dt \int\limits_0^\infty x(t') e^{-i\omega t'}\, dt' \right] = \int\limits_0^\infty \int\limits_0^\infty \mathbf{E}\left[x^*(t) x(t')\right] e^{i\omega (t-t')}\, dt\, dt$$

For example in Wiener-Khintchine theorem as far as I can see there is no time averaging: $$r_{xx} (\tau) = \int_{-\infty}^\infty S_{xx}(f) e^{2\pi i\tau f} df$$

And is there any difference when the signal is discrete?


1 Answer 1


The first definition of the power spectrum is the correct one (at least for causal signals, otherwise you need symmetrical integration limits). The second one assumes that the Fourier transform of $x(t)$ exists, which is generally not the case for random signals. The fact that the power spectrum and the autocorrelation function form a Fourier transform pair is no contradiction with the first definition of the power spectrum. This can be shown as follows:

$$S_x(\omega)=\lim_{T\rightarrow\infty}E\left\{ \frac{1}{T}\left| \int_{-T/2}^{T/2}x(t)e^{-j\omega t}dt \right|^2 \right\}=\\= \lim_{T\rightarrow\infty}E\left\{ \frac{1}{T} \int_{-T/2}^{T/2}x^*(t)e^{j\omega t}dt \int_{-T/2}^{T/2}x(t')e^{-j\omega t'}dt' \right\}$$

The inverse Fourier transform of the power spectrum $S_x(\omega)$ should equal the autocorrelation function $R_x(\tau)$:

$$\mathcal{F}^{-1}\{S_x(\omega)\}=\frac{1}{2\pi}\int_{-\infty}^{\infty}S_x(\omega)e^{j\omega \tau}d\omega=\\= \lim_{T\rightarrow\infty}E\left\{ \frac{1}{2\pi T} \int_{-\infty}^{\infty}\int_{-T/2}^{T/2}x^*(t)e^{j\omega t}dt \int_{-T/2}^{T/2}x(t')e^{-j\omega t'}dt'e^{j\omega \tau}d\omega \right\}=\\= \lim_{T\rightarrow\infty}E\left\{ \frac{1}{T} \int_{-T/2}^{T/2}x^*(t) \int_{-T/2}^{T/2}x(t')\left[\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{j\omega (t+ \tau-t')}d\omega\right]dtdt' \right\} $$ The expression in brackets equals $\delta(t+\tau-t')$, which gives

$$\mathcal{F}^{-1}\{S_x(\omega)\}= \lim_{T\rightarrow\infty}E\left\{ \frac{1}{T} \int_{-T/2}^{T/2}x^*(t) x(t+\tau)dt \right\}=\\ =\lim_{T\rightarrow\infty} \frac{1}{T} \int_{-T/2}^{T/2}E\{x^*(t) x(t+\tau)\}dt=\\ = \lim_{T\rightarrow\infty} \frac{1}{T} \int_{-T/2}^{T/2}R_x(\tau)dt= R_x(\tau)\lim_{T\rightarrow\infty} \frac{1}{T} \int_{-T/2}^{T/2}dt=R_x(\tau) $$

which shows that the chosen definition of the power spectrum is compatible with the fact that the power spectrum and the autocorrelation function are Fourier transform pairs. A similar derivation can be done for discrete-time signals.

  • $\begingroup$ I know this is an old thread but I wonder if you can answer a query for me. On your second to last equation, the expectation is brought inside the integral. I have been trying to read a little about ergodicity, and I have seen people say that it means you can replace the ensemble average with the time average - is that hidden in here somewhere? I assume this derivation can be done similarly for purely deterministic signals? Thank you! $\endgroup$
    – teeeeee
    Commented Apr 4, 2020 at 12:44
  • 1
    $\begingroup$ @teeeeee: Shifting the expectation operator to the integrand just means interchanging the order of integration (note that the expectation operator is defined by an integral), so this has nothing to do with ergodicity, just with conditions on the interchangeability of integrals, which we usually assume to be satisfied. $\endgroup$
    – Matt L.
    Commented Apr 4, 2020 at 14:03
  • $\begingroup$ Okay, thanks for the clarification. $\endgroup$
    – teeeeee
    Commented Apr 4, 2020 at 14:04
  • $\begingroup$ Sorry, I have a second question! When you write expectation $E\{\}$ (as is often done when dealing with power spectral density of stochastic signals), am I to understand that you refer to ensemble average, i.e the expected value that would be found at some time $t$ if you averaged over all realisations? Or are we speaking about temporal average - the average over time in a single realisation? Does it make sense to talk about expectation in the latter case, or is expectation always reserved for ensemble averaging? $\endgroup$
    – teeeeee
    Commented Apr 7, 2020 at 9:56
  • 2
    $\begingroup$ @teeeeee: It's an ensemble average. It's just that under certain conditions it can be exchanged for a time average, but by definition it's always an ensemble average. $\endgroup$
    – Matt L.
    Commented Apr 7, 2020 at 10:00

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