标签：系统 splay src cos row ref lis int http
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
\[
\frac{1}{t}\leftrightarrow -j\pi\text{sgn}(\omega)
\]
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
\[
\text{Sa}(\omega_0t)\leftrightarrow \frac{\pi}{\omega_0}G_{2\omega_0}(\omega)
\]
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
\[
\begin{align}
F(\omega)&=E\tau\text{Sa}(\frac{\omega\tau}{2})(1+e^{j\omega T}+e^{-j\omega T})\notag\&=E\tau\text{Sa}(\frac{\omega\tau}{2})[1+2\cos(\omega T)]\notag
\end{align}
\]
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
\[
F(\omega)=\frac{E\tau}{2}\{\text{Sa}[\frac{(\omega-\omega_0)\tau}{2}]+\text{Sa}[\frac{(\omega+\omega_0)\tau}{2}]\}
\]
已知\(f(t)\leftrightarrow F(\omega)\),求\(\mathscr{F}[(t-2)f(t)]\).
解:
\[
\begin{align}
\mathscr{F}[(t-2)f(t)]&=\mathscr{F}[tf(t)-2f(t)]\notag\&=jF^{'}(\omega)-2F(\omega)\notag
\end{align}
\]
求\(\mathscr{F}(t^n)\).
解:
\[
t^n=t^n\cdot1\leftrightarrow j^n\frac{d^n(2\pi\delta(\omega))}{d\omega^n}
\]
求门函数积分的傅氏变换.
解:当\(\omega=0\)时,\(F(0)\ne0\),可得
\[
\int_{-\infty}^{\tau}G_\tau(\tau)d\tau\leftrightarrow\pi\tau\delta(\omega)+\frac{\tau}{j\omega}\text{Sa}(\frac{\omega\tau}{2})
\]
据图写出\(f(t)\).
解:由图可知
\[
F(\omega)=AG_{2\omega_0}(\omega)e^{j\omega t_0}\leftrightarrow \frac{A\omega_0}{\pi}\text{Sa}[\omega_0(t-t_0)]
\]
标签：系统 splay src cos row ref lis int http
原文地址：https://www.cnblogs.com/Clouds42/p/11801551.html
