跳至主要內容

傅里叶变换的性质

Reflect314数学公式

基本性质

  • 线性性质：设 F(ω)=F[f(t)]F (\omega) = \mathcal{F}[f (t)] ， G(ω)=F[g(t)]G (\omega) = \mathcal{F}[g (t)] ， α,β\alpha ,\beta 为常数，则
    • F[αf(t)+βg(t)]=αF(ω)+βG(ω),\mathcal {F} [ \alpha f (t) + \beta g (t) ] = \alpha F (\omega) + \beta G (\omega),
    • F−1[αF(ω)+βG(ω)]=αf(t)+βg(t).\mathcal {F}^{- 1} \left[ \alpha F (\omega) + \beta G (\omega) \right] = \alpha f (t) + \beta g (t).
  • 位移性质：设 F(ω)=F[f(t)],t0,ω0F (\omega) = \mathcal{F}[f (t)], t_0, \omega_0 为实常数，则
    • F[f(t−t0)]=e−jωt0F(ω)\mathcal {F} \left[ f (t - t_{0}) \right] = \mathrm{e}^{- \mathrm{j} \omega t_{0}} F (\omega)
    • F−1[F(ω−ω0)]=ejω0tf(t)\mathcal {F}^{- 1} \left[ F (\omega - \omega_{0}) \right] = \mathrm{e}^{\mathrm{j} \omega_{0} t} f (t)
  • 相似性质：设 F(ω)=F[f(t)]F (\omega) = \mathcal{F}[f (t)] ， aa 为非零常数，则
    • \mathcal {F} [ f (a t) ] = \frac {1}{| a |} F \left (\frac {\omega}{a}\right).\tag{8.14}
  • 微分性质：
    • F[f′(t)]=jωF[f(t)]\mathcal {F} [ f^{\prime} (t) ] = \mathrm{j} \omega \mathcal {F} [ f (t)]
    • F[f(n)(t)]=(jω)nF[f(t)]\mathcal {F} [ f^{(n)} (t) ] = (\mathrm{j} \omega)^{n} \mathcal {F} [ f (t) ]
  • 像函数的导数公式：
    • ddωF(ω)=−jF[tf(t)]\frac {\mathrm{d}}{\mathrm{d} \omega} F (\omega) = - \mathrm{j} \mathcal {F} [ t f (t) ]
    • dnF(ω)dωn=(−j)nF[tnf(t)]\frac {\mathrm{d}^{n} F (\omega)}{\mathrm{d} \omega^ {n}} = (- \mathrm{j})^{n} \mathcal {F} [ t^{n} f (t) ]
  • 积分性质：设 g(t)=∫−∞tf(τ)dτg (t) = \int_{-\infty}^{t}f (\tau)\mathrm{d}\tau ，若 lim⁡t→+∞g(t)=0\lim_{t\to +\infty}g (t) = 0 ，则
    • F[g(t)]=1jωF[f(t)]\mathcal {F} [ g (t) ] = \frac {1}{\mathrm{j} \omega} \mathcal {F} [ f (t) ]
  • 帕塞瓦尔 (Parseval) 等式：设 F(ω)=F[f(t)]F (\omega) = \mathcal{F}[f (t)] ，则有
    • ∫−∞+∞f2(t)dt=12π∫−∞+∞∣F(ω)∣2dω\int_{- \infty}^{+ \infty} f^{2} (t) \mathrm{d} t = \frac {1}{2 \pi} \int_{- \infty}^{+ \infty} | F (\omega) |^{2} \mathrm{d} \omega

卷积

设实值函数 f1(t)f_{1}(t) 与 f2(t)f_{2}(t) 在 (−∞,+∞)(- \infty, +\infty) 内有定义. 若反常积分 ∫−∞+∞f1(τ)f2(t−τ)dτ\int_{-\infty}^{+\infty} f_{1}(\tau) f_{2}(t - \tau) \mathrm{d}\tau 对任何实数 tt 收敛, 则它定义了一个自变量为 tt 的函数, 称此函数为 f1(t)f_{1}(t) 与 f2(t)f_{2}(t) 的卷积, 记为 f1(t)∗f2(t)f_{1}(t) * f_{2}(t) , 即

f1(t)∗f2(t)=∫−∞+∞f1(τ)f2(t−τ)dτ.f_{1} (t) * f_{2} (t) = \int_{- \infty}^{+ \infty} f_{1} (\tau) f_{2} (t - \tau) \mathrm{d} \tau .

