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Laplace Transform

BIOE 4200

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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


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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


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What are Laplace transforms?


F(s)  L{f ( t )}   f ( t )e st dt
0
  j

1
1
st
f ( t )  L {F(s)} 
F
(
s
)
e
ds

2j  j


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
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Evaluating F(s) = L{f(t)}

let

Hard Way – do the integral
f (t)  1


1
1
F(s)   e st dt   (0  1) 
s
s
0

let

f ( t )  e at




0

0

F(s)   e at e st dt   e (s  a ) t dt 

let

1
sa

f ( t )  sin t


F(s)   e st sin( t )dt

0
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Integrate by parts

Evaluating F(s)=L{f(t)}- Hard Way
 udv  uv   vdu

remember
let

u  e st , du  se st dt
dv  sin( t )dt, v   cos( t ) 


st
st
  e sin( t )dt  [e cos( t ) ]  s  e st cos( t )dt 
0

0

0


 e (1)  s  e st cos( t )dt
st



 se



0

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e sin( t ) ]  s  e sin( t )dt  e
0

0

0

st
e
 sin( t )dt 

0
st

st
e
 sin( t )dt 

0

  e st cos( t )dt 


sin( t )dt  1  s

(1  s 2 )  e st sin( t )dt 1






2



u  e st , du  se st dt
dv  cos( t )dt, v  sin( t )

st

st

0

0

let

Substituting, we get:

st



(0)  s  e sin( t )dt
0

st

1
1  s2

It only gets worse…

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


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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

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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
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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

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st

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

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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

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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

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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

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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 )

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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



L{t n f ( t )}   t n f ( t )e st dt 
0

n

d 1
( )
n
ds s

n!
s n 1



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

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The “D” Operator
1.

2.

Differentiation shorthand

Integration shorthand
t

if

g( t )   f ( t )dt


then

Dg ( t )  f ( t )

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df ( t )
Df ( t ) 
dt
d2
2
D f (t)  2 f (t)
dt

t

if g( t )   f ( t )dt
a
1
then g( t )  D a f ( t )

Properties: Integrals

F(s)
L{D f ( t )} 
s
1
0

Proof :

g ( t )  D 01f ( t )


L{sin( t )}   g ( t )e st dt

Example : L{D 01 cos( t )} 

0

let u  g( t ), du  f ( t )dt

1
s
1
( )( 2 )  2
s s 1 s 1
L{sin( t )}

1
dv  e st dt , v   e st
s
1
1
F(s)
st 
st
 [ g( t )e ]0   f ( t )e dt 
s
s
s
t

g( t )   f ( t )dt If t=0, g(t)=0
0



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
0



for (t  )   f (t )e  st dt   so
0
f (t )dt  g (t )   slower than e st  0

Properties: Derivatives
(this is the big one)


L{Df (t )}  sF(s)  f (0 )
Example : L{D cos( t )} 
s2


f
(
0
)
2
s 1
s2
1 
2
s 1
s 2  (s 2  1)
s2  1
1
 L{ sin( t )}
2
s 1

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Proof :

let



d
f ( t )e st dt
dt
0

L{Df ( t )}  

u  e st , du  se st
d
dv  f ( t )dt , v  f ( t )
dt


[e st f ( t )]0  s  f ( t )e st dt 
0

 f (0  )  sF(s)





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

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Properties: Nth order derivatives

L{D f ( t )}  ?
2

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!
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Properties: Nth order derivatives
Start with L{Df (t )}  sF(s)  f (0)
Now apply again L{D2f (t )}
let g( t )  Df ( t ) and Dg ( t )  D 2f ( t )
then L{Dg ( t )}  sG (s)  g(0)
remember g( t )  Df ( t )
 g(0)  f ' (0)
G (s)  L{g( t )}  L{Df ( t )}  sF(s)  f (0)

 L{Dg (t )}  sG(s)  g(0)  s[sF(s)  f (0)]  f ' (0)  s 2 F(s)  sf (0)  f ' (0)

Can repeat for D3f (t ), D4f (t ), etc.
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)

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Relevant Book Sections
Modeling - 2.2
 Linear Systems - 2.3, page 38 only
 Laplace - 2.4
 Transfer functions – 2.5 thru ex 2.4


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