0
\$\begingroup\$

I always encounter random noise frequency domain representation as PSD(not FFT), something like in the below plot:

enter image description here

Without diving into math too much, practically speaking can we always use FFT instead of PSD to characterize a random signal? If so, what is the reason to use on to the other method in practice?

It even gets a bit more complicated because most of the time the signal can have both random and periodic components. Imagine I measure the constant pressure flow with a transducer's analog output. But those signal I sample will have both periodic and aperiodic components and random noise etc. In such case, if we have sampled the signal with enough sampling rate, what method between FFT and PSD would be preferred?

I before studied some Fourier series and transform and used many times FFT functions on MATLAB or Python for freq. domain view of a signal. Why would one need PSD if FFT is enough to for all types of signals?

(Digital scopes, for instance, show real-time FFT of a signal but not PSD. So not always PSD and not always FFT is used. How to decide which one to go for?)

\$\endgroup\$
8
  • 1
    \$\begingroup\$ The FFT is an algorithm, but PSD is a property of a stochastic process or signal, so "PSD, not FFT" makes no sense. What exactly do you calculate to get what you're calling "PSD"? There's many ways of estimating a PSD, and some use an FFT, others don't. \$\endgroup\$ Jul 21, 2019 at 19:02
  • \$\begingroup\$ Is PSD used more often for random signals? That was the main part of the questions' motivation. I need a practical example why using PSD us better than FFT for random signals. \$\endgroup\$
    – GNZ
    Jul 21, 2019 at 19:06
  • 1
    \$\begingroup\$ Oh so when calculating PSD we already use FFT algorithm(in programmin tools like MATLAB)? Is that what you mean? \$\endgroup\$
    – GNZ
    Jul 21, 2019 at 19:07
  • \$\begingroup\$ I think I understand what you mean we basically use FFT algorithm and then make some normalization and some math on the FFT results and then we call that PSD? \$\endgroup\$
    – GNZ
    Jul 21, 2019 at 19:09
  • \$\begingroup\$ no. As said, the PSD (power spectral density) is a property of a random process or signal. You can estimate it, as such. There's many methods of estimating it. So "PSD is better than FFT" makes no sense, because that's like "transportation is better than wheels": PSD and FFT are in two different categories and can't be compared. It's really not clear what you mean with PSD, because you certainly don't mean the abstract property, but some implementation of an estimation of that property. \$\endgroup\$ Jul 21, 2019 at 19:14

2 Answers 2

1
\$\begingroup\$

As Marcus Müller said in the comments, PSD is an output (an "answer"), while FFT is a calculation ("a way of getting an answer"). There are many ways to calculate the PSD, and they are all "estimates" since your input signal is noisy and uncertain.

More importantly, if you take the FFT of your raw signal, then you will get the amplitude spectral density. What you really want is the power spectral density -- the PSD. The most basic way to estimate a PSD is to simply take the FFT of the autocorrelation function of your signal. But that PSD estimate will be very noisy and uncertain (since your input signal is noisy and uncertain). If you add more sample points you just end up with another noisy and uncertain estimate, but one that has much higher frequency resolution.

There are a lot of ways to quiet down this PSD estimate, but spectral analysis of noisy signals is a tremendously involved subject, with entire books and graduate-level engineering courses devoted to it. You can find a good introduction in this tutorial on dsprelated.com. If you Google the phrase "spectral estimation of noisy signals" you will find a bunch of other resources including entire books. (I think they may be violating copyright law by putting those online, so I'm not sharing the link here.)

\$\endgroup\$
5
  • \$\begingroup\$ Thanks, Im wondering why in practice sometimes FFT result (amplitude spectral density) and why sometimes PSD is used. \$\endgroup\$
    – GNZ
    Jul 21, 2019 at 19:47
  • \$\begingroup\$ For instance, I came across this definition: "PSD is the power spectrum value normalized to the FFT's resolution bandwidth. It's unit of measure is dBm/Hz and it represents the power per unit bandwidth. PSD is useful for measuring broadband phenomena such as noise. The magnitude format shows the spectral magnitude in linear units which the oscilloscope is measuring like Volts or Amperes." It says PSD is useful for measuring broadband phenomena such as noise, but digital storage scopes have FFT function not PSD. I was wondering why. (?) \$\endgroup\$
    – GNZ
    Jul 21, 2019 at 19:47
  • \$\begingroup\$ I can see this is a very mathematics involved hard topic. I was just looking for practical aspects. \$\endgroup\$
    – GNZ
    Jul 21, 2019 at 19:52
  • \$\begingroup\$ Genzo, for the fourth time: You can't say "PSD is used". A PSD estimate is the result, whereas an FFT can be a method involved in getting that result. The definition your citing is simply wrong. \$\endgroup\$ Jul 21, 2019 at 20:12
  • \$\begingroup\$ @Genzo I think maybe your first comment here cuts to the heart of the problem. The result of "an FFT" is not the amplitude spectral density. The result of "an FFT of a voltage signal" is the amplitude spectral density. And the result of "an FFT of the autocorrelation of a voltage signal" is (an estimate of) the power spectral density. The FFT itself is just a mathematical operation, and the meaning of an FFT's output depends on what its input was. \$\endgroup\$
    – Mr. Snrub
    Jul 21, 2019 at 22:15
0
\$\begingroup\$

If you happen to know Parseval's rule for energy spectrum (=the square of the Fourier transform amplitudes calculated in each frequency point separately) and total energy of limited duration signal, you can calculate PSD for a signal sample set by dividing the energy spectrum by the duration of the signal. This really presents the signal which is stored in the memory except the phase information is totally lost.

Do not expect the PSD calculated from one sample set somehow reliably presents anything else than just that sample set. Those who use PSD want to know how accurately the calculated PSD present the output of the same noise source in the future. They use PSD in statistics based decision making or in designing that decision making. They are NOT interested one sample set of that noise. The sample set itself, or equivalently its FFT, shows that thing much better than PSD.

Unfortunately I cannot present any rule of thumb for validity estimations. That needs the theory of estimating the characteristics of stochastic processes. One thing is quite sure: You must observe a noise signal quite long time to get also in the future useful PSD estimate with the same frequency resolution as you get in a blink of eye for the FFT of one sample set of a signal.

Real time audio analyzers can well show PSD - it's the square of the amplitude spectrum and in dB scale it's the same, only scaled. But the calculation period is very long - it's a sliding window in time - and the shown resolution is low when compared to the internal FFT frequency resolution.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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