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+---
+title: "Fourier transform"
+firstLetter: "F"
+publishDate: 2021-02-22
+categories:
+- Mathematics
+- Physics
+
+date: 2021-02-22T21:35:54+01:00
+draft: false
+markup: pandoc
+---
+
+# Fourier transform
+
+The **Fourier transform** (FT) is an integral transform which converts a
+function $f(x)$ into its frequency representation $\tilde{f}(k)$.
+Great volumes have already been written about this subject,
+so let us focus on the aspects that are useful to physicists.
+
+The **forward** FT is defined as follows, where $A$, $B$, and $s$ are unspecified constants
+(for now):
+
+$$\begin{aligned}
+    \boxed{
+        \tilde{f}(k)
+        = \hat{\mathcal{F}}\{f(x)\}
+        = A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x}
+    }
+\end{aligned}$$
+
+The **inverse Fourier transform** (iFT) undoes the forward FT operation:
+
+$$\begin{aligned}
+    \boxed{
+        f(x)
+        = \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}
+        = B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k x) \dd{k}
+    }
+\end{aligned}$$
+
+Clearly, the inverse FT of the forward FT of $f(x)$ must equal $f(x)$
+again. Let us verify this, by rearranging the integrals to get the
+[Dirac delta function](/know/concept/dirac-delta-function/) $\delta(x)$:
+
+$$\begin{aligned}
+    \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{f(x)\}\}
+    &= A B \int_{-\infty}^\infty \exp(-i s k x) \int_{-\infty}^\infty f(x') \exp(i s k x') \dd{x'} \dd{k}
+    \\
+    &= 2 \pi A B \int_{-\infty}^\infty f(x') \Big(\frac{1}{2\pi} \int_{-\infty}^\infty \exp(i s k (x' - x)) \dd{k} \Big) \dd{x'}
+    \\
+    &= 2 \pi A B \int_{-\infty}^\infty f(x') \: \delta(s(x' - x)) \dd{x'}
+    = \frac{2 \pi A B}{|s|} f(x)
+\end{aligned}$$
+
+Therefore, the constants $A$, $B$, and $s$ are subject to the following
+constraint:
+
+$$\begin{aligned}
+    \boxed{\frac{2\pi A B}{|s|} = 1}
+\end{aligned}$$
+
+But that still gives a lot of freedom. The exact choices of $A$ and $B$
+are generally motivated by the [convolution theorem](/know/concept/convolution-theorem/)
+and [Parseval's theorem](/know/concept/parsevals-theorem/).
+
+The choice of $|s|$ depends on whether the frequency variable $k$
+represents the angular ($|s| = 1$) or the physical ($|s| = 2\pi$)
+frequency. The sign of $s$ is not so important, but is generally based
+on whether the analysis is for forward ($s > 0$) or backward-propagating
+($s < 0$) waves.
+
+
+## Derivatives
+
+The FT of a derivative has a very interesting property.
+Below, after integrating by parts, we remove the boundary term by
+assuming that $f(x)$ is localized, i.e. $f(x) \to 0$ for $x \to \pm \infty$:
+
+$$\begin{aligned}
+    \hat{\mathcal{F}}\{f'(x)\}
+    &= A \int_{-\infty}^\infty f'(x) \exp(i s k x) \dd{x}
+    \\
+    &= A \big[ f(x) \exp(i s k x) \big]_{-\infty}^\infty - i s k A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x}
+    \\
+    &= (- i s k) \tilde{f}(k)
+\end{aligned}$$
+
+Therefore, as long as $f(x)$ is localized, the FT eliminates derivatives
+of the transformed variable, which makes it useful against PDEs:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\mathcal{F}}\{f'(x)\} = (- i s k) \tilde{f}(k)
+    }
+\end{aligned}$$
+
+This generalizes to higher-order derivatives, as long as these
+derivatives are also localized in the $x$-domain, which is practically
+guaranteed if $f(x)$ itself is localized:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\mathcal{F}} \Big\{ \dv[n]{f}{x} \Big\}
+        = (- i s k)^n \tilde{f}(k)
+    }
+\end{aligned}$$
+
+Derivatives in the frequency domain have an analogous property:
+
+$$\begin{aligned}
+    \boxed{
+        \dv[n]{\tilde{f}}{k}
+        = A \int_{-\infty}^\infty (i s x)^n f(x) \exp(i s k x) \dd{x}
+        = \hat{\mathcal{F}}\{ (i s x)^n f(x) \}
+    }
+\end{aligned}$$