GATE 2020 EC Q.49: solution and answer

Official answer: 3.95 to 4.05Source: Official answer key, GATE 2020 EC (IIT Delhi)

1. Sinusoidal steady state (frequency response): the poles of $G(s)$ are at $s=-1$ and $s=-a$. Both lie in the left half-plane because $a>0$, so the system is stable and a sinusoidal steady state exists. For the input $A\cos \omega t$ the steady-state output is $A|G(j\omega )|\cos \left(\omega t+\angle G(j\omega )\right)$. Here $A=5$ and $\omega =3\text{rad/s}$.

2. Magnitude at $\omega =3$: $|G(j3)|=\frac{1}{|j3+1||j3+a|}=\frac{1}{\sqrt{3^2+1}\sqrt{3^2+a^2}}=\frac{1}{\sqrt{10}\sqrt{9+a^2}}$.

3. Match the output amplitudes: $$\frac{5}{\sqrt{10}\sqrt{9+a^2}}=\frac{1}{\sqrt{10}}\Rightarrow \sqrt{9+a^2}=5\Rightarrow a^2=16$$

4. Use $a>0$: this gives $a=4$.

5. Check with the phase (in radians): $\angle G(j3)=-{\tan }^{-1}(3)-{\tan }^{-1}\left(\frac{3}{4}\right)=-(1.2490+0.6435)=-1.8925\text{rad}$. This matches the printed $-1.892$, which confirms $a=4$.