零、预备知识
\[\def\e{\mathrm{e}} \def\d{\mathrm{d}} \def\g{\mathrm{g}} \def\j{\mathrm{j}} \def\Sa{\mathrm{Sa}} \def\fr{\mathscr{F}} \def\rect{\mathrm{rect}} \def\Fourier{\xrightarrow{\fr}} \def\dlt{\delta} \def\alp{\alpha} \def\dlt{\delta} \def\lmbd{\lambda} \def\omg{\omega} \newcommand{\intR}{\int_{-\infty}^{\infty}} \newcommand{\intT}{\int_{-T/2}^{T/2}} \newcommand{\sumZ}[1]{\sum_{#1=-\infty}^{+\infty}} \newcommand{\intP}[1]{\int_{-#1/2}^{#1/2}}\]常用函数
门函数
\[\g_{\tau}(x) = \rect(x/\tau) = \left\{ \begin{array}{} 1, & |x| \leq \tau/2 \\ 0, & \mathrm{else} \end{array} \right.\]门函数的卷积
\[(\g_{\tau} * \g_{\tau})(x) = \intP{\tau} \g_{\tau}(t) \g_{\tau}(x-t) =\left\{ \begin{array}{ll} \tau - |x|,& |x| \leq \tau \\ 0,& \mathrm{else} \end{array} \right.\]阶跃函数
\[u(x) = \left\{ \begin{array}{} 1, & x \geq 0 \\ 0, & x < 0 \end{array} \right.\]单位冲激函数
单位冲击函数$\dlt(x)$满足
\[\left\{ \begin{array}{ll} \dlt(x)=0 & x \neq 0 \\ \intR \dlt(x) \d x = 1 \\ \end{array} \right.\]$\dlt(x)$可以看作$u(x)$的导数。
放缩性
\[\begin{aligned} \intR \dlt(kx) \d x &= \frac{1}{|k|}\\ \dlt(k x) &= \frac{1}{|k|} \dlt(x) \\ \end{aligned}\]因此可知
\[\dlt(\omg) = \frac{1}{2\pi} \dlt(f)\]挑选性
\[\begin{aligned} f(x) \dlt(x - x_0) &= f(x_0) \dlt(x-x_0) \\ f(t) &= \intR f(x) \dlt(x - t) \d x \\ &= \intR f(x) \dlt(t - x) \d x \\ f(x - x_c) &= \intR f(t) \dlt(x - x_c - t) \d t \\ &= f(x) \ast \dlt(x-x_c) \end{aligned}\]周期脉冲
\[\dlt_{T}(t) = \sumZ{n} \dlt(t-nT)\]抽样函数
\[\Sa(x) = \frac{\sin x}{x}\]以下性质的推导使用了傅里叶变换，详见后文
\[\intR \Sa(t) \d t = \fr[\Sa(t)](0) = \pi \g_2(0) = \pi\] \[\intR \Sa^2(t) \d t = \fr[\Sa^2(t)](0) = \frac{ \pi}{2} (\g_2 * \g_2)(0) = \pi\]由以上性质可知
\[\begin{aligned} \dlt(x) &= \lim_{t\rightarrow \infty} \frac{t}{\pi} \Sa(tx) \end{aligned}\] \[\begin{aligned} \dlt(x) &= \lim_{t\rightarrow \infty} \frac{t}{\pi} \Sa^2(tx) \end{aligned}\]傅里叶级数
周期为$T= 1/f_0 = 2\pi/\omg_0$的信号$s(t)$可以表示为基波及其各次谐波之和
\[\begin{aligned} s(t) &= A_0 + \sum_{n=1}^{\infty} A_n \cos(n\omg_0 t + \theta_n) \\ &= \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(n\omg_0 t) + b_n \sin(n\omg_0 t) \right) \\ &= C_0 + \frac{1}{2}\sum_{n=1}^{\infty} \left( a_n (\e^{\j n \omg_0 t} + \e^{-\j n \omg_0 t}) - \j b_n (\e^{\j n \omg_0 t} - \e^{-\j n \omg_0 t}) \right) \\ &= C_0 + \frac{1}{2}\sum_{n=1}^{\infty} \left( (a_n - \j b_n) \e^{\j n \omg_0 t} + (a_n + \j b_n)\e^{-\j n \omg_0 t} \right)\\ &= \sumZ{n} C_n \e^{\j n \omg_0 t} \\ A_n &= \sqrt{a^2_n + b^2_n} \\ \theta_n &= - \arctan(\frac{b_n}{a_n}) \\ C_n &= \left\{ \begin{array}{l} (a_n - \j b_n)/2, & n=+1, +2, \dots \\ (a_n + \j b_n)/2, & n=-1, -2, \dots \\ a_0/2, & n=0 \end{array} \right. \end{aligned}\]其中$C_n$称为周期信号$s(t)$的频谱。
若$s(t)$是偶信号，则$b_n = 0$，$C_n$是实函数。
由于三角函数的正交性
\[\begin{aligned} & \intT \cos(n\omg_0 t) \cos(m \omg_0 t) \d t \\ &= \frac{1}{2} \intT (\cos((n+m)\omg_0 t) + \cos((n-m) \omg_0 t)) \d t \\ &= \frac{1}{2} \intT \cos((n-m) \omg_0 t) \d t \\ &= \left\{ \begin{array}{} \frac{T}{2}, & n = m \\ 0, & n \neq m \end{array} \right. \end{aligned}\] \[\begin{aligned} & \intT \cos(n\omg_0 t) \sin(m \omg_0 t) \d t \\ &= \frac{1}{2} \intT (\sin((n+m)\omg_0 t) - \sin((n-m) \omg_0 t)) \d t \\ &= 0 \end{aligned}\] \[\begin{aligned} & \intT \sin(n\omg_0 t) \sin(m \omg_0 t) \d t \\ &= \frac{1}{2} \intT (\cos((n-m)\omg_0 t) - \cos((n+m) \omg_0 t)) \d t \\ &= \frac{1}{2} \intT \cos((n-m) \omg_0 t)) \d t \\ &= \left\{ \begin{array}{} \frac{T}{2}, & n = m \\ 0, & n \neq m \end{array} \right. \end{aligned}\]可以推出傅里叶系数
\[\begin{aligned} & \frac{2}{T} \intT s(t) \cos(n \omg_0 t) \d t \\ &= \frac{2 a_n}{T} \intT \cos^2(n\omg_0 t) \d t \\ &= \frac{a_n}{T} \intT (1 + \cos(2n\omg_0 t)) \d t \\ &= a_n \\ \end{aligned}\] \[\begin{aligned} & \frac{2}{T} \intT s(t) \sin(n \omg_0 t) \d t \\ &= \frac{2 b_n}{T} \intT \sin^2(n\omg_0 t) \d t \\ &= \frac{b_n}{T} \intT (1 - \cos(2n\omg_0 t)) \d t \\ &= b_n \\ \end{aligned}\] \[\begin{aligned} & \frac{1}{T} \intT s(t) \e^{-\j n\omg_0 t} \d t \\ &= \frac{1}{T} \intT s(t) \left( \cos(n\omg_0 t) - \j \sin(n\omg_0 t) \right) \d t \\ &= C_n \\ \end{aligned}\]若$s(t)$为偶函数，则有
\[C_n = \frac{2}{T} \int_{0}^{T/2} s(t)\cos(n\omg_0 t) \d t\]傅里叶变换
当$s(t)$为任意信号时，需要使用傅里叶变换，$s(t)$的频谱密度为
\[S(\omg) = \intR s(t) \e^{-\j \omg t} \d t\]若$s(t)$为偶函数，则有
\[S(\omg) = 2\int_{0}^{+\infty} s(t) \cos(\omg t) \d t\]$S(\omg)$是$s(t)$的傅里叶变换。考虑周期信号$s(t)$的周期$T\rightarrow \infty$，有
