chuong 3 xu ly tin hieu so

7/30/2019 Chuong 3 Xu Ly Tin Hieu So

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Chng IIING DNG BIN I FOURIER PHN TCH TN

HIU S V H X L S

3.1. Bin i Fourier ca dy s

3.1.1. Bin i Fourier thun

a. nh ngha:Nu dy x(n) tha mn iu kin:1( )x n= [3.1-1]

th s tn ti php bin i fourier nh sau:

( ) ( ).j j n

n

X e x n e

=

= [3.1-2]

Bin i Fourier chuyn dy s x(n) thnh hm phc ( )jX e [3.1-2] l biu

thc bin i Fourier thun v c k hiu nh sau:

FT[x(n)] = ( )jX e [3.1-3 ]

Hay : x(n) FT ( )jX e [3.1.-4]

(FT l vit tt ca thut ng ting anhFourier Transform )

K hiu ( )jX e phn bit php bin i Fourier ca dy s x(n) vi php

bin i Fourier ca hm lin tc x(t): [ ]x(t) = ( ) ( ). .j tFT X x t e dt

= Biu thc bin i Fourier ca dy s x(n) [3.1-2] l xut pht t biu thc bin

i Fourier ca hm lin tc x(t), v khi hm di du tch phn l dy ri rc th phi

thay du tch phn bng du tng.

Do tnh cht tun hon ca hm m je nn ( )jX e l hm tun hon ca bin

vi chu k 2 :

( 2 ) ( 2 )

( ) ( ). ( ). ( )j k j k n j n j

X e x n e x n e X e

+ +

= = =

iu c ngha l ch cn nghin cu hm tn s ca cc dy ri rc x(n) vi

( , ) hoc (0,2 ) .

S dng bin i Fourier cho php nghin cu ph ca tn hiu s ca h x l

s. Nu x(n) l tn hiu s th FT[x(n)]= ( )jX e l ph ca x(n), cn nu h(n) l c

tnh xung ca h x l s th FT[h(n)]= ( )jH e l c tnh tn s ca h x l s.

b. S tn ti ca bin i Fourier

Theo nh ngha bin i Fourier thun [3.1.2] ch tn ti nu x(n) tha mn

in kin tng tuyt i [3.1-1]. iu c ngha l, nu dy x(n) tha mn iu kin

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[3.1-1] th chui [3.1-2] s hi t v hm ( )jX e , nn x(n) tn ti bin di Fourier.

Ngc li, nu dy x(n) khng tho mn iu kin [3.1-1] th chui [3.1-2] s phn

k, v th hm ( )jX e s khng tn ti v x(n) khng c bin i Fourier.

Cc tn hiu s x(n) c nng lng hu hn:2

) )xn

E x n

=

= <

lun tha mn

iu kin [3.1-1]], do lun tn ti bin i Fourier.

V d 3.1: Hy xt s tn ti v tm bin i Fourier ca dy sau:

a. u(n) b. 2 . ( )n u n c. 2 . ( )n u n

d. ( )n e. ( )n k f. ( )Nrect n

Gii: a.0

( ) 1n n

u n

= =

= =

Hm u(n) khng tha mn [3.1-1] nn khng tn ti bin i Fourier.b.

0

2 . ( ) 2n n

n n

u n

= =

= =

Hm 2 . ( )n u n khng tha mn [3.1-1] nn khng tn ti bin i Fourier

c. 10

12 . ( ) 2 2

1 2n n

n n

u n

= =

= = =

Hm 2 . ( )n u n tha mn [3.1-1] nn tn ti bin i Fourier

( )

1

0 02 . ( ) 2 . ( ) . 2 . 2 .

nn n j n n j n j

n n nFT u n u n e e e

= = = = = = Vy: 1

1 12 . ( )

1 2 1 0,5.n

j jFT u n

e e

= = [3.1-6]

d. ( ) 1n

n

=

=

Hm ( )n tha mn [3.1-1] nn tn ti bin i Fourier:

[ ] 0( ) ( ). 1. 1j n jn

FT n n e e

=

= = = [3.1-7]

e. Chui [3.1-1] l hi t vi ( )n k hi t nn n c bin i Fourier:

[ ]( ) ( ). 1.j n j kn

FT n k n k e e

=

= = [3.1-8]

f.1

0

( ) 1N

Nn n

rect n N

= =

= = <

Hm ( )Nrect n tha mn [3.1-1] nn n tn ti bin i Fourier:

[ ] ( )1

0

1( ) ( ).

1

j NN nj n jN N

jn n

eFT rect n rect n e e

e

= =

= = =

[3.1-9]

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C th thy rng cc dy c di hu hn lun lun tn ti php bin i

Fourier, cn cc dy c di v hn s tn ti bin i Fourier nu chui [3.1-1] ca

n hi t.

c. Cc dng biu din ca hm ( )jX e

V ( )j

X e

l hm phc, nn c th biu din n di dng,phn thc v phno, modul va argumen, ln v pha.

