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I'm a bit confused with the definition of the power spectral density (PSD). From Wiki https://en.wikipedia.org/wiki/Spectral_density , I found the definition is:

$$ S_{xx}(\omega) = \lim_{T\rightarrow \infty}\mathbb{E}[|x(\omega)|^2], $$ where $x(\omega)$ is the Fourier transform of the process $x(t)$.

I'm really confuses with that expectation $\mathbb{E}$ in the above equation. The expectation is taken w.r.t. $x(t)$, but $|x(\omega)|^2$ is a function of angular $\omega$. There is no $x(t)$ in the integrand at all, thus $$ \mathbb{E}[|x(\omega)|^2] = |x(\omega)|^2. $$ What is the point of this expectation?

However, in the other hand, $$ \mathbb{E} \left[ \left | x(\omega) \right |^2 \right] = \mathbb{E} \left[ \frac{1}{T} \int_0^T x^*(t) e^{i\omega t}\, dt \int_0^T x(t') e^{-i\omega t'}\, dt' \right] = \frac{1}{T} \int_0^T \int_0^T \mathbb{E}\left[x^*(t) x(t')\right] e^{i\omega (t-t')}\, dt\, dt' \neq |x(\omega)|^2 $$ the PSD is the Fourier transform of the cross-covariance (auto-correlation) of the process. This exectation is indeed needed.

Where am I wrong? Can't those integrals switch orders?

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  • $\begingroup$ It is only reasonable to put an expectation on a stochastic process. But $|x(\omega)|^2$ is a process of what? $\endgroup$ Commented Dec 31, 2019 at 13:12

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I think a better definition of the power spectrum is the following:

The power spectrum of $x(t)$ is the Fourier transform of the autocorrelation function of $x(t)$, where $x(t)$ can be either a deterministic power signal, or a wide-sense stationary (WSS) random process. The definition of the autocorrelation function depends on the model for $x(t)$.

If $x(t)$ is modeled as a WSS random process, then the autocorrelation function is defined by

$$R_x(\tau)=E\big\{x^*(t)x(t+\tau)\big\}\tag{1}$$

For deterministic power signals, the autocorrelation function is given by

$$R_x(\tau)=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}x^*(t)x(t+\tau)dt\tag{2}$$

In this answer it is shown that the following definition of the power spectrum for a WSS random process $x(t)$

$$S_x(\omega)=\lim_{T\rightarrow\infty}E\left\{ \frac{1}{T}\left| \int_{-T/2}^{T/2}x(t)e^{-j\omega t}dt \right|^2 \right\}\tag{3}$$

is equivalent to the definition of the power spectrum as the Fourier transform of $(1)$.

For deterministic power signals, the corresponding definition of the power spectrum is

$$S_x(\omega)=\lim_{T\rightarrow\infty}\frac{1}{T}\left| \int_{-T/2}^{T/2}x(t)e^{-j\omega t}dt \right|^2 \tag{4}$$

which can also be shown to be equivalent to the Fourier transform of $(2)$.

The definitions of autocorrelation and power spectrum of deterministic power signals is described in Chapter 12 of

Papoulis, A., The Fourier Integral and its Applications, McGraw Hill, 1962.

A good reference on random processes and the corresponding definitions of autocorrelation and power spectra is

Papoulis, A. and S.U. Pillai, Probability, random variables, and stochastic processes, Boston: McGraw-Hill, 2002.

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  • $\begingroup$ Yes, I know it make sense because it leads to the Fourier transform of the autocorrelation function. The proof is also sketched in the Wiki too. But my question is: observing that $|x(\omega)|^2$ is a function of $\omega$, and $T$, right?, If we expand this first, how to take expectation then? $\endgroup$ Commented Dec 31, 2019 at 7:46
  • $\begingroup$ @NathanExplosion: Doesn't the (truncated) Fourier transform of $x(t)$ quite clearly depend on $x(t)$? That's what is shown in Eq. (1) of my answer. $\endgroup$ Commented Dec 31, 2019 at 7:59
  • $\begingroup$ How? if you integrate out $t$ in $\int_{-T/2}^{T/2}x(t)e^{-j\omega t}dt$, then there is only $\omega$ and $T$ left. $\endgroup$ Commented Dec 31, 2019 at 8:02
  • $\begingroup$ @NathanExplosion: In that formulation, $x(t)$ is modeled as a random process, so you don't have $x(t)$ as a single realization for which you can compute that integral. You basically need an average of integrals over all possible realizations of $x(t)$, that's what the expectation operator indicates. In order to explicitly do that, you need to interchange the expectation operator with the integral. $\endgroup$ Commented Dec 31, 2019 at 12:28
  • $\begingroup$ thx, but still a bit confused. Do you mean |x(\omega)|^2 is a stochastic process? $\endgroup$ Commented Dec 31, 2019 at 12:45

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