Question 4 of 6: Partial fractions, stability and causality from the ROC

Question 4: Partial fractions, stability and causality from the ROC (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. Part (a): an all-pole second-order system $H(z) = 1/(1 - \tfrac14 z^{-2})$ whose impulse response is stated to be a sum of two one-sided geometric terms, which fixes the ROC as the causal one. Part (b): $H(z) = \bigl(1 - (\tfrac12)^{10}z^{-10}\bigr)/\bigl(1 - \tfrac12 z^{-1}\bigr)$ with the ROC printed as $|z| \gt 0$.

Find. (a)(i) the four constants $A_1$, $A_2$, $\alpha_1$, $\alpha_2$; (a)(ii) a justified stability verdict; (b) a justified causality verdict.

Approach. Factor the denominator of part (a) into two first-order terms, expand in partial fractions and read the constants off; then test stability by whether the ROC contains the unit circle. In part (b) let the printed ROC do the work: an ROC that reaches the whole plane except the origin is the signature of a finite-length causal sequence.

1. Factor the denominator of part (a). Treating $z^{-1}$ as the variable, $$1 - \tfrac14 z^{-2} = \left(1 - \tfrac12 z^{-1}\right)\left(1 + \tfrac12 z^{-1}\right),$$

   so the two poles are $z = +\tfrac12$ and $z = -\tfrac12$, and comparing with the stated form of $h[n]$ identifies the geometric ratios immediately as $\alpha_1 = \tfrac12$ and $\alpha_2 = -\tfrac12$.

2. Expand in partial fractions. Write $$H(z) = \frac{A_1}{1 - \tfrac12 z^{-1}} + \frac{A_2}{1 + \tfrac12 z^{-1}},$$

   and evaluate each residue at its own pole. Multiplying by $\left(1-\tfrac12 z^{-1}\right)$ and setting $z^{-1} = 2$ gives

   $$A_1 = \left.\frac{1}{1 + \tfrac12 z^{-1}}\right|_{z^{-1}=2} = \frac{1}{1+1} = \frac12,$$

   and the mirror calculation at $z^{-1} = -2$ gives $A_2 = \tfrac12$ as well. Hence

   $$\boxed{\ A_1 = A_2 = \tfrac12, \qquad \alpha_1 = \tfrac12, \qquad \alpha_2 = -\tfrac12 \ }$$
3. Check the expansion against the difference equation. The system function corresponds to $y[n] = \tfrac14 y[n-2] + x[n]$, whose impulse response is $1,\,0,\,\tfrac14,\,0,\,\tfrac1{16},\ldots$ — zero at every odd index. The closed form reproduces this exactly, because at odd $n$ the two terms $\tfrac12(\tfrac12)^n$ and $\tfrac12(-\tfrac12)^n$ cancel, and at even $n$ they add to $(\tfrac12)^n$. That parity check is the quickest way to catch a residue sign error.

Part (a): poles at z = +1/2 and z = -1/2 with the causal ROC |z| > 1/2, which contains the unit circle.
4. Decide stability, part (a)(ii). Because $h[n]$ is stated to be right-sided, the ROC is $|z| \gt \max|\alpha_i| = \tfrac12$, which contains the unit circle; equivalently, summing the absolute impulse response, $$\sum_{n=-\infty}^{\infty}|h[n]| = \tfrac12\sum_{n\ge0}\left(\tfrac12\right)^{n} + \tfrac12\sum_{n\ge0}\left(\tfrac12\right)^{n} = \tfrac12(2) + \tfrac12(2) = 2 \lt \infty.$$

   The impulse response is absolutely summable, so the system is BIBO stable. Both poles lie strictly inside the unit circle, which is the same statement for a causal system.

5. Read part (b) through its ROC. The printed region of convergence is $|z| \gt 0$: the entire $z$-plane except the origin. A rational $H(z)$ cannot have that ROC while retaining a pole at $z = \tfrac12$, so the numerator must cancel it — and it does, since $1 - (\tfrac12)^{10}z^{-10}$ vanishes at $z^{-1} = 2$. Cancelling leaves a polynomial in $z^{-1}$, i.e. a finite-length impulse response.

6. Recover $h[n]$ explicitly. This is exactly entry 13 of the supplied table of z-transform pairs, $$\begin{cases} a^{n}, & 0 \le n \le N-1\\ 0, & \text{otherwise}\end{cases} \;\longleftrightarrow\; \frac{1 - a^{N}z^{-N}}{1 - az^{-1}}, \qquad |z| \gt 0,$$

   read with $a = \tfrac12$ and $N = 10$. Therefore

   $$\boxed{\ h[n] = \left(\tfrac12\right)^{n}, \quad 0 \le n \le 9, \quad \text{and } h[n] = 0 \text{ otherwise} \ }$$

   a truncated geometric pulse of length 10. Its samples are $1,\ 0.5,\ 0.25,\ \ldots,\ (\tfrac12)^{9} = 1/512$, and they sum to $(1 - 2^{-10})/(1 - \tfrac12) = 2 - 1/512 = 1.998$, which is $H(z)$ evaluated at $z = 1$.

Part (b): the impulse response is a truncated geometric pulse of length 10, so h[n] = 0 for every n < 0.
7. Give the causality verdict. Since $h[n] = 0$ for every $n \lt 0$, the system is causal. The same conclusion follows from the ROC without computing $h[n]$: for a rational system function the ROC of a causal sequence is the exterior of a circle including $z = \infty$, and $|z| \gt 0$ certainly includes $z = \infty$. Because the system is also FIR it is stable as well — only nine delays and a finite sum can never diverge. The apparent pole at $z = \tfrac12$ is a removable one; what remains is a ninth-order pole at the origin, which is why the origin alone must be excluded from the ROC.