When given a discrete impulse response, let us say it is from an FIR filter so it is finite, how can I assume the filter type? The task is to recognize or to preclude one or more of the following types:

  • Low pass
  • High pass
  • Band stop
  • Band pass

For example, I read that a zero at $h(n = 0)$ (where h is the impulse response of my filter) means that the filter is a low pass but how can I say that with precluding the band stop immediately?

  • $\begingroup$ IIR filters can also have finite impulse responses. $\endgroup$ Feb 28, 2019 at 12:34

1 Answer 1


If you know for sure that the filter type is one of the four given ones, then the exercise is quite straightforward. Look at the zeros of the corresponding transfer function


where $N$ is the filter length. The zeros at DC ($z=1$) and at Nyquist ($z=-1$) are given by




It should be obvious that a (classic) low pass has a zero at Nyquist but no zero at DC. For a high pass filter the opposite is the case. A band pass filter has zeros at DC and at Nyquist, whereas a band stop filter has no zeros at DC or at Nyquist.

PS: What you read about the zero of the impulse response is either non-sense, or you misunderstood what they meant.

  • $\begingroup$ Thank you. I might have gotten it wrong (my example). $\endgroup$
    – Kutsubato
    Feb 27, 2019 at 11:01
  • $\begingroup$ So what does "Nyquist" and "DC" mean, in the sense of these "synonyms" that you have used? We have never used these expressions. What does it mean in this particular context? $\endgroup$
    – Kutsubato
    Feb 27, 2019 at 11:01
  • $\begingroup$ But how I can figure out the filter type purely based on the IR and not the corresponding transfer function? Just by looking at the plot of the IR! If I have got latter, I would appreciate to not having to create the transfer function additionally to solve the problem. $\endgroup$
    – Kutsubato
    Feb 27, 2019 at 11:05
  • $\begingroup$ For example, if the impulse response has a mean of zero, DC, i.e., $0\,\text{Hz}$ will be cancelled. In this case you have a high-pass filter. This becomes clear once you visualize the effects of convolution with the impulse response. $\endgroup$
    – applesoup
    Feb 27, 2019 at 14:14
  • $\begingroup$ If you look at Eqs (2) and (3) in my answer you see that it's simply a matter of adding the impulse response samples (in (3) with an alternating sign) in order to determine these two specific values of the transfer function. You can do that by inspection. $\endgroup$
    – Matt L.
    Feb 27, 2019 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.