フーリエ変換の定義による違い
フーリエ変換の定義による違い
フーリエ変換のよく使われる定義は3種類あり、それらの間の関係は次のようになります。
\[ \mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\xi\right)=\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi i\xi x}dx \] \[ \mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(k\right)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)e^{-ikx}dx \] \[ \mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(\nu\right)=\int_{-\infty}^{\infty}f\left(x\right)e^{-i\nu x}dx \] で定義すると、それぞれの間の関係は、
\[ \mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\xi\right)=\sqrt{2\pi}\mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(2\pi\xi\right) \] \[ \mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\xi\right)=\mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(2\pi\xi\right) \] \[ \mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(k\right)=\frac{1}{\sqrt{2\pi}}\mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\frac{k}{2\pi}\right) \] \[ \mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(k\right)=\frac{1}{\sqrt{2\pi}}\mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(k\right) \] \[ \mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(\nu\right)=\mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\frac{\nu}{2\pi}\right) \] \[ \mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(\nu\right)=\sqrt{2\pi}\mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(\nu\right) \] となる。
\[ \mathcal{F}_{1,\xi}^{\bullet}\left[f\left(\xi\right)\right]\left(x\right)=\int_{-\infty}^{\infty}f\left(\xi\right)e^{2\pi i\xi x}d\xi \] \[ \mathcal{F}_{2,k}^{\bullet}\left[f\left(k\right)\right]\left(x\right)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(k\right)e^{ikx}dk \] \[ \mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f\left(\nu\right)e^{i\nu x}d\nu \] となり、それぞれの間の関係は、
\[ \mathcal{F}_{1,x}^{\bullet}\left[f\left(x\right)\right]\left(\xi\right)=\sqrt{2\pi}\mathcal{F}_{2,x}^{\bullet}\left[f\left(x\right)\right]\left(2\pi\xi\right) \] \[ \mathcal{F}_{1,x}^{\bullet}\left[f\left(x\right)\right]\left(\xi\right)=2\pi\mathcal{F}_{3,x}^{\bullet}\left[f\left(x\right)\right]\left(2\pi\xi\right) \] \[ \mathcal{F}_{2,k}^{\bullet}\left[f\left(k\right)\right]\left(x\right)=\frac{1}{\sqrt{2\pi}}\mathcal{F}_{1,k}^{\bullet}\left[f\left(k\right)\right]\left(\frac{x}{2\pi}\right) \] \[ \mathcal{F}_{2,k}^{\bullet}\left[f\left(k\right)\right]\left(x\right)=\sqrt{2\pi}\mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right) \] \[ \mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right)=\frac{1}{2\pi}\mathcal{F}_{1,k}^{\bullet}\left[f\left(\nu\right)\right]\left(\frac{x}{2\pi}\right) \] \[ \mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right)=\frac{1}{\sqrt{2\pi}}\mathcal{F}_{2,k}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right) \] となる。
フーリエ変換のよく使われる定義は3種類あり、それらの間の関係は次のようになります。
(1)フーリエ変換
フーリエ変換を\[ \mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\xi\right)=\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi i\xi x}dx \] \[ \mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(k\right)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)e^{-ikx}dx \] \[ \mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(\nu\right)=\int_{-\infty}^{\infty}f\left(x\right)e^{-i\nu x}dx \] で定義すると、それぞれの間の関係は、
\[ \mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\xi\right)=\sqrt{2\pi}\mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(2\pi\xi\right) \] \[ \mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\xi\right)=\mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(2\pi\xi\right) \] \[ \mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(k\right)=\frac{1}{\sqrt{2\pi}}\mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\frac{k}{2\pi}\right) \] \[ \mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(k\right)=\frac{1}{\sqrt{2\pi}}\mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(k\right) \] \[ \mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(\nu\right)=\mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\frac{\nu}{2\pi}\right) \] \[ \mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(\nu\right)=\sqrt{2\pi}\mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(\nu\right) \] となる。
