拉普拉斯变换与拉普拉斯逆变换的常用结论与经典公式

2025-10-10 11:04:03

拉氏变换和拉氏逆变换常用结论和经典定理展示与证明

一、定义与概念1.1 拉普拉斯变换1.2 拉普拉斯逆变换

二、拉氏变换和拉氏逆变换常用结论和经典定理2.1 常用结论2.2 经典定理

三、常用结论证明3.1 Unit impluse function3.2 Unit step function3.3 Ramp function3.4 Exponential function3.5 Sine function3.6 Cosine function3.7 Power function

四、经典定理证明4.1 线性性质4.2 相似性质4.3 微分之导数的像函数4.4 微分之像函数的导数4.5 积分的像函数4.6 像函数的积分4.7 延迟性质4.8 位移性质4.9 终值定理4.10 初值定理

参考文献

本文适合于工科课程不过于要求过程严谨、侧重应用的特点,且拉普拉斯变换与拉普拉斯逆变换适用于工科课程中的信号与系统、复变函数与积分变换、电路理论、自动控制原理以及计算机控制原理的基础部分,因此本文提供拉氏变换与拉氏逆变换的重要结论与定理,同时,也为对相关证明感兴趣的同学提供了结论与定理的证明,如果你觉得本文对你有所帮助,可收藏本文,但转载不被允许

一、定义与概念

1.1 拉普拉斯变换

设函数

f

(

t

)

f(t)

f(t)是定义在

[

0

,

+

)

[0,+\infty)

[0,+∞)上的实值函数,如果对于复参数

s

=

β

+

j

w

s=\beta+\text{j}w

s=β+jw,积分

F

(

s

)

=

0

+

f

(

t

)

e

s

t

d

t

(1)

F(s)=\int_0^{+\infty}f(t)e^{-st}\,dt\tag{1}

F(s)=∫0+∞​f(t)e−stdt(1) 在复平面

s

s

s的某一个区域内收敛,则称

F

(

s

)

F(s)

F(s)为

f

(

t

)

f(t)

f(t)的拉普拉斯变换(简称拉氏变换),记为

F

(

s

)

=

L

[

f

(

t

)

]

F(s)=\mathscr{L}[f(t)]

F(s)=L[f(t)],该函数被称为像函数。

1.2 拉普拉斯逆变换

已知函数

f

(

t

)

f(t)

f(t)经过拉普拉斯变换后得到

F

(

s

)

F(s)

F(s),则原函数

f

(

x

)

f(x)

f(x)可由

F

(

s

)

F(s)

F(s)经过拉普拉斯逆变换得到:

f

(

t

)

=

L

1

[

F

(

s

)

]

=

1

2

π

j

β

j

β

+

j

F

(

s

)

e

s

t

d

s

(2)

f(t)=\mathscr{L}^{-1}[F(s)]=\frac{1}{2\pi j}\int_{\beta-j\infty}^{\beta+j\infty}F(s)e^{st}\,ds\tag{2}

f(t)=L−1[F(s)]=2πj1​∫β−j∞β+j∞​F(s)estds(2) 记

f

(

t

)

=

L

1

[

F

(

s

)

]

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

f(t)=L−1[F(s)],

f

(

t

)

f(t)

f(t)被称为原像函数,此过程称为拉普拉斯逆变换(简称拉氏逆变换)。

二、拉氏变换和拉氏逆变换常用结论和经典定理

2.1 常用结论

No.Name of items

f

(

t

)

f(t)

f(t)

F

(

s

)

\textit{F}(s)

F(s) 1unit impluse

δ

(

t

)

\delta(t)

δ(t)12unit step

u

(

t

)

u(t)

u(t)

1

s

\frac{1}{s}

s1​3ramp

tu

(

t

)

\textit{tu}(t)

tu(t)

1

s

2

\frac{1}{s^{2}}

s21​4exponential

e

at

u

(

t

)

e^{\textit{at}}\textit{u}(t)

eatu(t)

