Unit IV Laplace Transforms

7/29/2019 Unit IV Laplace Transforms

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Unit I Laplace Transform

Laplace Transform

The Laplace transform is an integral transform that transforms a real valued

function f of some non-negative variable, say t into a function F(s) for all s for

which the improper integral

0

st dt)t(fe

converges.

Definition 1.1 Let f (t) be a given real valued function that is

defined for

all t 0. The function defined by

0 )( dttfe

st

for all s for which this improper integral converges is

called the

Laplace transform of f (t) and will be denoted by

F (s) = L (f) =

0

)( dttfest

The operation which yields F (s) from a given real valued f (t) is also called the

Laplace transformof f (t) while f (t) is called

the inverse transformof F (s)

and will be denoted by

f (t) = )F(L 1

Example 1 Let f (t) = 1 for t 0. Then find F (s).

Solution Applying definition 1.1 we get:

L (f) = L (1) = b

0

st dteb

im =

0

bst

s

e

bim

=

+

s

e

bim

s

1 tb =s

1for s

> 0.

Therefore, L (1) =s1 for s > 0.

Remark: Let f (t) = k for any scalar k. Then L (f) =s

kfor s > 0.

Example 2 Let

>

=

1,0

10,4)(

tfor

tfortf . Then find F (s).

Prepared by Tekleyohannes Negussie 83


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Solution L (f) = F (s) = dteb

imb

tfst 0

)( = dtest

1

0

4 = ( )ses

14

.

Therefore, L (f) = ( )ses

14

for s > 0.

Example 3 Let f (t) = t for t 0. Then find F (s).


L (f) = F (s) = b

0

st dtetb

im

Now applying integration by parts we get:

b

0

st dtetb

im =0

2

bstst

s

ee

s

t

bim

=2s

1for s > 0.

Therefore, L (f) = 21

s

for s > 0.

Example 4 Prove that for any natural number n, L ( nt ) = 1!+n

s

nfor t 0.

Solution We need to proceed by applying the principle of mathematical

induction on n

i) For n = 1 it follows from example 3.

ii) Assume that it holds true for n = k, i.e.

L ( kt ) = 1!+k

s

kwhere t 0

iii) We need to show that it holds true for n = k + 1. i.e. L ( 1+k

t ) = 2!)1(

++

ks

k

for t 0.Now by the definition of the Laplace transform

L ( 1+k

t ) = +b

tskdtet

bim

0

1

Applying integration by parts we get:

u = 1+k

t and 'v = tse dt

then 'u = ktk )1( + dt and v =ts

es

1 .

Hence, +b

tskdtet

bim

0

1 =

++

+ btskts

k

dtets

kb

es

t

bim

0

11

0



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=

+ b tsk dtet

bim

s

k

0

1 =

+

+1!1

ks

k

s

k, by our

assumption

= 2!)1(

+

+k

s

kfor s > 0.

Thus, by the principle of mathematical induction it holds true for any naturalnumber n.

Therefore, L ( nt ) = 1!+n

s

nfor t 0.

Example 5 Let f (t) =t

e

for t 0. Then find F (s).


L (f) = L (t

e

) = b

0

t)s( dteb

im =

0

bt)s(

s

e

bim

=

+

s

1

s

e

bim

b)s( =

s1

for s > .

Therefore, L (t

e

) =s

1for s > .

Theorem 1.1 (linearity of the Laplace

transform)

The Laplace transform is a linear operation, that

is forany real valued functions f (t) and g (t) whose

Laplace

transform exists and any scalars and

L ( f + g) = L (f ) + L ( g)

ProofBy definition 1.1

L ( f + g) = { } +b

tsdte

bim tgtf

0

)()(

= b

tsdttfe

bim

0

)( + b

tsdttge

bim

0

)(

= h L (f) + k L ( g).

Therefore, Laplace transform is a linear operation.

Example 6 Find the Laplace transform of the hyperbolic functions f (t) = cosh

t and



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g (t) = sinh t for t 0.

Solution From theorem 1.1 and the result of example 2 we get:

L (f ) = L (cosh t) = b

tsdte

bim

0

)(

2

1 + +b

tsdte

bim

0

)(

2

1

=

02

1)()( b

s

e

s

e

bim

tsts

+

+

=

+

+

s

e

s

e

bim

bsbs

)()(

2

1

+

ss 11

2

1= 22 s

sfor s

> .

Similarly, L (g) = L (sinh t) = b

tsdte

bim

0

)(

2

1 +b

tsdte

bim

0

)(

2

1

=

021

)()( b

se

se

bim

tsts

+

+

+

=

++

+

s

e

s

e

bim

bsbs

)()(

2

1

++

ss 11

2

1= 22

sfor s

> .

