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Lecture : 1
(LAPLACE TRANSFORMS)
Course : B.Sc. (H) 2nd year, IV Sem Subject : Mathematical Physics III UPC : 32221401 Teacher : Ms. Bhavna Vidhani
(Deptt. of Physics & Electronics) Topics covered in this lecture:-
Laplace Transform of elementary functions Properties of LT: Change of Scale Theorem, Linearity property, Shifting Theorem
(First and Second), LT of Derivatives & Integrals of Functions, Multiplication by the powers of t, Division by t, LT of Periodic Functions.
Books to be referred: 1. Schaum's Outline of Theory and Problems of Laplace Transforms by Murray R. Spiegel. 2. Advanced Engineering Mathematics by Erwin Kreyzig. 3. Mathematical Physics by H. K. Dass. 1.1 Definition of Laplace Transformation: Let )(tf be a given function defined for all t 0, then the Laplace Transform of )(tf is defined as
L )}({ tf = dtetf st
0
)( = )(sF s may be real or complex
here, L is called Laplace Transform Operator. The function )(tf known as determining function, depends on t. The new function which is to be determined (i.e., )(sF ) is called generating function, depends on s. (t domain or (s domain or time domain) frequency domain) 1.2 Laplace Transform of some elementary functions:
1. If )(tf = 1, then )1(L = s1 ; s > 0
Proof: )1(L = dte st
0
)1( =
0
dte st = s1
)(tf )(sF L{f(t)}
2
2. If )(tf = t , then )(tL = 21s
; s > 0
Proof: )(tL = dtet st
0
= 2)2(
s = 2
1s
3. If )(tf = nt , then )( ntL = 1)1(
nsn ; s > 0
Proof: )( ntL = dtet stn
0
= 1)1(
nsn
10
)1(n
tsn
sndtet
4. If )(tf = 2/3t , then )( 2/3tL = 2/543
s ; s > 0
Proof: )( 2/3tL = dtet st
0
2/3 = 2/5)2/5(
s
= 2/5)2/1()2/3(
s = 2/54
3s
5. If )(tf = te , then )( teL = 1
1s
; s > 1
Proof: )( teL = dtee stt
0
= dte ts
0
)1( = 1
1s
0,1
0
cc
dte tc
6. If )(tf = tcos , then )(cos tL = 12 s
s ; s > 0
Proof: )(cos tL = dtet st
0
cos = dteee sttiti
0 2
= 21
0
)()( dtee istist
=
isis11
21 =
12 ss
0,1
0
cc
dte tc
7. If )(tf = tsin , then )(sin tL = 1
12 s
; s > 0
Proof: )(sin tL = dtet st
0
sin = dteiee st
titi
0 2
= i2
1
0
)()( dtee istist
=
isisi11
21 =
11
2 s
0,1
0
cc
dte tc
3
8. If )(tf = tcosh , then )(cosh tL = 12 s
s ; s > 1
Proof: )(cosh tL = dtet st
0
cosh = dteee sttt
0 2
= 21
0
)1()1( dtee stst
=
11
11
21
ss =
12 ss
0,1
0
cc
dte tc
9. If )(tf = tsinh , then )(sinh tL = 1
12 s
; s > 1
Proof: )(sinh tL = dtet st
0
sinh = dteee sttt
0 2
= 21
0
)1()1( dtee stst
=
11
11
21
ss =
11
2 s
0,1
0
cc
dte tc
1.3 Change of Scale Property:
If L{ )(tf } = )(sF then L{ )(atf } =
asF
a1
Proof: By definition, we have
L{ )(tf } = dtetf st
0
)( = )(sF
L{ )(atf } = dteatf st
0
)( Let a t = u dt = a
du
= a
dueuf aus /
0
)(
= a1 dueuf
uas
0
)(
= a1 dtetf
tas
0
)( =
asF
a1
Examples: 1. Find L( ate ) Sol. I Method: By Using Change of Scale Property
We know that L( te ) = 1
1s
= )(sF
4
L( ate ) =
asF
a1 =
1
11
asa
= as
1
OR (Direct Method)
L( ate ) = dtee stat
0
= dte tas
0
)( = as
1
0,1
0
cc
dte tc
2. Find L( atcos ) Sol. I Method: By Using Change of Scale Property
We know that L( tcos ) = 12 s
s = )(sF
L( atcos ) =
asF
a1 =
a1
12
as
as