1

s

a

\frac{1}{s-a}

s−a1​5sine

sin

w

t

\sin{wt}

sinwt

w

s

2

+

w

2

\frac{w}{s^{2}+w^{2}}

s2+w2w​6cosine

cos

w

t

\cos{wt}

coswt

s

s

2

+

w

2

\frac{s}{s^{2}+w^{2}}

s2+w2s​7power

t

m

t^m

tm

m

!

s

m

+

1

\frac{m!}{s^{m+1}}

sm+1m!​

2.2 经典定理

No.定理&性质名称 表达式 1线性性质

L

[

α

f

(

t

)

+

β

g

(

t

)

]

=

α

F

(

s

)

+

β

G

(

s

)

\mathscr{L}[\alpha f(t)+\beta g(t)]=\alpha F(s)+\beta G(s)

L[αf(t)+βg(t)]=αF(s)+βG(s),其中

α

\alpha

α和

β

\beta

β为常数2相似性质

L

[

f

(

a

t

)

]

=

1

a

F

(

s

a

)

\mathscr{L}[f(at)]=\frac{1}{a}F(\frac{s}{a})

L[f(at)]=a1​F(as​),其中

a

a

a为大于0的常数3微分之导数的像函数

L

[

f

(

n

)

(

t

)

]

=

s

n

F

(

s

)

s

n

1

f

(

0

)

s

n

2

f

(

0

)

f

(

n

1

)

(

0

)

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

L[f(n)(t)]=snF(s)−sn−1f(0)−sn−2f′(0)−⋯−f(n−1)(0)4微分之像函数的导数

F

(

n

)

(

s

)

=

(

1

)

n

L

[

t

n

f

(

t

)

]

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

F(n)(s)=(−1)nL[tnf(t)]5积分的像函数

L

[

0

t

d

t

0

t

d

t

0

t

n

f

(

t

)

d

t

]

=

1

s

n

F

(

s

)

\mathscr{L}[\underbrace{\int_0^tdt\int_0^tdt\cdots\int_0^t}_{n个}f(t)dt]=\frac{1}{s^n}F(s)

L[n个

∫0t​dt∫0t​dt⋯∫0t​​​f(t)dt]=sn1​F(s)6像函数的积分

s

d

s

s

d

s

s

n

F

(

s

)

d

s

=

L

[

f

(

t

)

t

n

]

\underbrace{\int_s^\infty ds\int_s^\infty ds\cdots\int_s^\infty}_{n个} F(s)ds=\mathscr{L}[\frac{f(t)}{t^n}]

n个

∫s∞​ds∫s∞​ds⋯∫s∞​​​F(s)ds=L[tnf(t)​]7延迟性质

L

[

f

(

t

τ

)

]

=

e

s

τ

F

(

s

)

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

L[f(t−τ)]=e−sτF(s)8位移性质

L

[

e

a

t

f

(

t

)

]

=

F

(

s

a

)

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

L[eatf(t)]=F(s−a)9终值定理

lim

t

+

f

(

t

)

=

lim

s

0

s

F

(

s

)

\lim\limits_{t \to +\infty}f(t)=\lim\limits_{s\to 0}sF(s)

t→+∞lim​f(t)=s→0lim​sF(s)10初值定理

lim

t

0

+

f

(

t

)

=

lim

s

+

s

F

(

s

)

\lim\limits_{t \to 0^+}f(t)=\lim\limits_{s\to +\infty}sF(s)

t→0+lim​f(t)=s→+∞lim​sF(s)

三、常用结论证明

3.1 Unit impluse function

已知函数

f

(

t

)

=

δ

(

t

)

f(t)=\delta (t)

f(t)=δ(t),因此该函数经过拉普拉斯变换将得到像函数:

已知

δ

(

t

)

\delta(t)

δ(t)的定义如下:

δ

(

t

)

=

{

A

τ

2

t

τ

2

0

t

>

τ

2

,

A

τ

=

1

,

τ

0

+

\delta(t)=\left\{ \begin{aligned} A && -\frac{\tau}{2}\leqslant t\leqslant\frac{\tau}{2} \\ 0 && t> \left|\frac{\tau}{2} \right | \\ \end{aligned}\,\,,且A\tau=1,\tau\to0^+ \right.