(2)フーリエ逆変換
フーリエ逆変換は\[ \mathcal{F}_{1,\xi}^{\bullet}\left[f\left(\xi\right)\right]\left(x\right)=\int_{-\infty}^{\infty}f\left(\xi\right)e^{2\pi i\xi x}d\xi \] \[ \mathcal{F}_{2,k}^{\bullet}\left[f\left(k\right)\right]\left(x\right)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(k\right)e^{ikx}dk \] \[ \mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f\left(\nu\right)e^{i\nu x}d\nu \] となり、それぞれの間の関係は、
\[ \mathcal{F}_{1,x}^{\bullet}\left[f\left(x\right)\right]\left(\xi\right)=\sqrt{2\pi}\mathcal{F}_{2,x}^{\bullet}\left[f\left(x\right)\right]\left(2\pi\xi\right) \] \[ \mathcal{F}_{1,x}^{\bullet}\left[f\left(x\right)\right]\left(\xi\right)=2\pi\mathcal{F}_{3,x}^{\bullet}\left[f\left(x\right)\right]\left(2\pi\xi\right) \] \[ \mathcal{F}_{2,k}^{\bullet}\left[f\left(k\right)\right]\left(x\right)=\frac{1}{\sqrt{2\pi}}\mathcal{F}_{1,k}^{\bullet}\left[f\left(k\right)\right]\left(\frac{x}{2\pi}\right) \] \[ \mathcal{F}_{2,k}^{\bullet}\left[f\left(k\right)\right]\left(x\right)=\sqrt{2\pi}\mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right) \] \[ \mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right)=\frac{1}{2\pi}\mathcal{F}_{1,k}^{\bullet}\left[f\left(\nu\right)\right]\left(\frac{x}{2\pi}\right) \] \[ \mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right)=\frac{1}{\sqrt{2\pi}}\mathcal{F}_{2,k}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right) \] となる。
(1)
\(\mathcal{F}_{1,x}\left[1\right]\left(\xi\right)=\delta\left(\xi\right)\)であることが分かっているとき、\(\mathcal{F}_{2,x}\left[1\right]\left(k\right)\)は\begin{align*} \mathcal{F}_{2,x}\left[1\right]\left(k\right) & =\frac{1}{\sqrt{2\pi}}\mathcal{F}_{1,x}\left[1\right]\left(\frac{k}{2\pi}\right)\\ & =\frac{1}{\sqrt{2\pi}}\delta\left(\frac{k}{2\pi}\right)\\ & =\sqrt{2\pi}\delta\left(k\right) \end{align*} となる。
(2)
\(\mathcal{F}_{1,x}\left[e^{iax}\right]\left(\xi\right)=\delta\left(\xi-\frac{a}{2\pi}\right)\)であることが分かっているとき、\(\mathcal{F}_{2,x}\left[1\right]\left(k\right)\)は\begin{align*} \mathcal{F}_{2,x}\left[e^{iax}\right]\left(k\right) & =\frac{1}{\sqrt{2\pi}}\mathcal{F}_{1,x}\left[e^{iax}\right]\left(\frac{k}{2\pi}\right)\\ & =\frac{1}{\sqrt{2\pi}}\delta\left(\frac{k}{2\pi}-\frac{a}{2\pi}\right)\\ & =\sqrt{2\pi}\delta\left(k-a\right) \end{align*} となる。
(1)
\begin{align*} \mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\xi\right) & =\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi i\xi x}dx\\ & =\sqrt{2\pi}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)e^{-i\left(2\pi\xi\right)x}dx\\ & =\sqrt{2\pi}\mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(2\pi\xi\right) \end{align*} \begin{align*} \mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\xi\right) & =\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi i\xi x}dx\\ & =\int_{-\infty}^{\infty}f\left(x\right)e^{-i\left(2\pi\xi\right)x}dx\\ & =\mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(2\pi\xi\right) \end{align*} \begin{align*} \mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(k\right) & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)e^{-ikx}dx\\ & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi i\frac{k}{2\pi}x}dx\\ & =\frac{1}{\sqrt{2\pi}}\mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\frac{k}{2\pi}\right) \end{align*} \begin{align*} \mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(k\right) & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)e^{-ikx}dx\\ & =\frac{1}{\sqrt{2\pi}}\mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(k\right) \end{align*} \begin{align*} \mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(\nu\right) & =\int_{-\infty}^{\infty}f\left(x\right)e^{-i\nu x}dx\\ & =\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi i\frac{\nu}{2\pi}x}dx\\ & =\mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\frac{\nu}{2\pi}\right) \end{align*} \begin{align*} \mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(\nu\right) & =\int_{-\infty}^{\infty}f\left(x\right)e^{-i\nu x}dx\\ & =\sqrt{2\pi}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)e^{-i\nu x}dx\\ & =\sqrt{2\pi}\mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(\nu\right) \end{align*}(1)-2
