GATE 2024 EE Q.59: solution and answer

Official answer: 1.11 to 1.13Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Open-loop transfer function (unity negative feedback): $L(s)=\frac{K_I}{s}G(s)$. On $s=j\omega$ this gives $\angle L(j\omega )=-90^{\circ }+\angle G(j\omega )$ and $|L(j\omega )|=\frac{K_I}{\omega }|G(j\omega )|$. In decibels: $|L{|}_{\text{dB}}=20{\log }_{10}{K}_I-20{\log }_{10}\omega +|G{|}_{\text{dB}}$.

2. Phase margin definition: $\text{PM}=180^{\circ }+\angle L(j{\omega }_{gc})$, where ${\omega }_{gc}$ is the gain crossover frequency ($|L(j{\omega }_{gc})|=1$, i.e. 0 dB). For $\text{PM}=50^{\circ }$ we need $\angle L(j{\omega }_{gc})=-130^{\circ }$.

3. So $\angle G(j{\omega }_{gc})=-130^{\circ }+90^{\circ }=-40^{\circ }$. The table gives exactly this phase at $\omega =0.5\text{rad/s}$, so no interpolation is needed: ${\omega }_{gc}=0.5\text{rad/s}$, where $|G|=-7\text{dB}$. Changing $K_I$ scales the magnitude only, not the phase, so this frequency is fixed and $K_I$ must be chosen to make it the crossover.

4. Gain crossover condition $|L(j{\omega }_{gc})|=1$: $$\frac{K_I}{0.5}\cdot 10^{-7/20}=1$$

5. Solve: $$K_I=0.5\times 10^{7/20}=0.5\times 10^{0.35}=0.5\times 2.2387\approx 1.1194$$

6. Check stability with this $K_I$ ($20{\log }_{10}1.1194\approx 0.98\text{dB}$). At $\omega =1\text{rad/s}$: $|L|\approx 0.98-0-10=-9.02\text{dB}$, $\angle L=-170^{\circ }$. At $\omega =2\text{rad/s}$: $|L|\approx 0.98-6.02-18=-23.04\text{dB}$, $\angle L=-220^{\circ }$. So the phase crossover ($\angle L=-180^{\circ }$) lies between 1 and 2 rad/s, where $|L|$ is already below $-9\text{dB}$. The gain margin is positive and the phase margin is $+50^{\circ }$, so the loop is stable, as the question states.

7. Rounded to 2 decimal places: $K_I\approx 1.12$.