ChewMA1506-14 Ch4
Short Description
asdf...
Description
MA1506 Mathematics II
Chapter 4 Laplace Transforms
Chew T S MA1506-14 Chapter 4
1
4.0 Introduction
4.0 Introduction In Chapter one, we have learnt how to solve nonhomogeneous 2nd order linear ODE (with or without initial conditions) by • Method of undetermined coefficients • Variation of parameters In this Chapter, we shall study how to use Laplace transforms to solve nonhomogeneous 2nd order linear ODE with initial conditions Chew T S MA1506-14 Chapter 4
2
4.0 Introduction
There are five parts in Chapter four. Part 1: Laplace transforms and inverse Laplace transforms Part 2: Finding Laplace transforms and inverse Laplace transforms using known results Part 3: Solving ODEs by Laplace transforms Part 4: Laplace transforms of piecewise continuous functions and ODEs Part 5: Laplace transforms of impulse functions and ODEs Chew T S MA1506-14 Chapter 4
3
Part 1 Laplace transforms and inverse Laplace transforms
4.1 Laplace transforms
Laplace transform L(f) is a mapping L which maps a function f(t) to a function F(s), where F(s) is given by
st e 0
L( f (t )) F ( s)
Chew T S MA1506-14 Chapter 4
f (t )dt 4
4.1 Laplace transform
Example
Let f(t)=1 for t>0. Then st e 0
L( f )
st e dt 0
f (t )dt
st t
e 1 s t 0 s
where s>0 Why? So Laplace transform maps constant function 1 to function F(s)=1/s, where s>0 Chew T S MA1506-14 Chapter 4
5
4.1 Laplace transform
where s>0 Why? ANS: In the above, if s0, then L(f)=1/s , s>0. Hence the inverse Laplace transform of 1/s is f(t)=1. We write
L (1/ s) 1 1
Chew T S MA1506-14 Chapter 4
7
4.3 Basic Laplace transforms
1 L (e ) ,s a sa at
L(t )
L (1/ ( s a)) e
s
at
1
n!
n
1
n 1
n=0,1,2…
n! n L n1 t s
Chew T S MA1506-14 Chapter 4
8
4.3 Basic Laplace transforms
L(sin t ) 2 s 2
L(cos t )
s s 2
2
L 2 sin t 2 s 1
1
s L 2 cos t 2 s
In the above, t>0, s>0 For the proofs, see appendix Chew T S MA1506-14 Chapter 4
9
4.4 Linearity We can apply L T term by term
L(af bg ) aL( f ) bL( g ) We can also apply inverse L T term by term 1
1
1
L (aF bG ) aL ( F ) bL (G) a, b are constants
Chew T S MA1506-14 Chapter 4
10
4.5 Operational properties
Let
L( f (t )) F (s) Then
L(tf (t )) F '( s) L(t f (t )) (1) F n
n
(n)
( s)
L e f (t ) F ( s a) at
where n is positive integer
called s-shifting
where s>a Chew T S MA1506-14 Chapter 4
11
Part 2: Finding Laplace transforms and inverse Laplace transforms using known results
4.6 Finding Laplace transforms
L(t f (t )) (1) F n
Example 1
L t sin 2t
d 2 4s ds 2 2 2 s 4 ( s 4)
n
(n)
L( f (t )) F (s) L(sin t ) 2 s 2
Example 2
L t sin t 2
d 2 1 2(1 3s 2 ) 2 2 ds s 1 ( s 2 1)3 Chew T S MA1506-14 Chapter 4
( s)
12
Example 3
1 L (e ) ,s a sa
4.6 Finding Laplace transforms
at
1 at 1 at 1 1 L(cosh at ) L (e e ) 2 2 s a s a
s L(cosh at ) 2 ,s a 0 2 s a Similarly,
1 at at a L sinh at L (e e ) 2 2 s a2 Chew T S MA1506-14 Chapter 4
13
Example 4
2
4.6 Finding Laplace transforms
Find L(sin t ) L(cos t )
1 L(sin t ) L (1 cos 2t ) 2 2
s s2 2
L(1) 1/ s
1 1 1 s L(1) L(cos 2t ) 2 2 2 s s 4 2 s ( s 2 4)
Chew T S MA1506-14 Chapter 4
14
4.6 Finding Laplace transforms
Example 5 s-shifting Recall
L(t )
n!
