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Diffstat (limited to 'content/know/concept/fourier-transform')
| -rw-r--r-- | content/know/concept/fourier-transform/index.pdc | 150 |
1 files changed, 138 insertions, 12 deletions
diff --git a/content/know/concept/fourier-transform/index.pdc b/content/know/concept/fourier-transform/index.pdc index 3be47ff..1d1e27d 100644 --- a/content/know/concept/fourier-transform/index.pdc +++ b/content/know/concept/fourier-transform/index.pdc @@ -25,8 +25,8 @@ The **forward** FT is defined as follows, where $A$, $B$, and $s$ are unspecifie $$\begin{aligned} \boxed{ \tilde{f}(k) - = \hat{\mathcal{F}}\{f(x)\} - = A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x} + \equiv \hat{\mathcal{F}}\{f(x)\} + \equiv A \int_{-\infty}^\infty f(x) \exp\!(i s k x) \dd{x} } \end{aligned}$$ @@ -35,8 +35,8 @@ The **inverse Fourier transform** (iFT) undoes the forward FT operation: $$\begin{aligned} \boxed{ f(x) - = \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\} - = B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k x) \dd{k} + \equiv \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\} + \equiv B \int_{-\infty}^\infty \tilde{f}(k) \exp\!(- i s k x) \dd{k} } \end{aligned}$$ @@ -46,9 +46,9 @@ again. Let us verify this, by rearranging the integrals to get the $$\begin{aligned} \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{f(x)\}\} - &= A B \int_{-\infty}^\infty \exp(-i s k x) \int_{-\infty}^\infty f(x') \exp(i s k x') \dd{x'} \dd{k} + &= A B \int_{-\infty}^\infty \exp\!(-i s k x) \int_{-\infty}^\infty f(x') \exp\!(i s k x') \dd{x'} \dd{k} \\ - &= 2 \pi A B \int_{-\infty}^\infty f(x') \Big(\frac{1}{2\pi} \int_{-\infty}^\infty \exp(i s k (x' - x)) \dd{k} \Big) \dd{x'} + &= 2 \pi A B \int_{-\infty}^\infty f(x') \Big(\frac{1}{2\pi} \int_{-\infty}^\infty \exp\!(i s k (x' - x)) \dd{k} \Big) \dd{x'} \\ &= 2 \pi A B \int_{-\infty}^\infty f(x') \: \delta(s(x' - x)) \dd{x'} = \frac{2 \pi A B}{|s|} f(x) @@ -74,15 +74,15 @@ on whether the analysis is for forward ($s > 0$) or backward-propagating ## Derivatives -The FT of a derivative has a very interesting property. +The FT of a derivative has a very useful property. Below, after integrating by parts, we remove the boundary term by assuming that $f(x)$ is localized, i.e. $f(x) \to 0$ for $x \to \pm \infty$: $$\begin{aligned} \hat{\mathcal{F}}\{f'(x)\} - &= A \int_{-\infty}^\infty f'(x) \exp(i s k x) \dd{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} + &= 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}$$ @@ -110,13 +110,139 @@ $$\begin{aligned} Derivatives in the frequency domain have an analogous property: $$\begin{aligned} + \dv[n]{\tilde{f}}{k} + &= A \dv[n]{k} \int_{-\infty}^\infty f(x) \exp\!(i s k x) \dd{x} + \\ + &= A \int_{-\infty}^\infty (i s x)^n f(x) \exp\!(i s k x) \dd{x} + = \hat{\mathcal{F}}\{ (i s x)^n f(x) \} +\end{aligned}$$ + + +## 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: + +$$\begin{aligned} + \boxed{ + \tilde{f}(\vb{k}) + \equiv \hat{\mathcal{F}}\{f(\vb{x})\} + \equiv A \int_{-\infty}^\infty f(\vb{x}) \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}} + } +\end{aligned}$$ + +Where the wavevector $\vb{k} = (k_1, ..., k_N)$. +Likewise, the inverse FT is given by: + +$$\begin{aligned} \boxed{ - \dv[n]{\tilde{f}}{k} - = A \int_{-\infty}^\infty (i s x)^n f(x) \exp(i s k x) \dd{x} - = \hat{\mathcal{F}}\{ (i s x)^n f(x) \} + f(\vb{x}) + \equiv \hat{\mathcal{F}}^{-1}\{\tilde{f}(\vb{k})\} + \equiv B \int_{-\infty}^\infty \tilde{f}(\vb{k}) \exp\!