Question 8 of 10: Regions of Convergence and Impulse Responses of a Discrete System

Question 8: Regions of Convergence and Impulse Responses of a Discrete System (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. $H(z) = \dfrac{z(z-1)}{(z-0.5)(z-2)}$, with zeros at $z = 0$ and $z = 1$ and poles at $z = 0.5$ and $z = 2$.

Find. All admissible ROCs; the causality and stability of each; the corresponding impulse responses with a proof that they exhibit those properties; and the difference equation.

Approach. The poles partition the plane into annuli; enumerate them, classify each by where it sits relative to the outermost pole and the unit circle, then invert by partial fractions choosing the right-sided or left-sided pair for each pole according to the annulus.

1. Part (a): enumerate the ROCs. An ROC is always an annulus bounded by poles and containing none. With poles at $|z| = 0.5$ and $|z| = 2$ there are exactly three: $$\boxed{\text{I: }|z|\lt 0.5;\qquad \text{II: }0.5\lt|z|\lt 2;\qquad \text{III: }|z|\gt 2}$$
2. Part (b): classify each ROC. A system is causal when its ROC is the exterior of the outermost pole (and $H(z)$ is proper, which it is here, numerator and denominator both being degree 2); it is stable when its ROC contains the unit circle $|z| = 1$. Applying both criteria:

Note that no ROC is both causal and stable — an unavoidable consequence of the pole at $z = 2$ lying outside the unit circle. Only one realisation can be built to run in real time, and it is the unstable one.

3. Part (c): partial-fraction expansion. Because each inverse pair in the table carries a factor $z$, expand $H(z)/z$ rather than $H(z)$: $$\frac{H(z)}{z}=\frac{z-1}{(z-0.5)(z-2)}=\frac{A}{z-0.5}+\frac{B}{z-2}$$ with residues $A = \dfrac{0.5-1}{0.5-2} = \dfrac{-0.5}{-1.5} = \dfrac13$ and $B = \dfrac{2-1}{2-0.5} = \dfrac{1}{1.5} = \dfrac23$. Multiplying back by $z$: $$\boxed{H(z)=\frac{1}{3}\cdot\frac{z}{z-0.5}+\frac{2}{3}\cdot\frac{z}{z-2}}$$ (As a check, $A+B = 1$, which must equal $h[0]$ for the causal branch by the initial-value theorem, and indeed $H(\infty) = 1$.)
4. Invert on ROC III ($|z| \gt 2$). Both poles are interior to the ROC, so both take the right-sided pair $\frac{z}{z-a}\leftrightarrow a^nu[n]$: $$h_{\text{III}}[n]=\left[\tfrac13(0.5)^n+\tfrac23(2)^n\right]u[n]$$ This is zero for $n \lt 0$, confirming causality directly from the sequence. It is not absolutely summable because the $\tfrac23 2^n$ term diverges geometrically, confirming instability — a bounded input can drive an unbounded output.
5. Invert on ROC II ($0.5 \lt |z| \lt 2$). The pole at 0.5 is inside the annulus boundary and takes the right-sided pair, while the pole at 2 is outside and must take the left-sided pair $\frac{z}{z-a}\leftrightarrow -a^nu[-n-1]$: $$h_{\text{II}}[n]=\tfrac13(0.5)^nu[n]-\tfrac23(2)^nu[-n-1]$$ This is non-zero for negative $n$, so the system is non-causal. It is absolutely summable: $$\sum_{n=0}^{\infty}\tfrac13(0.5)^n+\sum_{n=-\infty}^{-1}\tfrac23 2^{n}=\tfrac13\cdot 2+\tfrac23\cdot 1=\tfrac43\lt\infty$$ confirming BIBO stability. Note how each term decays in the direction it extends: the $(0.5)^n$ term dies away for large positive $n$, and the $2^n$ term dies away for large negative $n$.
6. Invert on ROC I ($|z| \lt 0.5$). Both poles now lie outside, so both take the left-sided pair: $$h_{\text{I}}[n]=-\left[\tfrac13(0.5)^n+\tfrac23(2)^n\right]u[-n-1]$$ It is zero for $n \ge 0$, hence purely anticausal, and unstable because as $n\to-\infty$ the term $\tfrac13(0.5)^n = \tfrac13 2^{|n|}$ grows without bound.
7. Part (d): the difference equation. Expand and divide through by $z^2$ to reach negative powers: $$H(z)=\frac{z^2-z}{z^2-2.5z+1}=\frac{1-z^{-1}}{1-2.5z^{-1}+z^{-2}}=\frac{Y(z)}{X(z)}$$ Cross-multiplying, $Y(z)[1-2.5z^{-1}+z^{-2}] = X(z)[1-z^{-1}]$, and each $z^{-k}$ is a delay of $k$ samples: $$\boxed{y[n]-2.5\,y[n-1]+y[n-2]=x[n]-x[n-1]}$$ Driving this recursion with a unit sample and zero initial conditions reproduces $h_{\text{III}}[n]$ term for term ($1,\ 1.5,\ 2.75,\ \ldots$), as it must, since the recursion run forward in time is the causal realisation.