= 22 ass
OR (Direct Method, using the definition of LT)
L( atcos ) = dteat st
0
cos = dteee sttaitai
0 2
= 21
0
)()( dtee iastiast
=
iasias11
21 = 22 as
s
0,1
0
cc
dte tc
3. Find L( atsin )
Sol. We know that L( tsin ) = 1
12 s
= )(sF
L( atsin ) =
asF
a1 =
a1
1
12
as
= 22 asa
Direct Method, using the definition of LT can also be used to get the required answer. 4. Find L( atcosh )
Sol. We know that L( tcosh ) = 12 s
s = )(sF
5
L( atcosh ) =
asF
a1 =
a1
12
as
as
= 22 ass
Direct Method, using the definition of LT can also be used to get the required answer. 5. Find L( atsinh )
Sol. We know that L( tsinh ) = 1
12 s
= )(sF
L( atsinh ) =
asF
a1 =
a1
1
12
as
= 22 asa
Direct Method, using the definition of LT can also be used to get the required answer. 1.4 Linearity Property: If C1 and C2 are any constants, while )(1 tf and )(2 tf are functions with Laplace Transforms )(1 sF and )(2 sF respectively, then L{C1 )(1 tf + C2 )(2 tf } = C1 L{ )(1 tf } + C2 L{ )(2 tf } = C1 )(1 sF + C2 )(2 sF Proof: By definition, we have
L{ )(tf } = dtetf st
0
)( = )(sF
L.H.S. = L{C1 )(1 tf + C2 )(2 tf }
= dtetfCtfC st
0
2211 )()(
= 1C dtetf st
0
1 )( + 2C dtetf st
0
2 )(
= C1 L{ )(1 tf } + C2 L{ )(2 tf } = RHS Examples: 1. Find L{4 te5 + 6 3t ‒ 3 t4sin + 2 t2cos } Sol: L{4 te5 + 6 3t ‒ 3 t4sin + 2 t2cos } = 4L( te5 ) + 6L( 3t ) ‒ 3L( t4sin ) + 2L( t2cos )
= 45
1s
+ 6 4!3
s ‒ 3
164
2 s + 2
42 ss
= 5
4s
+ 436s
‒ 16
122 s
+ 4
22 s
s
2. Find L{ )(tf } if )(tf =
3;030;5
tt
6
Sol: L{ )(tf } = dtetf st
0
)( = dtetf st
3
0
)( + dtetf st
3
)(
= dte st
3
0
5 + dte st
3
)0( = 5 3
0se st
+ 0
= s5 03 ee s =
s5
(1 ‒ se 3 )
1.5 First Translation or Shifting Property: If L{ )(tf } = )(sF then L{ )(tfebt } = )( bsF Proof: By definition, we have
L{ )(tf } = dtetf st
0
)( = )(sF
L{ )(tfebt } = dtetfe stbt
0
)( = dtetf tbs )(
0
)(
= )( bsF
Examples: 1. Find L{ tet 32 }
Sol. We know that L{ 2t } = 3
)3(s = 3
!2s
= )(sF
L{ tet 32 } = )3( sF = 33
!2s
= 33
2s
Direct Method, using the definition of LT can also be used to get the required answer. 2. Find L( te tb cos ) Sol. I Method: By using First Translation Property
We know that L( tcos ) = 12 s
s = )(sF
L( te tb cos ) = )( bsF = 1)( 2
bsbs
OR (Direct Method, using the definition of LT)
L( te tb cos ) = dtete sttb
0
cos = dteeee sttiti
tb
0 2
= 21
0
)()( dtee ibstibst
=
ibsibs11
21
0,1
0
cc
dte tc
7
= 21
22)( ibs
ibsibs =
1)( 2
bsbs
3. Find L( te tb sin )
Sol. We know that L( tsin ) = 1
12 s
= )(sF
L( te tb sin ) = )( bsF = 1)(
12 bs
Direct Method, using the definition of LT can also be used to get the required answer. 4. Find L( te tb cosh ) Sol. By using First Translation Property
We know that L( htcos ) = 12 s
s = )(sF
L( te tb cosh ) = )( bsF = 1)( 2
bsbs
Direct Method, using the definition of LT can also be used to get the required answer. 5. Find L( te tb sinh )
Sol. We know that L( tsinh ) = 1
12 s
= )(sF
L( te tb sinh ) = )( bsF = 1)(
12 bs
Direct Method, using the definition of LT can also be used to get the required answer. 1.6 Second Translation or Shifting Property:
If L { )(tf } = )(sF and )(tg =
atatatf
