Chapter 3 Laplace Transforms
MMP-1-3_Laplace_Transform_2014_DQLu_CoS_SHU
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.
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Laplace 变换的定义 ......
Laplace 变换的性质 ..
Laplace 变换的反演 .....
Laplace 变换法求解常微分方程时的应用 ....
例题[4]
♠ 给定像函数 f (p), 利用有理式反演法和 Laplace 变换的性质, 求其原函数 f (t).
习题: 1 . 2p + 5 p f (p) = 2 . p − 2p + 5 f (p) = √
[1]
请阅读教材 §§ 6.1–6.3 (90–104 页).
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Laplace 变换的定义 ......
Laplace 变换的性质 ..
Laplace 变换的反演 .....
Laplace 变换法求解常微分方程时的应用 ....
已知某个物理量在初始时刻 t = 0 时的值 f (0), 求解它在初始时刻之后的演化规律 f (t). 而对于 它在初始时刻之前的值, 我们并不感兴趣. 不妨设 f (t) = 0, (t < 0), 即 f (t) 理解为 f (t)H(t), H(t) 为 Heaviside 函数. 为了获得更宽 松的变换条件, 令 g (t) = exp(−σt)f (t), 其中 exp(−σt) 是收敛因子, 正实数 σ 选得足够大以保 证 g (t) 在 (−∞, ∞) 上绝对可积. 于是[2] ∫ ∞ 1 g (t) exp(−iωt)dt F [g (t)] = G(ω ) = 2π −∞ ∫ ∞ 1 f (t) exp[−(σ + iω )t]dt, = 2π 0
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Laplace 变换的定义 ......
Laplace 变换的性质 ..
拉普拉斯变换LAPLACETRANSFORM
F(s) L 1(t)
1
e
st
dt
1
e st
0
s
0
1 s
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2.1 拉氏变换的概念
• 例二：求单位脉冲函数(Unit Puise Fuction)的象函数。
• 设函数
0
(t 0)
(t
)
1
(0 t )
见图2-2（a）
f
(t)
e st d 0 dt
f (t)dt
estd f (t) f (t)est
0
0
f (t)(s)estdt
0
f (0) s f (t)estdt sF(s) f (0) 0
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2.2 拉氏变换的运算定理
重要的运算定理。
• 五、位移定理 L et f (t) F(s )
L et f (t)
et f (t)dt
0
f (t)etdt 0
F(s)
•
上中式的表用F(明ss代)，替原s函+f a(数t即) 可。乘也以就因是子将
e时t ，它的象函数只需把
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2.1 拉氏变换的概念
• 由于
f (t)est dt 0
是一个定积分，t 将在新函数中消失。因
此，F(s) 只取决于s，它是复变数s的函数。拉氏变换将原来
的实变量函数 f (t) 转化为复变量函数 F(s) 。
• 拉氏变换是一种单值变换。 f (t) 和 F(s)之间具有一一对应的
积分变换(Laplace)
根据拉氏变换的定义, 有
L[ f (t)] ektestd t e(sk )td t
0
0
这个积分在Re(s)>k时收敛, 而且有
e(sk )td t 1 e (sk )t 1
0
sk
0 sk
所以 L[ekt ] 1 (Re(s) k). sk
f (t )estdt.
0
20
例6 求单位脉冲函数 d (t ) 的Laplace变换.
解 因为
d (t) f (t)dt f (0),
所以
L[d (t)] L[d (t)]
d (t )estdt 0
d
(t )estdt
1.
d t 1
若令c e >0 (即 c+e = c1>c), 则
| f (t)est| Meet.
所以
f (t) est d t
M
eetd t
M
.
0
0
e
注1:大部分常用函数的Laplace变换都存在(常义下); 注2:存在定理的条件是充分但非必要条件.
14
例3 求 f (t)=sinkt (k为实数) 的拉氏变换
>L>L=f=[ afb(st()s]in(Lt1)[);fL(t1e=)]la1ps la1e0cseTe(f0T)stfs(ti)netstddtt.
1/(s1^2+1e1)*scoth1(s12/2e*1p si*s)
1 1
e s e s
s2
k
k2
Laplace_Transforms
A Useful Analogy.
To understand the Laplace transform, use of the Laplace to solve differential equations, and the relationship between the s-domain and the time doห้องสมุดไป่ตู้ain it is useful to consider the logarithm function.
