BME 333 Biomedical Signals and Systems - J.Schesser 31 Laplace Transforms Lesson 18 6CT.1-4
BME 333 Biomedical Signals and Systems - J.Schesser
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Laplace Transforms
Lesson 186CT.1-4
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Laplace Transform
• Up till now we have generally assumed that the quadratic content of a signal is finite (unless there are impulses) and that the signal may exist for t<0.
• More realistically, these assumptions are too limiting. In fact, f(t) = 0 for t<0 and we can not always guarantee the quadratic content. Therefore:
may not converge.
0)()()( dtetfdtetfjF tjtj
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Laplace Transform #2
• Let’s look at the following formulation such that the there is a real positive number σwhich makes the integral converge
even if is not finite.
0)( dtetf t
0)( dttf
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Laplace Transform #3 • Let’s then consider:
note that F(s) is not the FT of f(t)• F(s) is called the Laplace Transform of f(t)
£[f(t)]=F(s)• Note that for functions which are zero for t<0 and
have finite content F(s) F(s)|s=jω =F(jω)
( )
0
0
[ ( ) ] ( ) If ( ) 0, 0, then
[ ( ) ] ( ) ( )
Let's define ; Re( ) 0; ( ) 0, 0; then ( ) ( )
t t j t
t j t
st
f t e f t e e dt f t t
f t e f t e dt F j
s σ j s f t t F s f t e dt
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An example
se
dtedte
dteatutusF
atututf
saa
stst
st
1
)]()([)(
)()()(
0
0
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Inverse Laplace TransformRecall that
By analogy
1
-1
( ) ( )
0
( ) ( )
0
1[ ( )] ( ) ( )2
1£ [ ( )] ( ) ( ) ; since is considered a constant2
1 ( ) ( ) ( ) ( )2
1 { ( ) }2
1
j t
st
j t j
t
j j t
t
F j f t F j e d
F s f t F s e d
F j e d F s F j f e d
f e d e d
( ) ( )
0
( ) ( )
0
( )
0
( ) 2
1( ) { } 2
( ) { ( )}
( )
t j t
t
t j t
t
f e e d d
f e e d d
f e t d
f t
1Recall: [1] ( )1 1
2j t
t
e d
Recall: ( ) ( )
( )
f t t dt
f
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Inverse Laplace TransformAlso
0
1
1
£[ ( )] ( ) ( )
( ) ; since ( ) 0 for t 0
( ) [ ( ) ]
£[ ( )] ( ) [ ( ) ]And [ ( )] ( )
1[ ( )] ( ) ( )2
And
st
st
t j t t j t
t
t
t j t
f t F s f t e dt
f t e dt f t
f t e e dt f t e e dt
f t F s f t eF s f t e
F j f t e F j e d
f
( )1 1( ) ( ) ( ) 2 2
now substitute ;
1( ) ( )2
t j t j t
s jst
s j
t e F j e d F j e d
s j ds jd j s j
f t F s e dsj
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Some Properties• Superposition holds
£[f1(t)]=F1(s); £[f2(t)]=F2(s)£[f1(t)+ f2(t)]=F1(s)+F2(s)
• Differentiation
This seems clumsy – what is f(0) ?
0
00 0
0
( ) ( )]
( )
£
( )
[
( )( )
0 (0) ( ) ( ) (0)
st
st st st
st
df t df t e dtdt dt
df t e dt e f t f t se dtdt
f s f t e dt sF s f
Integration by parts, we get
( ) ;
( );
st
st
udv uv vdudf tdv dt u e
dtv f t du se dt
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Some Properties #2Let’s look at du(t)/dt = δ(t)
£[δ(t)] = £ [du(t)/dt]=s £ u(t)-u(0)
=
= 1- u(0)What is u(0)? If we choose u(0)=u(0+) =1, then £[δ(t)] =0;otherwise if we use u(0)=u(0-)=0, then £[δ(t)] =1.This is a better choice!!! Henceforth, we use:
£[df (t)/dt]=sF(s)-f (0-)
0)()( dtetfsF st
s est dt0
u(0)
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Some Properties #3
• Higher Order Derivatives
• Integration
11 2
0 01
( ) ( ) ( )[ ] ( ) (0) | |n n
n n nt tn n
d f t df t d f t£ s F s s f sdt dt dt
)0(1)()(
)()0()(
)()(
)()(
gsssFsG
sFgssG
tfdttdg
dftgt
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Initial Conditions
• To simplify this, we can assume that f(t)=fa(t)u(t) and all the initial conditions are zero.
• If we need to have systems which are not 0 at t=0-, then we can add special sources (called initial condition generators) to represent these conditions.