δ(t)=⎩

⎧​A0​​−2τ​⩽t⩽2τ​t>

​2τ​

​​,且Aτ=1,τ→0+

F

(

s

)

=

0

+

f

(

t

)

e

s

t

d

t

=

0

+

δ

(

t

)

e

s

t

d

t

=

+

δ

(

t

)

e

j

w

t

d

t

=

A

τ

2

τ

2

e

j

w

t

d

t

=

A

j

w

(

e

j

w

τ

2

e

j

w

τ

2

)

=

2

A

w

sin

w

τ

2

=

lim

τ

0

+

2

sin

w

τ

2

w

τ

=

1

\begin{aligned} F(s)&=\int_0^{+\infty}f(t)e^{-st}\,dt\\&=\int_0^{+\infty}\delta(t)e^{-st}dt\\ &=\int_{-\infty}^{+\infty}\delta(t)e^{-jwt}dt\\&=A\int_{-\frac{\tau}{2}}^{\frac{\tau}{2}}e^{-jwt}dt\\ &=\frac{A}{-jw}\left(e^{-jw\frac{\tau}{2}}-e^{jw\frac{\tau}{2}}\right)\\&=\frac{2A}{w}\sin\frac{w\tau}{2}\\ &=\lim\limits_{\tau\to0^+}\frac{2\sin\frac{w \tau}{2}}{w\tau}\\&=1 \end{aligned}

F(s)​=∫0+∞​f(t)e−stdt=∫0+∞​δ(t)e−stdt=∫−∞+∞​δ(t)e−jwtdt=A∫−2τ​2τ​​e−jwtdt=−jwA​(e−jw2τ​−ejw2τ​)=w2A​sin2wτ​=τ→0+lim​wτ2sin2wτ​​=1​ 证毕

3.2 Unit step function

已知函数

f

(

t

)

=

u

(

t

)

f(t)=u(t)

f(t)=u(t),因此该函数经过拉普拉斯变换将得到像函数:

F

(

s

)

=

0

+

f

(

t

)

e

s

t

d

t

=

0

+

u

(

t

)

e

s

t

d

t

=

1

s

e

s

t

0

+

=

1

s

F(s)=\int_0^{+\infty}f(t)e^{-st}\,dt=\int_0^{+\infty}u(t)e^{-st}dt=-\frac{1}{s}e^{-st}\big|_0^{+\infty}=\frac{1}{s}

F(s)=∫0+∞​f(t)e−stdt=∫0+∞​u(t)e−stdt=−s1​e−st

​0+∞​=s1​ 证毕

3.3 Ramp function

已知函数

f

(

t

)

=

t

u

(

t

)

f(t)=tu(t)

f(t)=tu(t),因此该函数经过拉普拉斯变换将得到像函数:

F

(

s

)

=

0

+

f

(

t

)

e

s

t

d

t

=

0

+

t

u

(

t

)

e

s

t

d

t

=

1

s

0

+

t

d

(

e

s

t

)

=

1

s

[

t

e

s

t

0

+

0

+

e

s

t

d

t

]

=

1

s

2

\begin{aligned} F(s)&=\int_0^{+\infty}f(t)e^{-st}\,dt\\ &=\int_0^{+\infty}tu(t)e^{-st}dt\\ &=-\frac{1}{s}\int_0^{+\infty}td(e^{-st})\\ &=-\frac{1}{s}\left[te^{-st}\big|_0^{+\infty}-\int_0^{+\infty}e^{-st}dt\right]\\&=\frac{1}{s^2} \end{aligned}

