Revision Inverse Laplace

7/17/2019 Revision Inverse Laplace

http://slidepdf.com/reader/full/revision-inverse-laplace 1/17

THEOREM : FIRST SHIFTING

If L{f(t)} = F(s), then L{eat

f t( )⋅ )} = F s a−( )

HENCE L

-1

(F s a−( )

) =e

atf t( )⋅

CLONE of EXM !"ESTION

Find L-1{

2

7s⋅

1

4−

s2

4 s⋅− 6+( ) } using the property of inverse Laplace transform

L 1−

2

7s⋅

1

4−

s2

4 s⋅− 6+( )

L 1−

2

7s⋅

1

4−

s 2−( )2

2+

L 1−

2

7s 2−( )⋅

4

7+

1

4−

s 2−( )2

2( ) 2+

=

L 1−

2

7s 2−( )⋅

9

28+

s 2−( )2

2+

L 1−

2

7s 2−( )⋅

s 2−( )2

2+

L 1−

9

28

2( )⋅

2 s 2−( )2

2( ) 2+⋅

+

=

2

7e

2 x⋅⋅ cos x 2⋅( )⋅ 9

28 2

e2 x⋅

sin x 2⋅( )⋅( )⋅+

CLONE of EXM !"ESTION (F#o$ Le%t&#e Note )

L-1{

s 5+

s2 s+ 2+ }=

L1− s 5+

s2

s+1

2

2

+

1

2

2

− 2+

⋅



=

L1− s 5+

s1

2+

27

4+

⋅ L1−

s1

2+

51

2−

+

s1

2+ 2 7

4

2+

⋅

=

L1−

s1

2+

s1

2+ 2 7

4

2+

⋅ L1−

9

2

s1

2+ 2 7

4

2+

⋅+

=

L1−

s1

2+

s1

2+ 2 7

4 2+

⋅9

2

L1−⋅

1

74

7

4

s1

2+ 2 7

4 2+

⋅

⋅+

=

L1−

s1

2+

s1

2+ 2 7

4

2+

⋅9

2

2

7

⋅ L1−⋅

7

4

s1

2+ 2 7

4

2+

⋅+

=

L1−

s 12

+

s1

2+ 2 7

2

2+

⋅9

7L

1−⋅

7

2

s1

2+ 2 7

2

2+

⋅+

=e

z−2

cos z7

2⋅

⋅9

7e

z−2

sin z7

2⋅

⋅

⋅+

=e

z−2

cos z7

2⋅

9

7sin z

7

2⋅

⋅+ ⋅



THEOREM : 'ERITIE OF TRNSFORM

If L{f(t)} = F(s), then L{tn

f t( )⋅ } =

1−( )n

Fn( )⋅ s( ) 1−( )

n dn

dsn

⋅ F s( )

F#o$ the *+en t.e of //en0* L{ tn

f t( ) } = 1−( )n

F n( )

s( )

L-1 {L{ tn

f t( ) }}= L-1 { 1−( )n

F n( )

s( ) } = 1−( )n

L-1 {F n( )

s( ) }

tn

f t( ) = 1−( )n

L-1 {F n( )

s( ) }

L-1 {F n( )

s( ) } =

tn

f ⋅ t( )⋅

1−( )n

1−( )n

tn⋅ f t( )⋅

L-1 {F n( )

s( ) } = 1−( )n

tn⋅ f t( )⋅

HENCE L-1 {F n( )

s( ) } = 1−( )n

tn⋅ f t( )⋅

THEOREM : L2LCE TRNSFORM FOR INTEGRL F"NCTION

If L{f(t)} = F(s), then L { 0

t

uf u( )⌠ ⌡

d

} =

1

s L{f(t)} =

F s( )

s

From the given appendix L{ 0

x

ug u( )⌠ ⌡

d

} =

G s( )

s

HENCE L-1 {

G s( )

