More Properties of Laplace Transforms - (4.4)
1. Derivatives of Laplace Transforms:
Let FŸs c£fŸt ¤. Then
n
c£t n fŸt ¤ Ÿ"1 n d n FŸs Ÿ"1 n F Ÿn Ÿs ds
n1
c£tfŸt ¤ Ÿ"1 F U Ÿs n2
c£t 2 fŸt ¤ F UU Ÿs It can be derived directly from the definition:
.
F U Ÿs d c£fŸt ¤ d ; fŸt e "st dt ds 0
ds
.
.
.
; 0 fŸt dsd Ÿe "st dt ; 0 fŸt Ÿ"te "st dt
" ; tfŸt e "st dt "c£tfŸt ¤,
0
U
c£tfŸt ¤ "F Ÿs Note that we can derive the following formula from this property:
U
F Ÿs c£"tfŸt ¤
Example Find c£gŸt ¤ where
a. gŸt te "2t
c. gŸt t 2 cosŸ2t b. gŸt t sinŸ=t a.
c£gŸt ¤ c£te "2t ¤ " d c£e "2t ¤ " d
ds
ds
1
s2
" "
1
Ÿs 2 2
1
Ÿs 2 2
b. c£gŸt ¤
c£gŸt ¤ c£t sinŸ=t ¤ " d c£sinŸ=t ¤ " d
ds
ds
"=
"2s
2
Ÿs = 2 2
=
s2 =2
2=s
2
Ÿs = 2 2
c. c£gŸt ¤
2
2
c£gŸt ¤ c£t 2 cosŸ2t ¤ d 2 c£cosŸ2t ¤ d 2
ds
ds
d
ds
4 " s2
2
Ÿs 2 4 s
s2 4
d
ds
"2sŸs 2 4 " Ÿ4 " s 2 Ÿ2Ÿs 2 4 Ÿ2s Ÿs 2 4 4
s 2 4 " sŸ2s 2
Ÿs 2 4 2
2s "9 2s 3
Ÿs 2 4 Example Find fŸt where
a. c£fŸt ¤ ln
s2
Ÿs " 5 Ÿs 2 4 b. c£fŸt ¤ arctanŸ4s Recall from the formula of the derivative of Laplace transform we can also derive the formula
U
U
F Ÿs c£"tfŸt ¤ or fŸt " 1 c "1 F Ÿs t
So if we know the inverse Laplace transform of F U Ÿs then we can find fŸt by above formula. Here are two
1
examples of this type.
a. Let
s2
lnŸs 2 " lnŸs " 5 " lnŸs 2 4 Ÿs " 5 Ÿs 2 4 F U Ÿs 1 " 1 " 22s
s"5
s2
s 4
1 " 1 " 2s
c "1 £F U Ÿs ¤ c "1
e "2t " e 5t " 2 cosŸ2t s"5
s2
s2 4
FŸs ln
Hence,
" tfŸt e "2t " e 5t " 2 cosŸ2t fŸt " 1
t
e "2t " e 5t " 2 cosŸ2t b. Let FŸs arctanŸ4s .
F U Ÿs 4
4
1
16
1 16s 2
1 Ÿ4s 2
"1
U
c £F Ÿs ¤ c
1
16
1
4
"1
1
16
" tfŸt sin 1 t ,
4
s
2
4
s2
1
4
1
16
s2
sin 1 t
4
fŸt " 1 sin 1 t
t
4
Example Use Laplace transform to solve y UU 16y cosŸ4t , yŸ0 1, y U Ÿ0 1
a. Find c£y¤.
c£y UU ¤ 16c£y¤ c£cosŸ4t ¤
s 2 c£y¤ " syŸ0 " y U Ÿ0 16c£y¤ Ÿs 2 16 c£y¤ s
s 1,
s2 6
c£y¤ s
s 16
2
1
s 2 16
s
s1
s2 6
b. Find y.