Therefore, L (cosh t) = 22 ss

and L (sinh t) = 22

sfor s > .

Example 7 Find the Laplace transform of functions f (t) = cos t and g (t) = sint for t 0.

Solution From theorem 1.1 and Eulers formula, where i = 1 we get:

tie = cos t + i sin t

L (f ) = L ( tie ) =is

1=

)()(

isis

is

++

= 2222

++

+ si

s

s

On the other hand, L (cos t + i sin t) = L (cos t) + i L (sin t).

Now equating the real and the imaginary parts we get:

L (cos t ) = 22 +ss

and L (sin t) = 22

+s.

Therefore, L (cos t ) = 22 +ss and L (sin t) = 22

+s

for s > .

We can also derive these by using definition 1. 1 and integration by parts with

out going into the operations in complex numbers.

Example 8 Find L (f), where



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i) f (t) = 3 cos 2t + 2 sin t 34 t ii) f (t) = 4 cosh 5t 3 sinh 2t +

te

5

iii) f (t) = sin 2t cos 4t

Solutions By the Linearity of the Laplace transform we have

i) L (f) = 3 L (cos 2t) + 2 L (sin t) )(4 3tL =4

32 +s

s+

1

22 +s

4)!3(4

s.

Therefore, L [3 cos 2t + 2 sin t 34 t ] =4

32 +s

s+

1

22 +s

424

sfor s > 0.

ii) L (f) = 4 L (cosh 5t) 3 L (sinh 2t) + )(5 teL =25

42 s

s

4

62 s

+s

5.

Therefore, L [4 cosh 5t 3 sinh 2t + te

5 ] =25

42

s

s

4

62

s

+s

5for

s > 0.iii) First observe that:

sin 2t cos 4t = tt )42(sin2

1)42(sin

2

1++ = tt 6sin

2

12sin

2

1+

Then by the Linearity of the Laplace transform we get:

L [sin 2t cos 4t] = ]6[sin2

1]2[sin

2

1tLtL + =

++

+

36

6

2

1

4

2

2

122 ss

.

Therefore, L [sin 2t cos 4t] =

++

+

36

3

4

122 ss

for s > 0.

Example 9 For any real valued functions f (t) and g (t) whose Laplace transform

exists and

any real numbers and , show that the inverse Laplace transform

is linear.

Solution From the linearity of the Laplace transform we get:

L ( f (t) + g (t)) = L (f ) + L ( g ) = F (s ) + F ( s )

Hence, ))()((1

sFsFL + = f (t) + g (t)) = )(1)(1 sGsF LL + .

Therefore, the inverse Laplace transform is linear.

Example 10 Find the inverse Laplace transforms of

i)s2

3ii) 3

40

siii)

1

22 ++

s

siv)

9

42 +

s

s

Solutions i) From the linearity of the inverse Laplace transform we get:



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sL

2

31=

s

L11

2

3=

2

3.

Therefore,

s

L2

31=

2

3.

ii) Now 340

s= 3

!220

sand 23

!21t

sL = .

Hence,

3

401

s

L =

3

!2120

s

L = 220 t .

Therefore,

3

401

s

L = 220 t .

iii)

++

1

22

1

s

sL =

+

121

s

sL +

+

1

12

2

1

sL = cos t + 2 sin t.

Therefore,

+

+1

2

2

1

s

s

L = cos t + 2 sin t.

iv)

+

9

42

1

s

sL =

921

s

sL +

9

3

3

42

1

sL = cosh 3t +

3

4sinh

3t.

Therefore,

+

9

42

1

s

sL = cosh 3t +

3

4sinh 3t.

Example 11 Find f (t) if

i) F (s) =ss 3

92 +

ii) F (s) =9

)1(42 +

s

s

Solution By partial fraction reduction we get:

i)ss 3

92 +

=s

A+

3+sB

=ss

sBsA

3

)3(2 +

++A + B = 0 and 3 A = 9 A = 3

and B = 3

Thus, F (s) =s

3

3

3

+s=

+

33

11113

ssLL = 3 3 te 3 .

Therefore, f (t) = 3 3 te 3 .

ii)9

)1(4

2

+

s

s=

3s

A+

3+s

B=

9

)3()3(

2

++

s

sBsA

A + B = 4 and 3 (A B) = 4 A =3

8and B =

3

4

Thus, F (s) =3

8

31

s+

3

4

+ 31

s=

+

+

33

4

3

1111

3

8

ssLL =

te 3

3

8+

3

4 te

3.



