From c705ac1d7dc74709835a8c48fae4a7dd70dc5c49 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 25 Feb 2021 16:14:20 +0100
Subject: Expand knowledge base

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