F(s)​=∫0+∞​f(t)e−stdt=∫0+∞​tu(t)e−stdt=−s1​∫0+∞​td(e−st)=−s1​[te−st

​0+∞​−∫0+∞​e−stdt]=s21​​ 证毕

3.4 Exponential function

已知函数

f

(

t

)

=

e

a

t

u

(

t

)

f(t)=e^{at}u(t)

f(t)=eatu(t),因此该函数经过拉普拉斯变换将得到像函数:

F

(

s

)

=

0

+

f

(

t

)

e

s

t

d

t

=

0

+

e

a

t

u

(

t

)

e

s

t

d

t

=

1

a

s

e

(

a

s

)

t

0

+

=

1

s

a

F(s)=\int_0^{+\infty}f(t)e^{-st}\,dt=\int_0^{+\infty}e^{at}u(t)e^{-st}dt=\frac{1}{a-s}e^{(a-s)t}\big|_0^{+\infty}=\frac{1}{s-a}

F(s)=∫0+∞​f(t)e−stdt=∫0+∞​eatu(t)e−stdt=a−s1​e(a−s)t

​0+∞​=s−a1​ 证毕

3.5 Sine function

已知函数

f

(

t

)

=

sin

w

t

f(t)=\sin wt

f(t)=sinwt,因此该函数经过拉普拉斯变换将得到像函数:

F

(

s

)

=

L

[

sin

w

t

]

=

1

2

j

(

L

[

e

j

w

t

]

L

[

e

j

w

t

]

)

=

1

2

j

(

1

s

j

w

1

s

+

j

w

)

=

w

s

2

+

w

2

\begin{aligned} F(s)&=\mathscr{L}[\sin wt]=\frac{1}{2j}\left(\mathscr{L}[e^{jwt}]-\mathscr{L}[e^{-jwt}]\right)\\&=\frac{1}{2j}\left(\frac{1}{s-jw}-\frac{1}{s+jw}\right)=\frac{w}{s^2+w^2} \end{aligned}

F(s)​=L[sinwt]=2j1​(L[ejwt]−L[e−jwt])=2j1​(s−jw1​−s+jw1​)=s2+w2w​​ 证毕

3.6 Cosine function

已知函数

f

(

t

)

=

cos

w

t

f(t)=\cos wt

f(t)=coswt,因此该函数经过拉普拉斯变换将得到像函数:

F

(

s

)

=

L

[

cos

w

t

]

=

1

2

(

L

[

e

j

w

t

]

+

L

[

e

j

w

t

]

)

=

1

2

(

1

s

j

w

+

1

s

+

j

w

)

=

s

s

2

+

w

2

\begin{aligned} F(s)&=\mathscr{L}[\cos wt]=\frac{1}{2}\left(\mathscr{L}[e^{jwt}]+\mathscr{L}[e^{-jwt}]\right)\\&=\frac{1}{2}\left(\frac{1}{s-jw}+\frac{1}{s+jw}\right)=\frac{s}{s^2+w^2} \end{aligned}

F(s)​=L[coswt]=21​(L[ejwt]+L[e−jwt])=21​(s−jw1​+s+jw1​)=s2+w2s​​ 证毕

3.7 Power function

已知函数

f

(

t

)

=

t

m

f(t)=t^m

f(t)=tm,因此该函数经过拉普拉斯变换将得到像函数:

F

(

s

)

=

L

[

t

m

]

=

0

+

t

m

e

s

t

d

t

=

1

s

0

+

t

m

d

(

e

s

t

)

=

1

s

t

m

e

s

t

0

+

+

m

s

0

+

t

m

1

e

s

t

d

t

=

m

s

L

[

t

m

1

]

=

m

(

m

1

)

s

2

L

[

t

m

2

]

=

m

!

