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 as follows, where $\pv{}$ is the Cauchy principal value, $A$ and $s$ are constants from the FT’s definition, and $\mathrm{sgn}$ is the signum function:

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

Proof

Proof. 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.