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Diffstat (limited to 'source/know/concept/fourier-transform/index.md')
| -rw-r--r-- | source/know/concept/fourier-transform/index.md | 23 |
1 files changed, 8 insertions, 15 deletions
diff --git a/source/know/concept/fourier-transform/index.md b/source/know/concept/fourier-transform/index.md index 0bc849b..c86d997 100644 --- a/source/know/concept/fourier-transform/index.md +++ b/source/know/concept/fourier-transform/index.md @@ -67,6 +67,7 @@ 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. @@ -113,6 +114,7 @@ $$\begin{aligned} \end{aligned}$$ + ## Multiple dimensions The Fourier transform is straightforward to generalize to $$N$$ dimensions. @@ -150,11 +152,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-constants-ND"/> -<label for="proof-constants-ND">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-constants-ND">Proof.</label> + +{% include proof/start.html id="proof-constants-ndim" -%} The inverse FT of the forward FT of $$f(\vb{x})$$ must be equal to $$f(\vb{x})$$ again, so: $$\begin{aligned} @@ -180,9 +179,8 @@ $$\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}$$ +{% include proof/end.html id="proof-constants-ndim" %} -</div> -</div> Differentiation is more complicated for $$N > 1$$, but the FT is still useful, @@ -197,11 +195,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-laplacian"/> -<label for="proof-laplacian">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-laplacian">Proof.</label> + +{% include proof/start.html id="proof-laplacian" -%} We insert $$\nabla^2 f$$ into the FT, decompose the exponential and the Laplacian, and then integrate by parts (limits $$\pm \infty$$ omitted): @@ -236,9 +231,7 @@ $$\begin{aligned} &= - 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> +{% include proof/end.html id="proof-laplacian" %} |