1. 单边指数函数
\[x(t)=\left \{
\begin{aligned}
& e^{-at}, & t \ge 0\\
& 0 , & t < 0
\end{aligned}
\right.
\]
其中\(a\)是实数。于是其傅里叶变换为:
\[\begin{aligned}
X(f) &= \int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}dt \\
&= \int_{0}^{\infty}e^{-at}e^{-j2\pi ft}dt \\
&= \int_{0}^{\infty}e^{-(a+j2\pi f)t}dt \\
&= [-\frac{e^{-(a+j2\pi f)t}}{a+j2\pi f}]_{0}^{\infty} \\
&= 0 - (-\frac{1}{a+j2\pi f}) = \frac{1}{a+j2\pi f}
\end{aligned}
\]
从而可以得到:
\[\left \{
\begin{aligned}
X(f) &= \frac{1}{a+j2\pi f} \\
|X(f)| &=\frac{1}{\sqrt{a^{2}+(2\pi f)^{2}}}\\
\varphi(f) &= -\text{arctan}(\frac{2\pi f}{a})
\end{aligned}
\right. \tag{1}
\]
2. 双边指数函数
\[x(t)=e^{-a|t|}\quad (-\infty<t<\infty)
\]
傅里叶变换为:
\[\begin{aligned}
X(f) &= \int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}dt = \int_{-\infty}^{\infty}e^{-a|t|}e^{-j2\pi ft}dt \\
&= \int_{-\infty}^{0}e^{at}e^{-j2 \pi ft}dt+\int_{0}^{\infty}e^{-at}e^{-j2\pi ft}dt \\
&= \int_{-\infty}^{0}e^{(a-j2\pi f)t}dt+\int_{0}^{\infty}e^{-(a+j2\pi f)t}dt \\
&= [\frac{e^{(a-j2\pi f)t}}{a-j2\pi f}]_{-\infty}^{0}+\frac{1}{a+j2\pi f} \\
&= \frac{1}{a-j2\pi f} + \frac{1}{a+j2\pi f} = \frac{2a}{a^{2}+(2\pi f)^{2}}
\end{aligned}
\]
从而可以得到:
\[\left \{
\begin{aligned}
X(f) &= \frac{1}{a^{2}+(2\pi f)^{2}} \\
|X(f)| &=\frac{1}{a^{2}+(2\pi f)^{2}}\\
\varphi(2\pi f) &= 0
\end{aligned}
\right. \tag{2}
\]
3. 矩形脉冲信号
矩形脉冲信号的表达式为:
\[x(t)=E[u(t+\frac{\tau}{2})-u(t-\frac{\tau}{2})]
\]
其中\(E\)为脉冲幅度,\(\tau\)为脉冲宽度。其傅里叶变换为:
\[\begin{aligned}
X(f) &= \int_{-\infty}^{\infty}x(t)e^{j2\pi ft}=\int_{-\frac{\tau}{2}}^{\frac{\tau}{2}}Ee^{-j2\pi ft}=[-\frac{Ee^{-j2\pi ft}}{j2\pi f}]_{-\frac{\tau}{2}}^{\frac{\tau}{2}}\\
&= -\frac{E}{j2\pi f}[e^{-j2\pi f \frac{\tau}{2}}-e^{j2\pi f\frac{\tau}{2}}] =-\frac{E}{j2\pi f}[(\text{cos}\pi f\tau-j\text{sin}\pi f\tau)-(\text{cos}\pi f\tau+j\text{sin}\pi f\tau)] \\
&= -\frac{E}{j2\pi f}[-2j\text{sin}\pi f\tau]=E\cdot\frac{sin(\pi f\tau)}{\pi f}
\end{aligned}
\]
定义抽样信号\(Sa(t)=\frac{\text{sin}t}{t}\),所以上式可以写为:
\[X(f)=E\tau[\frac{\text{sin}(\pi f\tau)}{\pi f\tau}]=E\tau Sa(\pi f\tau)
\]
从而得到
\[|X(f)| = E\tau \left|Sa(\pi f\tau)\right| \tag{3}
\]
矩形脉冲信号的频谱以\(Sa(\pi f\tau)\)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:
\[x(t) = \text{sgn}(t)= \left \{
\begin{aligned}
&+1\quad &(t>0) \\
&0 \quad &(t=0) \\
&-1 \quad &(t<0)
\end{aligned}
\right.
