ch 6.1: definition of laplace transform - dept of maths, nusmatysh/ma3220/chap6.pdf · ch 6.1:...

Ch 6.1: Definition of Laplace TransformMany practical engineering problems involve mechanical or electrical systems acted upon by discontinuous or impulsive forcing terms. For such problems the methods described in Chapter 3 are difficult to apply. In this chapter we use the Laplace transform to convert a problem for an unknown function f into a simpler problem for F, solve for F, and then recover f from its transform F. Given a known function K(s,t), an integral transform of a function f is a relation of the form

∞≤<≤∞= ∫ βαβ

α,)(),()( dttftsKsF

The Laplace TransformLet f be a function defined for t ≥ 0, and satisfies certain conditions to be named later. The Laplace Transform of f is defined as

Thus the kernel function is K(s,t) = e-st. Since solutions of linear differential equations with constant coefficients are based on the exponential function, the Laplace transform is particularly useful for such equations. Note that the Laplace Transform is defined by an improper integral, and thus must be checked for convergence. On the next few slides, we review examples of improper integrals and piecewise continuous functions.

{ } ∫∞ −==

0)()()( dttfesFtfL st

Example 1Consider the following improper integral.

We can evaluate this integral as follows:

Note that if s = 0, then est = 1. Thus the following two cases hold:

( )1lim1limlim0

00−===

∞→∞→∞→

∞

∫∫ sb

b

bst

b

b st

b

st ess

edtedte

∫∞

0dtest

0. if,diverges

and0; if,1

0

0

≥

<−=

∫

∫∞

∞

sdte

ss

dte

st

st

Example 2Consider the following improper integral.

We can evaluate this integral using integration by parts:

Since this limit diverges, so does the original integral.

[ ]( )[ ]1cossinlim

cossinlim

sinsinlim

coslimcos

00

00

00

−+=

+=

⎥⎦⎤

⎢⎣⎡ −=

=

∞→

∞→

∞→

∞→

∞

∫

∫∫

bsbsb

tstst

tdtstst

tdtsttdtst

b

bb

b

bb

b

b

b

∫∞

0cos tdtst

Piecewise Continuous FunctionsA function f is piecewise continuous on an interval [a, b] if this interval can be partitioned by a finite number of pointsa = t0 < t1 < … < tn = b such that

(1) f is continuous on each (tk, tk+1)

In other words, f is piecewise continuous on [a, b] if it is continuous there except for a finite number of jump discontinuities.

nktf

nktf

k

k

tt

tt

,,1,)(lim)3(

1,,0,)(lim)2(

1

K

K

=∞<

−=∞<

−+

+

→

→

Example 3Consider the following piecewise-defined function f.

From this definition of f, and from the graph of f below, we see that f is piecewise continuous on [0, 3].

⎪⎩

⎪⎨

⎧

≤<+≤<−≤≤

=32121,310,

)(

2

tttttt

tf

Example 4Consider the following piecewise-defined function f.

From this definition of f, and from the graph of f below, we see that f is not piecewise continuous on [0, 3].

( )⎪⎩

⎪⎨

⎧

≤<≤<−≤≤+

= −

32,421,210,1

)( 1

2

ttttt

tf

Theorem 6.1.2Suppose that f is a function for which the following hold:(1) f is piecewise continuous on [0, b] for all b > 0. (2) | f(t) | ≤ Keat when t ≥ M, for constants a, K, M, with K, M > 0.

Then the Laplace Transform of f exists for s > a.

Note: A function f that satisfies the conditions specified above is said to to have exponential order as t →∞.

{ } finite )()()(0∫∞ −== dttfesFtfL st

Example 5Let f (t) = 1 for t ≥ 0. Then the Laplace transform F(s) of f is:

{ }

0,1

lim

lim

1

0

0

0

>=

−=

=

=

−

∞→

−

∞→

∞ −

∫

∫

ss

se

dte

dteL

bst

b

b st

b

st

Example 6Let f (t) = eat for t ≥ 0. Then the Laplace transform F(s) of f is:

{ }

asas

ase

dte

dteeeL

btas

b

b tas

b

atstat

>−

=

−−=

=

=

−−

∞→

−−

∞→

∞ −

∫

∫

,1

lim

lim

0

)(

0

)(

0

Example 7Let f (t) = sin(at) for t ≥ 0. Using integration by parts twice, the Laplace transform F(s) of f is found as follows:

{ }

0,)()(1

sin/)sin(lim1

coslim1

cos/)cos(lim

sinlimsin)sin()(

222

2

00

0

00

00

>+

=⇒−=

⎥⎦⎤

⎢⎣⎡ +−=

⎥⎦⎤

⎢⎣⎡−=

⎥⎦⎤

⎢⎣⎡ −−=

===

∫

∫

∫

∫∫

−−

∞→

−

∞→

−−

∞→

−

∞→

∞ −

sas

asFsFas

a

ateasaate

as

a

ateas

a

ateasaate

atdteatdteatLsF

b stbst

b

b st

b

b stbst

b

b st

b

st

Linearity of the Laplace TransformSuppose f and g are functions whose Laplace transforms exist for s > a1 and s > a2, respectively.Then, for s greater than the maximum of a1 and a2, the Laplace transform of c1 f (t) + c2g(t) exists. That is,

with

{ } [ ] finite is )()()()(0 2121 ∫∞ − +=+ dttgctfcetgctfcL st

{ }{ } { })( )(

)( )()()(

21

020121

tgLctfLc

dttgecdttfectgctfcL stst

+=

+=+ ∫∫∞ −∞ −

Example 8Let f (t) = 5e-2t - 3sin(4t) for t ≥ 0. Then by linearity of the Laplace transform, and using results of previous examples, the Laplace transform F(s) of f is:

{ }{ } { }

0,16

122

5)4sin(35

)4sin(35)}({)(

2

2

2

>+

−+

=

−=

−=

=

−

−

sss

tLeLteL

tfLsF

t

t

Ch 6.2: Solution of Initial Value Problems The Laplace transform is named for the French mathematician Laplace, who studied this transform in 1782.The techniques described in this chapter were developed primarily by Oliver Heaviside (1850-1925), an English electrical engineer.In this section we see how the Laplace transform can be used to solve initial value problems for linear differential equations with constant coefficients. The Laplace transform is useful in solving these differential equations because the transform of f ' is related in a simple way to the transform of f, as stated in Theorem 6.2.1.

