ISRO EC 2023 Q.31: solution and answer

Official answer: BSource: ISRO's revised and final answer key of 16 February 2024

1. Take the z-transform with zero initial conditions: $Y(z)\left(1-1.8z^{-1}+0.72z^{-2}\right)=X(z)\left(1+0.5z^{-1}\right)$.

2. So $H(z)=\frac{1+0.5z^{-1}}{1-1.8z^{-1}+0.72z^{-2}}=\frac{z(z+0.5)}{z^2-1.8z+0.72}$. Stability depends on the poles, which are the roots of the denominator.

3. Poles: $z^2-1.8z+0.72=0$, so $z=\frac{1.8\pm \sqrt{3.24-2.88}}{2}=\frac{1.8\pm 0.6}{2}$.

4. This gives $z=1.2$ and $z=0.6$.

5. The difference equation is solved forward in time, so the system is causal. A causal system is BIBO stable only if all poles lie strictly inside the unit circle.

6. The pole at $z=1.2$ has $|z|>1$, so the impulse response contains a term $(1.2{)}^{n}u(n)$ that grows without bound.

7. The numerator zero at $z=-0.5$ (and the factor $z$) does not cancel this pole, so the system is unstable. Option (B).