Question

Let x(t) be a signal with Fourier transform $X\left(\omega \right)$. Meeting following conditions

1. $x\left(t\right)$ is real

2. $x\left(t\right)=0fort\le 0$

3. $\frac{1}{2\pi }{\int }_{-\infty }^{\infty }$ $Re\left\{X\right(\omega \left)\right\}$$e^{j\omega t}d\omega =|t|e^{-|t|}$

Re{p} means real part of complex quantity p. The expression for x(t) is

• A. $2|t|e^{-|t|}$ t < 0
• B. $2t{e}^{-t}u\left(t\right)$ t > 0
• C. $\frac{t{e}^{-t}-t{e}^t}{2}u\left(t\right)$ t > 0
• D. $\frac{t{e}^{-t}-2t{e}^t}{2}u\left(t\right)$ t > 0

Answer 1

The correct option is B $2t{e}^{-t}u\left(t\right)$ t > 0

$\frac{1}{2\pi }{\int }_{-\infty }^{\infty }Re\left\{X\right(\omega \left)\right\}$$e^{j\omega t}d\omega =|t|e^{-|t|}$

$F^{-1}\left[Re\left\{X\right(\omega \left)\right\}\right]=|t|e^{-|t|}$

We know that

$F\left[E\left\{x\right(t\left)\right\}\right]\leftrightarrow Re\left\{X\right(\omega \left)\right\}$

$x_e\left(t\right)=\frac{x\left(t\right)+x(-t)}{2}$

$\frac{x\left(t\right)+x(-t)}{2}=|t|e^{-|t|}$

Given that $x\left(t\right)=0fort<0$

$x(-t)=0fort>0$

$x\left(t\right)=2|t|e^{-|t|}$ $t\ge 0$

Therefore, $x\left(t\right)=2t{e}^{-t}u\left(t\right)$