;0;)(
then L { )(tg } = )(sFe sa Proof: By definition, we have
L{ )(tf } = dtetf st
0
)( = )(sF
L{ )(tg } = dtetg st
0
)( = dtetg sta
0
)( + dtetg st
a
)(
= dte sta
0
)0( + dteatf st
a
)(
Let t ‒ a = u dt = du
8
L{ )(tg } = 0 + dueuf aus )(
0
)(
= sae dueuf us
0
)(
= sae )(sF Examples: 1. Find L{ )(tf }, if
)(tf =
3/2;03/2;)3/2(cos
ttt
Sol. We know that L( tcos ) = 12 s
s = )(sF
L{ )3/2(cos t } = 3/22 1
ses
s
OR (Direct Method, using the definition of LT)
L{ )(tf } = dtetf st
0
)( = dtetf st
3/2
0
)(
+ dtetf st
3/2
)(
= dte st
3/2
0
)0(
+ dtet st
3/2
)3/2(cos
Let t ‒ 3
2 = u dt = du
= 0 + dueu us )3/2(
0
cos
= 3/2 se dueu us
0
cos
= 3/2 se dtet ts
0
cos
= 3/2 se L { tcos } = 3/2 se
11
2 s
2. Find L{ )(tf } if
)(tf =
10;01;)1( 2
ttt
Sol. We know that L( 2t ) = 32s
= )(sF
L{ 2)1( t } = 32s
se
Direct Method, using the definition of LT can also be used to get the required answer.
9
1.7 Combination of Scaling and Shifting Properties:
If L{ )(tf } = )(sF then L{ )(atfebt } =
absF
a1
Proof: By definition, we have
L{ )(tf } = dtetf st
0
)( = )(sF
L{ )(atfebt } = dteatfe stbt
0
)( = dteatf tbs )(
0
)(
Let a t = u dt = a
du
= adueuf aubs /)(
0
)(
= a1 dueuf
ua
bs
0
)(
=
absF
a1
Examples: 1. Find L( tae tb cos ) Sol. I Method: By using Combination of Scaling and Shifting Properties
We know that L( tcos ) = 12 s
s = )(sF
L( tae tb cos ) =
absF
a1 =
a1
12
absa
bs
= 22)( absbs
OR (using properties in steps)
We know that L( tcos ) = 12 s
s = )(sF
L( atcos ) =
asF
a1 =
a1
12
as
as
= 22 ass
= )(sG
and L( tae tb cos ) = )( bsG = 22)( absbs
Direct Method, by using the definition of LT can also be used to get the required answer. 2. Find L( tae tb sin ) Sol. I Method: By using Combination of Scaling and Shifting Properties
10
We know that L( tsin ) = 1
12 s
= )(sF
L( tae tb sin ) =
absF
a1 =
a1
1
12
abs
= 22)( absa
OR (using properties in steps)
We know that L( tsin ) = 1
12 s
= )(sF
L( atsin ) =
asF
a1 =
a1
1
12
as
= 22 asa
= )(sG
and L( tae tb sin ) = )( bsG = 22)( absa
Direct Method, by using the definition of LT can also be used to get the required answer. 3. Find L( tae tb cosh ) Sol. I Method: By using Combination of Scaling and Shifting Properties
We know that L( tcosh ) = 12 s
s = )(sF
L( tae tb cosh ) =
absF
a1 =
a1
12
absa
bs
= 22)( absbs
OR (using properties in steps)
We know that L( tcosh ) = 12 s
s = )(sF
L( atcosh ) =
asF
a1 =
a1
12
as
as
= 22 ass
= )(sG
and L( tae tb cosh ) = )( bsG = 22)( absbs
Direct Method, by using the definition of LT can also be used to get the required answer. 4. Find L( tae tb sinh ) Sol. I Method: By using Combination of Scaling and Shifting Properties
11
We know that L( tsinh ) = 1
12 s
= )(sF
L( tae tb sinh ) =
absF
a1 =
a1
1
12
abs
= 22)( absa
OR (using properties in steps)
We know that L( tsinh ) = 1
12 s
= )(sF
L( atsinh ) =
asF
a1 =
a1
1
12
as
= 22 asa
= )(sG
and L( tae tb sinh ) = )( bsG = 22)( absa
Direct Method, by using the definition of LT can also be used to get the required answer. 1.8 Laplace Transform of Derivatives: If L { )(tf } = )(sF , then (i) L { )(tf } = )0()( fsFs (ii) L { )(tf } = s2 )(sF ‒ s )0(f ‒ )0(f