Theorem 6.2.1Suppose that f is a function for which the following hold:(1) f is continuous and f ' is piecewise continuous on [0, b] for all b > 0. (2) | f(t) | ≤ Keat when t ≥ M, for constants a, K, M, with K, M > 0.

Then the Laplace Transform of f ' exists for s > a, with

Proof (outline): For f and f ' continuous on [0, b], we have

Similarly for f ' piecewise continuous on [0, b], see text.

{ } { } )0()()( ftfsLtfL −=′

⎥⎦⎤

⎢⎣⎡ +−=

⎥⎦⎤

⎢⎣⎡ −−=′

∫

∫∫−−

∞→

−−

∞→

−

∞→

b stsb

b

b stbst

b

b st

b

dttfesfbfe

dttfestfedttfe

0

000

)()0()(lim

)()()(lim)(lim

The Laplace Transform of f 'Thus if f and f ' satisfy the hypotheses of Theorem 6.2.1, then

Now suppose f ' and f '' satisfy the conditions specified for fand f ' of Theorem 6.2.1. We then obtain

Similarly, we can derive an expression for L{f (n)}, provided fand its derivatives satisfy suitable conditions. This result is given in Corollary 6.2.2

{ } { }{ }[ ]{ } )0()0()(

)0()0()()0()()(

2 fsftfLsfftfsLs

ftfsLtfL

′−−=

′−−=

′−′=′′

{ } { } )0()()( ftfsLtfL −=′

Corollary 6.2.2Suppose that f is a function for which the following hold:(1) f , f ', f '' ,…, f (n-1) are continuous, and f (n) piecewise continuous, on

[0, b] for all b > 0. (2) | f(t) | ≤ Keat, | f '(t) | ≤ Keat , …, | f (n-1)(t) | ≤ Keat for t ≥ M, for

constants a, K, M, with K, M > 0.

Then the Laplace Transform of f (n) exists for s > a, with

{ } { } )0()0()0()0()()( )1()2(21)( −−−− −−−′−−= nnnnnn fsffsfstfLstfL L

Example 1: Chapter 3 Method (1 of 4)

Consider the initial value problem

Recall from Section 3.1:

Thus r1 = -2 and r2 = -3, and general solution has the form

Using initial conditions:

Thus

We now solve this problem using Laplace Transforms.

( ) ( ) 30,20,065 =′==+′+′′ yyyyy

( )( ) 032065)( 2 =++⇔=++⇒= rrrrety rt

tt ececty 32

21)( −− +=

7,93322

2121

21 −==⇒⎭⎬⎫

=−−=+

cccccc

tt eety 32 79)( −− −=

Example 1: Laplace Tranform Method (2 of 4)Assume that our IVP has a solution φ and that φ'(t) and φ''(t) satisfy the conditions of Corollary 6.2.2. Then

and hence

Letting Y(s) = L{y}, we have

Substituting in the initial conditions, we obtain

Thus

( ) ( ) 30,20,065 =′==+′+′′ yyyyy

0}0{}{6}{5}{}65{ ==+′+′′=+′+′′ LyLyLyLyyyL

[ ] [ ] 0}{6)0(}{5)0()0(}{2 =+−+′−− yLyysLysyyLs

( ) ( ) 0)0()0(5)(652 =′−+−++ yyssYss

( ) ( ) 0352)(652 =−+−++ ssYss

( )( )23132)(}{++

+==

ssssYyL

Using partial fraction decomposition, Y(s) can be rewritten:

Thus

( )( ) ( ) ( )( ) ( )

9,71332,2

)32()(13232132

2323132

=−==+=+

+++=++++=+

++

+=

+++

BABABA

BAsBAssBsAs

sB

sA

sss

( ) ( )29

37)(}{

++

+−==

sssYyL

Example 1: Partial Fractions (3 of 4)

Recall from Section 6.1:

Thus

Recalling Y(s) = L{y}, we have

and hence

( ) ( ) ,2},{9}{72

93

7)( 23 −>+−=+

++

−= −− seLeLss

sY tt

{ } asas

dtedteesFeL tasatstat >−

==== ∫∫∞ −−∞ − ,1)(

0

)(

0

}97{}{ 23 tt eeLyL −− +−=

tt eety 23 97)( −− +−=

Example 1: Solution (4 of 4)

General Laplace Transform MethodConsider the constant coefficient equation

Assume that this equation has a solution y = φ(t), and that φ'(t) and φ''(t) satisfy the conditions of Corollary 6.2.2. Then

If we let Y(s) = L{y} and F(s) = L{ f }, then

)(tfcyybya =+′+′′

)}({}{}{}{}{ tfLycLybLyaLcyybyaL =+′+′′=+′+′′

[ ] [ ]( ) ( )

( )cbsas

sFcbsasyaybassY

sFyaybassYcbsassFycLyysLbysyyLsa

+++

++′++

=

=′−+−++

=+−+′−−

22

2

2

)()0()0()(

)()0()0()()(}{)0(}{)0()0(}{

Algebraic ProblemThus the differential equation has been transformed into the the algebraic equation

for which we seek y = φ(t) such that L{φ(t)} = Y(s).Note that we do not need to solve the homogeneous and nonhomogeneous equations separately, nor do we have a separate step for using the initial conditions to determine the values of the coefficients in the general solution.

( )cbsas

sFcbsasyaybassY

+++

++′++

= 22

)()0()0()(

Characteristic PolynomialUsing the Laplace transform, our initial value problem

becomes

The polynomial in the denominator is the characteristic polynomial associated with the differential equation. The partial fraction expansion of Y(s) used to determine φrequires us to find the roots of the characteristic equation. For higher order equations, this may be difficult, especially if the roots are irrational or complex.

