Question 3 of 5: Region of convergence, step response and the inverse system

Question 3: Region of convergence, step response and the inverse system (25 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. $H(z) = z^{-2}\big/\bigl[(1 - \tfrac12 z^{-1}) (1 - 2z^{-1})\bigr]$, whose poles are at $z = \tfrac12$ and $z = 2$ and whose only zeros are a double zero at $z = \infty$; equivalently $H(z) = 1/\bigl[(z - \tfrac12)(z-2)\bigr]$. In part (b) the input read from the figure is $x[n] = \{1,\,2,\,1,\,-1,\,\ldots\}$ for $n = 0,1,2,3,\ldots$ with $x[n] = 0$ for $n \lt 0$.

Find. (a) the step response of the stable version of the system; (b) the single output sample $y[2]$ of the version whose ROC contains $z = \infty$; (c) the impulse response of the inverse system and whether it depends on the ROC chosen for $H(z)$.

Approach. Two poles straddle the unit circle, so the ROC — not the algebra — decides the answer. Identify the annulus each part specifies, invert by partial fractions with the matching one-sided pairs, and obtain the inverse system by simply reciprocating $H(z)$.

1. Choose the ROC for part (a). The ROC of a rational transform is always bounded by poles, so the three candidates are $|z| \lt 1/2$, $1/2 \lt |z| \lt 2$ and $|z| \gt 2$. Stability requires the ROC to contain the unit circle, which selects $$\tfrac12 \lt |z| \lt 2 ,$$ a two-sided (and therefore non-causal) impulse response.
2. Form the output transform. The unit step has $X(z) = 1/(1 - z^{-1})$ with ROC $|z| \gt 1$; intersecting with the ROC of $H(z)$, $$Y(z) = \frac{z^{-2}}{\left(1 - \tfrac12 z^{-1}\right)\left(1 - 2z^{-1}\right) \left(1 - z^{-1}\right)}, \qquad 1 \lt |z| \lt 2 .$$
3. Expand in partial fractions. Writing $\alpha = z^{-1}$ and matching $\alpha^{2} = A(1-2\alpha)(1-\alpha) + B(1-\tfrac{\alpha}{2})(1-\alpha) + C(1-\tfrac{\alpha}{2})(1-2\alpha)$ at $\alpha = 2,\ \tfrac12,\ 1$ gives $A = \tfrac43$, $B = \tfrac23$, $C = -2$, so $$Y(z) = \frac{4/3}{1 - \tfrac12 z^{-1}} + \frac{2/3}{1 - 2z^{-1}} - \frac{2}{1 - z^{-1}} .$$ (The check $A + B + C = 0$ reproduces the vanishing constant term of $\alpha^{2}$.)
4. Invert each term with the ROC that applies to it. Inside $1 \lt |z| \lt 2$ the poles at $\tfrac12$ and $1$ lie inside the annulus boundary and give right-sided terms, while the pole at $2$ lies outside and gives a left-sided term: $$\frac{1}{1 - \tfrac12 z^{-1}} \to \left(\tfrac12\right)^{n} u[n], \qquad \frac{1}{1 - 2 z^{-1}} \to -2^{n} u[-n-1], \qquad \frac{1}{1 - z^{-1}} \to u[n].$$ Therefore $$\boxed{\;y[n] = \tfrac{4}{3}\left(\tfrac12\right)^{n} u[n] \;-\; \tfrac{2}{3}\,2^{\,n} u[-n-1] \;-\; 2\,u[n]\;}$$
5. Confirm the two limits. As $n \to +\infty$ the geometric term dies and $y[n] \to -2$, which must equal the dc gain $H(1) = 1/[(1-\tfrac12)(1-2)] = -2$; at $n = 0$ the expression gives $y[0] = \tfrac43 - 2 = -\tfrac23$, and for $n \lt 0$ the anticipatory tail $-\tfrac23 2^{n}$ decays backwards in time (for example $y[-1] = -\tfrac13$), as a two-sided stable response must. Both checks pass, so the ROC assignment was made correctly.
6. Re-select the ROC for part (b). Requiring the ROC to include $z = \infty$ forces the impulse response to have no samples at negative $n$: the system is causal, and the ROC is the outermost annulus $$|z| \gt 2 .$$ (That $H(\infty) = 0$ is finite is guaranteed by the $z^{-2}$ factor, so this choice is admissible.)
7. Get the causal impulse response. Expanding the pole part, $$\frac{1}{\left(1 - \tfrac12 z^{-1}\right)\left(1 - 2z^{-1}\right)} = \frac{-1/3}{1 - \tfrac12 z^{-1}} + \frac{4/3}{1 - 2z^{-1}} ,$$ which for $|z| \gt 2$ inverts to $g[n] = \bigl[-\tfrac13 (\tfrac12)^{n} + \tfrac43\,2^{n}\bigr]u[n]$. The factor $z^{-2}$ is a two-sample delay, so $$h[n] = g[n-2] = \left[-\tfrac13\left(\tfrac12\right)^{n-2} + \tfrac43\,2^{\,n-2}\right] u[n-2],$$ giving $h[0] = h[1] = 0$, $h[2] = 1$, $h[3] = 2.5$, $h[4] = 5.25$ — the same coefficients that long division of $1/(1 - 2.5 z^{-1} + z^{-2})$ produces.

