Trng i hc Khoa hc Hu Khoa CNTT

Gio trnh

Hu, thng 5 nm 2008

1

Li m u

X l tn hiu s l mt lnh vc c tnh nn tng cho nhiu lnh vc khoa hc khc nh x l m thanh, x l nh, truyn tin v vin thng, y hc, quan s v vin thm. N v ang c quan tm nhiu trn th gii. Trong khun kh gio trnh ny chng ti ch trnh by nhng khi nim v k nng c bn, nhng sinh vin chuyn ngnh cng ngh thng tin c c mt tm nhn v lnh vc x l tn hiu s, cng nh c c lng kin thc v k nng ti thiu c th bc u bc vo mt lnh vc x l tn hiu c th m th gii ang quan tm.

Do thi gian son tho gio trnh ny qu hn ch nn chng ti s khng th trnh khi c nhng sai st. Rt mong nhn c s thng cm ca nhng sinh vin hay c gi quan tm tm hiu n ti liu ny. V chng ti ha hn s c nhng bc sa i v nng cp cung cp thm khin thc v cp nht nhng kin thc mi cho kp ha nhp vi dng chy khoa hc.

Cui cng ti xin c gi li cm n chn thnh n Gio S L nh Chn Tm! Ngi Thy dn dt ti trn con ng hc tp v nghin cu trn lnh vc x l tn hiu s v x l nh.

Tc gi:

Nguyn Hu Ti

2

TI LIU THAM KHO Le Dinh Chon Tam, Notes de cours Ingnierie des systmes

numriques. Van Den Enden & N. Verhoeckx, Traitement numrique du

signal: Une introduction, Masson 1992, Ch 1, 3, 4, 5, 7, 8, 9. Sanjit . Mitra, Digital Signal Processing: A Computer-Based

Approach 3rd Edition, Mc Graw-Hill 2006, Ch 1-10, 12-15. S. Haykin, Adaptive Filter Theory, 4th Edition, Prentice Hall

2002, Ch 4-7, 17. J. G. Proakis & D. G. Manolakis, Introduction of Digital Signal

Processing, Macmillan Publishing Company, NewYork, 1988 J. Astola & P. Kuosmanen, Fundamentals of Nonlinear Digital

Filtering, CRC Press, 1997. R. Weeks, Jr., Fundamentals of Electronic Image Processing,

IEEE Press, 1996. Signal Processing ToolBox: MatLab Nguyn Quc Trung, X l tn hiu v lc s, Tp 1 v 2, Nh

xut bn Khoa hc K thut, H Ni, 2002. Tng Vn n, L thuyt v bi tp X l tn hiu s, Nh xut

bn Lao ng-X hi, 2002. Digital Signal Processing, Second Edition, www.DSPguide.com

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MC LC

Chng 1: PHP BIN I QUA LI GIA TN HIU LIN TC V TN HIU RI RC ........................................................................................................ 5

1.1 Tng quan v mt h thng x l tn hiu s .............................................. 5 1.2 Ly mu tn hiu (SAMPLING) .................................................................. 7 1.3 Bin i t tn hiu lin tc ra tn hiu s (A/D converter) ...................... 14 1.4 B i t tn hiu s ra lin tc (D/A) ....................................................... 16 1.5 Tn s chun ha (normalized frequence) ................................................ 19

Chng 2: H THNG V TN HIU RI RC .......................................... 21 2.1 Biu din tn hiu ri rc trong min thi gian ......................................... 21

2.1.1 Tn hiu xung n v: ......................................................................... 21 2.1.2 Tn hiu bc n v: ......................................................................... 22 2.1.3 Tn hiu sin ri rc: ............................................................................. 22

2.2 Biu din tn hiu ri rc trong min tn s .............................................. 24 2.2.1 nh ngha php bin i Fourier cho tn hiu ri rc (DTFT):......... 24