1. Dng phn thc v phn o

( ) ( ) ( )j R IX e X jX = + [3.1-10]

Theo cng thc Euler c:

[ ]( ) ( ). ( ). cos( ) sin( )j j nn n

X e x n e x n n j n

= =

= = [3.1-11]

Hm phn thc: ( ) Re ( ) ( ).cos( )j

RnX X e x n n

= = = [3.1-12]

Hm phn o: ( ) Im ( ) ( ).sin( )j

In

X X e x n n

=

= = [3.1-13]

2. Dng modul v argumen( )( ) ( ) .j j jX e X e e =

M un: 2 2( ) ( ) ( )j R IX e X X = + [3.1-15]

Argumen:2

2 ( )( ) arg ( ) arctg ( )j R

IXX e X

= =

[3.1-16]

( )jX e c gi l hm bin tn s, n l hm chn v i xng qua trc

tung: ( ) ( )j jX e X e =

( ) c gi l hm pha tn s, n l hm l v phn i xng qua gc ta :

( ) ( ) = .

3. D ng ln v pha

( )( ) ( )j j jA e A e e = [3.1-17]

Hm ln ( )jA e c th nhn gi tr dng hoc m: ( ) ( )j jX e A e = [3.1-18]

Cn: arg ( ) ( ) ( )jA e + = [3.1-19]

Hm pha: ( ) ( ) arg ( )jA e = [3.1-20]

Vi arg ( )jA e ph thuc vo du ca hm ( )

jA e nh sau:

arg ( ) 0 ( ) 0

arg ( ) ( ) 0

j j

j j

A e khi A e

A e khi A e

= =


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Mt cch tng qut, c th vit:

( )arg ( ) 1

2 ( )

jj

j

A eA e

A e

=

Theo [3.1-19], c th biu din hm pha ( ) di dng nh sau:

( )( )= ( )- 1

2 ( )

j

j

A e

A e

[3.1-21]

V d 3.2: Hy xc nh cc hm phn thc phn o, modul v argument, ln ca

hm tn s ( ) os(2 ).j jX e c e =

Gii: Theo [3.1-11] c: ( ) os(2 ). os( ) - j os(2 ).sin( )jX e c c c =

Hm phn thc: ( ) os(2 ). os( )RX c c =

Hm phn o:( ) - os(2 ).sin( )

I

X c =

Modul: 2 2 2 2( ) os (2 ). os ( ) + os (2 ).sin ( ) os(2 )jX e c c c c = =

Argumen:cos(2 ).sin( )

( )cos(2 ).cos( )

arctg

= =

Hm ln: ( ) os(2 )jA e c = )

Hm pha:os(2 )

( )=- - 12 os(2 )

c

c

d. Quan h gia bin i Fourier v bin i ZTheo biu thc nh ngha [2.1-1] ca bin i Z ta c:

[ ]( ) ( ) ( ). nn

ZT x n X z x n z

=

= = vi [ ]( ) : x xRC X z R z R +< 0 l x(n) b gi tr k, mu, nu k


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Tn hiu x(n) c ph tn l [ ]( ) ( )jX e FT x n = , hy tm ph tn s ca tn hiu

iu bin y(n) = x(n ).cos( 0 n)

Gii: C:0 0

0cos( ) 2

j n j ne en

=

Do : [ ] 0 00 1 1( )cos( ) ( ) ( )2 2j n j nFT x n n FT x n e FT x n e = +

Theo tnh cht dch ca hm tn s nhn c:

[ ] 0 0( ) ( )01 1

( )cos( ) ( ) ( )2 2

j jFT x n n x n e x n e += + [3.1-31]

Biu thc [3.1-31] chnh l ni dung ca nh l iu bin.

d. Tnh cht bin o:bin i Fourier ca cc dy thc c bin o x(n) v x(-n) l

hai hm lin hp phc.

Nu: [ ] ( )( ) ( ) ( ) .j j jFT x n X e X e e = =

Th: [ ] * ( )( ) ( ) ( ) ( ) .j j j jFT x n X e X e X e e = = = [3.1-32]

Chng minh: theo biu thc bin i Fourier thun [3.1-2] c:

[ ] ( )( ) ( ) ( ) ( )j n j n n jn n

FT x n x n e x n e X e

= =

= = =

V x(-n) l dy thc nn *( ) ( )j jX e X e = , do nhn c [3.1-32].

Nh vy, cc dy thc nhn qu v phn nhn qu tng ng c hm bin

tn s ging nhau, cn hm pha tn s ngc du.