n
s
L e f (t ) F ( s a)
n 1
at
Hence
L (e t ) ct n
n! ( s c)
n 1
Chew T S MA1506-14 Chapter 4
15
4.6 Finding Laplace transforms
Example 5 s-shifting Recall
L(sin t ) 2 2 s L(cos t ) Hence
s s2 2
L(e sin t ) ct
L e f (t ) F ( s a) at
( s c) sc ct L(e cos t ) 2 2 ( s c) 2
Chew T S MA1506-14 Chapter 4
2
16
4.6 Finding Laplace transforms
• The Laplace transform is named for the French mathematician Laplace, who studied this transform in 1782. Pierre-Simon de Laplace (1749-1827)
• The techniques described in this chapter were developed primarily by Oliver Heaviside (1850-1925), an English electrical engineer.
Oliver Heaviside (1850-1925)
Chew T S MA1506-14 Chapter 4
17
4.6 Finding Laplace transforms
Engineer Oliver Heaviside used Laplace transforms in the nineteenth century to solve ODE. However the mathematical community would not accept his work because he failed to justify his Methods. His reply: “ Should I refuse my dinner because I do not understand the process of digestion”. Heaviside’s method (using Laplace transforms to solve ODE) finally has been justified in the twentieth century. Chew T S MA1506-14 Chapter 4
18
4.7 Finding inverse Laplace transforms Recall
1 L (e ) sa at
L(sin t ) 2 s 2
s L cosh t 2 2 s
L(t )
n!
n
s
n 1
L(cos t )
s s2 2
L sinh t 2 2 s
Chew T S MA1506-14 Chapter 4
19
4.7 Finding inverse Laplace transforms
Recall
L (e t )
n!
ct n
( s c)
L(e sin t ) ct
n 1
( s c) sc ct L(e cos t ) 2 2 ( s c) 2
2
1 L(sin t ) L (1 cos 2t ) 2 2
Chew T S MA1506-14 Chapter 4
20
4.7 Finding inverse Laplace transforms
Example 1 From
L
L
L(t )
n!
n
s
n 1
get
1 3
! 3 4 t s
8 1 2 ! 1 2 ! 2 L 4 4 L 4 t 3 3 3 s s s
1
Chew T S MA1506-14 Chapter 4
21
4.7 Finding inverse Laplace transforms
Example 2 From
L(sin t ) 2 s 2
s L cosh t 2 s 2
s L(cos t ) 2 s 2 L sinh t 2 s 2
get 1 4 s 1 1 s L 2 4L 2
s 9
1 1 1 3 L 2 4cos3t sin 3t 3 s 9 3 s 9
s 1 1 3 1 4s 1 1 L 2 4L 2 L 2 4cosh 3t sinh 3t 3 s 9 s 9 3 s 9 1
Chew T S MA1506-14 Chapter 4
22
4.7 Finding inverse Laplace transforms
Example 3
L (e t )
n!