( )cbsas

sFcbsasyaybassY

+++

++′++

= 22

)()0()0()(

( ) ( ) 00 0,0),( yyyytfcyybya ′=′==+′+′′

Inverse ProblemThe main difficulty in using the Laplace transform method is determining the function y = φ(t) such that L{φ(t)} = Y(s). This is an inverse problem, in which we try to find φ such that φ(t) = L-1{Y(s)}. There is a general formula for L-1, but it requires knowledge of the theory of functions of a complex variable, and we do not consider it here.It can be shown that if f is continuous with L{f(t)} = F(s), then f is the unique continuous function with f (t) = L-1{F(s)}. Table 6.2.1 in the text lists many of the functions and their transforms that are encountered in this chapter.

Linearity of the Inverse TransformFrequently a Laplace transform F(s) can be expressed as

Let

Then the function

has the Laplace transform F(s), since L is linear. By the uniqueness result of the previous slide, no other continuous function f has the same transform F(s). Thus L-1 is a linear operator with

)()()()( 21 sFsFsFsF n+++= L

{ } { })()(,,)()( 11

11 sFLtfsFLtf nn

−− == K

)()()()( 21 tftftftf n+++= L

{ } { } { })()()()( 11

11 sFLsFLsFLtf n−−− ++== L

Example 2

Find the inverse Laplace Transform of the given function.

To find y(t) such that y(t) = L-1{Y(s)}, we first rewrite Y(s):

Using Table 6.2.1,

Thus

ssY 2)( =

⎟⎠⎞

⎜⎝⎛==

sssY 122)(

{ } ( ) 212122)( 111 ==⎭⎬⎫

⎩⎨⎧=

⎭⎬⎫

⎩⎨⎧= −−−

sL

sLsYL

2)( =ty

Example 3



Using Table 6.2.1,

Thus

53)(−

=s

sY

⎟⎠⎞

⎜⎝⎛

−=

−=

513

53)(

sssY

{ } tes

Ls

LsYL 5111 35

135

3)( =⎭⎬⎫

⎩⎨⎧

−=

⎭⎬⎫

⎩⎨⎧

−= −−−

tety 53)( =

Example 4



Using Table 6.2.1,

Thus

4

6)(s

sY =

44

!36)(ss

sY ==

{ } 34

11 !3)( ts

LsYL =⎭⎬⎫

⎩⎨⎧= −−

3)( tty =

Example 5



Using Table 6.2.1,

Thus

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛== 333

!24!2!2

88)(sss

sY

{ } 23

13

11 4!24!24)( ts

Ls

LsYL =⎭⎬⎫

⎩⎨⎧=

⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛= −−−

24)( tty =

3

8)(s

sY =

Example 6



Using Table 6.2.1,

Thus

⎥⎦⎤

⎢⎣⎡

++⎥⎦

⎤⎢⎣⎡

+=

++

=9

331

94

914)( 222 ss

ss

ssY

{ } tts

Ls

sLsYL 3sin313cos4

93

31

94)( 2

12

11 +=⎭⎬⎫

⎩⎨⎧

++

⎭⎬⎫

⎩⎨⎧

+= −−−

ttty 3sin313cos4)( +=

914)( 2 +

+=

sssY

Example 7



Using Table 6.2.1,

Thus

⎥⎦⎤

⎢⎣⎡

−+⎥⎦

⎤⎢⎣⎡

−=

−+

=9

331

94

914)( 222 ss

ss

ssY

{ } tts

Ls

sLsYL 3sinh313cosh4

93

31

94)( 2

12

11 +=⎭⎬⎫

⎩⎨⎧

−+

⎭⎬⎫

⎩⎨⎧

−= −−−

ttty 3sinh313cosh4)( +=

914)( 2 −

+=

sssY

Example 8



Using Table 6.2.1,

Thus

{ }( )

tets

LsYL −−− −=⎭⎬⎫

⎩⎨⎧

+−= 2

311 5

1!25)(

tetty −−= 25)(

( )3110)(+

−=s

sY

( ) ( ) ( ) ⎥⎦⎤

⎢⎣

⎡

+−=⎥

⎦

⎤⎢⎣

⎡

+−=

+−= 333 1

!251!2

!210

110)(

ssssY

Example 9

For the function Y(s) below, we find y(t) = L-1{Y(s)} by using a partial fraction expansion, as follows.

tt eetyss

sY

BAABBA

ABsBAssBsAs

sB

sA

sss

ssssY

34

2

710

711)(

31

710

41

711)(

7/10,7/11134,3

)34()(13)4()3(13

34)3)(4(13

1213)(

+=⇒⎥⎦⎤

⎢⎣⎡−

+⎥⎦⎤

⎢⎣⎡+

=

===−=+

−++=+++−=+

−+

+=

−++

=−++

=

−

Example 10

For the function Y(s) below, we find y(t) = L-1{Y(s)} by completing the square in the denominator and rearranging the numerator, as follows.

Using Table 6.1, we obtain

( ) ( )( )( ) ( ) ( ) ⎥

⎦

⎤⎢⎣

⎡

+−+⎥

⎦

⎤⎢⎣

⎡

+−−

=+−+−

=

+−+−

=++−

−=

+−−

=

1312

1334

13234

132124

196104

106104)(

222

222

sss

ss

ss

sss

ssssY

tetety tt sin2cos4)( 33 +=

Example 11: Initial Value Problem (1 of 2)


Taking the Laplace transform of the differential equation, and assuming the conditions of Corollary 6.2.2 are met, we have



Thus

( ) ( ) 60,00,0258 =′==+′−′′ yyyyy

[ ] [ ] 0}{25)0(}{8)0()0(}{2 =+−−′−− yLyysLysyyLs

( ) ( ) 0)0()0(8)(2582 =′−−−+− yyssYss

2586)(}{ 2 +−

==ss

sYyL

( ) 06)(2582 =−+− sYss

Completing the square, we obtain

Thus

Using Table 6.2.1, we have

Therefore our solution to the initial value problem is

( ) 91686

2586)( 22 ++−

=+−

=ssss

sY

( ) ⎥⎦

⎤⎢⎣

⎡

+−=

9432)( 2s

sY

{ } tesYL t 3sin2)( 41 =−

tety t 3sin2)( 4=


Example 12: Nonhomogeneous Problem (1 of 2)