V d 3.7: Hy tm ( ) 2 ( )j nX e FT u n =

Gii: Theo biu thc [3.1-6] v biu thc bin o c:1

2 ( )1 0,5

njFT u n e

=

e. Hm tn s ca tch chp hai dy:Hm tn s ca tch chp hai dy bng tch chp

ca hai dy thnh phn.

Nu: [ ]1 1( ) ( )jFT x n X e = v [ ]2 2( ) ( )

jFT x n X e =

Th : [ ]1 2 1 2( ) ( ) * ( ) ( ). ( )j j jY e FT x n x n X e X e = = [3.1-33]


[ ]1 2 1 2( ) ( ) * ( ) ( ) ( ) .j j n

n n

Y e FT x n x n x k x n k e

= =

= =

1 2( ) ( ) ( ). . .j j n j k j k

n k

Y e x k x n k e e e

= =

=

Hay:( )

1 2 1 2( ) ( ). ( ). ( ). ( )j j k j n k j j

n n

Y e x k e x n k e X e X e

= =

= =

V d 3.9: Hy tm ( ) 2 ( ) * ( 1)j nX e FT u n n =

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Gii: S dng biu thc [3.1-6], [3.1-8] vi k= 1, v [3.1-33], tm c:

12 ( )

1 0,5n

jFT u n

e

= v

[ ]( 1) jFT n e =

Vy:1

( )1 0,5 1 0,5

jj j

j j

eX e e

e e

= =

f. Hm tn s ca tch hai dy:Hm tn s ca tch hai dy bng tch chp ca hai

hm tn s thnh phn chia cho 2.

Nu [ ]1 1( ) ( )jFT x n X e = v [ ]2 2( ) ( )

jFT x n X e =

Th [ ]'( ) '

1 2 1 2

1( ). ( ) ( ). ( )

2j jFT x n x n X e X e d

= [3.1-34]

Hay: [ ]1 2 1 21

( ). ( ) ( )* ( )2

j jFT x n x n X e X e

= [3.1-35]


[ ] [ ]1 2 1 2( ). ( ) ( ). ( ) .j n

n

FT x n x n x n x n e

=

=

Khi thay 1( )x n bng biu thc bin i Fourier ngc ca n:

' ' '1 1

1( ) ( ). .

2j j nx n X e e d

=

Th: [ ]' ' '

1 2 1 2

1( ). ( ) ( ). . . ( ).

2j j n j n

n

FT x n x n X e e d x n e

=

= [3.1-36]

[ ]' '( ) '

1 2 1 21( ). ( ) ( ) ( ). .

2j j n

n

FT x n x n X e x n e d

=

=

[ ]' '( ) '

1 2 1 2 1 2

1 1( ). ( ) ( ) ( ). ( )* ( )

2 2j j j jFT x n x n X e X e d X e X e

= =

g. Cng thc Parseval tnh nng lng ca tn hiu theo hm ph.

22 1( ) ( )

2j

xn

E x n X e d

=

= = [3.1-37]

Chng minh: Vit li biu thc [3.1-36] di dng:

' ' '1 2 1 2

1( ). ( ). ( ). ( ). . .

2j n j j n j n

n n

x n x n e x n X e e d e

= =

=

Chia c 2 v ca biu thc trn cho j ne , nhn c:

' ' '1 2 1 2

1( ). ( ) ( ). . ( ).

2j n j

n n

x n x n x n e X e d

= =

=

Hay:' ' '

1 2 1 2

1( ). ( ) ( ). ( ).

2j j

n

x n x n X e X e d

=

=

Khi cho 1 2( ) ( ) ( )x n x n x n= = th theo [1.3-5], v tri ca biu thc trn chnh lnng lng xE ca tn hiu s ( )x n :

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22 1 1( ) ( ). ( ). ( ) .

2 2j j j

xn

E x n X e X e d X e d

=

= = =

Hay:2 1

( ) ( ).2x xn

E x n S d

=

= = [3.1-38]

Trong :

2

( ) ( )j

xS X e

= [3.1-39]( )xS c gi l hm mt ph nng lng ca tn hiu s x(n), n l hm

chn v i xng qua trc tung. V bn cht vt l, hm mt ph nng lng ( )xS

ca tn hiu trn trc tn s.

V d 3.9: Hy xc nh nng lng ca tn hiu s ( ) 2 ( )nx n u n= theo c hm thi

gian v hm ph, so snh kt qu nhn c.

Gii: Theo hm thi gian c:

2 21

0 0

1 4( ) (2 ) 41 4 3

n nx

n n n

E x n = = =

= = = = =

xc nh nng lng theo hm ph, trc ht tm:

1 1( ) 2 ( ).

1 0,5 1 0,5cos 0,5sinj n j n

jn

X e u n ee j

=

= = = +

Vy: 2 21 1

( )(1,25 )(1 0,5 ) (0,5sin )

jX ecoscos

= =

+

Tnh nng lng ca x(n) bng cng thc Parseval [3.1-37]:

2 2

(1, 25 1) ( )1 1 1 1 2.

