tranform laplace

8/12/2019 Tranform Laplace

1/34

Programme 6

Framesto 34

lntroduction

o

Laplace

ransforms

Learning

outcomes

\\'hen you have

completed

his Prograntnte

ou

will be able to:

o

Derive

he

Laplace ransform of an expression

y using the

integral

definition

o Obtain inverseLaplace ransformswith the help of a Tableof Laplace

transforms

o Derive

the

Laplace ransform of the derivative

of

an expression

o

Solve

irst-order,

constant-coefficient,

nhomogeneousdifferential

equations

using the

Laplace ransform

o

Derive

urther

Laplace ransforms

rom known transforms

o

Use

he

Laplace

ransform to obtain the solution

to linear, constant-

coefficient,

nhomogeneousdifferential equationsof second

and

higher

order


2/34

1118

Programme

2

The

Laplace

ransform

All the

differential equations

you

have looked

at so far have

had

solutions

containing a number of unknown integration constantsA, B, C etc.

fhe

values

of these

constants

have

then been found

by applying boundary

conditions

to

the solut ion, a

procedure

hat can often

prove

to be tedious. Fortunately,

or

a

certain type of differentiai

equation there is

a method of

obtaining

the

solution where these unknown

integration

constants are evaluated

during

ht

process

f solutiorr. urtherrnore,

ather than

employing integration

as

he r,var

of unravelling the differential

equation,

you

use straightforward

algebra.

The

method hinges on what

is called the Laploce

ronsform.If

f

(f

)

represents

some

expression

n

f defi ned for f

)

O, he

Laploce

ronsfbnn

of

f

(f

),

denoted

br

L{ f( . t ) ) , is def ined o be:

f \

L \ f ( t ) l :

I

e

" f i f ) d f

. l r o

where s is

a

variable whose

values

are chosen so as to

ensure that

the

seni-

infinite integral

converges.More will

be said about the variable.s

n

Frame.l.

For now, what would you

say s the Laplace

ransform

f

(t)

-Z

for

f

>

0?

Sttbstififte

br f

(t)

irt the integral

obove md then

perform

the integrotiutt

The

ctnswer s in the

next

ft'arttt

provided.s

> 0

Because:

f \

t { [ { t t }

-

|

r ,

" 1 r t r d I

.J -u

SO

L { 2 }

e

t t Z

d t

-0

[, '

l '

L l l , , ,

(0

-

( -1 l . s ) )

Notice that s > 0 is demanded because f s

s

-

0 then Ltz l is not

def ined

( in

both

diverges),so that

< 0 t h e n e 5 r - : c a s f + r . a n d 1 :

of these two

cases the integra.

I

- l

, | ,

a

a

Z

.s

L

;

12]i provided

s

>

0


3/34

-,ction

to Laplace

transforms

lltg

Bv

the same reasoning,

f k is

some

constant then

L'

t 1 ( |

- "

p r o v i d e d

0

_ s -

\ow,

how

about the I-aplace

transform

of

f(f)

:

e

kt,

t

>

0 where

k is a

ionstant?

Go bcrck

o tlte integrttldefinition

trtrd work

it ottt.

Again,

the artswer

s in the next

fiame

l { , '

^ ' }

- ,

l

O

p r ov i ded

(

f i

:

I

e ' t e

k t d t

.,

-0

l I

-

|

.

r s t ( 1 4 1

J :rt

[ , ,

' + ( r ] \

t - l

l

(s '+

)1,-,,

( , , (

|

\ \

. s + k > 0 i s d e r n a n d e d t o e n s u r e t h a t t h e

\-

(.t

+

k)//

integral

convergesat

both l imits

I

-

;: ^

provided .s+ k > 0, that is provided .s> k

\ J

|

^ l

1

I

Because

tk

r ' )

I hese

wo

exampleshave

demonstrated

hat

you

need to

be careful about

the

iinite

existence of

the Laplace

transform and not

just

take

the integral

clefinit ion

without

some thought.

For the Laplace

transform

to exist

the

integrand

e

't

f

(.t)

lnust

converge

o zero

as f

-

rc and

this

wil l

impose some

conditions

on the

values

of .s for

which

the integral

does converge and,

hence, the I-aplace

transform

exists. In

this Programme you

can

be assured hat

there are no

problems

concerning

the existence

of any of the Laplace

ransforms that you

ni l l

meet .

Move

on

The

inverse

Laplace

transform

-fhe

Laplace ransform

is an

expression n the variable

s which is

denoted by

F(s) . t is

sa id that

f ( f )

and F(s)

-LI f

( t ) j

form a t ransfonn ai r .This

means

that if

I(s) is

the Loplace ransfonn

of

f(t)

then

f(f)

is the lnverue

Lcrplace

trunsform

of

F(s).

We write:

f ( t ) - L ' { r ( r ) }

>

to the rrcxt

rame


4/34

1120

programme

_i

There

is no simple

integral definition

of the inverse

transform

so you

have

:

find

it

by

working

backwards.

For

example:

i f

f

( t )

:4

then

the Laplace

ransformL1f

@|:

f (s)

:1

.

S O

4

i f F(s)

:

:

then

the inverse

Laplace ransform

r

t{F(.r)}

-

f(t)

-

4

.s

It is this

abil ity to find

the Laplace

ransform

of an

expression

and

then

rever:.

it that

makes the Laplace

ransform

so useful

in the

solution

of

different:,

equations,

as

you

will soon

see.

Ior

now, what is

the inverse t,aplace

ransform

of F(s)

:

I

.,

. s - I

L

' { F ( s ) }

- f ( t ) - e '

know

that:

1

- Vou

CanSaVha t

) + K

sow henk :

r , r t {L }

,

l s

r

Becauseou

t { t

r ' }

:

1 ' {

1

} - . ^ ,

I S + K J

l ,

l l t

_ p t

To assist

n the

process

of

finding

Laplace

ransforms

and

their inverses

a

talr_.

is

used. n the next frame

is a

short table

containing

what

you

know

to

dat.

s

Table

of Laplace

transforms

f ( f ) : L ' { r ( r ) }

F(s)

L\f

( t ' ) j

k

e k t

k

J

I

s + k

s > 0

s > - k

Reading

he table from left

to right gives

the

Laplace

ransform

and

readinr

the table

from right to left gives

the lnverse Laplace

transform.

use

tlrcse,where

possible,

o answer

he

questions

n tlrc Revision

exercise

tlr,;.

follows. Otherwiseuse the basicdefinitictnbr Frtrnre

To answer

tlis, Iook

at the Laplace

ronsfonns

you

now

krtt..,.

Tlrc

nttswer

s in Ilrt

rtt 'xI

[r ',rt,


5/34

lntroduction

o

Laplace

ransforms

'l'l2l

SJ

nevision

ummary

1

The

Lttltloce ronsfonn

of

f(f

),

denoted

by L{f

(t)},

is defined

to

be:

f -

t { / 1 r r } | e " l r t r d t

. l t l

where

.s s a

variable

whose valuesare chosen

so as o

ensure hat

the senti-

infinite

integral converges.

