Fourier transform of unit step

The unit step $u(t)$ is not absolutely integrable, so its Fourier transform is found as a limit, using the signum function.

Step 1: Write u(t) using the signum function

$$u(t) = \frac{1}{2} + \frac{1}{2}\mathrm{sgn}(t), \qquad \mathrm{sgn}(t) = \begin{cases} 1, & t > 0 \\ -1, & t < 0 \end{cases}$$

Step 2: FT of the constant 1/2

The inverse Fourier transform of $\delta(\omega)$ is $\frac{1}{2\pi}$, so $1 \xleftrightarrow{F} 2\pi\delta(\omega)$ and

$$\frac{1}{2} \xleftrightarrow{F} \pi\delta(\omega)$$

Step 3: FT of sgn(t)

Write $\mathrm{sgn}(t) = \lim_{a\to 0}\left[e^{-at}u(t) - e^{at}u(-t)\right]$ with $a > 0$:

$$\begin{aligned} F\{\mathrm{sgn}(t)\} &= \lim_{a\to 0}\left[\int_0^{\infty}e^{-(a+j\omega)t}dt - \int_{-\infty}^{0}e^{(a-j\omega)t}dt\right] \\ &= \lim_{a\to 0}\left[\frac{1}{a+j\omega} - \frac{1}{a-j\omega}\right] = \lim_{a\to 0}\frac{-2j\omega}{a^2+\omega^2} = \frac{2}{j\omega} \end{aligned}$$

So $\frac{1}{2}\mathrm{sgn}(t) \xleftrightarrow{F} \frac{1}{j\omega}$.

Step 4: Combine

$$\boxed{U(\omega) = \pi\delta(\omega) + \frac{1}{j\omega}}$$

Spectrum

• The impulse $\pi\delta(\omega)$ represents the DC component (average value 1/2) of $u(t)$.
• $|U(\omega)| = \frac{1}{|\omega|}$ for $\omega \ne 0$.
• $\angle U(\omega) = -\pi/2$ for $\omega > 0$ and $+\pi/2$ for $\omega < 0$.

Check: $u(t) = \int_{-\infty}^{t}\delta(\tau)d\tau$, and the integration property gives $\frac{1}{j\omega}\cdot 1 + \pi\cdot 1\cdot\delta(\omega)$, which is the same result.