2 1, 25 2 1, 25 1 1, 25 1

tg

Ex d arctgcos

+

= =

1 1 33 (0)

0,75 2 2 0,75 0,75 4Ex arctg tg tg arctg

= = = =

h. o hm ca hm tn s

Nu: [ ]( ) ( )j

FT x n X e

=

Th: [ ]( )

. ( )jdX e

FT n x n jd

= [3.1-40]

Chng minh: Theo biu thc bin i Fourier thun ta c:

[ ]( )

( ) ( ) ( ). . . ( ).j

j j n j n

n n

dX eX e FT x n x n e j n x n e

d

= =

= = =

Nhn c hai v biu thc ny vi j, nhn c biu thc [3.1-40]

V d 3.10: Hy tm bin i Fourier ca dy ( ) 2 . . ( )nx n n u n=

Gii: Ta c:1

2 . ( )1 0,5.

nj

FT u ne

=

111


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Theo [3.1-40] c: 21 0,5. .

2 . . ( )1 0,5. (1 0,5. )

jn

j j

d eFT n u n j

d e e

= =

i. Ph tn s ca hm tng quan ( )xyr m

Nu: [ ]( ) ( )jFT x n X e = v [ ]( ) ( )jFT y n Y e =

Th: ( ) ( ) ( ). ( )j j jxy xyR e FT r m X e Y e = = [3.1.41]

Chng minh: Hm t tng quan ( )xyr m c xc nh nh sau:

( ) ( ). ( )xyn

r m x n y n m

=

=

Theo biu thc bin i Fourier thun ta c:

( )( )

( ) ( ). ( ). ( ) .

( ) ( ). ( ) . . .

( ) ( ). ( ). ( ). ( )

j m j mxy xym m n

j m j n j nxy

m n

j n j n m j jxy

n m

FT r m r m e x n y n m e

FT r m x n y n m e e e

FT r m x n e y n m e X e Y e

= = =

= =

= =

= = =

= =

V d 3.11: Cho cc tn hiu s ( ) 2 . ( )nx n u n= v ( ) ( 1)y n n= . Hy tm hm ph

( ) ( )jxy xyR e FT r m =

Gii: S dng [3.1-6], [3.1-8] vi k = 1, v [3.1-41], tm c:

1( ) ( ). ( ) .

1 0,5. 1 0,5.

jj j j j

xy j j

eR e X e Y e e

e e

= = =

k. Ph tn ca hm t tng quan ( )xr m : Ph tn s ( )j

xR e ca hm t tng quan

( )xr m chnh l hm mt ph nng lng ( )xS ca tn hiu s x(n).

Nu [ ]( ) ( )jFT x n X e =

Th [ ]( ) ( ) ( ). ( )j j j

x xR e FT r m X e X e = = [3.1-42]

Hay [ ]2

( ) ( ) ( ) ( )j jx x xR e FT r m X e S = = = [3.1-43]

chnh l ni dung nh l Winer - Khinchine.

Chng minh: Trong biu thc ca hm tng quan, khi thay y(n) = x(n) nhn c

hm t tng quan rx(m), v th ta c:

[ ]2

( ) ( ) ( ). ( ) ( ) ( )j j j jx x xR e FT r m X e X e X e S = = = =

V d 3.12: Hy tm hm ph ( )jxR e ca tn hiu s ( ) 2 . ( )nx n u n=

Gii: S dng [3.1-6] v [3.1-42] tm c:

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1 1 1( ) .

(1 0,5. ) (1 0,5. ) 1, 25 cosj

x j jR e

e e

= =

3.2 Ph ca tn hiu s

3.2.1 Cc c trng ph ca tn hiu s

Bin i Fourier ca tn hiu s ( )x n l hm ph ( )iX e ca n:

( )( ) ( ). ( ) . ( ) ( )jj j n j R In

X e x n e X e e X jX

=

= = = +

T xc nh c:

- Ph bin ( )iX e c tnh theo [3.1-15]:

2 2( ) ( ) ( )j R IX e X jX = +

- Ph pha ( ) ( )jArg X e = c tnh theo [3.1-16]:

( )( ) ( ) arc

( )j l

R

XArg X e tg

X

= =

- Nng lng Ex c tnh theo cng thc Parseval [3.1-37]:

21 1( ) ( )

2 2j

x xE X e d S d

= =

- Mt ph nng lng ( )xS c tnh theo [3.1-39]:2

( ) ( )jxS X e =

Hm ph ( )jX e , ph bin ( )jX e , ph pha )( , v hm mt nng

lng Sx() l c trng ph ca tn hiu s x(n)

V d 3.13: Cho tn hiu s x(n) = 2-n .u(n-2), hy xc nh cc c trng ph ca tn

hiu.