ct n
Let
( s c)
n 1
2! 10 10 2! Y ( s) 5 3 3 3 2! s 1 s 1 s 1 Then
2! t 2 L Y ( s ) 5L 5e t 3 s 1 1
1
Chew T S MA1506-14 Chapter 4
23
4.7 Finding inverse Laplace transforms
Example 4
First
2s 13 L 2 s 5s 6 2s 13 2s 13 2 s 5s 6 ( s 3)( s 2) Find
1
2s 13 A B s 3s 2 s 3 s 2 2s 13 As 2 Bs 3 2s 13 ( A B) s (2 A 3B) A B 2, 2 A 3B 13 A 7, B 9 Chew T S MA1506-14 Chapter 4
24
4.7 Finding inverse Laplace transforms
2s 13 7 9 ( s 3)( s 2) s 3 s 2 2s 13 7 9 1 L L ( s 3)( s 2) s 3 s 2 1
1 1 1 (7) L 9L s 3 s 2 1
(7)e
3t
9e
2 t
Chew T S MA1506-14 Chapter 4
Recall
1 L (e ) sa at
25
4.7 Finding inverse Laplace transforms
Example 5 Using partial fraction
2s s 8s 6 As B Cs D Y ( s) 2 2 2 2 ( s 1)(s 4) s 1 s 4 3
2
we obtain A = 2, B = 5/3, C = 0, and D = -2/3. Hence
2s 5/3 2/3 Y (s) 2 2 2 s 1 s 1 s 4 5 1 L (Y ( s)) 2cos t sin t sin 2t 3 3 1
Recall
L(sin t ) 2 s 2
s L(cos t ) 2 2 s Chew T S MA1506-14 Chapter 4
26
1 Example 6 Find 𝑠 𝑠+1 1 𝐴 𝐵 𝐶 = + + 2 𝑠 𝑠+1 𝑠 𝑠+1 𝑠+1 2 𝐿−1
2
𝐴 = 1, 𝐵 = 𝐶 = −1
1 𝑠 𝑠+1 −1
𝐿
2
1 1 1 = − − 𝑠 𝑠+1 𝑠+1
1 𝑠 𝑠+1 2
2
=
Chew T S MA1506-14 Chapter 4
27
4.7 Finding inverse Laplace transforms
Example 7
2s L( y ) ( s 1) 2 22
Find y
In order to use L(e sin t ) ct
( s c) 2
L(e cos t ) ct
2
sc ( s c) 2
2( s 1) 2 write L( y ) 2 2 ( s 1) 2 ( s 1) 2 22
So
y(t ) 2e cos 2t e sin 2t t
Chew T S MA1506-14 Chapter 4
t
28
2
4.7 Finding inverse Laplace transforms
Example 8 Find the inverse transform of
s 1 G( s) 2 s 2s 5
Note: We can’t factorize the denominator s sc ct Need to use L(e cos t ) ( s c) 2 2
2
2s 5
We first complete the square
s 1 G(s) 2 s 2s 5 s 2 2s 1 4 s 12 4 s 1
1
L
s 1
G ( s ) e
t
cos 2t
Chew T S MA1506-14 Chapter 4
29
Finding Laplace transform and inverse Laplace transform on line http://wims.unice.fr/wims/wims.cgi http://www.wolframalpha.com/widget s/view.jsp?id=3272010f63d3145699c a78bbe0db05a7
Solving ODE by Laplace transform animation http://www.atp.ruhr-unibochum.de/DynLAB/dynlab modules/Examples/Laplace transform/Rickeracke.html Chew T S MA1506-14 Chapter 4
30
Part 3 Solving ODE by Laplace transforms
4.8 Transform of derivatives
L( f ') sL( f ) f (0) L( f '') s 2 L( f ) sf (0) f '(0) L( f
(n)
) s L( f ) s n
n 1
f (0) s
n2
f '(0)
f
( n 1)
(0)
Note that after transform, derivatives disappear. So Laplace transform can be used to solve ODE and PDE. However, initial conditions should be given For the proofs of the above results, see appendix Chew T S MA1506-14 Chapter 4
31
4.9 Solving ODE Example 1
y '' y e
Take Laplace Transform
2t
y(0) 0, y '(0) 1
L( y '' y ) L(e ) 2t
L( y '') L( y ) L(e ) 2t
s L( y ) sy (0) 2
1 y '(0) L( y) s 2
1 1 s 1 L( y ) 2 1 2 s 1 s 2 ( s 2)( s 1) Chew T S MA1506-14 Chapter 4
32
1 1 1 s 3 L( y ) 2 5 s 2 5 ( s 1)
4.9 Solving ODE
We shall use
1 s L(sin t ) 2 L (e ) L(cos t ) 2 2 s sa s 2 1 1 1 s 1 3 Write L( y ) 2 5 s 2 5 ( s 1) 5 ( s 2 1) at
Remark: using Method of 1 2t 1 3 undetermined y (t ) e cos t sin t Coefficients is 5 5 5 Chew T S MA1506-14 Chapter 4 easier 33
Take inverse transform, get
Example 2 ANS:
y '' 2 y ' 5 y 0 y(0) 2, y '(0) 4 t t
4.9 Solving ODE
y (t ) 2e cos 2t e sin 2t
Example 3
y '' 2 y ' y et t y(0) 1, y '(0) 0 2
ANS:
t t t y (t ) e e t 2 2
We remark that these two examples are given in Lecture Note, They are solved by using Laplace Transforms. In fact it is easier if using method of undetermined coefficients. Chew T S MA1506-14 Chapter 4
34
4.9 Solving ODE
When we use Laplace transforms to solve Nonhomogeneous 2nd order linear ODE?