2

If f '(s)

is the

Ltrpluce ronsfbnn

of

[(r)

then

f(f)

is the inverse

Loploca

trotrsfitrrn

f

F(s). We

write:

l ' t t t

L r { F r s r }

There

s no simple

integral dcfinition

of thc

inverse ransform

so

you have

to

find

it

by

working

backwardsusing

a Tableof'Ltrpltce

ronsfbrnrs.

Z{

Revision

exercise

1

Hincl

he l.aplace

ransfclrm

of eacl'tof

the fcll lowing.

ln each

case

def ined

or I

- '0

:

(a)

( r )

-

3

( d ) f ( f )

-

5c

2

lrind

tftc

inversc

1

(a )

F (s )

s

:i

( ( l )

/ r ( r - )

-

L

a. )

( b )

l ( t )

-

c

(e)

f ( t )

-

2( '7t

2

transfr l rm

of cach

1

F ( s )

-

t

.s .)

l

1'(r")

2s

:3

(c)

l'(t)

-

e2t

of the fo l lowing:

.J

( c )

F ( . \ ) -

^

. \

r L

. l l

l.aplace

(b)

( c )

Solt t t i t rrr .s

tt

t rcx[

fr t r t r t '

(a)

f ( t )

-

. l

Because

{k }

(b)

f(t)

-

c

providcd ,t

>

0, l{ 3}

-

C

p r o v i d e d s > 0 , L { c } : -

provided .s

>

0

provided s >

0

_ {

.)

Because

U,)

-

|

(c)

f ( t )

-

ezt

Because

\t

u']1:

1

J

1

s + k

F

provideds

>

k, L\ezL1

provided

s

>

2


6/34

11?2

Programme

:

f

( t )

-se

' '

L l . 5e

" l

l '

e

" r

5 , ,

3 / rd r

- 5

|

e ' / c

i r c l t

_511e

1 1

. J r

, 1 , , ,

l { 5e

' r }

'

. , p rov ideds .3

. \ t J

f ( t ) - 2 e "

z

f -

L lZe" ' l - 1 e

) t -o

) , '

2

L { 2 e 7 '

2 }

-

=

F(s ) :

1

.s

Because ' { i

I s J

F ( s )

- 1

. s - . )

(

1 )

, , t (

I

I

B e c a u s e L

{ '

. I

- ( " , L ' - ( '

l \

K )

[ 5

s J

( 1 )

Because

'4= - l

:

e

2 tand

l { . 3e

z ' }

- . l t { e

2 t }

:

[ . )

| z J

( 2 )

t ' l - ' ' ; f

: 3c2 '

[ ) t 4 J

:l

. i ) s o

. ) T a

(d)

(e)

\ \ 2 e t ' , C , t -

2 1

2 |

e ' / e ' d t 2 c

) L 1 e - t 1

.J,-

provided

s

>

7

2

(a)

:k,Lr{+} . '{i} :- '

(b)

(c)

(d)

(e)

1

F ( s )

- .

4.S

-

3 t - 3 4 r

r ? )

, f

3

4 l

r ,

F r r -

-

4s s

|

+sJ

-

t

s

J

I

_

\ " /

a - a

z.\ .)

F(s)

2 s 3

so h a t

( t )

- L ' { " = }

1

;

Z

3

s - -

L

-, I

l

+ ,

-jl:''

Nert

fi

'r r

Beforeyou

can use the

l.aplace

ransform to solve a differential equation vo,:

need to know the Laplace ransform of a derivative. Given some expressi,r-

f ( f )

wi th

Laplace t ransform LI i( . t ) \ :F(s) , the Laplace t ransform

of

tnt

derivative

f'(f)

is:

f 1 .

L \ f ' ( t ) l

:

I

e ' t f ' ( . t ) d t

. . , /_o

l-aplace ransform of a derivative


7/34

lntroduction

to Laplace

transforms

'fhis

can be

integrated by

parts

as

follows:

l ' \

L l [ ' t t t l

-

|

.

' i 1 ' r t r d l

. ,

-0

f ^

ru ( f

dv{ r )

. l t o

t l ' I

( t h e P a r t s f o r m u l a - s e e

I t r { l

l t r ' ' . 1

, u

-

. 1 ,

, t ' " o t " "

P r . g r a m r n e 5 ,F r a m c

l t

where

u ( t ) :

c

' r

so du ( r )

. se

td f

and

where

dv ( f )

-

f l ( r ) d f

so

v ( f )

-

f

( t ) .

Therefore,

substitution

in the Parts ormula

gives:

I

l - f -

L{ f ' ( t ' t } -

] t

" f ( t ) l

o . t

I

c ' I f 1 r ;d r

L - l : t t

' l t

'o

-

( 0

f (0 ) )+

sF(s ) ssuming

' I f ( r )

-

0 as

* : r

That

is :

L { f ' ( t ) \

:

sF( . s )

f ( 0 )

So

the

Laplace

transform of the derivative

of

f(f)

is

given in terms of the

l.aplace

ransform

of

f

(r)

itself and the

value of

f

(t)

when f

-

0. Before

you

use

this

fact

just

consider

two

properties

of the Laplace transform

in the next

frame.

1129

Two

properties

of

Laplace transforms

Both the Laplace transform and its inverse are lirrcsr trcrrtsforms, y r.t'hich is

meant that:

(1)

The

transfornr

of a surn

(or

diffbretrceof expressiotts

s thc stun

(or

difference) f

the

individunl

rarrsfonns. hqt is:

r { r ( r )+s(r)}

r { r ( f ) } r {s(r) }

a n d

I

' 1 r 1 . s 1 + G ( s ) )

: l

1 { F ( . r ) } + L

{ { ; ( s ) }

(2)

The

trcrnsftinn

of on exprassionhat

is rruiltiplied by a constottt

s the cortstant

multiplied

by the tronsform

of tlte expressiort.

hctt is:

r { k f ( r ) } : kL \ f ( t ) } and l 11 ru1s ;1:kL |

{F (s ) }

wherek i s a cons tan t

'Ihese

are

easily

proved

using

the basic definition of the

Laplace ransform

in

Frame

1.

Armed

with

this information

let's try a simple differential

equation.

By using

Ll f ' ( ) \ :

sF( .s)

f (0)

take

the

Laplace ransfonn

of both sidesof

the equation

f '

( t )

+

f ( . t )

:

I

where

f(0)

:

O

and find an explession or the Laplace ransform I'(s).

