I have never really understood what the frequency axis meant when we plot the Power Spectral Density(PSD).

Does it correspond to frequency as we get after we take the Fourier Transform of a time domain signal. This doesn't make sense to me because the frequency here(when we do FT of a time domain signal) corresponds to the time variable "t".

To find the PSD we take the Fourier transform of the Autocorrelation function which is a function of either time difference $$\tau = t_2 - t_1$$ i.e. $$R_{xx}(\tau) $$ (for the Wide Sense Stationary case) or $$R_{xx}(t_1,t_2)$$ for other cases.

Here the autocorrelation is a function of either time difference or two time variables. If this is the case then how can we say that the PSD gives the power distributed in the frequency domain?

How do I understand this. If someone could give both a mathematical explanation and an intuitive explanation I would be really grateful.

Thank you.


2 Answers 2


The fact that the frequency variable of a power spectral density (PSD) equals the one of a Fourier transform of a "normal" time-domain signal can be seen more easily by considering the following definition of the PSD of a wide-sense stationary (WSS) random process $x(t)$:

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

Note that $(1)$ is just the limit of the expectation of the scaled squared magnitude of the Fourier transform of a truncated version of $x(t)$. So the frequency variable $\omega$ is just the frequency variable of the (squared magnitude of the) Fourier transform.

The Wiener-Khinchin theorem says that the autocorrelation function and the PSD of a WSS random process form a Fourier transform pair.

This answer shows that the inverse Fourier transform of $(1)$ indeed equals the autocorrelation function of $x(t)$.

  • $\begingroup$ Okay, this definition here makes sense. So they(ACF and PDS) are mathematically related (as fourier transform pairs) but this has no inherent physical meaning? $\endgroup$ Jul 27, 2016 at 17:59
  • $\begingroup$ @AnandKrishnadasNambisan: I'm not sure about the physical meaning of that relationship. In any case, even with deterministic signals you have the same: look at the relation between the squared magnitude of the Fourier transform of a signal, and the inverse transform of that (magnitude) squared Fourier transform. Its independent variable is also a time-shift (the corresponding time function is the deterministic auto-correlation). $\endgroup$
    – Matt L.
    Jul 27, 2016 at 19:47

I have churn over this question for a very long time. I never liked how answers to this question are again presented in mathematical form. Nevertheless, i continued looking at the same problem over and over again and whatever i have understood is represented below.

In this way i understood why frequency axis of PSD still represents the frequency of a time-series waveform.

enter image description here


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