[Complex Function Notes] Fourier Transform and Laplace Transform

1. Definition of Fourier transform and Laplace transform

1. Fourier integral

Fourier Integral Theorem If $f\left(t\right)$ at$(-\infty ,+\infty )$ satisfies the following conditions:
(1)$f\left(t\right)$ satisfies the Dirichlet condition on any finite interval (continuous or only a limited number of discontinuous points of the first type; only a limited number of extreme points);
(2)$f\left(t\right)$ in the infinite interval$(-\infty ,+\infty )$ is absolutely integrable (ie the integral${\int }_{-\infty }^{+\infty }|f(t)|dt$ equation), and if$$f(t)=\frac{1}{2p.m}{\int }_{-\infty }^{+\infty }\left[{\int }_{-\infty }^{+\infty }f\left(\tau \right)e^{-i\omega \tau }d\tau \right]e^{i\omega t}d\omega$$f ( t ) f(t)on the left$f\left(t\right)$ should be f ( t + 0 ) + f ( t − 0 ) 2 \frac{f(t+0)+f(t-0)}{2}at its discontinuity point$\frac{f(t+0)+f(t-0)}{2}$(left and right extreme mean values) instead.

2. Fourier transform

Fourier transform : if $f\left(t\right)$ at$(-\infty ,+\infty )$ satisfy the conditions of Fourier integral, then the function$$F\left(\omega \right)=\mathcal{F}[f(t)]={\int }_{-\infty }^{+\infty }f(t)e^{-i\omega t}dt$$is$The\; Fourier\; transform\; of\; f\left(t\right)$ ; and the function$$f(t)={\mathcal{F}}^{-1}\left[F\right(\omega )]=\frac{1}{2p.m}{\int }_{-\infty }^{+\infty }F\left(\omega \right)e^{i\omega t}d\omega \phi F$$(ω ) F(\omega)Inverse Fourier transform of$F$$($$\omega$$)\; .$

3. Unit impulse function and unit step function

Unit impulse function : $$d\left(t\right)=\{\begin{array}{ll}+\infty ,& t=0\\ 0,& t\ne 0\end{array}$$This is an imprecise statement, and a rigorous statement requires the use of weak limits. This function satisfies $${\int }_{-\infty }^{+\infty }\delta (t)\mathrm{d}t=1$$ If$f\left(t\right)$ is an infinitely differentiable function, then there is$${\int }_{-\infty }^{+\infty }\delta (t)f(t)\mathrm{d}t=f(0)$$$\delta \left(t\right)$ is an even function, satisfying$\delta (at)=\frac{1}{|a|}\delta \left(t\right)$（$a\ne 0$），且$${\int }_{-\infty }^{+\infty }{d}^{(n)}(t-t_0)f(t)\mathrm{d}t={(-1)}^n{f}^{\left(n\right)}\left(t_0\right)$$ (this is proved by integration by parts). In particular,$${\int }_{-\infty }^{+\infty }{d}^{\prime }(t)f(t)\mathrm{d}t=-f^{\prime }(0)$$

Unit step function : $$u(t)=\{\begin{array}{ll}1,& t>0\\ 0,& t<0\end{array}$$成动$$\begin{gathered} {\int }_{-\infty }^{t}d\left(t\right)dt=u\left(t\right) \\ \frac{\mathrm{d}}{\mathrm{d}t}u\left(t\right)=d\left(t\right) \end{gathered}$$

4. Laplace transform

Laplace transform : Let the function $f(t)$在$t\ge 0$ is defined, and the integral${\int }_0^{+\infty }f(t)e^{-st}dt$in the complex planessConvergence in a certain area of$s$$$F(s)=\mathcal{L}[f(t)]={\int }_0^{+\infty }f\left(t\right)e^{-st}dt$$isf ( t ) f(t)The Laplace transform (or image function) of$f$$($$t$$)$$f\left(t\right)$ is$The\; inverse\; Laplace\; transform\; of\; F\left(s\right)$ (or called the original function), recorded as$$f(t)={\mathcal{L}}^{-1}\left[F\right(s)]$$young age$s=b+i\omega$，则$\mathcal{L}[f(t)]=\mathcal{F}[f(t)u(t)e^{-\beta t}]$. This is the relationship between Laplace transform and Fourier transform.