s } = 0

x

ug u( )⌠ ⌡

d



CLONE of EXM !"ESTION

Find L-1{

2

s2

s 4+( )2⋅ } using the properties of the inverse Laplace transform

Consider L-1{

F s( )

s }= 0

t

uf u( )⌠

⌡

d

L 1− 1

s 4+( )2

t e 4− t⋅⋅

L 1− 4

s s 4+( )2⋅

L

1−

4

s 4+( )2

s

4

0

t

uu e 4− u⋅⋅

⌠ ⌡

d⋅ 4 u e

4− u⋅⋅4−

e 4− u⋅

16−

⋅ t

0

⋅

=

u− e 4− u⋅⋅

e 4− u⋅

4−

t

0

⋅ 1

4t e

4− t⋅⋅− e

4− t⋅

4−



L 1− 4

s2

s 4+( )2⋅

L

1−

1

s s 4+( )2⋅

s

0

t

u1

4u e

4− u⋅⋅− e

4− u⋅

4−

⌠ ⌡

d

=

1

4u⋅ 1

4u⋅ exp 4− u⋅( )⋅+ 1

8exp 4− u⋅( )⋅+

t

0

⋅

=

1

4t⋅

1

4t⋅ exp 4− t⋅( )⋅+

1

8exp 4− t⋅( )⋅+

1

8−

CLONE of MT343 5"NE 6711

Find the L-1{

s2+4¿¿¿2¿2

¿

using the properties of inverse Laplace transforms

Consider

s2+4¿

¿(¿2¿¿)=sin (2 x)=f ( x )2

¿ L

−1 ( F (s ) )= L−1 ¿

so f ( x )=sin (2 x)

s2+4

¿¿¿

F ' (s )=

d

ds ( 2

s2+4 )=−4 s

¿



L−1{ F (n )(s)}=(−1 )n ( x )n f ( x )

Hen%e L

−1

{

d

ds

( 2

s2+4

)}=(−1

)

n

( x )

n

f ( x )

L−1{ dds ( 2

s2+4 )}=(−1 )1 ( x)1 (sin (2 x ) )=− x (sin (2 x ))

L−1{ dds ( 2

s2+4 )}=− x (sin (2 x ))

s2+4

¿¿

¿=− x (sin (2 x))−4 s¿

L−1¿

s2

+4¿¿

¿=− x (sin (2 x))4 s¿

− L−1¿

s2+4

¿¿

¿= x (sin (2 x ))=g ( x)4 s¿

L−1¿



s2+4

¿¿

s2+4

¿¿

¿2

¿=1

2 L

−1 {G(s)s }=∫

0

x

g (u ) du

4 s¿¿

(¿2¿¿)=1

2 L

−1 ¿

2

¿ L

−1 ¿

here

s2+4

¿¿¿

G ( s )=4 s¿

s2+4¿

¿¿= x (sin (2 x ))4 s¿

g ( x )= L−1 {G(s)}= L

−1¿

!ences2+4

¿¿

g (u )du=¿

∫0

x

u(sin (2u ))du

(¿2¿¿)=∫0

x

¿

2

¿ L

−1¿



¿u (−cos (2u )2 )−(1)(

−sin (2u)4

)|0

x

= sin (2u )

4−

ucos (2u )2 |

0

x

s2+4¿¿

(¿2¿¿)=sin (2 x )

4−ucos (2 x )

2−

sin (0 )4

+ucos (0 )

2

2

¿ L

−1¿

s2+4

¿¿

(¿2¿¿)=sin (2 x )

4−

ucos (2 x )2

+u

2

2

¿ L

−1 ¿

MT343 5"NE 6718 (F#o$ Le%t&#e Note )



"s*n the /#o/e#t*es of the *n+e#se L/.%e t#nsfo#$, f*n0 the *n+e#se

L/.%e t#nsfo#$ of

5s

s2

16+( ) 2

F#o$ the *+en t.e of //en0* L{ tn

f t( ) } = 1−( )n

F n( )

s( )

