Question 3 of 6: Pole-zero transformations of a causal stable system (12 marks)

Question 3: Pole-zero transformations 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. A causal, stable system whose pole-zero constellation is fully specified, and two prescribed modifications of its impulse response.

Find. For each transformation: the new pole and zero locations with their rectangular coordinates, the derivation that produces them, the region of convergence, and a causality and stability verdict argued from the pole-zero plot.

Approach. Both transformations are single entries in the z-transform property table printed at the back of the paper — time reversal with a shift for (a), and multiplication by an exponential for (b). Apply each property to get $H_1$ and $H_2$ in closed form, map the ROC the same way, and then read causality from the shape of the ROC and stability from whether it contains the unit circle.

1. Fix the original system function and its ROC. Up to a gain constant $A$, the given constellation is $$H(z)=A\,\frac{z(z+1)}{\left(z-\tfrac12\right)\left(z-\tfrac{j}{2}\right)\left(z+\tfrac{j}{2}\right)}=A\,\frac{z(z+1)}{\left(z-\tfrac12\right)\left(z^{2}+\tfrac14\right)} .$$ A causal system has an ROC that is the exterior of a circle through the outermost pole, and all three poles have radius $\tfrac12$, so $$\text{ROC of } H(z):\quad |z| \gt \tfrac12 .$$ That region contains the unit circle, which is the pole-zero statement of the stability we were told to assume.

Part (a): $h_1[n]=h[1-n]$. Read the operation as a time reversal followed by a shift, since $h[1-n]=h[-(n-1)]$.

2. Apply the two properties in order (part a-ii). The appendix table gives $x[-n]\leftrightarrow X(1/z)$ with the ROC inverted, and $x[n-1]\leftrightarrow z^{-1}X(z)$. Composing them, $$H_1(z)=z^{-1}H(1/z) .$$ Substituting and clearing the reciprocal powers, $$H(1/z)=A\,\frac{\tfrac1z\left(\tfrac1z+1\right)}{\left(\tfrac1z-\tfrac12\right)\left(\tfrac{1}{z^{2}}+\tfrac14\right)} = -8A\,\frac{z(z+1)}{(z-2)(z^{2}+4)} ,$$ and therefore $$\boxed{\,H_1(z)=z^{-1}H(1/z)=-8A\,\frac{z+1}{(z-2)\left(z^{2}+4\right)}\,}$$
3. Read the new poles and zeros (part a-i). Reciprocation maps every finite pole $p$ to $1/p$, so the poles move outward by a factor of four in radius: $$\tfrac12 \mapsto 2,\qquad \tfrac{j}{2}\mapsto \frac{1}{j/2}=-2j,\qquad -\tfrac{j}{2}\mapsto +2j .$$ The pole pair is simply relabelled, so the pole set is $\{2,\,+2j,\,-2j\}$, i.e. rectangular coordinates $(2,0)$, $(0,2)$ and $(0,-2)$. The zeros need one extra step: reciprocation sends the zero at $z=0$ out to $z=\infty$ and brings the implicit zero at infinity in to $z=0$, while $z=-1$ maps to itself. The factor $z^{-1}$ from the shift then cancels that new zero at the origin, leaving a single finite zero at $(-1,0)$.
4. Map the ROC (part a-iii). Under $z\to 1/z$ an exterior region becomes an interior one: $|z| \gt \tfrac12$ becomes $|z| \lt 2$. Multiplying by $z^{-1}$ can only affect the point $z=0$, and $H_1(z)$ is finite there, so the origin stays inside: $$\boxed{\,\text{ROC of } H_1(z):\ |z| \lt 2\,}$$
5. Causality and stability from the plot (parts a-iv, a-v). The ROC is the interior of a circle, which is the signature of a left-sided (anti-causal) sequence — consistent with the direct reading that $h[1-n]$ is non-zero only for $n\le 1$, since $h$ is causal. So the system is not causal. The ROC $|z| \lt 2$ does contain the unit circle, so the system is stable; equivalently, all three poles lie outside the unit circle, which for a left-sided sequence is exactly the stability condition.

Part (b): $h_2[n]=\left(2e^{j\pi/4}\right)^{n}h[n]$. This is the exponential-modulation property, listed in the appendix as $z_0^{\,n}x[n]\leftrightarrow X(z/z_0)$ with the ROC scaled by $|z_0|$.

6. Apply the scaling property (part b-ii). With $z_0=2e^{j\pi/4}$, $$\boxed{\,H_2(z)=H\!\left(\frac{z}{z_0}\right),\qquad z_0=2e^{j\pi/4}\,}$$ Every pole and zero is multiplied by $z_0$, that is scaled in radius by $|z_0|=2$ and rotated counter-clockwise by $\arg z_0=\pi/4$.
7. Locate the new poles (part b-i). Each original pole has radius $\tfrac12$, and $\tfrac12\times 2 = 1$, so all three poles land exactly on the unit circle: $$\begin{aligned} \tfrac12 &\mapsto e^{j\pi/4} = \left(\tfrac{\sqrt2}{2},\ \tfrac{\sqrt2}{2}\right) \approx (0.707,\ 0.707),\\ \tfrac{j}{2} &\mapsto e^{j\pi/2}e^{j\pi/4}=e^{j3\pi/4}\approx (-0.707,\ 0.707),\\ -\tfrac{j}{2} &\mapsto e^{-j\pi/2}e^{j\pi/4}=e^{-j\pi/4}\approx (0.707,\ -0.707). \end{aligned}$$ The zeros scale the same way: $z=0$ stays at the origin, and $z=-1$ moves to $-2e^{j\pi/4}=2e^{j5\pi/4}\approx(-1.414,\ -1.414)$.
8. Map the ROC (part b-iii). The scaling property multiplies the ROC radii by $|z_0|$: $$|z| \gt \tfrac12 \ \longrightarrow\ |z| \gt \tfrac12\times 2 \quad\Longrightarrow\quad \boxed{\,\text{ROC of } H_2(z):\ |z| \gt 1\,}$$
9. Causality and stability from the plot (parts b-iv, b-v). The ROC is the exterior of a circle and extends to $z=\infty$, the signature of a right-sided causal sequence — which also follows directly, because multiplying a causal $h[n]$ by $z_0^{\,n}$ cannot create samples at negative $n$. The system therefore is causal. However the ROC $|z| \gt 1$ is an open region that excludes its boundary, and the unit circle is that boundary: the three poles sit on it. The ROC does not contain $|z|=1$, so the system is not stable. Physically, the modulation multiplied the impulse response by a factor of magnitude $2^{\,n}$, which exactly cancelled the $\left(\tfrac12\right)^{n}$ decay of the original modes and left non-decaying oscillations.