GATE 2018 IN Q.38: solution and answer

Official answer: 8.9 to 9.1Source: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Characteristic equation of the negative-feedback loop: $1+G(s)H(s)=0\Rightarrow 1+\frac{s+\alpha }{s(s+1)(s+2)}=0\Rightarrow s(s+1)(s+2)+(s+\alpha )=0$.

2. Expand: $s^3+3s^2+2s+s+\alpha =0\Rightarrow s^3+3s^2+3s+\alpha =0$.

3. Routh array: $$\begin{array}{rl}{s}^3& :13\\ s^2& :3\alpha \\ s^1& :\frac{3\cdot 3-1\cdot \alpha }{3}=\frac{9-\alpha }{3}\\ s^0& :\alpha \end{array}$$

4. Stability needs every first-column entry positive: $\alpha >0$ and $9-\alpha >0$, i.e. $0<\alpha <9$.

5. At the upper limit the $s^1$ row becomes entirely zero, which signals a pair of roots symmetric about the origin: $\frac{9-\alpha }{3}=0\Rightarrow \alpha =9$.

6. Auxiliary equation from the $s^2$ row with $\alpha =9$: $3s^2+9=0\Rightarrow s=\pm j\sqrt{3}$. Check by division: $s^3+3s^2+3s+9=(s^2+3)(s+3)$, so the closed-loop poles are $s=\pm j\sqrt{3}$ and $s=-3$. The pair lies on the $j\omega$ axis and the loop oscillates at $\omega =\sqrt{3}\approx 1.732\text{rad/s}$.

7. The other boundary, $\alpha =0$, is not the intended answer: it makes $H(s)=s/s$, so the zero of $H(s)$ cancels its own pole and the characteristic equation becomes $s(s^2+3s+3)=0$. That gives a single root at the origin (a cancelled mode), not the conjugate pair on the imaginary axis the question describes. Hence $\alpha =9.0$.