Question 5 of 6: Identifying filter families from pole-zero plots

Question 5: Identifying filter families from pole-zero plots (12 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Six pole-zero plots that all meet one common specification. Their contents are as follows.

Find. Which plots are recursive, what band type they all realise, which classical approximation each recursive plot corresponds to, and which designs have exactly linear phase.

[Figure not reproduced: Figure Q5.1 — the six pole-zero plots redrawn from source page 5. Crosses are poles, circles are zeros; the far-right open circle past the axis break in (e) and (f) is a real zero well outside the unit circle. See the official exam paper.]

Approach. Sort the plots first by whether they have poles at all, then use the location of the poles and zeros relative to $z = 1$ and $z = -1$ to fix the band type, then use the combination of zero placement and filter order to name each classical approximation, and finally test the zero constellation for reciprocal symmetry.

1. Part (a): separate recursive from non-recursive. A filter is IIR precisely when $H(z)$ has a non-trivial denominator, that is, poles somewhere other than the origin. Plots (a), (b), (c) and (d) all show crosses at finite non-zero locations, so $$\boxed{\text{(a), (b), (c) and (d) are IIR}}$$ Plots (e) and (f) show zeros only. Their poles all sit at $z = 0$, where they merely represent delay, so $H(z)$ is a polynomial in $z^{-1}$, the impulse response has finite length, and (e) and (f) are FIR.
2. Part (b): read the band type off the geometry. The magnitude at a frequency $\omega$ is the product of the distances from $e^{j\omega}$ to the zeros divided by the product of the distances to the poles. In every one of the six plots the poles (where there are any) cluster around $z = +1$, that is $\omega = 0$, which pushes the gain up at low frequency; the zeros sit at or near $z = -1$ and along the upper-left and lower-left arcs, that is $\omega \to \pi$, which pulls the gain down at high frequency. For (e) and (f) the same reasoning applies with zeros alone: every unit-circle zero lies on the left-hand arc. Hence $$\boxed{\text{all six are LOW-PASS filters}}$$ which is consistent with the statement that they meet one common specification.
3. Part (c): use the zeros to split the four IIR plots into two families. Plots (a) and (b) have all of their zeros piled at $z = -1$, so they are all-pole designs: the analogue prototype has all its zeros at $s = \infty$, and the bilinear transform maps $s = \infty$ to $z = -1$. Those are the Butterworth and Chebyshev type I approximations. Plots (c) and (d) instead show finite zeros distributed on the unit circle, i.e. exact transmission nulls in the stop band, which is the signature of the Chebyshev type II and elliptic approximations.
4. Break the first tie with the order. Since all six designs meet the same specification, the required order ranks the approximations. The maximally flat Butterworth response is the least efficient and needs the most poles, and (a) needs 18 against 8 for (b). Geometrically, (a) has its poles on a circular arc reaching well inside the unit circle while (b) has them pushed out onto an elliptical arc hugging $|z| = 1$, which is what produces pass-band ripple. Therefore (a) is Butterworth and (b) is Chebyshev type I.
5. Break the second tie with the pole radii and the order. Chebyshev type II has an equiripple stop band but a maximally flat pass band, so like the Butterworth its poles are spread over a wide range of radii rather than crowded against the rim, and it needs the same order as Chebyshev type I. Plot (c) shows exactly that: 8 poles with radii from about 0.24 out to 0.89. The elliptic design is equiripple in both bands, so its poles crowd near $|z| = 1$ like a Chebyshev I and it carries stop-band zeros, which is why it meets the specification with the fewest poles of all. Plot (d) achieves the job with just 5. Hence $$\boxed{\text{(a) Butterworth, (b) Chebyshev I, (c) Chebyshev II, (d) Elliptic}}$$
6. Sanity check on plot (d). An odd-order elliptic low-pass has one real pole and $(N-1)/2$ conjugate pole pairs, and its zero set is $(N-1)/2$ conjugate pairs on the unit circle plus the single bilinear zero at $z = -1$. With $N = 5$ that predicts one real pole plus two pairs, and four unit-circle zeros plus one at $z = -1$ — precisely the five crosses and five circles that the figure shows. The census closes, which confirms the identification.
7. Part (d): test for linear phase. A real causal FIR filter has exactly linear phase if and only if its impulse response is symmetric or antisymmetric, $h[n] = \pm h[M-n]$, and in the $z$-plane that condition reads $H(z) = \pm z^{-M}H(z^{-1})$: every zero at $z_0$ must be accompanied by one at $1/z_0$. Plots (e) and (f) display exactly that pattern — the labelled real zero at $9.38$ in (e) has its partner at $1/9.38 = 0.107$, the off-circle complex zeros come in reciprocal quadruples $\{z_0, 1/z_0, z_0^{*}, 1/z_0^{*}\}$, and the unit-circle zeros are self-reciprocal. Hence $$\boxed{\text{only (e) and (f) have linear phase}}$$
8. Explain why the IIR designs cannot. For a rational system, linear phase would demand the same mirror-image symmetry applied to the poles: every pole at $p$ would need a partner at $1/p$. One of each such pair necessarily lies outside the unit circle, so the system could not be simultaneously causal and stable. Plots (a) to (d) all have their poles strictly inside $|z| = 1$, which is exactly what makes them stable and causal — and exactly what rules out linear phase. This is the standard trade-off: the IIR designs reach the specification with far fewer coefficients, but only the FIR designs can deliver a constant group delay.