lesson #19 6ct.5-7joelsd/signals/classwork/bme314signalscw19.pdf · bme 333 biomedical signals and...

BME 333 Biomedical Signals and Systems - J.Schesser

53

Solving Systems using Laplace Transforms

Lesson #19 6CT.5-7


54

Inversion of the LT by Partial Fraction Expansion

F (s) N (s)D(s)

F0 (sm b1s

m1 bm )(sn a1s

n1 an )Like a System Function

V2 (s)V1(s)

H (s) B(s)A(s)

Simple Roots:

F (s) N (s)(s s1)(s s2 )(s sn )

;s1 s2 sn

F (s) K1

(s s1)

K2

(s s2 )

Kn

(s sn )

f (t) (K1es1t K2e

s2t Knesnt )u(t)


55

Finding the Ks

n

n

ssn

nssnn

ssssss

n

n

n

sssDsNsFssK

sssDsN

sDsNsssFssK

ssK

ssK

ssK

sssssssN

sDsNsF

|)()()(|)()(

|)()()(|)(

)()(|)()(

)()()(

)())(()(

)()()(

1

11

1

1111

2

2

1

1

21


56

Example

224

)13(17)3(13)3(2

326

)31(17)1(13)1(2)3()1(2

)3)(1()(2

)3)(1(17132)(

2

2

2

1

21

2

K

K

sK

sK

sssN

sssssF

)(]23[)(2)( 3 tueettf tt


57

Complex RootsF (s)

K1

s s1

K1

s s1

Kn

s sn

Kn

s sn

Roots are complex conjugates but the coefficients must also be complex conjugates so that f(t) is real.

Therefore, if

£-1[F (s)] £-1[K1

s s1

K1

s s1 ] £-1[

K1

s s1

] £-1[K1

s s1 ]

(K1es1t K1

*es1*t )u(t)

(K1ej e t j t K1e

j e t j t )u(t);

where K1 K1ej and s1 j

£-1[F (s)] K1e t[e j ( t ) e j (t ) ]u(t)

2K1e t cos(t )u(t)

2Re[K1es1t ]u(t)


58

Multiple Roots

1

1

1 1

1 1 2

0 1 11

1 1 1 1

1 20 11 1

0 1

1 1

( ) ( )( )( ) ( ) ( )

where ( ) ( )( ) ( )( )

( ) ( ) ( ) ( )

( ) { } ( ) ( )( 1)! ( 2)!

( ) ( ) |

{( ) ( )} |

P

n

PP P

s tP PP

Ps s

Ps s

N s N sF sD s s s D s

D s s s s s s sM M M R s

s s s s s s D sM Mf t t t M e u t f t

P PM s s F s

dM s s F sds

1

111 {( ) ( )} |!

kP

k s sk

dM s s F sk ds


59

An Example

15430

215

|})2(138393

2396{

|})2(138393{|)}()4{(

15230

)24(138)4(39)4(3|)()4(

18472

)42(138)2(39)2(3|)()2(

)2()4()4(

)2()4(138393)(

42

2

4

2

42

1

2

42

0

2

2

21

112

2

2

s

ss

s

s

o

sss

ss

sss

dsdsFsds

dM

sFsM

sFsK

sK

sM

sM

sssssF

)(]18)1(15[)( 24 tueettf tt


60

Another ExampleF(s) 3s2 17s 47

(s 2)(s2 4s 29)

3s2 17s 47(s 2)(s 2 j5)(s 2 j5)

K1

(s 2)

K2

(s 2 j5)

K2

(s 2 j5)

K1 (s 2)F(s) |s23(2)2 17(2) 47[(2)2 4(2) 29]

2525

1

K2 (s 2 j5)F(s) |s2 j5

3(2 j5)2 17(2 j5) 47(2 j5 2)(2 j5 2 j5)

(63 j60) (34 j85) 47

( j5)( j10)

(50 j25)

(50)1 j0.5

1.116 0.46K2

1 j.5 1.1160.46 f (t) [1 2.232cos(5t 0.46)]e2tu(t)


61

One More Example

44241

282

42

)(2

)(2|}

)(2

)(2{

|)(

|)()(

41

)2(1

)(|)()(

)()()()(

)()()1()(

3

2

23

2

2

2

22

1

22

22

1122

22

2

22

2

jjjj

jj

jjj

jjjj

jss

jss

jss

dsdsFjs

dsd

jjjjsFjs

jsjsjsjs

jsjss

sssF

js

jsjs

js

M

M

MMMM

o

oo

f (t) 2 14

[t cos t cos(t 2

)]u(t) 12

[t cos t cos(t 2

)]u(t)


62

Solving a System with LT

i(t)

)()(

)()(

)()(

])(

[])(

[][

)()(

)()()(

£££

LRs

RV

s

RVsI

LRssLV

LRssLV

sI

L

sVsI

LRssI

LtVu

LtV

iLR

dtdi

LtVu

LtV

iLR

dtdi

dttdi

LRtitV

Vs

1 2 R

L

)()1()( tutL

ReR

Vti


63

Replacing Circuit Element With Their LT Equivalents

0

R

V1

V2

L

C

0

R

V1

V2

sL

1/sC

Note based on the previous example, we can replace circuit elements with their LT equivalents

RR

L sL

C 1/(sC)

And we get, solving for V2 (s)/ V1 (s)

LCsLRs

LC

sCsLRsC

sVsV

1)(

1

1

1

)()(

2

1

2


64

Example Continued

Let’s choose R=5, L=1h, C=1/6f, and let’s solve for h(t), [i.e., v1(t)=δ(t)] and then v1(t)=u(t)

