200238 Mathematics for Engineers 2

LAPLACE TRANSFORMS

Things I MUST do before each teaching session:

• Listen to ‘Lectures on Line” • Print off lecture/workshop questions and tutorial questions • Switch off mobile phones during lectures and tutorials.

Things I CAN do:

• Read the relevant sections of the text(s) • Do the extra questions from the text(s) as indicated.

Student learning outcomes: At the end of this section I should be able to:

• Define the Laplace transform • Find the Laplace transform of basic functions using a transform table • Define and use the Unit step function and find its Laplace transform • Define the dirac-delta function and find its Laplace transform

1

Laplace transforms

The operations of differentiation and integration are examples of transform operations because they transform one function into another function, for example 2x is transformed to

x2 under the operation of differentiation. This section is concerned with an improper integral which transforms a function )(tf into another function )(sF . This transformation is useful, among other things, for solving certain ODEs commonly used in physics and engineering.

Let )(tf be defined for 0≥t , then the integral

∫∞

==

0 ))()}({ dtf(tesFtfL -st ( 0s > ).

is called the Laplace transform of the function )(tf (provided the integral converges).

NOTE: The Laplace transform is an improper integral which exists if ∫ −

∞→

b st

bdttfe

0 )(lim is

finite. Technically, functions that are piecewise continuous on every finite interval of 0≥t and of exponential order, i.e. ktMetf ≤)( for all 0≥t and some constants k and M, have

Laplace transforms for all ks > . Functions such as ttf /1)( = or 2

)( tetf = do not have Laplace transforms. NOTE: lowercase is used for the function of t and UPPER case for the transformed function.

NOTE: We are only interested in the function )(tf when 0≥t .

The Laplace transform as a linear operator

Assume that the functions )(tf and )(tg have Laplace transforms, then

)}({)}({

)()(

)()(

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

0

0

0

0

0

tgbLtfaL

dttgebdttfea

dttgbedttfae

dttgbtfaetgbtfaL

stst

stst

st

+=

+=

+=

+=+

∫∫∫∫

∫

∞ −∞ −

∞ −∞ −

∞ −

so L is a linear operator.

2

Basic transforms Examples: Find the Laplace transform of the following functions using the definition of the transform. 1. 1)( =tf

{ } { }0

0

1 1( ) 1 1 0st

st eL f t L e dts s s

∞−∞ − − = = × = = − = − ∫ .

2. ttf =)(

{ } 200

2 2

(using integration by parts)

1 10 .

st stst te eL t e t dt

s s

s s

∞− −∞ − = × = − −

= − − =

∫

Note: The Laplace transform of ,nt n a positive integer, may be found by repeated integration by parts :

1!}{ += n

n

sntL .

3. ( ) atf t e=

( )

( )

0 00

1{ } ( 0)( )

t s aat st at t s a eL e e e dt e dt s a

s a s a

∞− −∞ ∞− − − = × = = = − > − − − ∫ ∫ .

4. ( ) sinf t at=

( )2 2 2 200

{sin } sin sin cosst

st e aL at e at dt s at a ats a s a

∞−∞ − = × = − − = + + ∫ .

5. ( ) cosf t at= (There are at least three ways to find this transform.)

{ }

{ } { } { }

{ }

2 2 2 2

2 2 2 2

2 2

1( )( )

cos sin

cos sin cos sin

and by equating real and imaginary parts

cos

iat

iat

s ia s aL e is ia s ia s ia s a s a

e at i ats aL at i at L at iL at i

s a s a

sL ats a

+= = = +

− − + + +

= +

∴ + = + = ++ +

=+

3

Table of Laplace transforms Some common transforms: Function )(tf Transform )}({)( tfLsF = 1

s1

nt 1

!+ns

n

ate as −

1

sin at 2 2

as a+

cos at 2 2

ss a+

sinh at 2 2

as a−

cosh at 2 2

ss a−

)()(

)()(1

sFtf

sFtf

L

L

−

⇐

⇒

Some operational properties: Function Transform )(tfeat )( asF −

)()( atUatf −− )(sFe as−

)( atU − s

e as−

)( at −δ ase−

)()( tf n )0(...)0()( )1(1 −− −−− nnn ffssFs

0

( ) * ( ) ( ) ( )t

f t g t f g t dt t t= −∫

(the convolution of a function)

( ) ( )F s G s

4

Laplace transforms of many elementary functions can be obtained from the definition or from combinations of the basic formulae and use of the linearity property. First shifting theorem (s shifting: replacing s with as − in )(sF )

If )(tf has the transform )(sF (where ks > ) then )(tfeat has the transform )( asF − (where kas >− ).

