GATE 2025 EE Q.65: solution and answer

Official answer: 0.1Source: Official answer key, GATE 2025 EE (IIT Roorkee)

1. With unity negative feedback, the open-loop transfer function is $$L(s)=D(s)G(s)=\frac{1000\sqrt{2}(1+K_{D}s)}{s(s+10{)}^2}.$$

2. Plant magnitude at $\omega =10\text{rad/s}$: $|j10|=10$ and $|j10+10{|}^2=10^2+10^2=200$, so $$|G(j10)|=\frac{1000\sqrt{2}}{10\times 200}=\frac{1000\sqrt{2}}{2000}=\frac{1}{\sqrt{2}}.$$

3. Plant phase at $\omega =10\text{rad/s}$: the integrator gives $-90^{\circ }$ and each of the two poles at $s=-10$ gives $-{\tan }^{-1}(10/10)=-45^{\circ }$, so $$\angle G(j10)=-90^{\circ }-2(45^{\circ })=-180^{\circ }.$$

4. Phase-margin condition at the gain crossover: $\text{PM}=180^{\circ }+\angle L(j10)=45^{\circ }$, so $\angle L(j10)=-135^{\circ }$. Since the plant already gives $-180^{\circ }$, the controller has to add $+45^{\circ }$ at $\omega =10\text{rad/s}$.

5. Phase of the PD controller: $\angle D(j10)={\tan }^{-1}(10K_D)=45^{\circ }$, so $10K_D=1$ and $K_D=0.1$.

6. Check that $10\text{rad/s}$ really is the gain crossover: $|D(j10)|=|1+j1|=\sqrt{2}$, so $|L(j10)|=\sqrt{2}\cdot \frac{1}{\sqrt{2}}=1$. The magnitude and phase conditions both hold at the same frequency, so $K_D=0.1$ is consistent.