s

m

L

[

t

0

]

=

m

!

s

m

+

1

\begin{aligned} F(s)&=\mathscr{L}[t^m]\\&=\int_0^{+\infty}t^me^{-st}dt\\&=-\frac{1}{s}\int_0^{+\infty}t^md(e^{-st})\\ &=-\frac{1}{s}t^me^{-st}\big|_0^{+\infty}+\frac{m}{s}\int_0^{+\infty}t^{m-1}e^{-st}dt\\ &=\frac{m}{s}\mathscr{L}[t^{m-1}]\\&=\frac{m(m-1)}{s^2}\mathscr{L}[t^{m-2}]\\ &\,\,\,\,\,\,\,\,\,\,\,\vdots\\ &=\frac{m!}{s^m}\mathscr{L}[t^0]\\&=\frac{m!}{s^{m+1}} \end{aligned}

F(s)​=L[tm]=∫0+∞​tme−stdt=−s1​∫0+∞​tmd(e−st)=−s1​tme−st

​0+∞​+sm​∫0+∞​tm−1e−stdt=sm​L[tm−1]=s2m(m−1)​L[tm−2]⋮=smm!​L[t0]=sm+1m!​​ 证毕

四、经典定理证明

4.1 线性性质

α

,

β

\alpha,\beta

α,β为常数,且有

L

[

f

(

t

)

]

=

F

(

s

)

,

L

[

g

(

t

)

]

=

G

(

s

)

\mathscr{L}[f(t)]=F(s),\mathscr{L}[g(t)]=G(s)

L[f(t)]=F(s),L[g(t)]=G(s),则有

L

[

α

f

(

t

)

+

β

g

(

t

)

]

=

0

+

(

α

f

(

t

)

+

β

g

(

t

)

)

e

s

t

d

t

=

α

0

+

f

(

t

)

e

s

t

d

t

+

β

0

+

g

(

t

)

e

s

t

d

t

=

α

F

(

s

)

+

β

G

(

s

)

\begin{aligned} \mathscr{L}[\alpha f(t)+\beta g(t)]&=\int_0^{+\infty}\left(\alpha f(t)+\beta g(t)\right)e^{-st}dt\\ &=\alpha \int_0^{+\infty}f(t)e^{-st}dt+\beta \int_0^{+\infty}g(t)e^{-st}dt\\ &=\alpha F(s)+\beta G(s) \end{aligned}

L[αf(t)+βg(t)]​=∫0+∞​(αf(t)+βg(t))e−stdt=α∫0+∞​f(t)e−stdt+β∫0+∞​g(t)e−stdt=αF(s)+βG(s)​ 证毕

4.2 相似性质

L

[

f

(

t

)

]

=

F

(

s

)

\mathscr{L}\left[f(t)\right]=F(s)

L[f(t)]=F(s),则对任一常数

a

>

0

a>0

a>0有

L

[

f

(

a

t

)

]

=

0

+

f

(

a

t

)

e

s

t

d

t

=

x

=

a

t

1

a

0

+

f

(

x

)

e

(

s

a

)

x

d

x

=

1

a

F

(

s

a

)

\begin{aligned} \mathscr{L}[f(at)]&=\int_0^{+\infty}f(at)e^{-st}dt\\& \xlongequal{令x=at}\frac{1}{a}\int_0^{+\infty}f(x)e^{-(\frac{s}{a})x}dx\\&=\frac{1}{a}F\left({\frac{s}{a}}\right) \end{aligned}

L[f(at)]​=∫0+∞​f(at)e−stdt令x=at

a1​∫0+∞​f(x)e−(as​)xdx=a1​F(as​)​ 证毕

4.3 微分之导数的像函数

L

[

f

(

t

)

]

=

F

(

s

)

\mathscr{L}\left[f(t)\right]=F(s)

L[f(t)]=F(s),则有

L

[

f

(

t

)

]

