I would like to deconvolve the signal shown in the plot below from its linearly decaying tail and the exponential levelling. That means that after the deconvolutin, the tail should be at a constant value of ~410 mV. Does anyone have an idea how to approach this problem?

I uploaded the data here: https://nextcloud.mpp.mpg.de/nextcloud/index.php/s/RwXEzEa6bxHTByR

enter image description here

Thank you very much in advance!

  • $\begingroup$ not quite sure I get you correctly, but let me try to understand: you know your signal is 0 before time 4ms, and constant 400mV after, right? $\endgroup$ May 12 at 15:29
  • $\begingroup$ Hi Marcus! After the deconvolution, the signal up to ~4ms should look like before, but I want the tail (values >4ms) of the signal to become a constant at about 400mV. The background is an energy estimation - i.e. the pulse height is proportional to the energy - the maximum alone is not a good proxy. $\endgroup$
    – Frank
    May 12 at 15:47
  • $\begingroup$ What's the data you have? Is it synthetic? $\endgroup$
    – Royi
    May 12 at 18:09
  • $\begingroup$ do you, instead of a deconvolver, want a sliding maximum algorithm? $\endgroup$ May 13 at 2:40
  • $\begingroup$ @Royi The data come from an ASIC preamplifier (reset via capacitor instead of resistor. Therefore, the tail is mostly linearly decreasing instead of the usual exponential decay). $\endgroup$
    – Frank
    May 14 at 6:27

You probably can't. If I interpret your description correctly, this is the unit step response of your system. This looks very much like the unit step response of a high pass filter, that means the system is filtering out exactly the frequency that you want (which is DC or 0 Hz).

You may be better off creating a parametric model of your system response and try to do a weighted least square error fit of the data.


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.