EC3354 SIGNALS AND SYSTEM LAPLACE TRANSFORM

1. EC3354 SIGNALS AND SYSTEMS
UNIT III
Laplace Transform
Mrs.Mohanapriya S
AP/ECE
JCT CET
JCT COLLEGE OF ENGINEERING AND TECHNOLOGY
PICHANUR, COIMBATORE – 641105

2. Why use Laplace Transforms?
• Find solution to differential equation using
algebra
• Relationship to Fourier Transform allows easy
way to characterize systems
• No need for convolution of input and
differential equation solution
• Useful with multiple processes in system

3. How to use Laplace
• Find differential equations that describe
system
• Obtain Laplace transform
• Perform algebra to solve for output or variable
of interest
• Apply inverse transform to find solution

4. What are Laplace transforms?
















j
j
st
1
0
st
ds
e
)
s
(
F
j
2
1
)}
s
(
F
{
L
)
t
(
f
dt
e
)
t
(
f
)}
t
(
f
{
L
)
s
(
F
• t is real, s is complex!
• Inverse requires complex analysis to solve
• Note “transform”: f(t)  F(s), where t is integrated and s is
variable
• Conversely F(s)  f(t), t is variable and s is integrated
• Assumes f(t) = 0 for all t < 0

5. Evaluating F(s) = L{f(t)}
• Hard Way – do the integral

 



 




















0
st
0 0
t
)
a
s
(
st
at
at
0
st
dt
)
t
sin(
e
)
s
(
F
t
sin
)
t
(
f
a
s
1
dt
e
dt
e
e
)
s
(
F
e
)
t
(
f
s
1
)
1
0
(
s
1
dt
e
)
s
(
F
1
)
t
(
f
let
let
let
Integrate by parts

6. Evaluating F(s)=L{f(t)}- Hard Way
remember  

 vdu
uv
udv
)
t
cos(
v
,
dt
)
t
sin(
dv
dt
se
du
,
e
u st
st





 


 



 











0
st
st
0 0
st
0
st
st
dt
)
t
cos(
e
s
)
1
(
e
dt
)
t
cos(
e
s
)
t
cos(
e
[
dt
)
t
sin(
e ]
)
t
sin(
v
,
dt
)
t
cos(
dv
dt
se
du
,
e
u st
st




 




















0
st
st
0
st
0
st
0
st
dt
)
t
sin(
e
s
)
0
(
e
dt
)
t
sin(
e
s
)
t
sin(
e
[
dt
)
t
cos(
e
]
2
0
st
0
st
2
0 0
st
2
st
s
1
1
dt
)
t
sin(
e
1
dt
)
t
sin(
e
)
s
1
(
dt
)
t
sin(
e
s
1
dt
)
t
sin(
se









 




 


let
let
Substituting, we get:
It only gets worse…

7. Evaluating F(s) = L{f(t)}
• This is the easy way ...
• Recognize a few different transforms
• See table 2.3 on page 42 in textbook
• Or see handout ....
• Learn a few different properties
• Do a little math

8. Table of selected Laplace Transforms
1
s
1
)
s
(
F
)
t
(
u
)
t
sin(
)
t
(
f
1
s
s
)
s
(
F
)
t
(
u
)
t
cos(
)
t
(
f
a
s
1
)
s
(
F
)
t
(
u
e
)
t
(
f
s
1
)
s
(
F
)
t
(
u
)
t
(
f
2
2
at

















9. More transforms
1
n
n
s
!
n
)
s
(
F
)
t
(
u
t
)
t
(
f 



6
6
5
2
1
s
120
s
!
5
)
s
(
F
)
t
(
u
t
)
t
(
f
,
5
n
s
!
1
)
s
(
F
)
t
(
tu
)
t
(
f
,
1
n
s
1
s
!
0
)
s
(
F
)
t
(
u
)
t
(
f
,
0
n














1
)
s
(
F
)
t
(
)
t
(
f 




10. Note on step functions in Laplace




0
st
dt
e
)
t
(
f
)}
t
(
f
{
L
• Unit step function definition:
• Used in conjunction with f(t)  f(t)u(t)
because of Laplace integral limits:
0
t
,
0
)
t
(
u
0
t
,
1
)
t
(
u





