常用傅里叶变换对($t \rightarrow \omega$)

1. 单边指数函数

\[f(t)=\left \{ \begin{aligned} & e^{-at}, & t \ge 0\\ & 0 , & t < 0 \end{aligned} \right. \]

其中\(a\)是实数。于是其傅里叶变换为:

\[\begin{aligned} F(\omega) &= \int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt \\ &= \int_{0}^{\infty}e^{-at}e^{-j\omega t}dt \\ &= \int_{0}^{\infty}e^{-(a+j\omega)t}dt \\ &= [-\frac{e^{-(a+j\omega)t}}{a+j\omega}]_{0}^{\infty} \\ &= 0 - (-\frac{1}{a+j\omega}) = \frac{1}{a+j\omega} \end{aligned} \]

从而可以得到:

\[\left \{ \begin{aligned} F(\omega) &= \frac{1}{a+j\omega} \\ |F(\omega)| &=\frac{1}{\sqrt{a^{2}+\omega^{2}}}\\ \varphi(\omega) &= -\text{arctan}(\frac{\omega}{a}) \end{aligned} \right. \tag{1} \]

2. 双边指数函数

\[f(t)=e^{-a|t|}\quad (-\infty<t<\infty) \]

傅里叶变换为:

\[\begin{aligned} F(\omega) &= \int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt = \int_{-\infty}^{\infty}e^{-a|t|}e^{-j\omega t}dt \\ &= \int_{-\infty}^{0}e^{at}e^{-j\omega t}dt+\int_{0}^{\infty}e^{-at}e^{-j\omega t}dt \\ &= \int_{-\infty}^{0}e^{(a-j\omega)t}dt+\int_{0}^{\infty}e^{-(a+j\omega)t}dt \\ &= [\frac{e^{(a-j\omega)t}}{a-j\omega}]_{-\infty}^{0}+\frac{1}{a+j\omega} \\ &= \frac{1}{a-j\omega} + \frac{1}{a+j\omega} = \frac{2a}{a^{2}+\omega^{2}} \end{aligned} \]

从而可以得到:

\[\left \{ \begin{aligned} F(\omega) &= \frac{1}{a^{2}+\omega^{2}} \\ |F(\omega)| &=\frac{1}{a^{2}+\omega^{2}}\\ \varphi(\omega) &= 0 \end{aligned} \right. \tag{2} \]

3. 矩形脉冲信号

矩形脉冲信号的表达式为:

\[f(t)=E[u(t+\frac{\tau}{2})-u(t-\frac{\tau}{2})] \]

其中\(E\)为脉冲幅度,\(\tau\)为脉冲宽度。其傅里叶变换为:

\[\begin{aligned} F(\omega) &= \int_{-\infty}^{\infty}f(t)e^{i\omega t}=\int_{-\frac{\tau}{2}}^{\frac{\tau}{2}}Ee^{-j\omega t}=[-\frac{Ee^{-j\omega t}}{j\omega}]_{-\frac{\tau}{2}}^{\frac{\tau}{2}}\\ &= -\frac{E}{j\omega}[e^{-j\omega \frac{\tau}{2}}-e^{j\omega\frac{\tau}{2}}] =-\frac{E}{j\omega}[(cos\frac{\omega \tau}{2}-jsin\frac{\omega\tau}{2})-(cos\frac{\omega \tau}{2}+jsin\frac{\omega\tau}{2})] \\ &= -\frac{E}{j\omega}[-2jsin\frac{\omega\tau}{2}]=\frac{2E}{\omega}sin(\frac{\omega\tau}{2}) \end{aligned} \]

定义抽样信号\(Sa(t)=\frac{sint}{t}\),所以上式可以写为:

\[F(\omega)=E\tau[\frac{sin(\frac{\omega\tau}{2})}{\frac{\omega\tau}{2}}]=E\tau Sa(\frac{\omega\tau}{2}) \]

从而得到

\[|F(\omega)| = E\tau \left|Sa(\frac{\omega\tau}{2})\right| \tag{3} \]

矩形脉冲信号的频谱以\(Sa(\frac{\omega\tau}{2})\)的规律变化,分布在无限宽的频率范围上,但是其主要信号能量处于\(f=0\sim\frac{1}{\tau}\)的范围内。因而,通常认为这种信号的带宽为:

\[B\approx\frac{1}{\tau} \]

4. 符号函数

符号函数(或称正负号函数)以符号sgn记,其表达式为:

\[f(t) = \text{sgn}(t)= \left \{ \begin{aligned} &+1\quad &(t>0) \\ &0 \quad &(t=0) \\ &-1 \quad &(t<0) \end{aligned} \right. \]

