Fourier变换的积分性质及其证明过程

Fourier变换的积分性质及其证明过程

一、积分性质

如果当 t → + ∞ t \to +\infty t+时, g ( t ) = ∫ − ∞ t f ( t ) d t → 0 g(t) = \int_{ - \infty }^t {f(t){\rm{d}}t \to 0} g(t)=tf(t)dt0,则:
F [ ∫ − ∞ t f ( t ) d t ] = 1 j ω F [ f ( t ) ] \mathscr F\left[ {\int_{ - \infty }^t {f(t){\rm{d}}t} } \right] = \frac{1}{{{\rm{j}}\omega }} \mathscr F\left[ {f(t)} \right] F[tf(t)dt]=jω1F[f(t)]

该式表明一个函数的积分后的Fourier变换等于这个函数的Fourier变换除以因子 j ω j\omega .

注意:
lim ⁡ t → ∞ g ( t ) ≠ 0 \lim\limits_{t \to \infty } g(t) \ne 0 tlimg(t)=0时,积分性质应为:
F [ ∫ − ∞ t f ( t ) d t ] = 1 j ω F [ f ( t ) ] + π F ( 0 ) δ ( ω ) \mathscr F\left[ {\int_{ - \infty }^t {f(t){\rm{d}}t} } \right] = \frac{1}{{{\rm{j}}\omega }} \mathscr F\left[ {f(t)} \right]+ \pi F(0)\delta (\omega ) F[tf(t)dt]=jω1F[f(t)]+πF(0)δ(ω)
即:
F [ ∫ − ∞ t f ( t ) d t ] = 1 j ω F ( ω ) + π F ( 0 ) δ ( ω ) \mathscr F\left[ {\int_{ - \infty }^t {f(t){\rm{d}}t} } \right] = \frac{1}{{{\rm{j}}\omega }} F({\omega})+ \pi F(0)\delta (\omega ) F[tf(t)dt]=jω1F(ω)+πF(0)δ(ω)
其中: F ( ω ) = F [ f ( t ) ] F({\omega})=\mathscr F[f(t)] F(ω)=F[f(t)].

二、Fourier变换的积分性质的证明过程

证明:
根据高等数学理论,因为 d d t ∫ − ∞ t f ( t ) d t = f ( t ) \frac{d}{{dt}}\int_{ - \infty }^t {f(t){\rm{d}}t} = f(t) dtdtf(t)dt=f(t),所以
F [ d d t ∫ − ∞ t f ( t ) d t ] = F [ f ( t ) ] (1) \mathscr F\left[ {\frac{d}{{dt}}\int_{ - \infty }^t {f(t){\rm{d}}t} } \right] = \mathscr F\left[ {f(t)} \right] \tag 1 F[dtdtf(t)dt]=F[f(t)](1)

又根据Fourier变换的微分性质:
F [ g ′ ( t ) ] = j ω F [ g ( t ) ] \mathscr F[g'(t)]=j\omega \mathscr F[g(t)] F[g(t)]=F[g(t)]
可得:
F [ d d t ∫ − ∞ t f ( t ) d t ] = j ω F [ ∫ − ∞ t f ( t ) d t ] (2) \mathscr F\left[ {\frac{d}{{dt}}\int_{ - \infty }^t {f(t){\rm{d}}t} } \right] =j\omega \mathscr F[\int_{ - \infty }^t {f(t){\rm{d}}t}] \tag2 F[dtdtf(t)dt]=F[tf(t)dt](2)

对比公式(1)和(2)可知:
F [ d d t ∫ − ∞ t f ( t ) d t ] = F [ f ( t ) ] = j ω F [ ∫ − ∞ t f ( t ) d t ] (3) \mathscr F\left[ {\frac{d}{{dt}}\int_{ - \infty }^t {f(t){\rm{d}}t} } \right] = \mathscr F\left[ {f(t)} \right] \\=j\omega \mathscr F[\int_{ - \infty }^t {f(t){\rm{d}}t}] \tag3 F[dtdtf(t)dt]=F[f(t)]=F[tf(t)dt](3)

因此可得:
F [ f ( t ) ] = j ω F [ ∫ − ∞ t f ( t ) d t ] (4) \mathscr F\left[ {f(t)} \right]=j\omega \mathscr F[\int_{ - \infty }^t {f(t){\rm{d}}t}] \tag4 F[f(t)]=F[tf(t)dt](4)

对式子(4)两边除以 j ω j\omega 得到:
F [ ∫ − ∞ t f ( t ) d t ] = 1 j ω F [ f ( t ) ] (5) \mathscr F[\int_{ - \infty }^t {f(t){\rm{d}}t}] =\frac{1}{j\omega}\mathscr F\left[ {f(t)} \right] \tag5 F[tf(t)dt]=1F[f(t)](5)
证毕.

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