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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 sa at

L(t ) 

L (1/ ( s  a))  e

s

at

1 

n!

n

1

n 1

n=0,1,2…

n!  n L  n1   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 sa

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)   sc 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 )  sa 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)   sc 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  3s  2 s  3 s  2 2s  13  As  2  Bs  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

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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 )  sa 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

sc ( 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 sc 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  12  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

n2

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  sa 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  yL   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 11 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  e2t )  et 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

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