11. Properties of Laplace Transforms
• Linearity
• Scaling in time
• Time shift
• “frequency” or s-plane shift
• Multiplication by tn
• Integration
• Differentiation

12. Properties: Linearity
)
s
(
F
c
)
s
(
F
c
)}
t
(
f
c
)
t
(
f
c
{
L 2
2
1
1
2
2
1
1 


Example :
1
s
1
)
1
s
)
1
s
(
)
1
s
(
(
2
1
)
1
s
1
1
s
1
(
2
1
}
e
{
L
2
1
}
e
{
L
2
1
}
e
2
1
e
2
1
{
y
)}
t
{sinh(
L
2
2
t
t
t
t

















Proof :
)
s
(
F
c
)
s
(
F
c
dt
e
)
t
(
f
c
dt
e
)
t
(
f
c
dt
e
)]
t
(
f
c
)
t
(
f
c
[
)}
t
(
f
c
)
t
(
f
c
{
L
2
2
1
1
0
st
2
2
0
st
1
1
st
2
2
0
1
1
2
2
1
1

















13. )
a
s
(
F
a
1
)}
at
(
f
{
L 
Example :
2
2
2
2
2
2
s
)
s
(
1
)
1
)
s
(
1
(
1
)}
t
{sin(
L












 Proof :
)
a
s
(
F
a
1
du
e
)
u
(
f
a
1
du
a
1
dt
,
a
u
t
,
at
u
dt
e
)
at
(
f
)}
at
(
f
{
L
a
0
u
)
a
s
(
0
st












let
Properties: Scaling in Time

14. Properties: Time Shift
)
s
(
F
e
)}
t
t
(
u
)
t
t
(
f
{
L 0
st
0
0




Example :
a
s
e
)}
10
t
(
u
e
{
L
s
10
)
10
t
(
a






Proof :
)
s
(
F
e
du
e
)
u
(
f
e
du
e
)
u
(
f
t
u
t
,
t
t
u
dt
e
)
t
t
(
f
dt
e
)
t
t
(
u
)
t
t
(
f
)}
t
t
(
u
)
t
t
(
f
{
L
0
0
0
0
0
st
0
su
st
t
0
)
t
u
(
s
0
0
t
st
0
0
st
0
0
0
0






























let

15. Properties: S-plane (frequency) shift
)
a
s
(
F
)}
t
(
f
e
{
L at



Example :
2
2
at
)
a
s
(
)}
t
sin(
e
{
L






 Proof :
)
a
s
(
F
dt
e
)
t
(
f
dt
e
)
t
(
f
e
)}
t
(
f
e
{
L
0
t
)
a
s
(
0
st
at
at














16. Properties: Multiplication by tn
)
s
(
F
ds
d
)
1
(
)}
t
(
f
t
{
L n
n
n
n


Example :
1
n
n
n
n
n
s
!
n
)
s
1
(
ds
d
)
1
(
)}
t
(
u
t
{
L




Proof :
)
s
(
F
s
)
1
(
dt
e
)
t
(
f
s
)
1
(
dt
e
s
)
t
(
f
)
1
(
dt
e
t
)
t
(
f
dt
e
)
t
(
f
t
)}
t
(
f
t
{
L
n
n
n
0
st
n
n
n
0
st
n
n
n
0
st
n
0
st
n
n



























17. The “D” Operator
1. Differentiation shorthand
2. Integration shorthand
)
t
(
f
dt
d
)
t
(
f
D
dt
)
t
(
df
)
t
(
Df
2
2
2


)
t
(
f
)
t
(
Dg
dt
)
t
(
f
)
t
(
g
t

 

