Categories: Mathematics, Physics.

# Heaviside step function

The Heaviside step function $$\Theta(t)$$, is a discontinuous function used for enforcing causality or for representing a signal switched on at $$t = 0$$. It is defined as:

\begin{aligned} \boxed{ \Theta(t) = \begin{cases} 0 & \mathrm{if}\: t < 0 \\ 1 & \mathrm{if}\: t > 1 \end{cases} } \end{aligned}

The value of $$\Theta(t \!=\! 0)$$ varies between definitions; common choices are $$0$$, $$1$$ and $$1/2$$. In practice, this rarely matters, and some authors even change their definition on the fly for convenience. For physicists, $$\Theta(0) = 1$$ is generally best, such that:

\begin{aligned} \boxed{ \forall n \in \mathbb{R}: \Theta^n(t) = \Theta(t) } \end{aligned}

Unsurprisingly, the first-order derivative of $$\Theta(t)$$ is the Dirac delta function:

\begin{aligned} \boxed{ \Theta'(t) = \delta(t) } \end{aligned}

The Fourier transform of $$\Theta(t)$$ is noteworthy. In this case, it is easiest to use $$\Theta(0) = 1/2$$, such that the Heaviside step function can be expressed using the signum function $$\mathrm{sgn}(t)$$:

\begin{aligned} \Theta(t) = \frac{1}{2} + \frac{\mathrm{sgn}(t)}{2} \end{aligned}

We then take the Fourier transform, where $$A$$ and $$s$$ are constants from its definition:

\begin{aligned} \tilde{\Theta}(\omega) = \hat{\mathcal{F}}\{\Theta(t)\} = \frac{A}{2} \Big( \int_{-\infty}^\infty \exp(i s \omega t) \dd{t} + \int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t} \Big) \end{aligned}

The first term is proportional to the Dirac delta function. The second integral is problematic, so we take the Cauchy principal value $$\pv{}$$ and look up the integral:

\begin{aligned} \tilde{\Theta}(\omega) &= A \pi \delta(s \omega) + \frac{A}{2} \pv{\int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}} = \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \pv{\frac{1}{\omega}} \end{aligned}

The use of $$\pv{}$$ without an integral is an abuse of notation, and means that this result only makes sense when wrapped in an integral. Formally, $$\pv{\{1 / \omega\}}$$ is a Schwartz distribution. We thus have:

\begin{aligned} \boxed{ \tilde{\Theta}(\omega) = \frac{A}{|s|} \Big( \pi \delta(\omega) + i \: \mathrm{sgn}(s) \pv{\frac{1}{\omega}} \Big) } \end{aligned}