Lecture 25
Laplace Transform
Hung-yi Lee
Reference
• Textbook 13.1, 13.2
Laplace Transform
Motivation and Introduction
Laplace Transform

Time
domain
Fs     f (t )e
 st
0
 dt
Laplace Transform ( L[f(t)] )
s-domain
Fs 
f t 
Inverse Laplace Transform
  j
(L-1[F(s)])
1
st
f t  
F ( s )e  ds

2j   j
   c 
Res 
   c
Note
L f t     f (t )e  st  dt

0
0
   f (t )e
 st
0

 dt    f (t )e
0
 st
 dt
Always 0?
When f(0)=∞, it may not be zero
Note
1, u t ,
Time
domain
u t  1
u t 
s-domain
Fs 
=
When t≥0
f t 
Laplace
Transform (L)
fˆ t 
Inverse Laplace
Transform (L-1)
Domain
Different domains means view the same thing in
different perspectives
Position:
台北市羅斯福路
四段一號博理館
25°1'9"N 121°32'31"E
Domain
Different domains means view the same thing in
different perspectives
Linear Algebra:
y
1,1
y
x

2 ,0

x
Transform: switch between
different domains
Domain
Muggle
Signal
Engineer
Time domain
f t   1
Laplace
Transform
s-domain
Fs   1 / s
Inverse Laplace
Transform
Fourier Series
Fourier Transform
Time
Domain
Time
Domain
https://www.youtube.com/wat
ch?v=r4c9ojz6hJg
Frequency
Domain
Frequency
Domain
Fourier Transform
F   


1
f t  
2



f (t )e  jt  dt
F ( )e jt  d
Laplace Transform

Fs     f (t )e  st  dt
0
1   j
st
f t  
F ( s )e  ds

2j   j
Laplace Transform
v.s.
Fourier Transform
Laplace Transform
of f(t)
=
Fourier Transform:
F   


f (t )e
 j t
 dt
Laplace Transform:

Fs     f (t )e
 st
0



Fourier Transform
of f(t)e-σtu(t)

 dt    f (t )e t e  jt  dt
0
f (t )e u t e
t
 j t
 dt
Laplace Transform
v.s.
Fourier Transform
Multiply e-σt
Multiply u(t)
Do Fourier Transform

Fs     f (t )e  st  dt
0



f (t )e t u t e  jt  dt
Transformable Function
All Functions
e
t2
Laplace Transform
Fourier Transform
Fourier Series
e
t
1
0
Periodic Functions
D
Why we do Laplace transform?
Transfer Function
x(t )
H(s)
(circuit, filter …)
y (t )
Laplace transform can help us find y(t) easily
Transfer Function
The signal with complex
frequency s0 = σ0 + ω0
x(t )  Ae 0 t cos0t   
 jAe 0 t sin 0t   
 Ae j e s0 t
H(s)
(circuit, filter …)
y (t ) | H s0  | Ae 0 t cos0t    H s0 
 j | H s0  | Ae 0 t sin 0t    H s0 
| Hs0  | e jH  s0 Ae j e s0 t
Transfer Function
xt   Ae j e s0 t
xt   Ze s0 t
(z is complex)
 
H(s)
H(s)
| Hs0  | e jH  s0 Ae j e s0 t
y t   H s0 Ze s0 t
y t 
(H(s0) is complex)
Transfer Function
1   j
st
xt  
F ( s )e  ds

