Frequency derivative property; t·e^(−at)u(t)

Frequency derivative (differentiation in frequency) property

Statement: If $x(t) \leftrightarrow X(j\omega)$, then

$$-jt\,x(t) \;\longleftrightarrow\; \frac{dX(j\omega)}{d\omega} \qquad\text{or}\qquad t\,x(t) \;\longleftrightarrow\; j\frac{dX(j\omega)}{d\omega}$$

Proof: Start from the definition

$$X(j\omega) = \int_{-\infty}^{\infty} x(t)\,e^{-j\omega t}\,dt$$

Differentiate both sides with respect to $\omega$ (the derivative can be taken inside the integral):

$$\begin{aligned} \frac{dX(j\omega)}{d\omega} &= \int_{-\infty}^{\infty} x(t)\,\frac{d}{d\omega}\left(e^{-j\omega t}\right)dt \\ &= \int_{-\infty}^{\infty} \left[-jt\,x(t)\right]e^{-j\omega t}\,dt \end{aligned}$$

The right side is the FT of $-jt\,x(t)$. Multiplying both sides by $j$ gives $t\,x(t) \leftrightarrow j\,\frac{dX(j\omega)}{d\omega}$. Hence proved.

FT of $x(t) = t\,e^{-at}u(t)$, $a>0$

Let $x_1(t) = e^{-at}u(t)$. Its FT is

$$X_1(j\omega) = \int_0^\infty e^{-(a+j\omega)t}dt = \frac{1}{a+j\omega}$$

Now $x(t) = t\,x_1(t)$, so by the property:

$$\begin{aligned} X(j\omega) &= j\frac{d}{d\omega}\left[\frac{1}{a+j\omega}\right] \\ &= j\cdot\frac{-j}{(a+j\omega)^2} \\ &= \frac{1}{(a+j\omega)^2} \end{aligned}$$

Answer: $t\,e^{-at}u(t) \;\longleftrightarrow\; \dfrac{1}{(a+j\omega)^2}$, with $|X(j\omega)| = \dfrac{1}{a^2+\omega^2}$ and $\angle X(j\omega) = -2\tan^{-1}(\omega/a)$.

Check: $X(0) = 1/a^2$, and the area $\int_0^\infty t e^{-at}dt = 1/a^2$. They match.