信号処理 Tips

はじめに

信号処理でしばしば現れる事実について計算結果を記しておく。

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\[ \newcommand{\Ts}{T_\text{s}} \]

指数関数の無限積分

\[ \integrate{-\infty}{\infty}{e^{i\omega t}}{}{t} = 2\pi\delta(\omega) \quad (\omega\in\realNumbers) \]

指数関数とデルタ関数の無限級数

\[ \sum_{n=-\infty}^\infty\exp(i\omega n\Ts) = \frac{2\pi}{\Ts}\sum_{n=-\infty}^\infty\delta\parens*{\omega-n\frac{2\pi}{\Ts}} \quad (\omega\in\realNumbers,\;Ts>0) \]

Fourier 変換

定義

\[ \mathcal{F}(f) \coloneqq \frac{1}{\sqrt{2\pi}}\integrate{-\infty}{\infty}{f(t)\NapierE^{i\omega t}}{}{t} \]

積の Fourier 変換

\[ \mathcal{F}(fg)(\omega) = \frac{1}{\sqrt{2\pi}}(F*G)(\omega) \]

畳み込みの Fourier 変換

\[ \mathcal{F}(f*g)(\omega) = \sqrt{2\pi}F(\omega)G(\omega) \]

Gauss関数のFourier変換

$A,\sigma>0,\;f(t)=\exp\parens{-t^2/(2\sigma^2)}$ のとき $\mathcal{F}(f)(\omega) = A\sigma\exp\parens{-\sigma^2 \omega^2 /2}$

0次ホールド信号のFourier変換

\[ \newcommand{\Ts}{T_\text{s}} \]

連続時間信号 $f:\realNumbers\to\complexNumbers$ とサンプリング周期 $\Ts$,区間 $[0,\Ts]$ の単位ステップ関数 $u:\realNumbers\to\braces{0,1}$ を考える。

$f$ をサンプリング周期 $\Ts$ で0次ホールドした結果を $g$ すなわち

\[ g(t) = \sum_{n=-\infty}^\infty f(n\Ts)u(t-n\Ts) \]

とすると、その Fourier 変換 $G$ は次式である。

\[ G(\omega) = \frac{\Ts}{\sqrt{2\pi}}\exp\parens*{-i\omega\Ts/2}\parens*{\sinc \frac{\omega\Ts}{2}}\sum_{n=-\infty}^\infty f(n\Ts)\exp\parens*{-i\omega n\Ts} \]

投稿者: motchy

An embedded software and FPGA engineer for measuring instrument.

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