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'

---
 source/know/concept/convolution-theorem/index.md | 20 ++++++++++----------
 1 file changed, 10 insertions(+), 10 deletions(-)

(limited to 'source/know/concept/convolution-theorem')

diff --git a/source/know/concept/convolution-theorem/index.md b/source/know/concept/convolution-theorem/index.md
index d4655cf..742c8ff 100644
--- a/source/know/concept/convolution-theorem/index.md
+++ b/source/know/concept/convolution-theorem/index.md
@@ -9,14 +9,14 @@ layout: "concept"
 
 The **convolution theorem** states that a convolution in the direct domain
 is equal to a product in the frequency domain. This is especially useful
-for computation, replacing an $\mathcal{O}(n^2)$ convolution with an
-$\mathcal{O}(n \log(n))$ transform and product.
+for computation, replacing an $$\mathcal{O}(n^2)$$ convolution with an
+$$\mathcal{O}(n \log(n))$$ transform and product.
 
 ## Fourier transform
 
 The convolution theorem is usually expressed as follows, where
-$\hat{\mathcal{F}}$ is the [Fourier transform](/know/concept/fourier-transform/),
-and $A$ and $B$ are constants from its definition:
+$$\hat{\mathcal{F}}$$ is the [Fourier transform](/know/concept/fourier-transform/),
+and $$A$$ and $$B$$ are constants from its definition:
 
 $$\begin{aligned}
     \boxed{
@@ -46,7 +46,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Then we do the same again,
-this time starting from a product in the $x$-domain:
+this time starting from a product in the $$x$$-domain:
 
 $$\begin{aligned}
     \hat{\mathcal{F}}\{f(x) \: g(x)\}
@@ -63,7 +63,7 @@ $$\begin{aligned}
 
 ## Laplace transform
 
-For functions $f(t)$ and $g(t)$ which are only defined for $t \ge 0$,
+For functions $$f(t)$$ and $$g(t)$$ which are only defined for $$t \ge 0$$,
 the convolution theorem can also be stated using
 the [Laplace transform](/know/concept/laplace-transform/):
 
@@ -71,7 +71,7 @@ $$\begin{aligned}
     \boxed{(f * g)(t) = \hat{\mathcal{L}}{}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}}
 \end{aligned}$$
 
-Because the inverse Laplace transform $\hat{\mathcal{L}}{}^{-1}$ is
+Because the inverse Laplace transform $$\hat{\mathcal{L}}{}^{-1}$$ is
 unpleasant, the theorem is often stated using the forward transform
 instead:
 
@@ -85,8 +85,8 @@ $$\begin{aligned}
 <div class="hidden" markdown="1">
 <label for="proof-laplace">Proof.</label>
 We expand the left-hand side.
-Note that the lower integration limit is 0 instead of $-\infty$,
-because we set both $f(t)$ and $g(t)$ to zero for $t < 0$:
+Note that the lower integration limit is 0 instead of $$-\infty$$,
+because we set both $$f(t)$$ and $$g(t)$$ to zero for $$t < 0$$:
 
 $$\begin{aligned}
     \hat{\mathcal{L}}\{(f * g)(t)\}
@@ -95,7 +95,7 @@ $$\begin{aligned}
     &= \int_0^\infty \Big( \int_0^\infty f(t - t')  \exp(- s t) \dd{t} \Big) g(t') \dd{t'}
 \end{aligned}$$
 
-Then we define a new integration variable $\tau = t - t'$, yielding:
+Then we define a new integration variable $$\tau = t - t'$$, yielding:
 
 $$\begin{aligned}
     \hat{\mathcal{L}}\{(f * g)(t)\}
-- 
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