2j   j
Lxt   F ( s )
H(s)
1   j
st
y t  
H s F ( s )e  ds

2j   j
L y t   Hs F ( s )
We do not know y(t), but we
know its Laplace transform
Transfer Function
xt 
L
F (s )
H(s)
y t 
-1
L
Hs F ( s )
Laplace Transform
Laplace Transform Pairs
Laplace Transform Pairs (1/4)
f (t )  1
1  st t 
e
L f (t )    1 e dt 

t

0
0
s
1
1
If Re[s]=σ>0 
0  1  s
s

 st

Time domain: 1
Time domain: u(t)
(Only consider t>0)
s-domain: 1/s

ROC: σ > 0
ROC
Laplace Transform Pairs (2/4)
f (t )  e
 at
L f (t )    e

 at
0


 e dt    e a  s t dt
 st
0
1

e a  s t
 a  s 

t 
t 0
1
1
0  1 

sa
 a  s 
If Re[a+s]=a+σ>0

Time domain: e-at
s-domain: 1/(s+a)
ROC: σ > -a

-a
ROC
Laplace Transform Pairs (3/4)
f (t )  sin t 
e j  e  j
cos  
2
e j  e  j
sin  
2j
e j  cos   j sin 
e  j  cos   j sin 
L f (t )    sin  t e  st dt

0


0
e
j t
e
j2
 j t

1    s  j t
  s  j t

e
dt

e
dt
e dt 






0
0

j2
 st
If Re[s]=σ>0
1  1
1



j 2  s  j s  j
s  j  s  j

 
2
2
j
2
s







s 
2
2
Laplace Transform Pairs (4/4)
f (t )  cost 
e j  e  j
cos  
2
e j  e  j
sin  
2j
e j  cos   j sin 
e  j  cos   j sin 
L f (t )    cost e  st dt

0

1    s  j t
e jt  e  jt  st
  s  j t


e
dt

e
dt

e dt




0
0
0

2
2

If Re[s]=σ>0
1 1
1
 

2  s  j s  j
s
 s  j  s  j
 
 2
2
2
2
s


2
s





Summary of Transform Pairs
Time domain
1
e
 at
s-domain
1
s
sin t 
1
sa
cost 

s2   2
s
s2   2
The 4 transform pairs are sufficient to imply
all transform pairs in Table 13.2.
• More complete Transform Pairs:
http://www.vibrationdata.com/math/Laplace_Transforms.pdf
Note: Impulse function
• What is L-1[1]? L-1[1]=δ(t) (impulse function, Dirac
delta function)
 0 t0
 t   
" " t  0
 0 t0  0
 t  t0   
" " t0  0
1 / dt

  t dt  1
1
1

du t 
 t  
dt
0
t0
Laplace Transform
Properties
The six properties in Table 13.1 (P585)
Property 1: Linear Combination
• Let L[f(t)]=F(s) and L[g(t)]=G(s)
LAf (t )    Af (t )e  st dt  AF (s )

0
L f (t )  g (t )     f (t )  g (t )e  st dt

0


   f (t )e dt    g (t )e  st dt
 st
0
 F (s)  G (s)
0
Property 2: Multiplication by
• Let L[f(t)]=F(s)

Le
 at

-at
e

f (t )    e  at f (t )  e  st dt
0

   f (t )  e  s  a t dt  F ( s  a )
0
1

1
s
Multiplication
by e-at
e
 at


1
sa

-a
ROC
ROC
Property 3: Multiplication by t
• Let L[f(t)]=F(s)
d
Ltf (t )   F ( s )
ds
 st

de
d
d 
dt
 F ( s )     f t e  st dt     f t 
0
ds
ds
ds 0


    f t  t e dt    tf t e  st dt
0
 st
0
Property 4: Time Delay
g (t )  f (t  t0 ) u (t  t0 )
0
t  t0


 f (t  t0 ) t  t0
Delay by t0 and
zero-padding up to t0
t  t0
Lg (t )  
t 
t t0
f (t  t0 )  e dt  
 st
 0
  t  t0
e
 st 0
 
 

0
f ( )  e
t    t0
 s   t 0 
dt  d
f ( )  e  s d  e  st0 F s 
d
Property 5: Differentiation
Integration by parts:
• Let L[f(t)]=F(s)

a
v’
L f (t )   

0
b
v  udt v  u a   v  u dt
b
a
u
df (t )  st
 e dt
dt
 f (t ) e
v
b
 st 
0
u

   f (t )  ( s )e  st dt
0
v

u’
  f (0 )  s   f (t )e  st  dt  sF ( s )  f (0  )
0

Property 5: Differentiation
• Let L[f(t)]=F(s)
 
