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1 files changed, 33 insertions, 31 deletions
diff --git a/source/know/concept/fourier-transform/index.md b/source/know/concept/fourier-transform/index.md index f8b5b2a..0bc849b 100644 --- a/source/know/concept/fourier-transform/index.md +++ b/source/know/concept/fourier-transform/index.md @@ -10,11 +10,11 @@ layout: "concept" --- The **Fourier transform** (FT) is an integral transform which converts a -function $f(x)$ into its frequency representation $\tilde{f}(k)$. +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 +The **forward** FT is defined as follows, where $$A$$, $$B$$, and $$s$$ are unspecified constants (for now): $$\begin{aligned} @@ -35,9 +35,9 @@ $$\begin{aligned} } \end{aligned}$$ -Clearly, the inverse FT of the forward FT of $f(x)$ must equal $f(x)$ +Clearly, the inverse FT of the forward FT of $$f(x)$$ must equal $$f(x)$$ again. Let us verify this, by rearranging the integrals to get the -[Dirac delta function](/know/concept/dirac-delta-function/) $\delta(x)$: +[Dirac delta function](/know/concept/dirac-delta-function/) $$\delta(x)$$: $$\begin{aligned} \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{f(x)\}\} @@ -49,29 +49,29 @@ $$\begin{aligned} = \frac{2 \pi A B}{|s|} f(x) \end{aligned}$$ -Therefore, the constants $A$, $B$, and $s$ are subject to the following +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 $B$ +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. +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 useful 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 $x \to \pm \infty$: +assuming that $$f(x)$$ is localized, i.e. $$f(x) \to 0$$ for $$x \to \pm \infty$$: $$\begin{aligned} \hat{\mathcal{F}}\{f'(x)\} @@ -82,7 +82,7 @@ $$\begin{aligned} &= (- i s k) \tilde{f}(k) \end{aligned}$$ -Therefore, as long as $f(x)$ is localized, the FT eliminates derivatives +Therefore, as long as $$f(x)$$ is localized, the FT eliminates derivatives of the transformed variable, which makes it useful against PDEs: $$\begin{aligned} @@ -92,8 +92,8 @@ $$\begin{aligned} \end{aligned}$$ 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: +derivatives are also localized in the $$x$$-domain, which is practically +guaranteed if $$f(x)$$ itself is localized: $$\begin{aligned} \boxed{ @@ -115,9 +115,9 @@ $$\begin{aligned} ## Multiple dimensions -The Fourier transform is straightforward to generalize to $N$ dimensions. -Given a scalar field $f(\vb{x})$ with $\vb{x} = (x_1, ..., x_N)$, -its FT $\tilde{f}(\vb{k})$ is defined as follows: +The Fourier transform is straightforward to generalize to $$N$$ dimensions. +Given a scalar field $$f(\vb{x})$$ with $$\vb{x} = (x_1, ..., x_N)$$, +its FT $$\tilde{f}(\vb{k})$$ is defined as follows: $$\begin{aligned} \boxed{ @@ -127,7 +127,7 @@ $$\begin{aligned} } \end{aligned}$$ -Where the wavevector $\vb{k} = (k_1, ..., k_N)$. +Where the wavevector $$\vb{k} = (k_1, ..., k_N)$$. Likewise, the inverse FT is given by: $$\begin{aligned} @@ -138,9 +138,9 @@ $$\begin{aligned} } \end{aligned}$$ -In practice, in $N$D, there is not as much disagreement about -the constants $A$, $B$ and $s$ as in 1D: -typically $A = 1$ and $B = 1 / (2 \pi)^N$, with $s = \pm 1$. +In practice, in $$N$$D, there is not as much disagreement about +the constants $$A$$, $$B$$ and $$s$$ as in 1D: +typically $$A = 1$$ and $$B = 1 / (2 \pi)^N$$, with $$s = \pm 1$$. Any choice will do, as long as: $$\begin{aligned} @@ -155,7 +155,7 @@ $$\begin{aligned} <label for="proof-constants-ND">Proof</label> <div class="hidden" markdown="1"> <label for="proof-constants-ND">Proof.</label> -The inverse FT of the forward FT of $f(\vb{x})$ must be equal to $f(\vb{x})$ again, so: +The inverse FT of the forward FT of $$f(\vb{x})$$ must be equal to $$f(\vb{x})$$ again, so: $$\begin{aligned} \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\} @@ -180,14 +180,15 @@ $$\begin{aligned} &= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \ddn{N}{\vb{x}'} = \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x}) \end{aligned}$$ + </div> </div> -Differentiation is more complicated for $N > 1$, +Differentiation is more complicated for $$N > 1$$, but the FT is still useful, -notably for the Laplacian $\nabla^2 \equiv \idv{ {}^2}{x_1^2} + ... + \idv{ {}^2}{x_N^2}$. -Let $|\vb{k}|$ be the norm of $\vb{k}$, -then for a localized $f$: +notably for the Laplacian $$\nabla^2 \equiv \idv{ {}^2}{x_1^2} + ... + \idv{ {}^2}{x_N^2}$$. +Let $$|\vb{k}|$$ be the norm of $$\vb{k}$$, +then for a localized $$f$$: $$\begin{aligned} \boxed{ @@ -201,9 +202,9 @@ $$\begin{aligned} <label for="proof-laplacian">Proof</label> <div class="hidden" markdown="1"> <label for="proof-laplacian">Proof.</label> -We insert $\nabla^2 f$ into the FT, +We insert $$\nabla^2 f$$ into the FT, decompose the exponential and the Laplacian, -and then integrate by parts (limits $\pm \infty$ omitted): +and then integrate by parts (limits $$\pm \infty$$ omitted): $$\begin{aligned} \hat{\mathcal{F}}\{\nabla^2 f\} @@ -216,7 +217,7 @@ $$\begin{aligned} \end{aligned}$$ Just like in 1D, we get rid of the boundary term -by assuming that all derivatives $\idv{f}{x_n}$ are nicely localized. +by assuming that all derivatives $$\idv{f}{x_n}$$ are nicely localized. To proceed, we then integrate by parts again: $$\begin{aligned} @@ -228,13 +229,14 @@ $$\begin{aligned} \end{aligned}$$ Once again, we remove the boundary term -by assuming that $f$ is localized, yielding: +by assuming that $$f$$ is localized, yielding: $$\begin{aligned} \hat{\mathcal{F}}\{\nabla^2 f\} &= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}} = - s^2 \sum_{n = 1}^N k_n^2 \tilde{f} \end{aligned}$$ + </div> </div> |