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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