GATE 2024 EC Q.24: solution and answer

Official answer: A, C, DSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. Poles: $z=-\frac{1}{3}$ and $z=+\frac{1}{3}$. The system is causal, so the ROC is the exterior of the outermost pole, $|z|>\frac{1}{3}$, and it contains the unit circle.

2. Stability (causal system: every pole inside $|z|=1$): $\left|\pm \frac{1}{3}\right|<1$, so the system is BIBO stable. (A) is true.

3. Zeros: $2z^2+3=0\Rightarrow z^2=-\frac{3}{2}\Rightarrow z=\pm j\sqrt{\frac{3}{2}}$, so $|z|=\sqrt{1.5}\approx 1.22>1$.

4. Minimum phase (discrete time): a causal system is minimum phase when every pole and every zero lies inside the unit circle, so that its inverse is also causal and stable. Both zeros lie outside the unit circle, so the inverse system would have poles at $|z|\approx 1.22$ and could not be both causal and stable. The system is not minimum phase. (B) is false.

5. Initial value theorem (causal $h[n]$): $$h[0]=\underset{z\to \infty }{\lim }H(z)=\underset{z\to \infty }{\lim }\frac{2z^2+3}{z^2-\frac{1}{9}}=2$$ (C) is true.

6. Final value theorem: it applies here because every pole of $(z-1)H(z)$, namely $z=\pm \frac{1}{3}$, lies inside the unit circle. $$h[\infty ]=\underset{z\to 1}{\lim }(z-1)H(z)=0\cdot \frac{2+3}{\frac{4}{3}\cdot \frac{2}{3}}=0$$ This agrees with stability, since the impulse response of a stable system decays to zero. (D) is true.

7. Cross-check by division: $$H(z)=2+\frac{3+\frac{2}{9}}{z^2-\frac{1}{9}}=2+\frac{29}{9}\cdot \frac{z^{-2}}{1-\frac{1}{9}{z}^{-2}}$$ With $|z|>\frac{1}{3}$ this gives $h[n]=2\delta [n]+\frac{29}{9}{\left(\frac{1}{9}\right)}^{(n-2)/2}$ for even $n\ge 2$, and $h[n]=0$ for odd $n$. So $h[0]=2$ and $h[n]\to 0$, which confirms (C) and (D).

8. Answer: (A), (C), (D).