)(][6)(

6)32(6|)3(

6

6)23(6|)2(

623

)2)(3(6

656)(

32

22

31

21

2

tueethsK

sK

sK

sK

sssssH

tt

s

s

)(]321[)(

1)3)(2(6|)3)(2(

6

3)32)(2(6|)3(

6

2)23)(3(6|)2(

623

)2)(3(6)(

656)(

232

03

22

31

321

122

tueetvssK

ssK

ssK

sK

sK

sK

ssssVss

sV

tt

s

s

s


65

Poles of the Response

1

1

2

1

2 1

1 1

( ) ( )( ) ( )

( )( ) ( )( ) ( )( ) ( ) ( )

The poles of ( ) ( ) / ( ) (zeros of ( )) are associated with the solutions to the source free response.

The poles of ( ) (zeros of ( )

N

D

D

V s B sV s A s

V sB s B sV s V sA s A s V s

H s B s A s A s

V s V s

) are associated with the solutions to the source response.

Initial Condition GeneratorsInductors


66

L

a

b

i

i(0-)= i(0+)= IO

iL

a

bi(0-)= 0

LIOδ(t)+

-

L

a

b

i

i(0-)= 0 i’(0-)= 0

IOu(t)

( )

( ) ( )

( ) ( )

ab

ab O

ab O

div t Ldt

V s sLI s LIdiv t L LI tdt

0

( )

( ) ( )1 1( ) ( )

1( ) ( ) ( )

ab

ab O

ab O

t

ab O

div t Ldt

V s sLI s LI

I s V s IsL s

i t v t dt I u tL

i’ -

Initial Condition GeneratorsCapacitors


67

C

a

b

i

vab(0-)= vab (0+)= VO

iC

a

bv'ab (0-)= 0

CVOu(t)+

-

( )

( ) ( )

( ) ( )

ab

ab O

abO

dvi t Cdt

I s sCV s CVdvi t C CI tdt

0

( )

( ) ( )1 1( ) ( )

1( ) ( ) ( )

ab

ab O

ab O

t

ab O

dvi t Cdt

I s sCV s CV

V s I s VsC s

v t i t dt V u tC

C

a

b

i

vab(0-)= 0

CVOδ(t)i’

v'ab

+

-

Switching a motor


68

L

a

b

i

i(0-)= i(0+)= IO

iL

a

bi(0-)= 0

LIOδ(t)+

-

( )

( )

( ) ( )

( ) ( )

CC switch

switch CC

CCswitch O

switch CC O

diV v t Ldtdiv t V Ldt

VV s sLI s LIs

div t V L LI tdt

+VCC

--

t=0

vswitch+ -

+VCC

--

t=0

vswitch+ -

:1( ) ( )

1( )( )

( ) 1( )

switch

CCO

CCO

Assume

V s I ssC

VI s sL LIsC s

V LIsI s

sLsC

Switching a motorwith RC suppressor


69

2

1( ) ( )

1( ) ( ) ( ) ( )

( )( ) 1 1

CC

O

CC

O OO

diV Ri t i t dt LC dt

Vcc s RI s I s sLI s LIsC

VsVcc s LI LII s I RR sL s s

sC L LC

iL

a

bi(0-)= 0

LIOδ(t)+

-

+VCC

--

t=0vswitch+

-

R C

Vswitch (s) (R 1

sC)I (s) R(

s 1RCs

)I (s)

R(s 1

RCs

)IO

s VCC

LIO

s2 s RL

1LC

IOR(s 1

RC)(s

VCC

LIO

)

s(s2 s RL

1LC

)



70

iL

a

bi(0-)= 0

LIOδ(t)+

-

+VCC

--

t=0vswitch+

-

R CVswitch (s) IOR

(s 1RC

)(s VCC

LIO

)

s(s2 s RL

1LC

)

r1,2

RL ( R

L)2 4 1

LC2

Let's choose a 1, b RL 2n , and c 1

LCn

2

n 1

LC;2n 2 1

LC

RL

R2L

LC R2

CL

r1,2 n n 2 1



71

iL

a

bi(0-)= 0

LIOδ(t)+

-

+VCC

--

t=0vswitch+

-

R C

No suppressor

With a suppressor


72

Homework

• Problems: 3.3,

• 3.4b,

min100cycles/0.01for plot Bode theDraw)( :response impulse theFind

)1)(10(9.1263.5)(

as described is system LTIA

fth

ssssH

5 0.1

The unit impulse response is given as( ) (0.7 0.2 0.1 ) ( )

Find the response due to a unit step function.

t t th t e e e u t


73

Homework

• Problems: 3.12c

• 3.13c)()21( todue response theFind

)(3as described is system LTIA

tuetxyy

-t

)()( :stepunit a todue response theFind

)(5158as described is system LTIA

tutxtxyyy


74

Homework

• Problems: 3.18a&c

• 3.23 )( Find

system theof poles theFind)23(

2)(


2

ty

ssssY

)( Find

)2)(4(10)(


3

thss

sH


75

Homework• Find the inverse transforms for:

)4)(2(34 )(;

)106)(22(12188 )(

]1)1[(2 )(;

)4)(100(80090 )(

)5)(2(109 )(;)2(

1 )(

2

22

23

222

23

ssssf

ssssssse

ssd

sssc

sssssbss

eas

lesson #19 6ct.5-7joelsd/signals/classwork/bme314signalscw19.pdf · bme 333 biomedical signals and...

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