(- i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{k}} } \end{aligned}$$ +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 $s = \pm 1$. +Any choice will do, as long as: + +$$\begin{aligned} + \boxed{ + A B + = \bigg( \frac{|s|}{2 \pi} \bigg)^{\!N} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-constants-ND"/> +<label for="proof-constants-ND">Proof</label> +<div class="hidden"> +<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: + +$$\begin{aligned} + \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\} + &= A B \int \exp\!(- i s \vb{k} \cdot \vb{x}) + \int f(\vb{x}') \exp\!(i s \vb{k} \cdot \vb{x}') \dd[N]{\vb{x}'} \dd[N]{\vb{k}} + \\ + &= (2 \pi)^N A B \int f(\vb{x}') + \Big( \frac{1}{(2 \pi)^N} \int \exp\!(i s \vb{k} \cdot (\vb{x}' - \vb{x})) \dd[N]{\vb{k}} \Big) \dd[N]{\vb{x}'} + \\ + &= (2 \pi)^N A B \int f(\vb{x}') + \Big( \prod_{n = 1}^N \frac{1}{2 \pi} \int \exp\!(i s k_n (x_n' - x_n)) \dd{k_n} \Big) \dd[N]{\vb{x}'} +\end{aligned}$$ + +Here, we recognize the definition of the Dirac delta function again, +leading to: + +$$\begin{aligned} + \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\} + &= (2 \pi)^N A B \int f(\vb{x}') + \Big( \prod_{n = 1}^N \delta(s(x_n' - x_n)) \Big) \dd[N]{\vb{x}'} + \\ + &= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \dd[N]{\vb{x}'} + = \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x}) +\end{aligned}$$ +</div> +</div> + +Differentiation is more complicated for $N > 1$, +but the FT is still useful, +notably for the Laplacian $\nabla^2 \equiv \dv*[2]{x_1} + ... + \dv*[2]{x_N}$. +Let $|\vb{k}|$ be the norm of $\vb{k}$, +then for a localized $f$: + +$$\begin{aligned} + \boxed{ + \hat{\mathcal{F}}\{\nabla^2 f(\vb{x})\} + = - s^2 |\vb{k}|^2 \tilde{f}(\vb{k}) + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-laplacian"/> +<label for="proof-laplacian">Proof</label> +<div class="hidden"> +<label for="proof-laplacian">Proof.</label> +We insert $\nabla^2 f$ into the FT, +decompose the exponential and the Laplacian, +and then integrate by parts (limits $\pm \infty$ omitted): + +$$\begin{aligned} + \hat{\mathcal{F}}\{\nabla^2 f\} + &= A \int \big( \nabla^2 f \big) \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}} + \\ + &= A \int \Big( \sum_{n = 1}^N \pdv[2]{f}{x_n} \Big) \Big( \prod_{m = 1}^N \exp\!(i s k_m x_m) \Big) \dd[N]{\vb{x}} + \\ + &= A \sum_{n = 1}^N \bigg[ \pdv{f}{x_n} \exp\!(i s \vb{k} \cdot \vb{x}) \bigg] + - A \sum_{n = 1}^N i s k_n \int \pdv{f}{x_n} \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}} +\end{aligned}$$ + +Just like in 1D, we get rid of the boundary term +by assuming that all derivatives $\dv*{f}{x_n}$ are nicely localized. +To proceed, we then integrate by parts again: + +$$\begin{aligned} + \hat{\mathcal{F}}\{\nabla^2 f\} + &= - A \sum_{n = 1}^N i s k_n \int \pdv{f}{x_n} \Big( \prod_{m = 1}^N \exp\!(i s k_m x_m) \Big) \dd[N]{\vb{x}} + \\ + &= - A \sum_{n = 1}^N i s k_n \bigg[ f \exp\!(i s \vb{k} \cdot \vb{x}) \bigg] + + A \sum_{n = 1}^N (i s k_n)^2 \int f \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}} +\end{aligned}$$ + +Once again, we remove the boundary term +by assuming that $f$ is localized, yielding: + +$$\begin{aligned} + \hat{\mathcal{F}}\{\nabla^2 f\} + &= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}} + = - s^2 \sum_{n = 1}^N k_n^2 \tilde{f} +\end{aligned}$$ +</div> +</div> + ## References |