8. Convolve at the single instant $n = 2$ (part b). Causality of both $h$ and $x$ truncates the sum to three terms: $$y[2] = \sum_{k=0}^{2} h[k]\,x[2-k] = h[0]x[2] + h[1]x[1] + h[2]x[0] = 0 + 0 + 1\cdot 1 ,$$ so $$\boxed{\;y[2] = 1\;}$$ Only $x[0]$ contributes: the transfer function begins with a two-sample pure delay, so the output at $n = 2$ can only see the input at $n = 0$. Nothing else in the figure — the tall sample $x[1] = 2$, the negative sample $x[3] = -1$, or the continuation to the right — can influence this value.
9. Invert the system (part c). A cascade recovers the input when $H(z)H_i(z) = 1$, so $$H_i(z) = \frac{1}{H(z)} = \frac{\left(1 - \tfrac12 z^{-1}\right)\left(1 - 2z^{-1}\right)}{z^{-2}} = z^{2}\left(1 - \tfrac52 z^{-1} + z^{-2}\right) = z^{2} - \tfrac52 z + 1 .$$ Reading the coefficients as a sequence, $$\boxed{\;h_i[n] = \delta[n+2] - \tfrac52\,\delta[n+1] + \delta[n]\;}$$ a three-tap FIR filter that advances the signal by two samples. Its middle tap, $-5/2$, is $-(\tfrac12 + 2)$: the sum of the two poles it must cancel.
10. Answer the ROC question, with reasons. $h_i[n]$ is independent of which ROC was assigned to $H(z)$. The reason is structural: $H_i(z)$ is a polynomial in $z$, i.e. a finite-length sequence, and a finite-length sequence has only one possible region of convergence — the entire z-plane except the points $z = \infty$ (excluded by the positive powers of $z$) and, for a general two-sided polynomial, $z = 0$. There is no second inversion to choose between, so all three candidate ROCs of $H(z)$ lead to the same three-tap $h_i[n]$. Consistently, that ROC overlaps every candidate ROC of $H(z)$, so the cascade $h[n] * h_i[n] = \delta[n]$ converges in each case.
11. State the practical caveats. Two remarks complete the answer. First, $h_i[n]$ is non-causal (it has taps at $n = -1$ and $n = -2$), so it cannot be run in real time without inserting a compensating delay of two samples, after which the realisable inverse $z^{-2}H_i(z) = 1 - \tfrac52 z^{-1} + z^{-2}$ recovers $x[n-2]$ rather than $x[n]$. Second, the inverse is FIR and therefore unconditionally stable, which is the mirror image of the usual difficulty: it is the forward system, with a pole at $z = 2$, that cannot be both causal and stable.