eec4-4a-ss-lecture-10.pdf

Signals & Systems FEEE, HCMUT

Ch-6: Phn tch h thng lin tc dng bin i Laplace

Lecture-10

6.1. Bin i Laplace


6.1. Bin i Laplace

6.1.1. Bin i Laplace thun

6.1.2. Bin i Laplace ca mt s tn hiu thng dng

6.1.3. Bin i Laplace mt bn

6.1.4. Cc tnh cht ca bin i Laplace

6.1.5. Bin i Laplace ngc



Bin i Fourier cho php phn tch tn hiu thnh tng ca cc thnh phn tn s phn tch h thng n gin & trc quan hn trong min tn s.

| f(t)|dt & |h(t)|dt

Bin i Fourier l cng c ch yu phn tch TH & HT trong nhiu lnh vc (vin thng, x l nh, )

Mun p dng bin i Fourier th tn hiu phi suy gim & HT vi p ng xung h(t) phi n nh.

phn tch tn hiu tng theo thi gian (dn s, GDP,) v h thng khng n nh dng bin i Laplace (l dng tng qut ca bin i Fourier)



Xt tn hiu f(t) l hm tng theo thi gian to hm mi (t) t f(t) sao cho tn ti bin i Fourier: (t)=f(t).e- t; R

Bin i Fourier ca (t) nh sau:

t jt [ (t)] f(t)e e dt (+j)tf(t)e dt

t s= +j : st( ) f(t)e dt F(s)=()

Hay: stF(s)= f(t)e dt (Bin i Laplace thun)

t(t)=f(t)e

t

f(t)

t

F(s) f(t)]L[K hiu:



Min hi t (ROC) ca bin i Laplace: tp hp cc bin s trong mt phng phc c =Re{s} lm cho (t) tn ti bin i Fourier

V d: tm ROC tn ti F(s) ca cc tn hiu f(t) sau:

at(a) f(t)=e u(t); a>0 at(b) f(t)=e u( t); a>0 (c) f(t)=u(t)


6.1.2. Bin i Laplace ca mt s tn hiu thng dng

(a) f(t)=(t)

-at(b) f(t)=e u(t); a>0

-at(c) f(t)=-e u(-t); a>0

( ) 1; ROC: s-planeF s

1( ) ; : Re{ }F s ROC s a

s a

1( ) ; : Re{ }F s ROC s a

s a

(d) f(t)=u(t)1

( ) ; : Re{ } 0F s ROC ss



Kt qu phn trc cho ta cc tn hiu khc nhau c th c bin i Laplace ging nhau, nhng khc ROC. Do vy ROC phi c ch

r khi cn xc nh f(t) t F(s). V d: 1

( ) ; : Re{ }F s ROC s as a

( ) ( ); 0atf t e u t a

1( ) ; : Re{ }F s ROC s a

s a( ) ( ); 0atf t e u t a

gim s phc tp trn, ta nh ngha bin i Laplace 1 bn:

st

0F(s)= f(t)e dt 0

- c th dng khi f(t) l xung n v

0- c th kho st h thng c K 0-

Bin i Laplace 1 bn, ch c th dng kho st tn hiu & h thng nhn qu. Tuy nhin hn ch ny khng nh hng nhiu

n tn hiu v h thng thc.



Vy vi nh ngha bin i Laplace 1 bn, ta c th xc nh duy nht f(t) t F(s) m khng quan tm ti ROC. V d:

Trong chng ny ta ch tp trung vo dng bin i Laplace 1 bn phn tch h thng LTI. Do vy khi ni ti bin i Laplace, ta

ngm nh rng l bin i Laplace mt bn.

1( )F s

s a( ) ( )atf t e u t



Tnh cht tuyn tnh:

1 1( ) ( )f t F s

1 1 2 2 1 1 2 2( ) ( ) ( ) ( )a f t a f t a F s a F s2 2( ) ( )f t F s

Dch chuyn trong min thi gian:

( ) ( )f t F s 00( ) ( )

stf t t F s e

2 2 1: 2 ( ) ( ) ; : Re{ } 11 2

t tEx e u t e u t ROC ss s

3 54 1: ( 3) ( 5)2

s stVD rect u t u t e es



Dch chuyn trong min tn s:

( ) ( )f t F s 00( ) ( )

s tf t e F s s

2 2: cos ( )

sVD bt u t

s b 2 2cos ( )

( )

at s ae bt u ts a b

o hm trong min thi gian:

