Question

The Hibert transform of the signal $x\left(t\right)=\frac{1}{t}$ is

• A. $\pi \delta t$
• B. $-\pi \delta t$
• C. $\pi$
• D. $-\pi$

Answer 1

The correct option is B $-\pi \delta t$

Given, signal $x\left(t\right)=\frac{1}{t}$
We know that , $FT\left[\frac{1}{\pi t}\right]=-jsgn\left(\omega \right)$
$X\left(\omega \right)=FT\left[\frac{1}{t}\right]=-j\pi sgn\left(\omega \right)$
The fourier transform $\hat{X}\left(\omega \right)$ of the Hibert transform $\hat{x}\left(t\right)$ is
$\hat{X}\left(\omega \right)=-jsgn\left(\omega \right)X\left(\omega \right)$
$\hat{X}\left(\omega \right)=-jsgn\left(\omega \right)[-j\pi sgn(\omega \left)\right]$
$\hat{X}\left(\omega \right)=-\pi sg{n}^2\left(\omega \right)$
Since, $sg{n}^2\left(\omega \right)=1$ for all $\omega$, we obtain
$\hat{X}\left(\omega \right)=-\pi$
By taking inverse fourier transform,
$\hat{x}\left(t\right)=-\pi \delta \left(t\right)$