Taking the Laplace transform of the differential equation, and assuming the conditions of Corollary 6.2.2 are met, we have



Thus

( ) ( ) 10,20,2sin =′==+′′ yytyy

[ ] )4/(2}{)0()0(}{ 22 +=+′−− syLysyyLs

( ) )4/(2)0()0()(1 22 +=′−−+ sysysYs

)4)(1(682)( 22

23

+++++

=ss

ssssY

( ) )4/(212)(1 22 +=−−+ sssYs

Using partial fractions,

Then

Solving, we obtain A = 2, B = 5/3, C = 0, and D = -2/3. Thus

Hence


41)4)(1(682)( 2222

23

++

+++

=+++++

=s

DCss

BAsss

ssssY

( )( ) ( )( ))4()4()()(

1468223

2223

DBsCAsDBsCAsDCssBAssss

+++++++=

+++++=+++

43/2

13/5

12)( 222 +

−+

++

=sss

ssY

tttty 2sin31sin

35cos2)( −+=

Ch 6.3: Step Functions Some of the most interesting elementary applications of the Laplace Transform method occur in the solution of linear equations with discontinuous or impulsive forcing functions. In this section, we will assume that all functions considered are piecewise continuous and of exponential order, so that their Laplace Transforms all exist, for s large enough.

Step Function definitionLet c ≥ 0. The unit step function, or Heaviside function, is defined by

A negative step can be represented by

⎩⎨⎧

≥<

=ctct

tuc ,1,0

)(

⎩⎨⎧

≥<

=−=ctct

tuty c ,0,1

)(1)(

Example 1Sketch the graph of

Solution: Recall that uc(t) is defined by

Thus

and hence the graph of h(t) is a rectangular pulse.

0),()()( 2 ≥−= ttututh ππ

⎩⎨⎧

≥<

=ctct

tuc ,1,0

)(

⎪⎩

⎪⎨

⎧

∞<≤<≤<≤

=t

tt

thπ

πππ

202,1

0,0)(

Laplace Transform of Step FunctionThe Laplace Transform of uc(t) is

{ }

se

se

se

es

dte

dtedttuetuL

cs

csbs

b

b

c

st

b

b

c

st

b

c

stc

stc

−

−−

∞→

−

∞→

−

∞→

∞ −∞ −

=

⎥⎦

⎤⎢⎣

⎡+−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−==

==

∫

∫∫

lim

1limlim

)()(0

Translated FunctionsGiven a function f (t) defined for t ≥ 0, we will often want to consider the related function g(t) = uc(t) f (t - c):

Thus g represents a translation of f a distance c in the positive t direction.In the figure below, the graph of f is given on the left, and the graph of g on the right.

⎩⎨⎧

≥−<

=ctctfct

tg),(

,0)(

Example 2Sketch the graph of

Solution: Recall that uc(t) is defined by

Thus

and hence the graph of g(t) is a shifted parabola.

.0,)( where),()1()( 21 ≥=−= tttftutftg

⎩⎨⎧

≥<

=ctct

tuc ,1,0

)(

⎩⎨⎧

≥−<≤

=1,)1(

10,0)( 2 tt

ttg

Theorem 6.3.1If F(s) = L{f (t)} exists for s > a ≥ 0, and if c > 0, then

Conversely, if f (t) = L-1{F(s)}, then

Thus the translation of f (t) a distance c in the positive tdirection corresponds to a multiplication of F(s) by e-cs.

{ } { } )()()()( sFetfLectftuL cscsc

−− ==−

{ })()()( 1 sFeLctftu csc

−−=−

Theorem 6.3.1: Proof OutlineWe need to show

Using the definition of the Laplace Transform, we have

{ }

)(

)(

)(

)(

)()()()(

0

0

)(

0

sFe

duufee

duufe

dtctfe

dtctftuectftuL

cs

sucs

cusctu

c

st

cst

c

−

∞ −−

∞ +−−=

∞ −

∞ −

=

=

=

−=

−=−

∫∫

∫∫

{ } )()()( sFectftuL csc

−=−

Example 3Find the Laplace transform of

Solution: Note that

Thus

)()1()( 12 tuttf −=

⎩⎨⎧

≥−<≤

=1,)1(

10,0)( 2 tt

ttf

{ } { } { } 322

12)1)(()(setLettuLtfL

ss

−− ==−=

Example 4Find L{ f (t)}, where f is defined by

Note that f (t) = sin(t) + uπ/4(t) cos(t - π/4), and

⎩⎨⎧

≥−+<≤

=4/),4/cos(sin

4/0,sin)(

πππ

ttttt

tf

{ } { } { }{ } { }

11

111

cossin

)4/cos()(sin)(

2

4/

24/

2

4/4/

++

=

++

+=

+=

−+=

−

−

−

sse

sse

s

tLetL

ttuLtLtfL

s

s

s

π

π

π

π π

Example 5Find L-1{F(s)}, where

Solution: 4

73)(sesF

s−+=

( )373

471

41

4

71

41

7)(61

21

!361!3

21

3)(

−+=

⎭⎬⎫

⎩⎨⎧ ⋅+

⎭⎬⎫

⎩⎨⎧=

⎭⎬⎫

⎩⎨⎧

+⎭⎬⎫

⎩⎨⎧=

−−−

−−−

ttut

seL

sL

seL

sLtf

s

s

Theorem 6.3.2If F(s) = L{f (t)} exists for s > a ≥ 0, and if c is a constant, then

Conversely, if f (t) = L-1{F(s)}, then

Thus multiplication f (t) by ect results in translating F(s) a distance c in the positive t direction, and conversely.Proof Outline:

{ } cascsFtfeL ct +>−= ),()(

{ })()( 1 csFLtfect −= −

{ } )()()()(0

)(

0csFdttfedttfeetfeL tcsctstct −=== ∫∫

∞ −−∞ −

Example 4Find the inverse transform of

To solve, we first complete the square:

Since

it follows that

521)( 2 ++

+=

ssssG

( )( )

( ) 411

4121

521)( 222 ++

+=

++++

=++

+=

ss

sss

ssssG

{ } { } ( )tetfesFLsGL tt 2cos)()1()( 11 −−−− ==+=

{ } ( )ts

sLsFLtf 2cos4

)()( 211 =

⎭⎬⎫

⎩⎨⎧

+== −−

Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section focus on examples of nonhomogeneous initial value problems in which the forcing function is discontinuous.