Work through tltis steodily

using what

you

know;

vott will find the

onsv'er

n Frarne 12

by r.t'hich is


8/34

1124

Programme

26

Because,

that:

L{ f ' ( . t )

That is:

1

f

( s )

-

s (s r )

taking

Laplace

transforms of both sides of the

equation

you

have

r Frr \ rJ l l

The Laplace ransform

of the lef t -hand

sideequals

'

/

\ / '

-LLrr

the Laplace ransform of

the

r ight -hand

side

L{f '(L)l

+

LIf

G)}:

r{ 1}

l l :" tT: l"m

of a sum s hesum

of the

translorms.

From what

you

know about the Laplace transform

oti

f

(t)

and its

derivative

f'(f)

this

gives:

l sF(s)

f (o) l+r (s)

l

s

That

s:

(s

+

1)F(s)

f(0)

:1

and

you

are

given

hat

/(0)

:0

so

.t

l _ l

1s r l )F1s ) , t ha t i s F (s )

-

- - - -

's s ( s l )

Well done. Now, separate he ri ght-hand side into

partial

fractions.

Yotthsve

done

plenty

of this before n ProgrammeF.7;

the answer s in

Frame

1 1

i ( S ) - - - -

s s + l

Because

I A B

Assume hat

^

:

+

-

then, 1

:A(s+

1)

+Bs

fromwhichyou

s ( s + l )

s

s ,

I

f i n d t h a t A : 1

a n d B -

1

s o t h a t F ( s )

1 -

t ,

. s . r + 1

That was straightforward enough. Now take the inverse Laplace

transform

and

find

the solu tion to the differential eouation.

The znswer is in

Frame

11


9/34

I

ntroduction o Laplace

ransforms

1',25

f ( 1 )

: 1

c ' I

Because

f ( t ) : L ' { r ( . r ) }

, t [ 1

1

I

L ( - )

l s

s t J

_ ,

'11 \

_ r

,

|

\

r he inve rseLap lace t rans fo rmo f

d i f f e rence

-

|

rJ

"

[s

+

1J

is

the difference

of the inverse

ransforms

:

1

e

'

Using

the Table

of Laplace

ransforms

in Frame

6

You now have

a method

for

solving

a differential

equation

of the form:

af ' ( t )

+bf

( . t )

g( f )

g iven

that

f (0)

:

k

where a, b and

k are known

constants

and3(f)

is a known

expression

n

f:

(a)

Take the Laplace

ransform

of both

sidesof

the differential

equation

(b)

Find

the expression

F(.s)

L{f

(.t)}

n the

form

of an algebraic

raction

(c)

SeparateF(s)

nto

its

partial

fractions

(d)

Find the

inverse

Laplace

ransform

r

r{F(s)}

to find the

solution

f(f)

tcr

the

differential

equation.

Now yctu

ry sonrc

but beforeyou

do

jttst

ook dt the

Toble

of

Laploce

trartsfrtrmsn the next frame. You will neetl hem to solve

tlrc

equations

n fhe

Revision

exercise

that

fbllotus.

34

-".:

Table of Laplace

transforms

f ( t ) - L ' { F ( . s ) }

F(s)

L{ f

( t ) }

k

e k t

re

^'

r

s

1

. s + k

1

:

(.r

+

/()"

s > 0

s > - k

s > - k

We

will derive

this third

transform

later in

the

prograrnme.

For now, use lrcse

o answer

he

questions

hat

follow

f/le Revision

summary

in

the

next

ftcune

. $


10/34

1126

Programme

26

@

nevision

umm ry

I t ff

1 If F(s)

is the

Laplace transform of

f(f)

then the Laplace transform

oi

I l l J

I

t - ' r l r i s .

|

' t ' " '

L { f ' ( t ) }

- - s F ( s )

/ ( 0 )

2

(.a)

The Laplace ransform of a sum

(or

difference) of expressions s

the

sum

(or

difference)of

the individual

transforms.

That is:

t { f ( r )

+ . { ( t ) } t { f

( t t )

+ t { ( s ( f ) }

and

I

r {F ( s ) + t i ( s ) }

r

1 {F ( i ) } + r { c ( s ) }

(b)

1'he transform of an expression

multipl ied by

a constant is

the

constant

multipl ied by the transform of the expression.

I 'hat

is:

L l k l t t t l k L t t l ' t r l a r t d

r l k F n ' |

-

k L

r { F r r

w h e r e k i s a c o n s t a n t .

3

To solve a differential equation

of

the

form:

0f ' \ t )

t

bf

( l )

-

3( f )

given

that

f (0)

:

k

where

a,

b and k arc known constantsand

g(f

)

is

a

known

expression

n f :

(a)

Take

the

Laplace ransform of both sidesof the differential equation

(b)

Find the expression

(.s1

L{f

(t)}

in the form of an algebraic

ractior.t

(c)

Separate

(s) into

its partlal fractions

(d)

Find the

inverse Laplace ransform f

r{F1s)J

to f ind the solution

f

t

to

the

differential eouation.

ffi

Revision

xercise

I { ?

Solve each of the folloning differential equations:

,

'

'

( a ) f ' ( t )

f ( t ) : 2whe r e f ( o )

o

(b)

r ' ( r )

+

f ( t )

-e

t

where

(0)

0

(c )

f ' ( r )+ l ( r )

3

where

(0 )

2

(d)

f'(r) f'G)

e''

where

(0)

-

1

(e)

3r'(r) 2f

.t) 4e

t

-l-

where

(0)

0

Solutions n next

tidn:,


11/34

lntroduction

o Laplace

ransforms

1127

(a)

f ' ( t)

f

( t)

-

2 where

(0)

:

0

TakingLaplaceransforms

f both sides f this

equation

gives:

.sF(s)

/to1

F(s)

?

so hat F(.s)

- : ^

-

-2

+

-L

,

s \ ( 5 l ) . \ s I

The inverse transform

then

gives

the solution

as

f

( t ) :

- z

+

2 e t 2 ( t ' t 1 )

(b)

f ' ( f )

-

f

( t )

-

e '

t

where

f (0)

-

0

Taking

Laplace

ransforms of both sides

of this equation

gives:

sf ' (s)

f(0)

-

F(s)

- f

,o that F(.r)

1

=

J

r

r

r r

,

1 1 2

The

'l 'able

of inverse

ransforms then

gives

he solution as

f

(t)

:

7g

r

(c)

f ' ( f )

f

l ' ( r )

- 3 where

(0)

- 2

Taking Laplace ransforms

of both sidesof this

equation

gives:

.sF(s)

/ (0)

+

F(s)

9

so that

s

- 2 3 : J 2 s 3 5

s

I s r s l , 1

s '

l 1

s

s I

The

inverse

transform then

gives

the solution as

f(f)

-

.J 5e

I

(d)

f'(.t)

-

f ' \.t)

r '2rwhere

f(o)

-

1

Taking Laplace ransforms of both sidesof this equation gives:

. s F ( s )

f ( O ) - F ( . r )

j

S i u l r l g

- ,

l ) F ( r ) -

1 :

]

^

l l l

s o f l a t ' t '

-

i -

l

, ,

l

r r s

2 , . s 2

The inverse transform

then

gives

the soh,rtion

as

f

{.t1

szt

(e)

3f ' ( r )

-

2f

( t )

-

1e

t

+

2 where

f(0)

-

0

Taking Laplace ransforms of

both sidesof this equation

gives:

3 isF(s )(0) , 2F(s11* ? : ,1 ,1*? , o ha t

\

r

I

\ \ ( s

l )

6 s + 2 2 7 / 1

\

1 1 / 1

\

s r , s

l , ; t . i s 2 t

5

\ 3 s - 2 /

r

. 5 \ r | /

The inverse transform then

gives

he solution

as:

9 . . , +

I t l l - , " I

.) .)