L f (t )  sF s   f 0 
Example
f t   e
t
f t   e t
1
Fs   L f t  
s 1
1
L f t  
s 1
 
L f (t )  sF s   f 0

1
1
s
1 
s 1
s 1
Property 5: Differentiation
L f (t )  F s 
 
L f (t )  s sF s   f 0   f 0 
 s F s   sf 0   f 0 
L f (t )  sF s   f 0 


2

……



 
 
 
L f n  (t )  s n F s   s n 1 f 0   s n  2 f  0    f ( n 1) 0 
f ( )e  st dtd
Property 6: Integration
• Let L[f(t)]=F(s)
t
t
g (t )    f ( )d
0
t

Lg (t )      f ( )d  e  st dt

0 
0

(You can use integration by
parts as in P584)

t
t
 st

     f ( )e d  dt

0 
0



    f ( )e  st dt  d

0 



Property 6: Integration
• Let L[f(t)]=F(s)
t
g (t )    f ( )d
0


Lg t      f ( )e  st dt  d

0 




   f ( )  e  st dt  d
 

0



0
 1  st  
f (  )  e   d
 s



0
1  s
f (  ) e d
s
1 
 s
   f (  ) e d
s 0
1
 Fs 
s
Laplace Transform Properties
Table 13.1 Laplace Transform Properties (P585)
Operation
Time Function
Laplace Transform
Linear Combination
Af (t )  Bg (t )
AF ( s )  BG ( s )
Multiplication by e  at
e  at f (t )
F ( s  a)
Multiplication by t
tf (t )
 dF ( s ) ds
Time delay
f (t  t0 )u (t  t0 )
e  st0 F ( s )
Differentiation
f (t )
sF ( s )  f (0  )
Integration

t
0-
f (  ) d
1
F (s)
s
More properties (in Homework)
 Time-scaling property
 Integral of F(s)
L f t /    F s 

 F s ds  Lt f t 
1
s
 Periodic function
g t   f t   f t  T u t  T   f t  2T u t  2T   
f t 
f(t)=0 outside
0<t<T
g t 
…..
F s 
Lg t  
1  e  sT
Table 13.2 Laplace Transform Pairs (P585)
f(t)
A
u (t )  u (t  D )
t
tr
e  at
te at
t r e  at
sin  t
cos(  t   )
e  at cos(  t   )
F(s)
A
s
1  e  sD
s
1
s2
r!
s r 1
1
sa
1
s  a 2
r!
s  a r 1

s2   2
s cos    sin 
s2   2
s  a  cos    sin 
s  a 2   2
1
0
D
Lu t   u t - D 
1 1 -sD
 Lu t   Lu t - D    e
s s
Table 13.2 Laplace Transform Pairs (P585)
f(t)
A
u (t )  u (t  D )
t
tr
e  at
te at
t r e  at
sin  t
cos(  t   )
e
 at
cos(  t   )
F(s)
A
s
1  e  sD
s
1
s2
r!
s r 1
1
sa
1
s  a 2
r!
s  a r 1

s2   2
s cos    sin 
s2   2
s  a  cos    sin 
s  a 2   2
1
u t   u t - 1
0
t
1
1 e -s
Lu t   u t - 1 
s
tu t   u t - 1
1
0
1
1 e -s
d
s
Lt u t   u t - 1  
ds


e -s s  1  e -s 1 - e -s - e -s s


2
s
s2
Example for Periodic function
1
1 - e -s - e -s s
s2
L
0
1
1
L
0
1
2
3
4
-s
-s
1- e - e s 1
2
2 s
s
1 e
5
g t   f t   f t  T u t  T   f t  2T u t  2T   
F s 
Lg t  
 sT
1 e
Table 13.2 Laplace Transform Pairs (P585)
f(t)
A
u (t )  u (t  D )
t
tr
e  at
te at
t r e  at
cos(  t   )
e
 at
cos(  t   )
A
s
1  e  sD
s
1
s2
r!
s r 1
1
sa
1
s  a 2
r!
s  a r 1
1
L
t
t
2
L
t
3
L
t
r

