Chapter 4 Laplace Transforms

4-1 Laplace Transform F(s)=L[f(t)]=  e  st  f (t )dt
0
Eg. Evaluate L[cos(at)] and L[sin(at)].
(Sol.) e iat  cos( at )  i sin( at )


0
0
L[e iat ]   e st  e iat dt   e ( s ia)t dt 

1
1
 e ( s ia)t 
0 s  ia
s  ia
s
a
i 2
 L[cos( at )]  iL[sin( at )]
2
s a
s  a2
a
and L[sin( at )]  2
s  a2

∴ L[cos( at )] 

Eg Find

0
s
s  a2
2
2
e ax  cos(bx)dx , with a>0. [2005 台大電研] (Ans.) a/(a2+b2)
Basic theorems of Laplace transforms F(s)=L[f(t)] and G(s)=L[g(t)]:
1. L[c1f(t)+c2g(t)]=c1F(s)+c2G(s)
Eg. L[-2cos(2t)+3sin(2t)]. [2010台大光電所]
(Sol.) L[-2cos(2t)+3sin(2t)]= -2L[cos(2t)]+3L[sin(2t)]= 2 2s + 32  2 =  22 s  6 .
s 4
s 4 s 4
2. L[f(t)eat]=F(s-a), s>a


(Proof) For s>a, L[f(t)eat]=  e st  f (t )e at dt =  e ( s a )t  f (t )dt =F(s-a).
0
Eg. Find
L[eatcos(kt)],
(Sol.) L[cos( kt)] 
0
L[eatsin(kt)]
and
L[eat].
s
k
and L[sin( kt)]  2
2
s k
s  k2
2

According to L[ f (t )  e at ]  F (s  a) and L[1]   e  st dt 
0
∴ L[e at  cos( kt)] 
1
,
s
sa
k
1
, L[e at  sin( kt)] 
, and L[eat]=
2
2
2
2
sa
( s  a)  k
(s  a)  k
3. L[f’(t)]=sF(s)-f(0), L[f”(t)]=s2F(s)-sf(0)-f’(0), L[f’’’(t)]=s3F(s)-s2f(0)-sf’(0)-f”(0),
and L[f(n)(t)]=snF(s)-sn-1f(0)-sn-2f’(0)-sn-3f”(0)-…-sf(n-2)(0)-f(n-1)(0)



0
0
0
(Proof) L[f’(t)]=  e  st  f ' (t )dt =  e  st df (t ) =e-st f(t)| | 0 -  (s)e  st  f (t )dt

=-f(0)+s  e  st  f (t )dt =sF(s)-f(0)
0
Similarly, L[f”(t)]=s2F(s)-sf(0)-f’(0), and by mathematical induction, we have
L[f(n)(t)]=snF(s)-sn-1f(0)-sn-2f’(0)-sn-3f”(0)-…-sf(n-2)(0)-f(n-1)(0)
23
～～
t
F ( s)
4. L   f (u )du  
 0

s
 t
 
t

t
(Proof) L   f (u )du    e  st  f (u )dudt =   e st f (u)dudt =   e st f (u)dtdu
 0
 0
0
0 0
0 u


1 
1 
F (s)
=  f (u)du  e st dt =   f (u )[0  e  su ]du =  f (u )e  su du =
0
0
0
u
s
s
s

 f (t ) 
6. L 
  F (u )du

s
 t 
5. L[tnf(t)]=(-1)nF(n)(s)
de  st
d
 f (t )dt =0
0
ds
ds
n
n (n)
By mathematical induction, we have L[t f(t)]=(-1) F (s)
Eg. Find L[tn].


(Proof of (5)) L[tf(t)]=  e st  tf (t )dt =  


0
e  st  f (t )dt = -F’(s)

dn
1
(Sol.) According to L[t f (t )]  ( 1)
and
L
[
1
]

e  st dt 
F
(
s
)
n

0
s
ds
n
n
d n 1
n!
(n  1)
L[t ]  L[t  1]  (1)
( s )  n 1 
n
s
s n 1
ds
n
n
n
T
1
 e  st  f (t )dt if f(t+T)=f(t)
 sT 0
1 e
(Proof) Let t+T=u, t+2T=v, …
7. L[f(t)]=

