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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).

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Unit I Laplace Transform

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).

Solution Applying definition 1.1 we get:

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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Unit I Laplace Transform

=

+ 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).

Solution Applying definition 1.1 we get:

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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Unit I Laplace Transform

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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Unit I Laplace Transform

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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Unit I Laplace Transform

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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Unit I Laplace Transform

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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Unit I Laplace Transform

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

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Figure 1.1 a piecewise continuous function

90

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Unit I Laplace Transform

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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Unit I Laplace Transform

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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Unit I Laplace Transform

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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Unit I Laplace Transform

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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Unit I Laplace Transform

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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Unit I Laplace Transform

)()( 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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Unit I Laplace Transform

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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Unit I Laplace Transform

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.

Solution Applying definition 1.1 we get:

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)].

Prepared by Tekleyohannes Negussie

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)

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Unit I Laplace Transform

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

Prepared by Tekleyohannes Negussie

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

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Unit I Laplace Transform

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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Unit I Laplace Transform

( ) ( ) )()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

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Unit I Laplace Transform

( )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.

Example 28 (initial value problem)

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

Solution The subsidiary equation becomes

( ) ( ) 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.

Example 29 (initial value problem)

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

Solution The subsidiary equation becomes

( ) ( )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

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Unit I Laplace Transform

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.

Example 30 (initial value problem)

Solve tt

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

Solution The subsidiary equation becomes

( ) ( )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.

Example 31 (initial value problem)

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

Solution The subsidiary equation becomes

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Unit I Laplace Transform

( )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 .

Example 32 (initial value problem)

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

Solution The subsidiary equation becomes

( )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 .

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Unit I Laplace Transform

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

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Unit I Laplace Transform

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 .

Solution The subsidiary equations become

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

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

Solving for 1Y and 2Y algebraically we get

)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 .

Solution The subsidiary equations become

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

Solving for 1Y and 2Y algebraically we get

( )4

52

332

22

14

++=+

s

sssYsYs and

4

51

2122

++=

s

ssYYs

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Unit I Laplace Transform

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

Solving for 1Y and 2Y algebraically we get

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 )( = .