I'm trying to understand how the PSD is calculated. I've looked in a few of my Communication Engineering textbooks but to no avail. I've also looked online. Wikipedia seems to have the best explanation; however, I get lost at the part where they decide to make the CDF (Cumulative Distrubution Function) and then for some reason decide to related that to the autocorrelation function.

I guess what I don't understand is, how does autocorrelation having anything to do with calculating the PSD? I would've thought that the PSD simple be the Fourier Transform of $P(t)$ (where $P(t)$ is the power of the signal with respect to time).

  • $\begingroup$ How do you define $P(t)$? $\endgroup$
    – Phonon
    Commented Mar 9, 2013 at 6:31
  • $\begingroup$ I don't really define it as anything. It's just some power signal. I guess if I had to define it, it'd be $P(t) = v(t) \cdot i(t)$... I guess the point is that the PSD is not $\mathcal{F}\{P(t)\}$ and it has something to do with autocorrelation and I don't get what... $\endgroup$
    – user968243
    Commented Mar 9, 2013 at 6:36
  • $\begingroup$ You can't really define power like that for arbitrary signals. There are no voltage and current concepts. Power in this case is defined as in power of a wave (electromagnetic if you like). So it's $\frac{1}{T} \int_0^Tx^2(t)dt$, and it's a single number, not a time-varying quantity. $\endgroup$
    – Phonon
    Commented Mar 9, 2013 at 6:53
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    $\begingroup$ Read about the Wiener-Khinchin Theorem. You are refusing to understand what Phonon is pointing out to you that the limit that you are calculating is a constant and so its Fourier transform is just an impulse at $f=0$ in the frequency domain. If that floats your boat, go for it but it is not the power spectral density as everyone else understands it. $\endgroup$ Commented Mar 9, 2013 at 13:36
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    $\begingroup$ I have read about that theorem... And I get how it relates the Fourier transform to autocorrelation. And I'm not refusing to understand what Phonon said... I understand exactly what @Phonon said. What I don't understand is why the autocorrelation formula is used and I also don't understand why the fourier transform way is used (to get the PSD you can take the fourier transform, take the magnitude of it, square it etc.)... I have no idea why doing that would give a PSD and I haven't been able to find a decent derivation. $\endgroup$
    – user968243
    Commented Mar 9, 2013 at 14:13

2 Answers 2


You are right, PSD has to do with calculating the Fourier Transform of the power of the signal and guess what.....it does. But first let's look at the mathematical relationship between the PSD and the autocorrelation function.

  1. Notations:

    • Fourier Transform: $$ \mathcal{F}[ x(t)] = X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt $$
    • (Time) Auto-Correlation Function: $$ R(\tau) = x(\tau) * x(-\tau) = \int_{-\infty}^{\infty} x(t)x(t + \tau)dt $$
  2. Let's prove that the Fourier Transform of the Auto-Correlation function does indeed equal to the Power Spectral density of our stochastic signal signal $x(t)$.

$$ \mathcal{F}[ R(\tau)]= \int_{-\infty}^{\infty} R(\tau)e^{-j\omega \tau}d\tau $$ $$ = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x(t)x(t + \tau) e^{-j\omega \tau} dtd\tau $$ $$ = \int_{-\infty}^{\infty}x(t) \underbrace{\int_{-\infty}^{\infty} x(t + \tau) e^{-j\omega \tau} d\tau}_{\mathcal{F}[x(t + \tau) ] = X(\omega)e^{j\omega t}} dt $$ $$ = X(\omega) {\int_{-\infty}^{\infty}x(t)e^{j\omega t} dt} $$

$$ = X(\omega)X^*(\omega)= |X(\omega)|^2 $$

What does it all mean? Note: This explanation is a bit "hacky". But here it goes

The Fourier transform tells us the spectral components of a signal. In our case, the signal is Stochastic; So, trying to calculate the spectral components of the signal will be pointless because, for every realisation of the random process, you will have different expressions for $ \mathcal{F}[x(t)]$.

What if you take the Expected Value of the Fourier transform then? This wouldn't work. Let's take a zero mean signal for example.

$$ \mathbb{E}\{ \mathcal{F}[x(t)] \} = \mathcal{F}[\mathbb{E}\{ x(t) \}] = 0$$

Instead, what if you take the Fourier transform of the square of the signal. $$ \mathbb{E}\{ \mathcal{F}[x^2(t)] \} = \mathcal{F}[\underbrace{\mathbb{E}\{ x^2(t) \}}_{\text{Av. Power of the Signal}}] $$

The autocorrelation function is essentially the $P(t) $ which you were alluding to.


[1] Communications 1, P-L. Dragotti, Imperial College London

[2] White Noise and Estimation, F. Tobar [Unpublished Report]

  • $\begingroup$ Fantastic Explanation! A small calculus question - are you able to interchange the $dt$ and the $d\tau$ inside the double integrals, only because their limits are both from -$\infty$ to +$\infty$? $\endgroup$
    – Spacey
    Commented Mar 12, 2013 at 1:15
  • $\begingroup$ yes that's right. $\endgroup$
    – ssk08
    Commented Mar 12, 2013 at 1:30
  • $\begingroup$ Okay, I think that I kind of get it. I can see how the fourier transform is related to autocorrelation. I don't really understand, though, what the issue is with taking the fourier transform of $x(t)$ or $x^2(t)$. I don't really see why the expected value needs to be taken (I know it averages it, but I don't know why that is necessary) and I don't really understand what you mean by 'for every realisation of the random process, you will have different expressions for'. If you could elaborate a little, that'd be great! Thank you for your time! $\endgroup$
    – user968243
    Commented Mar 12, 2013 at 13:17
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    $\begingroup$ @user968243 As far as the "for every realization" part, think of it this way: Your original signal, lets say length $N$, that you want to find the PSD for, is a random vector. So it is a vector with $N$ components. Now, since this is a random vector, every time you 'roll the dice', you get different values for its components. One possibility might be [3 4 1 9 ...]. Another possibility might be [2.9 4.2 1.1 9.02...]. This is what he means when he says, "For every realization of a random process, (your vector), you get different expressions for" (the fourier transform. Make sense? $\endgroup$
    – Spacey
    Commented Mar 12, 2013 at 14:57
  • $\begingroup$ @Mohammad summed it up perfectly. $\endgroup$
    – ssk08
    Commented Mar 12, 2013 at 15:57

Nice derivation but I think you can do this even easier

Auto correlation $r(t) = x(t)*x(-t)$, it's the convolution of the signal with it's time flipped self.

Convolution in the time domain is multiplication in the frequency domain.

Time flip in the time domain is "complex conjugate" in the frequency domain.

Hence we get $$ R(\omega) = \mathcal{F}\{r(t)\} = \mathcal{F}\{x(t)\}\mathcal{F}\{x(-t)\} = X(\omega) X^*(\omega) = |X(\omega)|^2 = PSD $$

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    $\begingroup$ Isn't the auto-correlation the convolution of the signal with it's complex conjugate, time-flipped self? $\endgroup$
    – Jim Clay
    Commented Mar 12, 2013 at 13:47
  • $\begingroup$ I think he is assuming that the signal is real. $\endgroup$
    – ssk08
    Commented Mar 12, 2013 at 14:45
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    $\begingroup$ @Jim & ssk08: you are both correct, of course. Thanks for cleaning up the equations. $\endgroup$
    – Hilmar
    Commented Mar 12, 2013 at 15:04

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