s2   2
s cos    sin 
s2   2
s  a  cos    sin 
s  a 2   2
L
……
sin  t
F(s)
L
s 1
ds 1 2

s
ds
ds 2
3

2
s

ds
d 2s 3
4

 2  3s
ds
2  3 rs r 1
 r! s r 1
Table 13.2 Laplace Transform Pairs (P585)
f(t)
A
u (t )  u (t  D )
t
t
r
e  at
te at
t r e  at
sin  t
cos(  t   )
e  at cos(  t   )
F(s)
A
s
1  e  sD
s
1
s2
r!
s r 1
1
sa
1
s  a 2
r!
s  a r 1
cost   
 cost  cos   sin t sin  
Lcost   
s

 cos  2
 sin   2
2
s 
s 2

s2   2
s cos    sin 
s2   2
s  a  cos    sin 
s  a 2   2
Note: Euler’s formula
e
j 0 t
 cos0t   j sin 0t 
1
s  j 0
s  j 0
1

s  j 0 s  j 0
s  j 0
 2
2
s  0
s
s 2  02
0
j 2
s  02
Note: Multiplication
L f (t ) g (t )  F s G s 
Lsin t  

s 
2
2
1  cos 2 t
sin t 
2
2
  

L sin t   2
2 
s  

2

2


2
1 1
s

L sin t 
2 s 2 s 2  2  2
Laplace Transform
Inverse Laplace Transform
Partial-Fraction Expansions
• Rational Function
bm s m  bm 1s m 1    b1s  b0 N ( s )
F (s)  n

n 1
s  an 1s    a1s  a0
D( s)
N (s)

s  s1 s  s2  s  sn 
s1, s2, ……, sn are the
roots of D(s) ( poles
of F(s) )
An
A1
A2


  
One pole, one term
s  s1 s  s2
s  sn
We only consider the case that m < n.
(strictly proper rational function)
Partial-Fraction Expansions
 
f t   sF s   f 0 
• Rational Function
bm s m  bm 1s m 1    b1s  b0 N ( s )
F (s)  n

n 1
s  an 1s    a1s  a0
D( s)
If m = n
R( s)
Fs   bm 
D s 
If m = n + 1
R( s )
Fs   bm s  c 
D s 
δ(t)
1
differentiate
dδ(t)/dt ……
multiply s
s
Partial-Fraction Expansions
• Rational Function
An
A1
A2
F (s) 

  
s  s1 s  s2
s  sn
 1 
 at
L 

e

s  a
1
 Ai 
si t
L 

A
e
i

s

s
i

1
L1 F s   f (t )  A1e s1t  A2 e s2t    An e snt
t0
There are three tips you should know.
Tip 1: How to find A1,A2,∙∙∙∙∙∙∙,An
• Example 13.5
A3
 2 s 2  24 s  40 A1
A2
F( s ) 
 

s s  2 s  10 
s s  2 s  10
Panacea: reduce to a common denominator, and then
compare the coefficients
 2 s 2  24 s  40  A1 s  2 s  10   A2 s s  10   A3 s s  2 





 A1 s 2  12 s  20  A2 s 2  10 s  A3 s 2  2 s
A1  A2  A3  2
12 A1  10 A2  2 A3  24
20 A1  40

Tip 1: How to find A1,A2,∙∙∙∙∙∙∙,An
• cover-up rule
F (s) 
N (s)
A
A
 1  2
s  s1 s  s2  s  s1 s  s2

N (s)
s  s1 A2
 A1 
s  s2
s  s2
setting s  s1 , then
N ( s1 )
 A1
s1  s2
N ( s ) A1 s  s2 