If those ODE which can be solved by method of undetermined coefficients or variation of parameters, then we will not use Laplace transforms to solve them We need to use Laplace Transforms to solve those examples in next two sections, since we have difficulty to use the above two methods there Chew T S MA1506-14 Chapter 4
35
Part 4 Laplace transforms of piecewise continuous functions and ODEs
4.10 Piecewise Continuous Functions Jump discontinuity at a
lim f (t ) l finite value
t a
lim f (t ) l finite value
t a
Chew T S MA1506-14 Chapter 4
36
4.10 Piecewise Continuous Functions
piecewise continuous
Not piecewise cont since lim f (t ) t a
a Chew T S MA1506-14 Chapter 4
37
4.11 Unit Step (Heaviside) Function
u (t a) 0 t a u (t a) 1 t a
1
● ○ a
From now onwards, we will not draw ● ○
Chew T S MA1506-14 Chapter 4
38
4.11 Unit Step (Heaviside) Function
f (t ) u (t 1) u (t 3)
Example 1
u(t-1) 1
3 u(t-3)
1
3
f (t ) u (t 1) u (t 3)
1 Chew T S MA1506-14 Chapter 4
3 39
4.11 Unit Step (Heaviside) Function
Example 2
f (t ) u (t a) u (t b)
1 a
b
Chew T S MA1506-14 Chapter 4
40
4.11 Unit Step (Heaviside) Function
Example 3
Let g(t) be a function of t. Then
u(t-a)-u(t-b)
a
g(t)
b a Chew T S MA1506-14 Chapter 4
b 41
4.11 Unit Step (Heaviside) Function
Example 4
Piecewise cont. function in terms of u(t)
1
2
3
Chew T S MA1506-14 Chapter 4
4
42
4.12 t-Shifting
Suppose L( f (t )) F ( s)
Then
L( f (t a)u (t a)) e
as
F (s)
f (t a)u (t a) f (t )
a
a Chew T S MA1506-14 Chapter 4
43
4.12 t-Shifting
Suppose L( f (t )) F ( s)
Then
L( f (t a)u (t a)) e Hence
L(u (t a))
e
as
Recall
s
as
F (s)
1 L(1) s
u (t a) is also denoted by ua (t ) Chew T S MA1506-14 Chapter 4
44
4.12 t-Shifting
Example 1
L( f (t a)u (t a)) e
Use
to find
L t u (t 1) 2
as
L t u (t 1) L (t 1 1) u (t 1) 2
2
F (s)
L (t 1)2 u (t 1) L 2(t 1)u (t 1) L u (t 1)
Chew T S MA1506-14 Chapter 4
45
4.12 t-Shifting
Example 1
L (t 1) u (t 1) 2 L (t 1)u (t 1) L u (t 1) 2
s
2 2 1 e 3 2 s s s Recall
L( f (t a)u (t a)) e
2 2 L(t ) 3 s
1 L(t ) 2 s
as
F (s)
L(u (t a))
Chew T S MA1506-14 Chapter 4
e
as
s 46
4.12 t-Shifting
Example 2
L( f (t a)u (t a)) e
Use
to find
as