=

0

+

f

(

t

)

e

s

t

d

t

=

分部积分

f

(

t

)

e

s

t

0

+

+

s

0

+

f

(

t

)

e

s

t

d

t

=

s

F

(

s

)

f

(

0

)

\begin{aligned} \mathscr{L}[f'(t)]&=\int_0^{+\infty}f'(t)e^{-st}dt\\ &\xlongequal{分部积分}f(t)e^{-st}\big|_0^{+\infty}+s\int_0^{+\infty}f(t)e^{-st}dt\\ &=sF(s)-f(0) \end{aligned}

L[f′(t)]​=∫0+∞​f′(t)e−stdt分部积分

f(t)e−st

​0+∞​+s∫0+∞​f(t)e−stdt=sF(s)−f(0)​ 经过数学归纳法可得:

L

[

f

(

n

)

(

t

)

]

=

s

n

F

(

s

)

s

n

1

f

(

0

)

s

n

2

f

(

0

)

f

(

n

1

)

(

0

)

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

L[f(n)(t)]=snF(s)−sn−1f(0)−sn−2f′(0)−⋯−f(n−1)(0) 证毕

4.4 微分之像函数的导数

L

[

f

(

t

)

]

=

F

(

s

)

\mathscr{L}\left[f(t)\right]=F(s)

L[f(t)]=F(s),则有

F

(

s

)

=

d

d

s

0

+

f

(

t

)

e

s

t

d

t

=

0

+

s

[

f

(

t

)

e

s

t

]

d

t

=

0

+

t

f

(

t

)

e

s

t

d

t

=

L

[

t

f

(

t

)

]

\begin{aligned} F'(s)&=\frac{d}{ds}\int_0^{+\infty}f(t)e^{-st}dt\\&=\int_0^{+\infty}\frac{\partial}{\partial s}\left[f(t)e^{-st}\right]dt\\ &=-\int_0^{+\infty}tf(t)e^{-st}dt\\ &=-\mathscr{L}[tf(t)] \end{aligned}

F′(s)​=dsd​∫0+∞​f(t)e−stdt=∫0+∞​∂s∂​[f(t)e−st]dt=−∫0+∞​tf(t)e−stdt=−L[tf(t)]​ 对

F

(

s

)

F(s)

F(s)施行同样步骤,反复进行可得:

F

(

n

)

(

s

)

=

(

1

)

n

L

[

t

n

f

(

t

)

]

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

F(n)(s)=(−1)nL[tnf(t)] 证毕

4.5 积分的像函数

L

[

f

(

t

)

]

=

F

(

s

)

,

g

(

t

)

=

0

t

f

(

t

)

d

t

\mathscr{L}\left[f(t)\right]=F(s),g(t)=\int_0^tf(t)dt

L[f(t)]=F(s),g(t)=∫0t​f(t)dt,则

g

(

t

)

=

f

(

t

)

g'(t)=f(t)

g′(t)=f(t),且

g

(

0

)

=

0

g(0)=0

g(0)=0,利用微分之导数的像函数可得:

L

[

g

(

t

)

]

=

s

L

[

g

(

t

)

]

g

(

0

)

=

s

L

[

g

(

t

)

]

=

s

L

[

0

t

f

(

t

)

d

t

]

\mathscr{L}[g'(t)]=s\mathscr{L}[g(t)]-g(0)=s\mathscr{L}[g(t)]=s\mathscr{L}\left[\int_0^tf(t)dt\right]

L[g′(t)]=sL[g(t)]−g(0)=sL[g(t)]=sL[∫0t​f(t)dt] 即有

L

[

0

t

f

(

t

)

d

t

]

=

1

s

F

(

s

)

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

L[∫0t​f(t)dt]=s1​F(s),反复利用上式可得:

L

[

0

t

d

t

0

t

d

t

0

t

n

f

(

t

)

d

t

]

=

1

s

n

F

(

s

)