显然,这种信号不满足绝对可积条件,但是它却存在傅里叶变换(绝对可积是傅里叶变换存在的充分条件而非必要条件)。直接对该函数进行傅里叶变换:

\[\begin{aligned} F(\omega)&=\int_{-\infty}^{\infty}sng(t)e^{-j\omega t}=\int_{-\infty}^{0_{-}}-e^{-j\omega t}dt+\int_{0_{+}}^{\infty}e^{-j\omega t}dt \\ &=\frac{1}{j\omega}[\lim_{t\to0_{-}}e^{-j\omega t}-\lim_{t\to-\infty}e^{-j\omega t}]-\frac{1}{j\omega}[\lim_{t\to\infty}e^{-j\omega t}-\lim_{t\to0_{+}}e^{-j\omega t}] \end{aligned} \]

式中存在无穷项,不能收敛。所以无法直接求取。考虑到:

\[\lim_{a\to0}e^{-a|t|}=1 \]

我们将其与\(\text{sgn}(t)\)相乘,然后将乘积\(f_{1}(t)\)进行傅里叶变换,最后对结果取\(a\)趋近于0的极限即可得到符号函数的傅里叶变换。乘积信号的傅里叶变换为:

\[\begin{aligned} F_{1}(\omega)&=\int_{-\infty}^{\infty}f_{1}(t)e^{-j\omega t}dt=\int_{-\infty}^{0}(-e^{at})e^{-j\omega t}dt + \int_{0}^{\infty}e^{-at}e^{-j\omega t}dt \\ &=\int_{-\infty}^{0}(-e^{(a-j\omega)t})dt + \int_{0}^{\infty}e^{-at}e^{-j\omega t}dt \\ &=-[\frac{e^{(a-j\omega)t}}{a-j\omega}]_{-\infty}^{0} + \frac{1}{a+j\omega} \\ &= -[\frac{1}{a-j\omega}-0]+\frac{1}{a+j\omega}=\frac{-2j\omega}{a^{2}+\omega^{2}} \end{aligned} \]

从而符号函数\(sgn(t)\)的频谱函数为:

\[\left \{ \begin{aligned} F(\omega)&=\lim_{a\to0}\frac{-2j\omega}{a^{2}+\omega^{2}}=\frac{2}{j\omega} \\ \left| F(\omega) \right| &= \frac{2}{|\omega|} \\ \varphi(\omega)&=\left \{ \begin{aligned} &-\frac{\pi}{2}\quad (\omega >0) \\ &+\frac{\pi}{2}\quad(\omega<0) \end{aligned}\right. \end{aligned} \right. \tag{4} \]

5. 升余弦函数

升余弦脉冲函数的表达式为:

\[f(t)=\frac{E}{2}[1+cos(\frac{\pi t}{\tau})]\quad(0 \le |t| \le \tau) \]

其傅里叶变换为:

\[\begin{aligned} F(\omega)&=\int_{-\tau}^{\tau}\frac{E}{2}[1+cos(\frac{\pi t}{\tau})]e^{-j\omega t}dt \\ &= \frac{E}{2}\int_{-\tau}^{\tau}e^{-j\omega t}dt+\frac{E}{2}[\int_{-\tau}^{\tau}\frac{1}{2}(e^{\frac{j\pi t}{\tau}}+e^{-\frac{j\pi t}{\tau}})e^{-j\omega t}dt] \\ &=E\tau Sa(\omega \tau)+\frac{E\tau}{2}Sa[(\omega-\frac{\pi}{\tau})\tau]+\frac{E\tau}{2}Sa[(\omega+\frac{\pi}{\tau})\tau] \end{aligned} \tag{5a} \]

显然\(F(\omega)\)是由三项构成的,它们都是矩形脉冲的频谱,只是有两项沿频率轴左、右平移了\(\omega=\frac{\pi}{\tau}\)。把上式化简可以得到:

\[F(\omega)=\frac{Esin(\omega\tau)}{\omega[1-(\frac{\omega\tau}{\pi})^{2}]}=\frac{E\tau Sa(\omega\tau)}{1-(\frac{\omega\tau}{\pi})^{2}} \tag{5b} \]

升余弦脉冲信号的频谱比矩形脉冲的频谱更加集中。对于半幅度宽度为\(\tau\)的升余弦脉冲信号,它的绝大部分能量集中在\(\omega=0\sim\frac{2\pi}{\tau}\rightarrow f=0\sim\frac{1}{\tau}\)的范围内。

6. 冲激函数

单位冲激函数\(\delta(t)\)的傅里叶变换为:

\[F(\omega)=\int_{-\infty}^{\infty}\delta(t)e^{-j\omega t}dt=1 \tag{6a} \]