'i(+)

l-i(+)

\ .

1/

On now to Frante

19


12/34

1128

Programme

26

t * *

;

T

S

Generating ew transforms

Deriving

the Laplace ransform

of

f

(f)

otten requiresyou

to integrate

by

parts,

sometimes

repeatedly. However,

because LIf '(t)j

-

st{f(t)}

f(0.1

you

can

sometimesavoid this involved processwhen you know

the

transform

of

the

derivative

f'(f).

Take as an

example

the

problem

of finding

the Laplace

transform

of the expression

f

(t)

:

t. Now

f'(f)

:

1

and

f(0)

:

0

so that

substituting in

the equation:

LI f ' ( t ) | :

s r { f ( r ) }

f (0)

gives

t t 1 ] - s I - { r } - 0

that is

1

-

s r { f }

s

therefore

L { t }

: :

J '

That was

easyenough,

so

what

is

the

Laplace

ransform

of

f

(t)

-

t2?

The

snswer is in

the next

frorrtt

1

i

I

i

e*

Because

f

(t)

-

t,,

f '(.t)

2t and

(0)

:

o

Substituting in

L{f '

t)}

-

sr{f(t)}

f(0)

gives

L {z t } : . r r { f 2 } - O

that is

^ ?

2 L l t l

- s t { t 2 }

s o

;

-

s t t / 2 }

' s -

therefore

a

^ /

L { t ' }

:

=

Just

try another

one.

Verify

the third

entry in the Table

of Laplace

ransforms

in Frame 15 for k : 1, that is:

, - l

I ) t ) |

-

L t

|

-

( s +

1 ) '

This is

a

littler

harder but

jttst

follow

the

procedure

aicl

out in the

previorls

hlr)

frames

crnd ry it. The explattation

s irr the

next

frortrt


13/34

lntroduction

o Laplace ransforms

Because

f

( t ) :

t e

' ,

f ' ( t ) :

e

'

r e

I

and

f (o )

Substituting

in

L{f ' ( t ) } : sr { f ( t ) } f (0)

gives

L{e

'

te

t1

:

sL{te

t}

0

that

is

L { ,

' }

-

L { te

' }

-

s t { re

r }

therefore

t { t ' l :

( s +

I ) L { t e

t l

giving

- 1 ,

- ( s

*

I

) l t f e

' ]

a n d s o { f e

' }

s + 1

1129

1

:G+1f

On

now

to

Frame22

Laplace

ransforms of

higher

derivatives

The

Laplace transforms of derivatives higher than the

first are readily derived.

Let F(s) and

G(s) be the respectiveLaplace ransforms of

f(f)

and

J:(f).

That is

L \ f

( t ) l : F ( s )

so

tha t r { f ' ( f ) }

:

sF(s )

f

(0 )

and

r {s( t ) } :

G(s) nd {s ' ( t ) i sG(s)

s(0)

Now

let

.S(f)

f '( f

)

so that r{g(f)}

:

L{f'(t)} where

.s(0) f'(0)

andG(s)

sr(.s)

f(0)

Now,

because

(f

)

-

f '( t)

g ' ( t )

-

f "

( t )

This

means that

r {s ' ( r ) }

L{ f " ( t ) } : sG(s)

s(0)

:

s fsF(s)- f (O) l

f ' (o)

SO

L{f"

t)}

s2r(s)sr(O)

r'(0)

By a similar

argument

t

can be shown that

L{f" '(.t)}:

s3r1s;s',r(O)sr'(0)

f"

(0)

and

so on.

Can

you

see he

pattern

developing here?

The

Lap lacerans fo rm

f

f i ,

t , i s . . . .

Nex t

rome


14/34


15/34

1130

Programme

26

L{ fu

t ) }

-

s1F(s) s : r1 '101

21 ' tO1-r " (0)

f " ' , (O)

Now,

us i ng L { f " ( . t ) }

s 'F ( s ) s l ' ( 0 )

l ' ( 0 )

t he Lap lace

t r ans f o r m

o i

f ( f ) : s inkf where k is a constant s . .

Diffirentiate

f

(t)

twice

ord

follow

the

procedure

hat

yotr

used

n Franres

19 to

21.

Tqke it

corefully, he ttnswer

nrd tuorkingLtreht

tlte

following frarrrc.

r{sinkf}

"+"

Because

f( f )

:

s in kt ,

f '

( t )

-

k

coskf and

f"

( t )

-

k2

sinkf .

A lso

l ' (0)

:0

and

f ' (O)

-

f .

Substituting in

L I f " ( t ) )

: . s2 r 1 . s1

s l ( 0 )

f ' ( 0 )

wher eF( .s ) L l f

( . t ) j

gives

L { - k2s i nk f }

: . r z l { s i nk f }

s . 0 k

that is

k2Llsinkr l

:

.s2t{sin r }

k

S O

1.s2

i


16/34

lntroduction

o Laplace ransforms

1131

Tableof

Laplace

ransforms

f

( f )

t ,

' { r ( r ) }

F( .s ) L { f

( t ) }

k

e k '

te

kt

t

P

sin kf

coskf

t

s

1

. r + k -

1

-

(s

+

k ) '

1

5t

L

F

k

.)j

.r-

+

K'

J

U

.s-

+

K'

. s > 0

- s > - k

s > - k

s > 0

s > 0

. s 2 + k 2 > o

. s 2 + k 2 > O

Linear,

constant-coefficient, nhomogeneous

differentialequations

1'he Laplace ransform can be used

to solve equations of the form:

a, , f ( " ' )

. t )

a, ,

t f i "

t ' ( r )

+

. . .

+

a2f"

. t )

+

a1f '

t )

+

a\ f ( t )

-

s( t )

where al l ,011

, . . . ,

ez, a1,ag ?t? known constants,

( t )

is

a

known

expression

in f and the

values

of

f

(t)

and its

derivativesare known at f

:

0. This type of

equation

is called a linear, cotlstatlt-coeflicient,

nhr,tmogeneous iffbrential

equatiort nd

the values

of

f

(f)

and its derivatives

at

f

:

0 are

called

boundary

conditions.

