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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+
⋅
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 2/17
=
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⋅
⋅+ ⋅
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 3/17
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
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 4/17
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−
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 5/17
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
¿
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 6/17
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¿
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 7/17
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¿
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 8/17
¿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 )
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 9/17
"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
⋅
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 10/17
=
5−8
t− sin 4 t⋅( )⋅( )⋅ 5 t⋅ sin 4 t⋅( )⋅
8
CLONE of EXM !"ESTION (F#o$ Le%t&#e Note )
"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
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 11/17
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++
$here (s2
+1) is an irreduci%le &uadratic expression
s A s2
1+( )⋅ α s⋅ +( ) s 2+( )⋅+
Let s 2− ' 2− 5 A⋅ ⇔ A
2−5
s 0 ' 0 A 2 ⋅+ ⇔ 0
2−5
2 ⋅+ ⇔
1
5
s 1− ' 1− 2 A⋅ α− + ⇔ α
2
5
s
s 2+( ) s2
1+( )⋅
A
s 2+α s⋅ +
s2
1++
2−
5
s 2+
2
5s⋅
1
5+
s2
1++
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 12/17
=
1
5
2 s⋅ 1+
s2
1+
2
s 2+− ⋅
1
5
2 s⋅
s2
1+
1
s2
1++
2
s 2+− ⋅
L
1− s
s 2+( ) s2
1+( )⋅
⋅ L
1− 1
5
2 s⋅
s2
1+
1
s2
1++
2
s 2+−
⋅
⋅
=
1
5L
1−⋅ 2 s⋅
s2
1+
1
s2
1++
2
s 2+−
⋅ 1
52 cos x( )⋅ sin x( )+ 2 e
2− x⋅⋅−( )⋅
CLONE of EXM !"ESTION (F#o$ Le%t&#e Note )
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< &$&
'hus the (ethod of co()"etin* the s#uae +i"" !e a))"ied on 16s2
8 s⋅+ 5+
76 s⋅ 11−( )
16 s3⋅ 56 s
2⋅− 27 s⋅− 20−( )76 s⋅ 11−( )
s 4−( ) 16s2
8 s⋅+ 5+( )⋅
76 s⋅ 11−( )
s 4−( ) 16s2
8 s⋅+ 5+( )⋅
,
s 4−( )
s⋅ . +
16s2
8 s⋅+ 5+( )+
76 s⋅ 11−( ) , 16s2
8 s⋅+ 5+( )⋅ s⋅ . +( ) s 4−( )⋅+
su!stitute s 4 , 1
76 s⋅ 11− 16 s2⋅ 8 s⋅+ 5+ s⋅ . +( ) s 4−( )⋅+
76 s⋅ 11− 16+( ) s2⋅ 4− ⋅ 8+ . +( ) s⋅+ 5 4 / ⋅−+
co()ain* the coefficient of s2! 16+ 0 16−
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 13/17
co()ain* the coefficient of s0! 5 4 . ⋅− 11− . 4
ence
76 s⋅ 11−( )
s 4−( ) 16s2
8 s⋅+ 5+( )⋅
1
s 4−( )
4 16s−
16 s2⋅ 8 s⋅+ 5+
+
=
1
s 4−( )
4 16s−
16 s2 s
2+
1
4
2+1
4
2−
⋅ 5+
+1
s 4−( )
4 16s−
16 s1
4+ 2 1
16−
⋅ 5+
+
=
1
s 4−( )
4 16s−
16 s1
4
+ 2
⋅ 4+
+1
s 4−( )
161
4s−
16 s1
4
+ 2 1
4
+
⋅
+
=
1
s 4−( )
1
4s−
s1
4+ 2 1
4+
+1
s 4−( )
s−1
4−
1
2+
s1
4+ 2 1
4+
+1
s 4−( )
s1
4+
−1
2+
s1
4+ 2 1
4+
+
=
1
s 4−( )
1
2
s1
4+ 2 1
2 2+
+s
1
4+
s1
4+ 2 1
4+
−
L−1{
76 s−11
(16 s2+8 s+5)( s+4)
}
=
L1− 1
s 4−( )
1
2
s1
4+ 2 1
2
2+
+s
1
4+
s1
4+ 2 1
4+
−
=
L1− 1
s 4−( )
L1−
1
2
s1
4+ 2 1
2
2+
+ L1−
s1
4+
s1
4+ 2 1
2
2+
−
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 14/17
=e
4 ⋅e
−4
sin
2
⋅+ e
−4
cos
2
⋅−
CLONE of EXM !"ESTION (F#o$ Le%t&#e Note )
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 $ill %e applied on s2
2 s⋅− 2+
s2−2 s−4
(s2−2 s+2)(s+1)= A
(s+1)+ Bs+ K (s2−2 s+2)
s2
2 s⋅− 4− , s2
2 s⋅− 2+( )⋅ s⋅ . +( ) s 1+( )⋅+
su!stitute s 1− ,
1
5−
ences
2
2 s⋅− 4− 1
5− s2
⋅ 2
5+ . + s⋅2
5−+ . +
co()ain* the coefficient of s2!
1
5− 1
6
5
co()ain* the coefficient of s0!
. 2
5− 4−
.
18
5−
'hus
s2
2 s⋅− 4−( )
s 1+( ) s2 2 s⋅− 2+( )⋅
1−
5 s 1+( )⋅" #
6
5s⋅
18
5−
s2 2 s⋅− 2+( )
+
'heefoe
s2
2 s⋅− 4−( )
s 1+( ) s2
2 s⋅− 2+( )⋅
1−5 s 1+( )⋅" #
6
5s⋅
18
5−
s2
2 s⋅− 2+( )+
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 15/17
=
1−5 s 1+( )⋅" #
6
5s⋅
18
5−
s2
2 s⋅−2
2− 2+
2
2− 2−
2+
+
=
1−5 s 1+( )⋅" #
6
5s⋅
18
5−
s2
2 s⋅− 1−( )2+ 1− 2+
+
=
1−5 s 1+( )⋅" #
6
5s⋅
18
5−
s 1−( )2
1++
1−5 s 1+( )⋅" #
6
5s 1−( )⋅
6
5+
18
5−
s 1−( )2
1++
=
1−5 s 1+( )⋅" #
6
5s 1−( )⋅
12
5−
s 1−( )2
1++
1−5 s 1+( )⋅" #
6
5s 1−( )⋅
s 1−( )2
1++
12
5
s 1−( )2
1+−
L−1{
s2−2 s−4
(s2−2 s+2)( s+1)
}
L1− 1−
5 s 1+( )⋅" #
6
5s 1−( )⋅
s 1−( )2
1++
12
5
s 1−( )2
1+−
=
L1−
6
5s 1−( )⋅
s 1−( )2
1+
L1−
12
5
s 1−( )2
1+
− L
1−
1
5
s 1+
−
=
6
5
ez⋅ cos z( )⋅
12
5
ez⋅ sin z( )⋅−
ez−
5
−
MT343 5"N 6719
a) *ho$ that ℒ-1 {
1
(s2+2 s+5) s } =1
5−
1
5e−t (cos2 t +
1
5sin 2 t )
7/17/2019 Revision Inverse Laplace
http://slidepdf.com/reader/full/revision-inverse-laplace 16/17
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
¿