\]
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:
\[\begin{aligned}
X(f)&=\int_{-\infty}^{\infty}sng(t)e^{-j2\pi f t}=\int_{-\infty}^{0_{-}}-e^{-j2\pi f t}dt+\int_{0_{+}}^{\infty}e^{-j2\pi f t}dt \\
&=\frac{1}{j2\pi f}[\lim_{t\to0_{-}}e^{-j2\pi f t}-\lim_{t\to-\infty}e^{-j2\pi f t}]-\frac{1}{j2\pi f}[\lim_{t\to\infty}e^{-j2\pi f t}-\lim_{t\to0_{+}}e^{-j2\pi f t}]
\end{aligned}
\]
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\(x_{1}(t)\)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:
\[\begin{aligned}
X_{1}(f)&=\int_{-\infty}^{\infty}f_{1}(t)e^{-j2\pi f t}dt=\int_{-\infty}^{0}(-e^{at})e^{-j2\pi f t}dt + \int_{0}^{\infty}e^{-at}e^{-j2\pi f t}dt \\
&=\int_{-\infty}^{0}(-e^{(a-j2\pi f)t})dt + \int_{0}^{\infty}e^{-at}e^{-j2\pi f t}dt \\
&=-[\frac{e^{(a-j2\pi f)t}}{a-j2\pi f}]_{-\infty}^{0} + \frac{1}{a+j2\pi f} \\
&= -[\frac{1}{a-j2\pi f}-0]+\frac{1}{a+j2\pi f}=\frac{-j2(2\pi f)}{a^{2}+(2\pi f)^{2}}
\end{aligned}
\]
从而符号函数\(sgn(t)\)的频谱函数为:
\[\left \{
\begin{aligned}
X(f)&=\lim_{a\to0}\frac{-j2(2\pi f)}{a^{2}+(2\pi f)^{2}}=\frac{1}{j\pi f} \\
\left| F(f) \right| &= \frac{1}{|\pi f|} \\
\varphi(f)&=\left \{ \begin{aligned} &-\frac{\pi}{2}\quad (\pi f >0) \\ &+\frac{\pi}{2}\quad(\pi f<0) \end{aligned}\right.
\end{aligned}
\right. \tag{4}
\]
5. 升余弦函数
升余弦脉冲函数的表达式为:
\[x(t)=\frac{E}{2}[1+cos(\frac{\pi t}{\tau})]\quad(0 \le |t| \le \tau)
\]
其傅里叶变换为:
\[\begin{aligned}
X(f)&=\int_{-\tau}^{\tau}\frac{E}{2}[1+cos(\frac{\pi t}{\tau})]e^{-j2\pi f t}dt \\
&= \frac{E}{2}\int_{-\tau}^{\tau}e^{-j2\pi f t}dt+\frac{E}{2}[\int_{-\tau}^{\tau}\frac{1}{2}(e^{\frac{j\pi t}{\tau}}+e^{-\frac{j\pi t}{\tau}})e^{-j2\pi f t}dt] \\
&=E\tau Sa(2\pi f \tau)+\frac{E\tau}{2}Sa[(2\pi f-\frac{\pi}{\tau})\tau]+\frac{E\tau}{2}Sa[(2\pi f+\frac{\pi}{\tau})\tau]
\end{aligned} \tag{5a}
\]
显然\(X(f)\)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\(2\pi f=\frac{\pi}{\tau}\)。把上式化简可以得到:
\[X(f)=\frac{Esin(2\pi f\tau)}{2\pi f[1-(\frac{2\pi f\tau}{\pi})^{2}]}=\frac{E\tau Sa(2\pi f\tau)}{1-(\frac{2\pi f\tau}{\pi})^{2}} \tag{5b}
