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积分变换知识点整理1-3

作者:互联网

Fourier变换的性质

线性性质

设 F 1 ( ω ) = F [ f 1 ( t ) ] F_1(\omega)=\mathscr{F}[f_1(t)] F1​(ω)=F[f1​(t)], F 2 ( ω ) = F [ f 2 ( t ) ] F_2(\omega)=\mathscr{F}[f_2(t)] F2​(ω)=F[f2​(t)], α \alpha α, β \beta β是常数,则
F [ α f 1 ( t ) + β f 2 ( t ) ] = α F [ f 1 ( t ) ] + β F [ f 2 ( t ) ] = α F 1 ( ω ) + β F 2 ( ω ) \mathscr{F}[\alpha f_1(t)+\beta f_2(t)]=\alpha\mathscr{F}[f_1(t)]+\beta\mathscr{F}[f_2(t)]=\alpha F_1(\omega)+\beta F_2(\omega) F[αf1​(t)+βf2​(t)]=αF[f1​(t)]+βF[f2​(t)]=αF1​(ω)+βF2​(ω)

F − 1 [ α F 1 ( ω ) + β F 2 ( ω ) ] = α F − 1 [ F 1 ( ω ) ] + β F − 1 [ F 2 ( ω ) ] = α F 1 ( ω ) + β F 2 ( ω ) \mathscr{F}^{-1}[\alpha F_1(\omega)+\beta F_2(\omega)]=\alpha\mathscr{F}^{-1}[F_1(\omega)]+\beta\mathscr{F}^{-1}[F_2(\omega)]=\alpha F_1(\omega)+\beta F_2(\omega) F−1[αF1​(ω)+βF2​(ω)]=αF−1[F1​(ω)]+βF−1[F2​(ω)]=αF1​(ω)+βF2​(ω)

位移性质

F [ f ( t ± t 0 ) ] = e ± j ω t 0 F [ f ( t ) ] \mathscr{F}[f(t\pm t_0)]=e^{\pm j\omega t_0}\mathscr{F}[f(t)] F[f(t±t0​)]=e±jωt0​F[f(t)]
F − 1 [ F ( ω ∓ ω 0 ) ] = f ( t ) e ± j ω 0 t \mathscr{F}^{-1}[F(\omega \mp \omega_0)]=f(t)e^{\pm j\omega_0t} F−1[F(ω∓ω0​)]=f(t)e±jω0​t

微分性质

如果 f ( t ) f(t) f(t)在 ( − ∞ , + ∞ ) (-\infty,+\infty) (−∞,+∞)上连续或只有有限个可去间断点,且当 ∣ t ∣ → 0 |t|\to 0 ∣t∣→0时, f ( t ) → 0 f(t)\to 0 f(t)→0,则
F [ f ′ ( t ) ] = j ω F [ f ( t ) ] \mathscr{F}[f'(t)]=j\omega\mathscr{F}[f(t)] F[f′(t)]=jωF[f(t)]
推论
  若 f ( k ) ( t ) f^{(k)}(t) f(k)(t)在 ( − ∞ , + ∞ ) (-\infty,+\infty) (−∞,+∞)上连续或只有有限个可去间断点,且 lim ⁡ ∣ t ∣ → + ∞ f ( k ) ( t ) = 0 , k = 0 , 1 , 2 , ⋅ ⋅ ⋅ , n − 1 \lim\limits_{|t|\to +\infty}f^{(k)}(t)=0,k=0,1,2,···,n-1 ∣t∣→+∞lim​f(k)(t)=0,k=0,1,2,⋅⋅⋅,n−1,则有
F [ f ( n ) ( t ) ] = ( j ω ) n F [ f ( t ) ] \mathscr{F}[f^{(n)}(t)]=(j\omega)^n\mathscr{F}[f(t)] F[f(n)(t)]=(jω)nF[f(t)]
同样
d n d ω n F ( ω ) = ( − j ) n F [ t n f ( t ) ] \frac{d^n}{d\omega^n}F(\omega)=(-j)^n\mathscr{F}[t^nf(t)] dωndn​F(ω)=(−j)nF[tnf(t)]