The method of obtaining

the solution follows the

procedure

laid

down in Frame 14. For example:

To

find the solution of:

f "

( t )

+

3f ' ( t )

+

2f

( . t )

4f where

f

(0)

:

f ' (0)

:

0

(a)

Tcrke he

Loploce

ransform of both sidesof the etyntion

L{f"

. t)}

3L{f '(t) l

+

2L1f

rD

4L{t}

to

give

[s2r1s;

sf(0)

l"(0)]

+,3l.sF(.s)

(O)l

+

2r(.s)

-1

I

I

I

g6

4

s2

F


17/34

1132

Programme

):

(b)

Find

the exprcssitm (s)

-

L{f

(t)l

in the

fitrm

of an algebraic

fioction

Substituting

the

values

for

f(0)

and

f'(0)

and then rearranging

gives

1s2+ : l . s

z )F (s l -

52

so that

4

F t t :

-

- -

i '.;i

's

-)

(c)

Seporate

(s)

into

its

partiol

fructions

4 A B C I )

sL.s

16

+

z;

:

T- . t t

- . r

L '

t

+

2

Adding

the right-hand side partial

fractions

together and

then

equating

the left-hand side numerator

with

the

right-hand

side

numerator gives

4

:

As ( s

+

1) ( . s 2 )

+

B( s+

1) ( . s 21

+

Csz ( s 2 )

+Ds21s

1)

L e t s - 0 4 - Z B t h e r e f o r e B - 2

s :

- 1

4 : c ( 1 ) 2 ( - 1

z ) :

c

. s : 2 1 - D( . 2 )2(

2+ i )

-

. tD

t he r e f o r eD

-

-1

Equate

he coefficients

of s:

O

-

2A+

38

-

2A

+

6 therefore

A

-

-3

Consequentlv:

_ 3 2 4 1

s

s /

s

L \ 1 2

(d)

Usc he Tttbles o

firtd

the inverse

aplace ronsfonn

L

r{F(s)}

ctndso

furd

tlr,.

solutiorr

f

(.t)

to tlrc clifferential

eqtntiotr

f ( t ) :

3 + z t + 4 e t

e 2 t

So lnt wns

crllverystraightforwsrtl

even f it was involved.Now

try

your

hond

ot tlte

diff'erential

equations n

Frttnrc

29

Revision

ummary

:

#,t::'

1 If F(s)

is the Laplace ransform

of

f

(f)

then:

L{f"

t)1

s2r1.s;

f

0)

f '(0)

and

L{f"'(t) l :

.s:JF(s).s2r(0)

sf '(0)

f"(0)

2 Equations

of the

form:

t , , f i " ' ' ( t )

a , ,

t f ' ' "

t ' ( r ) +

. . . + e 2 f t t ( . t )

+ a 1 f t G )+

r i o f ( r )

. g ( f )

where

( t i l t

o i l

t , . . . ,

o2, oy, o11

t Constants re cal led l inear ,

coefficient, nhomogeneous

differential

equations.

constant-


18/34

lntroduction

o Laplace ransforms

3

The

l .aplace t ransform can be used to solve constant -coef f ic ient ,

inhomogeneous

di f ferent ia l equat ions

provided

a, , , a, ,

r , . . . ,

oz, or , Lto

are

known constants,

g(f

)

is a known expression

n f, and the vah-res f

f

(r)

and

its derivatir, 'es re known at f

:

0.

4

The

procedure

for solr. ing hese equations of

second and

higher

order

is

the same as that

for

solving the equations

of f irst order.

Namely:

(a)

Take the l-aplace ransform of both sidesof the

differential equation

(b)

Find the expression

(s) L{ f

( t ) l

in the form of an a lgebraic

ract ion

(c)

Separate (s) into its

part ial

fractions

(d)

Fincl he

inverseLaplace ransform L

r{} ' ( -s)} to

f ind the solut ion

f ( f )

to the differential e0uation.

1133

Zl Revisionexercise

Use ire

Laplaceransform o solveeachof

the fol lowing equations:

/fi={j;;

:

(a )

f ' l t )

r

f ( r )

:

- l

where

(0)

-

0

(b)

3r '( f)

+2f

( t)

-

t

n 'herc

(0)

-

2

(c)

r"(r)

t

5f

( f)

+

6f ' ( t) 2e

I

where

(0)

-

0 and

',(0)

0

@)

f '"( t)

4f

( t)

:

sin2t

where

(0)

-

1 and

'(0)

:

-2

An.sr,uerstr rrcxt t'orne

(a)

f ' ( t )

+f

t . t ) :3

where

(0)

o

: .3{ - t

' l 'aking

Laplaceransforms

f both sides f the equation

gives

L { f ' ( . t ) }

1 1 1 1 r ;1L { 3 } so h a t

s F f

s )

f ( O ) j

F ( t )

:

That is

(s

-r

1)i'(s)

i

,o

F(s)

--1=,

-=

*-3

-

5

s 1 r

I '

r

s

I

g iv ing

he so lu t ion s

( t )

-

3 3e

I

:3 (1

*e

t )

tr


19/34

1134

Programme

26

(b )

3 f ' ( f )

+2 f

( t ) :

f wh e re

(O ) :

- z

TakingLaplace

ransforms f both sides

f the equation

gives

L13f '(t) j

+

LIzf

( t ) l

:

r {r}

so hat 3lsF(.s)

(0)l

2F(,

:

\

.t '

That

s

(3s

2)F(s1

(-6)

:1

,o

Fisl

l : j*

\ r

s 2 r 3 s

2 1

The

partial

fraction breakdown gives

3

1 1 1

15 1 : l 1 1

1 5 1

, - r s r _

r

_ _

. t s

2 s 2

4

r 3 s l r -

+ r

t

j

+ r s - 1 '

giving

he solutionas

3 f

\ s z t " ' l

f I f \

t -

4 2 4

(c) f"( t ) + sf ' ( r ) + 6f ( t ) :2e t where f(0) - 0 and f ' (0) - 0

Taking

Laplace ransforms of

both sidesof the equation gives

L { f " ( . t ) }

r { s r ' ( f ) } L {6 f ( t ) l

L {2e

t }

so

hat

s2r'1.s1

f

0)

/ ' (0)l

+

sl.sF(.s)

(o)l

+

6F(s7:

?

-

5 + l

That is

(.s2

5.s 6)F(s)

-?

.

s + 1

so F / s )

2

J v

\ ' ' /

t t +

t [ s + 2 ) t t + 3 )

s + 1

-

s + 2 -

j ' + 3

giving

the solution as

f ( t ) : e t

2 1 2 t 1 1 3 t

(d)

f"( t )

4f

( t )

:

s in2f where

f(0)

:

1 and

f ' (0)

:

2

Taking

Laplace ransforms of both

sidesof the equation

gives

L\f"( . t ) | L{4f

( t ) l

-

L{s inzt l

so hat

.s2r1s;-

f(o)

f '(o)l

4F(s)

"+"Thats1s2+;r ' (s1s. l (_z) *L

. \ - T Z

2

s - 2

soF(s1

;: , ^

l ) -

-

+ , 1 s ' - 1 1 - ; ' -

1 5 1 1 1 1 2

: G

u r - r *

t

2

E s r + z ,

giving

he solutionas

-

1 5

, ,

I

, ,

s i n 2 l

l l l \ - - - p

' '

I

1 6 ' 1 6 ' 8

So, inally, the CanYou? checklist -ollowed y the Test exercise

and Further problems


20/34

Introduction

o Laplace ransforms

A

Can

You?