( ) ( ) 00 0,0),( yyyytgcyybya ′=′==+′+′′


Find the solution to the initial value problem

Such an initial value problem might model the response of a damped oscillator subject to g(t), or current in a circuit for a unit voltage pulse.

⎩⎨⎧

≥<≤<≤

=−=

=′==+′+′′

20 and 50,0205,1

)()()(

where0)0(,0)0(),(22

205 ttt

tututg

yytgyyy

Assume the conditions of Corollary 6.2.2 are met. Then

or

Letting Y(s) = L{y},


Thus

)}({)}({}{2}{}{2 205 tuLtuLyLyLyL −=+′+′′

[ ] [ ]seeyLyysLysyyLs

ss 2052 }{2)0(}{)0(2)0(2}{2

−− −=+−+′−−

( ) ( ) ( ) seeyyssYss ss 2052 )0(2)0(12)(22 −− −=′−+−++

Example 1: Laplace Transform (2 of 12)

( ) ( ) seesYss ss 2052 )(22 −− −=++

( )( )22

)( 2

205

++−

=−−

ssseesY

ss

0)0(,0)0(),()(22 205 =′=−=+′+′′ yytutuyyy

We have

where

If we let h(t) = L-1{H(s)}, then

by Theorem 6.3.1.

)20()()5()()( 205 −−−== thtuthtuty φ

Example 1: Factoring Y(s) (3 of 12)

( )( ) ( ) )(

22)( 205

2

205

sHeessseesY ss

ss−−

−−

−=++

−=

( )221)( 2 ++

=sss

sH

Thus we examine H(s), as follows.

This partial fraction expansion yields the equations

Thus


( ) 22221)( 22 ++

++=

++=

ssCBs

sA

ssssH

2/1,1,2/112)()2( 2

−=−==⇒=++++

CBAAsCAsBA

222/12/1)( 2 ++

+−=

sss

ssH

Completing the square,

Example 1: Completing the Square (5 of 12)

( )( )

( ) ⎥⎦

⎤⎢⎣

⎡

++++

−=

⎥⎦

⎤⎢⎣

⎡

+++

−=

⎥⎦⎤

⎢⎣⎡

++++

−=

⎥⎦⎤

⎢⎣⎡

+++

−=

+++

−=

16/154/14/14/1

212/1

16/154/12/1

212/1

16/1516/12/2/1

212/1

12/2/1

212/1

222/12/1)(

2

2

2

2

2

ss

s

ss

s

sss

s

sss

s

sss

ssH

Thus

and hence

For h(t) as given above, and recalling our previous results, the solution to the initial value problem is then


( )( )

( )( ) ( ) ⎥

⎦

⎤⎢⎣

⎡

++−⎥

⎦

⎤⎢⎣

⎡

+++

−=

⎥⎦

⎤⎢⎣

⎡

++++

−=

16/154/14/15

1521

16/154/14/1

212/1

16/154/14/14/1

212/1)(

22

2

sss

s

ss

ssH

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−== −−− tetesHLth tt

415sin

1521

415cos

21

21)}({)( 4/4/1

)20()()5()()( 205 −−−= thtuthtutφ

Thus the solution to the initial value problem is

The graph of this solution is given below.

Example 1: Solution Graph (7 of 12)

( ) ( )415sin152

1415cos21

21)(

where),20()()5()()(

4/4/

205

teteth

thtuthtut

tt −− −−=

−−−=φ

The solution to original IVP can be viewed as a composite of three separate solutions to three separate IVPs:

Example 1: Composite IVPs (8 of 12)

)20()20(),20()20(,022:200)5(,0)5(,122:205

0)0(,0)0(,022:50

2323333

22222

11111

yyyyyyytyyyyytyyyyyt

′=′==+′+′′>=′==+′+′′<<=′==+′+′′<≤

Consider the first initial value problem

From a physical point of view, the system is initially at rest, and since there is no external forcing, it remains at rest. Thus the solution over [0, 5) is y1 = 0, and this can be verified analytically as well. See graphs below.

Example 1: First IVP (9 of 12)

50;0)0(,0)0(,022 11111 <≤=′==+′+′′ tyyyyy

Consider the second initial value problem

Using methods of Chapter 3, the solution has the form

Physically, the system responds with the sum of a constant (the response to the constant forcing function) and a damped oscillation, over the time interval (5, 20). See graphs below.

Example 1: Second IVP (10 of 12)

205;0)5(,0)5(,122 22222 <<=′==+′+′′ tyyyyy

( ) ( ) 2/14/15sin4/15cos 4/2

4/12 ++= −− tectecy tt

Consider the third initial value problem


Physically, since there is no external forcing, the response is a damped oscillation about y = 0, for t > 20. See graphs below.

Example 1: Third IVP (11 of 12)

( ) ( )4/15sin4/15cos 4/2

4/13 tectecy tt −− +=

20);20()20(),20()20(,022 2323333 >′=′==+′+′′ tyyyyyyy

Our solution is

It can be shown that φ and φ' are continuous at t = 5 and t = 20, and φ'' has a jump of 1/2 at t = 5 and a jump of –1/2 at t = 20:

Thus jump in forcing term g(t) at these points is balanced by a corresponding jump in highest order term 2y'' in ODE.

