This is primarily a question about how to name my filter. Suppose that I have a photon detector, and that I want to measure the rate at which photons arrive, with a simple gated counter (because this is the only hardware I have). The photon rate is just the counter value divided by the gate interval.

It seems to me that the counter is an integrator: so by opening the gate for, say, 1 second, I have made a first-order low pass filter (LPF).

Now, suppose that I take the mean of, say, 10 such measurements. Have I now built a second-order LPF? Or merely a first-order LPF with a longer time-constant? Does the answer change if the measurements are not consecutive?


1 Answer 1


The easiest way to understand the concept of filter order is to look at the Laplace (continuous time) or Z-transform (discrete time) of the filter. The filter order is the order of the transform's denominator.

For example, the Laplace transform of an integrator is $\frac{1}{s}$. Extending the time of the integration does not change that it is an integrator, it just changes the bounds of the integration. If you were to make it a double integration, though (two integrators in series), then the Laplace transform would be $\frac{1}{s^2}$ and you would have a 2nd order filter.

Averaging can be considered a partial integration with a gain factor, so your integrator with averaging would be a variant of a double integrator, making it a second order filter.

  • $\begingroup$ Isn't summing the 10 measurements the same thing as integration, though? Or am I still missing something? Is the distinction whether I divide by the number of measurements? $\endgroup$
    – nick g
    Commented Aug 20, 2013 at 23:55
  • $\begingroup$ This is just an Integrator. A constant in front of any system doesn't affect the order of the system, it's just a Gain factor which appears as Magnitude shift across the Response of the system. Your 10 measurements are just the same filter used across 10 time intervals I guess. $\endgroup$
    – Sudarsan
    Commented Aug 21, 2013 at 0:56
  • $\begingroup$ @nickg Yes, I suppose that you are right. I will revise my answer. $\endgroup$
    – Jim Clay
    Commented Aug 21, 2013 at 1:23

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