T
0
0
2T
3T
L[f(t)]=  e  st  f (t )dt =  e st  f (t )dt +  e  st  f (t )dt +  e st  f (t )dt +…
T
2T
2T
T
3T
=  e st  f (t )dt +  e st  f (t  T )dt +  e  st  f (t  2T )dt +…
0
2T
T
T
2T
=  e st  f (t )dt + e-sT  e s (t T )  f (t  T )dt + e-2sT
0
T
T
T
0
0
=  e st  f (t )dt + e-sT  e  su  f (u )du + e-2sT
T
=(1+ e-sT+e-2sT+…)  e st  f (t )dt =
0
T

0

3T
2T
e s (t 2T )  f (t  2T )dt +…
e  sv  f (v)dv +…
T
1
 e  st  f (t )dt
 sT 0
1 e
 1 , 0  t  1
Eg. Find L[f(t)] if f(t+2)=f(t) and f(t)= 
.
 1, 1  t  2
T
1
 e  st  f (t )dt if f (t  T )  f (t ) and T=2,
(Sol.) According to L[ f (t )] 
 sT 0
1 e
L[ f (t )] 
2
1
1
1
  e  st dt   (1)e  st dt  
 2 s  0
 1  e 2 s
1

1 e
 e  st 1 e  st


0

s
s

1  e  s  e 2 s  e  s  1
1
(1  e  s ) 2
1 1  e s


  s  (1  e s )(1  e s )  s  1  e  s
1  e 2 s 
s

24
～～
2

1
8. L[f(at)]=[F(s/a)]/a and L[f(t/a)]=aF(as), a>0
(Proof) For a>0, let at=u

1
a

s
 ( ) at
a

e
 f (at )d (at ) =
1
a

s
( )u
a
1
s
F[( )]
0
0
0
a
a
1
t
1
s
1
t
Let b=  L[f( )]=L[f(bt)]= F[( )]=aF(as) and then L-1[F(as)]= [f( )]
a
a
b
b
a a
L[f(at)]=  e  st  f (at )dt =

e
 f (u )du =
9. L[f(t-a)u(t-a)]=e-asF(s)
(Proof) For t>a, let t-a=u


0
a
L[f(t-a)u(t-a)]=  e st  f (t  a)u(t  a)dt =  e  st  f (t  a)d (t  a)

=e-as  e s (t a )  f (t  a)d (t  a)
a

=e-as  e  su  f (u)du = e-asF(s), and then we have L-1[F(s)e-as]=f(t-a)u(t-a)
0
10. lim f (t )  lim sF ( s ) if (a) all nonzero roots of the denominator of F(s) must
t 
s 0
have negative real parts, or (b) F(s) must not have more than one pole at the origin.
11. lim f (t )  lim sF ( s )
t 0
s 
Eg. Find L[3t-5sin(2t)]. [2001台大電研]
(Sol.) L[3t-5sin(2t)]=3L[t]-5L[sin(2t)]= 32 - 210
s s 4
𝟏
Eg. Find L[𝟐te2tsin(2t)]. [2015中興電研乙丙丁組、中興光電所]
(Sol.) L[sin(2t)]=
2
2
, L[e2tsin(2t)]=
, ∵ L[tf(t)]= -F’(s),
s 4
( s  2) 2  4
2
1
2( s  2)
∴ L[te2tsin(2t)]= 2  2( s2  2) 2 , L[2te2tsin(2t)]=
[( s  2)  4]
[( s  2) 2  4] 2
Eg. Find L[e-tf(3t)] in case of L[f(t)]=e-1/s.
3
1
1
1
(Sol.) 1. L[ f (3t )]  e 1  s 3   e 3 s , L[e t  f (3t )]  e s 1 . The result is correct!
3
3
3
t
2. L[e  f (t )]  e

1
s 1
1
3
1
1
, L[e  f (3t )]  e ( s / 3)1  e s 3 . The result is wrong!
3
3
t


0
0
Another method: L[e t  f (3t )]   e  st  e t  f (3t )dt   e ( s 1)t  f (3t )dt
1 
  e
3 0
 s 1 

( 3t )
 3 
1 
f (3t )d (3t )   e
3 0
 s 1 

u
 3 
3
1  s  1  1 s 1
f (u )du  F 
 e
3  3  3
25
～～

Eg. Find

0
sin( x)
dx . [2003 中央光電所、1993 交大應數研]
x


 
1
 sin( t )    st sin( t )
1
(Sol.) L 

ds

tan
(
s
)
  tan 1 ( s)

e

dt

L
[sin(
t
)]
ds



2

0
s
s
s 2
t
s 1
 t 
Set s=0,


0
sin( x)