\[\begin{aligned} s(t) &= \lim_{T\rightarrow \infty} \sumZ{n} C_n \e^{\j n \omg_0 t} \\ &= \lim_{T\rightarrow \infty} \frac{1}{T} \sumZ{n} S(n \omg_0) \e^{\j n \omg_0 t} \\ &= \frac{1}{2\pi} \lim_{\omg_0\rightarrow 0} \sumZ{n} S(n \omg_0) \e^{\j n \omg_0 t} \omg_0 \\ &= \frac{1}{2\pi} \intR S(\omg) \e^{\j \omg t} \d \omg \end{aligned}\]$s(t)$为$S(\omg)$的傅里叶逆变换。
周期信号的频谱密度与频谱的关系
\[\begin{aligned} s(t) &= \sumZ{n} C_n \e^{\j n \omg_0 t} \\ &= \sumZ{n} C_n \intR \e^{\j \omg t} \dlt(n\omg_0 - \omg) \d \omg \\ &= \intR \e^{\j \omg t} \sumZ{n} C_n \dlt(n\omg_0 - \omg) \d \omg \\ &= \frac{1}{2\pi} \intR S(\omg) \e^{\j \omg t} \d \omg \\ S(\omg) &= 2\pi \sumZ{n} C_n \dlt(\omg - n\omg_0) \end{aligned}\]线性性
\[\begin{aligned} \fr[\alp f(t) + \beta g(t)] &= \alp F(\omg) + \beta G(\omg) \\ \fr^{-1}[\alp F(\omg) + \beta G(\omg)] &= \alp f(t) + \beta g(t) \\ \end{aligned}\]位移性
由时域卷积定理和$\fr[\dlt(t+\tau)]=\e^{\j \omg \tau}$（详见后文）可得
\[\begin{aligned} \fr[s(t + \tau)] &= \fr[s(t) \ast \dlt(t + \tau)] \\ &= S(\omg) \e^{\j \omg \tau} \end{aligned}\]频移性
由$\fr[1]=2\pi \dlt(\omg)$（详见后文）可得
\[\begin{aligned} \fr[\e^{\j \omg_0 t}] &= \intR \e^{-\j (\omg - \omg_0) t} \d t \\ &= 2\pi \dlt(\omg - \omg_0) \end{aligned}\]放缩性
\[\begin{aligned} \fr[s(\lmbd t)] &= \intR s(\lmbd t) \e^{-\j \omg t} \d t\\ &= \frac{1}{|\lmbd|} \intR s(\lmbd t) \e^{-\j \frac{\omg}{\lmbd} \lmbd t} \d (\lmbd t)\\ &= \frac{1}{|\lmbd|}S(\frac{\omg}{\lmbd}) \end{aligned}\]对称性
设$\fr[f(t)]=F(\omg)$，由$\fr[\e^{\j \omg_0 t}]=2\pi\dlt(\omg-\omg_0)$，$\fr[2\pi\dlt(t-\tau)]=2\pi \e^{-\j \omg \tau}$可知
\[\fr[F(t)] = 2\pi f(-\omg)\]微分关系
由$\frac{\d}{\d t}\e^{\j \omg t} = \j \omg \e^{\j \omg t}$可知
\[\begin{aligned} \fr \left[ \frac{\d^n f(t)}{\d t^n} \right] &= (\j \omg)^n F(\omg) \\ \fr^{-1} \left[\frac{\d^n F(\omg)}{\d \omg^n} \right] &= (-\j t)^n f(t) \\ \end{aligned}\]积分关系
由微分关系可知，设$g’(t)=f(t)$，有
\[G(\omg) = \frac{1}{\j \omg} F(\omg)\]上式不考虑$\omg=0$的情况，设$g(t)=\int f(t) \d t + C$，则有
\[G(\omg) = \frac{1}{\j \omg} F(\omg) + 2\pi C \dlt(\omg)\]频域卷积定理
\[\begin{aligned} \fr[s(t) h(t)] &= \intR s(t)h(t) \e^{-\j \omg t} \d t\\ &= \frac{1}{2\pi} \intR s(t) \left( \intR H(\alp)\e^{\j \alp t} \d \alp \right) \e^{-\j \omg t} \d t \\ &= \frac{1}{2\pi} \intR H(\alp) \left( \intR s(t) \e^{-\j (\omg - \alp) t} \d t \right) \d \alp \\ &= \frac{1}{2\pi} \intR H(\alp) S(\omg-\alp) \d \alp \\ &= \frac{1}{2\pi} \left(S(\omg) \ast H(\omg)\right) \\ \end{aligned}\]时域卷积定理