s
1
s1
s 2 16 s 2 6
2 1
9 2 s
10 s 16
s 16
1
9 2 s
1 3s
10 s 16
10 s 6
s 2 16
FŸs y c "1
9s 1
s
1 10
10 s 2 16
10 s 2 6
1
s
10 s 3 6
1 sinŸ4t 9 cosŸ4t 1 cos
4
10
10
6t
2. Convolution:
a. Definition:
Let f and g be piecewise continuous functions. The convolution of f and g denoted by f ' g is defined
as
Ÿf ' g Ÿt Example Find
Ÿi 2
Ÿi t ' e 2t ;
Ÿii e "t ' cosŸ=t t
; 0 fŸA gŸt " A dA.
t ' e 2t t
; 0 Ae 2Ÿt"A dA " 12 t "
1 1 e 2t
4
4
Ÿii e "t ' cosŸ=t t
"e "t cos t= = sin t=
1 =2
; 0 e "A cosŸ=Ÿt " A dA b. Laplace Transform of a convolution:
Let c£f¤ FŸs and c£g¤ GŸs . Then
c£f ' g¤ c£f¤c£g¤ FŸs GŸs c "1 £FŸs GŸs ¤ f ' g
Example Find
Ÿii c£e "t ' cosŸ=t ¤
Ÿi c£t ' e 2t ¤
t
; 0 e A sinŸt " A dA
Ÿiii c
Ÿi 1
s2
c£t ' e 2t ¤ c£t¤c£e 2t ¤ 1
s"2
1
s 2 Ÿs " 2 Ÿii 1
s1
c£e "t ' cosŸ=t ¤ c£e "t ¤c£cosŸ=t ¤ s
s2 =2
s
Ÿs 1 Ÿs 2 = 2 Ÿiii c
t
; 0 e A sinŸt " A dA
c£e t ' sinŸt ¤ c£e t ¤c£sin t¤
1
s"1
1
s2 1
1
Ÿs " 1 Ÿs 2 1 Example Find hŸt where c£h¤ HŸs Ÿi HŸs Ÿi HŸs 1
sŸs 2 2 1
2
Ÿs 4 2
1
sŸs 2 2 1
sŸs 2 2 1
s
fŸt 1 ' 1 sin
2
1
Ÿii HŸs 2
2
Ÿs 4 1
1
2
2
2
s
4
Ÿs 4 1
s2 2
2t
1
s2 4
c£1¤c
1
2
t
; 0 sin
c
hŸt 1 sinŸ2t ' sinŸ2t 1
4
4
3
Ÿii HŸs 1 sin
2
c 1 ' 1 sin
2
2 A dA 1 2 " 1 cos t 2
2
2
1 sinŸ2t c
2
t
2t
1 sinŸ2t 2
; 0 sinŸ2A sinŸ2Ÿt " A dA c
2t
2 1 2
2
1 sinŸ2t ' 1 sinŸ2t 2
2
1 sin 2t " t Ÿcos 2t 16
8
Example Solve the initial value problem: y UU 4y 2 sin 2t, yŸ0 1, y U Ÿ0 "1
a. Find c£y¤.
c£y UU ¤ 4c£y¤ 2c£sinŸ2t ¤
s 2 c£y¤ " sŸ1 " Ÿ"1 4c£y¤ 2
4 s " 1,
s2 4
Ÿs 2 4 c£y¤ 2
s2 4
1
s2 4
c£y¤ 4 s"1
s2 4
b. Find y.
