6
$\begingroup$

What's the difference between these? Both are measurements of some form of signal power, but surely there's some difference between the power they are measuring?

$\endgroup$
  • $\begingroup$ Can you give more detail about what you don't understand? Please edit your question to add more details. $\endgroup$ – Peter K. Jul 22 '15 at 8:26
7
$\begingroup$

The fast Fourier transform ($\textrm{FFT}$) algorithms are fast algorithms for computing the discrete Fourier transform ($\textrm{DFT}$). This is achieved by successive decomposition of the $N$-point $\textrm{DFT}$ into smaller-block $\textrm{DFT}$, and taking advantage of periodicity and symmetry.

Now, the $N$-point $\textrm{DFT}$ of a sequence $\{x[0], x[1],\cdots, x[N-1]\}$ is: \begin{equation} X(f_k) = \displaystyle \sum_{n = 0}^{N - 1} x[n]\exp\left(-j2\pi f_kn\right) \tag{1} \end{equation} For $f_k = k/N$ and $k = 0, 1, \cdots, N - 1$. And the $\textrm{FFT}$ magnitude at bin $k$ is the $\textrm{DFT}$magnitude at bin $k$. For a given $N$ that is:

\begin{equation} \left|X(f_k)\right| = \left|X\left(\frac{k}{N}\right)\right| = \left|X(k)\right| = \displaystyle \left|\sum_{n = 0}^{N - 1} x[n]\exp\left(-j2\pi nk/N\right)\right| \tag{2} \end{equation}

You talk about power spectral estimation when the signals being analyzed are characterized as random processes. With random fluctuations in such signals, statistical characteristics and average characteristics are normally adopted. For a wide sense stationary $(\textrm{WSS})$ discrete random process, the PSD is defined as:

\begin{equation} P(f) = \displaystyle \sum_{m = -\infty}^{\infty} r_{xx}[m]\exp\left(-j2\pi fm\right) \tag{3} \end{equation}

For reasons you can find in this answer, you see that the squared magnitude of the signal's $\textrm{DFT}$ is taken as the estimate of the PSD in most practical situations. One form, among other variations/methods, is:

\begin{equation} P(f_k) = \frac{1}{N} \displaystyle \left| \sum_{n = 0}^{N - 1} x[n]\exp\left(-j2\pi f_kn\right) \right|^2 \tag{4} \end{equation}

What's the difference between these?

Comparing $(2)$ and $(4)$, you have:

\begin{equation} P(f_k) = \frac{1}{N} \left| X(f_k)\right|^2 \end{equation}

From the bin number $k$ to frequency in $\textrm{Hz}$, $F = \frac{F_s}{N}k$

For more reading on PSD estimation check this question, that question, and this question.

EDIT:

The power spectral density, $\textrm{PSD}$, describes how the power of your signal is distributed over frequency whilst the $\textrm{DFT}$ shows the spectral content of your signal, the amplitude and phase of harmonics in your signal. You pick one or the other depending on what you want to observe/analyze. And no they're not the same as you can see from the equations above and links given. Their spectra are generally not the same. One is estimated as the squared magnitude of the other.

$\endgroup$
  • $\begingroup$ Do the two measurements produce a spectra that has the same relative amplitudes (i.e. the spectras look the same)? Or if they do have different spectras, then when/why pick one over another? $\endgroup$ – mavavilj Jul 24 '15 at 10:10
  • $\begingroup$ @mavavilj See EDIT in my answer above. $\endgroup$ – Gilles Jul 24 '15 at 10:50
  • $\begingroup$ So for e.g. audio, which one would it be? I'm trying to understand which one of those measurements corresponds to what I want to observe in a FFT plot (of an audio signal) that must be interpretable to be used in sound mixing. $\endgroup$ – mavavilj Jul 24 '15 at 11:00
  • $\begingroup$ @mavavilj As I said, it all depends on what you're looking for. One feature might describe better what you want than the other. Speech signals for instance are non stationary, even the usual DFT won't do the job, the short-time Fourier transform (STFT) is the way to go. That said, I'm not an expert in sound mixing. You may consider asking a separate new question on sound mixing. $\endgroup$ – Gilles Jul 24 '15 at 14:45
  • $\begingroup$ @mavavilj Also, consider doing the tour on the site. This will help a lot when asking, answering, commenting, accepting answers, and voting in a more informed way. Plus you get a bonus for this. $\endgroup$ – Gilles Jul 24 '15 at 15:00
0
$\begingroup$

The FFT is the Fast Fourier Transform. It is a special case of a Discrete Fourier Transform (DFT), where the spectrum is sampled at a number of points equal to a power of 2. This allows the matrix algebra to be sped up. The FFT samples the signal energy at discrete frequencies.

The Power Spectral Density (PSD) comes into play when dealing with stochastic signals, or signals that are generated by a common underlying process, but may be different each time the signal is measured. Given just one "realization" of a stochastic process--a stochastic signal--you can only estimate what the underlying Power Spectral Density is. You can make this estimate poorly with the Periodogram, which involves squaring the FFT (amplitude squared yields power). The periodogram suffers from very high variance and is not a good estimator. You are better off using Welch's method of periodogram averaging, or better yet, the Blackman-Tukey method of periodogram smoothing.

$\endgroup$
  • 2
    $\begingroup$ i didn't down-arrow you, but there are Fast Fourier Transforms that are not limited to number of points being $N=2^p$ where $p$ is an integer. $\endgroup$ – robert bristow-johnson Jul 21 '15 at 16:45

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.