Gii: 2 ( 2)( ) 2 .2 . ( 2)j nX e FT u n =

Theo [3.1-6] v tnh cht tr ca bin i Fourier c:

X(ej ) = FT[2-nu(n-2)]= 2-2 e 2j).5,01(

1je

Hm ph: X (ej ) =)5,01(4

2

j

j

e

e

=

)sin5,0cos5,01(

.25,0 2

j

e j

+

Hm ph bin : )(ejX = 22 )sin5,0()cos5,01(

25,0

+=

cos25,1

25,0

Hm ph pha: )( = -2- arctg[ )cos5,01(

sin5,0

]

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Hm mt ph nng lng: Sx() =2

)( jeX =)cos25,1(

0625,0

V ngha vt l, th ca hm ph bin )(ejX v hm mt ph nng

lng Sx() chnh l bc tranh cho bit s phn b nng lng ca tn hiu s x(n)

trn trc tn s. th ph pha )( cho bit quan h v pha gia cc thnh phn tns ca ph tn hiu.

Phng php phn tch tn hiu s x(n) da trn cc c trng ph ca n c

gi l phng php tn s, hay phng php phn tch ph, n thng c s dng

x l s tn hiu m thanh.

3.2.2 Ph ca tn hiu lin tc x(t) v tn hiu ly mu x(n.T)

a. nh l ly mu

nh l ly mu l c s ri rc ho tn hiu lin tc m khng lm mt

thng tin ca n, v v th c th khi phc tn hiu lin tc t tn hiu ly mu.

Gio trnh l thuyt mch trnh by v chng minh nh l ly mu, do

y ch nhc li ni dung ca nh l.

nh l ly mu:Mi tn hiu lin tc x(t) c ph hu hnf fc u hon ton

c xc nh bi cc gi tr tc thi ri rc ca n ti cc thi im cch u nhau

mt khong thi gian T 1/2fc (Tng ng T / c )

nh l ly mu nu ln hai iu kin bt buc phi c m bo vic ly

mu khng lm mt thng tin ca tn hiu lin tc:1. Tn hiu lin tc x(t) phi c ph hu hnf fc

2. Chu k ly mu T phi tho mn iu kin T 1/2 fc

b. Ph ca tn hiu lin tc x(t) v ph ca tn hiu ly mu x(n.T)

thy c bn cht vt l ca nh l ly mu, chng ta s xc nh quan h

gia hm ph X() ca tn hiu lin tc x(t) v hm ph X( je ) ca tn hiu ly

x(n.T) tng ng.

Xt tn hiu lin tc x(t) c ph hu hnf< cf (hay c< ), quan h gia x(t) v

ph ca n l cp tch phn Fourier:

Bin i Fourier thun: ( ) ( ). .j tX x t e dt

= & [3.2-1]

Bin i Fourier ngc:1

( ) ( ). .2

j tx t X e d

= & [3.2-2]

Khi ri rc ho tn hiu lin tc x(t) vi chu k ly mu T, nhn c tn hiu

ly mux(n.T). Quan h gia x(nT) v hm ph X(e j ) ca n l cp bin i Fourier

ca tn hiu s [3.1-23] v [3.1-24], khi thay bin n bng bin (n.T):

Bin i Fourier thun: ( ) ( ).j j nTn

X e x nT e

=

= [3.2-3]

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Bin i Fourier ngc: ( ) ( ). .2

T

T

j j nTTx nT X e e d

= [3.2-4]Khi thc hin ri rc ho tn hiu lin tc x(t) theo nh l ly mu th x(n.T) =

x(t)| Tnt .= nn ta c th vit li [3.2-2] di dng:

x(n.T)=

deXT nTjc

c

)(2

[3.2-5]

Khi gi tr ca x(n.T) ti thi im n=k c xc nh l:

x(n.T)| kn= =

deXT nTj

Tk

Tk

+

)12(

)12(

)(2

= x(n.T)=

dekT

XT nTj

T

T

+ )2

(2

Biu thc trn nhn c do tnh cht tun hon ca hm m nTje .

Khi cho k bin thin t - n + nhn c:

x(n.T) =

=k

kTx )( Hay: x(n.T)=

dekT

X nTjT

T

k

).2

(21

+

=[3.2-6]

Nhn v chia [3.2-6] cho chu k trch mu t, ng thi i th t ca du tng

v du tch phn, nhn c biu thc:

x(n.T)=

dekT

XT

T nTjT

T

k

).2

(1

2+

+

=

[3.2-7]

So snh biu thc di du tch phn [3.2-7] nhn c:

x(n.T)= )2

(1

kT

XT k

+

=

[3.2-8]

Biu thc [3.2-8] cho thy rng hm ph X(e j ) ca tn hiu ly mu x(n.T) l

hm tun hon ca bin tn s gc vi chu k T/21 = v l tng v s cc hm

ph )(

X ca tn hiu lin tc x(t).