dx 
x
2
Eg. Determine f(0) if the Laplace transform F(s) of f(t) is given as below:
𝒔
F(s)=(𝒔−𝟏)(𝒔+𝟐). [2017 台師大電研]
s2
=1
s  ( s  1)( s  2)
(Sol.) f(0)= lim f (t )  lim sF ( s ) = lim
t 0
s 
4-2 Inverse Laplace Transform L-1[F(s)]=f(t)
Basic theorems of the inverse Laplace Transforms:
1. L-1[c1F(s)+c2G(s)]=c1f(t)+c2g(t)
2. L-1[F(s+a)] = f(t)e-at
 f (t  a), t  a
3. L1 [ F ( s)  e  as ]  f (t  a)  u (t  a)  
0, t  a
4. L-1[F(as)]= [f(t/a)]/a

f (t )
6. L1 [  F (u )du ] 
s
t
 F (s)  t
8. L1 
   f (u )du
 s  0
Heaviside’s formulae: L1 [
5. L-1[F(n)(s)]=(-1)ntnf(t)
7. L-1[sF(s)]=f’(t)+f(0)δ(t)
9. L-1[1/sn]=tn-1/(n-1)!=tn-1/Γ(n)
B
A
Bt m1e at
]
] =Aeat, L1 [
,

sa
( s  a) m
(m  1)!
Cs
D
] =C cos(t ) , L1 [ 2
] =D sin( t )
2
s 
s 2
 (s  r )

L1 [
] =  cos(t )  e rt , L1 [
] =  sin( t )  e rt
2
2
2
2
(s  r )  
(s  r )  
L1 [
2
1 
1
1
1
Eg. Find L1   , L1  2  , L1  3  , L1  4  , and L1 1 .
s
s 
s 
s 
0
2
t2
1  t
1 t
1 t
(Sol.) L-1[1/sn]=tn-1/(n-1)!, L1   = =1, L1  2  = =t, L1  3  = = ,
 s  0!
 s  1!
 s  2! 2
3
t3
1 t
L1  4  = = .
 s  3! 6
 1
By L-1[sF(s)]=f’(t)+f(0)δ(t), L1 1 = L1  s   =0+1∙δ(t)=δ(t).
 s
26
～～
 1 
Eg. Find L1 
. [2013 成大電研]
3
 ( s  2) 
t
1 t
(Sol.) L-1[1/sn]=tn-1/(n-1)!, L1  3  = = , and L-1[F(s+a)] = f(t)e-at,
 s  2! 2
2
2
 1  t 2 e 2t
∵ a=-2, ∴ L1 
=
3
2
 ( s  2) 
 e 2 s 
Eg. Find L1 
. [1993 中山電研]
4 
(s

2
)


(Sol.) According to L1[ F (s)  e  as ]  f (t  a)  u(t  a) , L1 [
L1 [
1
e 2t  t 3
]

3!
( s  2) 4
e 2 s
e 2(t-2)  (t - 2) 3
]

 u t - 2
3!
( s  2) 4
 2s  1 
Eg. Find L1 
.
 s ( s  1) 
(Sol.)
2s  1 A B
 
 A(s - 1)  Bs  2s - 1 
s( s  1) s s - 1
A  B  2
 s 1 B 1
or 

  A  1
s  0   A  1
 2s - 1 
1
1
 A=1, B=1  L-1 
 L1 [ ]  L-1[
]  1 et



s
s
1
s
s

1


 2s 2  9s  19 
Eg. Find L1 
.
2
 ( s  1) ( s  3) 
(Sol.)
A1
A2
2s 2  9s  19
B



2
2
s3
( s  1) ( s  3) ( s  1) ( s  1)
 A1  2
 A1  B  2
 2s 2  9s  19 


t
t
3t
 2 A1  A2  2 B  9   A2  3 , L1 
 =  2e  3te  4e
2
 ( s  1) ( s  3) 
 3 A  3 A  B  19  B  4

1
2

  s  2 
Eg. Find L1 n
 . [2005 北科大電研]
  s  1 

f (t )
(Sol.) According to L1 [  F (u )du ] 
and L1[ F (s  a)]  e  at  f (t )
s
t
 1
  s  2 
1   1 t
1 
 2t
L1 n

  L  s 
ds    [e  e ]
t
s

1
s

1
s

2

 
 
 
27
～～


s 1
Eg. Find L1  2
.
2
 (s  2s  2) 
(Sol.) According to L1[ F (s  a)]  e  at  f (t ) and L1 [ F ( n ) (s)]  (1) n t n f (t )



d  1  e  t
s 1
s
t
1  1
t
1 

e

L

(

1
)

 t  sin( t )
L1 

e

L


 
 2
2
2
2 
2
ds  s 2  1 

 ( s  1) 
 ( s  1) 2  1 




s2
Eg. Find L1  2
 . [2015 台大電研]
 s  2s  3 
(Sol.) According to L1[ F (s  a)]  e at  f (t ) and L1 1 =δ(t)