积分性质

如果当 t → + ∞ t\to +\infty t→+∞时, g ( t ) = ∫ − ∞ t f ( t )   d t → 0 g(t)=\displaystyle\int_{-\infty}^{t}f(t)\,dt\to0 g(t)=∫−∞t​f(t)dt→0,那么 F [ ∫ − ∞ t f ( t ) ] = 1 j ω F [ f ( t ) ] \mathscr{F}\left[\int_{-\infty}^{t}f(t)\right]=\frac{1}{j\omega}\mathscr{F}[f(t)] F[∫−∞t​f(t)]=jω1​F[f(t)]

乘积定理

若 F 1 ( ω ) = F [ f 1 ( t ) ] F_1(\omega)=\mathscr{F}[f_1(t)] F1​(ω)=F[f1​(t)], F 2 ( ω ) = F [ f 2 ( t ) ] F_2(\omega)=\mathscr{F}[f_2(t)] F2​(ω)=F[f2​(t)],则
∫ − ∞ + ∞ f 1 ( t ) ‾ f 2 ( t )   d t = 1 2 π ∫ − ∞ + ∞ F 1 ( ω ) ‾ F 2 ( ω ) d ω \int_{-\infty}^{+\infty}\overline{f_1(t)}f_2(t)\,dt=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\overline{F_1(\omega)}F_2(\omega)d\omega ∫−∞+∞​f1​(t)​f2​(t)dt=2π1​∫−∞+∞​F1​(ω)​F2​(ω)dω
∫ − ∞ + ∞ f 2 ( t ) ‾ f 1 ( t )   d t = 1 2 π ∫ − ∞ + ∞ F 2 ( ω ) ‾ F 1 ( ω ) d ω \int_{-\infty}^{+\infty}\overline{f_2(t)}f_1(t)\,dt=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\overline{F_2(\omega)}F_1(\omega)d\omega ∫−∞+∞​f2​(t)​f1​(t)dt=2π1​∫−∞+∞​F2​(ω)​F1​(ω)dω
其中 f 1 ( t ) ‾ \overline{f_1(t)} f1​(t)​, f 2 ( t ) ‾ \overline{f_2(t)} f2​(t)​, F 1 ( ω ) ‾ \overline{F_1(\omega)} F1​(ω)​及 F 2 ( ω ) ‾ \overline{F_2(\omega)} F2​(ω)​分别为 f 1 ( t ) f_1(t) f1​(t), f 2 ( t ) f_2(t) f2​(t), F 1 ( ω ) F_1(\omega) F1​(ω)及 F 2 ( ω ) F_2(\omega) F2​(ω)的共轭函数

能量积分

  若 F ( ω ) = F [ f ( t ) ] F(\omega)=\mathscr{F}[f(t)] F(ω)=F[f(t)],则有
∫ − ∞ + ∞ [ f ( t ) ] 2 d t = 1 2 π ∫ − ∞ + ∞ ∣ F ( ω ) ∣ 2 d ω \int_{-\infty}^{+\infty}[f(t)]^2dt=\frac{1}{2\pi}\int_{-\infty}^{+\infty}|F(\omega)|^2d\omega ∫−∞+∞​[f(t)]2dt=2π1​∫−∞+∞​∣F(ω)∣2dω
称为Parseval等式
能量密度函数(或称能量谱密度
S ( ω ) = ∣ F ( ω ) ∣ 2 S(\omega)=|F(\omega)|^2 S(ω)=∣F(ω)∣2
显然 S ( ω ) = S ( − ω ) S(\omega)=S(-\omega) S(ω)=S(−ω)
—————————————————————————————————
若 F ( ω ) = F [ f ( t ) ] F(\omega)=\mathscr{F}[f(t)] F(ω)=F[f(t)]
F [ f ( t ) cos ⁡ ω 0 t ] = 1 2 [ F ( ω ) + F ( ω 0 ) ] \mathscr{F}[f(t)\cos\omega_0t]=\frac{1}{2}[F(\omega)+F(\omega_0)] F[f(t)cosω0​t]=21​[F(ω)+F(ω0​)]
F [ f ( t ) sin ⁡ ω 0 t ] = 1 2 j [ F ( ω − ω 0 ) − F ( ω + ω 0 ) ] \mathscr{F}[f(t)\sin\omega_0t]=\frac{1}{2j}[F(\omega-\omega_0)-F(\omega+\omega_0)] F[f(t)sinω0​t]=2j1​[F(ω−ω0​)−F(ω+ω0​)]

标签:知识点,F2,infty,变换,mathscr,f2,F1,积分,omega
来源: https://blog.csdn.net/qq_37278781/article/details/120319156