Question 8 of 10: Regions of Convergence, Causality/Stability, and Inversion of a Rational z-Transform

Question 8: Regions of Convergence, Causality/Stability, and Inversion of a Rational z-Transform (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 $z=0,1$ and poles at $z=0.5,2$.

Find. (a) all valid ROCs; (b) causality/stability of each; (c) $h[n]$ for each ROC; (d) the difference equation.

Approach. Since there are two poles, three annular ROCs are possible (interior of the inner pole, the annulus between them, exterior of the outer pole). Causality follows from whether the ROC is the exterior of the outermost pole; stability follows from whether the ROC contains the unit circle $|z|=1$. Partial-fraction $H(z)$ and read $h[n]$ off each term via the standard $a^nu[n]\leftrightarrow1/(1-az^{-1})$ (right-sided) and $-a^nu[-n-1]\leftrightarrow1/(1-az^{-1})$ (left-sided) pairs, matched to the chosen ROC.

1. Part (a) — possible ROCs. The two poles at $z=0.5$ and $z=2$ split the $z$-plane into three annular regions: $$\boxed{\text{(i) } |z|<0.5,\quad\text{(ii) } 0.5<|z|<2,\quad\text{(iii) } |z|>2.}$$

2. Part (b) — causality and stability per ROC. Causality requires the ROC to be the EXTERIOR of the outermost pole; stability requires the ROC to CONTAIN the unit circle $|z|=1$:
   • (i) $|z|<0.5$: interior of the inner pole — the sequence is purely left-sided (anticausal); the region excludes $|z|=1$ (since $0.5<1$), so unstable.
   • (ii) $0.5<|z|<2$: an annulus — the sequence is two-sided (non-causal); this is the ONLY one of the three regions that contains $|z|=1$, so it is the unique stable choice.
   • (iii) $|z|>2$: exterior of the outer pole — the sequence is right-sided (causal); the region excludes $|z|=1$ (since $2>1$), so unstable.
   No ROC is simultaneously causal AND stable here, because that would require every pole strictly inside the unit circle, and $z=2$ is not.
3. Part (c) — impulse response per ROC. Polynomial division plus partial fractions give $$H(z)=1+\frac{1/6}{z-0.5}+\frac{4/3}{z-2}=1+\frac{(1/6)z^{-1}}{1-0.5z^{-1}}+\frac{(4/3)z^{-1}}{1-2z^{-1}}$$ (: $H(z)$ minus this expansion simplifies to exactly 0). Applying $z^{-1}/(1-az^{-1})\leftrightarrow a^{n-1}u[n-1]$ (right-sided, ROC $|z|>|a|$) or $\leftrightarrow-a^{n-1}u[-n]$ (left-sided, ROC $|z|<|a|$) to each term, matched to the chosen ROC: $$\text{(i) } |z|<0.5:\quad h[n]=\boxed{\delta[n]-\tfrac16(0.5)^{n-1}u[-n]-\tfrac43(2)^{n-1}u[-n]}$$ $$\text{(ii) } 0.5<|z|<2:\quad h[n]=\boxed{\delta[n]+\tfrac16(0.5)^{n-1}u[n-1]-\tfrac43(2)^{n-1}u[-n]}$$ $$\text{(iii) } |z|>2:\quad h[n]=\boxed{\delta[n]+\tfrac16(0.5)^{n-1}u[n-1]+\tfrac43(2)^{n-1}u[n-1]}$$ Each sequence's own causality/stability confirms part (b) directly: in (i) both geometric pieces run over $n\le0$ (left-sided, anticausal) and the $2^{n-1}$ piece grows without bound as $n\to-\infty$ (unstable, since $\sum|h[n]|$ diverges); in (ii) the $0.5^{n-1}$ piece is right-sided and decaying while the $2^{n-1}$ piece is left-sided and decaying as $n\to-\infty$ — both halves are absolutely summable, so $\sum_n|h[n]|<\infty$; in (iii) both pieces are right-sided (causal) but the $2^{n-1}$ piece grows without bound as $n\to+\infty$ (unstable).
4. Part (d) — difference equation. Cross-multiplying $H(z)=Y(z)/X(z)$: $(z-0.5)(z-2)Y(z)=z(z-1)X(z)$, i.e. $(z^2-2.5z+1)Y(z)=(z^2-z)X(z)$. Dividing by $z^2$ (delay form): $$(1-2.5z^{-1}+z^{-2})Y(z)=(1-z^{-1})X(z)\ \Longrightarrow\ \boxed{y[n]=2.5y[n-1]-y[n-2]+x[n]-x[n-1]}$$ This equation is the SAME for all three ROCs — the ROC only selects which particular sequence (which set of initial/boundary conditions) realizes it, not the recursion itself (confirmed: recursively evaluating this equation for a unit-impulse input reproduces the causal $h[n]$ of part (iii) exactly for the first 15 samples).