Question 1 of 7: Pole-Zero Diagrams, ROC, Stability and Causality

Question 1: Pole-Zero Diagrams, ROC, Stability and Causality (20 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. Four transfer functions specified only by their pole/zero locations and their regions of convergence:

Find. For each system: whether it is FIR or IIR, whether it is BIBO stable, whether it is causal, and what kind of sequence its impulse response is — each conclusion justified from the diagram rather than asserted.

[Figure not reproduced: Figure 1 redrawn. Crosses are poles, circles are zeros; the hatched region is the stated ROC and the dashed circles are its boundaries. The solid circle is the unit circle in every panel. See the official exam paper.]

Approach. Every answer follows from three readings of the diagram: where the poles are (all at the origin means FIR), whether the ROC contains the unit circle (stability), and whether the ROC is the exterior region including $z = \infty$ (causality) — and the shape of the ROC then fixes the sidedness of $h[n]$.

1. State the three tests once, then apply them mechanically. A rational $H(z)$ has a finite-length impulse response exactly when all of its poles lie at $z = 0$ or $z = \infty$, because then $H(z)$ is a polynomial in $z^{-1}$ (or in $z$) and its coefficients are the taps — that is the FIR/IIR test. BIBO stability requires $\sum_n |h[n]| \lt \infty$, which for a rational transfer function is equivalent to
   $$\text{ROC} \supset \{z : |z| = 1\} \quad \Longleftrightarrow \quad \text{the system is BIBO stable.}$$

   Causality requires $h[n] = 0$ for $n \lt 0$, which for a rational $H(z)$ means the ROC is the exterior of the outermost pole and includes $z = \infty$. Note the two tests are independent: the unit circle can lie inside an ROC that does not reach infinity.

2. Systems A and B — poles only at the origin, so both are FIR. With a double pole at $z = 0$ and the conjugate zero pair, system A is
   $$H_A(z) = \frac{(z - z_1)(z - z_1^{*})}{z^{2}} = 1 - 2\,\mathrm{Re}\{z_1\} z^{-1} + |z_1|^{2} z^{-2},$$

   a three-tap polynomial in $z^{-1}$ only, so $h_A[n]$ lives on $n = 0, 1, 2$. Its ROC excludes only $z = 0$, which does include both the unit circle and $z = \infty$: A is stable and causal. System B has only a single pole at the origin, so

   $$H_B(z) = \frac{(z - z_1)(z - z_1^{*})}{z} = z - 2\,\mathrm{Re}\{z_1\} + |z_1|^{2} z^{-1},$$

   and the leading $+z$ term is exactly why its ROC must also exclude $z = \infty$. That single excluded point is the whole answer to part (a) for B: the term $z$ inverts to $\delta[n+1]$, so $h_B[-1] \ne 0$ and B is not causal. It is still FIR and still stable — a finite sum of finite taps is always absolutely summable.

3. System C — outermost-pole exterior ROC lying inside the unit circle. The ROC $|z| \gt |p_1|$ is the exterior of the largest pole radius and therefore contains $z = \infty$, so C is causal and $h_C[n]$ is right-sided. Because the dashed boundary circle sits inside the unit circle ($|p_1| \lt 1$), the unit circle lies in the ROC and C is stable. The poles are not at the origin, so C is IIR. Note that the zeros lie outside the unit circle: this makes C a non-minimum-phase system, which affects the phase response but has no bearing on stability or causality.
4. System D — annular ROC straddling the unit circle. Here the ROC is a ring bounded inside by $|p_2| \lt 1$ and outside by $|p_1| \gt 1$. An annulus never contains $z = \infty$, so D cannot be causal. Splitting $H_D(z)$ by partial fractions, the pole $p_2$ (inside the inner boundary) contributes a right-sided term and the pair $p_1, p_1^{*}$ (outside the outer boundary) contributes a left-sided term:
   $$h_D[n] = \underbrace{c\,p_2^{\,n} u[n]}_{\text{right-sided}} \; - \; \underbrace{\left(a\,p_1^{\,n} + a^{*} (p_1^{*})^{n}\right) u[-n-1]}_{\text{left-sided}},$$

   so $h_D[n]$ is genuinely double-sided. Since $|p_2| \lt 1 \lt |p_1|$, the ring contains $|z| = 1$ and D is stable — both one-sided pieces decay away from $n = 0$ in their own directions. This is the case worth dwelling on: poles outside the unit circle do not by themselves make a system unstable; they only forbid it from being causal and stable at the same time.

   $$\boxed{\text{Stability} \Leftrightarrow |z| = 1 \in \text{ROC}; \qquad \text{Causality} \Leftrightarrow z = \infty \in \text{ROC}.}$$