Trng hp tn hiu lin tc x(t) c ph hu hn v chu k ly mu T tho mn

iu kin ca nh l ly mu: T c/ , th ph X(e j ) ca tn hiu ly mu x(n.T) c

chu k cT 2 . Khi ph X(e j ) l tng ca cc ph )(

X hu hn tch bit nhau

nh trn cc th hnh 3.4 v hnh 3.5, nn ng vi mi gi tr k, ph ca tn hiu ly

mu x(n.T) c dng ng vi ph ca tn hiu lin tc x(t) nhng bin b gim T

ln: x(e j )= )(1

XT

. V th khi cho tn hiu ly mu x(n.T) i qua b lc thng thp

ly thnh phn ph ca X(ej

) ng vi k=0, s nhn c ng ph )(

X . Do khi phc c tn hiu x(t).

115


15/21

Trng hp tn hiu lin tc x(t) c ph hu hn nhng chu k ly mu khng

tho mn iu kin ca nh l ly mu: T> c/ , th ph X(e j ) ca tn hiu ly mu

x(n.T) s c chu k 1 < c . Khi ph X(e j ) l tng ca cc ph )(

X hu hn c

cc bin tn trm ln nhau nh hnh 3.6. S trm ph lm cho X(e j ) b mo dng so

vi ph )(

X ca tn hiu lin tc x(t), v th khng th khi phc tn hiu lin tc x(t)

nh tn hiu ly mu x(n.T).

Trng hp tn hiu lin tc x(t) c ph hu hn nh trn hnh 3.7 th chc chn

xy ra hin tng trm ph, nn ph ca tn hiu ly mu x(n.T) khng th c dng

ging vi ph ca tn hiu lin tc x(t), do khng th khi phc c tn hiu lin

tc x(t) t tn hiu ly mu x(n.T).

Nh vy bn cht vt l ca vic ri rc ho tn hiu lin tc x(t) m khng lm

mt thng tin trong n l ch, khi m bo cc iu kin ca nh l ly mu x(n.T)

ca ph X(e j ) tun hon v mi chu k ca ph X(e j ) hon ton ging vi ph

)( jeX

ca tn hiu lin tc x(t), do thng tin ca tn hiu lin tc x(t) c bo

ton trong tn hiu ly mu x(n.T).

Nh vy, khi c ri rc ho theo ng iu kin ca nh l ly mu, th

rng ph ca mt chu k ph tn hiu s ng bng rng ph ca tn hiu lin tc.

Do khng gy mo tn hiu s th di thng ca h x l s phi rng ph

ca tn hiu lin tc tng ng.

116

X )( je

)(

X

c

c

Hnh 3.2:Tn hiu lin tc x(t) c ph X(hu hn: /th


16/21

3.3 c tnh tn s v hm truyn t phc ca h x l s TTBBNQ

3.3.1 c tnh tn s v hm truyn t phc H )( j

ea. nh ngha: c tnh tn s H )( je ca h x l s TTBBNQ l bin i Fourier

ca c tnh xung h(n)

H )( je =FT[h(n)] =

=

n

njenh ).( [3.3-1]

c tnh tn s H )( je cho bit tnh cht ca tn s ca h x l s TTBBNQ.

Xt h x l s c c tnh xung h(n), tc ng x(n), v phn ng y(n)

c tnh tn s:H )(j

e =FT [h(n)]Tc ng: X( )je = FT[x(n)]

Phn ng: Y( )je = FT[y(n)] = FT[x(n)*h(n)]

Theo tnh cht chp ca bin i Fourier nhn c:

Y( )je =X( )je .H( )je [3.3-2]

Suy ra: H( )je =)(

)Y(ej

jeX[3.3-3]

Nh vy, c tnh tn s ( )j

H e

ca h x l s TTBBNQ bng t s gia hmph ca phn ng ( )jY e v hm ph tc ng ( )jX e , v th ( )jH e cng chnh l

hm truyn t phc ca h x l s TTBBNQ.

C th tm c c tnh xung h(n) ca h x l s t c tnh tn s H( )je

bng bin i Fourier ngc:

h(n)= IFT[H( )je ] =

deeH njj .).(2

1[3.3-4]

b. c tnh bin tn s v c tnh pha tn s

117

c

c

X

Hnh 3.6:Tn hiu lin tc x(t), c ph X( v hn


17/21

T [3.3-3] ta c:)(

)()(

j

j

j

eX

eYeH = [3.3-5]

V: Arg [H( )je ] = Arg [Y( )je ]- Arg[X( )je ] [3.3-6]

Vy modul hm truyn t phc ( )jH e ca h x l s bng t s gia ph

bin ca phn ng v ph bin ca tc ng, cn Argument ca hm truyn t

phc Arg [H( )je ] bng hiu ph pha ca phn ng v ph pha ca tc ng.