The existence theorem of Laplace transform If the function $f\left(t\right)$ satisfies the following conditions:
(1) At$t\ge 0$ continuous or piecewise continuous on any finite interval;
(2) When$t\to +\infty$ ,$The\; growth\; rate\; of\; f\left(t\right)$ does not exceed a certain exponential function, that is, there exists$M>0$和$c\ge 0$使得$$|f(t)|\le M{e}^{ct},\forall t\in [0,+\infty )$$ thenf ( t ) f(t)Laplace transform of$f$$($$t$$)$$F(s)={\int }_0^{+\infty }f(t)e^{-st}dt$in the semi-plane$\mathrm{Re}(s)>There\; must\; exist\; on\; c$ , and the integral on the right is at$\mathrm{Re}\left(s\right)\ge c_1>$Absolutely convergent and uniformly convergent on$c$$\mathrm{Re}\left(s\right)>In\; the\; half-plane\; of\; c$ ,$F\left(s\right)$ is an analytic function. $c$ is called$The\; growth\; exponent\; of\; f\left(t\right)$ .

2. Fourier transform and Laplace transform of common functions

Like primitive function $f(t)$	Fourier transform	Laplace transform
$1$	$2pd\left(o\right)$	-
$d\left(t\right)$	$1$	$1$
$u(t)$	$\frac{1}{i\omega }+pd\left(o\right)$	$\frac{1}{s}$
$t^n$	$2\pi i^n{d}^{\left(n\right)}\left(o\right)$	$\frac{C(n+1)}{s^{(n+1)}}(n>-1)$ (whennnWhen$n$$\frac{n!}{s^{n+1}}$）
$\cos {oh}_{0}t$	$p\left[d\right(o+{oh}_0)+d(o-{oh}_0)]$	$\frac{s}{s^2+{oh}_0^2}$（requirementsω ${oh}_0\in \mathbb{R}$）
$\sin {oh}_{0}t$	$i\pi \left[d\right(o+{oh}_0)-d(o-{oh}_0)]$	$\frac{{oh}_0}{s^2+{oh}_0^2}$（requirementsω ${oh}_0\in R$）
$e^{-\beta t}u(t)$	$\frac{1}{b+i\omega }(b>0)$	$\frac{1}{s+b}$

