GATE 2018 EE Q.39: solution and answer

Official answer: DSource: Official answer key, GATE 2018 EE (IIT Guwahati)

1. Write each term as a shifted unit ramp $r(t)=tu(t)$: $$x(t)=r(t-1)-r(t-2)-r(t-3)+r(t-4)$$

2. Build the waveform piece by piece. Each ramp switches on at its breakpoint, so the slope changes by $+1,-1,-1,+1$ at $t=1,2,3,4$.

3. $t<1$: $x(t)=0$. For $1\le t<2$: $x(t)=t-1$, rising from $0$ to $1$ (slope $+1$).

4. $2\le t<3$: $x(t)=(t-1)-(t-2)=1$ (slope $0$). For $3\le t<4$: $x(t)=1-(t-3)=4-t$, falling from $1$ to $0$ (slope $-1$).

5. $t\ge 4$: $x(t)=(4-t)+(t-4)=0$ (slope $0$). So $x(t)$ is a trapezoid: $0$ at $t=1$, up to $1$ at $t=2$, flat until $t=3$, back to $0$ at $t=4$.

6. Energy definition: $E={\int }_{-\infty }^{\infty }|x(t){|}^{2}dt$. Split the integral over the three non-zero pieces: $$E={\int }_1^2(t-1{)}^{2}dt+{\int }_2^3{1}^{2}dt+{\int }_3^4(4-t{)}^{2}dt$$

7. Substituting $\tau =t-1$ in the first integral and $\tau =4-t$ in the third, each ramp edge gives ${\int }_0^1{\tau }^{2}d\tau =\frac{1}{3}$, and the flat top gives $1$: $$E=\frac{1}{3}+1+\frac{1}{3}=\frac{5}{3}$$

8. This matches option (D).