Introduction to Laplace
Transforms
Definition of the Laplace Transform
One-sided Laplace transform:

F ( s)  L f (t )  0 f (t )e
 st

l Some functions may not have Laplace
transforms but we do not use them in circuit
analysis.
Choose 0- as the lower limit (to capture
discontinuity in f(t) due to an event such as
closing a switch).
dt
The Step Function
?
Laplace Transform of the Step Function

Lu( t )  0 u( t )e
 st

1  st
 - e
s

0
  st
e
dt
0
dt  
1

s
The Impulse Function
The sifting property:


f (t ) (t  a )dt  f (a )
The impulse function is a derivative of the step function:
du(t )
 (t ) 
dt
t
  ( x )dx  u(t )
?
Laplace transform of the impulse function:

L ( t  a )  0  ( t  a )e

For a  0,
L ( t )  1
 st
dt e
 as
Laplace Transform Features
1) Multiplication by a constant
L f ( t )  F ( s )
 L Kf ( t )  KF ( s )
2) Addition (subtraction)
L f1 ( t )  F1 ( s ),
L f 2 ( t )  F2 ( s ),
 L f1 ( t )  f 2 ( t )  F1 ( s )  F2 ( s )
3) Differentiation
 df ( t ) 

L
  sF ( s )  f (0 ),
 dt 
 d 2 f ( t )  2



L

s
F
(
s
)

sf
(0
)

f
(0
)

2
 dt 
 d n f ( t )  n
n 1

n 2


L

s
F
(
s
)

s
f
(0
)

s
f
(0
)

n
 dt 
 sf ( n  2) (0 )  f ( n 1) (0
Laplace Transform Features (cont.)
3) Differentiation
 df ( t ) 

L
  sF ( s )  f (0 ),
 dt 
 d 2 f ( t )  2



)
(0
f

)
(0
sf

)
s
(
F
s

L

2
 dt 
.
.
.
 d n f ( t )  n

n 2

n 1

)
(0
f
s

)
(0
f
s

)
s
(
F
s

L
n 
 dt 
 sf ( n  2) (0 )  f ( n 1) (0 )
Laplace Transform Features (cont.)
4) Integration
L

t
0

F ( s)
f ( x )dx 
s
5) Translation in the Time Domain
L f (t  a )u(t  a )  e
 as
F ( s ),
a
0
Laplace Transform Features (cont.)
6) Translation in the Frequency Domain

L e
 at

f (t )  F ( s  a )
7) Scale Changing
1 s
L f (at )  F   ,
a a
a
0
Laplace Transform of Cos and Sine
?
Laplace Transform of Cos and Sine

Lcos  t .u( t )  0 cos  t .u( t )e
 st

e
 0
j t

dt  0 cos  te  st dt
 e  j t  st
1   ( s  j )t
  ( s  j ) t
e dt  [ 0 e
dt  0 e
dt ]
2
2
1 1