An Important Limitation.
The Laplace transform and techniques related to it are only applicable to systems described by linear constant-coefficient models.
Using the Laplace Transform.
In order to apply the technique described above, it is necessary to be able to do the forward and inverse Laplace transforms. Although in principle, you could do the necessary integrals, people have been doing those integrals for centuries. They have them pretty well figured out by now, and you’ve got better things to do. So a better plan is to use a table of Laplace transforms. The book has a pretty good table, but we’ll provide a larger table. The table is used primarily for the inverse transform, and for transforming inputs. For other parts of your equation(s), it is only necessary to know a few properties of the Laplace transform. Some of these are derived in Appendix II, but you only need to be able to use them, not derive them. Linearity: (both superposition and homogeneity):
[数学]拉普拉斯变换 Laplace Transform
![[数学]拉普拉斯变换 Laplace Transform](https://img.taocdn.com/s3/m/7db5dd64783e0912a2162a99.png)
0
∫ F(s ′) ds ′
s T − st
∞
15.
f (t) periodic
∫e L { f (t )} =
f (t ) dt f (t ) = f (t + T )
1 − e − sT
2.2.2 Methods of Finding the Laplace Transform
1. 2. 3. 4. Direct method by solving (2.1.1). Expand f (t) in power series if such an expansion exists. Differentiation with respect to a parameter. Use of tables.
F( s) =
∫ f (t ) e
0
∞
− st
dt
s = σ + jω
f (t) = piecewise continuous and of exponential order
2 Transform
f (t ) = 1 2 πj
σ+ j ∞
σ− j ∞
∫ F( s) e
st
ds
where the integration is within the regions of convergence which is a vertical strip σ1 < Re{s} < σ2.
Laplace transform
∞
Unit-step function ( 记为 us(t) or 1(t) )(单位阶跃 函数)
0
1 = s +α
1, t > 0 1(t ) = 0, t < 0 ∞ 1 − st − st L(1(t )) = ∫ 1(t )e dt = − e s 0
∞
0
1 = s
PDF 文件使用 "pdfFactory Pro" 试用版本创建
−1
c + i∞
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Laplace 变换的例子 Exponential function(指数函数)
f (t ) = e
那么
−α t
,t ≥ 0
− ( s +α ) t ∞
L (e
−αt
e ) = ∫ e e dt = s +α 0
dn f (t) n L[ n ] = s L[ f (t)]−sn−1 f (0) −sn−2 f (1) (0) −sn−3 f (2) (0) −L− f (n−1) (0) dt
特别地， 如果有零初始条件，即
f (0) = f (1) (0) = f (2) (0) = f (n−1) (0) = 0
( s 2 + 3s + 8)Y = (2 + s ) X
Transfer function
Y 2+ s = 2 X s + 3s + 8
Block diagram
X
2+ s s 2 + 3s + 8
Y
chapter15 Laplace transform(英文版课件 拉普拉斯变换)
closed. Calculate iK(t) for t>0.
＋ uC(－t)
1H iL(t)
1F
5A
2Ω
2Ω
iK(t)
S(t=0)
3uC(t) (a)
Solution: t=0-，in Fig. (b).
＋uC(0－-) iL(0-)
5A
2Ω
2Ω
(b)
uC(0-)=2×5+2iL(0-) iL(0-)=3uC(0-)RFra biblioteksC 1
1
sC
1
IS (s) 1 1
IS(s) sC
C
(b)
s 1
RC
u(t)
L1[U (s)]
1
1t
e RC (t)V
C
R＋
U(s)
－
Example: The switch in the circuit of Fig. (a) has
been opened for a long time. At t=0, the switch is
Example: R 500 k, C 1F, iS (t) 2et A
If uC(0-)=0, find uC(t).
iC
Solution:
If iS (t) (t)
iS(t)
+
RC
uC
h(t) uC (t)
1 C
t
e RC
10 6 e2t
t
uC (t) 0 iS (t )h()d
denotes convolution.
Equation (1) or (2) states that the output is equal to the input convolved with the unit impulse response.