3. Properties of Fourier transform and Laplace transform

令$F\left(\omega \right)=\mathcal{F}[f(t)]$，$F(s)=\mathcal{L}[f(t)]$。

nature	Fourier transform	Laplace transform
linear properties	$F[a{f}_1(t)+b{f}_2(t)]=aF\left[f_1\right(t)]+b\mathcal{F}[f_2(t)]$	$L[a{f}_1(t)+b{f}_2(t)]=aL\left[f_1\right(t)]+bL\left[f_2\right(t)]$
Similarity	$\mathcal{F}[f(at)]=\frac{1}{|a|}F\left(\frac{}{a}\right)$	$\mathcal{L}[f(at)]=\frac{1}{a}F\left(\frac{s}{a}\right)$（$a>0$）
Symmetric properties	$F\left[F\right(\omega )]=2\pi f(-t)$
Differential properties	$\mathcal{F}[f^{\prime }(t)]=i\omega \mathcal{F}[f(t)]$ $F[f^{(n)}(t)]={\left(i\omega \right)}^n\mathcal{F}[f(t)]$	$\mathcal{L}[f^{\prime }(t)]=sF(s)-f(0)$ $L\left[f^{\left(n\right)}\right(t)]=s^{nF}\left(s\right)\_-s^{n-1}f(0)-\cdots -f^{(n-1)}(0)$
象函数的微分性质	$\mathcal{F}[tf(t)]=i{F}^{\prime }(\omega )$ $\mathcal{F}[t^{n}f(t)]=i^n{F}^{(n)}(\omega )$	$\mathcal{L}[tf(t)]=-F^{\prime }(s)$ $\mathcal{L}[t^{n}f(t)]={(-1)}^n{F}^{(n)}(s)$
位移性质	$\mathcal{F}[f(t+t_0)]=e^{i\omega t_0}\mathcal{F}[f(t)]$	$\mathcal{L}[f(t+t_0)]=e^{s{t}_0}\mathcal{L}[f(t)]$（要求$t_0<0$）
象函数的位移性质	$\mathcal{F}[e^{i{\omega }_{0}t}f(t)]=F(\omega -{\omega }_0)$	$\mathcal{L}[e^{at}f(t)]=F(s-a)$（要求$\mathrm{Re}(s-a)>s_0$，其中$s_0$是$f(t)$的增长指数）
积分性质	$\mathcal{F}\left[{\int }_{-\infty }^{t}f(\tau )\mathrm{d}\tau \right]=\frac{1}{i\omega }F(\omega )+\pi F(0)\delta (\omega )$	$\mathcal{L}[{\int }_0^{t}f(\tau )\mathrm{d}\tau ]=\frac{1}{s}F(s)$
象函数的积分性质		$\mathcal{L}\left[\frac{f(t)}{t}\right]={\int }_s^{\infty }F(s)\mathrm{d}s$
卷积定理	$\mathcal{F}[f_1(t)\ast f_2(t)]=\mathcal{F}[f_1(t)]\cdot \mathcal{F}[f_2(t)]$ $\mathcal{F}[f_1(t)\cdot f_2(t)]=\frac{1}{2\pi }\mathcal{F}[f_1(t)]\ast \mathcal{F}[f_2(t)]$	$\mathcal{L}[f_1(t)\ast f_2(t)]=\mathcal{L}[f_1(t)]\cdot \mathcal{L}[f_2(t)]$
乘积定理	令$F_1(\omega )=\mathcal{F}[f_1(t)]$，$F_2(\omega )=\mathcal{F}[f_2(t)]$，则 ${\int }_{-\infty }^{+\infty }\overline{f_1(t)}{f}_2(t)\mathrm{d}t=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }\overline{F_1(\omega )}{F}_2(\omega )\mathrm{d}\omega$ ${\int }_{-\infty }^{+\infty }{f}_1(t)\overline{f_2(t)}\mathrm{d}t=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{F}_1(\omega )\overline{F_2(\omega )}\mathrm{d}\omega$
能量积分	帕西瓦尔等式： ${\int }_{-\infty }^{+\infty }{[f(t)]}^2\mathrm{d}t=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{|F(\omega )|}^2\mathrm{d}\omega$
初值定理		$f(0)=\underset{s\to \infty }{\lim }sF(s)$（若这个极限存在）
终值定理		$\underset{t\to +\infty }{\lim }f(t)=\underset{s\to 0}{\lim }sF(s)$ （要求$sF(s)$的所有奇点都在$s$平面的左半部）

四、拉普拉斯逆变换

由于$F(s)=\mathcal{L}[f(t)]=\mathcal{F}[f(t)u(t)e^{-\beta t}]$（$s=\beta +i\omega$），我们有$$\begin{gathered} \frac{1}{2\pi }{\int }_{-\infty }^{+\infty }F(\beta +i\omega )e^{i\omega t}\mathrm{d}\omega =f(t)u(t)e^{-\beta t} \\ f(t)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }F(\beta +i\omega )e^{(\beta +i\omega )t}\mathrm{d}\omega \end{gathered}$$注意到$\mathrm{d}s=i\mathrm{d}\omega$，我们把积分写出复变函数的积分：$$f(t)=\frac{1}{2\pi i}{\int }_{\beta -i\infty }^{\beta +i\infty }F(s)e^{st}\mathrm{d}s$$下面我们通过留数来计算这个积分。

定理 设$s_1,s_2,\cdots ,s_l$是$F(s)$的所有孤立奇点（有限个），除这些点外，$F(s)$处处解析。且存在$R_0>0$，使得当$|s|>R_0$时，$|F(s)|\le M(r)$，其中$M(r)$是$r$的实函数，满足$\underset{r\to +\infty }{\lim }M(r)=0$（即当$s\to \infty$时，$F(s)\to 0$）。选取$\beta$，使所有孤立奇点的实部都小于$\beta$，则当$t>0$时，$$f(t)=\frac{1}{2\pi i}{\int }_{\beta -i\infty }^{\beta +i\infty }F(s)e^{st}\mathrm{d}s=\sum _{k=1}^l\mathrm{Res}[F(s)e^{st},s_k]$$