Chuong 1 - Tín hiệu và hệ thống.pdf

7/21/2019 Chuong 1 - Tn hiu v h thng.pdf

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CHNG 1:GII THIU TN HIU V

H THNG


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1.1. TN HIU

nh ngha tn hiu Phn loi tn hiu

Cc php ton c bn trn tn hiu

Cc tn hiu c bn


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nh ngha tn hiu:

Mt i lng vt l truyn ti thng ti v bncht ca mt hin tng vt l

C th biu din di dng hm thi gian lin tchoc ri rc Hm ca mt hay nhiu bin:

Tn hiu m thanh: hm ca thi gian (tn hiu mtchiu) nh ng: (php chiu ca ca mt cnh ng vo mtphng nh): hm ca 3 bin x,y,t (tn hiu nhiu chiu)


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V d v tn hiu:


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Phn loi tn hiu

Tn hiu lin tc v ri rc

Tn hiu tng t v s

Tn hiu tun hon v khng tun hon Tn hiu nhn qu v khng nhn qu

Tn hiu chn v l

Tn hiu xc nh v ngu nhin

Tn hiu a knh v a chiu

Tn hiu bn tri v phi

Tn hiu hu hn v v hn

Tn hiu nng lng v cng sut


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Phn loi tn hiu:Tn hiu lin tc v ri rc Thi gian lin tc:

Gi tr hay bin thay i lin tc theo thi gian C bn cht t nhin

Thi gian ri rc: Gi tr ch thay i ti nhng thi im nht nh C th to ra t tn hiu lin tc bng vic ly mu tn hiulin tc ti nhng thi im nht nh Thng lin quan n cc h thng nhn to

x(t) x[n]=x(nTs) n=0,1,2,


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Phn loi tn hiu: Gi tr lin tc: Gi tr ca tn hiu thay i mt cchlin tc Gi tr ri rc: gi tr ca tn hiu thay i khng lin

tcTn hiu tng t v s Tn hiu tng t: tn hiu lin tc theo thi gian v cgi tr lin tc

Tn hiu s: tn hiu ri rc theo thi gian v c gi trc lng t ha hay c gi tr ri rc


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Phn loi tn hiu:Tn hiu tun hon v khng tun hon Tn hiu tun hon: lp li chnh bn thn tn hiu sau mtkhong thi gian nht nh

x(t)=x(t+T) vi mi T>0hay x[n]=x[n+N] viN nguyn dng Chu k c bn ca tn hiu tun hon l gi tr nh nht ca Ttha mn iu kin trn (T hay N)

Tn s c bn = 1/chu k c bn (f=1/T hay f=1/N) Tn s gc c bn = 2*tn s c bn ( = 2/T rad/s hay = 2/Nrad)

Tn hiu khng tun hon: khng c gi tr no ca T thamn iu kin trn hay gi tr ca tn hiu khng c lpli mt cch c chu k


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Phn loi tn hiu:

T=? =?

N=? =?

Tn hiu tun hon

Tn hiu khng tun hon

V d:2

( ) os (2 ) [n] ( 1)n

x t c t x


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Phn loi tn hiu:

Tn hiu nhn qu v khng nhn qu Tn hiu nhn qu: gi tr ca tn hiu lun bng khngtrn phn m ca trc thi gian

Tn hiu phn nhn qu: gi tr ca tn hiu lun bngkhng trn phn dng ca trc thi gian

Tn hiu phi (khng) nhn qu: tn hiu c gi tr khckhng trn c phn m v phn dng ca trc thi gian


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Phn loi tn hiu: V d:

0 : ( ) 0t f t : ( ) 0t f t

Tn hiu nhn qu Tn hiu phi nhn qu Tn hiu phn nhn qu

0 : ( ) 0t f t


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Phn loi tn hiu:Tn hiu chn v l Tn hiu chn: x(t) = x(-t) hay x[n]=x[-n] Tn hiu l: x(-t) = -x(t) hay x[-n]=-x[n] Bt k mt tn hiu no c th c biu din nh l tngca mt tn hiu chn v mt tn hiu l

x(t)=xe(t)+xo(t)Trong :

sin( ),( )

0, c

( ) cos( ) sin( ) sin( ) os( )

tT t T

x t Tt kh

x t t t t c t

Tn hiu chn

Tn hiu l

V d

1( ) ( ) ( )21

( ) ( ) ( )2

x t x t x te

x t x t x to


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Phn loi tn hiu:Tn hiu xc nh v ngu nhin Tn hiu xc nh: c m hnh nh l mt hm ca thigian, v th gi tr ca tn hiu ti bt k thi im no u c

th tnh trc c bng biu thc ton hc hoc bng gi tr Tn hiu ngu nhin: nhiu yu t khng chc chn xuthin trc khi tn hiu xut hin, do khng xc nh cchnh xc gi tr ti mt thi im trong tng lai.