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Therefore, f (t) =te 3

3

8+

3

4 te

3.

Example 12 For any 0, the Gamma function () is defined by:

() =

0

1dtte

t

Show that )( atL = 1)1(

++

as

afor a > 0 and in particular (n + 1) = n!,

for any non-negative integer n.

Solution Put x = st, so that dt =s

dx.

Thus, )( atL =s

dx

s

xe

bim

b ax

0

= dxxeb

im

s

bax

a

+ 01

1 =

1

)1(

+

+a

s

a

.

Therefore, )( atL = 1)1(

++

as

afor a > 0.

Further more, for any natural number n

(n + 1) = dteb

imb

nt

t 0 =

0

bnt

te

bim

+ n

dteb

imb

nt

t 01

= n dteb

imb

nt

t 01 = n (n)

Thus, (n + 1) = n (n) = n (n 1) (n 1) = n (n 1) (n 2) (n2) = . . . =

n! (1) = n!

We summarize the Laplace Transforms of some functions for future reference in

the table below.

N

o

Function f

(t)

Laplace

transform

No Function f

(t)

Laplace

transform

1 1s

16

te

s1

2 k for k s

k7 cos t

22 +ss

3 t2

1

s8 sin t

22

+s



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4nt for

n = 0, 1,2, . . .

1

!+n

s

n9 cosh t

22 ss

5 at for a

01

)1(++

as

a10

sinh t22

s

Existence of Laplace Transforms

Before we define the theorem that guarantees the existence of Laplace

transform, let us see the definition of piecewise continuity.

Definition 1.2 A function f (t) ispiecewise continuous on a

finite

interval a t b if f (t) is defined on the interval and is

such that

the interval can be divided into finitely many subintervals ,

in

each of which f (t) is continuous and has finite limits as t

approaches either endpoint of the subintervals from the

interior.

By definition 1.2, it follows that finite jumps are the only discontinuities that a

piecewise

continuous function may have.

Theorem 1.2 (Existence of Laplace Transforms)

Let f (t) be a real valued function that is piecewise

continuous

on every finite interval in [0, ) and satisfies the

Prepared by Tekleyohannes Negussie

x

y

Figure 1.1 a piecewise continuous function

90


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inequality

)t(f teM

for all t 0 and for some constants and M.

Then there

is a unique Laplace transform of f (t) for all s >

.

Proof Since f (t) is piecewise continuous, )t(fe st is integrable over any finite

interval on

the t-axis. Thus, for s > we have

)f(L = b

0

)t(fst dte

bim

b

0

)t(fst dte

bim

b

sttdteeM

bim

0

=s

Mfor s > .

Therefore, )( fL is finite, and hence it exists for s > .

Example 13 Let f (t) =t

1for t > 0. Then show that L [f] doesnt exist.

Solution We prove this by showing that dtet

ts

0

1diverges. For some c > 0,

dte

t

ts

0

1= dte

tb

imc

b

ts

+

1

0

+ dte

td

imd

c

ts

1

To this end, let

u =ts

e

and 'v = dtt

1. Then du = dt

tses and v = tn .

Hence, dtetb

imc

b

ts +1

0

= { }

b

ctstne

b

im

+0

+ dttnesb

imc

b

ts

+0 .

But, { }b

ctstne

b

im

+0

= { }bnecneb

im bscs

+0

=

cse

cn { }bnbse

b

im

+

0= for any c > 0.

Consequently, dtet

ts

0

1diverges.

Therefore, the Laplace transform of f (t) =t

1for t > 0 doesnt exist.

Remark: If the Laplace transform of a given real valued function exists, then it

is unique.



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Conversely, if two functions have the same Laplace transform, then these

functions

are equal over any interval of positive length (i.e. they differ at various

isolated points).

Exercises 1.1

In exercises 1-8, find the Laplace transforms of the following functions by

showing the details of your steps. (a, b, c, and are constants) .

1. f (t) = 2tctba ++ 2. f (t) = sin t cos t 3. f (t) = tcos 2 4. f (t) =

t3coshe t

5. f (t) = )t(sin + 6.

=1x0for1

1x0for0)t(f

7. 2x0for1x1)t(f = 8.

=1x0for11x0fort)t(f

In exercises 9 13, given F (s) = L (f), find f (t) by showing the details of your

steps. (, , and are constants) .

9.4s

4s

2

10.25s

s5

2 11.

4s

112.

14

3

s

s2

13.

2ss

10s

2

In exercises 14 17, find the Laplace transforms of the following functions by

showing the details of your steps. (a, b, c, and are constants) .14. sinh t cos t 15. cosh t sin t 16. t2sinhe5 t2 17. tet 2)1( +

In exercises 18 20, find the inverse Laplace transforms of the following by

showing the details of your steps.