F (s)
L((e 1)u (t 2)) t
L((e 1)u (t 2)) L((et 2e2 1)u (t 2)) t
t 2
e L (e 2
u (t 2)) L(u (t 2))
Chew T S MA1506-14 Chapter 4
47
4.12 t-Shifting
Example 2
e L(e u (t 2)) L(u (t 2)) 2
t 2
2 2 s 1 2 s 1 e e e s 1 s Recall
L( f (t a)u (t a)) e
1 t L (e ) s 1
as
L(u (t a)) Chew T S MA1506-14 Chapter 4
e
F (s) as
s 48
4.13 t-Shifting and ODE
4.13 t-Shifting and ODE
Example
where
L( y '') L(3 y ') L(2 y) L( g )
s 2 L( y ) sy (0) y '(0) 3( sL( y) y(0)) 2 L( y) L( g )
( s 3s 2) L( y) 1 L( g ) 2
Chew T S MA1506-14 Chapter 4
49
4.13 t-Shifting and ODE
Now compute
L( g )
Recall
So
g (t ) u (t ) u (t 1) L( g (t )) L(u (t )) L(u (t 1))
s
1 e L( g (t )) s s
Chew T S MA1506-14 Chapter 4
50
4.13 t-Shifting and ODE
s
1 e ( s 3s 2) L( y) 1 L( g (t )) s s 2
1 1 s L( y ) e s( s 2) s ( s 1)( s 2) 1
1 1 s 1 yL L e s( s 1)( s 2) s( s 2) Now find
1 1 and 1 s L L e s ( s 2) s ( s 1)( s 2) 1
Chew T S MA1506-14 Chapter 4
51
4.13 t-Shifting and ODE
1 1 1 1 1 L L s ( s 2) 2 s s 2 1
1 1 1 1 1 1 L L 2 s 2 s 2
1 2t (1 e ) 2
Chew T S MA1506-14 Chapter 4
52
To find
1 s
L e
1 s ( s 1)( s 2)
4.13 t-Shifting and ODE
Recall
L( f (t a)u (t a)) e So
as
F (s)
1 F (s) Now need to find f s( s 1)( s 2)
i.e. to find
1 L s ( s 1)( s 2) 1
Chew T S MA1506-14 Chapter 4
53
4.13 t-Shifting and ODE
1 11 1 1 s ( s 1)( s 2) 2 s s 2 s 1
1 1 2t t L (1 e ) e s ( s 1)( s 2) 2 1
Recall
L( f (t a)u (t a)) e
as
F (s)
1 1 2t t F (s) s( s 1)( s 2) So f (t ) 2 (1 e ) e Chew T S MA1506-14 Chapter 4
54
1 s
L e
4.13 t-Shifting and ODE
1 s ( s 1)( s 2)
f (t 1)u (t 1)
1 from f (t ) (1 e2t ) et get 2
1 2(t 1) (t 1) (1 e )e 2
u (t 1)
1 1 1 s L e Hence y L s( s 2) s( s 1)( s 2) 1
1 1 2t 2(t 1) (t 1) (1 e ) (1 e )e u ( t 1) 2 2 Chew T S MA1506-14 Chapter 4
55
Part 5: Laplace transforms of impulse functions and ODEs
• In some applications, it is necessary to deal with an impulsive force. • For example, a mechanical system subject to a sudden large force g(t) over a short time interval about t0. • The differential equation will then have the form
Chew T S MA1506-14 Chapter 4
56
(Cont.)
ay by cy g (t ),
where big , t0 t t0 g (t ) otherwise 0, and 0 is small.
