Improve knowledge base

1 files changed, 44 insertions, 33 deletions

diff --git a/source/know/concept/convolution-theorem/index.md b/source/know/concept/convolution-theorem/index.md
index d10d85d..3f9eafb 100644
--- a/source/know/concept/convolution-theorem/index.md
+++ b/source/know/concept/convolution-theorem/index.md
@@ -7,54 +7,59 @@ categories:
 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.
+The **convolution theorem** states that a convolution in the real domain
+is equal to a product in the frequency domain.
+This fact is especially useful for computation,
+as it allows 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:
+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 the constants from its definition:
 
 $$\begin{aligned}
     \boxed{
         \begin{aligned}
-            A \cdot (f * g)(x) &= \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\
-            B \cdot (\tilde{f} * \tilde{g})(k) &= \hat{\mathcal{F}}\{f(x) \: g(x)\}
+            A \cdot (f * g)(x)
+            &= \hat{\mathcal{F}}{}^{-1}\Big\{ \tilde{f}(k) \: \tilde{g}(k) \Big\}
+            \\
+            B \cdot (\tilde{f} * \tilde{g})(k)
+            &= \hat{\mathcal{F}}\Big\{ f(x) \: g(x) \Big\}
         \end{aligned}
     }
 \end{aligned}$$
 
 
 {% include proof/start.html id="proof-fourier" -%}
-We expand the right-hand side of the theorem and
-rearrange the integrals:
+We expand the right-hand side of the theorem and rearrange the integrals:
 
 $$\begin{aligned}
-    \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\}
-    &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \: e^{i s k x'} \dd{x'} \Big) e^{-i s k x} \dd{k}
+    \hat{\mathcal{F}}{}^{-1}\Big\{ \tilde{f}(k) \: \tilde{g}(k) \Big\}
+    &= B \int_{-\infty}^\infty \tilde{f}(k) \bigg( A \int_{-\infty}^\infty g(x') \: e^{i s k x'} \dd{x'} \bigg) e^{-i s k x} \dd{k}
     \\
-    &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \: e^{-i s k (x - x')} \dd{k} \Big) \dd{x'}
+    &= A \int_{-\infty}^\infty g(x') \bigg( B \int_{-\infty}^\infty \tilde{f}(k) \: e^{-i s k (x - x')} \dd{k} \bigg) \dd{x'}
     \\
     &= A \int_{-\infty}^\infty g(x') \: f(x - x') \dd{x'}
-    = A \cdot (f * g)(x)
+    \\
+    &= A \cdot (f * g)(x)
 \end{aligned}$$
 
 Then we do the same again,
 this time starting from a product in the $$x$$-domain:
 
 $$\begin{aligned}
-    \hat{\mathcal{F}}\{f(x) \: g(x)\}
-    &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \: e^{-i s x k'} \dd{k'} \Big) e^{i s k x} \dd{x}
+    \hat{\mathcal{F}}\Big\{ f(x) \: g(x) \Big\}
+    &= A \int_{-\infty}^\infty f(x) \bigg( B \int_{-\infty}^\infty \tilde{g}(k') \: e^{-i s x k'} \dd{k'} \bigg) e^{i s k x} \dd{x}
     \\
-    &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \: e^{i s x (k - k')} \dd{x} \Big) \dd{k'}
+    &= B \int_{-\infty}^\infty \tilde{g}(k') \bigg( A \int_{-\infty}^\infty f(x) \: e^{i s x (k - k')} \dd{x} \bigg) \dd{k'}
     \\
     &= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'}
-    = B \cdot (\tilde{f} * \tilde{g})(k)
+    \\
+    &= B \cdot (\tilde{f} * \tilde{g})(k)
 \end{aligned}$$
 {% include proof/end.html id="proof-fourier" %}
 
@@ -62,45 +67,51 @@ $$\begin{aligned}
 
 ## Laplace transform
 
-For functions $$f(t)$$ and $$g(t)$$ which are only defined for $$t \ge 0$$,
+For functions $$f(t)$$ and $$g(t)$$ that are only defined for $$t \ge 0$$,
 the convolution theorem can also be stated using
 the [Laplace transform](/know/concept/laplace-transform/):
 
 $$\begin{aligned}
-    \boxed{(f * g)(t) = \hat{\mathcal{L}}{}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}}
+    \boxed{
+        (f * g)(t)
+        = \hat{\mathcal{L}}{}^{-1}\Big\{ \tilde{f}(s) \: \tilde{g}(s) \Big\}
+    }
 \end{aligned}$$
 
-Because the inverse Laplace transform $$\hat{\mathcal{L}}{}^{-1}$$ is
-unpleasant, the theorem is often stated using the forward transform
-instead:
+Because the inverse Laplace transform $$\hat{\mathcal{L}}{}^{-1}$$ is usually difficult,
+the theorem is often stated using the forward transform instead:
 
 $$\begin{aligned}
-    \boxed{\hat{\mathcal{L}}\{(f * g)(t)\} = \tilde{f}(s) \: \tilde{g}(s)}
+    \boxed{
+        \hat{\mathcal{L}}\Big\{ (f * g)(t) \Big\}
+        = \tilde{f}(s) \: \tilde{g}(s)
+    }
 \end{aligned}$$
 
 
 {% include proof/start.html id="proof-laplace" -%}
 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 $$t < 0$$:
+because we choose to set both $$f(t)$$ and $$g(t)$$ to zero for $$t < 0$$:
 
 $$\begin{aligned}
-    \hat{\mathcal{L}}\{(f * g)(t)\}
-    &= \int_0^\infty \Big( \int_0^\infty g(t') \: f(t - t') \dd{t'} \Big) e^{-s t} \dd{t}
+    \hat{\mathcal{L}}\Big\{ (f * g)(t) \Big\}
+    &= \int_0^\infty \bigg( \int_0^\infty g(t') \: f(t - t') \dd{t'} \bigg) e^{-s t} \dd{t}
     \\
-    &= \int_0^\infty \Big( \int_0^\infty f(t - t') \: e^{-s t} \dd{t} \Big) g(t') \dd{t'}
+    &= \int_0^\infty \bigg( \int_0^\infty f(t - t') \: e^{-s t} \dd{t} \bigg) \: g(t') \dd{t'}
 \end{aligned}$$
 
 Then we define a new integration variable $$\tau = t - t'$$, yielding:
 
 $$\begin{aligned}
-    \hat{\mathcal{L}}\{(f * g)(t)\}
-    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \: e^{-s (\tau + t')} \dd{\tau} \Big) g(t') \dd{t'}
+    \hat{\mathcal{L}}\Big\{ (f * g)(t) \Big\}
+    &= \int_0^\infty \bigg( \int_0^\infty f(\tau) \: e^{-s (\tau + t')} \dd{\tau} \bigg) \: g(t') \dd{t'}
     \\
-    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \: e^{-s \tau} \dd{\tau} \Big) g(t') \: e^{-s t'} \dd{t'}
+    &= \int_0^\infty \bigg( \int_0^\infty f(\tau) \: e^{-s \tau} \dd{\tau} \bigg) \: g(t') \: e^{-s t'} \dd{t'}
     \\
     &= \int_0^\infty \tilde{f}(s) \: g(t') \: e^{-s t'} \dd{t'}
-    = \tilde{f}(s) \: \tilde{g}(s)
+    \\
+    &= \tilde{f}(s) \: \tilde{g}(s)
 \end{aligned}$$
 {% include proof/end.html id="proof-laplace" %}