18.2)1s(

1

+19.

4)3s(

12

20.

18s6s

3

2 ++

1.2 More on Transforms of Functions

1.2.1 Laplace Transforms of Derivatives

Theorem 1.3 ( Laplace Transform of the Derivative of f

(t))

Suppose that f (t) is a piecewise continuous real valued



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function

for all t 0 and )t(f teM

for some constants

and M,

and has a derivative )t(f' that is piecewise continuouson every

finite interval on [0, ). Then the Laplace transform of

)t(f'

exists when s >, and

)f(L ' = sL (f) f (0), for all s >.

ProofApplying integration by parts we get:

)]([ tfL =

0

)( dttfe st =

+

00)()( dtetfsetf

b

im stbst

= [ ] ].[)0()( fLsfebfb

im sb +

Sinceb

Mebf)( , we have sbebf )( bsbeMebf )( = bsMe )(

But,bs

Meb

im )(

= bse

M

b

im)(

= 0, for >s .

This implies thatsb

ebfb

im

)(

= 0, and hencesb

ebfb

im

)(

= 0.

Consequently,

)0()]([)]([ ftfsLtfL =

Therefore, )f(L ' = sL (f) f (0), for all s >.

Theorem 1.4 (Laplace Transforms of Higher order

Derivatives)

Let )(tf and its derivatives )(tf , )(tf,

,

)()1(

tfn

be real

valued

continuousfunctionsfor all 0t and satisfying )()( tf k

teM

for some and M, and k 0 and let the derivative )()( tf n be



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piecewise

continuous on every finite interval contained in ),0[ . Then the

Laplace

transform of )(

)(

tf

n

exists when>s

, and is given by

))(( )( tfL n = )0()0()0())(( )1(21 nnnn ffsfstfLs

Proof( by the principle of mathematical induction on n)

For n = 1, it holds by theorem 1.4 i.e

)0()]([)]([ ftfsLtfL = .

Assume that it holds true for n = k

][)(k

fL = )0(...)0(')0(][)1(21 kkkk ffsfsfLs

Now we need to show that it holds true for n = k + 1

][)1( +k

fL = )0(][)()( kk

ffLs

= )0(...)0(')0(][)1(21 kkkk ffsfsfLss

= )0(...)0(')0(][)(11 kkkk ffsfsfLs +

Therefore, by the principle of mathematical induction

))(()(

tfLn = )0()0()0())((

)1(21 nnnn ffsfstfLs for any natural number n.

Example 14 Let .)(3

ttf = Then find ].[ fLSolution )(tf = 23t , 0)0( =f , )(tf = t6 , 0)0( =f , 6)( = tf and .6)0( =f

Thus, )]([ tfL = )0()0()0()]([ 23 ffsfstfLs , and also, )]([ tfL =s

L6

)6( = .

Equating these values we get:

][3 fLs =s

6 4

6][

sfL =

Therefore, ][ 3tL = 46

s.

Remark: It can be shown by induction on n that

)( ntL = 1!+n

s

nfor n = 0, 1, 2, and s > 0.

Example 15Find ).(cos tL

SolutionLet )(tf = .cos t Then )0(f = 1, )(tf = tsin , )0(f = 0,

)(tf = t cos2 = ).(2 tf



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Now we have ))(( tfL = )0()0())((2 fsftfLs and )]([ tfL = ][2 fL

)]([ tfL = 22 +ss

.

Therefore, )(cos tL = .22

+s

s

Example 16 Find L [ t2

cos ].

Solution Let )(tf = t2

cos . Then f (0) = 1, )(tf = ttsincos2 , )0(f = 0 and

)(tf = tt 22 cos2sin2 = ]cos[sin2 22 tt

= ]coscos1[2 22 tt = t2cos42 = ).(42 tf

Now, since ))(( tfL = ),0()0())((2 fsftfLs

we have

]2)(4[ + tfL = )0()0())((2 fsftfLs and ]2)(4[ + tfL = )2()]([4 LtfL +

sfLs )(2 = )2(][4 LfL + L[f] =)4(

22

2

++

ss

s.

Therefore, )(cos2 tL =)4(

22

2

++

ss

sfor s > 0.

Example 17 Show that )(sin2

tL =)4(

22 +ss

.

Solution Let f (t) = )(sin 2 tL . Then f (0) = 0, )(tf = ttsincos2 , )0(f = 0 and

)(tf = tt 22 sin2cos2 = ]sin21[2 2 t = )(42sin42 2 tft = .

