Laplace Transform

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Transcript Laplace Transform

Laplace
Transform
Prepared By :





Akshay Gandhi
Kalpesh kale
Jatin Patel
Prashant Dhobi
Azad Hudani
: 130460119029
: 130460119038
: 130460119036
: 130460119026
: 130460119031
The French Newton
Pierre-Simon Laplace

Developed mathematics in
astronomy, physics, and
statistics

Began work in calculus which
led to the Laplace Transform

Focused later on celestial
mechanics

One of the first scientists to
suggest the existence of black
holes
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

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

What are Laplace
transforms?

F ( s)  L{ f (t )}   f (t )e  st dt
0
t is real, s is complex!
II.
Note “transform”: f(t)  F(s), where t is
integrated and s is variable
III. Conversely F(s)  f(t), t is variable and s is
integrated
IV. Assumes f(t) = 0 for all t < 0
I.
Laplace Transform Theory
•General Theory
•Example
Laplace Transform for ODEs
•Equation with initial conditions
•Laplace transform is linear
•Apply derivative formula
Table of selected Laplace
Transforms
1
f ( t )  u ( t )  F(s) 
s
1
f ( t )  e u ( t )  F(s) 
sa
 at
s
f ( t )  cos( t )u ( t )  F(s)  2
s 1
1
f ( t )  sin( t )u ( t )  F(s)  2
s 1
More transforms
n!
f ( t )  t u ( t )  F(s)  n 1
s
n
0! 1
n  0, f ( t )  u ( t )  F(s)  1 
s s
1!
n  1, f ( t )  tu ( t )  F(s)  2
s
5! 120
n  5, f ( t )  t 5 u ( t )  F(s)  6  6
s
s
f ( t )  ( t )  F(s)  1
Note on step functions in
Laplace

Unit step function definition:
u ( t )  1, t  0
u ( t )  0, t  0

Used in conjunction with f(t) 
f(t)u(t) because of Laplace integral
limits:

L{f ( t )}   f ( t )e dt
0
st
Properties of Laplace
Transforms
Linearity
 Scaling in time
 Time shift
 “frequency” or s-plane shift
 Multiplication by tn
 Integration
 Differentiation

Properties: Linearity
L{c1f1 (t )  c2f 2 (t )}  c1F1 (s)  c2 F2 (s)
Example : L{sinh( t )} 
Proof :
1 t 1 t
y{ e  e } 
2
2
1
1
L{e t }  L{e  t } 
2
2
1 1
1
(

)
2 s 1 s 1
1 (s  1)  (s  1)
1
(
)

2
s2 1
s2 1
L{c1f1 ( t )  c 2 f 2 ( t )} 

st
[
c
f
(
t
)

c
f
(
t
)]
e
dt 
2 2
 11
0


0
0
c1  f1 ( t )e st dt  c 2  f 2 ( t )e st dt 
c1F1 (s)  c 2 F2 (s)
Properties: Scaling in Time
1 s
L{f (at )}  F( )
a a
Example : L{sin( t )}
1
1
(
 1) 
2
s
 ( )

1
2
( 2
)
2
 s 

s 2  2
Proof :
L{f (at )} 

st
f
(
at
)
e
dt 

0
let
u  at , t 

a
u
1
, dt  du
a
a
s
( ) u
1
f (u )e a du 

a0
1 s
F( )
a a
Properties: Time Shift
L{f ( t  t 0 )u ( t  t 0 )}  e
 a ( t 10)
u ( t  10)} 
Example : L{e
e 10s
sa
Proof :
 st 0
F(s)
L{f ( t  t 0 )u ( t  t 0 )} 

 st
f
(
t

t
)
u
(
t

t
)
e
dt 
0
0

0

 st
f
(
t

t
)
e
dt 
0

t0
let
u  t  t0, t  u  t0
t0
s ( u  t 0 )
f
(
u
)
e
du 

0
e
 st 0

 st 0
 su
f
(
u
)
e
du

e
F(s)

0
Properties: S-plane
(frequency) shift
 at
L{e f (t )}  F(s  a )
Example : L{e  at sin( t )} 

(s  a ) 2  2
Proof :
L{e  at f ( t )} 

 at
st
e
f
(
t
)
e
dt 

0

(s  a ) t
f
(
t
)
e
dt 

0
F(s  a )
Properties: Multiplication
by tn n
n
n d
L{t f ( t )}  (1)
F
(
s
)
n
ds
Example :
Proof :
L{t n u ( t )} 
(1) n
n!
s n 1

L{t n f ( t )}   t n f ( t )e st dt 
0
n
d 1
( )
n
ds s

n st
f
(
t
)
t
e dt 

0

n

(1) n  f ( t ) n e st dt 
s
0

n
n
st
n 
(1)
f ( t )e dt (1)
F(s)
n 
n
s 0
s
n
The “D” Operator
1.
2.
Differentiation shorthand
Integration
t
if
g( t )   f ( t )dt

then
Dg ( t )  f ( t )
df ( t )
Df ( t ) 
dt
d2
2
D f (t)  2 f (t)
dt
shorthand
t
if g( t )   f ( t )dt
a
1
then g( t )  D a f ( t )
Difference in 

f (0 ), f (0 ) & f (0)
The values are only different if f(t)
is not continuous @ t=0
 Example of discontinuous function:
u(t)

f (0  )  lim u ( t )  0
t 0
f (0  )  lim u ( t )  1
t 0
f (0)  u (0)  1
Properties: Nth order
derivatives
2
L{D f ( t )}  ?
let
g( t )  Df ( t ), g(0)  Df (0)  f ' (0)
 L{D 2 g( t )}  sG (s)  g(0)  sL{Df ( t )}  f ' (0)
 s(sF(s)  f (0))  f ' (0)  s 2 F(s)  sF(0)  f ' (0)
L{Dn f (t )}  s n F(s)  s( n 1) f (0)  s( n 2) f ' (0)    sf ( n 2)' (0)  f ( n 1)' (0)
NOTE: to take L{D n f ( t )}
you need the value @ t=0 for
Dn 1f (t ), Dn 2f (t ),...Df (t ), f (t )  called initial conditions!
We will use this to solve differential equations!
Real-Life Applications
Semiconductor
mobility
Call completion in
wireless networks
Vehicle vibrations on
compressed rails
Behavior of magnetic
and electric fields
above the atmosphere
THANK YOU