根据定义, 很容易知道卷积满足：

  • 交换律：f1(t)∗f2(t)=f2(t)∗f1(t)f_{1} (t) * f_{2} (t) = f_{2} (t) * f_{1} (t)
  • 结合律：f1(t)∗[f2(t)∗f3(t)]=[f1(t)∗f2(t)]∗f3(t)f_{1} (t) * [ f_{2} (t) * f_{3} (t) ] = [ f_{1} (t) * f_{2} (t) ] * f_{3} (t)
  • 分配律：f1(t)∗[f2(t)+f3(t)]=f1(t)∗f2(t)+f1(t)∗f3(t)f_{1} (t) * [ f_{2} (t) + f_{3} (t) ] = f_{1} (t) * f_{2} (t) + f_{1} (t) * f_{3} (t)
例题：求下列函数的卷积：$$f (t) = \left{ \begin{array}{l l} \mathrm{e}^{- \alpha t}, & t \geqslant 0, \ 0, & t < 0; \end{array} \right. g (t) = \left{ \begin{array}{l l} \mathrm{e}^{- \beta t}, & t \geqslant 0, \ 0, & t < 0. \end{array} \right.$$

其中 α>0,β>0\alpha>0,\beta>0 且 α≠β.\alpha\neq\beta.

(a)

(b)

解：

由定义有：$$f (t) * g (t) = \int_{- \infty}^{+ \infty} f (\tau) g (t - \tau) \mathrm{d} \tau ,$$

由图可得：当 t<0t < 0 时， f(t)∗g(t)=0f (t) * g (t) = 0 ；当 t⩾0t \geqslant 0 时，$$\begin{array}{r l} f (t) * g (t) & = \int_{0}^{t} f (\tau) g (t - \tau) \mathrm{d} \tau = \int_{0}^{t} \mathrm{e}^{- \alpha \tau} \mathrm{e}^{- \beta (t - \tau)} \mathrm{d} \tau \ & = \mathrm{e}^{- \beta t} \int_{0}^{t} \mathrm{e}^{- (\alpha - \beta) \tau} \mathrm{d} \tau = \frac {1}{\alpha - \beta} (\mathrm{e}^{- \beta t} - \mathrm{e}^{- \alpha t}). \end{array}$$

综合得：$$f (t) * g (t) = \left{ \begin{array}{l l} 0, & t < 0, \ \frac {1}{\alpha - \beta} (\mathrm{e}^{- \beta t} - \mathrm{e}^{- \alpha t}), & t \geqslant 0. \end{array} \right.$$

例题：求下列函数的卷积：$$f (t) = t^{2} u (t), \quad g (t) = \left{ \begin{array}{l l} 1, & | t | \leqslant 1, \ 0, & | t | > 1. \end{array} \right.$$

(a)

(b)

解：

由定义有$$f (t) * g (t) = \int_{- \infty}^{+ \infty} f (\tau) g (t - \tau) \mathrm{d} \tau = \int_{- \infty}^{+ \infty} g (\tau) f (t - \tau) \mathrm{d} \tau ,$$

由图可得: 当 t<−1t<-1 时, $$f (t) * g (t) = 0;$$

当 −1⩽t⩽1-1 \leqslant t \leqslant 1 时，$$f (t) * g (t) = \int_{- 1}^{t} 1 \cdot (t - \tau)^{2} \mathrm{d} \tau = \frac {1}{3} (t + 1)^{3};$$

当 t>1t > 1 时，$$f (t) * g (t) = \int_{- 1}^{1} 1 \cdot (t - \tau)^{2} \mathrm{d} \tau = \frac {1}{3} (6 t^{2} + 2).$$

综合得$$f (t) * g (t) = \left{ \begin{array}{l l} 0, & t < - 1, \ (t + 1)^{3} / 3, & - 1 \leqslant t \leqslant 1, \ (6 t^{2} + 2) / 3, & t > 1. \end{array} \right.$$

通过上述例子可知，卷积由反褶、平移、相乘、积分几个部分组成。即将 g(τ)g (\tau) 反褶平移得 g(t−τ)=g(−(τ−t))g (t - \tau) = g (-( \tau - t)) ，再与 f(τ)f (\tau) 相乘求积分，因此卷积又称为褶积或卷乘。若采用图解方式，则很容易确定积分限。

卷积定理

设 F(ω)=F[f(t)]F(\omega) = \mathcal{F}[f(t)] ， G(ω)=F[g(t)]G(\omega) = \mathcal{F}[g(t)] ，则有

  • 时域卷积定理：F[f(t)∗g(t)]=F(ω)⋅G(ω)\mathcal {F} \left[ f (t) * g (t) \right] = F (\omega) \cdot G (\omega)
  • 频域卷积定理（对偶性）：F[f(t)⋅g(t)]=12πF(ω)∗G(ω)\mathcal {F} \left[ f (t) \cdot g (t) \right] = \frac {1}{2 \pi} F (\omega) * G (\omega)

