Determining which Filter from a Z-Plane Plots?

How do i determine which FIR filter (LP, HP, BS, BP) it is from looking at it's z-plane plot?

just to add a little to Matt's answer. if you know where your poles, $p_n$, and zeros, $q_m$, are your transfer function is

$$H(z) = H_0 \frac{(z-q_1)(z-q_2)(z-q_3)...(z-q_M)}{(z-p_1)(z-p_2)(z-p_3)...(z-p_N)} \quad \quad M \le N$$

the complex frequency response (which contains both magnitude and phase information) is

$$H(e^{j\omega}) = H_0 \frac{(e^{j\omega}-q_1)(e^{j\omega}-q_2)(e^{j\omega}-q_3)...(e^{j\omega}-q_M)}{(e^{j\omega}-p_1)(e^{j\omega}-p_2)(e^{j\omega}-p_3)...(e^{j\omega}-p_N)}$$

and the magnitude response (which is where you determine if it's a LPF or BPF or HPF or something else like that) is:

$$|H(e^{j\omega})| = |H_0| \frac{|e^{j\omega}-q_1| \ |e^{j\omega}-q_2| \ |e^{j\omega}-q_3|...|e^{j\omega}-q_M|}{|e^{j\omega}-p_1| \ |e^{j\omega}-p_2| \ |e^{j\omega}-p_3|...|e^{j\omega}-p_N|}$$

and $e^{j\omega}$ is a point on the unit circle. what this means is, as your frequency increases from DC (or $\omega=0$), that corresponds to moving along the unit circle from $z=e^{j0}=1$ to other points on the unit circle. then as your point on the unit circle gets close to a zero $q_m$, a factor of $|e^{j\omega}-q_m|$ gets small. $|e^{j\omega}-q_m|$ is literally the distance your point on the unit circle is from that particular zero.

likewise for the poles, but your poles are in the denominator, so as your point on the unit circle gets close to a pole $p_n$, a factor of $|e^{j\omega}-p_n|$ gets small, but since you're dividing by that small number, the effect on the gain magnitude is to increase it.

so a LPF will have poles closer to the DC point of $z=e^{j0}=1$ and, perhaps, zeros closer to Nyquist $z=e^{j\pi}=-1$ so when you're close to DC you're multiplying by big numbers and when you're close to Nyquist, you're multiplying by small numbers.

likewise for a HPF, you'll have poles closer to Nyquist and zeros closer to DC.

for a BPF, you might have zeros close to both DC and Nyquist and poles close to the unit circle at other places where they will boost the gain for those bandpass frequencies.

that's how you look at the pole-zero plots on the $z$-plane to judge if it's a LPF or whatever.

• I am interested in the posted question. I hope you won't mind if I ask for a few details. 1) LPF will have [...] zeros closer to Nyquist - If I'm correct, this gets rid of the frequency $\frac{f_s}{2}$ where $f_s$ is the sampling frequency. What about a frequency $f$ slightly larger than $f_s$? An ideal low-pass filter needs to get rid of it completely. Would this Z transform get rid of it? 2) LPF will have poles closer to the DC point - what would go wrong if I put a pole somewhere inside the unit circle not at the origin? Commented May 22, 2017 at 15:15
• geez, i never noticed this question before. Commented Nov 14, 2017 at 3:45

You obtain the filter's frequency response by evaluating its transfer function $H(z)$ on the unit circle $z=e^{j\omega}$, where $\omega=2\pi f/f_s$ is the normalized angular frequency, and $f_s$ is the sampling frequency. So DC corresponds to $z=1$, and Nyquist (the maximum frequency) corresponds to $z=\pm\pi$. Since for real-valued filters the magnitude response is symmetric, you only need to consider the upper half of the unit circle, i.e. you look at frequencies $\omega\in[0,\pi]$.

Looking at the pole-zero plot in the $z$-plane, you need to identify the pass bands and the stop bands of the filter. Since you're considering FIR filters, you only have to look at the zeros (the poles are at the origin, or at infinity if the filter is not causal). In the stop band(s), there must be zeros on the unit circle or at least very close to the unit circle to guarantee a reasonable stop band attenuation. In the pass band(s) there mustn't be any zeros very close to the unit circle because you want the filter to pass all frequency components in the pass band. You do get zeros at angles corresponding to the pass band(s), but they are much further away from the unit circle. They are there to guarantee a more or less flat pass band response, and they also determine the phase response in the pass band. If you have a linear phase FIR filter, each zero inside the unit circle must have a mirror image outside the unit circle.

In sum, identify the stop band(s) by finding regions on the unit circle where there are zeros on (or very close to) the circle. All other regions are pass bands. Obviously, a low pass filter must have one stop band extending from some given band edge to Nyquist. A high pass filter has its stop band between DC and some given cut-off frequency. I guess you can figure it out yourself for band pass and band stop filters.

As an example, the figure below shows a pole-zero plot of a typical linear-phase low pass FIR filter. Note the zeros on the unit circle in the stop band, and the zeros symmetrical to the unit circle in the pass band. Also note that all poles are at the origin since the filter is a causal FIR filter.