\[\begin{aligned} \fr^{-1}[S(\omg) H(\omg)] &= \frac{1}{2\pi} \intR S(\omg) H(\omg) \e^{\j \omg t} \d \omg\\ &= \frac{1}{2\pi} \intR S(\omg) \left( \intR h(x) \e^{-\j \omg x} \d x \right) \e^{\j \omg t} \d \omg \\ &= \frac{1}{2\pi} \intR h(x) \left( \intR S(\omg) \e^{\j \omg (t-x)} \d \omg \right) \d x \\ &= \intR h(x) s(t-x) \d x \\ &= s(t) \ast h(t) \\ \end{aligned}\]常用傅里叶变换
\[\begin{aligned} 1 & \Fourier 2\pi\dlt(\omg) \\ \dlt(t) & \Fourier 1 \\ \e^{\j \omg_0 t} & \Fourier 2 \pi \dlt(\omg - \omg_0) \\ \dlt(t - t_0) & \Fourier \e^{-\j \omg t_0}\\ \g_{\tau}(t) & \Fourier \tau\Sa(\omg \tau /2) \\ \omg_H \Sa(\omg_H t/2) & \Fourier 2\pi g_{\omg_H}(\omg) \\ \cos(\omg_0 t) & \Fourier \pi [\dlt(\omg - \omg_0) + \dlt(\omg + \omg_0)] \\ \sin(\omg_0 t) & \Fourier \frac{\pi}{\j} [\dlt(\omg - \omg_0) - \dlt(\omg + \omg_0)] \\ s(t + \tau) & \Fourier S(\omg) \e^{\j \omg \tau}\\ s(\lmbd t) & \Fourier \frac{1}{|\lmbd|}S(\frac{\omg}{\lmbd}) \\ s(t)\cos(\omg_0 t) & \Fourier \frac{1}{2} [S(\omg - \omg_0) + S(\omg + \omg_0)] \\ s(t) h(t) & \Fourier \frac{1}{2\pi} S(\omg) \ast H(\omg) \\ s(t) \ast h(t) & \Fourier S(\omg) H(\omg) \\ \dlt_T(t)=\sumZ{n} \dlt(t-nT) & \Fourier \omg_0 \sumZ{n} \dlt(\omg - n \omg_0) \\ \e^{-\lmbd |t|} &\Fourier \frac{2\lmbd}{\lmbd^2 + \omg^2} \\ u(t) &\Fourier \pi \dlt(\omg) + \frac{1}{\j \omg} \\ u(t)\e^{-\alp t} &\Fourier \frac{1}{\alp + \j \omg} \end{aligned}\]门函数
\[\begin{aligned} \fr[\g_\tau(t)] &= \intP{\tau} \e^{-\j \omg t} \d t \\ &= \frac{\e^{-\j \omg \tau / 2} - \e^{\j \omg \tau / 2}}{-\j \omg} \\ &= \frac{2 \sin(\omg \tau / 2)}{\omg} \\ &= \frac{\tau \sin(\omg \tau / 2)}{\omg \tau / 2} \\ &= \tau \Sa(\omg \tau / 2) \end{aligned}\]常函数
\[\begin{aligned} \fr[1] &= \intR \e^{-\j \omg t} \d t \\ &= \lim_{\tau \rightarrow \infty} \tau \Sa(\omg \tau / 2) \\ &= 2\pi \lim_{\tau/2 \rightarrow \infty} \frac{\tau/2}{\pi} \Sa(\omg \tau / 2) \\ &= 2\pi \dlt(\omg) \end{aligned}\]抽样函数
\[\begin{aligned} \fr[\omg_H \Sa(\omg_H t/2)] &= \intR \omg_H \Sa(\omg_H t/2) \e^{-\j \omg t} \d t \\ &= \intR \left( \int_{-\omg_H /2}^{\omg_H /2} \e^{-\j tx} \d x \right) \e^{-\j \omg t} \d t \\ &= \int_{-\omg_H /2}^{\omg_H /2} \left( \intR \e^{-\j (\omg + x) t} \d t \right) \d x \\ &= 2\pi \int_{-\omg_H /2}^{\omg_H /2} \dlt(\omg + x) \d x \\ &= 2\pi g_{\omg_H}(\omg) \end{aligned}\]正弦波