Checklist

26

(.lrcck his list befbreond afteryou try the end of Programtneest.

On a scale

of

1 to 5 how confident are

you that you can:

o

Derive

he Laplace ransform of an expression

y using the

integral

definition?

Yes

o

Obtain

inverse

Laplace ransforms

with

the

help

of

a Table of

Laplace

ransforms?

. l

Nr.r

1135

Frames

r"*rJ.

-r-J

[:*].

ilC

ilil

Yes

r

Derive

he

Laplace ransform of the derivative

of an

expression?

Y e s l ' . .

f No

o

Solve

irst-order,


nhomogeneous

differential

equations

using the Laplace ransform?

ffi.

{-'qJ

No

Derive

further

Laplace ransforms from

known

transforms?

ilpl

.

ilr,uJ

Ycs -- i

Use he

Laplace ransform to obtain the solution to

linear,


nhomogeneousdifferential equations

of

higher

order

than the first?

i'_Zl

.

il t

Y e s r [ - N o

S

Test

exercise

26

1 Using

the integral definit ion,

f ind the Laplace ransforms

for eac h of the

following:

(a)

f ( r )

-

8

(b)

l ' ( f )

-

esr

(c)

f ( r )

:

-4e2t t3

2 Using the

l'able of Laplace ransforms,

ind

the

inverseLaplace ransforms

of each

of the fol lowing:

( a )

l ( 5 )

-

.

l s ) \ '

\ ' -

-

/

3

l

3fri

I

c , , 1

tb t

F ls t

1 l

) .

q

( d l

F l s t :

- #

J ' t J

ts

( c )

F ( s )

s2 + 9


21/34

1136 Programme

26

yHllll

3

Ciiven

that

the Laplace ransform of fe

kt

is F(s)

-

,+

derive

the

[,,-..i

(s

+

/()-

Lapiaceransformof t2c:'1 ithout using he integral

definition.

4 Use he Laplaceransform o solveeachof the following

equations:

@)

f' '

.t)+ 2f ft) : I where (0) 0

( b )

f ' ( r )

f ( . t ) - a

r w he r e f ( 0 )

:

1

(c)

f"( f)

+

Lf ' ( t)

+

4f

( t)

-

e

21

where

(0)

:0

and

'(0)

:

t )

(d)

4f " ( t )

9 f

( t ) :

18where

(0)

-

0 and

' (0)

0

&

Further

problems

26

#"3

I

Find

the

being de

(a)

(r)

(d)

(r)

Find

the

(a)

F(.s)

(d )

F (s )

3s 4

7s2 27

s r + 9 \

T1Tl

(b)

Laplace ransfo

f i n e d f o r f > 0 ) :

: ( l k l , ( t > o

( X

f o r 0 < f

- )

f

o

fo r t>s

inverse

Laplace

2

of each of the following expressions

each

f( r)

:

sinhkt

0

( c l

f ( r )

- c os hk r

transform of each of the

(c)

-s2 fl

fn l l o r r r i n o .

3 . s - 4

F t t t - _

-s'+ ro

- )

-

.\-

()5

+

I'+

s r

-

J z

,

4 r

-

- l

3 Show that

i f F(s)

(b)

F(.s)

-+s

( e )

F r s )

( t )

( \ 2

7 ) '

L l l r t t l

|

.

" / ' , l r d /

t h e n :

.,

:o

(a)

( i )

F ' (s)

-L{ t f

( t ) }

( i i )

} ' "1.s1 L\ f f

( t ) }

Use

part

(a)

to find

(b)

( i )

r {r s inZt l

( i i )

t { f2cos3f}

(c) What would you say the nth derivative of I (.s) s equal to?

4 Show

that

if L{f

ft)}

-

F(s) then L{ektf

(f)}

:

F(.s k) where k is

a constant.

Hence find:

(a)

L\e"I s inbf

)

(b)

l{e"rcosbf}

where

s and

b

are constants

n

both cases.


22/34

lntroduction

o

Laplace ransforms

5 Solve

each of

the following differential equations:

(a)

f"

( t )

-

sf ' ( f )

+

6f(t)

-

0

(b)

f"( t )

sr '( f

+

6f

(t) :

I

(c)

f"( t )

5f '( t )

+

6f(t1

gzr

(ci) 2f" ( t ) - f ' ( t ) f ( t1 : ,

t t

(e)

f( t \

+

f '( t )

2f"

1t |

7g

( f

f " ( t )

+

16r(r) 0

1137

wherc

(0)

:0

aqd

'(0)

-

1

where

(0)

-

0 and

' ' (0)

0

where

(0)

=

0

and

' (0 )

0

where (0) - 2 antl '(O)- |

where

(0)

-

0 and

'(0)

-

1

where

(0)

=

1

and

'(0)

:

4

-:lrr

It,W$l

( S )

Z f " ( t )

f ' ( t ) f ( f )

- s i n r - c o s t

w h e r e f ( 0 )

0 a n d l ' ' ( 0 ) - 0

Now

visit the companlon

website for this book at

www.palglave.com/stroud

#4,

for more

questions applying this

mathematics to science

and engineering.


23/34


24/34

/(s)

F{t)

t4.

f-

f

(u)

d.u

F(t)

T

t5.

i=+=t

fo'

"-*"u'ou

F(t;

=

P1;a1,

t6.

/(/t)

a

hf"""

uz/4t

pt(ul

6.r,

17 .

Iliu'l

{o*

tn{'lfrt'{u1 ou

t 8 .

#lv'l

p,,

f

o

u-nrz 1n12,,1ffi)

fu)

du

t9.

f(s

* 1/s)

- 7 +

1

f,'

rrtzt/ if t=AtF@)

ilu

20.

#1,'

-atz a-s?lau (ul il u

r'(f,)

2r

/(ln

s)

s l n s f,-ryn"

27.

P(e)

=

suku banyak

dari bilangan

yang

lebih

kecil dari n,

Q ( s )

=

( s - c 1 ) ( s - a 2 )

" '

( s - a n )

di

manacl

,Q2, , , , ,o,

semuanYa

aPat

dibedakan

g

p("r)

oo*t

rar

Q'@x)

-


25/34

' t \ t

^lpendiks

TABEL

TRANSFORMASI

APLACE

HUSUS

/(s)

r(r)

I

I

;

1

2.

1

8,

t

3.

t

S"

n

=

1 , 2 , 3 , .

tn -

L

(;-ill

0 = 1

4.

