2
$\begingroup$

My background is in signal processing, and never took any course related to functional analysis or even advanced algebra. But I have a strong conviction (may be wrong) that we may be do better then zero padding as given below.

Consider we have a $N$ complex time domain samples (with unknown noise) and we want to estimate the spectral density. Spectral density is nothing but square of the Fourier Transform of the signal. Let $N$ be the length of the original signal $x_n=x_0, x_1 \cdots x_{N-1}$ and let $X(\omega)$ be its DTFT, $X(\omega)= \sum x_n \exp(-i \omega n)$. It is easily seen that the number of DTFT bins in this case is also $N$. In our problem we want to have a better resolution in frequency axis. In other words, we want to get more DTFT bins than $N$. A common approach here is to zero pad $x_n$ as much as you want and then perform Fourier Transform as before to get more frequency bins.

One can show that by zero padding we are doing some sort of interpolation (trigonometric). Can we do something better (else)? Something better means retain the original roughness if the roughness is correct? Equivalently, can't we built some sort of topological structure from $x_n$, assuming that $x_n$ has come from a structure, and then do the interpolation?

I do not know what else can we do, so the examples of "something else" may not be relevant. I am looking for answers related to the limitation/restriction applicable of this problem. A perfect answer may be explain why we cannot do anything other than zero padding?

$\endgroup$
2
  • 1
    $\begingroup$ With all respect to the OP, can someone rewrite the question into more standard mathematical terminology so that we can tell what it is about? $\endgroup$ Commented Feb 9, 2016 at 23:27
  • $\begingroup$ The zero-padding gives a faithful sinc-function-type interpolation of the DFT. I think Bracewell in The Fourier Transform and Its Applications pretty much covers the theory. Dealing with distortions of the PSD from aliasing induced by noise as well as achieving super-resolution of an undersampled signal are other matters. $\endgroup$ Commented Feb 10, 2016 at 13:15

0

You must log in to answer this question.