Example 1: Solution Smoothness (12 of 12)

)20()()5()()( 205 −−−= thtuthtutφ

-.5072)(lim,-.0072)(lim

2/1)(lim,0)(lim

2020

55

≅′′≅′′

=′′=′′

+−

+−

→→

→→

tt

tt

tt

tt

φφ

φφ

Consider a general second order linear equation

where p and q are continuous on some interval (a, b) but g is only piecewise continuous there. If y = ψ(t) is a solution, then ψ and ψ ' are continuous on (a, b) but ψ '' has jump discontinuities at the same points as g. Similarly for higher order equations, where the highest derivative of the solution has jump discontinuities at the same points as the forcing function, but the solution itself and its lower derivatives are continuous over (a, b).

Smoothness of Solution in General

)()()( tgytqytpy =+′+′′


The graph of forcing function g(t) is given on right, and isknown as ramp loading.

( )⎪⎩

⎪⎨

⎧

≥<≤−<≤

=−

−−

=

=′==+′′

10,11055550,0

510)(

55)()(

where0)0(,0)0(),(4

105

tttt

ttuttutg

yytgyy


Assume that this ODE has a solution y = φ(t) and that φ'(t) and φ''(t) satisfy the conditions of Corollary 6.2.2. Then

or

Letting Y(s) = L{y}, and substituting in initial conditions,

Thus

( )[ ] ( )[ ] 5}10)({5}5)({}{4}{ 105 −−−=+′′ ttuLttuLyLyL

[ ] 2

1052

5}{4)0()0(}{

seeyLysyyLs

ss −− −=+′−−

Example 2: Laplace Transform (2 of 12)

( ) ( ) 21052 5)(4 seesYs ss −− −=+

( )( )45

)( 22

105

+−

=−−

sseesY

ss

0)0(,0)0(,510)(

55)(4 105 =′=

−−

−=+′′ yyttuttuyy

We have

where

If we let h(t) = L-1{H(s)}, then

by Theorem 6.3.1.

[ ])10()()5()(51)( 105 −−−== thtuthtuty φ

Example 2: Factoring Y(s) (3 of 12)

( )( ) )(

545)(

105

22

105

sHeess

eesYssss −−−− −

=+

−=

( )41)( 22 +

=ss

sH

Thus we examine H(s), as follows.

This partial fraction expansion yields the equations

Thus


( ) 441)( 2222 +

+++=

+=

sDCs

sB

sA

sssH

4/1,0,4/1,0144)()( 23

−====⇒=+++++

DCBABAssDBsCA

44/14/1)( 22 +

−=ss

sH

Thus

and hence

For h(t) as given above, and recalling our previous results, the solution to the initial value problem is then


( )ttsHLth 2sin81

41)}({)( 1 −== −

⎥⎦⎤

⎢⎣⎡

+−⎥⎦

⎤⎢⎣⎡=

+−=

42

811

41

44/14/1)(

22

22

ss

sssH

[ ])10()()5()(51)( 105 −−−== thtuthtuty φ

Thus the solution to the initial value problem is

The graph of this solution is given below.

Example 2: Graph of Solution (6 of 12)

[ ]

( )ttth

thtuthtut

2sin81

41)(

where,)10()()5()(51)( 105

−=

−−−=φ

The solution to original IVP can be viewed as a composite of three separate solutions to three separate IVPs (discuss):

Example 2: Composite IVPs (7 of 12)

)10()10(),10()10(,14:100)5(,0)5(,5/)5(4:105

0)0(,0)0(,04:50

232333

2222

1111

yyyyyytyytyytyyyyt

′=′==+′′>=′=−=+′′<<=′==+′′<≤

Consider the first initial value problem

From a physical point of view, the system is initially at rest, and since there is no external forcing, it remains at rest. Thus the solution over [0, 5) is y1 = 0, and this can be verified analytically as well. See graphs below.

Example 2: First IVP (8 of 12)

50;0)0(,0)0(,04 1111 <≤=′==+′′ tyyyy

Consider the second initial value problem


Thus the solution is an oscillation about the line (t – 5)/20, over the time interval (5, 10). See graphs below.

Example 2: Second IVP (9 of 12)

105;0)5(,0)5(,5/)5(4 2222 <<=′=−=+′′ tyytyy

( ) ( ) 4/120/2sin2cos 212 −++= ttctcy

Consider the third initial value problem


Thus the solution is an oscillation about y = 1/4, for t > 10. See graphs below.

Example 2: Third IVP (10 of 12)

10);10()10(),10()10(,14 232333 >′=′==+′′ tyyyyyy

( ) ( ) 4/12sin2cos 213 ++= tctcy

Recall that the solution to the initial value problem is

To find the amplitude of the eventual steady oscillation, we locate one of the maximum or minimum points for t > 10. Solving y' = 0, the first maximum is (10.642, 0.2979). Thus the amplitude of the oscillation is about 0.0479.

Example 2: Amplitude (11 of 12)

[ ] ( )ttththtuthtuty 2sin81

41)( ,)10()()5()(

51)( 105 −=−−−==φ

Our solution is

In this example, the forcing function g is continuous but g' is discontinuous at t = 5 and t = 10.It follows that φ and its first two derivatives are continuous everywhere, but φ''' has discontinuities at t = 5 and t = 10 that match the discontinuities of g' at t = 5 and t = 10.

Example 2: Solution Smoothness (12 of 12)

[ ] ( )ttththtuthtuty 2sin81

41)( ,)10()()5()(

51)( 105 −=−−−== φ

Ch 6.5: Impulse Functions In some applications, it is necessary to deal with phenomena of an impulsive nature. For example, an electrical circuit or mechanical system subject to a sudden voltage or force g(t) of large magnitude that acts over a short time interval about t0. The differential equation will then have the form

small. is 0 and otherwise,0

,big)(

where),(

00

>⎩⎨⎧ +<<−

=

=+′+′′

τ

ττ ttttg

tgcyybya

Measuring ImpulseIn a mechanical system, where g(t) is a force, the total impulseof this force is measured by the integral

Note that if g(t) has the form

then

In particular, if c = 1/(2τ), then I(τ) = 1 (independent of τ ).