V ngha vt l, modul hm truyn t phc ( )jH e c trng cho tnh cht

chn lc tn hiu theo tn s ca h x l s TTBBNQ, v th ( )jH e cn c gi l

c tnh bin tn s.

Cn Argument ca hm truyn t phc )( cho bit dch pha ca cc thnh

phn tn s tn hiu khi truyn qua h x l s TTBBNQ, v th )( = Arg[H )(j

e ]cn c gi l c tnh pha tn s.

tn hiu s khng b mo ph khi truyn qua h x l s TTBBNQ th c

tnh bin tn s ca h x l s phi m bo cho qua tt c cc thnh phn, tn s

ca tn hiu vi h s truyn t l nh nhau. Tc l, v l tng h x l s phi c

c tnh bin tn s dng hnh ch nht nh hnh 3.7a. Tuy nhin, cc h x l s

thc t c c tnh bin tn s vi s nhp nh v hai sn dc, v d nh hnh

3.7b.

Khi nim v di thng v di chn: Di thng l di tn s m h x l s cho tn

hiu i qua, di chn l di tn s m h x l s khng cho tn hiu i qua.

i vi h x l s l tng: Di thng 2 l vng c gi tr ca c tnh bin

tn s )( jeH (hnh 3.7a), cn tn s gii hn di thng v di chn gi l tn s ct

v thng c k hiu l c .

i vi h x l s thc t: Tn s gii hn ca di thng l c

, tn s gii hnca di chn l p . gia di thng v di chn c mt di qu cpp = (hnh

118

1 2( ) ( ) ( )x n x n x n=)( jeH

c c

1 2( ) ( ) ( )x n x n x n=

p

c

0.00 (jeH

2 2

Hnh 3.7:c tnh bin tn s ca h x l sa. H x l s l tng b. H x l s thc t


18/21

3.4b). Nu rng di qu p cng nh th dc hai bin tn ca c tnh bin

tn s )(jeH cng ln, nn tnh cht chn lc tn hiu theo tn s ca h x l s

cng tt.

Cc tn hiu s c ph nm trn trong di thng ca c tnh bin tn s s

i qua c h x l s v khng b mo dng ph. Cc tn hiu s c b rng ph ln

hn di thng s b mt cc thnh phn ph nm ngoi di thng. Cc tn hiu s c

ph nm ngoi di thng ca h x l s s hu nh b gim hon ton khi i qua h x

l s. T cc hiu ng ngi ta xy dng cc h x l s c tnh chn lc tn hiu

theo tn s, l b lc s.

V d 3.14: H x l s c phn ng y(n)=6. n2 .u(n-1)-2. (n-1) ng vi tc ng

x(n) = 2 n u(n). Hy xc nh hm truyn t phc )( jeH ,c tnh xung h(n),v cc

c tnh tn s )( jeH , )(

Gii: C: X( je )=FT [2 )(nun ]= )5.01(1

je

V y(n)= 6. n2 .u(n-1)-2. (n-1)=6.2 )1(2)1(2 )1(1 nnun

Nn: Y( je )=FT[6.2 )1(2)1(2 )1(1 nnun ]=

jj

j

ee

e

2)5.01(

3

Y(

j

e )=

j

jj

j

jji

e

ee

e

eee

+

=

+

5.015.01

23 22

Theo[3.3-3] c: H ( je )= =)(

)(

j

j

eX

eY

j

jjj

e

eee

+

5,01

5,01)(( 2

Hm truyn t phc:H( je )= 2jj ee +

c tnh xung: h(n) = IFT[H( je )]= )1()2()1( 2 =+ nrectnn

tm c tnh bin tn s v pha tn s,bin i H( je ) nh sau:

H( je )= 2jj ee + =(cos )2sin(sin)2cos ++ j

H(j

e )=2cos )5,0cos()5,1sin(2)5.0cos().5,1( j)5,0(cos)5,1(sin4)5,0(cos)5,1(cos4( 2222) +=jeH

c tnh bin tn s: )5,0cos(2( ) =jeH

])5,1cos(

)5,1sin([]

)5,0cos()5,1cos(2

)5,0cos()5,1sin(2[)(

arctgarctg =

=

c tnh pha tn s 5,1)( =

Vy hm truyn t phc l: 5,1)5,0cos(2)( jj eeH =

V d 3.15: Cho h x l s c c tnh xung h(n) =rect 2 (n-1) v tc ng x(n) = 2

)(nun , hy tm hm ph Y( je ) v phn ng y(n).