( ) ( )f t F s

1 2 (1) ( 1)( ) ( ) (0 ) (0 ) ... (0 )n

n n n n

n

d f ts F s s f s f f

dt

(1) ( )t s( ) 1t ( ) ( )n nt s

4( )

2

tf t rect

2

2

( )?

d f t

dt



Tch phn min thi gian:

( ) ( )f t F s0

( )( )

t F sf d

s

0

( ) ( )( )

t f d F sf d

s s

T l thi gian:

( ) ( )f t F s1

( ) ; 0s

f at F aa a



Tch chp min thi gian:

1 1 2 2( ) ( ); ( ) ( )f t F s f t F s 1 2 1 2( ) ( ) ( ) ( )f t f t F s F s

Tch chp min tn s:

1 1 2 2( ) ( ); ( ) ( )f t F s f t F s1

21 2 1 2( ) ( ) ( ) ( )jf t f t F s F s

o hm trong min tn s:

( ) ( )f t F s( )

( )dF s

tf tds

1( )

1

te u ts 2

1( )

1

tte u ts

2 ( ) ?t u t



Tn hiu f(t) c tng hp nh sau: ( ) ( ).tf t t e

1 12( ) [ ( )]. ( ) .

t j t tf t e F s e d e

12( ) ( )

jst

jj

f t F s e ds (Bin i Laplace ngc)

Chng ta khng tp trung vo vic tnh trc tip tch phn trn!!!

M t F(s) v cc hm n gin m c kt qu trong bng cc cp bin i Laplace. Thc t ta quan tm ti cc hm hu t!!!

Zero ca F(s): cc gi tr ca s F(s)=0

Pole ca F(s): cc gi tr ca s F(s)

Nu F(s)=P(s)/Q(s) Nghim ca P(s)=0 l cc zero & nghim ca Q(s)=0 l cc pole

K hiu: -1f(t) ( )F sL



Dng bng

Dng ? V d:

2

3 2

2 1 1 1

3 2 1 2

s

s s s s s s

2-1 -1 2

3 2

2 1 1 11 ( )

3 2 1 2

t ts e e u ts s s s s s

L L



start

m



( ) ( ) / ( )F s P s Q s

Xc nh zero & pole ca F(s); zero & pole phi khc nhau

Khai trin cc hm proper:

Gi s cc pole l: s= 1, 2, 3,

Khai trin F(s) dng quy lut sau:

Cc pole khng lp li:

31 2

1 2 3

( ) ...( ) ( ) ( )

kk kF s

s s s

Cc pole lp li, gi s 2 lp li r ln

12 31

01 2 3

( ) ...( ) ( ) ( )

rj

r jj

k kkF s

s s s



Phng php hm tng minh xc nh cc h s:

Nhn 2 v vi Q(s); sau cn bng thu c h phng trnh

theo cc h s cn tm

Its easy to understand and perform, but it needs so much work and time!!!

Gii h phng trnh tm cc h s

2

31 2

3 2

2

3 2 1 2

kk ks

s s s s s s

2

1 2 32 ( 1)( 2) ( 2) ( 1)s k s s k s s k s s

v d:

1 2 3

1 2 3

1

1

3 2 0

2 2

k k k

k k k

k

1

2

3

1

1

1

k

k

k



Phng Heaviside xc nh cc h s:

Cc pole khng lp li: ( ) ( )i

i i sk s F s

Cc pole lp li:

0 ( ) ( )

1( ) ( ) ; 0

!

i

i

r

i i s

jr

ij ij s

k s F s

dk s F s j

j ds

3

8 10( )

( 1)( 2)

sF s

s s V d: 201 21 22

3 2( 1) ( 2) ( 2) ( 2)

kk k k

s s s s



Phng hn hp: phng php thng dng

3

8 10( )

( 1)( 2)

sF s

s s V d: 201 21 22

3 2( 1) ( 2) ( 2) ( 2)

kk k k

s s s s

1 22 220 2k k k( ); :sF s s

0:s20 21 22

1

5

8 4 2 4

k k kk

1 3

1

8 102

2s

sk

s20

2

8 106

1s

sk

s

1 20 2221

10 8 4

2

k k kk

21

10 16 6 82

2k



V d: tm bin i Laplace ngc ca cc hm sau:

2

7 - 6( ) F(s)=

6

sa

s s

2

2

2 5( ) F(s)=

3 2

sb

s s

2

6( 34)( ) F(s)=

( 10 34)

sc

s s s

eec4-4a-ss-lecture-10.pdf

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