GATE 2020 EC Q.52: solution and answer

Official answer: 58.50 to 58.80Source: Official answer key, GATE 2020 EC (IIT Delhi)

1. Parseval's theorem (with $\omega$ in rad/s): ${\int }_{-\infty }^{\infty }|X(\omega ){|}^{2}d\omega =2\pi {\int }_{-\infty }^{\infty }|x(t){|}^{2}dt=2\pi E_x$. So we only need the energy of $x(t)$; $X(\omega )$ never has to be found.

2. Read the segments off the graph (corners at $(-1,0)$, $(0,1)$, $(1,3)$, $(2,1)$, $(3,0)$): $x(t)=t+1$ on $[-1,0]$, $x(t)=1+2t$ on $[0,1]$, $x(t)=5-2t$ on $[1,2]$, $x(t)=3-t$ on $[2,3]$, and zero elsewhere. The signal is symmetric about $t=1$ (for example $x(1-\tau )=x(1+\tau )=3-2\tau$ for $0\le \tau \le 1$), so the energy of the right half equals the energy of the left half.

3. Outer ramp, $[-1,0]$: ${\int }_{-1}^0(t+1{)}^{2}dt={\int }_0^1{u}^{2}du=\frac{1}{3}$.

4. Inner ramp, $[0,1]$: ${\int }_0^1(1+2t{)}^{2}dt={\int }_0^1(1+4t+4t^2)dt=1+2+\frac{4}{3}=\frac{13}{3}$.

5. By symmetry, $[1,2]$ gives $\frac{13}{3}$ and $[2,3]$ gives $\frac{1}{3}$. Total: $$E_x=2\left(\frac{1}{3}+\frac{13}{3}\right)=\frac{28}{3}\approx 9.333$$

6. Apply Parseval's theorem: $${\int }_{-\infty }^{\infty }|X(\omega ){|}^{2}d\omega =2\pi \cdot \frac{28}{3}=\frac{56\pi }{3}\approx 58.64$$