 A2
s  s1
s  s1
setting s  s2 , then
N ( s2 )
 A2
s2  s1
Tip 1: How to find A1,A2,∙∙∙∙∙∙∙,An
• Example 13.5: cover-up rule
A3
 2 s 2  24 s  40 A1
A2
F( s ) 
 

s s  2 s  10 
s s  2 s  10
Take A2 as example. Multiplying s+2 at both sides
A3 
 2 s 2  24 s  40
A2
 A1
s  2 
 s  2  


s s  2 s  10 
 s s  2 s  10 

s  2 A3
 2 s 2  24 s  40 s  2 A1

 A2 
s s  10 
s
s  10
2
 2 2   24 2   40
Set s=-2:
A2  5
 A2
 2 2  10
Tip 2: For Complex Poles
15s 2  16 s  7
-1[F(s)]
• Example 13.6: F ( s) 
Find
f(t)
=
L
s  2 s 2  6s  25


s  2,  3  j 4
A1
A21
A22
F (s) 


s  2 s  3  j4 s  3  j4
It is not easy to
find A21 and A22.


s  3  j 4 A1
s  3  j 4 A22
s  3  j 4F ( s) 
 A21 
s2
15s 2  16 s  7

s  2s  3  j 4
s  3  j4
Set s=-3+j4 to find A21…….
Tip 2: For Complex Poles
15s 2  16 s  7
-1[F(s)]
• Example 13.6: F ( s) 
Find
f(t)
=
L
s  2 s 2  6s  25
5
Bs  C
F (s) 
 2
s  2 s  6 s  25


Do not split the complex poles
Find B and C
5s 2  30 s  125  Bs 2  2 Bs  Cs  2C
15s 2  16 s  7

2
s  2 s  6s  25
s  2 s 2  6s  25

5  B  15 B  10
125  2C  7 C  66


5
10 s  66
F (s) 
 2
s  2 s  6 s  25

Tip 2: For Complex Poles
15s 2  16 s  7
-1[F(s)]
• Example 13.6: F ( s) 
Find
f(t)
=
L
s  2 s 2  6s  25
5
Bs  C
F (s) 
 2
s  2 s  6 s  25


Do not split the complex poles
Find B and C (another approach)
lim sF s 
s 
15s 3 
5s
Bs 2  Cs
sF ( s )  3

 2
s   s  2 s  6 s  25
lim sF s   15  5  B
s 
7
5 C
 
F (0) F (0) 
225 2 25
B  10
C  66
Tip 2: For Complex Poles
15s 2  16 s  7
-1[F(s)]
• Example 13.6: F ( s) 
Find
f(t)
=
L
s  2 s 2  6s  25


5
10 s  66
1  10 s  66 
F (s) 
 2
Find L
 s 2  6 s  25  (Refer to P593)
s  2 s  6 s  25





s 
1 
t
1 
t
L

e
sin t
L 

e
cos

t

2
2
2
2
 s      
 s      
 10 s  66 
1 10s  3  96 
L  2
L 
2
2 

s

6
s

25


 s  3  4 
1



s3
4
1 
 10L 
 24L 
2
2
2
2




s

3

4
s

3

4




1
Tip 3: Repeated Poles
• Exercise 13.31
A2  A3
A1
4 s 2  12 s  10
A3
A1
A2


F (s) 



2
s  4s  3 s  4 s  3 s  3 s  4 s  3
order=3
order=2
4 s 2  12 s  10
A1
A2
F (s) 


2
s  4s  3 s  4 s  32
4 s 2  12 s  10
 s  3 A1  s  4 A2
2
4 s 2  12 s  10


 s  6 s  9 A1  s  4 A2
2
A1  4
6 A1  A2  12
9 A1  4 A2  10
?
Tip 3: Repeated Poles
• Exercise 13.31
A3
4 s 2  12 s  10
A1
A2
F (s) 