{ } )()()()(0

)(

0asFdttfedttfeetfeL tasatstat −=== ∫∫

∞ −−∞ − so

)()}({ asFtfeL at −= .

The steps are )()()}({)}({ asFsFtfLtfeLassass

at −===−→−→

.

Examples:

Find the Laplace transform of the following functions:

1. ( ) costf t e t−= .

{ } 2 2( 1)( 1)

1{ cos } cos .1 ( 1) 1

ts s

s s

s sL e t L ts s

−

→ +→ +

+= = =

+ + +

2. 5 3( ) .tf t e t=

{ }5 3 34 4( 5)

( 5)

3! 6{ } .( 5)

t

s ss s

L e t L ts s→ −

→ −

= = =−

Second shifting theorem (t-shifting: replacing t with at − in )(tf )

In many practical applications there are functions that can essentially be either “on” or “off”. For example, an external force acting on a mechanical system or a voltage impressed on a circuit can be turned on or off at some point in time. It is useful to define a special function to model these situations.

The unit step function is denoted by )( atU − and is defined as

><

=−atat

atU10

)(

5

NOTE: For convenience, we can leave the function undefined at at = .

Examples:

1. Graph the following:

a. )3( −tU

2. Given

><<<<

=31

312100

)(t

tt

tf , write )(tf in terms of the unit step function.

( ) 2 ( 1) ( 3).f t U t U t= − − −

A function )(tf can be shifted to the right and then “switched on” after time at = as

follows:

>−

<=−−

.)(

0)()(

atatf

atatUatf

The Laplace transform of this shifted function is

)(

)(

) so , (where )(

)(0

)()()()(

)()()}()({

0

0

)(

0

0

sFe

dvvfee

dtdvatvdvvfe

dtatfe

dtatUatfedtatUatfe

dtatUatfeatUatfL

as

svas

avs

a

st

a

sta st

st

−

∞ −−

∞ +−

∞ −

∞ −−

−∞

=

=

=−==

−+=

−−+−−=

−−=−−

∫

∫

∫

∫∫

∫

so { } )()()( sFeatUatfL as−=−− .

1

3

2

1

3

1

6

In particular, if 1)( =− atf then 1)( =tf so s

sF 1)( = and { }s

eatULas−

=− )( .

Examples:

1. Find { })(tfL if

><<

=.12

100)(

tt

tf

( ) 2 ( 1)f t U t= −

{ }1 2( ) 2 .

s se eL f ts s

− × −

∴ = × =

2. Find ( ){ }2 2 ( 3)L t t U t+ − .

To match the theorem let 2( 3) 2f t t t− = + so that

( ) ( )

( ){ }

2 23 2

2 33 2

2 8 15( ) 3 2 3 8 15 and ( )

2 8 152 ( 3) .s

f t t t t t F ss s s

L t t U t es s s

−

= + + + = + + = + +

∴ + − = + +

Dirac delta (unit impulse) function Phenomena of an impulse nature, such as the action of very large forces or voltages over very short intervals of time are of great practical interest since they arise in various applications.

For example, a vibrating airplane wing could be struck by lightning, a mass on a spring could be given a sharp blow or a ball could be struck by a bat.

This type of external force acting on a system for a brief period can been modelled by a (generalized) function called the Dirac delta or unit impulse function. It is named after Paul Dirac, a physicist awarded the Nobel Prize in 1933 for his work in quantum mechanics.

7

The impulse of a force )(tf acting over a time interval kata +≤≤ is given by the integral of )(tf from a to ka + . Now consider the function shown below.

+≤≤=

otherwise.0

1)(

kataktf k

The impulse is given by no matter how small k is. The Dirac delta function is the limit of 0 as )( →ktf k and is denoted by )( at −δ .

)(lim)(0

tfat kk→=−δ .

To find the Laplace transform of the delta function, the function )(tf k can be represented in terms of the unit step function as

[ ]))(()(1)( katUatUk

tf k +−−−=

so that

{ } { } .11))(()(1)()(

−=

−=+−−−=

−−

+−−

ksee

se

se

kkatUatUL

ktfL

ksas

skaas

k

[ ]

rule). sHopital'l' (using

1lim

)}({lim)}({

0

0

as

ksas

k

kk

e

ksee

tfLatL

−

−−

→

→

=

−=

=−∴ δ

{ } aseatL −=− )(δ .

