Is this filter a BPF? $$\dfrac{z}{z-a}$$

where $a$ is some complex number?

If we put a pole somewhere on the unit circle it will emphasise a certain frequency, is that right?


1 Answer 1


Yes, if you put a pole approaching the unit circle (but not on the unit circle, as this will create an unstable filter) , you will have a bandpass. The center frequency of this bandpass filter depends on how many radians along the unit circle where you have placed the pole (the pole's 'degree').

Moving counter-clockwise from zero will give you an increasing center frequency of the bandpass filter.

Once you start moving towards zero on the Z-plane and away from the unit circle, you will decrease the sharpness Q and consequently increase the bandwidth.

  • 1
    $\begingroup$ If you have a pole on the unit circle you have infinite Q and an unstable filter. So it's not really useful as a bandpass filter. $\endgroup$
    – Matt L.
    May 16, 2015 at 19:10
  • $\begingroup$ Good point, will edit $\endgroup$
    – panthyon
    May 16, 2015 at 19:27
  • 1
    $\begingroup$ also, keep in mind that with purely real coefficients you can't do a 1st-order BPF. so you really need two of these with complex conjugate a to do this for real. and then it will be 2nd-order. $\endgroup$ May 16, 2015 at 22:35

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.