\begin{align*} \mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\xi\right) & =\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi i\xi x}dx\\ & =\frac{1}{2\pi}\int_{-\infty}^{\infty}f\left(\frac{x}{2\pi}\right)e^{-i\xi x}dx\\ & =\frac{1}{\sqrt{2\pi}}\mathcal{F}_{2,x}\left[f\left(\frac{x}{2\pi}\right)\right]\left(\xi\right)\\ & =\frac{\left|2\pi\right|}{\sqrt{2\pi}}\mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(2\pi\xi\right)\\ & =\sqrt{2\pi}\mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(2\pi\xi\right) \end{align*} \begin{align*} \mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\xi\right) & =\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi i\xi x}dx\\ & =\frac{1}{2\pi}\int_{-\infty}^{\infty}f\left(\frac{x}{2\pi}\right)e^{-i\xi x}dx\\ & =\frac{1}{2\pi}\mathcal{F}_{3,x}\left[f\left(\frac{x}{2\pi}\right)\right]\left(\xi\right)\\ & =\frac{\left|2\pi\right|}{2\pi}\mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(2\pi\xi\right)\\ & =\mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(2\pi\xi\right) \end{align*} \begin{align*} \mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(k\right) & =\frac{2\pi}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(2\pi x\right)e^{-2\pi ikx}dx\\ & =\sqrt{2\pi}\mathcal{F}_{1,x}\left[f\left(2\pi x\right)\right]\left(k\right)\\ & =\frac{\sqrt{2\pi}}{\left|2\pi\right|}\mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\frac{k}{2\pi}\right)\\ & =\frac{1}{\sqrt{2\pi}}\mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\frac{k}{2\pi}\right) \end{align*} \begin{align*} \mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(k\right) & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)e^{-ikx}dx\\ & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)e^{-ikx}dx\\ & =\frac{1}{\sqrt{2\pi}}\mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(\xi\right) \end{align*} \begin{align*} \mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(\nu\right) & =\int_{-\infty}^{\infty}f\left(x\right)e^{-i\nu x}dx\\ & =2\pi\int_{-\infty}^{\infty}f\left(2\pi x\right)e^{-2\pi i\nu x}dx\\ & =2\pi\mathcal{F}_{1,x}\left[f\left(2\pi x\right)\right]\left(\nu\right)\\ & =\frac{2\pi}{\left|2\pi\right|}\mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\frac{\nu}{2\pi}\right)\\ & =\mathcal{F}_{1,x}\left[f\left(x\right)\right]\left(\frac{\nu}{2\pi}\right) \end{align*} \begin{align*} \mathcal{F}_{3,x}\left[f\left(x\right)\right]\left(\nu\right) & =\int_{-\infty}^{\infty}f\left(x\right)e^{-i\nu x}dx\\ & =\sqrt{2\pi}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)e^{-i\nu x}dx\\ & =\sqrt{2}\mathcal{F}_{2,x}\left[f\left(x\right)\right]\left(\nu\right) \end{align*}(2)
\begin{align*} \mathcal{F}_{1,x}^{\bullet}\left[f\left(x\right)\right]\left(\xi\right) & =\int_{-\infty}^{\infty}f\left(x\right)e^{2\pi i\xi x}dx\\ & =\sqrt{2\pi}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)e^{i\left(2\pi\xi\right)x}dx\\ & =\sqrt{2\pi}\mathcal{F}_{2,x}^{\bullet}\left[f\left(x\right)\right]\left(2\pi\xi\right) \end{align*} \begin{align*} \mathcal{F}_{1,x}^{\bullet}\left[f\left(x\right)\right]\left(\xi\right) & =\int_{-\infty}^{\infty}f\left(x\right)e^{2\pi i\xi x}dx\\ & =2\pi\frac{1}{2\pi}\int_{-\infty}^{\infty}f\left(x\right)e^{i\left(2\pi\xi\right)x}dx\\ & =2\pi\mathcal{F}_{3,x}^{\bullet}\left[f\left(x\right)\right]\left(2\pi\xi\right) \end{align*} \begin{align*} \mathcal{F}_{2,k}^{\bullet}\left[f\left(k\right)\right]\left(x\right) & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(k\right)e^{ikx}dk\\ & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(k\right)e^{2\pi i\frac{k}{2\pi}x}dk\\ & =\frac{1}{\sqrt{2\pi}}\mathcal{F}_{1,k}^{\bullet}\left[f\left(k\right)\right]\left(\frac{x}{2\pi}\right) \end{align*} \begin{align*} \mathcal{F}_{2,k}^{\bullet}\left[f\left(k\right)\right]\left(x\right) & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(k\right)e^{ikx}dk\\ & =\frac{\sqrt{2\pi}}{2\pi}\int_{-\infty}^{\infty}f\left(k\right)e^{ikx}dk\\ & =\sqrt{2\pi}\mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right) \end{align*} \begin{align*} \mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right) & =\frac{1}{2\pi}\int_{-\infty}^{\infty}f\left(\nu\right)e^{i\nu x}d\nu\\ & =\frac{1}{2\pi}\int_{-\infty}^{\infty}f\left(\nu\right)e^{2\pi i\frac{\nu}{2\pi}x}d\nu\\ & =\frac{1}{2\pi}\mathcal{F}_{1,k}^{\bullet}\left[f\left(\nu\right)\right]\left(\frac{x}{2\pi}\right) \end{align*} \begin{align*} \mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right) & =\frac{1}{2\pi}\int_{-\infty}^{\infty}f\left(\nu\right)e^{i\nu x}d\nu\\ & =\frac{1}{\sqrt{2\pi}}\cdot\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(\nu\right)e^{i\nu x}d\nu\\ & =\frac{1}{\sqrt{2\pi}}\mathcal{F}_{2,k}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right) \end{align*}(2)-2
\begin{align*} \mathcal{F}_{1,x}^{\bullet}\left[f\left(x\right)\right]\left(\xi\right) & =\int_{-\infty}^{\infty}f\left(x\right)e^{2\pi i\xi x}dx\\ & =\frac{1}{2\pi}\int_{-\infty}^{\infty}f\left(\frac{x}{2\pi}\right)e^{i\xi x}dx\\ & =\frac{1}{\sqrt{2\pi}}\mathcal{F}_{2,x}^{\bullet}\left[f\left(\frac{x}{2\pi}\right)\right]\left(\xi\right)\\ & =\sqrt{2\pi}\mathcal{F}_{2,x}^{\bullet}\left[f\left(x\right)\right]\left(2\pi\xi\right) \end{align*} \begin{align*} \mathcal{F}_{1,x}^{\bullet}\left[f\left(x\right)\right]\left(\xi\right) & =\int_{-\infty}^{\infty}f\left(x\right)e^{2\pi i\xi x}dx\\ & =\frac{1}{2\pi}\int_{-\infty}^{\infty}f\left(\frac{x}{2\pi}\right)e^{i\xi x}dx\\ & =\mathcal{F}_{3,x}^{\bullet}\left[f\left(\frac{x}{2\pi}\right)\right]\left(\xi\right)\\ & =2\pi\mathcal{F}_{3,x}^{\bullet}\left[f\left(x\right)\right]\left(2\pi\xi\right) \end{align*} \begin{align*} \mathcal{F}_{2,k}^{\bullet}\left[f\left(k\right)\right]\left(x\right) & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(k\right)e^{ikx}dk\\ & =\frac{2\pi}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(2\pi k\right)e^{2\pi ikx}dk\\ & =\sqrt{2\pi}\mathcal{F}_{1,k}^{\bullet}\left[f\left(2\pi k\right)\right]\left(x\right)\\ & =\frac{\sqrt{2\pi}}{\left|2\pi\right|}\mathcal{F}_{1,k}^{\bullet}\left[f\left(k\right)\right]\left(\frac{x}{2\pi}\right)\\ & =\frac{1}{\sqrt{2\pi}}\mathcal{F}_{1,k}^{\bullet}\left[f\left(k\right)\right]\left(\frac{x}{2\pi}\right) \end{align*} \begin{align*} \mathcal{F}_{2,k}^{\bullet}\left[f\left(k\right)\right]\left(x\right) & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(k\right)e^{ikx}dk\\ & =\frac{\sqrt{2\pi}}{2\pi}\int_{-\infty}^{\infty}f\left(k\right)e^{ikx}dk\\ & =\sqrt{2\pi}\mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right) \end{align*} \begin{align*} \mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right) & =\frac{1}{2\pi}\int_{-\infty}^{\infty}f\left(\nu\right)e^{i\nu x}d\nu\\ & =\int_{-\infty}^{\infty}f\left(2\pi\nu\right)e^{2\pi i\nu x}d\nu\\ & =\mathcal{F}_{1,k}^{\bullet}\left[f\left(2\pi\nu\right)\right]\left(x\right)\\ & =\frac{1}{2\pi}\mathcal{F}_{1,k}^{\bullet}\left[f\left(\nu\right)\right]\left(\frac{x}{2\pi}\right) \end{align*} \begin{align*} \mathcal{F}_{3,\nu}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right) & =\frac{1}{2\pi}\int_{-\infty}^{\infty}f\left(\nu\right)e^{i\nu x}d\nu\\ & =\frac{1}{\sqrt{2\pi}}\cdot\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(\nu\right)e^{i\nu x}d\nu\\ & =\frac{1}{\sqrt{2\pi}}\mathcal{F}_{2,k}^{\bullet}\left[f\left(\nu\right)\right]\left(x\right) \end{align*}
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フーリエ変換の性質
\[
\mathcal{F}_{x}\left[f\left(x\right)*g\left(x\right)\right]\left(\xi\right)=\mathcal{F}_{x}\left[f\left(x\right)\right]\left(\xi\right)\mathcal{F}_{x}\left[g\left(x\right)\right]\left(\xi\right)
\]
フーリエ変換でのパーセバルの等式
\[
\int_{-\infty}^{\infty}\overline{f\left(x\right)}g\left(x\right)dx=\int_{-\infty}^{\infty}\overline{F\left(\xi\right)}G\left(\xi\right)d\xi
\]
フーリエ変換の定義とフーリエ逆変換
\[
\mathcal{F}_{x}\left[f\left(x\right)\right]\left(\xi\right)=\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi i\xi x}dx
\]