Tn hiu xc nh Tn hiu ngu nhin


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Phn loi tn hiu:

Tn hiu a knh v a chiu Tn hiu a knh: thng c biu din di dng mt

vct trong cc thnh phn ca vct l cc tn hiu nknh:F(t) = [ f1(t) f2(t).fN(t)]

Tn hiu a chiu: thng c biu din di dng hmca nhiu bin c lp:

f(x1,x2,xN)


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Phn loi tn hiu:

Tn hiu thun v nghch Tn hiu thun (bn phi): gi tr ca tn hiu lun bng 0

k t mt thi im v trc, ngha l

Tn hiu nghch (bn tri): gi tr ca tn hiu lun bng 0k t mt thi im tr v sau, ngha l

0 : ( ) 0t t x t

0 : ( ) 0t t x t


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Phn loi tn hiu:

Tn hiu hu hn v v hn Tn hiu hu hn: tt c cc gi tr khc khng ca tn hiu

u nm trong mt khong hu hn, ngoi khong gi trca tn hiu lun bng 0, ngha l tn ti mt khong huhn sao cho

Tn hiu v hn: khng tn ti khong hu hn tha mniu kin trn hay min cc gi tr khc khng ca tn hiul v hn

1 2 1 2: ( ) 0 [t ,t ]t t f t khi t


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Tn hiu nng lng v tn hiu cng sut Nng lng ca mt tn hiu lin tc x(t) v ri rc x[n]c nh ngha bi:

Mt tn hiu l tn hiu nng lng khi n c nng lnghu hn, ngha l tha mn:

0


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Tn hiu nng lng v tn hiu cng sut Cng sut ca mt tn hiu c nh ngha l nng lngtrung bnh ca tn hiu trong mt n v thi gian Cng sut ca tn hiu lin tc x(t) v ri rc x[n] c

nh ngha nh sau:

/22

/2

1lim ( )

T

x TT

P x t dtT

21lim [ ]2 1

N

xN

n N

P x nN


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Tn hiu nng lng v tn hiu cng sut

2

0

1( )

T

xP x t dtT

12

0

1[ ]

N

xn

P x nN

Cng sut ca tn hiu lin tc tun hon x(t) vi chu k Tv thi gian ri rc x[n] vi chu k N l tng ng vinng lng trung bnh trong mt chu k nn cng suttrung bnh ca tn hiu tun hon c nh ngha l:

Mt tn hiu l tn hiu cng sut khi n c cng sut trungbnh hu hn, ngha l tha mn:

0


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Tn hiu nng lng v tn hiu cng sut Mt tn hiu nu l tn hiu nng lng th khng th l tnhiu cng sut do cng sut ca tn hiu nng lng lunbng 0

Mt tn hiu nu l tn hiu cng sut th khng th l tnhiu nng lng do nng lng ca tn hiu cng sut lunv hn, v d i vi tn hiu tun hon

V d: xc nh nng lng v cng sut trung bnh ca tnhiu sau:

0 1os( ) 0

( ) 2 1 2 [ ]0 c

0 c

t tc n n

x t t t x nkh

kh


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Cc php ton c bn trn tn hiu

Php ton trn bin ph thuc

Php ton trn bin c lp


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Php ton trn cc bin c lp T l thi gian y(t) = x(at)

a>1: nn tn hiu 01: mt gi tr


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Php ton trn cc bin c lp nh x: thay t = -t hoc n = -n

y(t) = x(-t) nh x tn hiu chn l chnh n

nh x tn hiu l l gi tr m ca chnh n

V d: xc nh tn hiu hp y[n]=x[n]+x[-n] ca

1 11 1& 1

[ ] 1 1 [ ]0 0& | | 1

0 0&|n|>1

nn n

x n n x nn n

n


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Php ton trn cc bin c lp Dch thi gian: y(t) = x(t - t0)

t0>0: dch sang phi (tr) t0


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Mt s tn hiu c bn Tn hiu xung

Tn hiu xung n v thi gian lin tc, k hiu l (t),c nh ngha bng hm delat Dirac nh sau:

Tn hiu xung n v thi gian ri rc, k hiu [n],c nh ngha bi

0 0( )0 0

ttt

( ) 1t dt

0 0[ ]

1 0

nn

n


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Tn hiu nhy bc Tn hiu nhy bc n v thi gian lin tc, k hiuu(t), c nh ngha nh sau:

Tn hiu nhy bc n v thi gian ri rc, k hiu u[n],c nh ngha nh sau:

0 0( )1 0

tu tt

0 0[ ]

1 0

nu n

n

Mt s tn hiu c bn


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Tn hiu dc Tn hiu dc thi gian lin tc, k hiu r(t), c nhngha nh sau:

Tn hiu dc thi gian ri rc, k hiu r[n], c nhngha nh sau:

tng ng r(t)=tu(t), r[n]=tu[n]

0 0( )0

tr tt t

0 0[ ]

0

nr n

n n

Mt s tn hiu c bn


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Tn hiu sin Tn hiu sin gi tr thc thi gian lin tc c biudin di dng sau:

s(t) = Acos(t+)A: bin , : tn s gc (rad/s), : gc pha (rad). Chuk ca tn hiu tun hon ny l T=2/ C th biu din di dng l hm ca bin tn sf=1/T (Hz):

s(t) = Acos(2ft+)

