Question

The Fourier transform $X(j\omega)$ of the signal \[ x(t)=\frac{t}{(1+t^{2})^{2}} \] is ________________.

• A. $\displaystyle \frac{\pi}{2j}\,\omega e^{-|\omega|}$
• B. $\displaystyle \frac{\pi}{2}\,\omega e^{-|\omega|}$
• C. $\displaystyle \frac{\pi}{2j}\, e^{-|\omega|}$
• D. $\displaystyle \frac{\pi}{2}\, e^{-|\omega|}$

Hint:

Use time-domain differentiation when a function resembles a derivative of a known transform. Many rational functions involving $t$ and $(1+t^2)$ reduce easily.

Answer 1

The Correct Option is A

Step 1: Recall a known Fourier transform pair.
A standard transform pair is \[ \mathcal{F}\left\{\frac{1}{1+t^{2}}\right\}=\pi e^{-|\omega|} \] under the convention \[ X(j\omega)=\int_{-\infty}^{\infty} x(t)e^{-j\omega t}\,dt. \] Step 2: Use differentiation property in the time domain.
The given signal \[ x(t)=\frac{t}{(1+t^{2})^{2}} \] can be recognized by observing the derivative: \[ \frac{d}{dt}\left( \frac{1}{1+t^{2}} \right)=\frac{-2t}{(1+t^{2})^{2}}. \] Therefore, \[ \frac{t}{(1+t^{2})^{2}} = -\frac{1}{2}\frac{d}{dt}\left( \frac{1}{1+t^{2}} \right). \] Step 3: Apply the Fourier transform differentiation property.
If \[ \mathcal{F}\{f(t)\}=F(\omega), \] then \[ \mathcal{F}\{f'(t)\}=j\omega F(\omega). \] Thus, \[ X(j\omega)=\mathcal{F}\left\{-\frac{1}{2}\frac{d}{dt}\left( \frac{1}{1+t^{2}} \right)\right\} = -\frac{1}{2} (j\omega) \cdot \pi e^{-|\omega|}. \] Step 4: Simplify.
\[ X(j\omega)= -\frac{\pi j\omega}{2} e^{-|\omega|} = \frac{\pi}{2j} \omega e^{-|\omega|}. \] This matches option (A).
Final Answer: $\displaystyle \frac{\pi}{2j}\, \omega e^{-|\omega|}$