∫∫+

−

∞

∞−==

τ

ττ 0

0

)()()(t

tdttgdttgI

⎩⎨⎧ +<<−

= otherwise,0

,)( 00 ττ tttc

tg

0,2)()()( 0

0

>=== ∫∫+

−

∞

∞−τττ

τ

τcdttgdttgI

t

t

Unit Impulse FunctionSuppose the forcing function dτ(t) has the form

Then as we have seen, I(τ) = 1.

We are interested dτ(t) acting over shorter and shorter time intervals (i.e., τ → 0). See graph on right.

Note that dτ(t) gets taller and narrower as τ → 0. Thus for t ≠ 0, we have

⎩⎨⎧ <<−

= otherwise,0

,21)(

ττττ

ttd

1)(limand,0)(lim00

==→→

ττττ

Itd

Dirac Delta FunctionThus for t ≠ 0, we have

The unit impulse function δ is defined to have the properties

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,

1)(limand,0)(lim00

==→→

ττττ

Itd

1)(and,0for 0)( =≠= ∫∞

∞−dtttt δδ

1)(and,for 0)( 000 =−≠=− ∫∞

∞−dttttttt δδ

Laplace Transform of δ (1 of 2)

The Laplace Transform of δ is defined by

and thus { } { } 0,)(lim)( 0000 >−=−

→tttdLttL ττ

δ

{ }

( ) ( )[ ]

00

00

00

0

0

0

0

)cosh(lim

)sinh(lim2

lim

21lim

2lim

21lim)(lim)(

0

00

00

00 000

stst

stssst

tstst

t

st

t

t

stst

es

sse

sseee

se

eess

e

dtedtttdettL

−

→

−

→

−−−

→

−−+−

→

+

−

−

→

+

−

−

→

∞ −

→

=⎥⎦⎤

⎢⎣⎡=

⎥⎦⎤

⎢⎣⎡=⎥

⎦

⎤⎢⎣

⎡ −=

+−=−

=

=−=− ∫∫

τ

ττ

τ

ττ

τδ

τ

τ

ττ

τ

ττ

τ

τ

ττ

τ

ττττ

Laplace Transform of δ (2 of 2)

Thus the Laplace Transform of δ is

For Laplace Transform of δ at t0= 0, take limit as follows:

For example, when t0= 10, we have L{δ(t -10)} = e-10s.

{ } 0,)( 000 >=− − tettL stδ

{ } { } 1lim)(lim)( 0

00 000==−= −

→→

st

tettdLtL

ττδ

Product of Continuous Functions and δThe product of the delta function and a continuous function fcan be integrated, using the mean value theorem for integrals:

Thus

[ ]

)(

*)(lim

)* where( *)(221lim

)(21lim

)()(lim)()(

0

0

000

0

000

0

0

tf

tf

ttttf

dttf

dttfttddttftt

t

t

=

=

+<<−=

=

−=−

→

→

+

−→

∞

∞−→

∞

∞−

∫

∫∫

τ

τ

τ

ττ

ττ

ττττ

τ

δ

)()()( 00 tfdttftt =−∫∞

∞−δ

Consider the solution to the initial value problem

Then

Letting Y(s) = L{y},


or

)}7({}{2}{}{2 −=+′+′′ tLyLyLyL δ

[ ] [ ] sesYyssYysysYs 72 )(2)0()()0(2)0(2)(2 −=+−+′−−


( ) sesYss 72 )(22 −=++

22)( 2

7

++=

−

ssesY

s

0)0(,0)0(),7(22 =′=−=+′+′′ yytyyy δ

We have

The partial fraction expansion of Y(s) yields

and hence


22)( 2

7

++=

−

ssesY

s

( ) ⎥⎦

⎤⎢⎣

⎡

++=

−

16/154/14/15

152)( 2

7

sesY

s

( ) ( )7415sin)(

1521)( 4/7

7 −= −− tetuty t

With homogeneous initial conditions at t = 0 and no external excitation until t = 7, there is no response on (0, 7). The impulse at t = 7 produces a decaying oscillation that persists indefinitely. Response is continuous at t = 7 despite singularity in forcing function. Since y' has a jump discontinuity at t = 7, y'' has an infinite discontinuity there. Thus singularity in forcing function is balanced by a corresponding singularity in y''.

Example 1: Solution Behavior (3 of 3)

Ch 6.6: The Convolution Integral Sometimes it is possible to write a Laplace transform H(s) as H(s) = F(s)G(s), where F(s) and G(s) are the transforms of known functions f and g, respectively.In this case we might expect H(s) to be the transform of the product of f and g. That is, does

H(s) = F(s)G(s) = L{f }L{g} = L{f g}?On the next slide we give an example that shows that this equality does not hold, and hence the Laplace transform cannot in general be commuted with ordinary multiplication. In this section we examine the convolution of f and g, which can be viewed as a generalized product, and one for which the Laplace transform does commute.

Example 1Let f (t) = 1 and g(t) = sin(t). Recall that the Laplace Transforms of f and g are

Thus

and

Therefore for these functions it follows that

{ } { }1

1sin)()( 2 +==

stLtgtfL

{ } { } { } { }1

1sin)(,11)( 2 +====

stLtgL

sLtfL

{ } { } { })()()()( tgLtfLtgtfL ≠

{ } { } ( )11)()( 2 +

=ss

tgLtfL

Theorem 6.6.1Suppose F(s) = L{f (t)} and G(s) = L{g(t)} both exist for s > a ≥ 0. Then H(s) = F(s)G(s) = L{h(t)} for s > a, where

The function h(t) is known as the convolution of f and g and the integrals above are known as convolution integrals.

Note that the equality of the two convolution integrals can be seen by making the substitution u = t - τ. The convolution integral defines a “generalized product” and can be written as h(t) = ( f *g)(t). See text for more details.

∫∫ −=−=tt

dtgtfdgtfth00

)()()()()( τττττ

Theorem 6.6.1 Proof Outline

{ })(

)()(

)()(

)()(

)()()(

)()(

)()()()(

00

0 0

0

0

0 0

)(

0 0

thL

dtdgtfe

dtdgtfe

ddttfge

utdttfedg

duufedg

dgeduufesGsF

tst

t st

st

st

us

ssu

=⎥⎦⎤

⎢⎣⎡ −=

−=

−=

+=−=

=

=

∫∫

∫ ∫

∫ ∫

∫ ∫

∫ ∫

∫ ∫

∞ −

∞ −

∞ ∞ −

∞ ∞ −

∞ ∞ +−

∞ ∞ −−

τττ

τττ

τττ

ττττ

ττ

ττ

τ

τ

τ

τ

Example 2Find the Laplace Transform of the function h given below.