证：由卷积与傅氏变换定义有

F[f(t)∗g(t)]=∫−∞+∞f(t)∗g(t)e−jωtdt=∫−∞+∞[∫−∞+∞f(τ)g(t−τ)dτ]e−jωtdt=∫−∞+∞f(τ)[∫−∞+∞g(t−τ)e−jωtdt]dτ=∫−∞+∞f(τ)e−jωτ[∫−∞+∞g(t−τ)e−jω(t−τ)dt]dτ=F(ω)∗G(ω).\begin{array}{r l} \mathcal {F} [ f (t) * g (t) ] & = \int_{- \infty}^{+ \infty} f (t) * g (t) \mathrm{e}^{- \mathrm{j} \omega t} \mathrm{d} t \\ & = \int_{- \infty}^{+ \infty} \left[ \int_{- \infty}^{+ \infty} f (\tau) g (t - \tau) \mathrm{d} \tau \right] \mathrm{e}^{- \mathrm{j} \omega t} \mathrm{d} t \\ & = \int_{- \infty}^{+ \infty} f (\tau) \left[ \int_{- \infty}^{+ \infty} g (t - \tau) \mathrm{e}^{- \mathrm{j} \omega t} \mathrm{d} t \right] \mathrm{d} \tau \\ & = \int_{- \infty}^{+ \infty} f (\tau) \mathrm{e}^{- \mathrm{j} \omega \tau} \left[ \int_{- \infty}^{+ \infty} g (t - \tau) \mathrm{e}^{- \mathrm{j} \omega (t - \tau)} \mathrm{d} t \right] \mathrm{d} \tau \\ & = F (\omega) * G (\omega). \end{array}

同理可证频域卷积定理.

利用卷积定理可以简化卷积计算及某些函数的傅氏变换.

例题：已知 α>0,β>0\alpha>0,\beta>0，求下列函数的卷积$$f (t) = \frac {\sin \alpha t}{\pi t}, \quad g (t) = \frac {\sin \beta t}{\pi t}$$

解：

设 F(ω)=F[f(t)]F (\omega) = \mathcal{F}[f (t)] ， G(ω)=F[g(t)]G (\omega) = \mathcal{F}[g (t)] ，由例题知：$$F (\omega) = \left{ \begin{array}{l l} 1, & | \omega | \leqslant \alpha , \ 0, & | \omega | > \alpha ; \end{array} \right. \quad G (\omega) = \left{ \begin{array}{l l} 1, & | \omega | \leqslant \beta , \ 0, & | \omega | > \beta . \end{array} \right.$$
因此有F(ω)⋅G(ω)={1,∣ω∣⩽γ,0,∣ω∣>γ(其中γ=min⁡(α,β)),F (\omega) \cdot G (\omega) = \left\{ \begin{array}{l l} 1, & | \omega | \leqslant \gamma , \\ 0, & | \omega | > \gamma \end{array} \right. (\text {其中} \gamma = \min (\alpha , \beta)),

由卷积定理有f(t)∗g(t)=F−1[F(ω)⋅G(ω)]=sin⁡γtπt.f (t) * g (t) = \mathcal {F}^{- 1} [ F (\omega) \cdot G (\omega) ] = \frac {\sin \gamma t}{\pi t}.

例题：设 f(t)=e−βtu(t)cos⁡ω0t(β>0)f (t)=\mathrm{e}^{-\beta t}u (t)\cos\omega_{0}t (\beta>0) ，求 F[f(t)]\mathcal{F}[f (t)]

解：

由频域卷积定理得$$\mathcal {F} [ f (t) ] = \frac {1}{2 \pi} \mathcal {F} [ \mathrm{e}^{- \beta t} u (t) ] * \mathcal {F} [ \cos \omega_{0} t ].$$

又由例题a与例题b可知 $$\begin{align}&\mathcal {F} \left[ \mathrm{e}^{- \beta t} u (t) \right] = \frac {1}{\beta + \mathrm{j} \omega}, \ & \mathcal {F} \left[ \cos \omega_{0} t \right] = \pi \left[ \delta (\omega + \omega_{0}) + \delta (\omega - \omega_{0}) \right].\end{align}$$

因此有$$\begin{array}{r l} & {\mathcal {F} [ f (t) ]} \ & {= \frac {1}{2 \pi} \int_{- \infty}^{+ \infty} \frac {\pi}{\beta + j \tau} [ \delta (\omega + \omega_{0} - \tau) + \delta (\omega - \omega_{0} - \tau) ] d \tau} \ & {= \frac {1}{2} \left[ \frac {1}{\beta + j (\omega + \omega_{0})} + \frac {1}{\beta + j (\omega - \omega_{0})} \right] = \frac {\beta + j \omega}{(\beta + j \omega)^{2} + \omega_{0}^{2}}.} \end{array}$$

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