在Sun Jian等学者的序阻抗建模论文中,采用的Fourier分析以\(f\)为自变量,相关的公式有所变化。以下给出以\(f\)为自变量的Fourier分析的相关公式推导过程。
\(\mathscr{F}\)为Fourier算子,定义为:
\[\mathscr{F} \left\{ f\left( t \right) \right\} =F\left( \omega \right) =\int_{-\infty}^{\infty}{f\left( t \right) e^{-j\omega t}\mathrm{d}t}
\]
类似地,定义算子\(\mathscr{H}\):
\[\mathscr{H} \left\{ f\left( t \right) \right\} =H\left( f \right) =\int_{-\infty}^{\infty}{f\left( t \right) e^{-j2\pi ft}\mathrm{d}t}
\]
则:
\[\begin{aligned}
H\left( f \right) &=\int_{-\infty}^{\infty}{f\left( t \right) e^{-j2\pi ft}\mathrm{d}t}
\\
&=\frac{1}{2\pi}\int_{-\infty}^{\infty}{f\left( t \right) e^{-j2\pi ft}\mathrm{d}2\pi t}
\\
&\xlongequal{\tau =2\pi t}\frac{1}{2\pi}\int_{-\infty}^{\infty}{f\left( t \right) e^{-jf\tau}\mathrm{d}\tau}
\\
&=\frac{1}{2\pi}F\left( f \right)
\end{aligned}
\]
故:
\[\mathscr{H} \left\{ f\left( t \right) \right\} =\frac{1}{2\pi}\mathscr{F} \left\{ f\left( t \right) \right\} \Leftrightarrow \mathscr{F} \left\{ f\left( t \right) \right\} =2\pi \mathscr{H} \left\{ f\left( t \right) \right\}
\]
又(“\(*\)”为卷积运算符):
\[\mathscr{F} \left\{ f_1\left( t \right) \cdot f_2\left( t \right) \right\} =\frac{1}{2\pi}\left[ \mathscr{F} \left\{ f_1\left( t \right) \right\} *\mathscr{F} \left\{ f_1\left( t \right) \right\} \right]
\]
因此:
\[\begin{aligned}
\mathscr{H} \left\{ f_1\left( t \right) \cdot f_2\left( t \right) \right\} &=\frac{1}{2\pi}\mathscr{F} \left\{ f_1\left( t \right) \cdot f_2\left( t \right) \right\}
\\
&=\left( \frac{1}{2\pi} \right) ^2\left[ \mathscr{F} \left\{ f_1\left( t \right) \right\} *\mathscr{F} \left\{ f_1\left( t \right) \right\} \right]
\\
&=\left( \frac{1}{2\pi} \right) ^2\left\{ \left[ 2\pi \mathscr{H} \left\{ f_1\left( t \right) \right\} \right] *\left[ 2\pi \mathscr{H} \left\{ f_1\left( t \right) \right\} \right] \right\}
\end{aligned}
\]
设:
\[H_1\left( f \right) =\mathscr{H} \left\{ f_1\left( t \right) \right\} \\
H_2\left( f \right) =\mathscr{H} \left\{ f_1\left( t \right) \right\}
\]
则:
\[\begin{aligned}
\mathscr{H} \left\{ f_1\left( t \right) \cdot f_2\left( t \right) \right\} &=\left( \frac{1}{2\pi} \right) ^2\left\{ \left[ 2\pi \mathscr{H} \left\{ f_1\left( t \right) \right\} \right] *\left[ 2\pi \mathscr{H} \left\{ f_1\left( t \right) \right\} \right] \right\}
\\
&=\left( \frac{1}{2\pi} \right) ^2\left\{ \left[ 2\pi H_1\left( f \right) \right] *\left[ 2\pi H_2\left( f \right) \right] \right\}
\\
&=\left( \frac{1}{2\pi} \right) ^2\int_{-\infty}^{\infty}{\left\{ \left[ 2\pi H_1\left( x \right) \right] \cdot \left[ 2\pi H_2\left( f-x \right) \right] \right\} \mathrm{d}x}
\\
&=\int_{-\infty}^{\infty}{\left[ H_1\left( x \right) \cdot H_2\left( f-x \right) \right] \mathrm{d}x}
\\
&=H_1\left( f \right) *H_2\left( f \right)
\\
&=\mathscr{H} \left\{ f_1\left( t \right) \right\} *\mathscr{H} \left\{ f_1\left( t \right) \right\}
\end{aligned}
\]
因此:
\[\mathscr{H} \left\{ f_1\left( t \right) \cdot f_2\left( t \right) \right\} = \mathscr{H} \left\{ f_1\left( t \right) \right\} *\mathscr{H} \left\{ f_1\left( t \right) \right\}
\]
Provided by 昨夜三更雨, see https://www.cnblogs.com/zysgy/p/16182312.html
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