System function, all admissible ROCs and the value of $h[0]$: 22-Elec-B1 Digital Signal Processing, May 2018, Q5

Question 5: System function, all admissible ROCs and the value of $h[0]$ (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. The difference equation $y[n] - \tfrac52 y[n-1] + y[n-2] = x[n] - x[n-1]$, with no statement about causality or stability — which is precisely why part (b) has more than one answer.

Find. $H(z)$; every region of convergence consistent with it; and $h[0]$ for each.

Approach. Transform the difference equation, factor the denominator to locate the poles, enumerate the annuli they bound, and obtain $h[0]$ from a partial-fraction expansion by asking which terms are right-sided in each case.

Transform the difference equation. The z-transform is linear and a delay of $k$ samples multiplies by $z^{-k}$, so $$Y(z)\left(1 - \tfrac{5}{2}z^{-1} + z^{-2}\right) = X(z)\left(1 - z^{-1}\right),$$ giving $$\boxed{\;H(z) = \frac{Y(z)}{X(z)} = \frac{1 - z^{-1}}{1 - \tfrac{5}{2}z^{-1} + z^{-2}}\;}$$
Factor the denominator. Treating $z^{-1}$ as the variable, $1 - \tfrac52 z^{-1} + z^{-2}$ factors as $$\left(1 - 2z^{-1}\right)\left(1 - \tfrac{1}{2}z^{-1}\right),$$ which is confirmed by the coefficient checks $2 + \tfrac12 = \tfrac52$ and $2 \times \tfrac12 = 1$. The system therefore has poles at $$z = 2 \qquad\text{and}\qquad z = \tfrac{1}{2},$$ and a single zero at $z = 1$ (with a second zero at $z = 0$ if one counts the difference in polynomial orders).
Enumerate the regions of convergence. A region of convergence is always an annulus centred on the origin that is bounded by poles and contains none. With poles of magnitude $\tfrac12$ and $2$, exactly three such annuli exist: $$\boxed{\;|z| \lt \tfrac12, \qquad \tfrac12 \lt |z| \lt 2, \qquad |z| \gt 2.\;}$$ Their interpretations follow from where each ROC sits relative to the poles: $|z| \gt 2$ is outside the outermost pole, so $h[n]$ is right-sided (causal) but unstable; $|z| \lt \tfrac12$ is inside the innermost pole, so $h[n]$ is left-sided (anticausal) and also unstable; the annulus $\tfrac12 \lt |z| \lt 2$ is the only one containing the unit circle, so it is the two-sided, stable (and therefore non-causal) system.
Expand in partial fractions. Writing $$H(z) = \frac{A}{1 - 2z^{-1}} + \frac{B}{1 - \tfrac12 z^{-1}},$$ the residues follow from covering up each factor: $$A = \left.\frac{1 - z^{-1}}{1 - \tfrac12 z^{-1}}\right|_{z^{-1} = 1/2} = \frac{1 - \tfrac12}{1 - \tfrac14} = \frac{2}{3}, \qquad B = \left.\frac{1 - z^{-1}}{1 - 2z^{-1}}\right|_{z^{-1} = 2} = \frac{1 - 2}{1 - 4} = \frac{1}{3}.$$ As a check, $A + B = 1$, which must equal the value of $H(z)$ as $z \to \infty$.
Assign each term its sidedness and read off $h[0]$. A term $\dfrac{C}{1 - p z^{-1}}$ contributes $C p^{n}u[n]$ when the ROC lies outside $|z| = |p|$, and $-C p^{n}u[-n-1]$ when it lies inside. Applying this pole by pole:
Case $|z| \gt 2$ (causal). Both terms are right-sided, $h[n] = \left(\tfrac23 2^{n} + \tfrac13 (\tfrac12)^{n}\right)u[n]$, so $$h[0] = A + B = \boxed{1}.$$
Case $\tfrac12 \lt |z| \lt 2$ (stable). The pole at 2 lies outside the annulus and so becomes left-sided, while the pole at $\tfrac12$ stays right-sided: $$h[n] = -\tfrac{2}{3}\,2^{n}u[-n-1] + \tfrac{1}{3}\left(\tfrac12\right)^{n}u[n] \;\Longrightarrow\; h[0] = B = \boxed{\tfrac{1}{3}}.$$
Case $|z| \lt \tfrac12$ (anticausal). Both terms are left-sided, both carry $u[-n-1]$, and neither contributes at $n = 0$: $$h[0] = \boxed{0}.$$
Cross-check the causal case two independent ways. The initial-value theorem quoted on page 8 of the exam gives, for a causal sequence, $h[0] = \lim_{z\to\infty} H(z) = 1/1 = 1$, matching. Running the recursion $h[n] = \tfrac52 h[n-1] - h[n-2] + \delta[n] - \delta[n-1]$ forward from rest gives $h[0] = 1$, $h[1] = \tfrac32$, $h[2] = \tfrac{11}{4}$, which agree term by term with $\tfrac23 2^{n} + \tfrac13 (\tfrac12)^{n}$. Note also that the correct answers differ across the three cases, which is the whole point of the question: a difference equation alone does not define a system.