e  ( s  j )t
2 s  j

0
1 1

e  ( s  j )t
2 s  j
1 1
1 1
s


 2
2 s  j 2 s  j s   2

Lsin  t .u(t )  2
2
s 

0
Name
Time function f(t) Laplace Transform
Unit Impulse
(t)
u(t)
t
tn
e-at
t n e-at
sin(bt)
cos(bt)
e-at sin(bt)
e-at cos(bt)
t sin(bt)
t cos(bt)
Unit Step
Unit ramp
nth-Order ramp
Exponential
nth-Order exponential
Sine
Cosine
Damped sine
Damped cosine
Diverging sine
Diverging cosine
1
1/s
1/s2
n!/sn+1
1/(s+a)
n!/(s+a)n+1
b/(s2+b2)
s/(s2+b2)
b/((s+a)2+b2)
(s+a)/((s+a)2+b2)
2bs/(s2+b2)2
(s2-b2) /(s2+b2)2
Some
Laplace
Transform
Properties
(t)
F(s)
A (t)
A F(s)
1(t) ± 2(t)
F1(s) ± F2(s)
Property
Scaling
Linearity
1 s
F 
a a
a0
Time Scaling
(a·t)
Time Shifting
(delay)
(t–t0) u(t–t0)
e–s·t0 F(s) t00
Frequency Shifting
e–a·t (t)
F(s+a)
Time Domain
Differentiation
d f (t )
dt
s F(s) – (0)
Frequency Domain
Differentiation
t (t)
t
0
Time Domain
Integration
t
Convolution
0
f ( ) d
f1 ( ) f 2 ( t   ) d
d F ( s)
ds
1
F ( s)
s
F1(s) F2(s)
Inverse Laplace Transform
Partial Fraction Expansion
Step1: Expand F(s) as a sum of partial fractions.
Step 2: Compute the expansion constants (four
cases)
Step 3: Write the inverse transform
Example:
s6
2
s( s  3)( s  1)
K3
K1
K2
K4
s6




2
2
s
s

3
s1
s( s  3)( s  1)
( s  1)
1 

s6
L 
2
 s( s  3)( s  1) 


 K1  K 2e 3 t  K 3te  t  K 4e  t u( t )
Distinct Real Roots
K3
K2
96( s  5)( s  12) K1
F ( s) 



s( s  8)( s  6)
s
s8 s6
96( s  5)( s  12)
96(5)(12)
K1  sF ( s ) s  0  s

 120
s( s  8)( s  6) s  0
8(6)
96( s  5)( s  12)
96( 3)(4)
K 2  ( s  8)F ( s ) s 8  ( s  8)

 72
s( s  8)( s  6) s 8
8( 2)
K 3  ( s  6)F ( s ) s 6  ( s  6)
96( s  5)( s  12)
96( 1)(6)

 48
s( s  8)( s  6) s 6
6(2)
120
48
72
F ( s) 


s
s6 s8
f ( t )  (120  48e 6 t  72e 8 t )u( t )
Distinct Complex Roots
100( s  3)
100( s  3)
F ( s) 

2
( s  6)( s  6 s  25) ( s  6)( s  3  j 4)( s  3  j 4)
K3
K1
K2



s  6 s  3  j4 s  3  j4
100( s  3)
100( 3)
K1  2

 12
25
s  6 s  25 s 6
K2 
100( s  3)
100( j 4)

( s  6)( s  3  j 4) s 3  j 4 (3  j 4)( j 8)
 6  j 8  10e
 j 53.130
100( s  3)
100(  j 4)
K3 

( s  6)( s  3  j 4) s 3  j 4 (3  j 4)(  j 8)
 6  j 8  10e
j 53.130
Distinct Complex Roots
(cont.)
12 10  53.130 10 53.130
F ( s) 


s6
s  3  j4
s  3  j4
f ( t )  ( 12e
10e
6 t
 10e
 j 53.130  (3  j 4) t
 10e
e
3 t
(e
 j 53.130  (3  j 4) t
e
 10e
j (4 t  53.130 )
e
 10e
j 53.130  (3  j 4) t
j 53.130  (3  j 4) t
e
 j (4 t  53.130 )
)
 20e 3 t cos(4t  53.130 )
f ( t )  [12e 6 t  20e 3 t cos(4t  53.130 )]u( t )
e
)u( t )
Whenever F(s) contains distinct complex roots at the denominator
as (s+-j)(s++j), a pair of terms of the form
K
K*

s    j s    j
appears in the partial fraction.
Where K is a complex number in polar form K=|K|ej=|K| 0
and K* is the complex conjugate of K.
The inverse Laplace transform of the complex-conjugate pair is
*


K
K

1 
L 



 s    j s    j 

2K e
 t
cos(  t   )
Repeated Real Roots
K3
K2
K4
100( s  25) K 1
F ( s) 