2.3 H thng ri rc tuyn tnh bt bin theo thi gian(LTI system) .............. 35 2.3.1 Cc nh ngha c bn: ....................................................................... 36 2.3.2 Biu din h thng bng phng trnh sai bit: .................................. 38 2.3.3 Biu din h thng bng p ng xung .............................................. 38

2.4 Tch chp ................................................................................................... 41 2.4.1 Php tch trc tip ............................................................................... 41

2.5 Biu din h thng LTI bng p ng tn s ............................................ 44 2.6 Tng quan (Correlation) .......................................................................... 46 2.7 Bin i Z .................................................................................................. 52 2.8 Biu din h thng LTI trong min Z........................................................ 59

Chng 3: BIN I FOURIER RI RC .................................................... 64 3.1 DFT cho tn hiu ri rc tun hon ........................................................... 64 3.2 DFT CHO TN HIU RI RC C CHIU DI HU HN ............... 66 3.3 Cc tnh cht ca bin i DFT ................................................................. 67 3.4 Cch chn N trong tnh ton DFT ............................................................. 67 Trong trng hp ny, ch c bin i DTFT l c th biu din chnh xc tn hiu trong min tn s. i vi bin i DFT, ta phi chn N ln. .............. 72 3.5 TCH CHP VNG (CIRCULAR CONVOLUTION) ........................... 72 Hnh 3.11 Tnh tch chp thng bng cch phn on mt tn hiu di ........ 76 3.6 ng dng ca DFT trn tn hiu lin tc ................................................... 76 3.7 Bin i Fourier nhanh (Fast Fourier Transform, FFT) ............................ 83

Chng 4: CU TRC B LC ..................................................................... 89 4.1 Cu trc b lc FIR ................................................................................... 89 4.2 CU TRC B LC IIR ......................................................................... 94 4.3 NHNG CU TRC B LC C BIT ............................................ 98 4.4 HIN TNG PHN LNG T CC H S: ................................ 107

4.4.1 NHY (SENSIBILITY): ........................................................... 107 4.4.2 Cu trc kt hp (coupled structure) : .............................................. 109

Chng 5: THIT K B LC S ............................................................... 112

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5.1 Thit k b lc FIR ................................................................................. 112 5.1.1 C TNH CA B LC FIR C PHA TUYN TNH .............. 113 5.1.2 PHNG PHP CA S .............................................................. 115 5.1.3 THIT K DA VO DFT - PHNG PHP LY MU T MIN TN S ............................................................................................. 127 5.1.4 THIT K VI SAI GN SNG U (EQUIRIPPLE): THUT TON REMEZ .............................................................................. 134 5.1.5 MT THIT K KHC .................................................................. 143

5.2 BI TP ................................................................................................. 146 5.3 B LC DA TRN THNG K XP TH T ............................... 147

5.3.1 B lc trung v (median): ................................................................. 147 Chng 6: H THNG A NHP & B LC A PHA ............................. 151

6.1 Php gim nhp vi h s nguyn D (Decimation) ................................. 152 6.2 Php ni suy tng nhp vi h s nguyn U ............................................ 156 6.3 Php bin nhp vi h s hu t .............................................................. 159 6.4 B lc nhiu pha (hay a pha, Polyphase filter) ..................................... 160 6.5 Thc hin b phn chia gim nhp .......................................................... 161 6.6 Thc hin php ni suy tng nhp ........................................................... 163

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Chng 1: PHP BIN I QUA LI GIA TN HIU LIN TC V TN HIU RI RC

1.1 Tng quan v mt h thng x l tn hiu s

Hnh 1.1 H thng x l tn hiu s tng ng yd(t) y(t)

Hnh 1.1 m t kin trc mt h thng x l tn hiu s. V mt l thuyt chng ta c th thit k mt h thng x l tn hiu s tng ng vi mt h thng x l tn hiu lin tc mong c hay sn c no .