FŸs y c "1
4 s"1
s2 4
1
s2 4
4
" 21
2s
2
s 4
s
4
4 Ÿs
4
" 21
2s
2
s 4
s
4
Ÿs 4 2
sinŸ2t ' sinŸ2t cosŸ2t " 1 sinŸ2t 2
2
t
; 0 sinŸ2A sinŸ2Ÿt " A dA cosŸ2t "
1 sinŸ2t 1 sin 2t " 1 tŸcos 2t cosŸ2t " 1 sinŸ2t 2
2
4
2
3. Laplace Transform of an Integral:
t
; 0 fŸu du
c
1s FŸs where c£f¤ FŸs c "1
FŸs s
t
; 0 fŸt dt
Example Find
a. c
a. fŸu ue
t
t
; 1 ue "2u du
b. c t ; e "u cosŸ=u du ,
0
"2u
FŸs c£fŸt ¤ c£te "2t ¤ t
; 1 ue "2u du
c
1s
1
Ÿs 2 2
1
Ÿs 2 2
1
sŸs 2 2
b. fŸu e "u cosŸ=u s"1
"
Ÿs 1 2 = 2
FŸs c£fŸt ¤ c£e "t cosŸ=t ¤ t
c t ; e "u cosŸ=u du
0
" d c
ds
t
; 0 e "u cosŸ=u du
" d
ds
3
2
2
" " 2s " 5s 4s " 1 " =2
2 2
2
s Ÿs " 2s 1 = 1
s
s"1
Ÿs " 1 2 = 2
3
2
2
2s " 5s 4s " 1 " =2
2 2
2
s Ÿs " 2s 1 = Example Find hŸt where HŸs is
a. HŸs 1
sŸs 2 2 b. HŸs 1
s 2 Ÿs " 2 a.
HŸs 4
1
1s
sŸs 2 2
1
s 2
2
1s c
1 sin
2
2t
hŸt t
;0
2 A dA " 1 cos t 2 1
2
2
1 sin
2
b.
HŸs 1
1s
s 2 Ÿs " 2 hŸt ; 0 ; 0 e 2z dz du t
u
1
s
1s c£1 ' e 2t ¤
1
s"2
1 e 2t " 1 t " 1
4
4
2
t
Example Solve fŸt using Laplace transform where fŸt 2t " 4 ; sin A fŸt " A dA
0
a. Find c£f¤ :
t
c£f¤ c 2t " 4 ; sin A fŸt " A dA
0
1
4
c£f¤ 22 ,
s2 1
s
2c£t¤ " 4c£sin t¤c£f¤ 22 " 4 2 1 c£f¤
s
s 1
2
2Ÿs 1 s 2 5 c£f¤ 2 ,
c£f¤ 2 2
2
2
s
s 1
s Ÿs 5 b. Find f :
fŸt c "1
2Ÿs 2 1 s 2 Ÿs 2 5 c "1
8
2 5Ÿs 2 5 5s 2
Example LC-circuit: Solve the initial value problem:
L dI RI 1
C
dt
2 t 8 sin
5
5 5
5t
t
; 0 IŸA dA EŸt where L 0. 1, R 2, C 0. 1, E 120t " 120tUŸt " 1 , IŸ0 0.
a. Find c£I¤.
c£EŸt ¤ c£120t " 120tUŸt " 1 ¤ c£120t " 120Ÿt " 1 1 UŸt " 1 ¤
1 " 120 1 1 e "s
s
s2
s2
t
c dI 20I 100 ; IŸA dA
dt
0
120
10c£EŸt ¤
1 "s
1
1
Ÿsc£I¤ " IŸ0 20c£I¤ 100
s c£I¤ 1200 s 2 " 1200 s 2 s e
s 20 100
c£I¤ 1200 12 " 1200 12 1s e "s
s
s
s
2
s 20s 100 c£I¤ 1200 1 " 1200 1 1 e "s
s
s
s2
s2
s
1200 12 " 1200 12 1s e "s
c£I¤ 2
s 20s 100
s
s
1200
1200
1
"s
"
1 e
sŸs 10 2
Ÿs 10 2 s
b. Find I :
c "1
5
1200
sŸs 10 2
c "1
120
12 "
" 12
2
s
s 10
Ÿs 10 12 " 120te "10t " 12e "10t
1200
Ÿs 10 2
c "1
c "1
1200
Ÿs 10 2
1200 1 1200
Ÿs 10 2
Ÿs 10 2 s
12 " 120te "10t " 12e "10t 1200te "10t
1 1
s
1 1 e "s
s
c "1
Ÿ12 " 12e "10Ÿt"1 1080Ÿt " 1 e "10Ÿt"1 UŸt " 1 IŸt 12 " 120te "10t " 12e "10t Ÿ12 " 12e "10Ÿt"1 1080Ÿt " 1 e "10Ÿt"1 UŸt " 1 4. Laplace Transform of a Periodic Function:
Let f be a periodic function with period T. Then
c£f¤ T
; 0 e "st fŸt dt
1
1 " e "Ts
Derivation
c£f¤ .
