傅里叶变换——以f为自变量
\(\mathscr{F}\) is the Fourier operator, which is defined as:
\[\mathscr{F} \left\{ f\left( t \right) \right\} =F\left( \omega \right) =\int_{-\infty}^{\infty}{f\left( t \right) e^{-j\omega t}\mathrm{d}t} \]Similarly, define the operator \(\mathscr{H}\) as:
\[\mathscr{H} \left\{ f\left( t \right) \right\} =H\left( f \right) =\int_{-\infty}^{\infty}{f\left( t \right) e^{-j2\pi ft}\mathrm{d}t} \]Then,
\[\begin{aligned} H\left( f \right) &=\int_{-\infty}^{\infty}{f\left( t \right) e^{-j2\pi ft}\mathrm{d}t} \\ &=\frac{1}{2\pi}\int_{-\infty}^{\infty}{f\left( t \right) e^{-j2\pi ft}\mathrm{d}2\pi t} \\ &\xlongequal{\tau =2\pi t}\frac{1}{2\pi}\int_{-\infty}^{\infty}{f\left( t \right) e^{-jf\tau}\mathrm{d}\tau} \\ &=\frac{1}{2\pi}F\left( f \right) \end{aligned} \]Thus,
\[\mathscr{H} \left\{ f\left( t \right) \right\} =\frac{1}{2\pi}\mathscr{F} \left\{ f\left( t \right) \right\} \Leftrightarrow \mathscr{F} \left\{ f\left( t \right) \right\} =2\pi \mathscr{H} \left\{ f\left( t \right) \right\} \]Because,
\[\mathscr{F} \left\{ f_1\left( t \right) \cdot f_2\left( t \right) \right\} =\frac{1}{2\pi}\left[ \mathscr{F} \left\{ f_1\left( t \right) \right\} *\mathscr{F} \left\{ f_1\left( t \right) \right\} \right] \]Hence,
\[\begin{aligned} \mathscr{H} \left\{ f_1\left( t \right) \cdot f_2\left( t \right) \right\} &=\frac{1}{2\pi}\mathscr{F} \left\{ f_1\left( t \right) \cdot f_2\left( t \right) \right\} \\ &=\left( \frac{1}{2\pi} \right) ^2\left[ \mathscr{F} \left\{ f_1\left( t \right) \right\} *\mathscr{F} \left\{ f_1\left( t \right) \right\} \right] \\ &=\left( \frac{1}{2\pi} \right) ^2\left\{ \left[ 2\pi \mathscr{H} \left\{ f_1\left( t \right) \right\} \right] *\left[ 2\pi \mathscr{H} \left\{ f_1\left( t \right) \right\} \right] \right\} \end{aligned} \]Let:
\[H_1\left( f \right) =\mathscr{H} \left\{ f_1\left( t \right) \right\} \\ H_2\left( f \right) =\mathscr{H} \left\{ f_1\left( t \right) \right\} \]There is:
\[\begin{aligned} \mathscr{H} \left\{ f_1\left( t \right) \cdot f_2\left( t \right) \right\} &=\left( \frac{1}{2\pi} \right) ^2\left\{ \left[ 2\pi \mathscr{H} \left\{ f_1\left( t \right) \right\} \right] *\left[ 2\pi \mathscr{H} \left\{ f_1\left( t \right) \right\} \right] \right\} \\ &=\left( \frac{1}{2\pi} \right) ^2\left\{ \left[ 2\pi H_1\left( f \right) \right] *\left[ 2\pi H_2\left( f \right) \right] \right\} \\ &=\left( \frac{1}{2\pi} \right) ^2\int_{-\infty}^{\infty}{\left\{ \left[ 2\pi H_1\left( x \right) \right] \cdot \left[ 2\pi H_2\left( f-x \right) \right] \right\} \mathrm{d}x} \\ &=\int_{-\infty}^{\infty}{\left[ H_1\left( x \right) \cdot H_2\left( f-x \right) \right] \mathrm{d}x} \\ &=H_1\left( f \right) *H_2\left( f \right) \\ &=\mathscr{H} \left\{ f_1\left( t \right) \right\} *\mathscr{H} \left\{ f_1\left( t \right) \right\} \end{aligned} \]Therefore,
\[\mathscr{H} \left\{ f_1\left( t \right) \cdot f_2\left( t \right) \right\} = \mathscr{H} \left\{ f_1\left( t \right) \right\} *\mathscr{H} \left\{ f_1\left( t \right) \right\} \]