\mathscr{L}[\underbrace{\int_0^tdt\int_0^tdt\cdots\int_0^t}_{n个}f(t)dt]=\frac{1}{s^n}F(s)

L[n个

∫0t​dt∫0t​dt⋯∫0t​​​f(t)dt]=sn1​F(s) 证毕

4.6 像函数的积分

L

[

f

(

t

)

]

=

F

(

s

)

\mathscr{L}\left[f(t)\right]=F(s)

L[f(t)]=F(s),则有

s

+

F

(

s

)

d

s

=

s

+

[

0

+

f

(

t

)

e

s

t

d

t

]

d

s

=

0

+

f

(

t

)

[

s

+

e

s

t

d

s

]

d

t

=

0

+

f

(

t

)

[

1

t

e

s

t

]

s

+

d

t

=

0

+

f

(

t

)

t

e

s

t

=

L

[

f

(

t

)

t

]

\begin{aligned} \int_s^{+\infty}F(s)ds&=\int_s^{+\infty}\left[\int_0^{+\infty}f(t)e^{-st}dt\right]ds\\ &=\int_0^{+\infty}f(t)\left[\int_s^{+\infty}e^{-st}ds\right]dt\\ &=\int_0^{+\infty}f(t)\cdot\left[-\frac{1}{t}e^{-st}\right]\big|_s^{+\infty}dt\\ &=\int_0^{+\infty}\frac{f(t)}{t}e^{-st}\\ &=\mathscr{L}\left[\frac{f(t)}{t}\right] \end{aligned}

∫s+∞​F(s)ds​=∫s+∞​[∫0+∞​f(t)e−stdt]ds=∫0+∞​f(t)[∫s+∞​e−stds]dt=∫0+∞​f(t)⋅[−t1​e−st]

​s+∞​dt=∫0+∞​tf(t)​e−st=L[tf(t)​]​ 反复利用上式可得:

s

d

s

s

d

s

s

n

F

(

s

)

d

s

=

L

[

f

(

t

)

t

n

]

\underbrace{\int_s^\infty ds\int_s^\infty ds\cdots\int_s^\infty}_{n个} F(s)ds=\mathscr{L}[\frac{f(t)}{t^n}]

n个

∫s∞​ds∫s∞​ds⋯∫s∞​​​F(s)ds=L[tnf(t)​] 证毕

4.7 延迟性质

L

[

f

(

t

)

]

=

F

(

s

)

\mathscr{L}\left[f(t)\right]=F(s)

L[f(t)]=F(s),当

t

<

0

t<0

t<0时,

f

(

t

)

=

0

f(t)=0

f(t)=0,则对任一非负实数

τ

\tau

τ有

L

[

f

(

t

τ

)

]

=

0

+

f

(

t

τ

)

e

s

t

d

t

=

τ

+

f

(

t

τ

)

e

s

t

d

t

\mathscr{L}[f(t-\tau)]=\int_0^{+\infty}f(t-\tau)e^{-st}dt=\int_\tau^{+\infty}f(t-\tau)e^{-st}dt

L[f(t−τ)]=∫0+∞​f(t−τ)e−stdt=∫τ+∞​f(t−τ)e−stdt 令

t

1

=

t

τ

t_1=t-\tau

t1​=t−τ有

L

=

0

+

f

(

t

1

)

e

s

(

t

1

+

τ

)

d

t

1

=

e

s

τ

0

+

f

(

t

1

)

e

s

t

1

d

t

1

=

e

s

τ

F

(

s

)

\mathscr{L}=\int_0^{+\infty}f(t_1)e^{-s(t_1+\tau)}dt_1=e^{-s\tau}\int_0^{+\infty}f(t_1)e^{-st_1}dt_1=e^{-s\tau}F(s)

L=∫0+∞​f(t1​)e−s(t1​+τ)dt1​=e−sτ∫0+∞​f(t1​)e−st1​dt1​=e−sτF(s) 证毕

4.8 位移性质

L

[

f

(

t

)

