拉普拉斯变换入门

Introduction to the Laplace Transform

定义

对于一个给定的函数 $f(t)$，其拉普拉斯变换为：

$$\mathcal{L}[f(t)]=F(s)={\int }_{0^{-}}^{\mathrm{\infty }}f(t)e^{-st}\mathrm{d}t$$

其中， $s$ 是一个复变量，其定义为

$$s=\sigma +j\omega$$

The Laplace transform is an integral transformation of a function $f(t)$ from the time domain into the complex frequency domain, given by $F(s)$.

性质

Linearity

$$\mathcal{L}[a_1{f}_1(t)+a_2{f}_2(t)]=a_1{F}_1(s)+a_2{F}_2(s)$$

Scaling

$$\mathcal{L}[f(at)]=\frac{1}{a}F\left(\frac{s}{a}\right)$$

公式推导

$$\begin{array}{rl}\mathcal{L}[f(at)]& ={\int }_{0^{-}}^{\mathrm{\infty }}f(at)e^{-st}\mathrm{d}t\\ & ={\int }_{0^{-}}^{\mathrm{\infty }}f(at)e^{(-s/a)(at)}\frac{\mathrm{d}(at)}{a}\\ & =\frac{1}{a}{\int }_{0^{-}}^{\mathrm{\infty }}f(u)e^{-(s/a)u}\mathrm{d}u,\text{ where }u=at\\ & =\frac{1}{a}F\left(\frac{s}{a}\right)\end{array}$$

Time Shift

$$\mathcal{L}[f(t-a)u(t-a)]=e^{-as}F(s),\text{ where }a\ge 0$$

公式推导

$$\begin{array}{rl}\mathcal{L}[f(t-a)u(t-a)]& ={\int }_{0^{-}}^{\mathrm{\infty }}f(t-a)u(t-a)e^{-st}\mathrm{d}t\\ & ={\int }_{t=a^{-}}^{\mathrm{\infty }}f(t-a)e^{-st}\mathrm{d}t\\ & ={\int }_{(t-a)=0^{-}}^{\mathrm{\infty }}f(t-a)e^{-s(t-a)}{e}^{-as}\mathrm{d}(t-a)\\ & =e^{-as}{\int }_{u=0^{-}}^{\mathrm{\infty }}f(u)e^{-su}\mathrm{d}u,\text{ where }u=t-a\\ & =e^{-as}F(s)\end{array}$$

Frequency Shift

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

公式推导

$$\begin{array}{rl}\mathcal{L}[e^{-at}f(t)]& ={\int }_{0^{-}}^{\mathrm{\infty }}{e}^{-at}f(t)e^{-st}\mathrm{d}t\\ & ={\int }_{0^{-}}^{\mathrm{\infty }}f(t)e^{-(s+a)t}\mathrm{d}t\\ & =F(s+a)\end{array}$$

Time Differentiation

$$\mathcal{L}[f^{\prime }(t)]=sF(s)-f(0^{-})$$

公式推导

$$\begin{array}{rl}\mathcal{L}[f^{\prime }(t)]& =\mathcal{L}[\frac{\mathrm{d}}{\mathrm{d}t}f(t)]\\ & ={\int }_{0^{-}}^{\mathrm{\infty }}\frac{\mathrm{d}}{\mathrm{d}t}f(t)e^{-st}\mathrm{d}t\\ & ={\int }_{0^{-}}^{\mathrm{\infty }}{e}^{-st}\mathrm{d}f(t)\\ \text{分部积分法 }& =e^{-st}f(t){|}_{0^{-}}^{\mathrm{\infty }}-{\int }_{0^{-}}^{\mathrm{\infty }}f(t)\mathrm{d}(e^{-st})\\ & =-f(0^{-})-{\int }_{0^{-}}^{\mathrm{\infty }}f(t)(-s{e}^{-st})\mathrm{d}t\\ & =-f(0^{-})+s{\int }_{0^{-}}^{\mathrm{\infty }}f(t)e^{-st}\mathrm{d}t\\ & =sF(s)-f(0^{-})\end{array}$$