I

8t

n ) 0

t tu* t

ilD-

5.

8 - A

eo t

6.

I

G-A;

n

=

1 , 2 1 3 , .

f f i . '

o =1

7.

- l -

n ) O

(8

-

a) "

Xn-I

ed t

riil-

8.

'|

TT7'

sin ol

a

9.

s

F 4 " ; ;

cos 6t,

lo.

I

G

*-bFT?

ebt

sin ot

a

t l

s - b

G:O?T?

ebt

cosot

12.

I

a*a

sinh

dt

.L

t3 .

E

a=at

cosh

ot

t4.

1

G=W=A

ct,

slnh

4

&


26/34

246

TABEL TRANSFORMASI

LAPLACE

KHUSUS

IApendiks

B

/(s)

F(r)

t5.

e - b

G=6?

-7

eb,

cosh

oi

t6.

ebt

-

eot

b - a

17.

a * b

bebt

-

aeat

b - o

18 .

k,

+W

g i n a t

-

a t co sa t

-__,4-_

t9.

a

@+W

I

sin g.i

2a

20.

a2

@+W

s i n c t

* o t co se t

2a

2 l

cos

ot

-

\a't

sin

at

22,

s 2 - a 2

;--:--;i;

ls. t

d'1.

t cos o,t

,3.

1

Gr=w

at cosh

ot

-

sinh

at

--w-*

2*.

I

Gt=w

t

sinh

ot

---2;*

25.

92

@-;rF

sinh

ot

* at

cosh ot

2a

26.

83

@-w

eoshot *

$atsinhat

27.

e 2 * a 2

;..;-.-M-

18'

-

d ' l '

t cosh

ol

28.

1

t--;1-d5

| 8 " - t o ' . ) "

-

$ot cosat

29.

s

Gt+ory

30.

82

@Ti-ozlt

(1

* aztz) sin

at

-

at cos

at

8o3

3 l

$3

@T;'F

3t

sin

of *

atz

cosat

-__T;-


27/34

: x J i k s

B

TABEL

TRANSFORMASI

LAPLACE

KHUSUS

247

i(s)

F(t)

32.

8{

(F+7p

(3

-

a2tzl

sin

at

|

\at

eosot

8o

33 .

G'+"43

(8

-

art2)

cos ot

-

?ot

sin of

8

34.

lsz

-

az

G'+"ry

t2

sin

af

_T;

35.

s3

-

3a2s

Gt+;ry

$tz

easat

36.

$t3 cosc.t

37.

s3

-

o2s

Grfiry

t3

sin ot

24"*

38.

I

Grj45

($

*

o2lz)

sinhot

-

Sot

coshot

8o5

39.

a

@=w

oi2

cosh at

-

t

einh

ot

8a3

40.

a2

@=ry

ot

cosh of

*

(azts

1)

ginh

ot

8a3

4 l

a3

Gt_"ry

3t

sinh ot

* ot2

eoshol

---_

6d,

42.

3{

G';T

(3

* az1z1

inh

ot

*

5oi

cosh of

8a

43.

g5

G':ZF

(8

* azfe)

cosh

at

* Tat

sinhs,t

8

44.

t2

sinh ot

-Ta

45.

sB

* 3o2s

tr=ry

{t2

cosh

ot

46.

a 4 + 6 @ 2 s 2 + a 4

-*lp

-;4a-

*ts

cosh ot

47.

ss

*

a2s

@-w

13

sinh ot

-ffi-*

48.

s 3 * o 3

#{lu"t"ry-"*ry*"-*,,)


28/34

/(s)

F(0

49.

a

s i * a s

#{*++y'

sin+-'- ' ' "}

s0.

gt

FT;6

+

(,-"'

+

zea'tz*"4t)

5 l

I

' s3

*

c3

ffi

{**,,

"o"

r*

/F"'"

}

52.

a

s 3 - c 3

u#{{"'"+-"'"ry*"*"'

53.

PTJg

*

(*'

*

z"-*,'

ro"{L"t)

54.

1

TT?;|

frtrtn

a.t cosh t - cosdt sinhot )

55.

*4

* 4aa

gin

of sinh

at

----zF-*

56.

a3

FT 4A

l(sin

ot

cosh

ct

* cos

at

sinh

at )

2e

s7.

s3

{77&

cos al

cosh

ot

58.

t

F*7

1|-{rittt

ot

-

sin

ot)

59.

s

s 4 * 4 4

,|'t"ost

oi

-

eos

at)

60.

sz

F-7

fi-{.introt

*

sinol)

6l

83

il-:;a

$(cosh

at

* cos

af )

62.

1

V s * o

*

V s * b

e-bt

-

e-a t

-_-.-*....*:

2(b

-

a){rF

63.

I

* { r * o

erf

dat

na

u.

Vr (s -o )

eet

e$l&,

{a

..'

'

I

6:?+

0

*'{+-

a b""rrc

6r/l)}

LVrt


29/34

/(s)

r(r)

1 6 . -

I

I

I

{ e z * d

Jn@tl

I

cr.

1

1/62

-

q'z

Is(at)

58.

1r/Ploz

-

t1n

n >

* l

{st

* az

*n

Jn{atJ

69.

( s

-

\ / 8 2 = e 2 ) "

n )

- 1

,ls'z

e'

ei

I"(et\

70.

,uts

-

y'rETE

I

{sz

|

6z

to@,/&TdEtt

7l

"-vt{FiF

\F+e

J o o r / t r - t )

t > b

f o

t < b

72.

(82

+

a2)s/2

t

J1{at)

a

73.

. a

(az

+

a2lx/2

t

J

o(ot)

74.

g2

FTTF

J o f u t ) o t J l @ t \

75.

lsz

-

oz)slr

t

lr{at\

(r,

76,

a

@-wn

t Is{o,tl

77.

a2

@:wn

Ie(ot)

*

o,tl1@t\

78,

1 - ___3::_

e(et

-

1) a( l

-

e-s1

Lihat

juga

entri

141,

Halsmttl 264

F ( t 1 n , n S t < - n * L , n = 0 , 1 , 2 , .

79.

1

=

____g:_

e(es

r)

e(l

*

re-s;

Iu

r ' ( 0 =

E f k

k = l

di

mana

[fl

=

bilangan

bulat

terbeear{

t

80.

e s - 1

1 - e - t

o(et

-

r )

s( l

-

re-r ;

Lihat

juga

entti 143, Halaman 254

f ( t J = q ' * ,

n = t 1 n * I ,

n

= 0 ,

1 ' 2 , .

8 l

g - a l s

Va

eos2{A

t/

rt


30/34

f(s)

F(r)

82.

e -a /8

FiT

sin

2lat

{"a

83.

*T1-

n

)

-1

/ r \ n / 2

( ; )

r^Q\/at)

84.

e - o G

V s

e*

a2/4t

-.--.:

'l

"t

85.

*o

Vi

--3-

e-a2/4t

2Gts

86.

