From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

---
 .../know/concept/heaviside-step-function/index.md  | 33 +++++++++++-----------
 1 file changed, 17 insertions(+), 16 deletions(-)

(limited to 'source/know/concept/heaviside-step-function')

diff --git a/source/know/concept/heaviside-step-function/index.md b/source/know/concept/heaviside-step-function/index.md
index 1933037..15d1729 100644
--- a/source/know/concept/heaviside-step-function/index.md
+++ b/source/know/concept/heaviside-step-function/index.md
@@ -8,9 +8,9 @@ categories:
 layout: "concept"
 ---
 
-The **Heaviside step function** $\Theta(t)$,
+The **Heaviside step function** $$\Theta(t)$$,
 is a discontinuous function used for enforcing causality
-or for representing a signal switched on at $t = 0$.
+or for representing a signal switched on at $$t = 0$$.
 It is defined as:
 
 $$\begin{aligned}
@@ -23,11 +23,11 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The value of $\Theta(t \!=\! 0)$ varies between definitions;
-common choices are $0$, $1$ and $1/2$.
+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:
+For physicists, $$\Theta(0) = 1$$ is generally best, such that:
 
 $$\begin{aligned}
     \boxed{
@@ -35,7 +35,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Unsurprisingly, the first-order derivative of $\Theta(t)$ is
+Unsurprisingly, the first-order derivative of $$\Theta(t)$$ is
 the [Dirac delta function](/know/concept/dirac-delta-function/):
 
 $$\begin{aligned}
@@ -45,10 +45,10 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The [Fourier transform](/know/concept/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:
+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:
 
 $$\begin{aligned}
     \boxed{
@@ -62,16 +62,16 @@ $$\begin{aligned}
 <label for="proof-fourier">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-fourier">Proof.</label>
-In this case, it is easiest to use $\Theta(0) = 1/2$,
+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 $\mathrm{sgn}(t)$:
+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:
+where $$A$$ and $$s$$ are constants from its definition:
 
 $$\begin{aligned}
     \tilde{\Theta}(\omega)
@@ -80,7 +80,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The first term is proportional to the Dirac delta function.
-The second integral is problematic, so we take the Cauchy principal value $\pv{}$
+The second integral is problematic, so we take the Cauchy principal value $$\pv{}$$
 and look up the integral:
 
 $$\begin{aligned}
@@ -88,10 +88,11 @@ $$\begin{aligned}
     &= 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}$$
+
 </div>
 </div>
 
-The use of $\pv{}$ without an integral is an abuse of notation,
+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/).
+Formally, $$\pv{\{1 / \omega\}}$$ is a [Schwartz distribution](/know/concept/schwartz-distribution/).
 
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