Categories: Mathematics, Physics.

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:

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

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.

© Marcus R.A. Newman, a.k.a. "Prefetch".
Available under CC BY-SA 4.0.