GATE 2018 IN Q.33: solution and answer

Official answer: CSource: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Read the spectrum from the figure: $X(j\omega )=1+\omega$ for $-1\le \omega \le 0$ (the ramp from $0$ at $\omega =-1$ to $1$ at $\omega =0$), $X(j\omega )=1$ for $0<\omega <1$ (the flat part), and $X(j\omega )=0$ elsewhere.

2. Frequency-shift property: multiplying a signal by $e^{j{\omega }_{0}t}$ shifts its spectrum to the right by ${\omega }_0$, i.e. $e^{j{\omega }_{0}t}x(t)\leftrightarrow X(j(\omega -{\omega }_0))$. Here $e^{jt}=e^{j(1)t}$, so ${\omega }_0=1\text{rad/s}$.

3. By linearity, $$Y(j\omega )=X(j\omega )+X(j(\omega -1))$$

4. First term at $\omega =0.5$: $\omega =0.5$ lies on the flat part, so $X(j0.5)=1$.

5. Second term at $\omega =0.5$: $X(j(0.5-1))=X(j(-0.5))$. The frequency $-0.5$ lies on the ramp, so $X(j(-0.5))=1+(-0.5)=0.5$.

6. Add the two: $$Y(j0.5)=1+0.5=1.5$$ which is option (C).