第三章典型例题
[对称性]
例一
已知\(\text{sgn}(t)\leftrightarrow \frac{2}{j\omega}\),求\(\mathscr{F}(\frac{1}{t})\).
解:由对称性可知
\[\frac{2}{jt}\leftrightarrow 2\pi\text{sgn}(-\omega) \]
可得
\[\frac{1}{t}\leftrightarrow -j\pi\text{sgn}(\omega) \]
例二
已知\(EG_\tau(t)\leftrightarrow E\tau\text{Sa}(\frac{\omega\tau}{2})\),求\(\mathscr{F}(\text{Sa}(\omega_0t))\).
解:由对称性可知
\[\text{Sa}(\omega_0t)\leftrightarrow \frac{\pi}{\omega_0}G_{2\omega_0}(\omega) \]
[时移特性]
例三
图为三矩形脉冲信号,求其傅氏变换.
解:相当于矩形脉冲信号进行了两次时移后的叠加,可得
\[\begin{align} F(\omega)&=E\tau\text{Sa}(\frac{\omega\tau}{2})(1+e^{j\omega T}+e^{-j\omega T})\notag\\ &=E\tau\text{Sa}(\frac{\omega\tau}{2})[1+2\cos(\omega T)]\notag \end{align} \]
[频移特性]
例四
图为矩形调幅信号的波形,表达式为\(f(t)=EG_\tau(t)\cos(\omega_0t)\),求其傅氏变换.
解:由欧拉公式,可得
\[f(t)=\frac{1}{2}EG_\tau(t)(e^{j\omega_0t}+e^{-j\omega_0t}) \]
根据频移特性,时域中信号乘以\(e^{j\omega_0t}\)在频域表现为进行频移,得
\[F(\omega)=\frac{E\tau}{2}\{\text{Sa}[\frac{(\omega-\omega_0)\tau}{2}]+\text{Sa}[\frac{(\omega+\omega_0)\tau}{2}]\} \]
[频域微分特性]
例五
已知\(f(t)\leftrightarrow F(\omega)\),求\(\mathscr{F}[(t-2)f(t)]\).
解:
\[\begin{align}\mathscr{F}[(t-2)f(t)]&=\mathscr{F}[tf(t)-2f(t)]\notag\\&=jF^{'}(\omega)-2F(\omega)\notag\end{align} \]
例六
求\(\mathscr{F}(t^n)\).
解:
\[t^n=t^n\cdot1\leftrightarrow j^n\frac{d^n(2\pi\delta(\omega))}{d\omega^n} \]
[时域积分特性]
例七
求门函数积分的傅氏变换.
解:当\(\omega=0\)时,\(F(0)\ne0\),可得
\[\int_{-\infty}^{\tau}G_\tau(\tau)d\tau\leftrightarrow\pi\tau\delta(\omega)+\frac{\tau}{j\omega}\text{Sa}(\frac{\omega\tau}{2}) \]
[综合]
例八
据图写出\(f(t)\).
解:由图可知
\[F(\omega)=AG_{2\omega_0}(\omega)e^{j\omega t_0}\leftrightarrow \frac{A\omega_0}{\pi}\text{Sa}[\omega_0(t-t_0)] \]
例九
已知\(f(t)=\text{Sa}(t)=\frac{\sin t}{t}\),求其带宽(\(Hz\)).
解:首先运用对称性进行傅氏变换,即为
\[\text{Sa}(t)\leftrightarrow \pi G_2(\omega) \]
频谱是一个门宽为\(2\)的门函数,则其带宽为\(1(Rad/s)\).因题中所给单位是\(Hz\),通过
\[\omega=2\pi f \]
可得带宽为
\[f=\frac{1}{2\pi} \]
单位为\(Hz\).
[卷积定理]
例十
求升余弦脉冲的傅氏变换.
解:升余弦脉冲的表达式为
\[f(t)=(1+\cos t)G_{2\pi}(t) \]
根据频域卷积定理可得
\[\begin{align}F(\omega)&=\frac{1}{2\pi}[2\pi\delta(\omega)+\pi\delta(w-1)+\pi\delta(\omega+1)]*2\pi\text{Sa}(\pi\omega)\notag\\&=-\frac{2\sin\pi\omega}{\omega(\omega^2-1)}\notag\end{align} \]
[周期函数傅氏变换]
例十一
求如图周期矩形脉冲序列的傅氏变换.
解:先看单个脉冲的傅氏变换
\[F_0(\omega)=E\tau\text{Sa}(\frac{\omega\tau}{2}) \]
再根据脉冲变周期的傅氏变换关系得
\[F(n\omega_1)=\frac{1}{T_1}F_0(\omega)|_{\omega=n\omega_1} \]
所以可得该周期函数的傅氏变换为
\[\begin{align} F(\omega)&=2\pi\sum_{n=-\infty}^{\infty}\frac{E\tau}{T_1}\text{Sa}(\frac{n\omega_1\tau}{2})\delta(\omega-n\omega_1)\notag\\ &=E\tau\omega_1\sum_{n=-\infty}^{\infty}\text{Sa}(\frac{n\omega_1\tau}{2})\delta(\omega-n\omega_1)\notag \end{align} \]
也可以利用时域卷积定理求解,将信号写为
\[f(t)=f_0(t)*\delta_T(t) \]
根据时域卷积定理,得
\[F(\omega)=F_0(\omega)\omega_1\sum_{n=-\infty}^{\infty}\delta(\omega-n\omega_1) \]
利用冲激信号的抽样性质
\[F(\omega)=\omega_1\sum_{n=-\infty}^{\infty}F_0(n\omega_1)\delta(\omega-n\omega_1) \]
其中
\[F_0(n\omega_1)=E\tau\text{Sa}(\frac{n\omega_1\tau}{2}) \]
带入上式,得
\[\begin{align} F(\omega)&=\omega_1\sum_{n=-\infty}^{\infty}E\tau\text{Sa}(\frac{n\omega_1\tau}{2})\delta(\omega-n\omega_1)\notag\\ &=E\tau\omega_1\sum_{n=-\infty}^{\infty}\text{Sa}(\frac{n\omega_1\tau}{2})\delta(\omega-n\omega_1)\notag \end{align} \]
结果和之前的相同.