T
; 0 fŸt e "st dt ; 0
fŸt e "st dt ;
2T
T
fŸt e "st dt ;
3T
2T
fŸt e "st dt . . .
Observe that let u t " T
2T
;T
fŸt e "st dt T
T
; 0 fŸu T e "sŸuT du e "sT ; 0 fŸu e "su du
c£f¤ Ÿ1 e "sT s "s2T . . . ;
T
0
fŸt e "st dt 1
1 " e "sT
T
;0
fŸt e "st dt
T
Note that the integral ; e "st fŸt dt is equivalent to c£gŸt ¤ where
0
gŸt fŸt , 0 t t T
0, t u T
.
Since gŸt fŸt " fŸt UŸt " T ,
c£gŸt ¤ c£fŸt " fŸt UŸt " T ¤.
Example Find c£f¤ where
a. fŸt t, 0 t t 1, with period 1
b. fŸt t 0tt1
0 1tt2
, with period 2
c. fŸt sin t, 0 t t =, with period =
a.
1 c£t " Ÿt " 1 1 UŸt " 1 ¤
1 c£t " tUŸt " 1 ¤ 1 " e "s
1 " e "s
1
1 " 1 1 e "s 1 " 1 e "s
s 1 " e "s
s
1 " e "s s 2
s2
s2
c£f¤ 6
1.8
2
1.6
1.4
1.5
1.2
1
1
0.8
0.6
0.4
0.5
0.2
0
0.5
1
1.5
t
2
2.5
3
0
1
2
fŸt s
3
4
5
FŸs b.
1
1
c£t " tUŸt " 1 ¤ c£t " Ÿt " 1 1 UŸt " 1 ¤
1 " e "2s
1 " e "2s
1
1 " 1 1 e "s
s
1 " e "2s s 2
s2
c£f¤ 2
1.8
1.2
1.6
1
1.4
1.2
0.8
1
0.6
0.8
0.6
0.4
0.4
0.2
0.2
0
1
2t
fŸt 3
4
1
2
s
3
4
5
FŸs c.
1
1
c£sin t " sin tUŸt " = ¤ c£sin t " sinŸt " = = UŸt " = ¤
1 " e "=s
1 " e "=s
1
1
1 1 e "=s
c£sin t sinŸt " = UŸt " = ¤ 1 " e "=s
1 " e "=s s 2 1
s2 1
c£f¤ 7
3
1.8
1.6
2.5
1.4
2
1.2
1
1.5
0.8
1
0.6
0.4
0.5
0.2
0
2
4
6
t
0
8
1
2
s
3
4
5
FŸs fŸt 1 0tt1
Example Solve the initial value problem: dI I EŸt , IŸ0 0, where EŸt dt
0 1tt2
period 2.
a. Find c£I¤.
1
c£1 " UŸt " 1 ¤
1 " e "2s
1
1 " 1 e "s
s
s
Ÿs 1 Ÿ1 " e "2s sc£I¤ " 0 c£I¤ c£EŸt ¤, Ÿs 1 c£I¤ 1
s1
1
1 " e "2s
1
sŸs 1 Ÿ1 e "s c£I¤ 1 Ÿ1 " e "s s
b. Find I.
c "1
IŸt c "1
1
1 e "s
1
sŸs 1 1
sŸs 1 c
1 " 1
s
s1
c "1
1
sŸs 1 1 " e "t
Ÿ1 " e "s e "2s " e "3s . . . 1 " e "t " Ÿ1 " e "Ÿt"1 UŸt " 1 Ÿ1 " e "Ÿt"2 UŸt " 2 " Ÿ1 " e "Ÿt"3 UŸt " 3 . . .
2
1
1.8
0.8
1.6
1.4
0.6
1.2
1
0.4
0.8
0.6
0.2
0.4
0.2
d
0
1
2x
Input EŸt 8
3
4
0
1
2
t
IŸt 3
4
5
with