Solution: Note that f (t) = t and g(t) = sin2t, with

Thus by Theorem 6.6.1,

42}2{sin)}({)(

1}{)}({)(

2

2

+===

===

stLtgLsG

stLtfLsF

∫ −=t

tdtth0

2sin)()( ττ

{ } ( )42)()()()( 22 +

===ss

sGsFsHthL

Example 3: Find Inverse Transform (1 of 2)

Find the inverse Laplace Transform of H(s), given below.

Solution: Let F(s) = 2/s2 and G(s) = 1/(s - 2), with

Thus by Theorem 6.6.1,

{ }{ } tesGLtg

tsFLtf21

1

)()(2)()(

==

==−

−

)2(2)( 2 −

=ss

sH

{ } ∫ −==− tdetthsHL

0

21 )(2)()( ττ τ

Example 3: Solution h(t) (2 of 2)

We can integrate to simplify h(t), as follows.

[ ] [ ]

21

21

21

21

1211

22)(2)(

2

222

222

0

2

0

2

0

2

0

2

0

2

0

2

−−=

−+−−=

⎥⎦⎤

⎢⎣⎡ −−−−=

⎥⎦⎤

⎢⎣⎡ −−=

−=−=

∫

∫∫∫

te

eettte

eetet

deete

dedetdetth

t

ttt

ttt

ttt

ttt

ττ

τττττ

τττ

τττ


Solution:

or

Letting Y(s) = L{y}, and substituting in initial conditions,

Thus

)}({}{4}{ tgLyLyL =+′′

[ ] )(}{4)0()0(}{2 sGyLysyyLs =+′−−


( ) )(13)(42 sGssYs +−=+

4)(

413)( 22 ++

+−

=s

sGs

ssY

1)0(,3)0(),(4 −=′==+′′ yytgyy

We have

Thus

Note that if g(t) is given, then the convolution integral can be evaluated.

τττ dgttttyt

)()(2sin212sin

212cos3)(

0∫ −+−=


)(4

221

42

21

43

4)(

413)(

222

22

sGsss

ss

sGs

ssY

⎥⎦⎤

⎢⎣⎡

++⎥⎦

⎤⎢⎣⎡

+−⎥⎦

⎤⎢⎣⎡

+=

++

+−

=

Recall that the Laplace Transform of the solution y is

Note Φ (s) depends only on system coefficients and initial conditions, while Ψ (s) depends only on system coefficients and forcing function g(t).

Further, φ(t) = L-1{Φ (s)} solves the homogeneous IVP

while ψ(t) = L-1{Ψ (s)} solves the nonhomogeneous IVP

Example 4: Laplace Transform of Solution (3 of 4)

)()(4)(

413)( 22 sΨsΦ

ssG

sssY +=

++

+−

=

1)0(,3)0(),(4 −=′==+′′ yytgyy

1)0(,3)0(,04 −=′==+′′ yyyy

0)0(,0)0(),(4 =′==+′′ yytgyy

Examining Ψ (s) more closely,

The function H(s) is known as the transfer function, and depends only on system coefficients. The function G(s) depends only on external excitation g(t) applied to system.If G(s) = 1, then g(t) = δ(t) and hence h(t) = L-1{H(s)} solves the nonhomogeneous initial value problem

Thus h(t) is response of system to unit impulse applied at t = 0, and hence h(t) is called the impulse response of system.

Example 4: Transfer Function (4 of 4)

41)( where),()(

4)()( 22 +

==+

=s

sHsGsHs

sGsΨ

0)0(,0)0(),(4 =′==+′′ yytyy δ

Consider the general initial value problem

This IVP is often called an input-output problem. The coefficients a, b, c describe properties of physical system, and g(t) is the input to system. The values y0 and y0' describe initial state, and solution y is the output at time t. Using the Laplace transform, we obtain

or

Input-Output Problem (1 of 3)

00 )0(,)0(),( yyyytgcyybya ′=′==+′+′′

[ ] [ ] )()()0()()0()0()(2 sGscYyssYbysysYsa =+−+′−−

)()()()()( 2200 sΨsΦ

cbsassG

cbsasyaybassY +=

+++

++′++

=

We have

As before, Φ (s) depends only on system coefficients and initial conditions, while Ψ (s) depends only on system coefficients and forcing function g(t).

Further, φ(t) = L-1{Φ (s)} solves the homogeneous IVP

while ψ(t) = L-1{Ψ (s)} solves the nonhomogeneous IVP

Laplace Transform of Solution (2 of 3)

00 )0(,)0(,0 yyyycyybya ′=′==+′+′′

0)0(,0)0(),( =′==+′+′′ yytgcyybya

00 )0(,)0(),( yyyytgcyybya ′=′==+′+′′

)()()()()( 2200 sΨsΦ

cbsassG

cbsasyaybassY +=

+++

++′++

=

Examining Ψ (s) more closely,

As before, H(s) is the transfer function, and depends only on system coefficients, while G(s) depends only on external excitation g(t) applied to system.Thus if G(s) = 1, then g(t) = δ(t) and hence h(t) = L-1{H(s)} solves the nonhomogeneous IVP

Thus h(t) is response of system to unit impulse applied at t = 0, and hence h(t) is called the impulse response of system, with

Transfer Function (3 of 3)

0)0(,0)0(),( =′==+′+′′ yytcyybya δ

cbsassHsGsH

cbsassGsΨ

++==

++= 22

1)( where),()()()(

{ } τττψ dgthsGsHLtt

)()( )()()(0

1 ∫ −== −

ch 6.1: definition of laplace transform - dept of maths, nusmatysh/ma3220/chap6.pdf · ch 6.1:...

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