Now, since ))(( tfL = ),0()0())((2 fsftfLs

we have: ]2)(4[ + tfL = )0()0())((2 fsftfLs and ]2)(4[ + tfL = )2()]([4 LtfL +

)(2 fLs = )2(][4 LfL + L[f] =)4(

22 +ss .

Therefore, )(sin 2 tL =)4(

22 +ss

for s > 0.

Example 18 Find )cos( ttL .

Solution Let )(tf = tt cos . Then f (0) = 0, )(tf = ttt sincos , )0(f = 1and )(tf = tttt cossinsin 2 = ).(sin2 2 tft

Thus, )]([ tfL = )](sin2[ 2 tftL

)0()0())((2 fsftfLs = )].([)(sin2 2 tfLtL

1)(2 fLs = ].[2 222 fLs

+



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)()( 22 fLs + = 12

22

2

++

s

= .

22

22

+

s

s

Therefore, )cos( ttL = )]([ tfL =22

22

)( 2

+

s

s

Similarly, it can be shown that

)cos( ttL =222

22

)(

+

s

sand )sin( ttL = 222 )(

2

+ss

.

1.2.2 Laplace Transforms of the Integral

Theorem 1.5 (Integration of f (t))

Let F (s) be the Laplace transform of the real valued

function

f (t). If f (t) is piecewise continuous and satisfies the

inequality

teMtf)(

for some constants and M, then

)(1

)(0

sFs

dfLt

=

for s > 0 and s > .

Proof Suppose that f (t) is piecewise continuous andt

eMtf

)( for some

constants

and M . Then the integral

g (x) = t

df0

)(

is continuous and for any positive number t

=t

dtftg0

)()( t

deM0

= )1( ek

M

e

k

Mfor s > .

This shows that g (t) also satisfies an inequality of the form

)(tg

ek

Mfor s > .

Also, )(' tg = f (t), except for points at which f (t) is discontinuous. Hence,

)(' tg is

piecewise continuous on each finite interval, and by theorem 1.4,



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L[f (t)] = )]([ ' tgL = s L [g (t)] g (0) for s > .

Hence, g (0) = 0, so that L [f (t)] = s L [g (t)] for s > 0 and s > .

Therefore, )(1

)(0

sFs

dfLt

=

for s > 0 and s > .

Example 19 Let F (s) =)(

122 +ss

. Find f (t).

Solution From the result of example 7 we get

+

)(

122

1

ssL = t

sin

1

Again, using theorem 1.7 we have

+

)(

122

1

ssL =

d

t

0

sin1

= )cos1(12

t

Therefore, f (t) = )cos1(1

2t

.

Example 20 Let F (s) =)(

1222 +ss

. Find f (t).

Solution From the result of example 19 we get

+

)(

1222

1

ssL =

d

t

0

2)cos1(

1= )

sin(

12

tt

Therefore, f (t) = )sin

(12

tt .

1.2.3 S-shifting, Unit step functions and t-shiftingThe first shifting theorem

If we replace s by s a, in the definition of the Laplace transform we get the

following important result.

Theorem 1.6 (First shifting theorem)

If f (t) has the transform F (s) where s > k, then )t(feat



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has the

transform F (s a) where s a > k. Symbolically,

L { )t(feat } = F (s a) or )t(feat = )}as(F{L 1

ProofF (s a) can be obtained by replacing s by s a in definition 1.1 as

follows:

F (s a) = b

0

)t(ft)as( dte

bim = [ ]

b

0

)t(ftaets dte

bim = L { )t(feat }.

Now if F (s) exists for s greater than some k, then our integral exists for

s a > k.

Example 21 Find the Laplace transform of f (t) = ate cos t and g (t) = ate sin t

for t 0.

Solution Applying theorem 1.6 on the result of example 5 we get:

L (f ) = L ( ate cos t) =22)as(

as

+

and L (g ) = L ( ate sin t) =

22)as( +

Unit step functions and t-shifting Theorem

The unit step function defined below is a typical engineering function made to

measure for engineering applications, which often involve function that are off

and on.

Definition 1.3The function u

defined by

> 0.


L [u (t a)] =

b

tsdtatue

b

im

0

)(

=

b

a

tsdte

b

im=

b

im

a

b

s

esa

=

s

esa

.

Therefore, L [u (t a)] =s

esa

for s > 0.

Theorem 1.7 (The second shifting theorem, t-shifting

theorem)

If f (t) has the Laplace transform F (s), then the shiftedfunction

)(~

tf = f (t a) u (t a) =

atfor

atfor

1

0

has the Laplace transform )(sFeas . That is

L[f (t a) u (t a)] = )(sFe as .

