0
$\begingroup$

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?

Improve this question
$\endgroup$
1
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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 Aug 20 '13 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 Aug 21 '13 at 0:56
  • $\begingroup$ @nickg Yes, I suppose that you are right. I will revise my answer. $\endgroup$ – Jim Clay Aug 21 '13 at 1:23

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.