1
$\begingroup$

I am struggling to understand the effects that partially undersampling a signal would have on its FFT. I have a complex signal consisting on a sum of sinusoids of different frequencies, which I am sampling at the Nyquist rate, except for a couple of samples that are skipped at some point. Assuming I interpolate the missing samples, how would this affect the resulting FFT?

$\endgroup$
2
  • $\begingroup$ I believe there’s quite some information missing. Let alone the complexity of the issue, different interpolation schemes have different characteristics. Without even knowing what interpolation method you will use I believe it is very hard to answer that. Even then, I am not sure there would be a “trivial” answer (I haven’t put extensive thought into it so I may very well be mistaken though). $\endgroup$
    – ZaellixA
    Oct 4 at 12:57
  • 1
    $\begingroup$ Hey @ZaellixA, you are right that my question lacked information. However, I find Hilmar's answer to fit my issue. Thanks $\endgroup$ Oct 4 at 13:23

1 Answer 1

1
$\begingroup$

You can model the resulting signal, $y[n]$ as

$$y[n] = x[n] + e[n]$$

where $e[n]$ is the error signal given by

$$e[n] = \sum_m a_k\cdot \delta[n-n_m] $$

$a_k$ is the residual interpolation error at the sample time $n_m$ where there are missing samples.

The DFT spectrum of the result would be

$$Y[k] = X[k] + E[k]$$

The DFT $E[k]$ of the error is

$$E[k] = \sum_m a_m \cdot e^{-j2\pi\frac{km}{N}} $$

In most cases this will simply be white noise, i.e. the total error energy ($\sum a_m^2$) will fairly equally distributed over all frequencies. That's kind of handy: you can use the zeros in $X[k]$ to estimate the noise floor.

The error will NOT be white if the missing samples follow some regular pattern in either amplitude or time index and/or if the error is correlated with the signal itself.

$\endgroup$
1
  • $\begingroup$ Thanks! This is precisely what I needed to understand the scenario. $\endgroup$ Oct 4 at 13:21

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.