分部积分法

$$\int u\mathrm{d}v=uv-\int v\mathrm{d}u$$

推广

$$\begin{array}{rl}\mathcal{L}[f^{\prime \prime }(t)]& =s\mathcal{L}[f^{\prime }(t)]-f^{\prime }(0^{-})\\ & =s[sF(s)-f(0^{-})]-f^{\prime }(0^{-})\\ & =s^{2}F(s)-sf(0^{-})-f^{\prime }(0^{-})\\ \mathcal{L}[f^n(t)]& =s\cdot \mathcal{L}[f^{n-1}(t)]-f^{n-1}(0^{-})\end{array}$$

Time Integration

$$\mathcal{L}[{\int }_0^{t}f(\tau )\mathrm{d}\tau ]=\frac{1}{s}F(s)$$

公式推导

使用分部积分法，令

$$\begin{array}{rlr}& u={\int }_0^{t}f(\tau )\mathrm{d}\tau ,& \mathrm{d}u=f(t)\mathrm{d}t\\ & \mathrm{d}v=e^{-st}\mathrm{d}t,& v=-\frac{1}{s}{e}^{-st}\end{array}$$

于是

$$\begin{array}{rl}\mathcal{L}[{\int }_0^{t}f(\tau )\mathrm{d}\tau ]& ={\int }_{0^{-}}^{\mathrm{\infty }}[{\int }_0^{t}f(\tau )\mathrm{d}\tau ]e^{-st}\mathrm{d}t\\ & =[{\int }_0^{t}f(\tau )\mathrm{d}\tau ](-\frac{1}{s}{e}^{-st}){|}_{0^{-}}^{\mathrm{\infty }}-{\int }_{0^{-}}^{\mathrm{\infty }}(-\frac{1}{s}{e}^{-st})f(t)\mathrm{d}t\\ & =0+\frac{1}{s}{\int }_{0^{-}}^{\mathrm{\infty }}f(t)e^{-st}\mathrm{d}t\\ & =\frac{1}{s}F(s)\end{array}$$

Frequency Differentiation

$$\mathcal{L}[tf(t)]=-\frac{\mathrm{d}F(s)}{\mathrm{d}s}$$

公式推导

$$\begin{array}{rl}\frac{\mathrm{d}F(s)}{\mathrm{d}s}& ={\int }_{0^{-}}^{\mathrm{\infty }}f(t)(-t{e}^{-st})\mathrm{d}t\\ & ={\int }_{0^{-}}^{\mathrm{\infty }}(-tf(t))e^{-st}\mathrm{d}t\\ & =\mathcal{L}[-tf(t)]\end{array}$$

Hence,

$$\mathcal{L}[tf(t)]=-\frac{\mathrm{d}F(s)}{\mathrm{d}s}$$

推广

$$\mathcal{L}[t^{n}f(t)]=(-1{)}^n\frac{{\mathrm{d}}^{n}F(s)}{\mathrm{d}{s}^n}$$

Time Periodicity

$$F(s)=\frac{F_1(s)}{1-e^{-Ts}}$$

公式推导

对于满足以下性质的函数 $f(t)$

$$f(t)=f(t-nT),\text{ where }n\in \mathbb{N}$$

被称作周期函数。函数 $f(t)$ 可以表示为一组和

$$\begin{array}{rl}f(t)& =f(t)[u(t)-u(t-T)]\\ & \phantom{=}+f(t)[u(t-T)-u(t-2T)]\\ & \phantom{=}+f(t)[u(t-2T)-u(t-3T)]\\ & \phantom{=}+\cdots \\ & =f_1(t)+f_2(t)+f_3(t)+\cdots \\ & =f_1(t)+f_1(t-T)u(t-T)+f_1(t-2T)u(t-2T)+\cdots \end{array}$$

其中，$f_1(t)=f(t)[u(t)-u(t-T)]$。

于是

$$\begin{array}{rl}F(s)& =\mathcal{L}[f_1(t)]+\mathcal{L}[f_1(t-T)u(t-T)]+\mathcal{L}[f_1(t-2T)u(t-2T)]+\cdots \\ & =F_1(s)+e^{-Ts}{F}_1(s)+e^{-2Ts}{F}_1(s)+\cdots \\ & =F_1(s)[1+e^{-Ts}+e^{-2Ts}+\cdots ]\end{array}$$

