---
title: "Heaviside step function"
firstLetter: "H"
publishDate: 2021-02-25
categories:
- Mathematics
- Physics

date: 2021-02-25T11:28:02+01:00
draft: false
markup: pandoc
---

# 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](/know/concept/dirac-delta-function/):

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

The [Fourier transform](/know/concept/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](/know/concept/schwartz-distribution/).
We thus have:

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