0
$\begingroup$

I have some noisy signal. When I substract real (for calibration it is known) or filtered signal I get residuals. When I plot distribution of these residuals I see that it is sum of exponential and Gaussian distributions. Is it possible to speculate about what does it means? Maybe it is common situation? Could you help me with explanation?

UPDATE:

Blue is RAW signal, red- filtered. enter image description here Distribution of residuals:enter image description here

Distribution of abs(residuals):enter image description here

Distribution near 0:enter image description here

$\endgroup$
  • $\begingroup$ How can you tell for sure that it's the sum of an exponential and Gaussian rather than (say) a gamma distribution or a Weibull distribution or some other moderately right skew distribution? $\endgroup$ – Glen_b -Reinstate Monica May 4 '17 at 2:46
  • $\begingroup$ It has maximum on 0. In reality it is symmetrical over 0. I just use absolute values. $\endgroup$ – zlon May 4 '17 at 5:49
  • $\begingroup$ I don't follow this at all. Why would having a mode at 0 imply exponential + Gaussian? If its symmetrical about 0, how does that imply exponential + Gaussian? $\endgroup$ – Glen_b -Reinstate Monica May 4 '17 at 8:56
  • $\begingroup$ Ok, it is not exponential it is something very skewed. Is it important for the explanation which 2 sources of noise I have? $\endgroup$ – zlon May 4 '17 at 8:58
  • $\begingroup$ A. I'm just trying to understand (i) what your actual claim about your data actually is, and (ii) what the basis for the claim is. B. Thanks for the picture; that's potentially quite useful. If that's the absolute value of residuals, how did you decide the residuals were actually symmetric about 0? $\endgroup$ – Glen_b -Reinstate Monica May 4 '17 at 11:21
2
$\begingroup$

Your side-lobes appear to be approximately at +/- twice your signal amplitude, i.e. when the two signals are 180 degrees out of phase (peak of one = trough of the other). The prevalence likely arises because the blue signal is much higher frequency than the red, so the blue signal goes through several cycles over any interval where the slower red signal is near a peak/trough.

$\endgroup$

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.