高等数学英文教材书名
高等数学英文教材书名Title: Advanced Calculus - An English TextbookIntroduction:Advanced Calculus is a comprehensive English textbook designed to provide a solid foundation in the principles and applications of higher-level mathematics. This textbook aims to cater to the needs of undergraduate and graduate students studying mathematics, engineering, and other related fields. Its well-structured content, precise explanations, and engaging examples make it an indispensable resource for anyone seeking to deepen their understanding of advanced calculus.Chapter 1: Limits and ContinuityIn this chapter, readers are introduced to the fundamental concepts of limits and continuity. Starting with an overview of the concept of a limit, the textbook gradually delves into the intricacies of one-sided and infinite limits. Through clear explanations and illustrative examples, students will develop a solid understanding of continuity and its various properties, such as the intermediate value theorem and the extreme value theorem.Chapter 2: DifferentiationThis chapter explores the concept of differentiation - a crucial tool in calculus. The textbook provides a step-by-step approach to calculating derivatives of various functions, including polynomials, exponential and logarithmic functions, and trigonometric functions. Additionally, the concept of higher-order derivatives and their applications are extensively covered,allowing students to grasp the intricacies of curve sketching and optimization problems.Chapter 3: IntegrationThe textbook presents integration as a powerful technique for finding areas, volumes, and cumulative sums. Starting with the definite integral and its properties, the chapter progresses to explore various integration methods, such as substitution, integration by parts, and partial fractions. Furthermore, the textbook covers improper integrals and numerical integration methods, equipping students with the necessary tools to solve a wide range of mathematical problems.Chapter 4: Sequences and SeriesThis chapter focuses on the concepts of sequences and series, which play a fundamental role in advanced calculus. Beginning with an introduction to arithmetic and geometric sequences, the textbook moves on to explore the convergence and divergence of sequences. Furthermore, different types of series, including geometric, Taylor, and Fourier series, are presented, along with their convergence properties and applications.Chapter 5: Multivariable CalculusIn this chapter, readers are introduced to the complexities of multivariable calculus. The textbook covers topics such as partial derivatives, directional derivatives, multiple integrals, and vector calculus. Through practical examples and applications, students will develop a deep understanding of how calculus applies to functions of multiple variables, preparing them for more advanced mathematical concepts.Chapter 6: Differential EquationsThe final chapter of this textbook focuses on differential equations, which are essential in modeling real-world phenomena. Starting with first-order differential equations, the textbook progressively introduces higher-order differential equations, separable equations, and systems of linear differential equations. Different solution methods, such as integrating factors and Laplace transforms, are also presented, allowing students to solve a wide range of differential equation problems.Conclusion:Advanced Calculus - An English Textbook provides a comprehensive and rigorous study of advanced mathematical concepts, serving as an invaluable resource for students pursuing mathematics and related disciplines. With its clear explanations, numerous examples, and well-organized content, this textbook equips students with the necessary tools to tackle complex mathematical problems and fosters a deeper understanding of the principles underlying advanced calculus.。
常见函数的拉普拉斯变换
s2 (s2 − a2 )2
s3 (s2 − a2 )2
s2 (s2 − a2)3/2
1 (s2 + a2 )3
s (s2 + a2 )3
s2 (s2 + a2 )3
s3 (s2 + a2 )3
F(t)
ebt cosh at
ebt − eat b− a
bebt − aeat b− a
sin at − at cos at 2a3ቤተ መጻሕፍቲ ባይዱ
1 s−a +b
F(t)
eat /2 ⎧
3a
⎨cos ⎩
3at 2
+
3 sin
3at 2
−
e−3at
/
2
⎫ ⎬
⎭
1 3
⎛ ⎝⎜e
−
at
+
2eat /2
cos
3at⎞ 2 ⎠⎟
e−at / 2 3a2
⎧ ⎨e3at /2 ⎩
−
cos
3at 2
−
3 sin
3at ⎫
2
⎬ ⎭
e−at /2 ⎧
3a
⎨ ⎩
3 sin
F(n)(t)
33.10.
f ′(s)
–tF(t)
33.11. 33.12. 33.13.
33.14.
f ′′(s)
f (n)(s)
f (s) s
f (s) sn
t2F(t)
(–1)nt nF(t)
t
∫0 F(u)du
∫ ∫ ∫ t 0
t
F(u)dun =
0
t (t − u)n−1 0 (n − 1)! F(u)du
信号与系统奥本海姆英文版课后答案chapter9
s1 = −3
1 3 s2 = − + j 2 2
1 3 s3 = − − j 2 2
Based on the locations of the locations of these poles , we my choose form the following regions of convergence: (i) Re{s}>- 1
( s − 1)( s + 1)
s −1
Clearly ,X(s) has no zero in the finite s-plane .Since in X(s) the of the denominator polynomial exceeds the order the numerator polynomial by 1,X(s) has a zero at ∞ .therefore X(s) has no zero in the finite s-plane and one zero at infinity. (c) The given Laplace transform may be written as
s+5 s+β
Ae( s +5)t0 s+5
β.