ProofFrom the definition of the Laplace transform we have

)(sFe as = b

dfeb

ime sas

0

)( = +b

dfeb

im as

0

)( )(

Substituting + a = t in the integral we get:

)(sFeas =

+

ba

dtatfeb

im

a

ts)(

=

b

dtatuatfeb

im ts

0

)()(

= L[f (t a) u (t a)].


t

f(t)

t

f (t)

1 1

1

f (t) = u (t 1)

1

f (t) = t [u (t) u (t 1)] + u (t 1)

99


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Therefore, )(sFeas = L[f (t a) u (t a)].

Example 24 Find L [f], where

=

2sin

20

02

)(

tift

tif

tif

tf

Solution We write f (t) in terms of unit step functions. For 0 < t < , we take 2 u

(t).

For < t < 2 we want 0, so we must subtract the step function 2 u (t

) and for

t > 2 we need to add 2 u (t ) sin t.

Hence, f (t) = 2 u (t) 2 u (t ) 2 u (t 2) sin t. f (t) = 2 u (t) 2 u (t ) 2 u

(t 2) sin (t 2). (since sine is a periodic function with period 2)

Therefore, L [f] =s

e

s

s

22

+12

2

+

s

es

.

Example 25Find the inverse Laplace transform of

F (s) =2

2

2

22

s

e

s

s

s

es2

4

+1

2

+

s

ess

.

Solution Without the exponential functions the four terms of F (s) would have

the inverses

2t, 2t, 4 and cos t. Hence, by theorem 1.7

f (t) = 2t 2(t 2) u(t 2) 4 u(t 2) + u (t ) cos (t )= 2t 2t u (t 2) u (t

)cos t.

Therefore,

=

tift

tif

tift

tf

cos

20

202

)( .

1.2.3 LaplaceTransforms ofPeriodic Functions

ProofBy definition 1.1


Theorem 1.8 (Laplace Transforms of Periodic

Functions)

The Laplace transform of a piecewise

continuous

function f (t) with period p is

L [f] =

p

tsps dttfee 0

)(1

1

100


19/26


L [f] = =

+

m

n

pn

np

tsdttfe

m

im

0

)1(

)(

= ( )

=

0 0)(

n

ptsnps

dttfeem

im(put t = t + np since f (t) is periodic with

period p)

=

pts

ts dttfee 0

)(1

1, because it is a geometric series.

Therefore, L [f] =

pts

ts dttfee 0

)(1

1.

Example 26 Find the Laplace transform of the saw-tooth wave given by

p

ktf =)(

where 0 t p and f (t + p) = f (t) for t 0.

Solution From theorem 1.8 we have

L [f] =

p tsts

dteep

k

01

1=

01

1p

s

e

ep

kts

ts = ps

k

Therefore,

t

p

kL =

ps

k.

1.3 Differential Equations

1.3.1 Ordinary Linear Differential Equations

We shall now discuss how the Laplace transform method solves differential

equations. We began with an initial value problem

)(trbyyay =++ , y (0) = 0k and 1)0( ky =

(1)

with constants a and b. Here r (t) is the input (driving force) applied to the

mechanical system and y (t) is the output (response of the system). In Laplace

method we do three steps:

Step 1 We transfer (1) by means of theorem 1.3 and 1.4, writing Y = L (y) and R

= L (r). This gives

[ ] )()0()0()0(2 sRbYysYayysYs =++

(2)

This is called the subsidiary equation.

Collecting Y terms we have



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( ) ( ) )()0()0(2 sRyyasYbass +++=++

(3)

Step II We solve the subsidiary equation algebraically for Y. Division by

bass ++2 and use

of the so- called the transfer function

basssQ

++=

2

1)(

(4)gives the solution

( )[ ] )()()0()0( )( sQsQyyasY sR+++=

(5)

If y (0) = )0(y = 0, then this implies Y = QR; thus Q is the quotient

R

YsQ =)( =

)(

(

inputL

outputL

(6)and this explains the name of Q.

Note that: Q depends only on a and b, but neither on r (t) nor on the

initial conditions.

Step III We reduce (5) to a sum of terms whose inverse can be found from the

table, so that

the solution

y (t) = ][1

YL

.

Example 27 (initial value problem)

Solve tyy = , y (0) = 1 and )0(y = 1

Solution The subsidiary equation becomes

21)0()0(2

sYyysYs =

Thus,2

11)1( 2

ssYs += ( )1

1

1

122 2

+

=sss

sY



21/26


( )11

1

122

++

=sss

Y =

+

+

2

1

1

1

1

12

sss

Nowt

es

L =

+111

, ts

L sinh1

12

1 =

and ts

L =

2

1 1.

