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Chapt er 5 Laplace Tr ansf or ms

5 . 1 I nt r oductionI t was shown ear lier in t he course t hat 1st order ODEs of t he f orm

x = f ( x,t ) x(0) = a

can be solved by int egrat ion one t ime wrt t . This reduces t he ODE

t o an algebr aic equat ion t hat can be solved f or x(t ) which sat isf ies t he

I C, x(0) = a. However, as t he order of t he ODE is 2 or higher, mor e

t han one int egrat ion is requir ed t o r educe t he ODE t o an algebraic

equat ion. Theref or e, more wor k is r equir ed.

Anot her met hod of solving linear ODEs is t he so-called M et hod of

Laplace Tr ansf or ms which r educes t he ODE t o an algebraic equat ion

in one st ep and sat isf ies t he init ial condit ions as par t of t he same

process. Laplace (1749-1827) and Cauchy (1789-1857) are normally

credit ed wit h developing t he mat h t heory, and Heaviside, an English

elect r ical engineer (1850-1925), f ir st applied t he met hod t o cir cuit

problems. The met hod was originally called Heaviside Transf orms, but

t he name Laplace Tr ansf orms is used t oday.

Def init ion: The Laplace Tr ansf orm of a f unct ion f (t ) is def ined by

t he int egr al

L{ f (t ) } = F(s) = dtef ( t )0

st-

(2)

where s is a paramet er .

Note: I n or dinar y dif f er ent ial and int egral calculus t he int egral is t he

inverse operat ion of a der ivat ive and can be t hought of as a

t ype of t r ansf orm.

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5 . 2 Calculat ion of Some Simple T r ansf or ms

Example 3 : f (t ) = 1

F(s) = dtef ( t )0

st-

= dte10

st-

= -s1 (-s)dte

0

st-

= -s

1e

-st= -

s

1{Lim [ e

-st] - 1 } =

s

1

Theref or e: L{1} =s

1, s 0 (5)

Example 4 : f (t ) = e

at

F(s) = dtee0

st-at

= dte0

a)t-(s-

= -a)-(s

1dta)](s[e

0

a)t-(s-

= -a)-(s

1 e

a)t--(s= -

a)-(s

1{Lim [ e

a)t--(s] - 1 }

=a)-(s

1(6)

Ther ef or e: L{ eat

} =a)-(s

1, s a

Example: f (t ) = t

Shor t Cut

F(s) = dtet0

st-

D I

t + e-st

1 - -s

1e

-st

0 +s

12

e-st

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t

0

t

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Example (Cont inued )

F(s) = [ -s

1t e

-st-

s

12

e-st

]

= Lim [ -s

1t e

-st-

s

12

e-st

] - Lim [ -s

1t e

-st-

s

12

e-st

]

t = 0

=s

12

Theref or e: L{ t } =s

1

2, s 0

Since t he Laplace Transf or m int egral dtef ( t )0

st-

must exist in order t o be

used, cer t ain requir ement s on f (t ) must , in general, be met :

(1) f (t ) must be at least piecewise ( in sect ions) cont inuous.

f ( t )

t

(2) f (t ) e-st

must be bounded as t

Ex: f (t ) = t , Lim t e-st

= Lime

tst

=

= Limes

1st

= 0

Theref or e: f (t ) = t = O( est

), s 0 Page 3

0

t

t t

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Ex: f (t ) = et2

Lime

est

2t

=

t

Lim e stt2

= NOT BOUNDED!

t

Theref or e: et2

O( est

), s 0.

5. 3 T ransf orms of Derivat ives

Consider L{ f ' (t )} = dte(t )'f0

st-

{ I nt egr at e by par t s }

u = e-st

v = f (t )

u = - s e-st

v = f (t )

dte(t )'f0

st-

= e-st

f ( t ) - dte(t )(-s)f0

st-

= Lim e-st

f (t) - f (0) + s dte(t )f0

st-

t

Since f (t ) = O( est

) f or s 0, t hen Lim e-st

f (t ) = 0 as t .

Theref or e: L{ f (t ) } = = sL{ f (t ) } - f (0) = sF(s) - f (0) (8)

I n a similar mannerL{ f (t ) } = sL{ f (t ) } - f (0) = s[sL{ f (t ) } - f (0) ] f (0)

= s2L{ f (t) } - sf (0) - f (0) (13)

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I nverse T ransf orms Let t he symbol L-1

{F(s)} = f (t ) denot e a f unct ion

whose Laplace Tr ansf orm is F(s). Then, if

L{f (t )} = F(s) t hen L-1

{F(s)} = f (t )

Example: I f L{ eat

} =a)-(s

1, t hen L

-1{

a)-(s

1} = e

at

As a pract ical mat t er , ever y given f unct ion F(s) cannot have more t han one

inverse t r ansf orm f (t ) t hat is cont inuous f or each t 0.