利用几何级数公式

$$1+x+x^2+\cdots =\frac{1}{1-x},\text{ for }|x|<1$$

可知

$$F(s)=\frac{F_1(s)}{1-e^{-Ts}}$$

Initial and Final Values

$$\begin{array}{rl}f(0)& =\underset{s\to \mathrm{\infty }}{lim}sF(s)\\ f(\mathrm{\infty })& =\underset{s\to 0}{lim}sF(s)\end{array}$$

公式推导 1

利用 the differentiation property，可知

$$sF(s)-f(0)=\mathcal{L}\left[\frac{\mathrm{d}f}{\mathrm{d}t}\right]={\int }_{0^{-}}^{\mathrm{\infty }}\frac{\mathrm{d}f}{\mathrm{d}t}{e}^{-st}\mathrm{d}t$$

因为 $\underset{s\to \mathrm{\infty }}{lim}{e}^{-st}=0$，所以

$$\underset{s\to \mathrm{\infty }}{lim}[sF(s)-f(0)]=0$$

因为 $f(0)$ 是常数，所以

$$f(0)=\underset{s\to \mathrm{\infty }}{lim}sF(s)$$

公式推导 2

$$\underset{s\to 0}{lim}[sF(s)-f(0^{-})]={\int }_{0^{-}}^{\mathrm{\infty }}\frac{\mathrm{d}f}{\mathrm{d}t}{e}^{0t}\mathrm{d}t={\int }_{0^{-}}^{\mathrm{\infty }}\mathrm{d}f=f(\mathrm{\infty })-f(0^{-})$$

于是

$$f(\mathrm{\infty })=\underset{s\to 0}{lim}sF(s)$$

拉普拉斯逆变换

The Inverse Laplace Transform

$${\mathcal{L}}^{-1}[F(s)]=f(t)=\frac{1}{2\pi j}{\int }_{{\sigma }_1-j\mathrm{\infty }}^{{\sigma }_1+j\mathrm{\infty }}F(s)e^{st}\mathrm{d}s$$$$F(s)=\frac{N(s)}{D(s)}$$

import sympy as sp
import IPython.display as display
s = sp.symbols('s', complex=True)
t = sp.symbols('t', real=True)
F = 1 / s
# 输出：
# $\displaystyle \theta\left(t\right)$
f = sp.inverse_laplace_transform(F, s, t)
display(f)

Simple Poles

$$\begin{array}{rl}F(s)& =\frac{N(s)}{(s+p_1)(s+p_2)\cdots (s+p_n)}\\ & =\frac{k_1}{s+p_1}+\frac{k_2}{s+p_2}+\cdots +\frac{k_n}{s+p_n}\end{array}$$

Partial Fractions

$$k_i=(s+p_i)F(s){|}_{s=-p_i}$$$$f(t)=(k_1{e}^{-p_{1}t}+k_2{e}^{-p_{2}t}+\cdots +k_n{e}^{-p_{n}t})u(t)$$

Repeated Poles

$$F(s)=\frac{k_n}{(s+p{)}^n}+\frac{k_{n-1}}{(s+p{)}^{n-1}}+\cdots +\frac{k_2}{(s+p{)}^2}+\frac{k_1}{s+p}+F_1(s)$$$$\begin{array}{rl}{k}_n& =(s+p{)}^{n}F(s){|}_{s=-p}\\ k_{n-1}& =\frac{\mathrm{d}}{\mathrm{d}s}[(s+p{)}^{n}F(s)]{|}_{s=-p}\\ k_{n-2}& =\frac{1}{2!}\frac{{\mathrm{d}}^2}{\mathrm{d}{s}^2}[(s+p{)}^{n}F(s)]{|}_{s=-p}\\ k_{n-m}& =\frac{1}{m!}\frac{{\mathrm{d}}^m}{\mathrm{d}{s}^m}[(s+p{)}^{n}F(s)]{|}_{s=-p}\end{array}$$$${\mathcal{L}}^{-1}\left[\frac{1}{(s+a{)}^n}\right]=\frac{t^{n-1}{e}^{-at}}{(n-1)!}u(t)$$$$f(t)=(k_1{e}^{-pt}+k_{2}t{e}^{-pt}+\frac{k_3}{2!}{t}^2{e}^{-pt}+\cdots +\frac{k_n}{(n-1)!}{t}^{n-1}{e}^{-pt})u(t)+f_1(t)$$

