Question 3 of 6: ROC, realness and z-domain scaling of a causal stable system

Question 3: ROC, realness and z-domain scaling of a causal stable system (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. From the printed pole-zero plot: a double zero at the origin and a conjugate pole pair at $z = \tfrac{1}{2}e^{\pm j\pi/4}$, i.e. pole radius $r = 1/2$ and pole angle $\theta = \pi/4$. The system is stated to be causal and stable, and $H(1) = 4/(5-2\sqrt{2})$ is supplied.

Find. (a) the ROC with justification, (b) whether $h[n]$ is real with justification, and (c), (d) the pole-zero plots of the two exponentially weighted sequences.

Approach. Identify $H(z)$ from the plot and the supplied $H(1)$, then answer (a) from the ROC rules for causal and stable systems, (b) from conjugate symmetry of the pole-zero pattern, and (c), (d) from the single z-transform property $z_{0}^{\,n}x[n] \leftrightarrow X(z/z_{0})$ with $\text{ROC} = |z_{0}|R_{x}$ — the property printed on page 8 of the paper.

1. Read $H(z)$ off the plot. Two poles and a double zero at the origin give $$H(z) = \frac{A\,z^{2}}{(z - re^{j\theta})(z - re^{-j\theta})} = \frac{A}{1 - 2r\cos\theta\,z^{-1} + r^{2}z^{-2}} .$$ With $r = 1/2$ and $\theta = \pi/4$, $2r\cos\theta = 2(\tfrac12) (\tfrac{\sqrt2}{2}) = \tfrac{\sqrt2}{2}$ and $r^{2} = \tfrac14$, so $$H(z) = \frac{A}{1 - \frac{\sqrt{2}}{2}z^{-1} + \frac{1}{4}z^{-2}} .$$ Evaluating the denominator at $z = 1$ gives $1 - \tfrac{\sqrt2}{2} + \tfrac14 = \tfrac{5 - 2\sqrt2}{4}$, so $H(1) = 4A/(5-2\sqrt2)$. Matching the supplied value fixes $$\boxed{A = 1,\qquad H(z) = \frac{1}{1 - \frac{\sqrt{2}}{2}z^{-1} + \frac{1}{4}z^{-2}}}$$ The double zero at the origin is what makes the transfer function exactly second order in $z^{-1}$ with no numerator delay.
2. (a) The ROC. For a causal sequence the ROC is the exterior of a circle, $|z| \gt R$; for a stable system the ROC must contain the unit circle. The only pole radius is $|{\tfrac12}e^{\pm j\pi/4}| = \tfrac12$, and an ROC can never contain a pole, so the exterior region must start outside the outermost pole: $$\boxed{\text{ROC}:\quad \tfrac{1}{2} \lt |z| \le \infty }$$ This region does contain $|z| = 1$, so the two requirements are mutually consistent — as they must be, since both poles lie strictly inside the unit circle. (Had a pole sat outside $|z|=1$, no ROC could have been both causal and stable.) Note that the two zeros at $z=0$ place no restriction: a zero may lie anywhere, including inside the excluded disc.
3. (b) Is $h[n]$ real? Yes. A sequence is real if and only if its transform satisfies $H^{*}(z^{*}) = H(z)$, and equivalently if and only if every pole and zero is either real or occurs in a complex-conjugate pair. Here the two poles $\tfrac12 e^{+j\pi/4}$ and $\tfrac12 e^{-j\pi/4}$ are conjugates of each other and the double zero at the origin is real, so the coefficients of $H(z)$ are real — visibly so above, $-\tfrac{\sqrt2}{2}$ and $+\tfrac14$. A real rational system function driven by a real impulse produces a real output, so $h[n]$ is real. Explicitly, the standard pair for this denominator gives $$\boxed{\,h[n] = \frac{r^{n}\sin\!\big((n+1)\theta\big)}{\sin\theta}\,u[n] = \left(\tfrac{1}{2}\right)^{n} \frac{\sin\!\big((n+1)\pi/4\big)}{\sin(\pi/4)}\,u[n]\,}$$ whose first samples $1,\ \tfrac{\sqrt2}{2},\ \tfrac14,\ 0,\ -\tfrac{1}{16},\dots$ are manifestly real, and which the difference equation $h[n] = \tfrac{\sqrt2}{2}h[n-1] - \tfrac14 h[n-2] + \delta[n]$ reproduces term by term.
4. (c) Pole-zero plot of $\left(\tfrac12\right)^{n}h[n]$. The exponential-weighting property is $$z_{0}^{\,n}h[n] \;\longleftrightarrow\; H\!\left(\frac{z}{z_{0}}\right), \qquad \text{ROC} = |z_{0}|\,R_{h},$$ so with $z_{0} = \tfrac12$ the transform is $H(2z)$. Substituting $z/z_{0}$ for $z$ moves every pole and zero from $p$ to $z_{0}p$: each pole radius is multiplied by $\tfrac12$ while its angle is unchanged, and the double zero at the origin stays at the origin (scaling zero gives zero). Hence $$\boxed{\text{poles at } \tfrac{1}{4}e^{\pm j\pi/4},\quad \text{double zero at } z=0,\quad \text{ROC } |z| \gt \tfrac{1}{4}}$$ Because $z_{0}$ is real and positive the conjugate symmetry survives, so this sequence is still real — and it is "more stable" than the original, its poles having moved further inside the unit circle.

5. (d) Pole-zero plot of $\left(\tfrac{j}{2}\right)^{n}h[n]$. Now $z_{0} = \tfrac{j}{2} = \tfrac12 e^{j\pi/2}$, so the transform is $H\!\left(z/(j/2)\right) = H(-2jz)$ and each pole moves to $z_{0}p$: the radius is again halved, but the angle is now rotated by $+\pi/2$. Therefore $$\tfrac{j}{2}\cdot\tfrac12 e^{+j\pi/4} = \tfrac14 e^{j3\pi/4}, \qquad \tfrac{j}{2}\cdot\tfrac12 e^{-j\pi/4} = \tfrac14 e^{j\pi/4},$$ and $$\boxed{\text{poles at } \tfrac{1}{4}e^{j\pi/4}\ \text{and}\ \tfrac{1}{4}e^{j3\pi/4},\quad \text{double zero at } z=0,\quad \text{ROC } |z| \gt \tfrac{1}{4}}$$ The ROC radius is $|z_{0}|\,R_{h} = \tfrac12\cdot\tfrac12 = \tfrac14$, the same as in (c), because only $|z_{0}|$ enters the ROC. The important structural difference is that both poles now sit in the upper half-plane and are no longer conjugates of one another, so the pole-zero pattern has lost its conjugate symmetry and $\left(\tfrac{j}{2}\right)^{n}h[n]$ is complex. That is exactly what one expects: multiplying a real sequence by $j^{n}$, which cycles through $1, j, -1, -j$, cannot leave it real.

Parts (c) and (d) are the same theorem applied twice; the pedagogical point is that the ROC responds only to $|z_{0}|$ while the realness of the sequence responds only to $\arg z_{0}$. A real positive $z_{0}$ rescales radii and preserves realness; a complex $z_{0}$ rotates the pattern and destroys it.