T he Use of Part ial Fr act ions For I nver se T r ansf or ms.

Example: Find L-1

{2ss

1s2

} .

Let2ss

1s2

=

2)s(s

1s

=

s

A+

2s

B

Mult iply bot h sides of t he equat ion by s(s+2).

s + 1 = (s + 2)A + sB = (A + B)s + 2A

Equat e coef f icient s of like power s of s:

A + B = 1 and 2A = 1

or

A =2

1and B =

2

1

Theref or e, L-1

{2ss

1s2

} = L

-1{

2

1

s

1} + L

-1{

2

1

2)(s

1

}

=2

1+

2

1e

-2t

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PARTI AL FRACTI ON SHORT CUT:

A =2)(s

1s

at s = 0 A =

2

1

B = s

1s at s = - 2 B = 2

1

Example: Repeat ed Linear Fact or s.

Find L-1

{

a)s(s

a2

2

} .

Assume:

a)s(s

a

2

2

= s

A

+ as

B

+ a)(s

C

2

Convent ional M et hod: Mult iply bot h sides by s a)(s2

.

a2

= A a)(s2

+ Bs(s + a) + Cs

Mult iply out and collect like power s of s

a2

= A( s2

+ 2as + a2

) + B( s2

+ as) + Cs

or

a2

= ( A + B)s2

+ ( 2aA + aB + C)s + a2

A

Ther ef ore: A = 1, A + B = 0 B = - 1

2aA + aB + C = 0 C = - a

and

a)s(s

a

2

2

= s

1

- as

1

- a)(s

a

2

L-1

{a)s(s

a2

2

} = L

-1{

s

1} - L

-1{

as

1

} - aL

-1{

a)(s

12

}

= 1 - e-at

- at e-at

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PARTI AL FRACTI ON SHORT CUT:

A =

a)(s

a2

2

at s = 0 A = 1

C =s

a2

at s = - a C = - a

B = )ds

dC(

a-s

= -s

a2

2

at s = - a B = - 1

These are t he same result s as wit h t he convent ional met hod!

Example 2 (Text ): Find L-1

{

10-3ss

32

}.

10-3ss

32

=2)-5)(s(s

3

=

5s

A

+

2-s

B

A = )2-s

3(

5-s

= -7

3

B = )5s

3(

2s =

7

3

L-1

{10-3ss

32

} = -7

3L

-1{

5s

1

} +

7

3L

-1{

2-s

1}

Using L-1

{a-s

1} = e

at,

L-1

{10-3ss

32

} = -7

3e

-5t+

7

3e

2t=

7

3( e

2t- e

-5t)

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Convolut ion ( Pr oduct of T r ansf or ms)

Let F(s) = L{f (t )} and G(s) = L{g(t )}

Then consider t he product

F(s)G(s) = dxef (x )0

xs-

dyeg(y)0

ys-

=

0

dydxg(y)f (x )e0

y)x(s-

(1)

Consider t he f ollowing region in t he 1st Quadrant :

y

(0,a)

(0,0) (a,0) x

I n t he limit as a the triangle cover s t he 1st quadrant . Now consider

t he f ollowing change in variables:

Let t = t (x,y) = x + y and = (x,y) = y

The reverse t r ansf ormat ion is given by

x = x(t , ) = t - and y = y(t , ) =

Eq(1) can be wri t t en

F(s)G(s) = Lim dtdJ)g()-f (test-

(2)

where t he limit s of int egrat ion will depend on a , and J is t he J acobian of

t he t r ansf ormat ion.

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J is given by J (t ,

yx,) = = = 1

Theref or e, eq(2) becomes

F(s)G(s) = Lim dtd)g()-f (test- 1

Limit s of int egr at ion are f ound as f ollows: (0,0) (0,0) , (a,0) (a,0) ,

and (0,a) (a,a)

a (a,a)

= t

0 dt a t

F(s)G(s) = Lim a

0

dtd)g()-f (tet

0

st-

F(s)G(s) =

0

st-e [ dtd)g()-f (tt

0 ] (3)

Let f (t)*g(t ) = t

0

d)g()-f (t (4)

Then, F(s)G(s) = L{ f (t )*g(t ) } or f (t )*g(t ) = L-1 { F(s)G(s) } (5)

or L-1

{ F(s)G(s) } = f (t )*g(t ) = t

0

d)g()-f (t

Page 7b

t

x

t

y

x

y

1 0

- 1 1

a

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Example: L-1

{s

14

} =6

t 3f r om Laplace Transf orm Table.