Complex Poles

$$F(s)=\frac{A_{1}s+A_2}{s^2+as+b}+F_1(s)$$$$s^2+as+b=s^2+2\alpha s+{\alpha }^2+{\beta }^2=(s+\alpha {)}^2+{\beta }^2$$$$A_{1}s+A_2=A_1(s+\alpha )+B_1\beta$$$$F(s)=\frac{A_1(s+\alpha )}{(s+\alpha {)}^2+{\beta }^2}+\frac{B_1\beta }{(s+\alpha {)}^2+{\beta }^2}+F_1(s)$$$$f(t)=(A_1{e}^{-\alpha t}\cos (\beta t)+B_1{e}^{-\alpha t}\sin (\beta t))u(t)+f_1(t)$$

卷积积分

The Convolution Integral

定义

$$y(t)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(\lambda )h(t-\lambda )\mathrm{d}\lambda$$

或

$$y(t)=x(t)\ast h(t)$$

如果

1. 当 $t<0$ 时，$x(t)=0$
2. 当 $t<\lambda$ 时，$h(t-\lambda )=0$

那么

$$y(t)=h(t)\ast x(t)={\int }_0^{t}x(\lambda )h(t-\lambda )\mathrm{d}\lambda$$

性质

1. $x(t)\ast h(t)=h(t)\ast x(t)$

2. $f(t)\ast [x(t)+y(t)]=f(t)\ast x(t)+f(t)\ast y(t)$

3. $f(t)\ast [x(t)\ast y(t)]=[f(t)\ast x(t)]\ast y(t)$

4. $f(t)\ast \delta (t)=f(t)$

5. $f(t)\ast \delta (t-t_0)=f(t-t_0)$

6. $f(t)\ast {\delta }^{\prime }(t)=f^{\prime }(t)$

7. $f(t)\ast u(t)={\int }_{-\mathrm{\infty }}^{t}f(\lambda )\mathrm{d}\lambda$

与拉普拉斯变换的关系

假设 $f_1(t)$ 和 $f_2(t)$ 的拉普拉斯变换分别为 $F_1(s)$ 和 $F_2(s)$，那么

$$\boxed{{\displaystyle F_1(s)F_2(s)=\mathcal{L}[f_1(t)\ast f_2(t)]}}$$

Python 示例

# Example 15.12
# Page 698

import sympy as sp
# 我们要计算 x_1(t) 和 x_2(t) 的卷积:
# y(t) = ∫ x_1(λ) * x_2(t-λ) dλ, λ 从 0 到 t

t = sp.symbols('t', real=True)
lam = sp.symbols('\\lambda', real=True)

x_1 = 2 * (sp.Heaviside(t) - sp.Heaviside(t-1))
x_2 = (sp.Heaviside(t-1) - sp.Heaviside(t-3))

# 卷积表达式，注意 x_2 用 t-lam
y_conv = sp.integrate(x_1.subs(t, lam) * x_2.subs(t, t - lam), (lam, 0, t))
display(y_conv.simplify())

# 作图验证
import matplotlib.pyplot as plt
import numpy as np

# 将 y_conv 转换为可用于数值计算的函数
conv_func = sp.lambdify(t, y_conv, modules=['numpy'])

ts = np.linspace(0, 5, 301)
ys = conv_func(ts)

plt.figure(figsize=(7,4))
plt.plot(ts, ys, label=r'$y(t)$')
plt.xlabel('$t$')
plt.ylabel('卷积值')
plt.title('x_1(t) 和 x_2(t) 的卷积图像, $t \\in [0,5]$')
plt.rcParams['font.sans-serif'] = ['SimHei']  # 设置中文字体为黑体，避免中文乱码
plt.rcParams['axes.unicode_minus'] = False    # 正常显示负号
plt.grid(True)
plt.legend()
plt.show()

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