Therefore, y (t) = t te

+ sinh t.


Solve teyyy=++ 2 , y (0) = 1 and )0(y = 1


( ) ( ) 11

121

2

+=++++ sYsYsYs

Thus,1

11)12( 2

++=++

ssYss 3)1(

1

)1(

12 ++

++

=ss

sY

3)1(

1

1

1

++

+=

ssY

Nowt

es

L =

+111

,2

3

1

2

1

)1(

1t

sL =

+ t

e

.

Therefore, y (t) =t

et

12

1 2.


Solve tetyyy 3423 +=+ , y (0) = 1 and )0(y = 1


( ) ( )3

142131

2

2

+=++

ssYsYsYs

Thus,3

144)23(

2

2

++=+

sssYss

( ) ( )23)3(1

23

4

23

42222 +++++

= ssssssss

s

Y

Now2123

42

+

=+

=

s

B

s

A

ss

sY A + B = 1 and 2A + B = 4 A = 3 and

B = 2

( )234

22 + sss=

s

A+ 2s

B+

1sC

+2s

DA = 3, B = 2, C = 4 and D =

1



22/26


and ( )23)3(12 + sss

=3s

A+

1sB

+2s

CA = 0.5, B = 0.5 and C = 1.

Hence, ( ) ( )23)3(1

23

4

23

42222 +

++

++

ssssssss

s

=s3 + 2

2s

11

21

s

22s

+

31

21

s

Now 331 =

s

L , ts

L 222

1 =

,

1

1

2

11

sL =

te2

1and

3

1

2

11

sL =

te 3

2

1.

Therefore, y (t) = 3 + 2t te2

1 te 22 +

te 3

2

1.


Solve tt

eyyy sin52=++ , y (0) = 0 and )0(y = 1.


( ) ( )1)1(

1521

2

2

++=++

sYsYYs

Thus, ( )1)1(

114)1(

2

2

+++=++

sYs ( )( )4)1(1)1(

1

4)1(

1222 ++++

+++

=sss

Y

Now ( )( )4)1(1)1(1

22 ++++ ss=

4)1(2 +++

s

BAs+

1)1(2 ++

+s

DCs

1)52()2522()22()( 23 =+++++++++++ DBsDCBAsDCBAsCA

A = C = 0, B =3

1 and D =

3

1.

Hence, Y =

++

++

+

++ 4)1(1

1)1(

1

3

1

4)1(

1222 sss

=

++

+++

4)1(

2

1)1(

1

3

122 ss

Now

++

42)1(

21

sL = tte 2sin and

++

12)1(

11

sL = tte sin

Therefore, y (t) =t

ett+ )2sinsin(

3

1.


Solve tyyy sin1022 =++ , y (0) = 0 and )0(y = 1.




23/26


( )1

10221

22

+=++

sYsYYs

Thus, ( )1

1011)1(

2

2

++=++

sYs ( )( )11)1(

10

1)1(

1222 +++

+++

=sss

Y

Now ( )( )11)1( 10 22 +++ ss = 1)1( 2 ++ +s BAs + 12 ++s DCs

10)2()22()2()( 23 =++++++++ + DBsDCAsDCBsCA

A = 4, B = 6, C = 4 and D = 2.

Hence, Y =

+

+

+

++

++

++ 11

211)1(

14

1)1(

32222 ss

s

s

s

s

Now

+

12

11

s

L = tsin and

++

+

12)1(

11

s

sL = t

te sin

Therefore, y (t) = tetttt ++ )cos4sin3(cos4sin2 .


Solve tyy cosh189 =+ , y (0) = 2 and )0(y = 0.


( )1

1892

22

=+

s

sYsYs

Thus, ( )1

1829

2

2

+=+

s

ssYs ( )( )19

18

9

2222 +

++

=ss

s

s

sY

Now ( )( )1918

22 + sss

=92 +

+s

BAs+

12 +

s

DCs

sDBsCAsDBsCA 18)()9()()( 23 =++++++ +

B = D = 0, A =5

9 and B =

5

9.

Hence, Y =

+

+

+ 9159

9

2222 s

s

s

s

s

s=

+

+

91

9

5

122 s

s

s

s

Now

121

s

sL = tshco and

+

921

s

sL = t3cos

Therefore, y (t) = tt 3cos5

1cosh

5

9+ .

Example 33 (shifted data problem)

Solve tyy 2''1

=+ ,24

=

y and 224

' =

y .



24/26


Solution set t = 0~

tt + , so that4

0

=t .