Figure 5.1 — the pole–zero map of $H(z)$ (poles $\times$ at $z = \tfrac12, 2$; zero $\circ$ at $z = 1$) with the three admissible regions of convergence shaded. Only the middle annulus contains the unit circle, so only that choice gives a stable system.

Quantity	Result
(a) System function	$H(z) = \dfrac{1 - z^{-1}}{1 - \tfrac52 z^{-1} + z^{-2}} = \dfrac{1-z^{-1}}{(1-2z^{-1})(1-\tfrac12 z^{-1})}$
Poles and zero	poles $z = \tfrac12$ and $z = 2$; zero $z = 1$
Partial-fraction residues	$A = \tfrac23$ (pole 2), $B = \tfrac13$ (pole $\tfrac12$)
(b) ROC 1: $\|z\| \gt 2$	causal, unstable
(b) ROC 2: $\tfrac12 \lt \|z\| \lt 2$	two-sided, stable (contains $\|z\| = 1$)
(b) ROC 3: $\|z\| \lt \tfrac12$	anticausal, unstable
(c) $h[0]$ for the three ROCs	$1$, $\tfrac13$, $0$ respectively

Topic. The region of convergence as the missing half of a z-domain system description.

Key relations. $x[n-k] \leftrightarrow z^{-k}X(z)$. A rational $H(z)$ admits one ROC per annulus bounded by distinct pole magnitudes: for $P$ distinct magnitudes there are $P+1$ choices. Causal $\Leftrightarrow$ ROC outside the outermost pole and $H(\infty)$ finite; stable $\Leftrightarrow$ ROC contains $|z| = 1$. Standard pairs: $a^n u[n] \leftrightarrow 1/(1-az^{-1})$, $|z| \gt |a|$; $-a^n u[-n-1] \leftrightarrow 1/(1-az^{-1})$, $|z| \lt |a|$. Initial-value theorem: $h[0] = \lim_{z\to\infty}H(z)$ for causal $h$.

Why this works. The algebra of $H(z)$ fixes only the poles and zeros; the ROC records which of the several sequences sharing that algebra is actually meant. Since a pole at $|p|$ can be expanded either as a decaying right-sided geometric or as a growing left-sided one, each pole offers a binary choice, and the requirement that the ROC be a connected annulus reduces the combinations to $P+1$. Here that yields three systems with the same difference equation and three genuinely different impulse responses — and because both poles are real and straddle the unit circle, no single choice is both causal and stable. That trade-off is the standard motivation for non-causal implementations such as zero-phase filtering, where a recorded signal is filtered forwards and then backwards.

Common pitfall. Assuming causality because the difference equation "looks recursive" and reporting only $|z| \gt 2$ with $h[0] = 1$. The exam deliberately omits any causality statement, and part (c) is designed so that a candidate who missed the point produces one number instead of three. The second common slip is to make the pole at $z = 2$ right-sided inside the annulus; the rule is that a pole contributes a left-sided term whenever the ROC lies inside that pole's circle, which for the middle annulus applies to the pole at 2.

Source. Oppenheim & Schafer, Discrete-Time Signal Processing, 3rd ed., §3.2–3.3 (ROC properties), §3.5 (inverse z-transform by partial fractions) and Table 3.1, reproduced on page 8 of this exam.

Question 5 of 6: System function, all admissible ROCs and the value of $h[0]$

Question 5: System function, all admissible ROCs and the value of $h[0]$ (12 marks)