GATE 2024 EC Q.37: solution and answer

Official answer: ASource: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. Closed-loop characteristic equation (unity negative feedback, $1+G(s)=0$): $$(s+1)(s+2)(s+3)+K=s^3+6s^2+11s+6+K=0$$

2. Translate the requirement. The impulse response is a sum of terms of the form $e^{p_{i}t}$ (times polynomials or sinusoids in $t$). It decays faster than $e^{-t}$ only if every closed-loop pole satisfies $\text{Re}(p_i)<-1$, i.e. all poles lie strictly to the left of the line $s=-1$.

3. Shift the axis with $s=z-1$, so that the line $s=-1$ becomes the imaginary axis of the $z$-plane: $$(z-1+1)(z-1+2)(z-1+3)+K=z(z+1)(z+2)+K=z^3+3z^2+2z+K=0$$

4. Routh array in $z$: $$\begin{array}{rl}{z}^3& :12\\ z^2& :3K\\ z^1& :\frac{3\cdot 2-1\cdot K}{3}=\frac{6-K}{3}\\ z^0& :K\end{array}$$

5. No sign change in the first column (all $z$-roots in the left half plane, hence all $s$-roots left of $s=-1$): $\frac{6-K}{3}>0$ and $K>0$, giving $0<K<6$.

6. Check the boundaries. At $K=0$ a pole sits exactly at $s=-1$, so that mode decays like $e^{-t}$, not faster. At $K=6$ the $z^1$ row vanishes; the auxiliary equation $3z^2+6=0$ gives $z=\pm j\sqrt{2}$ (indeed $z^3+3z^2+2z+6=(z+3)(z^2+2)$), i.e. $s=-1\pm j\sqrt{2}$, again not faster than $e^{-t}$. So the interval is open: $0<K<6$.

7. Match the options. Only $1\le K\le 5$ lies entirely inside $0<K<6$. For option (B), $K\ge 7$ makes $\frac{6-K}{3}<0$, giving two sign changes, so two poles lie to the right of $s=-1$. For options (C) and (D), $K<0$ makes the last first-column entry negative: one sign change, i.e. a real pole to the right of $s=-1$ (for $K\le -6$ the constant term $6+K\le 0$ puts it at $s=0$ or beyond, so the loop is not even stable). Answer: (A).