L-1 {L{ tn

f t( ) }}= L-1 { 1−( )n

F n( )

s( ) } = 1−( )n

L-1 {F n( )

s( ) }

tn

f t( ) = 1−( )n

L-1 {F n( )

s( ) }

L-1 {F

n( )s( ) } =

tn

f ⋅ t( )⋅

1

−( )

n1−( )

nt

n⋅ f t( )⋅

L-1 {F n( )

s( ) } = 1−( )n

tn⋅ f t( )⋅

Cons*0e#

F s( ) 4

s2

16+( )

L-1 {F s( ) } = L-1 {

4

s2

16+( ) } = sin 4 t⋅( ) f t( )

So f t( ) sin 4 t⋅( )

F n( )

s( )s

4

s2

16+( )

d

d

8− s⋅

s2

16+( ) 2

L-1 {F n( )

s( ) } = 1−( )n

tn⋅ f t( )⋅

L-1 {

8− s⋅

s2 16+( ) 2 } = 1−( )1 t1( )⋅ sin 4t( )⋅ t− sin 4t( )⋅

L 1− 5s

s2

16+( ) 2

⋅ L

1− 5−8

8− s⋅

s2

16+( ) 2

⋅

⋅

5−8

L 1−⋅

8− s⋅

s2

16+( ) 2

⋅



=

5−8

t− sin 4 t⋅( )⋅( )⋅ 5 t⋅ sin 4 t⋅( )⋅

8


"s*n the /#o/e#t*es of the *n+e#se L/.%e t#nsfo#$,

Find L-1{

s

s2

1+( )2

}

L-1{

1

s2

1+( )}= sin t( )

s1

s2

1+( )

dd

2− s⋅s

21+( )

2

L-1{

2− s⋅

s2

1+( )2

}= 1−( )1

t1( )⋅ sin t( )⋅ t− sin t( )⋅

L1− s

s2

1+( )2

⋅ L

1− 1−2

2− s⋅

s2

1+( )2

⋅

⋅

1−2

L1−⋅

2− s⋅

s2

1+( )2

⋅

=

1−2

t− sin t( )⋅( )⋅t sin t( )⋅

2

MT343 5N 6716

"etermine ℒ-1 {

1

s (s−1)5 } using the property of inverse Laplace transform#

Consider

L-1{

1

s 1−( )5

} =

eu

4!u

4 u4

eu⋅

24



L-1{

1

s s 1−( )5

} = L-1{

1

s 1−( )5

s } = 0

t

uu

4e

u⋅24

⌠ ⌡

d

0

t

uu

4e

u⋅24

⌠ ⌡

d et t

4

24

t3

6−

t2

2+ t− 1+

⋅ 1−

L-1{

1

s s 1−( )5

} = 0

t

uu

4e

u⋅24

⌠ ⌡

d et t

4

24

t3

6−

t2

2+ t− 1+

⋅ 1−

MT343 'EC 6718

Find L

−1{ s

(s2+1)( s+2)