Mt s tn hiu c bn


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Tn hiu m thc Tn hiu m thc thi gian lin tc c nh nghanh sau:

f(t) = Aet

A v l cc gi tr thc Nu >0: f(t) m tng,


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Tn hiu m phc Tn hiu m phc thi gian lin tc c nh nghanh sau:

f(t) = Ae(+j)t

Mi lin quan gia tn hiu m v tn hiu sin: s dngbiu thc Euler cho ejt, ta thu c biu thc sau cho tnhiu m phc:

f(t) = Aet

[cos(t)+jsin(t)]

Mt s tn hiu c bn


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f(t) l mt hm gi tr phc trong phn thc v phno c tnh nh sau:

Re[f(t)] = Aetcos(t)

Im[f(t)] = Ae

t

sin(t) f(t) cng c gi l tn hiu sin phc vi ln phcl Aet v tn s gc

f(t) = Ae(+j)t

ln thc ca f(t) l |A|et v pha l :

2 2 Im( )| | Re( ) Im( ) , arctanRe( )

AA A A

A

Mt s tn hiu c bn


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1.2. H THNG

nh ngha h thng

M hnh ton hc ca h thng

Mt s v d v h thng

Phn loi v c im ca h thng


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nh ngha h thng

Mt h thng l mt thc th hot ng khi c tn hiu livo (kch thch) v sinh ra tn hiu li ra (p ng)

Theo biu diu ton hc, h thng c c trng bi miquan h gia tn hiu li vo v tn hiu li ra

y(t) = T[x(t)]y[n] = T{x[n]}

T:php bin i c trng cho h thng

x(t) T y(t) x[n] T y[n]


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M hnh ton hc ca h thng

Mi quan h gia li vo ca h thng v li ra ca hthng, cn gi l hnh vi ca h thng, c biu din bngmt m hnh ton hc

M hnh ton hc cho php xc nh h thng: xc nhtn hiu li ra khi bit tn hiu li vo

M hnh ton hc c s dng trong vic phn tch vthit k h thng


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V d v h thng

H thng truyn thng tng t


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V d v h thng

H thng truyn thng s


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V d v h thng

H thng iu khin phn hi


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Phn loi h thng

H thng lin tc v ri rc

H thng lin tc: cc tn hiu vo, tn hiu ra v cc tnhiu s dng trong h thng u l cc tn hiu thi gian lintc

H thng thi gian ri rc: tn hiu vo v tn hiu ra l cctn hiu thi gian ri rc


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Phn loi h thng

H thng n bin v h thng a bin

SISO: (Single-input Single output): mt bin vo-mt bin ra SIMO: (Single input Multiple ouput): mt bin vo-nhiu bin ra

MISO: (Multiple input Single output): nhiu bin vo-mt bin ra

MIMO: (Multiple input Multiple output): nhiu bin vo- nhiubin ra


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Phn loi h thng H thng tnh v ng (nh v khng nh)

H thng tnh hay h thng khng nh: l h thng trong gitr ca tn hiu ra ch ph thuc gi tr ca tn hiu vo cng

thi im H thng ng hay h thng c nh: l h thng trong gitr ca tn hiu ra ph thuc c vo gi tr trong qu kh hoctng lai ca tn hiu vo

2

1( ) ( )

[ ] [ ]

i t v t R

y n x n

1( ) ( )

1[ ] ( [ ] [ 1] [ 2])

3

t

i t v d L

y n x n x n x n


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Phn loi h thng H thng nhn qu v phi nhn qu

Mt h thng nu tn hiu li ra ca h thng ch c th ph

thuc cc gi tr ca tn hiu vo hin ti hoc trong qu kh chkhng th ph thuc vo cc gi tr tng lai ca tn hiu cgi lh thng nhn qu.

Mt h thng khi tn hiu ra ca h thng c th ph thuc voc cc gi tr tng lai ca tn hiu vo c gi lh thng

khng nhn qu


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Phn loi h thng H thng tuyn tnh

H thng c xem l tuyn tnh khi v ch khi tha mnnguyn l ng nht v nguyn l xp chng:

H thng khng tuyn tnh nu khng tha mn iu kintrn

1 2 1 2, : [ ( ) ( )]= [ ( )] [ ( )]x t x t x t x t R T T T


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Phn loi h thng H thng bt bin thi gian

H thng bt bin thi gian: mt s dch chuyn thi gianca tn hiu li vo dn n s dch chuyn thi gian tng

ng tn hiu li ra quan h vo/ra khng ph thuc vothi im bt u:

H thng thay i theo thi gian khi quan h vo/ra ph

thuc vo thi im bt u

0 0 0( ) [ ( )] : ( ) [ ( )]y t x t t y t t x t t


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Phn loi h thng H thng n nh

H thng c gi l n nh gii hn BIBO (BoundedInput Bounded Output) khi v ch khi tn hiu ra lun c giihn hu hn khi tn hiu vo c gii hn hu hn

Nu tn hiu vo c gii hn hu hn to ra mt tn hiu ra

gii hn khng hu hn th h thng s khng n nh

| ( ) | | ( ) |x t y t

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Chuong 1 - Tín hiệu và hệ thống.pdf