上述结果也可由矩形脉冲取极限得到,当脉宽\(\tau\)逐渐变窄时,其频谱必然展宽。可以想象,若\(\tau\rightarrow0\),而\(E\tau=1\),这时矩形脉冲就变成了\(\delta(t)\),其相应的频谱等于常数1。

单位冲激函数的频谱等于常数,也就是说,在整个频率范围内频谱时均匀分布的。显然,在时域中变化异常剧烈的冲激函数包含幅度相等的所有频率分量。因此,这种频谱常称为“均匀谱”或“白色谱”。

怎么样的函数其频谱为冲激函数呢?也就是求\(\delta(\omega)\)的傅里叶逆变换。

\[\mathscr{F}^{-1}[\delta(\omega)]=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\delta(\omega)e^{j\omega t}d\omega = \frac{1}{2\pi} \tag{6b} \]

此结果说明,直流信号的傅里叶变换是冲激函数。由\((6b)\)进一步可得:

\[\mathscr{F}(\frac{1}{2\pi})=\delta(\omega) \tag{6c} \]

\[\mathscr{F}(1)=2\pi\delta(\omega) \tag{6d} \]

7. 阶跃函数

从波形容易看出阶跃函数\(u(t)\)不满足绝对可积条件,即使如此,它仍然存在傅里叶变换。由于

\[u(t)=\frac{1}{2}+\frac{1}{2}sgn(t) \]

两边进行傅里叶变换可以得到:

\[\begin{aligned} \mathscr{F}[u(t)]&=\mathscr{F}(\frac{1}{2})+\frac{1}{2}\mathscr{F}[sng(t)]\\ &= \pi\delta(\omega)+\frac{1}{j\omega} \end{aligned} \tag{7} \]

8. 正弦、余弦信号的傅里叶变换

\(\mathscr{F}[f_{0}(t)]=F_{0}(\omega)\),由频移特性可知

\[\mathscr{F}[f_{0}(t)e^{j\omega_{1}t}]=F_{0}(\omega-\omega_{1}) \]

上式中,令\(f_{0}(t)=1\),其傅里叶变换为\(\mathscr{F}[1]=2\pi\delta(\omega)\),那么上式变为:

\[\mathscr{F}[e^{j\omega_{1}t}]=2\pi\delta(\omega-\omega_{1}) \]

同理:

\[\mathscr{F}[e^{-j\omega_{1}t}]=2\pi\delta(\omega+\omega_{1}) \]

由两式并结合欧拉公式,可以得到:

\[\left \{ \begin{aligned} \mathscr{F}[cos(\omega_{1}t)]&=\pi[\delta(\omega+\omega_{1})+\delta(\omega-\omega_{1})] \\ \mathscr{F}[sin(\omega_{1}t)]&=j\pi[\delta(\omega+\omega_{1})-\delta(\omega-\omega_{1})] \end{aligned} \tag{8} \right. \]

9. 一般周期信号的傅里叶变换

令周期信号\(f(t)\)的周期为\(T_{1}\),角频率为\(\omega_{1}(=2\pi f_{1}=\frac{2\pi}{T_{1}})\),可以将\(f(t)\)展成傅里叶级数,为:

\[f(t)=\sum_{n=-\infty}^{\infty}F_{n}e^{jn\omega_{1}t} \]

将上式两边取傅里叶变换

\[\begin{aligned} \mathscr{F}[f(t)]&=\mathscr{F}[\sum_{n=-\infty}^{\infty}F_{n}e^{jn\omega_{1}t}] \\ &= \sum_{n=-\infty}^{\infty}F_{n}\mathscr{F}[e^{jn\omega_{1}t}] \\ &= 2\pi\sum_{n=-\infty}^{\infty}F_{n}\delta(\omega-n\omega_{1}) \end{aligned} \tag{9} \]

其中\(F_{n}\)\(f(t)\)的傅里叶级数的系数,已经知道它等于

\[F_{n}=\frac{1}{T_{1}}\int_{-\frac{T_{1}}{2}}^{\frac{T_{1}}{2}}f(t)e^{-jn\omega_{1}t}dt \tag{9a} \]

\((9)\)表明,周期信号\(f(t)\)的傅里叶变换是由一些冲激函数组成,这些冲激位于信号的谐频(\(0,\pm\omega_{1},\pm\omega_{2},\cdots\))处,每个冲激的强度等于\(f(t)\)的傅里叶级数相应系数\(F_{n}\)\(2\pi\)倍。显然,周期信号的频谱是离散的。然而,由于傅里叶变换反应频谱密度的概念,因此周期函数的傅里叶变换不同于傅里叶级数,这里不是有限值,而是冲激函数,它表明在无穷小的频带范围内(即谐频点)取得了无限大的频谱值。

posted @ 2022-02-09 13:28  Vinson88  阅读(8694)  评论(1编辑  收藏  举报