10
$\begingroup$

I read the following on Wikipedia:

Power spectral density:

The above definition of energy spectral density is most suitable for transients, i.e., pulse-like signals, for which the Fourier transforms of the signals exist. For continued signals that describe, for example, stationary physical processes, it makes more sense to define a power spectral density (PSD), which describes how the power of a signal or time series is distributed over the different frequencies, as in the simple example given previously.

I don't quite understand that paragraph. The first part says that "for some signals .. the Fourier transform does not exist".

  • For which signals (in the context that we are discussing) does the Fourier transform not exist, and we therefore need to resort to the PSD rather than using the energy spectral density?

  • When obtaining the power spectral density, why can't we compute it directly? Why do we need to estimate it?

  • Finally, on this topic, I have read about methods that use Kayser-windows when computing the PSD over time. What is the purpose of these windows in PSD estimation?

$\endgroup$
  • $\begingroup$ A short answer to one of your questions: for a deterministic signal $x(t)$, you can compute its power spectral density. However, the power spectral density is also defined for wide-sense stationary random processes. In this context, the PSD is defined as the Fourier transform of the process's autocorrelation function. In that scenario, you typically don't know the exact autocorrelation function of a particular random process that you might be observing, so you try to estimate its PSD from your observations. $\endgroup$ – Jason R Jul 31 '13 at 0:04
  • 5
    $\begingroup$ A deterministic signal $x(t)$ for which $$\lim_{T\to \infty}\int_{-T}^T |x(t)|^2\,\mathrm dt$$ exists is called a (finite) energy signal and its Fourier transform exists. But if the limit does not exist, the Fourier transform need not exist in the sense that $\int_{-\infty}^\infty x(t)e^{-j2\pi ft}\,\mathrm dt$ is a divergent integral. If $$\lim_{T\to \infty}\frac{1}{2T}\int_{-T}^T |x(t)|^2\,\mathrm dt$$ exists, the signal is called a power signal and its Fourier transform exists in a generalized sense (meaning that impulses are usually involved). $\endgroup$ – Dilip Sarwate Jul 31 '13 at 1:42
2
$\begingroup$

Random process is never ending, non-periodic phenomenon, so taking Fourier transform of its realizations makes no sense, not possible either. However if random process is stationary, then it is for sure that it has some finite power over some band of frequencies. Now, here the question arises that how to compute the power of this stationary random process, (fourier tranform is not possible to be taken directly)? So, what to do? we find the auto-correlation function of the given random process, whose fourier transform always exist. Finally, we take the fourier tranform of this autocorrelation function to get the power spectral density of the given stationary process.

If you integrate the power spectral density of a given stationary process over the interval from -$\infty$ to $\infty$ you ll get the total power contained in the given random process.

$\endgroup$
  • $\begingroup$ When you said: "However if random process is stationary, then it is for sure that it has some finite power over some band of frequencies."- why is that? And does it necessarily have to be stationary to have finite power over some band of frequencies? $\endgroup$ – Amelio Vazquez-Reina Oct 27 '13 at 2:34
  • $\begingroup$ Staionary processes have always finite mean and finite variance. It means that staionary process has always finite power. Since, power is finite, this means that power spectral density of staionary process is finite over some band of frequencies. ( frequency band may be infinite). $\endgroup$ – kaka Oct 27 '13 at 7:56
  • 2
    $\begingroup$ Staionary processes have always finite mean and finite variance. It means that staionary process has always finite power. This is incorrect. See the second paragraph of this answer for a counter-example. $\endgroup$ – Dilip Sarwate Oct 28 '13 at 2:12

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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