3种积分变换笔记

Fourier 变换

若函数 $f(t)$ 在 $(-\infty, \infty)$ 上满足条件:

则:

F(\omega) = \int_{-\infty}^{+\infty}f(t)e^{-j\omega t}dt

f(t) = \frac 1 {2\pi} \int_{-\infty}^{+\infty}F(\omega)e^{j\omega t}d\omega

其中, $F(\omega)$ 称为 $f(t)$ 的 $Fourier$ 变换, 记为 $F(\omega) = \mathscr{F}[f(t)]$ , 逆变换记为 $f(t) = \mathscr{F}^{-1}[F(\omega)]$ .

\begin{array}{ll} \displaystyle \mathscr{F}[1] = 2\pi\delta(\omega) \\\\ \displaystyle \mathscr{F}[\delta(t)] = 1 \\\\ \displaystyle \mathscr{F}[\cos \omega_0 t] = \pi [\delta(\omega + \omega_0) + \delta(\omega - \omega_0)] \\\\ \displaystyle \mathscr{F}[\sin \omega_0 t] = \pi j [\delta(\omega + \omega_0) - \delta(\omega - \omega_0)] \end{array}

\mathscr{F}[\alpha f_1(t) + \beta f_2(t)] = \alpha F_1(\omega) + \beta F_2(\omega)

\mathscr{F}^{-1}[\alpha F_1(\omega) + \beta F_2(\omega)] = \alpha f_1(t) + \beta f_2(t)

\mathscr{F}[f(t \pm t_0)] = F(\omega) e^{\pm j \omega_0 t}

\mathscr{F}^{-1}[F(\omega \pm \omega_0)] = f(t)e^{\mp j \omega_0 t}

\mathscr{F}[f_1(t) * f_2(t)] = F_1(\omega) \cdot F_2(\omega)

\mathscr{F}[f_1(t) \cdot f_2(t)] = \frac1 {2\pi} F_1(\omega) * F_2(\omega)

\int_{-\infty}^{+\infty} \left[ f(t) \right]^2 dt = \frac 1{2\pi} \int_{-\infty}^{+\infty} \left[ F(\omega) \right]^2 d\omega

设 $f(t)$ 在 $t \geq 0$ 上有定义, 且积分 $\displaystyle \int_0^{+\infty}f(t)e^{-st}dt$ ( $s$ 是复参数) 关于某一范围内的 $s$ 收敛, 则由这个积分确定的函数

F(s) = \int_0^{+\infty}f(t)e^{-st}dt

称为 $f(t)$ 的 $Laplace$ 变换, 记为 $F(s) = \mathscr{L}[f(t)]$ , 逆变换记为 $f(t) = \mathscr{L}^{-1}[F(s)]$ .

\begin{array}{ll} \displaystyle \mathscr{L}[1] = \frac1 {s} \\\\ \displaystyle \mathscr{L}[e^{kt}] = \frac{1}{s-k} \\\\ \displaystyle \mathscr{L}[\cos \omega t] = \frac{s}{s^2 + \omega^2} \\\\ \displaystyle \mathscr{L}[\sin \omega t] = \frac{\omega}{s^2 + \omega^2} \end{array}

\mathscr{L}[\alpha f_1(t) + \beta f_2(t)] = \alpha F_1(s) + \beta F_2(s)

\mathscr{L}^{-1}[\alpha F_1(s) + \beta F_2(s)] = \alpha f_1(t) + \beta f_2(t)

\mathscr{L}[f'(t)] = s F(s) - f(0)

\mathscr{L}[f^{(n)}(t)] = s^n F(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0)

F'(s) = -\mathscr{L}[tf(t)]

F^{(n)}(s) = (-1)^n \mathscr{L}[t^n f(t)]

\mathscr{L}\left [\int_0^t f(t) dt \right] = \frac 1s F(s)

\mathscr{L}\left [\int_0^t dt \int_0^t dt \cdots \int_0^t f(t) dt \right] = \frac 1{s^n} F(s)

\int_s^{+\infty} F(u) du = \mathscr{L}\left [ \frac{f(t)}{t} \right ]

\mathscr{L}\left[ \frac{f(t)}{t^n} \right] = \int_s^{+\infty} ds \int_s^{+\infty} ds \cdots \int_s^{+\infty} F(s) ds

\mathscr{L}[e^{at}f(t)] = F(s-a)

\mathscr{L}[f_1(t) * f_2(t)] = F_1(s) \cdot F_2(s)

\mathscr{L}^{-1}[F_1(s) \cdot F_2(s)] = f_1(t) * f_2(t)

若 $s_1, s_2, \cdots, s_n$ 是函数 $F(s)$ 的所有孤立奇点(有限个), 除这些奇点外处处解析, 则

f(t) = \sum\limits_{k=1}^{n} \text{Res}[F(s)e^{st}, s_k]

即 $f(t)$ 是 $F(s)e^{st}$ 的所有留数之和.

设 $x(n)$ 是无限序列, 则和式

X(z) = \sum\limits_{n=-\infty}^{+\infty} x(n) z^{-n}

是序列 $x(n)$ 的 $Z$ 变换, 记为 $x(z) = Z[x(n)]$ , 其逆变换为

x(n) = \frac{1}{2\pi j}\oint_C X(z) z^{n-1} dz

记为 $x(n) = Z^{-1}[X(z)]$ .

\begin{array}{ll} \displaystyle Z[u(n)] = \frac{z}{z - 1} &, |z| > 1 \\\\ \displaystyle Z[\alpha^n u(n)] = \frac{z}{z-\alpha} &, |z| > |\alpha| \\\\ \displaystyle Z[n u(n)] = \frac{z}{(z-1)^2} &, |z| > 1 \\\\ \displaystyle Z[n \alpha^n u(n)] = \frac{\alpha z}{(z - \alpha)^2} &, |z| > 1 \\\\ \displaystyle Z[\sin \alpha n \cdot u(n)] = \frac{z \sin \alpha }{z^2 -2z \cos \alpha +1} &, |z| > 1 \\\\ \displaystyle Z[\cos \alpha n \cdot u(n)] = \frac{z^2-z\cos \alpha}{z^2 - 2 z \cos \alpha + 1} &, |z| > 1 \\\\ \displaystyle Z[\frac 1{n!}] = e^{\frac 1z} &, |z| \neq 0 \end{array}

Z[\alpha x_1(n) + \beta x_2(n)] = \alpha Z[x_1(n)] + \beta Z[x_2(n)]

Z[x(n \pm m)] = z^{\pm m} Z[x(n)]

Z[nx(n)] = -zX'(z)

Z[\alpha^nx(n)] = X\left( \frac z\alpha \right)

Z[x_1(n) * x_2(n)] = X_1(z) \cdot X_2(z)