Chew T S MA1506-14 Chapter 4
57
4.14 Measuring Impulse
In a mechanical system, where g(t) is a force, the total impulse of this force is measured by the integral
t0
t0
I ( ) g (t )dt
Chew T S MA1506-14 Chapter 4
g (t )dt 58
4.14 Measuring Impulse
Note that if g(t) has the form
c, t0 t t0 g (t ) 0, otherwise then
t0
t0
I ( ) g (t )dt
g (t )dt 2 c, 0
In particular, if c = 1/(2), then I ( ) 1 (independent of ). Chew T S MA1506-14 Chapter 4
59
4.14 Measuring Impulse
Unit Impulse Function Suppose the forcing function d(t) has the form
1 2 , t d (t ) otherwise 0,
Then as we have seen, I ( ) 1
Chew T S MA1506-14 Chapter 4
60
4.14 Measuring Impulse
We are interested in d(t) acting over shorter and shorter time intervals (i.e., 0). See graph on right. Note that d(t) gets taller and narrower as 0. Thus for t 0, we have
lim d (t ) 0, and lim I ( ) 1 0
0
lim d (0) 0
Chew T S MA1506-14 Chapter 4
t≠0 why? 61
4.14 Measuring Impulse
Dirac Delta Function Thus for t 0, we have
lim d (t ) 0, and lim I ( ) 1 0
0
The unit impulse function is defined to have the properties
(t ) 0 for t 0, and
Chew T S MA1506-14 Chapter 4
(t )dt 1 62
4.14 Measuring Impulse
The unit impulse function is an example of a generalized function and is usually called the Dirac delta function. In general, for a unit impulse at an arbitrary point t0,
(t t0 ) 0 for t t0 , and
Chew T S MA1506-14 Chapter 4
(t t0 )dt 1 63
4.15 Laplace Transform of
The Laplace Transform of is defined by
L (t t0 ) lim L d ( t t ) 0 0 We can proved that
L (t t0 ) e
st0
For example, we have L( (t 10)) e
10 s
L( (t )) 1 Chew T S MA1506-14 Chapter 4
64
4.15 Laplace Transform of
Paul A. M. Dirac (1902-1984), an English physicist, received the Nobel prize at age 31 for his work in quantum theory When Dirac introduced his delta functions in early 1930s, again mathematical community was somewhat suspect Finally his delta functions have been Justified. Today, delta functions have a solid place.
Chew T S MA1506-14 Chapter 4
65
4.16 impulse functions and ODE Example 1 Solve
2 y y 2 y (t 7), y(0) 0, y(0) 0
Apply L T, get
2 L( y) L( y) 2 L( y) L( (t 7)) 2s 2 L( y ) 2sy (0) 2 y(0) sL( y ) y (0) 2 L( y ) e 7 s
L (t t0 ) e Chew T S MA1506-14 Chapter 4
st0 66
4.16 impulse functions and ODE
2s 2 L( y ) 2sy (0) 2 y(0) sL( y ) y (0) 2 L( y ) e 7 s Substituting in the initial conditions, we obtain
2s
2
s 2 L( y ) e
L( y )
e
7 s
7 s
2s s 2 2
Chew T S MA1506-14 Chapter 4
67
4.16 impulse functions and ODE
7 s
e L( y ) 2 2s s 2 Recall
L( f (t a)u (t a)) e Then F ( s )
as
F (s)
1 2s 2 s 2
2 15 / 4 2 15 s 1/ 4 15/16 Chew T S MA1506-14 Chapter 4
68
4.16 impulse functions and ODE
2 15 / 4 F ( s) 15 s 1/ 4 2 15/16 By
L(e sin t )
ct
( s c)
2 f (t ) 15
2
2
1 t e 4 sin
Chew T S MA1506-14 Chapter 4
get
15 t 4 69