here, )(tf )(tfdtd
Proof: (i) By definition, we have
L{ )(tf } = dtetf st
0
)( = )(sF
L{ )(tf } = dtetf st
0
)(
=
0)(tfe ts ‒ dtestf st
0
)()(
= )0(0 f + s dtetf st
0
)( [ e = 0]
= ‒ )0(f + s )(sF = s )(sF ‒ )0(f L{ )(tf } = s )(sF ‒ )0(f (1.1) (ii) L{ )(tf } = s2 )(sF ‒ s )0(f ‒ )0(f Proof: Let )(tg = )(tf then L{ )(tg } = s L{ )(tg } ‒ )0(g [using eq. (1.1)] L{ )(tf } = s L{ )(tf } ‒ )0(f
12
= s [s )(sF ‒ )0(f ] ‒ )0(f = s2 )(sF ‒ s )0(f ‒ )0(f Examples: 1. Find L{ t } by using derivatives method. Sol. Here, )(tf = t )(tf = 1 and )0(f = 0 We know that L{ )(tf } = s )(sF ‒ )0(f L{1} = s L{ t } ‒ 0
s1 = s L{ t }
L{ t } = 21s
2. Find L{ tasin } by using derivatives method. Sol. Here, )(tf = tasin )(tf = taa cos and )(tf = taa sin2 Also, )0(f = 0 and )0(f = a We know that L{ )(tf } = s2 )(sF ‒ s )0(f ‒ )0(f L{ taa sin2 } = s2 L{ tasin } ‒ 0 ‒ a 22 as L{ tasin } = a
L{ tasin } = 22 asa
1.9 Laplace Transform of Integrals:
If L{ )(tf } = )(sF then L
t
duuf0
)( = ssF )(
Proof: Let )(tg = t
duuf0
)( (1.2)
)(tg = )(tf with the condition )0(g = 0 Taking Laplace Transform on both sides, we get L{ )(tg } = L{ )(tf } s L{ )(tg } ‒ )0(g = L{ )(tf } s L{ )(tg } ‒ 0 = L{ )(tf }
s L
t
duuf0
)( = )(sF [using (1.2)]
L
t
duuf0
)( = ssF )(
Examples:
1. Find L
tu dueuu
0
2 )(
13
Sol. Let )(tf = tett 2
L ( tett 2 ) = 1
11223
sss = )(sF
L
tu dueuu
0
2 )( = s1 )(sF =
s1
1112
23 sss
3. Verify directly that L
t
duua0
cos = s1 L ( tacos )
Sol. Let )(tf = tacos
then L { )(tf } = L ( tacos ) = 22 ass
= )(sF (1.3)
and t
duuf0
)( = t
duua0
cos = t
aua
0
sin = a
tasin
L
t
duuf0
)( = L
atasin
= a1
22 asa
= 22
1as
= s1
22 ass
= s1 )(sF [using eq. (1.3)]
hence, verified. 1.10 Multiplication by the powers of t: If L { )(tf } = )(sF , then
(i) L { t )(tf } = ‒ ds
sFd )(
(ii) L { 2t )(tf } = 2
2 )(ds
sFd
Proof: (i) By definition, we have
L { )(tf } = )(sF = dtetf st
0
)(
Then by Leibnitz Rule for differentiation under the integral sign,
ds
sdF )( = )(sF =
0
)(tfedsd st = dttfet st )(
0
= dttfte st )]([0
(1.4)
14
= ‒ L { )(tft }
i. e., L { )(tft } = ‒ ds
sdF )( = ‒ )(sF
(ii) Differentiate both sides of eq. (1.4) w. r. t. s , we get
2
2 )(ds
sFd = dttftet st )]([0
= 2)1( dttfte st )]([ 2
0
= 2)1( L { 2t )(tf }
i. e., L { 2t )(tf } = 2)1( 2
2 )(ds
sFd = 2
2 )(ds
sFd
Generalizing, L { nt )(tf } = n)1( n
n
dssFd )(
Examples: 1. Find L{ tatcos } Sol. We know that
L{ tacos } = 22 ass
L{ tatcos } =
22 as
sdsd
=
222
222 2as
sas = 222
22
asas
2. Find L{ ttt 3sin)23( 2 } Sol. L { ttt 3sin)23( 2 } = L )3sin( 2 tt ‒ 3 L ( tt 3sin ) + 2 L ( t3sin )
= )(2
2
sFdsd + 3 )(sF
dsd + 2 )(sF
here, )(sF = L ( t3sin ) = 9
32 s
=
93
22
2
sdsd + 3
93
2sdsd + 2
93
2s
=
22 )9()2(3
ss
dsd + 3
22 )9(
)2(3s
s + 9
62 s
= ‒ 6
42
222
)9()2()9(21.)9(
sssss ‒ 22 )9(
18s
s + 22
2
)9()9(6
ss
= ‒6
42
222
)9()49()9(
ssss ‒ 6 22
2
)9(93
sss
15
= 6 32
2
)9(93
ss + 6 32
22
)9()9()93(
ssss
= 6 32
2
)9(93
ss + 6 32
2324
)9(8192739
ssssss
= 6 32
234
)9(7227213
sssss
1.11 Division by t:
If L{ )(tf } = )(sF then L
ttf )( = duuF
s
)( provided 0t
Limttf )( exists.