 1 


s2
2( s  1)
1
2s  3 
= L1 1 

L  2
 = L 1  2

2
2

 s  2s  3 
 ( s  1)  2 ( s  1)  2 
 s  2s  3 
1


1
2( s  1)
1
2
sin(√2t)et.
= L1 1 
=δ(t)+2cos(√2t)et 

2
2
2
2 
2 ( s  1)  ( 2 ) 
2
 ( s  1)  ( 2 )
Eg. Determine f(0) and f(∞) if the Laplace transform F(s) of f(t) is given as below:
𝒔
F(s)=(𝒔−𝟏)(𝒔+𝟐). [2017 台師大電研]
1
3
𝑠
2
3
𝑒 𝑡 2𝑒 −2𝑡
(Sol.) F(s)=(𝑠−1)(𝑠+2)= 𝑠−1 + 𝑠+2 , f(t)= 3 +
3
, ∴ f(0)=1 and f(∞)=∞
t
Eg. Solve y’+y+  y(u)du =1, y(0) =0. [2011中正電研]
0
(Sol.) sY(s)-y(0)+Y(s)+
1
Y(s)= 2
=
s  s 1
Y (s) 1
= ,
s
s
1
1
3
(s  ) 2  ( ) 2
2
2
=
2
3

3
2
1
3
(s  ) 2  ( ) 2
2
2
 y(t)=
t
2
3
t
sin(
Eg. Solve y’+2y+  y(u)du =u(t-1), y(0) =0. [2013台聯大系統電機類聯招]
0
e s
1
]=te-t and L-1[F(s)e-as]=f(t-a)u(t-a),
, L-1[
2
2
( s  1)
(s  1)
-(t-1)
∴ y(t)=(t-1)e ．u(t-1)
(Sol.) Y(s) =
28
～～
3t  2
)e
2
4-3 Laplace Transform Solutions of Differential Equations with Polynomial
Coefficients
Eg. Solve xy”-xy’-y=0, y(0)=0 and y’(0)=3. [1991 成大電研]

(Sol.) L[y(x)]=  y( x)e sx dx  Y (s) , L[ x n y ( x )]  ( 1) n
0
dn
Y ( s ) , and
ds n
L[ y ( n ) ( x)]  s nY ( s)  s n1  y(0)  s n2  y' (0)    s  y ( n2) (0)  y ( n1) (0)

d 2
d
[ s Y ( s )  sy (0)  y (0)]  ( )[ sY ( s)  y (0)]  Y ( s )  0
ds
ds
 2sY (s)  s 2Y (s)  Y (s)  sY (s)  Y (s)  0
(s 2  s)Y (s)  2sY (s)  0 , Y ( s ) 
 Y ( s) 
2
Y (s)  0
s 1
A
 y ( x )  Axex , y’(0)=3  A=3, ∴ y(x)=3xex
( s  1) 2
Eg. Solve 2y”+ty’-2y=10, y(0)=y’(0)=0. [2011 台大電子所甲組]
10
,
s
d
10
,
2[ s 2Y ( s )  sy (0)  y (0)]  (1) [ sY ( s )  y (0)]  2Y ( s ) 
ds
s
10
3
10
3
-sY’(s)+(2s2-3)Y(s)  , Y’(s)+(-2s+ )Y(s)=  2 , ∫(-2s+ )ds=-s2+3ln(s),
s
s
s
s
(Sol). L[2y”+ty’-2y]=L(10)=
exp[-s2+3ln(s)]= s 3 e  s , s 3 e  s Y’(s)+[-2 s 4 e  s +3 s 2 e  s ]Y(s)=-10s e  s ,
2
2
2
2
2
5
[ s 3 e  s Y(s)]’ =-10s e  s , s 3 e  s Y(s) =5 e  s +C, Y(s)= 3 +C s 3 e s ,
s
5
lim y (t )  lim sY ( s) =0  y(t)= t 2
t 0
s 
2
2
2
2
2
2
4-4 Convolution and Dirac Delta Function
t
Convolution in Laplace transform: f(t)*g(t)=  f (t   ) g ( )d
0
t
Theorem L[f(t)*g(t)]= L[ f (t   ) g ( )d ] =F(s)G(s)
0
Eg. Solve