]

=

F

(

s

)

\mathscr{L}\left[f(t)\right]=F(s)

L[f(t)]=F(s),则有

L

[

e

a

t

f

(

t

)

]

=

0

+

e

a

t

f

(

t

)

e

s

t

d

t

=

0

+

f

(

t

)

e

(

s

a

)

t

d

t

=

F

(

s

a

)

\mathscr{L}[e^{at}f(t)]=\int_0^{+\infty}e^{at}f(t)e^{-st}dt=\int_0^{+\infty}f(t)e^{-(s-a)t}dt=F(s-a)

L[eatf(t)]=∫0+∞​eatf(t)e−stdt=∫0+∞​f(t)e−(s−a)tdt=F(s−a) 证毕

4.9 终值定理

从拉普拉斯变换的微分性质我们知道以下一个简单的等式:

L

[

f

(

t

)

]

=

0

+

f

(

t

)

e

s

t

d

t

=

s

F

(

s

)

f

(

0

)

\mathscr{L}[f'(t)]=\int_0^{+\infty}f'(t)e^{-st}dt=sF(s)-f(0)

L[f′(t)]=∫0+∞​f′(t)e−stdt=sF(s)−f(0) 我们将等式两边取极限

s

0

s\to0

s→0,可得

lim

s

0

0

+

f

(

t

)

e

s

t

d

t

=

0

+

f

(

t

)

d

t

=

lim

t

+

f

(

t

)

f

(

0

)

=

lim

s

0

s

F

(

s

)

f

(

0

)

\lim\limits_{s\to0}\int_0^{+\infty}f'(t)e^{-st}dt=\int_0^{+\infty}f'(t)dt=\lim\limits_{t\to+\infty}f(t)-f(0)=\lim\limits_{s\to0}sF(s)-f(0)

s→0lim​∫0+∞​f′(t)e−stdt=∫0+∞​f′(t)dt=t→+∞lim​f(t)−f(0)=s→0lim​sF(s)−f(0) 化简可得:

f

(

+

)

=

lim

t

+

f

(

t

)

=

lim

s

0

s

F

(

s

)

f(+\infty)=\lim\limits_{t\to+\infty}f(t)=\lim\limits_{s\to0}sF(s)

f(+∞)=t→+∞lim​f(t)=s→0lim​sF(s) 证毕

4.10 初值定理

从拉普拉斯变换的微分性质我们知道以下一个简单的等式:

L

[

f

(

t

)

]

=

0

+

f

(

t

)

e

s

t

d

t

=

s

F

(

s

)

f

(

0

)

\mathscr{L}[f'(t)]=\int_0^{+\infty}f'(t)e^{-st}dt=sF(s)-f(0)

L[f′(t)]=∫0+∞​f′(t)e−stdt=sF(s)−f(0) 我们将等式两边取极限

s

+

s\to+\infty

s→+∞,可得

lim

s

+

0

+

f

(

t

)

e

s

t

d

t

=

0

=

lim

s

+

s

F

(

s

)

f

(

0

)

\lim\limits_{s\to+\infty}\int_0^{+\infty}f'(t)e^{-st}dt=0=\lim\limits_{s\to+\infty}sF(s)-f(0)

s→+∞lim​∫0+∞​f′(t)e−stdt=0=s→+∞lim​sF(s)−f(0) 因此,我们可以得到:

lim

t

0

+

f

(

t

)

=

f

(

0

)

=

lim

s

+

s

F

(

s

)

\lim\limits_{t\to0^+}f(t)=f(0)=\lim\limits_{s\to+\infty}sF(s)

t→0+lim​f(t)=f(0)=s→+∞lim​sF(s) 证毕

参考文献

[1]李红, 谢松法. 复变函数与积分变换.第4版[M]. 高等教育出版社, 2013.