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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?

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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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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