t
0
f ( ) f (t   )d  6t 3 . [2006 台大電研]
(Sol.) According to L[t n ] 
F(s)=
(n  1)
36
and L-1[1/sn]=tn-1/Γ(n), [F(s)]2=6L[t3]= 4 ,
n 1
s
s
6
 f(t)=6t
s2
29
～～
t
Eg. Solve y”+y  4 y( ) sin( t   )d  e 2t , y(0)=1 and y’(0)=0. [1990 交大電信所]
0
(Sol.) s 2Y ( s )  sy (0)  y (0)  Y ( s )  4Y ( s )
1
1

s 1 s  2
2
( s 2  1)( s  1)
A
B
Cs  D


 2
 Y ( s) 
2
( s  2)( s  3)( s  1) s  1 s  2 s  3
(s2+1)(s+1)=A(s+2)(s2+3)+B(s-1)(s2+3)+(Cs+D)(s-1)(s+2)
1
Let s=1  4=A∙3∙4  A= .
3
5
Let s=-2  -5=B(-3)∙7  B= .
21
5
1
Let s=0  1=2- -2D  D= .
7
7
1
5
3
1
Let s=-1  0= ∙1∙4+ ∙(-2)∙4+(-C+ )(-1)(2)  C=
7
3
7
21
3s
1
5
1
3
 Y (s)  3  21  2 7 
 2
s 1 s  2 s  3 7 3 s  3
 
1
5
3
1
 y (t )  e t  e 2t  cos 3t 
sin
3
21
7
7 3
 3t 
t
Eg. Solve f(t)=3t2-e-t-  f (  )e t   d . [2011 台大電研]
0
6
1
1
s
6
1
6 6 1 2
 F (s) ,
 F ( s) = 3 , F(s)= 3 - 4 + ,
3
s 1
s s 1
s s  1 s 1
s s 1
s s
 f(t)=3t2-t3+1-2e-t
(Sol.) F(s)=
t
Eg. Solve y+ e t  y( )e  d  cos(t ) . [2015 中興電研乙丙丁組、中興光電所]
0
t
(Sol.) y+  y( )e t  d  cos(t ) , y+y*et=cos(t), Y ( s )  Y ( s )
0
 Y ( s) 
s 1
s
1
 2
 2
 y(t )  cost   sin t 
2
s 1 s 1 s 1
t
Eg. Solve f(t)=e-t+ 2 e 3 f (t   )d .
0
1
2

 F (s)
s 1 s  3
s3
1
2
 f (t )  e t  2te t .
F ( s) 


2
2
s

1
( s  1)
( s  1)
(Sol.) F ( s ) 
30
～～
1
s
 2
s 1 s 1
Eg. Solve
t
 x(u) sin( t  u)du  x(t )  sin( t )  cos(t ) .
0
【1991 交大電信所】
t
Eg. Solve y(t )  sin( 2t )   y( ) sin[ 2(t   )]d . 【1991 淡江機研】
0
Dirac delta function:
1

,a  t  a  
lim


0
 (t  a)   a (t )  lim [u (t  a)  u (t  a   )]   
 0 
0, elsewhere
1
1, a  b
Kronecker delta:  ab  
0, a  b
Characteristics of Dirac’s delta function:
1.



 (t  a)dt  1 . 2. L[ (t  a)]  e  as . 3.


0
f (t ) (t  a)dt  f (a) .

4. f (t ) *  (t )   f ( x) (t  x)dx  f (t )
0
5. f (t ) (t  a)  f (a) (t  a)
Eg. Solve y(4)=δ(x-a), y(0)=y”(0)=y(1)=0, y(3)(0)=1, 0<a<1. [1990 交大電子所]
(Sol.) L( y ( 4) )  s 4Y (s)  s 3 y(0)  s 2 y (0)  sy (0)  y (3) (0)  s 4Y ( s)  s 2 c  1
L[ ( x  a)]  e  as  Y ( s ) 
( x  a) 3
x3
e  as c 1



y
(
x
)


u
(
x

a
)

cx

s4 s2 s4
6
6
y (1)  0 
(1  a) 3
1
 (1  a) 3 1
c c 

6
6
6
6
∴ y ( x) 
 (1  a) 3 1 
( x  a) 3
x3
 u ( x  a)  
 x
6
6
6
 6
  (1  a) 3 1 
x3
 x
, 0 xa
 
6
6
  6

3
3
3
 ( x  a)   (1  a)  1   x  x , a  x  1


 6
6
6
 6

Eg. Solve y”+5y’+4y=3+2δ(t).【北科大土木所】
31
～～