Proof: By definition, we know that
L{ )(tf } = dtetf st
0
)( = )(sF
duuFs
)( = dudttfes
ut
)(
0
= dtduetfs
ut
0
)(
= dtt
etfs
ut
0
)(
= dtetft
ts
0
0)(1
= dtettf ts
0
)( = L
ttf )(
Examples:
1. Find L
ttsin
Sol. We know that
L{ tsin } = 1
12 s
= )(sF
L
ttsin =
s
duuF )( =
s
duu 1
12
=
su1tan = s11 tantan
= s1tan2
= s1cot
= s1tan 1
16
Additional:
L
ttsin =
s1tan 1
0
sin dtet
t ts = s1tan 1
0
sin dtt
t = 2 [putting s = 0 on both sides]
2. Solve
0
3
dtt
ee tt
Sol. Let )(tf = tt ee 3
L{ )(tf } = 1
1s
‒ 3
1s
= )(sF
L
ttf )( =
s
duuF )( = duuus
31
11
0
3
dtet
ee sttt
= suu )3(ln)1(ln
0
3
dtet
ee sttt
=
suu
31ln =
31ln)1(ln
ss =
13ln
ss
0
3
dtt
ee tt
= 3ln [putting s = 0 on both sides]
1.12 Laplace Transform of Periodic Function: Let )(tf have period T > 0 so that )( Ttf = )(tf
then L{ )(tf } = Tse11 dttfe
Tst )(
0
Proof: By definition, we have
L{ )(tf } = dtetf st
0
)(
= dtetf stT
0
)( + dtetf stT
T
2
)( + dtetf stT
T
3
2
)( + .........
In the second integral on RHS, Let t = u + T, In the third integral on RHS, Let t = u + 2T, and so on.
Then, L{ )(tf } = dtetf stT
0
)( + dueTuf TusT
)(
0
)(
+ dueTuf TusT
)2(
0
)2(
+ .........
= dueuf suT
0
)( + dueufe suT
sT
0
)( + dueufe suT
sT
0
2 )( + .........
[ )( Tuf = )2( Tuf = .......... = )(uf ]
= ......)1( 2 sTsT ee dueuf suT
0
)(
17
= Tse11 duufe
Tsu )(
0
= Tse11 dttfe
Tst )(
0
Example:
1. If )(tf =
2;02;sin
ttt
Find L{ )(tf } where )(tf extends periodically with period 2 . Sol. We know that
L{ )(tf } = Ts
Tst
e
dttfe
1
)(0 =
2
2
0
1
)(
s
st
e
dttfe
[ T = 2 ]
= se 211
dttfe st )(0
+ se 211
dttfe st )(2
= se 211
dtte st sin0
+ se 211
dte st )0(2
= se 211
dtte st sin0
(1.5)
Consider, I = dtte st sin = tsin
se st
‒
dts
etst
cos
= s
te st sin
+ s1
dts
ets
etstst
)sin(cos
= s
te st sin
‒ 2
coss
te st
‒ 21s
I
2
11s
I = ‒ 2se st
(s tsin + tcos )
I = ‒ 12
se st
(s tsin + tcos ) (1.6)
Put eq. (1.6) in eq. (1.5), we get
L{ )(tf } = se 211
02 )cossin(
1
ttss
e st
= se 211 1
12
s )10()}1(0{ 0 ee s
= se 211 1
12 s
)1( se
= )1()1()1(
12
seee
ss
s
= )1()1(
12 se s
