GATE 2024 IN Q.55: solution and answer

Official answer: 2Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. Read the map: poles at $0$, $-1$ and $-4$, no zeros. With unity negative feedback, $G(s)H(s)=\frac{K}{s(s+1)(s+4)}$.

2. Characteristic equation: $1+G(s)=0\Rightarrow s(s+1)(s+4)+K=0\Rightarrow s^3+5s^2+4s+K=0$.

3. Routh array: $s^3:1,4$; $s^2:5,K$; $s^1:\frac{5\cdot 4-K}{5}=\frac{20-K}{5}$; $s^0:K$.

4. The locus crosses the $j\omega$-axis when the $s^1$ row becomes zero: $20-K=0\Rightarrow K=20$.

5. Auxiliary equation from the $s^2$ row: $5s^2+20=0\Rightarrow s^2=-4\Rightarrow s=\pm j2$.

6. Check by substituting $s=j\omega$: $$-j{\omega }^3-5{\omega }^2+j4\omega +K=0$$ Imaginary part: $\omega (4-{\omega }^2)=0\Rightarrow \omega =2\text{rad/s}$ (taking $\omega >0$). Real part: $K-5{\omega }^2=0\Rightarrow K=20>0$, so the point lies on the positive-$K$ root locus.

7. Angle check at $s=j2$: the pole angles are $90^{\circ }$ (from $0$), ${\tan }^{-1}(2)=63.43^{\circ }$ (from $-1$) and ${\tan }^{-1}(0.5)=26.57^{\circ }$ (from $-4$). They sum to $180^{\circ }$, so $\angle G(j2)=-180^{\circ }$ and the angle condition holds.

8. Hence $s=j\alpha =j2$, giving $\alpha =2$.