1

-

e * o G

I

efi(a/hfi)

87.

e * o G

a

er{.i(allfi)

88.

e - o G

v 8

( v s

+

D )

eb(br+a)

rtc

Utft

+

-i=)

\

2\/t

/

89.

s-at

{i

?TT-

n )

- 7

#*,

I

r-

"n

c*u2/442tz.(^t/ilt

du

90.

r , / "19 )

\ 8 + D /

e -b t

-

e -a l

t

9 t .

ln

l(sz

*

az)/q,21

-----?;--

Ci

(at)

92.

ln

[(s

+ oilcr.]

s

Ei

(ot)

93.

*

(y

* lns)

a

y

=

tetapan Euler

=

0,5772156.

l n t

94.

, , 1 s 2 * a 2 \

--

\sz

+ bz/

2

(cos

ot

-

cos

bt )

95.

, 2

,

( v * l n s ) 2

6 " * -

8 -

y =

te tapanEuler=

0,5??2156. .

lnz

t

96.

l n s

I

* ( l n t * y )

7

=

tetapanEuler

=

0,57?2156

97.

ln2

s

a

( l n t * Y ) z

*

{ 7 2

7

=

ttapanEuler=

0,5?72156


31/34

tF)

r(r)

9 8

r ' ( n * 1 )

- r ( a * 1 ) l n e

n > _ L

5 t * l

t n l n ,

99

,un_r

(als)

sin ai

t

1 0 0 .

,rr.,_r

(o/s)

s

Si

(ot)

l 0 l

? a / s

U

erfc

(Vcls

)

e-2'l;i

---=-

V T t

I 02.

"s2/aaz

s;fc.

(s/Za\

?L

s*aztz

r03.

e"'/ao'

erfc

(s/Zol

s erf (ot)

I

104.

eos

erfc

Vii

Vs

I

t["Q

+

",

I 05.

eds Ei

{os)

1

t - t d

r06.

*

[.o*

""

{;

-

tt

(*)}

-

sin

as

ci

(cs)]

t

+ d............t

lo7.

"1","

{i

-

si

(as)}

+ cos

as

ci

(cs)

t

FT|t

r08.

f - r

cos

as

{z

*

Si

(as)}

-

sin

os

Ci

(as)

I

tan_

|

(t/a)

t09.

,in

o*

i

-

si

tas)I

*

cos s ci

(c.s)

l o )

I

1 ,

/ t 2 + o 2 \

, '"

\__F_

t to.

[u

-

.t,*,]' * 6iz

(as)

I

1r,

t'

+=cr'''

t \ a z /

i l l

0

a((0

l t2 .

I

8(r )

I t 3 .

e -

as

6(t

-

o)

I t4 .

e-

as

a

Lihat

juga

entri 139, Halaman 254

u(t

-

a,


32/34

f(a)

r(a

| 15 .

sinh s*

s

sinh sa

i

*

1

, I

.in

i

cosT:

l l 6 .

sinhsr

il;E;;

4

S

(-1) '

. ' -

r f r 1 2 n - 1 " "

-

l ) ru

- Slll

I 17.

eosh.so

s sinh so

i *?2, ( - l ln"o , rys in 'Y

I t8.

cosh

sr

il;fr;

1

l 19 .

sinh

so

ez

sinh so

r t , 2 a ( - l ) o . n T t . n r t

" - 7 , 4 r V s r n - s r n -

120.

sinh eo

;%;ffi;

sln n

-

llrn

-

fOS

t 2 l

cosh

so

p;frfi;

*.

#" ,

#

"o,ff( '-

"Y)

122.

eoshso

p;;fr;

,+gi

x

n = l

( -1 ) " (2n

l \ r r

(2"

-TF

cos

t;-

srn

t23.

cosh

slr

F;;fr;

+(t2

xz a2)

+F "i, #+ "ou? n

(2n

-

l )nt

cos

za

124.

sinh

rfi'

;;h

"v?

:1 s

f,z

,3t

(- lt

n

p'*zrttto,

rinT

r25.

coshcVF

cosh

a.y'3

126.

sinh

oy'T

VE-cosh

,V?

a # t

lZn

*

llrr

(*1)"-

t

e-

t

2n t)2t2t

/

4a2

sin

-;:r*:

127.

eoshoVF

{T

sinh oy'a

+

-

i, i,

{-t)"

e-n2r'ttoz

"o"W

t28.

sinh

oV?

e sinh

oy'?

g

+

?

3

( -1 \n

" -no r t / o r

s i n ,o "

e

o G t

n

-

- " ' a

r29.

coshsVF

o eosh

ay'?

t

*

;

, : ,

t

"- ,rn

t)zr2t/4o'

"o"@; l

t30.

sinh

oV?

er

sinh

oV

r t

* 2 a 2 $

& ' , g # t

/ - 1 \ n

* tf

_

e-n2n2t/s2\

sin

,."-

' a

t 3 t .

cosh

cV;

su cosh

oV '

$(x2-o2l

+

t

-

Y.i,

#+

e-.2^-t,znztt+oz

oa


33/34

f(e)

r(4

r32.

Jsli*,r/l)

ffi

1

-

2

3

e-*t/dJo'(^nrla)

n=1

I,

J1(trn)

mana r1

\2,

... adalah

akar-akar

positif

dari J6(l)

=

i 0

t33.

Jo$sr/i)

ezJo6.a\/t) l (az-oz l

4

t

*

mana1,1,

r2,

...

adalah

,ou#,.##y,

akar-akar

positif

dari Je(1.)i

0

r34.

#**

(?)

I

0

Fungsi gelombang

segitiga

t35.

r"''h

(?)

tr'ungsi

gelombang

segi+mpat

t36.

Fungsi

penyearahan

gelombang

sinu*

rs7.

v(L

@FTAG4

Fungsi

penyearahan

separuh

gelombang

sinus

r38.

1

e - 6

d,8z

s(1

_

e-os)

Fungsi

gelombang gigi gergaji

a 2 a g 6 4 a


34/34

/{s)

F(t',

t39.

e-as

I

Lihat juga entri 144, Halaman 251

Fungri

satuan Heaviside u( t

-

a)

F.(.)

140.

a - o s

( 1

-

e - s )

..-

a

Fungsi

pulsa

t 4 l .

I

A 1 -

' = E

Lihat

juga

entri

?8, Halaman 249

2

I

0

Fungsi tangga

142.

C- s

+

e -zs

8(1

-

-s)2

F(t,

=

n2,

n < t < n * 7 , n = 0 , 1 , 2 , .

r(o

4 .

0

t43.

1 * e - s

Ai_?,-q

Lihat

juga

entri 80, Halarnan

249

F(t)

*

,4,

n = t 1 n * L , n = 0 , 1 , 2 , .

I

rft)

141.

ro {1

* e *os )

d"282

+ T2

(

sin

Gt/a\

= a ,

r(0

tranform laplace

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