\[\begin{aligned} \fr[\cos(\omg_0 t)] &= \frac{1}{2} \fr[\e^{\j \omg_0 t} + \e^{-\j \omg_0 t}] \\ &= \pi [\dlt(\omg - \omg_0) + \dlt(\omg + \omg_0)] \\ \fr[s(t)\cos(\omg_0 t)] &= \frac{1}{2\pi} S(\omg) \ast \pi [\dlt(\omg - \omg_0) + \dlt(\omg + \omg_0)] \\ &= \frac{1}{2} [S(\omg - \omg_0) + S(\omg + \omg_0)] \end{aligned}\]周期脉冲
\[\begin{aligned} \dlt_T(\omg) &= \fr[\dlt_T(t)]\\ C_n &= \frac{1}{T} \intT \dlt(t) \e^{-\j n \omg_0 t} \d t \\ &= \frac{1}{T} \\ \end{aligned}\]根据周期信号的频谱密度与频谱的关系
\[\begin{aligned} \dlt_T(\omg) &= 2\pi \sumZ{n} \frac{1}{T} \dlt(\omg - n \omg_0) \\ &= \omg_0 \sumZ{n} \dlt(\omg - n \omg_0) \end{aligned}\]冲击函数
\[\begin{aligned} \fr[u(t)]&= \frac{1}{\j \omg}\fr[\dlt(t)] + 2\pi C \dlt(\omg)\\ &= \frac{1}{\j \omg} + 2\pi C \dlt(\omg) \\ u(t) &= \fr^{-1}[\frac{1}{\j \omg}] + C \\ C &= u(1) - \frac{1}{2\pi} \intR \frac{1}{\j \omg} \e^{\j \omg} \d \omg \\ &= 1 - \frac{1}{2\pi} \intR \frac{\sin( \omg)}{ \omg} \d \omg \\ &= \frac{1}{2} \\ \fr[u(t)] &= \frac{1}{\j \omg} + \pi \dlt(\omg) \end{aligned}\]其他
\[\begin{aligned} \fr[\e^{-\lmbd |t|}] &= \intR \e^{-\lmbd |t|} \e^{-\j \omg t} \d t \\ &= 2\int_0^{+\infty} \e^{-\lmbd t} \cos(\omg t) \d t\\ &= \frac{2}{\lmbd} - \frac{2\omg^2}{\lmbd^2} \int_0^{+\infty} \e^{-\lmbd t} \cos(\omg t) \d t\\ &= \frac{2\lmbd}{\lmbd^2 + \omg^2} \end{aligned}\]\[\begin{aligned} s(t) &= u(t)\e^{-\alp t}\\ s'(t) &= \dlt(t) - \alp s(t)\\ \fr[s'(t)] &= \fr[\dlt(t) - \alp s(t)] \\ \j \omg S(\omg) &= 1 - \alp S(\omg)\\ S(\omg) &= \frac{1}{\alp + \j \omg} + 2\pi C \dlt(\omg) \\ \end{aligned}\]
由于
\[\int_{0}^{+\infty} \e^{-\alp t} \d t = \frac{1}{\alp} = S(0)\]可知
\[S(\omg) = \frac{1}{\alp + \j \omg}\]\[\begin{aligned} s(t) &= u(t)t\e^{-\alp t}\\ s'(t) &= u(t)\e^{-\alp t} - \alp s(t)\\ \fr[s'(t)] &= \fr[u(t)\e^{-\alp t}] - \alp\fr[s(t)] \\ \j \omg S(\omg) &= \frac{1}{\alp + \j \omg} - \alp S(\omg)\\ S(\omg) &= \frac{1}{(\alp + \j \omg)^2} \end{aligned}\]
由于
\[\int_{0}^{+\infty} t\e^{-\alp t} \d t = \frac{1}{\alp^2} = S(0)\]可知
\[S(\omg) = \frac{1}{(\alp + \j \omg)^2}\]