s  4s  32 s  4 s  3 s  32

4 s 2  12 s  10
s  4 A2 s  4 A3
 A1 

2
s3
s  3 
s  32
A1  6
4 s 2  12 s  10 s  3 A1

 s  3A2  A3
s  4 
s4
A3  10
2
A
4 s  12 s  10 s  3A1

 A2  3
s  4s  3
s4
s3
2
We cannot find A2 by
multiplying (s+3)
(Refer to P596 – 597)
Tip 3: Repeated Poles
• Exercise 13.31
4 s 2  12 s  10
6
A2
 10
F (s) 



s  4s  32 s  4 s  3 s  32
4 s 2  12 s  10
6s
A2 s
 10 s
sF ( s )  s



2
s  4s  3 s  4 s  3 s  32
s
4
6
A2
0
A 2  2
Time delay
Exercise 13.10
8 s




5 s  3  2s  4 e
Fs  
2
s  3 

5
2 s  4  8 s


e
2
s  3 s  3
L-1
5e
3t
L-1
?
Delay by 8 and
zero-padding up to 8
f (t  t0 )u (t  t0 )
e
 st 0
F (s)
2s  4
Fs  
s  32
A1
A2


s  3 s  32
2 s  4  s  3A1  A 2
s  3
A 2  10
Time delay
Exercise 13.10
8 s




5 s  3  2s  4 e
Fs  
2
s  3 

5
2 s  4  8 s


e
2
s  3 s  3
L-1
5e
3t
L-1
?
Delay by 8 and
zero-padding up to 8
f (t  t0 )u (t  t0 )
e
 st 0
F (s)
2s  4
Fs  
s  32
A1
 10


s  3 s  32
2s  4
sFs   s
2
s  3
A1
 10
s
s
2
s3
s  3
s
2  A1
Time delay
Exercise 13.10
8 s




5 s  3  2s  4 e
Fs  
2
s  3 

5
2 s  4  8 s


e
2
s  3 s  3
L-1
5e
3t
L-1
?
Delay by 8 and
zero-padding up to 8
f (t  t0 )u (t  t0 )
e
 st 0
F (s)
2s  4
Fs  
s  32
2
 10


s  3 s  32
L-1
L-1
2e 3t  10te 3t
f t   2e 3t  10te 3t
f t   5e 3t
 f t  8u t  8
Initial and Final Values
• We can find the value f(0+) and f(∞) from F(s)
We know
L f t   F s 
Initial-value

sF (s)
Theorem f (0 )  lim
s 
Because F(s) is strictly proper,
lim sF ( s ) is defined.
s 
Example 13.9
• Find the initial value f(0+)
5s 3  1600
F( s )  3
s s  18s 2  90 s  800 
 
f 0  5

f (0 )  lim sF (s)
s 
5s 3  1600
5
 lim 3
2
s  s  18s  90 s  800
Example 13.9
• Find the initial slope f’(0+)
5s 3  1600

F( s )  3


f
0
5
2
s s  18s  90 s  800 
Lf s   F( s ) f (0  )  lim sF ( s )
 
s 5s  1600 
Lf s   sF( s )  f 0 
f (0  )  lim
s 
s 
5s 3  1600
 3
5
2
s  18s  90 s  800
3
s  18s  90 s  800
3

2


 5s
∞-∞?

s 5s 3  1600  5s s 3  18s 2  90 s  800
 lim
 90
3
2
s 
s  18s  90 s  800
Initial and Final Values
• We can find the value f(0+) and f(∞) from F(s)
We know
L f t   F s 
Initial-value

sF (s)
Theorem f (0 )  lim
s 
Because F(s) is strictly proper,
lim sF ( s ) is defined.
s 
If the final value exists
(Can be known from the poles)
Final-value
f
(

)

lim
sF
(
s
)
Theorem
s 0
Final Values
4 regions

Region B
Region A
Region D

Region C

Final Values
Region A

A
F( s ) 
sa
f (t )  e at
a>0
No final value
A1
A2
Ar
F( s ) 

  
2
s  a s  a 
s  a r
a>0
Bs  C
Bs     C 
F( s )  2

2
s  bs  c
s      2
b<0
α<0
f (t )  te at
No final value
f (t )  Be t cos t
 C e t sin t
No final value

