GATE 2025 IN Q.47: solution and answer

Official answer: BSource: Official answer key, GATE 2025 IN (IIT Roorkee)

1. Desired closed-loop pole: $s_d=-1.5+j\frac{\sqrt{27}}{2}=-1.5+j2.598$. Since ${\left(\frac{\sqrt{27}}{2}\right)}^2=6.75$, its distance from the origin is $|s_d|=\sqrt{1.5^2+6.75}=\sqrt{9}=3$, so ${\omega }_n=3\text{rad/s}$ and $\zeta =1.5/3=0.5$.

2. Angle of the vector drawn from a pole at the origin to $s_d$, which lies in the second quadrant: $\angle s_d=180^{\circ }-{\tan }^{-1}\left(\frac{2.598}{1.5}\right)=180^{\circ }-{\tan }^{-1}\left(\sqrt{3}\right)=180^{\circ }-60^{\circ }=120^{\circ }$.

3. Plant phase at $s_d$: $G(s)=1/s^2$ has a double pole at the origin, and each pole contributes $-\angle s_d$, so $\angle G(s_d)=-2\times 120^{\circ }=-240^{\circ }$.

4. Root-locus angle condition: for $s_d$ to be a closed-loop pole of the unity-feedback loop, $\angle C(s_d)+\angle G(s_d)=\pm 180^{\circ }(2k+1)$.

5. Required compensator phase: $\angle C(s_d)=-180^{\circ }-(-240^{\circ })=+60^{\circ }$. (Using $+180^{\circ }$ instead gives $420^{\circ }$, which is the same direction, $60^{\circ }$.) The compensator must supply a phase lead of $60^{\circ }$.

6. Check: without compensation, $1+K/s^2=0$ gives $s=\pm j\sqrt{K}$, so the locus lies only on the $j\omega$ axis. The locus has to be pulled into the left half-plane, and a positive (lead) angle at $s_d$ is what does that. This agrees with option (B).