158
The ROC of X(s) is such that if s0 was in the ROC of X 1 ( s) , then - s0 will be in the ROC of X(s). Putting the two above equations together ,we have L −t → X(s)= X 1 ( − s ) =2 Re {s} <1 x(t)= x1 (-t) = e sin(2t )u ( −t ) ←⎯ ,
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1
Laplace Transform
L(f (t))= f (t)e-st dt
0
Example 1:
Chapter 3
a st a a L(a)= ae dt e 0 0 s s s 0
-st
L(e )= e e dt e
= s2F(s ) s f(0) f (0)
etc. for
d nf dt n
e-jt e jt L(cosωt) = L 2
= = 1 1 1 2 s jω s jω
e jt - e jt L(sin ωt) = L 2j ω = 2 s ω2
Note:
1 s jω s jω 2 s2 ω2 s2 ω2 s = 2 s ω2
3
Chapter 3
4
Table 3.1 Laplace Transforms for Various Time-Domain Functionsa
f(t) F(s)
12
f(t) F(s)
7
Example 3.1
Solve the ODE,
dy 5 4y 2 y 0 1 dt First, take L of both sides of (3-26), 2 5 sY s 1 4Y s s Rearrange, 5s 2 Y s s 5s 4 (3-26)
Multiply by s, set s = 0
α α 4 α α1 s 2 3 4 (s +1 )(s + 2 )(s + 3 ) s 0 s 1 s 2 s 3 s 0 4 2 α1 1 2 3 3
10
For a2, multiply by (s+1), set s=-1 (same procedure for a3, a4) 2 α2 2, α3 2, α4 3
Chapter 3
5
Table 3.1 Laplace Transforms for Various Time-Domain Functionsa
f(t) F(s)
Chapter 3
6
Table 3.1 Laplace Transforms for Various Time-Domain Functionsa (continued)
11
Laplace transforms can be used in process control for: 1. Solution of differential equations (linear)
Chapter 3
2. Analysis of linear control systems (frequency response) 3. Prediction of transient response for different inputs
Laplace Transforms
Chapter 3
1. Standard notation in dynamics and control (shorthand notation)
2. Converts mathematics to algebraic operations
3. Advantageous for block diagram analysis
You can use this method on any order of ODE, limited only by factoring of denominator polynomial (characteristic equation) Must use modified procedure for repeated roots, imaginary roots
Chapter 3
s(s3+6s 2+11s+6 )=s(s+ 1 )(s+2 )(s+3 )
Step 2b. Use partial fraction decomposition
α α α α 4 1 2 3 4 s(s +1 )(s + 2 )(s + 3 ) s s 1 s 2 s 3
-bt -bt -st 0 0
-(b+s)t
1 1 ( b s)t dt -e 0 s+b b+s
df L(f ) L dt
df -st e dt sL(f) f(0) 0 dt
Usually define
f(0) = 0
Chapter 3
Step 1
Take L.T. (note zero initial conditions) 4 3 2 s Y(s)+ 6s Y(s)+11sY(s) 6Y ( s ) = s
9
Rearranging,
Y(s)= 4 ( s 3 6s 2 11s 6) s
Step 2a. Factor denominator of Y(s)
Chapter 3
Step 3. Take inverse of L.T.
y(t)=
(Y(s)=
2 2 2 2/3 + ) 3s s 1 s 2 s3
2 2 2e t 2e 2t e 3t 3 3 2 t y(t) t 0 y (0) 0. (check original ODE) 3
Chapter 3
(3-34)
Take L-1,
y t L
1
From Table 3.1 (line 11),
5s 2 s 5s 4
(3-37)
8
y t 0.5 0.5e0.8t
Example:
d3y d2y dy 6 2 11 6 y 4 3 dt dt dt y( 0 )= y( 0 )= y ( 0 )= 0
(e.g., the error)
2
Other Transforms
d2 f df dφ L 2 = L w he re φ = dt dt dt sφ (s )- φ (0) = ss F(s )- f(0) f (0)
Chapter 3