4.16 impulse functions and ODE
From above we have 7 s
e L( y ) 2 2s s 2
L( f (t a)u (t a)) e 2 f (t ) 15
1 t e 4 sin
as
F (s)
15 t 4
So
2 t 7 / 4 15 y (t ) e sin t 7 u (t 7) 4 15 Chew T S MA1506-14 Chapter 4
70
4.16 impulse functions and ODE
Example 2
A mass m=1 is attached to a spring with constant k=4 and released from rest 3 units below its equilibrium position. Hence we have
x " 4 x 0, x(0) 3, x '(0) 0 equilibrium position Chew T S MA1506-14 Chapter 4
x 71
4.16 impulse functions and ODE
Example 2
Now at the instant t=2π the mass is struck with a Hammer in the downward direction, exerting an impulse of 8 units 8 (t 2 ) on the mass. So we have x " 4 x 8 2 (t ) 8 (t 2 ), x(0) 3, x '(0) 0
Chew T S MA1506-14 Chapter 4
72
4.16 impulse functions and ODE
Solution
Apply L T to the above ODE, get
L( x ") 4 L( x) 8L( 2 (t )), x(0) 3, x '(0) 0
s L( x) 3s 4 L( x) 8e 2
2 s
2 s
3s 8e L( x) 2 2 s 4 s 4
L (t t0 ) e st0
2 s 3 s 8 e 1 1 x(t ) L 2 L 2 s 4 s 4 Chew T S MA1506-14 Chapter 4
73
4.16 impulse functions and ODE
Solution
2 s 3 s 8 e 1 1 x(t ) L 2 L 2 s 4 s 4
x(t ) 3cos 2t 4u(t 2 )sin 2(t 2 )
Chew T S MA1506-14 Chapter 4
74
Example 3 : Injections
It is known that the body removes morphine from the bloodstream at the rate proportional to the amount of morphine present y(t) where unit of t is day Hence So
dy ky dt
y (t )
y (0) Ae
k 0
kt Let t=0, get Ae So
y (t ) y (0)e
Chew T S MA1506-14 Chapter 4
kt 75
y (t ) y (0)e
kt
Example 3 : Injections
Now shall find k It is given that half-life of morphine = 18 hours = 0.75 days
i.e.,
(1/ 2) y(0) y(0.75) y(0)e
1/ 2 e
k (0.75)
So the soln is and the ODE is
k (0.75)
ln 2 k 0.924 0.75
y (t ) y (0)e
(0.924) t
dy (0.924) y dt
Chew T S MA1506-14 Chapter 4
76
Example 3 : Injections
Assume that no morphine was in the body before t=0
Now at t=0, 100mg morphine was injected almost instantly (impulsive injection) into a patient, dy 100 can be written as 100 (t ) dt t 0
at t=1, another 100mg was injected Hence we have
100 (t 1) Chew T S MA1506-14 Chapter 4
77
Example 3 : Injections
dy (0.924) y 100 (t ) 100 (t 1) dt Here y(0)=0
sL( y) y(0) 0.924 L( y) 100 L( (t )) 100 L( (t 1))
sL( y ) y (0) 0.924 L( y) 100 100e
sL( y ) 0.924 L( y ) 100 100e Chew T S MA1506-14 Chapter 4
s
s 78
s
Example 3 : Injections
100 100e L( y ) s 0.924 s 0.924
y 100e 2nd
0.924t
method
100e
0.924(t 1)
u (t 1)
dy (0.924) y 100 (t 1) y (0) 100 dt
y 200
150
100
50
0 0
1
2
3
4
5
6
Chew T S MA1506-14 Chapter 4
7
8
9
x
79
4.17 Parameter Reconstruction Sometimes it happens that you have a system whose nature you understand (e.g it is a damped harmonic oscillator), but you don’t know the values of the parameters (e.g the spring constant, the mass, the friction coefficient) In such case, what you can do is to poke the system with a sudden, sharp force and watch how it behaves For example, if it is a damped harmonic oscillator , after poking,at t=1, we observe the displacement of the mass, say Chew T S MA1506-14 Chapter 4
