Topic: Signals and Systems
Fourier transform of $te^{-\alpha t}u(t)$, (where, $\alpha>0$, $u(t)$ is the Unit step function) is:
- A. $\frac{\alpha}{(\alpha+j2\pi f)^2}$
- B. $j\frac{dX(\omega)}{d\omega}$
- C. $\frac{\alpha}{(\alpha+j2\pi f)}$
- D. $\frac{1}{(\alpha+j2\pi f)^2}$
Answer: D. $\frac{1}{(\alpha+j2\pi f)^2}$
Explanation: The Fourier transform of $e^{-\alpha t}u(t)$ is $\frac{1}{\alpha + j\omega}$. Multiplication by $t$ in the time domain corresponds to differentiation in the frequency domain multiplied by $j$: $F\{t x(t)\} = j \frac{d}{d\omega} X(\omega)$. So, $j \frac{d}{d\omega} (\alpha + j\omega)^{-1} = j (-1)(\alpha + j\omega)^{-2} (j) = \frac{1}{(\alpha + j\omega)^2}$. Since $\omega = 2\pi f$, the result is $\frac{1}{(\alpha+j2\pi f)^2}$.