Final Values
Region B

A
F( s ) 
sa
f (t )  e at
a<0
Final value = 0
A1
A2
Ar
F( s ) 

  
2
s  a s  a 
s  a r
a<0
Bs  C
Bs     C 
F( s )  2

2
s  bs  c
s      2
b>0
α>0
f (t )  te at
Final value = 0
f (t )  Be t cos t
 C e t sin t
Final value = 0

Region C
Final Values

Bs  C
F( s )  2
s 2
s

 B 2
 C 2
2
s 
s 2
f (t )  B cos t  C  sin t
No final value
Region D
Final Values
A
F( s ) 
s


f (t )  A
final value = constant
A1 A2
Ar
F( s )   2    r
s s
s
f (t )  A 2t No final value
Summary for Final Values (P601)
Final value exists
single pole at the origin
(1) Poles on the left plane, or
(2) single pole at the origin
Non zero final value
Final-value
f
(

)

lim
sF
(
s
)
Theorem
s 0
Final Values
The final-value theorem gives the wrong
answer when the final value does not exist.

Only one pole

s
Fs   2
s 2
lim sF ( s )  0
s 0
f s   cos t The final value
not exists
1
Fs  
sa
The final value exists
iff the poles are in
this region
f s   e  at
lim sF ( s )  0
a0
s 0
The final value
not exists
Final Values

Only one pole

Final-value
f
(

)

lim
sF
(
s
)
Theorem
s 0
The final value is not zero iff there is
only one pole at the origin
A
A1
A2
F( s )  


s s  a1 s  a2
The final value is 0
The final value is A
The final value is clearly A
The final value exists
iff the poles are in
this region
A  f ()  lim sF ( s )
s0
Example 13.9
• Find the final value
5s 3  1600
F( s )  3
s s  18s 2  90 s  800 
Four poles: 0, -10, -4+8j, -4-8j
 
f 0  5
The final value exists.
The final value is not zero.
f ()  lim sF ( s )
s0
5s 3  1600
 2
 lim 3
2
s 0 s  18s  90 s  800
Laplace Transform
Application
Differential Equation
Find v(t)
8
v (0 ) 
15V  6V
12  8

Ri (t )  v(t )  VB
i (t )  Cv(t )
RCv(t )  v(t )  VB
1
12  v(t )  v(t )  15
60
0.2v(t )  v(t )  15
0.2Lv(t )  Lv(t )  L15
L15 
15
s
V ( s )  Lv(t )
Lv(t )  sLv(t )  v(0  )
 sV ( s )  6
15
0.2sV ( s )  6  V ( s ) 
s
Differential Equation
15
0.2sV ( s )  6  V ( s ) 
s
15
1 .2 
6 s  75 A
B
s
 V (s) 

 
0 .2 s  1 s s  5  s s  5
6 s  75
sB
sV ( s ) 
 A
s5
s5
6 s  75
A
s  5V ( s) 
 s  5  B
s
s
15  9
V (s)  
s s5
L 1 V ( s )
 v(t )  15  9e 5t
t0
s  0  A  15
s  5  B  9
Homework
• 13.6, 13.9, 13.10, 13.16, 13.25, 13.28, 13.35, 13.38.
13.46
Thank you!
Answer
• 13.6: derive by yourself
• 13.9: proof by yourself
• 13.10: proof by yourself
• 13.16: F(s)=2(1-3se-2s-e-3s)/s2
• 13.25: f(t)=-2+5e-2t-3e-4t-e-6t
• 13.28: f(t)=5-5e-4t+10e-3tcos(t-36.9。)
• 13.35: f(t)=2te-tcos(2t-180 。)
• 13.46: f(0+)=2, f’(0+)=-5, f(∞) not exist
Appendix
Fourier Series
• Periodic Function: f(t) = f(t+nT)
Fourier Series:
Laplace Transform Pairs (1/4)
1
L1 
s
s0
1  1  e st
σ=0
0
?
ω