)
t
(
f
D
)
t
(
g
dt
)
t
(
f
)
t
(
g
1
a
t
a


 
if
then then
if

18. Properties: Integrals
s
)
s
(
F
)}
t
(
f
D
{
L 1
0 

Example :
)}
t
{sin(
L
1
s
1
)
1
s
s
)(
s
1
(
)}
t
cos(
D
{
L
2
2
1
0





Proof :
let
st
st
0
st
1
0
e
s
1
v
,
dt
e
dv
dt
)
t
(
f
du
),
t
(
g
u
dt
e
)
t
(
g
)}
t
{sin(
L
)
t
(
f
D
)
t
(
g



















 


t
0
st
0
st
dt
)
t
(
f
)
t
(
g
s
)
s
(
F
dt
e
)
t
(
f
s
1
]
e
)
t
(
g
s
1
[








0
)
(
)
( dt
e
t
f
t st
If t=0, g(t)=0
for so
slower than





0
)
(
)
( t
g
dt
t
f 0

st
e

19. Properties: Derivatives
(this is the big one)
)
0
(
f
)
s
(
sF
)}
t
(
Df
{
L 


Example :
)}
t
sin(
{
L
1
s
1
1
s
)
1
s
(
s
1
1
s
s
)
0
(
f
1
s
s
)}
t
cos(
D
{
L
2
2
2
2
2
2
2
2















Proof :
)
s
(
sF
)
0
(
f
dt
e
)
t
(
f
s
)]
t
(
f
e
[
)
t
(
f
v
,
dt
)
t
(
f
dt
d
dv
se
du
,
e
u
dt
e
)
t
(
f
dt
d
)}
t
(
Df
{
L
0
st
0
st
st
st
0
st





















let

20. Difference in
• The values are only different if f(t) is not
continuous @ t=0
• Example of discontinuous function: u(t)
)
0
(
f
&
)
0
(
f
),
0
(
f 

1
)
0
(
u
)
0
(
f
1
)
t
(
u
lim
)
0
(
f
0
)
t
(
u
lim
)
0
(
f
0
t
0
t













21. ?
)}
t
(
f
D
{
L 2

)
0
(
'
f
)
0
(
sF
)
s
(
F
s
)
0
(
'
f
))
0
(
f
)
s
(
sF
(
s
)
0
(
'
f
)}
t
(
Df
{
sL
)
0
(
g
)
s
(
sG
)}
t
(
g
D
{
L
)
0
(
'
f
)
0
(
Df
)
0
(
g
),
t
(
Df
)
t
(
g
2
2














let
)
0
(
f
)
0
(
sf
)
0
(
'
f
s
)
0
(
f
s
)
s
(
F
s
)}
t
(
f
D
{
L )'
1
n
(
)'
2
n
(
)
2
n
(
)
1
n
(
n
n 








 
NOTE: to take
you need the value @ t=0 for
called initial conditions!
We will use this to solve differential equations!



)
t
(
f
),
t
(
Df
),...
t
(
f
D
),
t
(
f
D 2
n
1
n
)}
t
(
f
D
{
L n
Properties: Nth order derivatives

22. Properties: Nth order derivatives
)
0
(
f
)
s
(
sF
)}
t
(
Df
{
L 

)}
t
(
f
D
{
L 2
)
0
(
f
)
s
(
sF
)}
t
(
Df
{
L
)}
t
(
g
{
L
)
s
(
G
)
0
(
'
f
)
0
(
g
)
t
(
Df
)
t
(
g
)
0
(
g
)
s
(
sG
)}
t
(
Dg
{
L
)
t
(
f
D
)
t
(
Dg
and
)
t
(
Df
)
t
(
g 2











)
0
(
'
f
)
0
(
sf
)
s
(
F
s
)
0
(
'
f
)]
0
(
f
)
s
(
sF
[
s
)
0
(
g
)
s
(
sG
)}
t
(
Dg
{
L 2









.
etc
),
t
(
f
D
),
t
(
f
D 4
3
Start with
Now apply again
let
then
remember
Can repeat for
)
0
(
f
)
0
(
sf
)
0
(
'
f
s
)
0
(
f
s
)
s
(
F
s
)}
t
(
f
D
{
L )'
1
n
(
)'
2
n
(
)
2
n
(
)
1
n
(
n
n 








 