80
4.17 Parameter Reconstruction
x(t ) u (t 1)e(t 1) sin(t 1) How to get the above equation? ANS: We can use Graphmatica to guess the equation of the curve
Now with the above equation (soln) , we can find the mass, the spring constant, the friction coefficient as follows Chew T S MA1506-14 Chapter 4
81
4.17 Parameter Reconstruction
It is known that
mx bx kx (t 1)
Apply L T, get
m( s L( x) sx(0) x(0)) 2
b( sL( x) x(0)) kL( x) e
L( x)
e
s
s
ms bs k 2
Chew T S MA1506-14 Chapter 4
82
4.17 Parameter Reconstruction
On the other hand, the L T of
x(t ) u (t 1)e is
L( x)
e
sin(t 1)
s
( s 1) 1 2
Compare with
L( x)
(t 1)
e
s 2s 2 2
s
ms bs k 2
e
s
We have
L( f (t a)u (t a)) e Chew T S MA1506-14 Chapter 4
as
F (s) 83
4.17 Parameter Reconstruction
e s ms bs k 2
e s s 2s 2 2
ms bs k s 2s 2 2
2
Hence
m 1, b 2, k 2
Chew T S MA1506-14 Chapter 4
84
APPENDIX Laplace transforms
Chew T S MA1506-14 Chapter 4
85
APPENDIX
Example
Chew T S MA1506-14 Chapter 4
86
APPENDIX
Example Set a = iw
Chew T S MA1506-14 Chapter 4
87
APPENDIX
Example
Chew T S MA1506-14 Chapter 4
88
APPENDIX
Transform of Derivatives and Integrals
Pf
Chew T S MA1506-14 Chapter 4
89
APPENDIX
Transform of Derivatives and Integrals
Chew T S MA1506-14 Chapter 4
90
APPENDIX
Transform of Derivatives and Integrals
Chew T S MA1506-14 Chapter 4
91
APPENDIX
t-Shifting
If L(f(t)) = F(s) then
Chew T S MA1506-14 Chapter 4
92
APPENDIX
Additional Operational properties
L
t
0
1 f ( x)dx F ( s) s
f (t ) L( ) F ( x)dx s t
1 sinh t L s L(sinh t )dx s 2 dx x 1 t
1 1 1 1 x 1 1 s 1 1 s 1 0 ln ln dx ln s 2 x 1 x 1 2 x 1 s 2 s 1 2 s 1
Apply L’ Hopital’s rule to
1 1 L sin xdx L(sin t ) 0 s s (s 2 2 ) t
Chew T S MA1506-14 Chapter 4
form
93
APPENDIX
Laplace Transform of
The Laplace Transform of is defined by
L (t t0 ) lim L d ( t t ) 0 0 L (t t0 ) lim 0
st t0
e lim 0 2s
t0
0
1 e d (t t0 )dt lim 0 2 st
t0 t0
st
e dt
1 s t0 s t0 e lim e 0 2s Chew T S MA1506-14 Chapter 4
94
APPENDIX
Cont.
e e e lim 0 s 2 st0
s
s
sinh( s ) st0 e lim 0 s o/o form
e
st0
s cosh( s ) st0 lim e 0 s Chew T S MA1506-14 Chapter 4
95
APPENDIX
Product of Continuous Functions and The product of the delta function and a continuous function f can be integrated, using the mean value theorem for integrals: (t t0 ) f (t )dt lim d (t t0 ) f (t )dt
0
1 t0 lim f (t )dt 0 2 t 0 1 lim 2 f (t*) ( where t0 t* t0 ) 0 2 lim f (t*) 0
f (t0 ) Chew T S MA1506-14 Chapter 4
96
APPENDIX
Cont.
(t t0 ) f (t )dt f (t0 )
Thus
(t t0 )sin tdt sin(t0 ) (t t0 )cos tdt cos(t0 ) End Chapter 4 Chew T S MA1506-14 Chapter 4
97
View more...
Comments
