拉普拉斯变换与拉普拉斯逆变换的常用结论与经典公式
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。.........
拉氏变换和拉氏逆变换常用结论和经典定理展示与证明
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) |
---|---|---|---|
1 | unit impluse | δ ( t ) \delta(t) δ(t) | 1 |
2 | unit step | u ( t ) u(t) u(t) | 1 s \frac{1}{s} s1 |
3 | ramp | tu ( t ) \textit{tu}(t) tu(t) | 1 s 2 \frac{1}{s^{2}} s21 |
4 | exponential | e at u ( t ) e^{\textit{at}}\textit{u}(t) eatu(t) | 1 s − a \frac{1}{s-a} s−a1 |
5 | sine | sin w t \sin{wt} sinwt | w s 2 + w 2 \frac{w}{s^{2}+w^{2}} s2+w2w |
6 | cosine | cos w t \cos{wt} coswt | s s 2 + w 2 \frac{s}{s^{2}+w^{2}} s2+w2s |
7 | power | 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)]=a1F(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 个 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 ∫0tdt∫0tdt⋯∫0tf(t)dt]=sn1F(s) |
6 | 像函数的积分 | ∫ s ∞ d s ∫ s ∞ d s ⋯ ∫ s ∞ ⏟ n 个 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→+∞limf(t)=s→0limsF(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+limf(t)=s→+∞limsF(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
∣
,
且
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⩽2τt>
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∫−2τ2τe−jwtdt=−jwA(e−jw2τ−ejw2τ)=w2Asin2wτ=τ→0+limwτ2sin2wτ=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=−s1e−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−s1e(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)=−s1tme−st
0+∞+sm∫0+∞tm−1e−stdt=smL[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
设
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
设
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
=
令
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=ata1∫0+∞f(x)e−(as)xdx=a1F(as)
证毕
4.3 微分之导数的像函数
设
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
设
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
设
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)=∫0tf(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[∫0tf(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[∫0tf(t)dt]=s1F(s),反复利用上式可得:
L
[
∫
0
t
d
t
∫
0
t
d
t
⋯
∫
0
t
⏟
n
个
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
∫0tdt∫0tdt⋯∫0tf(t)dt]=sn1F(s)
证毕
4.6 像函数的积分
设
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)⋅[−t1e−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
个
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
设
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−st1dt1=e−sτF(s)
证毕
4.8 位移性质
设
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→+∞limf(t)−f(0)=s→0limsF(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→+∞limf(t)=s→0limsF(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→+∞limsF(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+limf(t)=f(0)=s→+∞limsF(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.
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