I get confused when there are a lot of zeros/poles in the zero-pole diagram and I find difficulty understanding their frequency response.

I know the following: 1. Complex conjugates cause double peaks at their respective angles. The larger their magnitude, the sharper their peaks.

  1. A single pole causes a peak at its respective angle. The closer it is to the unit circle the sharper the peak in the frequency response.

  2. A single zero causes a dip at its respective angle. The closer it is to the unit circle the sharper the dip.

I am not sure how to deal with zero pole diagrams when they come in bundles as shown below where it is asked to match each zero-pole diagram to its frequency response.

My trials:

I matched:

Frequency Response 3 to Zero-Pole Diagram 2

Frequency Response 6 to Zero-Pole Diagram 3

enter image description here


1 Answer 1


These are all frequency selective filters. Note the zeros on the unit circle, which correspond to zeros of the frequency response. The filter's passband is determined by range of the angles of the poles, and the stopband is determined by the range of the angles of the zeros.

As an example, let's consider PZ-diagram $\#1$: since the stopband is around DC (i.e., angle zero), and there are no more zeros, just poles, at angles greater than approximately $\pi/3$ (and, by symmetry, smaller than approximately $-\pi/3$), this must be a high pass filter. The cut-off frequency must be close to $\pi/3$, so all frequency responses can be eliminated except for $\#4$ and $\#5$. However, frequency response $\#5$ has a zero at DC, which is not present in the PZ-diagram. Consequently, we can match PZ-diagram $\#1$ to frequency response $\#4$. I'm sure you can match the other diagrams yourself.

  • $\begingroup$ Thanks! That was very helpful. I was able to map all the cases. However, can you help me understand what is the role of the pole at the z=0 in PZ-diagram #6 ? $\endgroup$
    – HaneenSu
    Commented Mar 8, 2020 at 20:28
  • $\begingroup$ @HaneenSu: It's not exactly at $z=0$, but at a small negative value. Without that pole the frequency response would have a slight dip at Nyquist. $\endgroup$
    – Matt L.
    Commented Mar 8, 2020 at 21:17

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