11)(0

=== ∫∫+∞

dtk

dttfIka

akk

8

The delta “function” has the properties:

(i)

=∞

=−otherwise0

)(at

atδ

(ii) 1)(

0 =−∫

∞dtatδ

(iii) ).()(

0 atUda

t−=−∫ ttδ

NOTE: This is different to the sort of function we normally deal with, since it would be expected that a function that is zero everywhere except at a single point would have an integral equal to zero. Nevertheless, it is convenient to use )( at −δ as though it were an ordinary function. It is an example of a generalized function.

There are many other operational properties of the Laplace transforms that would enable you to find the transform of many more functions.

9

Questions for Lecture/ Workshop Find the Laplace transforms of the following functions:

1. ttf 6)( =

2. tetf 54)( −=

3. ttf 2sin)( =

4. tetf t cosh)( −=

5. 2( ) sin .tf t e t=

6. Write the function

>−

<<=

1)1(

10 0)(

tt

ttf in terms of the unit step function.

7. Graph the function ( ) 2 ( 1) 3 ( 2)f t U t U t= − − − and find its Laplace Transform

8. Write the function

>−

<<=

21

203)(

t

ttf in terms of the unit step function and hence

find its Laplace Transform.

9. Find the Laplace transform of the function having the following graph.

2

1

-1 1 2 3

10

Answers:

2 2

2

2 2

2 3

6 4 2 11. 2. 3. 4.5 ( 2)( 4)

25. 6. ( ) ( 1) ( 1)( 1)[( 1) 4]

2 3 47. { ( )} 3 8. { ( )}

2 29. { ( )} .

s s s

s s

ss s ss s s

f t t U ts s

e e eL f t L f ts s s s

e eL f ts s s

− − −

− −

++ ++

= − −− − +

= − = −

= − −

11

Tutorial Questions 1. Find the Laplace Transform of the following functions:

32

2

4 3 2 2 2

3 1

1(i) ( ) 3 (ii) ( ) cos5 sinh 3 (iii) ( )

(iv) ( ) 2 5 8 (v) cosh (vi) ( ) ( 1)(vii) ( ) sin cos (viii) ( ) cosh (ix) ( )

tt

t

t t

ef t e f t t t f te

f t t t t f(t) t f t t tf t t t f t e t f t e −

+= − = + =

= − + − = = −

= = =

2. Evaluate: 3 2 2(i) { sin } (ii) { } (iii) { sin }t t tL e t L t e L e t− −

2(iv) { sinh }L t t

3. Let

≥

<≤=

.10

101)(

t

ttf

(i) Sketch )(tf . (ii) Write )(tf in terms of the unit step function and hence find )}({ tfL .

4. Given 0 0 1

( ) 1 1 22 2

tf t t

t

< <= < < >

find { ( )}L f t using the unit step function.

5. Given 1 0 1

( )0 1

te tf t

t − < <

= >

find { ( )}L f t using the unit step function.

Extra questions 1. James: Page 892 Questions 1 and 3, Page 897 Question 4, Page 906 Question 5 2. Kreyszig: Page 210 Set 6.1 Q1-8, 33-36, Page 223 Set 6.3 Q2-11

Answers: 2

2 2 5 4 2 2

1

6 4 2 2

2 4 2

3 1 3 2 1 24 12 5 8 2. ( ) ( ) ( ) ( ) ( )2 ( 2)( 1)25 9 ( 4)

5! 2 3! 1 1 1( ) ( ) ( ) ( )( 2) 34

1 3! 1 1 1 1. ( ) ( ) ( ) ( )2 1( 1) 1 ( 2) ( 1) 4 ( 1)

s s si ii iii iv vs s s s ss s s s s s s

s evi vii viii ixs s ss s s s

si ii iii ivss s s s

−

+ −− + − + −

− + −+ − −

× −− +

− −+

+− +− + + + + −

1

2 3 3

1 .( 1)s

−+

2

( 1)

1. (i) (ii) (iii) ( ) 1 ( 1) .

1 1.1 1

s s s

s s

e e ef t U ts s s s

e es s s s

− − −

− − −

− + = − − +

− − +− −

3 4

5

12