3
3
2
s ( s  5)
s5
s( s  5)
( s  5)
100( s  25)
100(25)
K1 

 20
3
125
( s  5) s  0
100( s  25)
100(20)
K2 

 400
s
5
s 5
d 
 s  ( s  25) 
3

K 3  ( s  5) F ( s ) 
 100 
 100

2
s 5
ds
s

 s 5
1 d2 
1
 2 s(25) 
3

K4 
( s  5) F ( s ) 
 100 
 20

2
4
s

5
2 ds
2
 s
 s 5
20
400
100
20
F ( s) 



3
2
s ( s  5)
s5
( s  5)
f ( t )  [20  200t 2e 5 t  100te 5 t  20e 5 t ]u( t )
Repeated Complex Roots
768
768
F ( s)  2

2
( s  6 s  25)
( s  3  j 4)2 ( s  3  j 4)2
K1
K1*
K2
K 2*




2
2
( s  3  j 4) ( s  3  j 4)
( s  3  j 4)
( s  3  j 4)
768
K1 
( s  3  j 4)2
768

 12
2
( j 8)
s 3  j 4

d 
768
2(768)
K2  

2
ds  ( s  3  j 4)  s 3  j 4
( s  3  j 4)3
2(768)
0



j
3

3
-90
( j 8)3
s 3  j 4


12
12
F ( s)  

2
2
( s  3  j 4) 
 ( s  3  j 4)
 3 -900
3 900 
 


 s  3  j4 s  3  j4 
f ( t )  [ 24te
3 t
cos 4t  6e
3 t
cos(4t  90 )]u( t )
0
Improper Transfer Functions
An improper transfer function can always be expanded into a
polynomial plus a proper transfer function.
s 4  13 s 3  66 s 2  200 s  300
F ( s) 
s 2  9 s  20
30 s  100
2
 s  4 s  10  2
s  9 s  20
20
50
2
 s  4 s  10 

s4 s5
d  (t )
d ( t )
4 t
5 t
f (t ) 

4

10

(
t
)

(

20
e

50
e
)u( t )
2
dt
dt
2
POLES AND ZEROS OF F(s)
The rational function F(s) may be expressed as
the ratio of two factored polynomials as
K ( s  z1 )( s  z2 ) ( s  zn )
F ( s) 
( s  p1 )( s  p2 ) ( s  pm )
The roots of the denominator polynomial –p1, -p2,
..., -pm are called the poles of F(s). At these values
of s, F(s) becomes infinitely large. The roots of the
numerator polynomial -z1, -z2, ..., -zn are called the
zeros of F(s). At these values of s, F(s) becomes
zero.
10( s  5)( s  3  j 4)( s  3  j 4)
F ( s) 
s( s  10)( s  6  j 8)( s  6  j 8)
The poles of F(s) are at
0, -10, -6+j8, and –6-j8.
The zeros of F(s)
are at –5, -3+j4, -3-j4
num=conv([1 5],[1 6 25]);
den=conv([1 10 0],[1 12 100]);
pzmap(num,den)
Initial-Value Theorem
The initial-value theorem enables us to determine the
value of f(t) at t=0 from F(s). This theorem assumes that
f(t) contains no impulse functions and poles of F(s),
except for a first-order pole at the origin, lie in the left
half of the s plane.
lim f (t )  lim sF ( s)
t 0

s 
100( s  3)
F ( s) 
2
( s  6)( s  6 s  25)
f ( t )  [12e 6 t  20e 3 t cos(4t  53.130 )]u( t )
Final-Value Theorem
The final-value theorem enables us to determine the behavior of
f(t) at infinity using F(s).
lim f ( t )  lim sF ( s )
t 
s 0
The final-value theorem is useful only if f(∞) exists. This condition is
true only if all the poles of F(s), except for a simple pole at the
origin, lie in the left half of the s plane.
100( s  3)
lim F ( s )  lim s
0
2
s 0
s  0 ( s  6)( s  6 s  25)
lim f ( t )  lim[12e 6t  20e 3t cos(4t  53.130 )]u( t )  0
t 
t 