Lesson 18 6CT.1-4 - Information Services and...

BME 333 Biomedical Signals and Systems - J.Schesser

31

Laplace Transforms

Lesson 186CT.1-4


32

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


33

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


34

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


35

An example

se

dtedte

dteatutusF

atututf

saa

stst

st

1

)]()([)(

)()()(

0

0


36

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


37

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


38

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


39

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)


40

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


41

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


42

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)]


43

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


44

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


45

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


46

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


47

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


48

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


49

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


50

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


51

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


52

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

Lesson 18 6CT.1-4 - Information Services and...

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