ISRO EC 2023 Q.58: solution and answer

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

1. The transfer function is $G(s)=\frac{4s+1}{4s^2+1}$. The numerator has no common factor with the denominator, so the poles are the roots of $4s^2+1=0$.

2. Routh array for $4s^2+0s+1$: the $s^2$ row is $4,1$ and the $s^1$ row is $0$ (an entire row of zeros), so the auxiliary polynomial is $A(s)=4s^2+1$ with $\frac{dA}{ds}=8s$.

3. Replace the zero row with the coefficient $8$. The $s^0$ row is $1$. The first column is $4,8,1$: no sign change, so no root lies in the right half-plane.

4. Solve directly: $s^2=-\frac{1}{4}$, so $s=\pm j\frac{1}{2}$. These are two simple, distinct poles on the imaginary axis, and they are the roots of the auxiliary polynomial.

5. A system whose poles are all in the left half-plane or simple on the $j\omega$-axis, with at least one on the axis, is marginally stable. Here the impulse response contains sustained oscillation at $\omega =0.5\text{rad/s}$ that neither grows nor decays. It is not asymptotically stable, and it is not unstable because the imaginary-axis poles are not repeated. Stability is fully decided by the pole locations, so the answer is marginally stable.