The Laplace Transform
The University of Tennessee
Electrical and Computer Engineering Department
Knoxville, Tennessee
wlg
The Laplace Transform
The Laplace Transform of a function, f(t), is defined as;

L[ f (t )]  F ( s)   f (t )e dt
 st
Eq A
0
The Inverse Laplace Transform is defined by
1
L [ F ( s)]  f (t ) 
*notes
1
  j
Eq B
F
(
s
)
e
ds

2 j   j
ts
The Laplace Transform
We generally do not use Eq B to take the inverse Laplace. However,
this is the formal way that one would take the inverse. To use
Eq B requires a background in the use of complex variables and
the theory of residues. Fortunately, we can accomplish the same
goal (that of taking the inverse Laplace) by using partial fraction
expansion and recognizing transform pairs.
*notes
The Laplace Transform
Laplace Transform of the unit step.

 1  st 
L[u(t )]   1e dt  e |
0
s
0
 st
1
L[u(t )] 
s
The Laplace Transform of a unit step is:
*notes
1
s
The Laplace Transform
The Laplace transform of a unit impulse:
Pictorially, the unit impulse appears as follows:
(t – t0)
f(t)
t0
0
Mathematically:
(t – t0) = 0 t
*note

t 0 
0
 (t  t )dt 1
0
t0 
 0
The Laplace Transform
The Laplace transform of a unit impulse:
An important property of the unit impulse is a sifting
or sampling property. The following is an important.
t2

t1
t1  t 0  t 2
 f (t 0 )
f (t ) (t  t 0 )dt  
t 0  t1 , t 0  t 2
0
The Laplace Transform
The Laplace transform of a unit impulse:
In particular, if we let f(t) = (t) and take the Laplace

L[ (t )]    (t )e dt  e
0
 st
0 s
1
The Laplace Transform
An important point to remember:
f (t )  F ( s)
The above is a statement that f(t) and F(s) are
transform pairs. What this means is that for
each f(t) there is a unique F(s) and for each F(s)
there is a unique f(t). If we can remember the
Pair relationships between approximately 10 of the
Laplace transform pairs we can go a long way.
The Laplace Transform
Building transform pairs:

e
L[e u(t )]   e e dt   e
at
0
at  st
L(e
( s  a ) t
0
 st
e
1

L[e u( t )] 
|0 
( s  a)
sa
 at
A transform
pair
 at
e u( t )

1
sa
dt
The Laplace Transform
Building transform pairs:

L[tu(t )]   te dt
 st
0



 udv  uv |   vdu
0
0
u=t
dv = e-stdt
0
tu(t )

1
2
s
A transform
pair
The Laplace Transform
Building transform pairs:

(e jwt  e  jwt )  st
L[cos( wt )]  
e dt
2
0
1 1
1 
 


2  s  jw s  jw 
s
 2
s  w2
cos( wt )u(t )

s
s2  w2
A transform
pair
The Laplace Transform
Time Shift

L[ f (t  a )u(t  a )]   f (t  a )e  st
a
Let x  t  a , then dx  dt and t  x  a
As t  a , x  0 and as t  , x  . So,


0

f ( x )e  s ( x  a ) dx  e as  f ( x )e  sx dx
0
L[ f (t  a)u(t  a)]  e
 as
F ( s)
The Laplace Transform
Frequency Shift
L[e
 at

f (t )]   [e
 at
 st
f (t )]e dt
0


 f ( t )e
( s  a ) t
dt  F ( s  a )
0
L[e
 at
f (t )]  F ( s  a)
The Laplace Transform
Example: Using Frequency Shift
Find the L[e-atcos(wt)]
In this case, f(t) = cos(wt) so,
s
F ( s)  2
s  w2
(s  a)
and F ( s  a ) 
(s  a)2  w 2
L[e
 at
( s  a)
cos( wt )] 
( s  a)2  (w)2
The Laplace Transform
Time Integration:
The property is:

 t
  st
L  f (t )dt      f ( x )dx e dt
0
 0 0

Integrate by parts :
t
Let u   f ( x )dx , du  f (t )dt
0
and
 st
dv  e dt ,
1  st
v  e
s
The Laplace Transform
Time Integration:
Making these substitutions and carrying out
The integration shows that

