GATE 2020 IN Q.19: solution and answer

Official answer: 2Source: Official answer key, GATE 2020 IN (IIT Delhi)

1. Put $\omega =0$ in the definition. Then $e^{-j\omega t}=e^0=1$, so the transform at zero frequency is the total area under the signal: $$X(j0)={\int }_{-\infty }^{\infty }x(t)dt={\int }_{-\infty }^{\infty }{e}^{-|t|}dt$$

2. $e^{-|t|}$ is an even function of $t$, so the integral over the whole line is twice the integral over $t\ge 0$. There $|t|=t$: $$X(j0)=2{\int }_0^{\infty }{e}^{-t}dt$$

3. Evaluate: $$2{\int }_0^{\infty }{e}^{-t}dt=2[-e^{-t}{]}_0^{\infty }=2\left(0-(-1)\right)=2$$

4. Cross-check with the standard pair $e^{-a|t|}\leftrightarrow \frac{2a}{a^2+{\omega }^2}$ for $a>0$. With $a=1$, $X(j\omega )=\frac{2}{1+{\omega }^2}$, and at $\omega =0$ this gives $X(j0)=2$.

5. Therefore $X(j0)=2$.