GATE 2018 IN Q.12: solution and answer

Official answer: 90, or -270Source: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Sinusoidal steady state: $G(s)$ has its only pole at $s=-1$, so it is stable and the steady state exists. An input $\sin (\omega t)$ then gives the output $|G(j\omega )|\sin \left(\omega t+\angle G(j\omega )\right)$. Here $\omega =1\text{rad/s}$.

2. Substitute $s=j$: $$G(j1)=\frac{-1+j}{1+j}$$

3. Rationalise: $$G(j1)=\frac{(-1+j)(1-j)}{(1+j)(1-j)}=\frac{-1+j+j-j^2}{1-j^2}=\frac{-1+2j+1}{2}=\frac{2j}{2}=j$$

4. Magnitude: $|G(j1)|=1$. This is an all-pass network ($|G(j\omega )|=1$ at every $\omega$), which matches the unit amplitude of $y(t)$ given in the question.

5. Phase: $\angle G(j1)=\angle j=90^{\circ }$. Check by quadrants: the numerator $-1+j$ lies in the second quadrant, so its angle is $180^{\circ }-45^{\circ }=135^{\circ }$. The denominator $1+j$ has angle $45^{\circ }$. Difference: $135^{\circ }-45^{\circ }=90^{\circ }$. In general, $$\angle G(j\omega )=\left(180^{\circ }-{\tan }^{-1}\omega \right)-{\tan }^{-1}\omega =180^{\circ }-2{\tan }^{-1}\omega ,$$ which is $180^{\circ }-90^{\circ }=90^{\circ }$ at $\omega =1\text{rad/s}$.

6. So $y(t)=\sin (t+90^{\circ })$ and $\phi =90^{\circ }$. This already lies in $[0^{\circ },360^{\circ }]$; the equivalent angle $-270^{\circ }$ is the same phase.