Question 3 of 6: System function, ROC, impulse response and difference equation

Question 3: System function, ROC, impulse response and difference equation (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. A two-sided input built from a right-sided geometric term and a left-sided one, and a purely right-sided output.

Find. $H(z)$ with its pole-zero plot and ROC, the impulse response, a difference equation, and the stability and causality verdicts.

Approach. Transform both signals with the standard pairs on page 9, form $H(z) = Y(z)/X(z)$ and cancel the common factor, then choose the one ROC for $H$ that is consistent with $\mathcal{R}_x \cap \mathcal{R}_h \subseteq \mathcal{R}_y$.

1. Transform the input. Pairs 5 and 6 of the exam table give $a^n u[n] \leftrightarrow 1/(1-az^{-1})$ for $|z| \gt |a|$ and $-a^n u[-n-1] \leftrightarrow 1/(1-az^{-1})$ for $|z| \lt |a|$. Hence $$X(z) = \frac{1}{1-\tfrac{1}{3}z^{-1}} - \frac{1}{1-2z^{-1}}, \qquad \tfrac{1}{3} \lt |z| \lt 2.$$ The two half-planes must overlap, which is what makes the ROC an annulus.
2. Put the input over a common denominator. Combining the two terms, $$X(z) = \frac{\left(1-2z^{-1}\right) - \left(1-\tfrac{1}{3}z^{-1}\right)}{\left(1-\tfrac{1}{3}z^{-1}\right)\left(1-2z^{-1}\right)} = \frac{-\tfrac{5}{3}z^{-1}}{\left(1-\tfrac{1}{3}z^{-1}\right)\left(1-2z^{-1}\right)}.$$
3. Transform the output the same way. Both terms are right-sided, so $$Y(z) = \frac{5}{1-\tfrac{1}{3}z^{-1}} - \frac{5}{1-\tfrac{2}{3}z^{-1}} = \frac{-\tfrac{5}{3}z^{-1}}{\left(1-\tfrac{1}{3}z^{-1}\right)\left(1-\tfrac{2}{3}z^{-1}\right)}, \qquad |z| \gt \tfrac{2}{3}.$$ Notice that the numerators of $X$ and $Y$ are identical — the examiner chose the leading 5 to make them so.
4. Divide and cancel. Forming the ratio, the factor $-\tfrac{5}{3}z^{-1}$ and the factor $(1-\tfrac{1}{3}z^{-1})$ both cancel, leaving $$H(z) = \frac{Y(z)}{X(z)} = \frac{1-2z^{-1}}{1-\tfrac{2}{3}z^{-1}} = \frac{z-2}{z-\tfrac{2}{3}}.$$ So there is a single pole at $z = 2/3$ and a single zero at $z = 2$. The cancelled pole at $z = 1/3$ belonged to the input, not to the system.
5. Select the ROC. A single pole leaves only two candidate regions, $|z| \gt 2/3$ or $|z| \lt 2/3$. Convolution requires $\mathcal{R}_x \cap \mathcal{R}_h \subseteq \mathcal{R}_y = \{|z| \gt 2/3\}$. Taking $\mathcal{R}_h = \{|z| \gt 2/3\}$ gives the intersection $\tfrac{2}{3} \lt |z| \lt 2$, which lies inside $\mathcal{R}_y$; taking $\mathcal{R}_h = \{|z| \lt 2/3\}$ gives $\tfrac{1}{3} \lt |z| \lt \tfrac{2}{3}$, which does not. Therefore $$\boxed{\,H(z) = \frac{1-2z^{-1}}{1-\tfrac{2}{3}z^{-1}}, \qquad \text{ROC: } |z| \gt \tfrac{2}{3}\,}$$
6. Part (b): invert. Split the transform into a constant plus a delayed term, $$H(z) = \frac{1}{1-\tfrac{2}{3}z^{-1}} - \frac{2z^{-1}}{1-\tfrac{2}{3}z^{-1}},$$ and use pair 5 with the shift property. This gives $h[n] = (2/3)^n u[n] - 2(2/3)^{n-1}u[n-1]$. Collecting the two terms for $n \ge 1$, where $2(2/3)^{n-1} = 3(2/3)^{n}$, $$\boxed{\,h[n] = \delta[n] - 2\left(\tfrac{2}{3}\right)^{n}u[n-1]\,}$$ so $h[0] = 1$, $h[1] = -4/3$, $h[2] = -8/9$, and the tail decays geometrically.
7. Part (c): read off the difference equation. Cross-multiplying $H(z) = Y(z)/X(z)$ gives $\left(1-\tfrac{2}{3}z^{-1}\right)Y(z) = \left(1-2z^{-1}\right)X(z)$, and each $z^{-1}$ is one sample of delay: $$\boxed{\,y[n] - \tfrac{2}{3}\,y[n-1] = x[n] - 2\,x[n-1]\,}$$ Substituting the printed $x[n]$ and $y[n]$ satisfies this identity at every $n$, positive and negative, which is the check the question is really asking for.
8. Part (d): stability and causality. The ROC $|z| \gt 2/3$ is the exterior of the outermost pole and it includes the unit circle. Including $|z| = 1$ is equivalent to $\sum_n |h[n]| \lt \infty$, so the system is stable; directly, $\sum_{n\ge1} 2(2/3)^n = 4$ converges. An exterior ROC together with a system function that is proper in $z^{-1}$ means $h[n] = 0$ for $n \lt 0$, so the system is also causal. The zero at $z = 2$ lies outside the unit circle, so the system is stable but not minimum phase — its inverse could not be both causal and stable.