Categories: Mathematics, Physics.

Heaviside step function

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

Θ(t)={0ift<01ift>1\begin{aligned} \boxed{ \Theta(t) = \begin{cases} 0 & \mathrm{if}\: t < 0 \\ 1 & \mathrm{if}\: t > 1 \end{cases} } \end{aligned}

The value of Θ(t ⁣= ⁣0)\Theta(t \!=\! 0) varies between definitions; common choices are 00, 11 and 1/21/2. In practice, this rarely matters, and some authors even change their definition on the fly for convenience. For physicists, Θ(0)=1\Theta(0) = 1 is generally best, such that:

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

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

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

The Fourier transform of Θ(t)\Theta(t) is as follows, where P\pv{} is the Cauchy principal value, AA and ss are constants from the FT’s definition, and sgn\mathrm{sgn} is the signum function:

Θ~(ω)=As(πδ(ω)+isgn(s)P1ω)\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 Θ(0)=1/2\Theta(0) = 1/2, such that the Heaviside step function can be expressed using the signum function sgn(t)\mathrm{sgn}(t):

Θ(t)=12+sgn(t)2\begin{aligned} \Theta(t) = \frac{1}{2} + \frac{\mathrm{sgn}(t)}{2} \end{aligned}

We then take the Fourier transform, where AA and ss are constants from its definition:

Θ~(ω)=F^{Θ(t)}=A2(exp(isωt)dt+sgn(t)exp(isωt)dt)\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 P\pv{} and look up the integral:

Θ~(ω)=Aπδ(sω)+A2Psgn(t)exp(isωt)dt=Asπδ(ω)+iAsP1ω\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 P\pv{} without an integral is an abuse of notation, and means that this result only makes sense when wrapped in an integral. Formally, P{1/ω}\pv{\{1 / \omega\}} is a Schwartz distribution.