 1
L  f (t )dt    f (t )e  st dt
0
 s0
1
 F ( s)
s
The Laplace Transform
Time Differentiation:
If the L[f(t)] = F(s), we want to show:
df (t )
L[
]  sF ( s )  f (0)
dt
Integrate by parts:
u  e , du   se dt and
df ( t )
dv 
dt  df ( t ), so v  f ( t )
dt
 st
*note
 st
The Laplace Transform
Time Differentiation:
Making the previous substitutions gives,
 df 
L    f ( t )e
 dt 

|  f (t ) se dt
 st 
0
0

 0  f (0)  s  f (t )e  st dt
0
So we have shown:
 df (t ) 
L
 sF ( s)  f (0)

 dt 
 st
The Laplace Transform
Time Differentiation:
We can extend the previous to show;
 df (t ) 2  2
L
 s F ( s )  sf (0)  f ' (0)
2 
 dt 
 df (t )3 
3
2
L

s
F
(
s
)

s
f (0)  sf ' (0)  f ' ' (0)
3 
 dt 
general case
 df (t ) n 
n
n 1
n2
L

s
F
(
s
)

s
f
(
0
)

s
f ' (0)
n 
 dt 
 ...  f ( n1) (0)
The Laplace Transform
Transform Pairs:
f(t)
 (t )
F(s)
1
1
f
(
t
)
F
(
s
)
u( t ) ____________________________________
s
1
 st
e
sa
1
t
s2
n!
n
t
n 1
s
The Laplace Transform
Transform Pairs:
f(t)
te
 at
n  at
t e
sin( wt )
cos( wt )
F(s)
1
s  a 2
n!
( s  a )n 1
w
s2  w2
s
s2  w2
The Laplace Transform
Transform Pairs:
f(t)
e at sin( wt )
e
 at
cos( wt )
sin( wt   )
cos( wt   )
F(s)
w
(s  a)2  w 2
sa
2
2
(s  a)  w
s sin   w cos 
s2  w2
s cos   w sin 
s2  w2
Yes !
The Laplace Transform
Common Transform Properties:
f(t)
F(s)
f ( t  t 0 )u( t  t 0 ), t 0  0
f ( t )u( t  t 0 ), t  0
e  at f ( t )
d n f (t )
dt n
e
e
 to s
 to s
F ( s)
L[ f ( t  t 0 )
F (s  a)
s n F ( s )  s n  1 f ( 0)  s n  2 f ' ( 0)  . . .  s 0 f n  1 f ( 0)
dF ( s )
ds
tf ( t )

t
1
F ( s)
s
 f ( )d
0
The Laplace Transform
Using Matlab with Laplace transform:
Example
Use Matlab to find the transform of
te  4 t
The following is written in italic to indicate Matlab code
syms t,s
laplace(t*exp(-4*t),t,s)
ans =
1/(s+4)^2
The Laplace Transform
Using Matlab with Laplace transform:
Example
F ( s) 
Use Matlab to find the inverse transform of
s ( s  6)
( s  3)( s  6 s  18)
2
prob .12.19
syms s t
ilaplace(s*(s+6)/((s+3)*(s^2+6*s+18)))
ans =
-exp(-3*t)+2*exp(-3*t)*cos(3*t)
The Laplace Transform
Theorem:
Initial Value Theorem:
If the function f(t) and its first derivative are Laplace transformable and f(t)
Has the Laplace transform F(s), and the lim sF ( s ) exists, then
s
lim sF ( s )  lim f ( t )  f (0)
s
t 0
Initial Value
Theorem
The utility of this theorem lies in not having to take the inverse of F(s)
in order to find out the initial condition in the time domain. This is
particularly useful in circuits and systems.
The Laplace Transform
Example: Initial Value Theorem:
Given;
F ( s) 
( s  2)
( s  1)2  5 2
Find f(0)


s 2  2s
f (0)  lim sF ( s )  lim s
 lim  2

2
2
s
s
s
( s  1)  5
 s  2 s  1  25 
( s  2)
s2 s2  2 s s2
 lim
s
s
2
s  2 s s  ( 26 s )
2
2
2
1
The Laplace Transform
Theorem:
Final Value Theorem:
If the function f(t) and its first derivative are Laplace transformable and f(t)
has the Laplace transform F(s), and the lim sF ( s ) exists, then
s
lim sF ( s )  lim f ( t )  f ( )
s0
t 
Final Value
Theorem
Again, the utility of this theorem lies in not having to take the inverse
of F(s) in order to find out the final value of f(t) in the time domain.
This is particularly useful in circuits and systems.
The Laplace Transform
Example: Final Value Theorem:
Given:
F ( s) 
( s  2) 2  3 2
( s  2)
2
 32

note F 1 ( s )  te  2 t cos 3t
Find f ( ) .
f ( )  lim sF ( s )  lim s
s0
s0
( s  2) 2  3 2
( s  2)
2
3
2

0