} using partial fraction

s

s 2+( ) s2

1+( )⋅

A

s 2+α s⋅ +

s2

1++

$here (s2

+1) is an irreduci%le &uadratic expression

s A s2

1+( )⋅ α s⋅ +( ) s 2+( )⋅+

Let s 2− ' 2− 5 A⋅ ⇔ A

2−5

s 0 ' 0 A 2 ⋅+ ⇔ 0

2−5

2 ⋅+ ⇔

1

5

s 1− ' 1− 2 A⋅ α− + ⇔ α

2

5

s

s 2+( ) s2

1+( )⋅

A

s 2+α s⋅ +

s2

1++

2−

5

s 2+

2

5s⋅

1

5+

s2

1++



=

1

5

2 s⋅ 1+

s2

1+

2

s 2+− ⋅

1

5

2 s⋅

s2

1+

1

s2

1++

2

s 2+− ⋅

L

1− s

s 2+( ) s2

1+( )⋅

⋅ L

1− 1

5

2 s⋅

s2

1+

1

s2

1++

2

s 2+−

⋅

⋅

=

1

5L

1−⋅ 2 s⋅

s2

1+

1

s2

1++

2

s 2+−

⋅ 1

52 cos x( )⋅ sin x( )+ 2 e

2− x⋅⋅−( )⋅


F*n0 L

−1{ 76 s−11

(16 s2+8 s+5)( s+4)}

&s*n /#t*. f#%t*on

Notice that 16s2

8 s⋅+ 5+( ) is an ieduci!"e #uadatic e#uation$

%an ieduci!"e #uadatic e#uation % !2

4 ac⋅− 0< &$&

'hus the (ethod of co()"etin* the s#uae +i"" !e a))"ied on 16s2

8 s⋅+ 5+

76 s⋅ 11−( )

16 s3⋅ 56 s

2⋅− 27 s⋅− 20−( )76 s⋅ 11−( )

s 4−( ) 16s2

8 s⋅+ 5+( )⋅

76 s⋅ 11−( )

s 4−( ) 16s2

8 s⋅+ 5+( )⋅

,

s 4−( )

s⋅ . +

16s2

8 s⋅+ 5+( )+

76 s⋅ 11−( ) , 16s2

8 s⋅+ 5+( )⋅ s⋅ . +( ) s 4−( )⋅+

su!stitute s 4 , 1

76 s⋅ 11− 16 s2⋅ 8 s⋅+ 5+ s⋅ . +( ) s 4−( )⋅+

76 s⋅ 11− 16+( ) s2⋅ 4− ⋅ 8+ . +( ) s⋅+ 5 4 / ⋅−+

co()ain* the coefficient of s2! 16+ 0 16−



co()ain* the coefficient of s0! 5 4 . ⋅− 11− . 4

ence

76 s⋅ 11−( )

s 4−( ) 16s2

8 s⋅+ 5+( )⋅

1

s 4−( )

4 16s−

16 s2⋅ 8 s⋅+ 5+

+

=

1

s 4−( )

4 16s−

16 s2 s

2+

1

4

2+1

4

2−

⋅ 5+

+1

s 4−( )

4 16s−

16 s1

4+ 2 1

16−

⋅ 5+

+

=

1

s 4−( )

4 16s−

16 s1

4

+ 2

⋅ 4+

+1

s 4−( )

161

4s−

16 s1

4

+ 2 1

4

+

⋅

+

=

1

s 4−( )

1

4s−

s1

4+ 2 1

4+

+1

s 4−( )

s−1

4−

1

2+

s1

4+ 2 1

4+

+1

s 4−( )

s1

4+

−1

2+

s1

4+ 2 1

4+

+

=

1

s 4−( )

1

2

s1

4+ 2 1

2 2+

+s

1

4+

s1

4+ 2 1

4+

−

L−1{

76 s−11

(16 s2+8 s+5)( s+4)

}

=

L1− 1

s 4−( )

1

2

s1

4+ 2 1

2

2+

+s

1

4+

s1

4+ 2 1

4+

−

=

L1− 1

s 4−( )

L1−

1

2

s1

4+ 2 1

2

2+

+ L1−

s1

4+

s1

4+ 2 1

2

2+

−



=e

4 ⋅e

−4

sin

2

⋅+ e

−4

cos

2

⋅−


Find L

−1{ s

2−2 s−4

(s2−2 s+2)( s+1)

} using partial fraction

otice that s2

2 s⋅− 2+( ) is an irreduci%le &uadratic e&uation (!2

4 ac⋅− 0< )#

hus the method of completing the s&uare $ill %e applied on s2

2 s⋅− 2+

s2−2 s−4

(s2−2 s+2)(s+1)= A

(s+1)+ Bs+ K (s2−2 s+2)

s2

2 s⋅− 4− , s2

2 s⋅− 2+( )⋅ s⋅ . +( ) s 1+( )⋅+

su!stitute s 1− ,

1

5−

ences

2

2 s⋅− 4− 1

5− s2

⋅ 2

5+ . + s⋅2

5−+ . +

co()ain* the coefficient of s2!

