1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ It is only reasonable to put an expectation on a stochastic process. But $|x(\omega)|^2$ is a process of what? $\endgroup$ – Nathan Explosion Dec 31 '19 at 13:12
1
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Nathan Explosion Dec 31 '19 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. $\endgroup$ – Matt L. Dec 31 '19 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. $\endgroup$ – Nathan Explosion Dec 31 '19 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. $\endgroup$ – Matt L. Dec 31 '19 at 12:28
  • $\begingroup$ thx, but still a bit confused. Do you mean |x(\omega)|^2 is a stochastic process? $\endgroup$ – Nathan Explosion Dec 31 '19 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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