\]
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\(2\pi f=0\sim\frac{2\pi}{\tau}\rightarrow f=0\sim\frac{1}{\tau}\)的范围内。
6. 冲激函数
单位冲激函数\(\delta(t)\)的傅里叶变换为:
\[X(f)=\int_{-\infty}^{\infty}\delta(t)e^{-j2\pi f t}dt=1 \tag{6a}
\]
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\(\delta(2\pi f)\)的傅里叶逆变换。
\[\mathscr{F}^{-1}[\delta(f)]=\int_{-\infty}^{+\infty}\delta(f)e^{j2\pi f t}df = 1 \tag{6b}
\]
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:
\[\mathscr{F}(1)=\delta(f) \tag{6c}
\]
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:
\[\begin{aligned}
\mathscr{F}[u(t)]&=\mathscr{F}(\frac{1}{2})+\frac{1}{2}\mathscr{F}[sng(t)]\\
&= \frac{1}{2}\delta(f)+\frac{1}{j2\pi f}
\end{aligned} \tag{7}
\]
8. 正弦、余弦信号的傅里叶变换
若\(\mathscr{F}[x_{0}(t)]=X_{0}(f)\),由频移特性可知
\[\mathscr{F}[f_{0}(t)e^{j2\pi f_{1}t}]=F_{0}(f-f_{1})
\]
上式中,令\(f_{0}(t)=1\),其傅里叶变换为\(\mathscr{F}[1]=\delta(f)\),那么上式变为:
\[\mathscr{F}[e^{j2\pi f_{1}t}]=\delta(f-f_{1})
\]
同理:
\[\mathscr{F}[e^{-j2\pi f_{1}t}]=\delta(f+f_{1})
\]
由两式并结合欧拉公式,可以得到:
\[\left \{
\begin{aligned}
\mathscr{F}[cos(2\pi f_{1}t)]&=\frac{1}{2}[\delta(f+f_{1})+\delta(f-f_{1})] \\
\mathscr{F}[sin(2\pi f_{1}t)]&=\frac{j}{2}[\delta(f+f_{1})-\delta(f-f_{1})]
\end{aligned} \tag{8}
\right.
\]
9. 一般周期信号的傅里叶变换
令周期信号\(x(t)\)的周期为\(T_{1}\),角频率为\(2\pi f_{1}(=2\pi f_{1}=\frac{2\pi}{T_{1}})\),可以将\(x(t)\)展成傅里叶级数,为:
\[x(t)=\sum_{n=-\infty}^{\infty}X_{n}e^{jn2\pi f_{1}t}
\]
将上式两边取傅里叶变换
\[\begin{aligned}
\mathscr{F}[x(t)]&=\mathscr{F}[\sum_{n=-\infty}^{\infty}X_{n}e^{jn2\pi f_{1}t}] \\
&= \sum_{n=-\infty}^{\infty}X_{n}\mathscr{F}[e^{jn2\pi f_{1}t}] \\
&= \sum_{n=-\infty}^{\infty}X_{n}\delta(f-nf_{1})
\end{aligned} \tag{9} \]
其中\(F_{n}\)是\(x(t)\)的傅里叶级数的系数,已经知道它等于
\[X_{n}=\frac{1}{T_{1}}\int_{-\frac{T_{1}}{2}}^{\frac{T_{1}}{2}}x(t)e^{-jn2\pi f_{1}t}dt \tag{9a}
\]
式\((9)\)表明,周期信号\(x(t)\)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(\(0,\pm f_{1},\pm f_{2},\cdots\))处,每个冲激的强度等于\(x(t)\)的傅里叶级数相应系数\(X_{n}\)的\(2\pi\)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。