1

5− 1

6

5

co()ain* the coefficient of s0!

. 2

5− 4−

.

18

5−

'hus

s2

2 s⋅− 4−( )

s 1+( ) s2 2 s⋅− 2+( )⋅

1−

5 s 1+( )⋅" #

6

5s⋅

18

5−

s2 2 s⋅− 2+( )

+

'heefoe

s2

2 s⋅− 4−( )

s 1+( ) s2

2 s⋅− 2+( )⋅

1−5 s 1+( )⋅" #

6

5s⋅

18

5−

s2

2 s⋅− 2+( )+



=

1−5 s 1+( )⋅" #

6

5s⋅

18

5−

s2

2 s⋅−2

2− 2+

2

2− 2−

2+

+

=

1−5 s 1+( )⋅" #

6

5s⋅

18

5−

s2

2 s⋅− 1−( )2+ 1− 2+

+

=

1−5 s 1+( )⋅" #

6

5s⋅

18

5−

s 1−( )2

1++

1−5 s 1+( )⋅" #

6

5s 1−( )⋅

6

5+

18

5−

s 1−( )2

1++

=

1−5 s 1+( )⋅" #

6

5s 1−( )⋅

12

5−

s 1−( )2

1++

1−5 s 1+( )⋅" #

6

5s 1−( )⋅

s 1−( )2

1++

12

5

s 1−( )2

1+−

L−1{

s2−2 s−4

(s2−2 s+2)( s+1)

}

L1− 1−

5 s 1+( )⋅" #

6

5s 1−( )⋅

s 1−( )2

1++

12

5

s 1−( )2

1+−

=

L1−

6

5s 1−( )⋅

s 1−( )2

1+

L1−

12

5

s 1−( )2

1+

− L

1−

1

5

s 1+

−

=

6

5

ez⋅ cos z( )⋅

12

5

ez⋅ sin z( )⋅−

ez−

5

−

MT343 5"N 6719

a) *ho$ that ℒ-1 {

1

(s2+2 s+5) s } =1

5−

1

5e−t (cos2 t +

1

5sin 2 t )



Let H (s )= 1

(s2+2 s+5 ) s and h ( t )=¿ ℒ-1 { H (s )}

H (s )=

1

(s2+2 s+5 ) s=

A

s +

Bs+C

(s2+2 s+5 )

1= A ( s2+2 s+5)+( Bs+C ) s

s=0 ;1=5 A → A=1

5

s=−1 ;1=1

5(4 )+B−C →B−C =

1

5

s=−2 ;1=1

5(5 )+4B−2C →2 B−C =0

B=−1

5;C =

−2

5

s+1¿¿

(¿2

− (1

)

2

+5

¿)5¿1

( s2+2 s+5 ) s=

1

5 s−

s+2

5 (s2+2 s+5)=

1

5 s−

(s+1 )+1¿

s+1¿¿

(¿2+4¿)5¿

1

( s2+2 s+5 ) s= 1

5 s−( s+1 )+1

¿



s+1¿¿

(¿2+4¿)¿

s+1

¿¿(¿2+4¿)

5¿5¿

¿ 1

5 s−

s+1¿

h (t )= L−1{ H ( s) }= L−1{ 1

(s2+2 s+5)s }

¿ ℒ-1 {

s+1¿¿

(¿2+22¿)¿

s+1¿¿

(¿2+22¿)¿

5(2)¿

5¿1

5 s−

s+1¿

¿1

5−

1

5e−t cos (2t )−

1

10e−t sin (2t )

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