119


19/21

Gii: y l bi ton ngc ca v d 3.13, trc ht cn xc nh:

Hm truyn t phc:H( je )=FT[h(n)] =

=

n

njenrect )1(2

Hay: H( je )= 2

2

1

jj

n

nj eee

=

+=

Ph ca tc ng: X( )je = FT[2 )(nun ] =)5,01(

1je

Theo biu thc [3.3-2] tm c ph phn ng:

Y(e )().() jjj eHeX= =)5,01(

1je .

)( 2 jj ee +

Y( je )= e j )5,01(1

je + )5,01(1

je e2j

Phn ng:y(n) = IFT[Y(j

e )] = 0,5 )1( n u(n-1) + 0,5 )2( n u(n-2)Y(n) =2.2 n u(n-1)+22.2-n[u(n-1)- (n-1)

y(n) = 2.2-nu(n-1)+4.2-n[u(n-1)- (n-1)]

y(n) = 6.2-nu(n-1)-4.2-n (n-1)

V (n-1) ch c 1 mu ti n= 1, nn 4.2-n (n-1) = 2 (n-1), do kt qu l:

y(n) = 6.2-nu(n-1)-2 (n-1)

Kt qu nhn c ph hp vi phn ng y(n) cho v d 3.13

c. Tm hm truyn t phc H( je ) theo phng trnh sai phn

C th tm c hm truyn t phc ca h x l s TTBBNQ khi bit phng

trnh sai phn ca n. Xt h x l s TTBBNQ c m t bng phng trnh sai

phn bc N:

=

N

rr rnya

0

)( ==

M

kk knxb

0

)(

Ly bin i Fourier ca hai v ca phng trnh trn ta c:

=

N

r

jrjr eYea

0

)( = =

M

k

jkj

k eXeb0

)(

Suy ra: H( je )=)(

)(

j

j

eX

eY=

=

=

N

r

rjr

M

k

kjk

ea

eb

0

0

[3.3-7]

V d 3.16: Hy xc nh hm truyn t phc v cc c tnh tn s ca h x l s

c phng trnh sai phny(n) = x(n) +x(n-1)+y(n-2)

Gii: Ly bin i Fourier c hai v ca phng trnh trn ta nhn c:

Y(

j

e )=X(

j

e )+

j

e

X(

j

e )+

2j

e

Y(

j

e )Hay: Y( je ).(1- 2je )= X( je ).(1+ je )

120


20/21

Vy: H( je ) =)(

)(

j

j

eX

eY=

)1(

)1(2

j

j

e

e

+

=)1(

1je

Hm truyn t phc: H( je ) =)1(

1je

=)sincos1(

1

j+

c tnh bin tn s: )( jeH = 22 sin)cos1(

1 + = )5,0sin(1

c tnh tn s: Arg[H( je )] = arctg[)cos1(

sin

] =

2

Vy hm truyn t phc l: H( je ) =)5,0sin(

5,0

je

3.3.2 Phn tch h x l s theo hm truyn t phc H( je )

a. S khi, s cu trc trong min tn s ca h x l s

Theo quan h vo ra [3-51]: Y( je ) = X( je ).H( je ), c th m t h x l s TTBBbng s khi theo hm truyn t phc nh hnh 3.8

Cc h x l s phc tp c th c m t bng s khi gm nhiu khi,

mi khi c hm truyn t phc H( je ). Khi , hm truyn t phc H( je ) ca h

x l s cng c th c xc nh theo cc hm truyn t phc Hi(j

e ) ca cckhi thnh phn.

V H( je ) = H(z) jez= , nn t cc phn t cu trc v s khi theo hm h

thng H(z) c th nhn c cc phn t cu trc v s khi theo hm truyn t

phc H( je ) theo s khi cng tng t nh cch xc nh hm h thng H(z). S

khi v s cu trc trong min tn s ca h x l s thng c s dng khi

phn tch v tng hp cc b lc s.

V d3.17: Hy xy dng s khi v s cu trc trong min tn s ca h x l

s cho v d 3.16: y(n)= x(n)+x(n-1)+y(n-2).

Gii: Theo kt qu v d 3.14, c s khi nh trn hnh 3.9

Hnh 3.9:S khi trong min tn s ca h x l s v d 3.16

Ly bin i Fourier c hai v ca phng trnh sai phn cho, nhn c

quan h vo ra: Y(j

e ) = X(j

e )+j

e

X(j

e )+2j

e

Y(j

e )

121

)5,0s i n (

5,0

jeX() Y()

H()X() Y()

Hnh 3.8:S khi trong min tn s ca h x l s


21/21

Theo quan h vo ra trn, xy dng c s cu trc ca h x l s cho

trn hnh 3.10:

b. Xt tnh n nh ca h x l s theo H( je )

Theo nh ngha ca bin i Fourier, ch c cc h x l s c c tnh xung

h(n) tho mn iu kin [3.1-1]:

=

chuong 3 xu ly tin hieu so

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