GATE 2023 EC Q.25: solution and answer

Official answer: ASource: Official answer key, GATE 2023 EC (IIT Kanpur)

1. Phase crossover definition: the phase crossover frequency ${\omega }_{pc}$ is the frequency at which $\angle G(j\omega )=-180^{\circ }$. The gain $k>0$ adds no phase, so it drops out and the answer cannot depend on $k$.

2. Phase of each factor: the pole at the origin contributes $-90^{\circ }$, and each first-order pole $(1+j\omega T)$ with $T>0$ contributes $-{\tan }^{-1}(\omega T)$:
   $$\angle G(j\omega )=-90^{\circ }-{\tan }^{-1}(\omega T_1)-{\tan }^{-1}(\omega T_2)$$
   As $\omega$ rises from $0$ to $\infty$, this phase falls steadily from $-90^{\circ }$ to $-270^{\circ }$, so it passes through $-180^{\circ }$ exactly once.

3. Set the phase to $-180^{\circ }$:
   $${\tan }^{-1}(\omega T_1)+{\tan }^{-1}(\omega T_2)=90^{\circ }$$

4. Solve the arctangent equation: write $a=\omega T_1>0$ and $b=\omega T_2>0$. Then ${\tan }^{-1}b=90^{\circ }-{\tan }^{-1}a={\tan }^{-1}\left(\frac{1}{a}\right)$, so $b=\frac{1}{a}$, which means $ab=1$. (Equivalently, the denominator of ${\tan }^{-1}\frac{a+b}{1-ab}$ goes to zero.)
   $${\omega }^2{T}_1{T}_2=1$$

5. Solve for $\omega$:
   $${\omega }_{pc}=\frac{1}{\sqrt{T_1{T}_2}}\text{rad/s}$$

6. Cross-check with real and imaginary parts: the denominator of $G(j\omega )$ is
   $$j\omega (1+j\omega T_1)(1+j\omega T_2)=-{\omega }^2(T_1+T_2)+j\omega \left(1-{\omega }^2{T}_1{T}_2\right)$$
   At ${\omega }^2{T}_1{T}_2=1$ the imaginary part is zero and the real part $-{\omega }^2(T_1+T_2)$ is negative. So $G(j\omega )$ is a negative real number there, which is a phase of $-180^{\circ }$. This matches option (A).