Rework using Convolut ion t r ansf orm. Let F(s) =s

1

2

and G(s) =s

1

2

.

L-1

{ F(s)G(s) } = f (t )*g(t ) = t * t

f (t ) = t , g(t ) = t or g() =

Then, using eqs(4) and (5)

L-1 { F(s)G(s) } = f (t )*g(t ) = t

0

d)g()-f (t =

= t

0

2 d)(t = [ t2

2

-3

3

]

=2

t 3-

3

t 3=

6

t 3Same Result !

Example: L-1

{s

13

} =2

t 2f r om Laplace Tr ansf orm t able.

L-1

{s

13

} = L-1

{s

1s

12

} = f (t )*g(t )

where F(s) =s

1or f (t ) = 1

G(s) = s

12 or g(t ) = t

L-1

{s

13

} = 1* t = t

0

d)( 1 =2

2

=2

t 2

Page 7c

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0

d)-(t

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t

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Sect ion 5. 3, Example 3: Part ial Fract ion (See Ex 2) Versus Convolut ion

Met hod

L-1

{

10-3ss

32

} = L-1

{

2)-5)(s(s

3

} = L

-1{

5)(s

3

2)-(s

1}

Then F(s) =5)(s

3

or f (t ) = 3 e

-5t

and G(s) =2)-(s

1or g(t ) = e

2t

L-1

{ F(s)G(s) } = f (t )*g(t ) = t

0

2)-5(t- dee3 = 3 t

0

255t- deee

= 3 e-5t

t

0

7 de =7

3e

-5t

t

0

7 7de

=7

3e

-5t[ e

7t- 1 ] =

7

3[ e

2t- e

-5t]

This is t he same result as Example 2!

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5 . 4 Solut ions of OD Es Using Laplace T ransf or m.

Example: ODE y (t ) + a2y = 0

I C y(0) = A

y (0) = B

L{ y (t ) } + a2L{ y(t ) } = 0 (1)

From eq(13)

L{ y (t ) } = s2L{ y(t ) } - sy(0) - y(0)

= s2y(s) - A s - B

Eq(1) becomes an algebr aic equat ion in y(s)

s2y(s) - A s - B + a

2y(s) = 0

Solve f or y(s).

y(s)( s2

+ a2

) = As + B

or

y(s) = A (as

s22

) + B(

as

122

) = A(

as

s22

) +

a

B(

as

a22

)

Using t he inver se t r ansf orm t able

L-1

{y(s)} = A L-1

{as

s22

} +

a

BL

-1{

as

a22

}

y(t ) = Acos(at ) +a

Bsin(at )

or t he solut ion t o eq(1) becomes

y(t ) = Acos(at ) + Bsin(at )

where A and B ar e arbit rar y constant s based on t he I C.

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Example: ODE y (t ) - y (t ) - 6y(t ) = 2

I C y(0) = 1

y (0) = 0

Take t he Laplace Tr ansf orm of both sides of t he ODE.

s2y(s) - sy(0) - y (0) - [ sy(s) - y(0) ] - 6y(s) =

s

2

s2y(s) - s - sy(s) + 1 - 6y(s) =

s

2

Solve f or y(s).

y(s) [ s2

- s - 6 ] = s - 1 +s

2=

s

2s-s2

y(s) ( s 3)( s + 2) =s

2s-s2

or

y(s) =2)s3)(-s(s

2s-s2

To f ind t he inverse t r ansf orm, use par t ial f r act ions.

y(s) =2)s3)(-s(s

2s-s2

=

s

A+

3s

B

+

2s

C

Using t he shor t cut ,

A =2)s3)(-s(

2s-s2

at s = 0 A =

2)3)(-(

2= -

3

1

B =2)s(s

2s-s2

at s = 3 B =

2)3(3

23-)3(2

=

15

8

C =3)-s(s

2s-s2

at s = - 2 C =

3)-2-(-2)(

2(-2)-(-2)2

=

10

8=

5

4

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Theref or e, y(s) = - (3

1)

s

1+ (

15

8)

3s

1

+ (

5

4)

2s

1

Taking t he inver se t r ansf orm,

y(t ) = = -3

1+ (

15

8) e

3t+ (

5

4) e

-2t

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Example 2 (T ext ) Solve t he I nit ial Value Pr oblem:

ODE y (t ) - y(t ) = 0 (16)

I C y(0) = 1

y (0) = y (0) = y (0) = 0

Tale t he Laplace Tr ansf or m of each t erm of t he ODE:

L{ y (t ) } - L{ y(t ) } = 0

From Appendix C Tr ansf orm No. 26

s4Y(s) - s

3y (0) - s

2y (0) - sy (0) - y (0) - Y(s) = 0 (17)

s4Y(s) - s

3(1) - s

2(0) - s(0) - y (0) - Y(s) = 0

Y(s) [ s4 - 1 ] = s3

Y(s) =1s

s

4

3

(18)

To f ind t he I nverse Transf or m use t he Met hod of Par t ial Fract ions

wit h t he shor t cut .

Y(s) =1s

s

4

3

= 1)s1)(-s(

s

22

3

= i)-i)(s1)(s-1)(s(s

s3

= 1s

A

+ 1s

B

+ is

C

+ is

D

(19)

A =1)s1)(-(s

s

2

3

=

1)1)(1-(-1

(-1)3

=

(-2)(2)

1-=

4

1

B =1)s1)((s

s

2

3

=

1)1)(1(1

(1)3

=

(2)(2)

1=

4

1

C =i)-1)(ss(

s

2

3

==

i)-1)(-i-(-1

(-i)3

=(-2)(-2i)

i)(-i)(-i)(-=

4i

i=

4

1

D =i)1)(ss(

s

2

3

=

i)1)(i-(-1

(i)3

=

(-2)(2i)

(-1)(i)=

4i

i=

4

1Page 11

s = -1

s = 1

s = -i

s = +i

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Theref or e, t he t r ansf orm in par t ial f r act ion f orm is

Y(s) =4

1

1)(s

1

+

4

1

1)(s

1

+

4

1

i)(s

1

+

4

1

i)(s

1

y(t ) =4

1L

-1{

1s

1

} +

4

1L

-1{

1s

1

} +

4

1L

-1{

is

1

} +

4

1L

-1{

is

1

}

Using t he I nver se Tr ansf or m L-1

{a-s

1} = e

at

y(t ) =4

1e

-t+

4

1e

t+

4

1e

-i t+

4

1e

it

Recall: cosh(t ) =21 ( et + e-t )

and eit

= cos(t ) + isin(t )

e-i t

= cos(t ) - isin(t )

Addit ion gives 2cos(t ) = eit

+ e-i t

or

cos(t ) =21 ( eit + e-i t )

Theref or e, t he f inal solut ion t o eq(16) is

y(t ) =2

1cosh(t ) +

2

1cos(t ) (20)

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Example 3 (Text) Solve the 1 st or der I nit ial Value Pr oblem:

ODE x (t ) + px(t ) = q(t ) (21)

I C x(0) = xo

where p is a constant and q(t ) is a given f or cing f unct ion of t .

Take t he Laplace Transf or m of both sides of eq(21).

L{ x (t ) } + pL{ x(t ) } = L{ q(t ) }

sX(s) - x(0) + pX(s) = Q(s)

or X(s)( s + p ) = xo + Q(s)

or X(s) = xo(

ps

1

) + (

ps

1

)Q(s)

Taking t he I nver se Transf or m,

x(t ) = xoe

-pt+ L

-1{ (

ps

1

)Q(s) }

Consider t he 2nd

t erm as a Convolut ion wit h

F(s) =ps

1

and G(s) = Q(s)

f ( t ) = e-pt

and g(t ) = q(t )

or f (t - ) = e)-p(t -

and g() = q()

Theref or e,

x(t) = xoe

-pt+

t

0

d)g()-f (t = xoe

-pt+

t

0

)-p(t- d)q(e

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x(t) = xoe

-pt+ e

-pt

t

0

p d)q(e

= e-pt

[ xo

+ t

0

p d)q(e

] (23)

Let q(t ) = t . Then, eq(23) represent s a closed f orm solut ion since

I nt egrat ion by par t s using t he shor t cut D I

+ ep

1 -p

1e

p

0 +p

12

ep

t

0

p d)q(e

= t

0

p de

= [p

e

p-

p

12

ep

]

=p

te

pt-

p

12

ept

- [ 0 -p

12

]

=p

te

pt-

p

12

ept

+p

12

Theref or e, eq(23) becomes

x(t) = e-pt

[ xo

+p

te

pt-

p

12

ept

+p

12

]

x(t ) = ( xo

+p

12

) e-pt

+ (p

t-

p

12

)

Page 14

t

0