• So £[df (t)/dt]=sF(s)
£[dnf (t)/dtn]=snF(s)£[ ]=F(s)/sf (t)dt
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Impulse Response
• Recall:A(p)y(t)=B(p)x(t)A(p)h(t)=B(p)δ(t)
£[A(p)h(t)]= £[ B(p)δ(t)]A(s)H(s) = B(s)1H(s) = B(s) / A(s)h(t) = £-1[H(s)]
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Some Transforms£[ (t)] 1
£[u(t)] 1s
£[tu(t)] 1s2
£[ 12
t 2u(t)] 1s3 ;£[t 2u(t)] 2
s3
£[ 1n!
tnu(t)] 1sn1 ;£[t nu(t)] n!
sn1
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Frequency Displacement
1 1 1( )1
0 0
-
2 2
2 2
£[ ( )] ( )
£[ ( ) ] ( ) ( ) ( )
1£[ ( )]
1£[ ( )]
1 1 1£[cos ( )] { }2
1 1 1£[sin ( )] { }2
s t s t s s tst
t
j t
f t F s
f t e f t e e dt f t e dt F s s
then
e u ts
e u ts j
t u ts j s js
s
t u tj s j s j
s
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Examples
-4 -4
2 2 2 2
2
2
Example #1 £[ cos (10 -30 ) ( )] £[ {.866cos 10 .5sin10 } ( )]
.866( 4) .5 10( 4) 10 ( 4) 10
Example #22 4
2 52( 1) 2( 1) 4
( ) (2cos 2 sin 2 ) ( )
t t
t
e t u t e t t u ts
s s
sF(s)s s
ss
f t e t t u t
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Time Displacement
)(
)(
)(
Let
)()(0)()(
:Proof)()]()(£[)()]()(£[
0
0
)(
00
sFe
dxexfe
dxexf
t-Tx
dtet-Tut-Tfdtt-Tut-Tf
esFt-Tut-TfsFtutf
sT
sxsT
Txs
T
-stT
sT
]1[1)(
)()()()(
22 se
sTsV
TtuTTttuT
ttv
Example
sT
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Convolution1 2 1 2
0
1 20 0
1 20 0
( )1 2
0 0
1 2
£[ ( ) ( ) ] £[ ( ) ( ) ]
[ ( ) ( ) ]
since 0 ; 0 0
( )[ ( ) ]
Let -
( )[ ( ) ]
( ) [ ( )
t t
st
t
st
t
s x
t
s
t
f f t d f f t d
f f t d e dt
t t
f f t e dt d
x t
f f x e dx d
f e f x
1 20 0 0
1 2
] ( ) [ ( )]
( ) ( )
sx se dx d f e F s d
F s F s
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Convolution
1 2 1 20
1 20
( )1 2
0
1 2
£[ ( ) ( )] ( ) ( )
1[ ( ) ] ( )2
1 [ ( )][ ( ) ]2
1 [ ( )][ ( )]2
tst
w jtwt st
w j
w j ts w t
w j
w j
w j
f t f t f t f t e dt
F w e dw f t e dtj
F w f t e dt dwj
F w F s w dwj
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De-Convolution Example( ) ( )
0
2
sin( ) ( ) ( ) ( ) ( )
What is ( ) ? Note the above equation is equivalent to theconvolution of ( ) with ( ).
1 1£[sin ( ) ( )] and £[ ( )]=1 1
an
t tt t
-t
-t
t u t f e u t d f e d
f tf t e u t
t u t e u ts s
2
2
d the LT of the convolution is:1 ( )
11 1 ( )
1 11( )1
( ) {cos sin } ( )
F ss
F ss s
sF ss
f t t t u t
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Homework
• Problems: 3.1a,b,
)()( stepunit an todue response Plot the))()( i.e., response, impulse Plot the )
min5.0min,75.0,0.1Let
.constant time theis and tominutesin delay time theis
constant, response spatient' theis K ight/min.patient we mg/kgin rateinfusion IV SNP specific theis Q MAP,in change induced-SMP theis where
)(1
)(
us Sheppard) and Slateby (developed (SNP) idenitorpruss sodium of infusion IVan following pressure aterialmean for model following The b a, 3.1
p
tutUbttQa
K
MAP
MAP
sQseK
sMAP
ppp
pp
p
sp
p
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Homework
• Problems: 3.4a,
• 3.12a,b
5 0.1
The unit impulse response is given as( ) (0.7 0.2 0.1 ) ( )
Find the transfer function ( ) and its poles and zeroes.
t t th t e e e u tH s
)( :response impulse and )(function transfer theFind)(3
as described is system LTIA
thsHtxyy
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Homework
• Problems: 13a,b
• Sketch f (t) = tu(t)u(1-t) and find F(s) • Find f (t), if F(s) = (1-e-2s)/s2. Repeat
for F(s) = (1-s+e-2s)/s3
• 6CT.2.1• 6CT.2.2
)( :response impulse and )(function transfer theFind
)(5158as described is system LTIA
thsHtxyyy