Now the subsidiary equation in terms of the new variable t~ becomes

ssYyysYs

2

2~)0('~)0(~

~

2

2

+=+

ss

YsYs2

2~22

2

~2

2 +=++

ss

sYs2

222

2

~)1(

22 +++=+

)1(2)1(

2

1

2

1

2

)1(2

~222222 +

++

++

+

++

=sssssss

sY

Now)1(

222 +ss=

+ 1

11222 ss

and)1(2 2 +ss

=

+

11

2 2ss

s

Hence,1

2

1

2

)1(2

~222 +

+

++

=sss

sY

+

+

1

112

22 ss+

+

1

1

2 2s

s

s

=1

22 +

s

+ 22

s+

s2

Therefore, y~ ( t~ ) = t

~sin2 + 2 t~ +

2

.

Now in terms of the original variable t we get:

y(t) = 2t + cos t sin t.

Therefore, y(t) = 2t + cos t sin t.

1.3.2 Systems of Linear Differential Equations

Example 34 Solve211

' yyy += ,212

' yyy = , 1)0(1

=y and 0)0(2

=y .

Solution The subsidiary equations become

2111 )0( YYysY += and 2122 )0( YYysY = 1)1( 21 =+ YYs and0)1(

12 =++YYs

Solving for 1Y and 2Y algebraically we get

)1()1()1( 212 +=++ sYsYs and 0)1( 12 =++ YYs

)1(1)1( 12 +=++ sYs and 12

1

1Y

sY

+

=

1)1(

121 ++

+=

s

sY and

1)1(

122 ++

=s

Y



25/26


Now,

+++

1)1(

12

1

s

sL =

te

cos t and

++

1)1(

12

1

sL =

te

sin t

Therefore, tt

ety cos1 )(= and ttety sin2 )(

= .

Example 35 Solve211

5' yyy += ,212

5' yyy += , 3)0(1

=y and 7)0(2

=y .


2111 5)0( YYysY += and 2122 5)0( YYysY +=

3)5( 21 = YYs and 7)5( 12 = YYs


)5(3)5()5( 212 = sYsYs and 7)5( 12 = YYs

)5(371)5( 12 = sYs and 12

5

1

5

7Y

ssY

+

=

= 1)5(5

31)5(

7221 s

s

sY and

( )

+

=

1)5(

3

5

7

1)5()5(

7222 ssss

Y

Now( ) 5

7

1)5(

57

1)5()5(

722

= ss

s

ss.

Hence,

=1)5(

3

1)5(

57

222 ss

sY .

Now,

1)5(52

1

s

sL = te 5 cosh t and

1)5(1

2

1

sL = te 5 sinh t

Therefore, )cosh3sinh7(5

1 )( ttt

ety = and )sinh3cosh7(52 )( ttt

ety = .

Example 36 Solve tyy 2cos5''21

=+ , tyy 2cos5''12=+ , 1)0(

1=y , 1)0('

1=y , 1)0(

2=y

and 1)0('2

=y .


4

5)0(')0(

22111

2

+=+

s

sYyysYs and

4

5)0(')0(

21222

2

+=+

s

sYyysYs

( )45

1221

2

++=+

s

ssYYs and

4

51

2122

++=+

s

ssYYs


( )4

52

332

22

14

++=+

s

sssYsYs and

4

51

2122

++=

s

ssYYs



26/26


4

5

4

51)1(

22

332

14

+

++++=

s

s

s

ssssYs and ( ) 1

222 4

51 Ys

s

ssY

++=

+

+

+=

4

5

)1(

1

11

122221 s

s

ss

s

sY

Now41)4)(1(

52222 +++

+=

+ sDCs

s

BAs

ss

s 41)4)(1(

52222 +

=+ s

s

s

s

ss

s

Hence,41

1221 +

++

=s

s

sY and

41

1222 +

+

=s

s

sY

Therefore, ttty 2cossin)(1 += and ttty 2cossin)(2 = .

Example 37 Solve211

3'' yyy += , teyy 44''12= , 2)0(

1=y , 3)0('

1=y , 1)0(

2=y and

2)0('2

=y . Solution The subsidiary equations become

211112 3)0(')0( YYyysYs += and

144)0(')0( 1222

2

=

sYyysYs

sYYs 233)1( 212 += and

1

424 12

2

+=

ssYYs


sYYs 81212)1(4 212 += and

)1(4)1()2()1(4)1( 212

222 ++= sssYsYss

2

12 =

sY and

2

1

1

11

+

=ss

Y

Therefore, tt eety 21 )( += andtety 22 )( = .

Documents

Unit IV Laplace Transforms