y: o x(t): tn hin lin tc (analog signal) u vo cn x l o y(t): tn hiu lin tc u ra mong mun o x(n): tn hiu s (digtal signal) nhn c t tn hiu lin tc

u vo x(t) sau qu trnh tin x l o y(n): tn hin s u ra ca qu trnh x l tn hiu s o yd(t): tn hiu lin tc nhn c t tn hiu ri rc y(n) qua

qu trnh hu x l. Qu trnh tin x l bao gm giai on lc chng chng ph

ri tip n l giai on bin i A/D (Analog to Digital) nhm bin i tn hiu lin tc sang tn hiu s, kt qu thu c mt tn hiu c biu din bi chui s (c tnh cht ri rc nn cn c th gi l tn hiu s ri rc)

Qu trnh hu x l bao gm php bin i ngc li D/A ri tip n l lc khi phc tn hiu lin tc.

X l tn hiu lin tc

X l tn hiu s

Lc chng

chng ph

Lc khi

phc

Bin i

A/D

Bin i

D/A

tn hiu i vo tn hiu i ra x(t) y(t)

x(n) y(n)

Tin x l Hu x l

yd(t)

ng h ly mu

H thng x l tn hiu s

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Mt h thng x l tn hiu s, vi c trng tn hiu c s ha, nn c nhng thun li khi x l trn my tnh hay cc vi mch s, iu m mt h thng x l tn hiu lin tc khng th c c.

T khi khoa hc my tnh ra i m ra mt cnh ca mi vi s pht trin vt bc ca lnh vc x l tn hiu s vi nhiu ng dng mi v cht lng mi. X l tn hiu s l nn tng ca cng ngh truyn tin v vin thng hin i ang bng n trn th gii hin nay cng nh cng ngh x l nh, x l m thanh v rt nhiu lnh vc khc na.

Mt cch hnh thc c th m t mt qu trnh x l tn hiu s nh l mt php bin i T bin i u vo x(n) thnh u ra y(n) bi:

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

cu to ca mt h thng x l tn hiu s gm c nhng phn t sau

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

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Hnh 1.3 Cc phn t thng dng trong mt h thng s

u im ca H thng x l tn hiu s: tn s thp, chnh xc cao, tnh n nh ca nhng phn t cu to, kch thc nh, v tnh hp nht (integration) c.

Nhc im ca H thng x l tn hiu s: Kh thc hin c nhng tn s cao. V tc ca h thng quyt nh mc tn s c th x l c.

ng h: c chc nng nh nhp ly mu. N quyt nh xung nhp x l ca c h thng x l tn hiu s, nn c mt vai tr rt quan trng. Song n gin ha trong cc s chng ta s khng v n tr nhng trng hp cn thit.

1.2 Ly mu tn hiu (SAMPLING)

nh l ly mu:

Nu tn hiu lin tc x(t) c tn s gii hn M, ngha l:

|X()|=0, vi ||> M (1.1)

V nu x(t) b ly mu bi tn s s = 2/T th

s 2M (1.2) (y c gi l iu kin Nyquist)

tn hiu lin tc x(t) c th c khi phc hon ho t nhng tn hiu

ri rc x(nT), cn p dng mt b lc vi tn s gii hn c nm gia M

v s - M:

M c s - M (1.3)

Hnh 1.4 nh l ly mu v khi phc

Trc khi xem xt chng minh nh l ly mu chng ta cn xem xt khi nim xung Delta

0n khi 0

0n ,1)(

khin

8

Hay

in khi 0

in ,1)(

khiin

Hnh 1.5 H thng xung n v

Theo hnh trn th tn hiu lin tc xs(t) sau ly mu s bng:

xs(t) = x(t).(t-nT)

xs(t) = x(t).(t-nT) = x(nT).(t-nT)

(v nhng gi tr tnT th x(t).(t-nT)=x(t).0=0) Hn na, theo php bin i Fourier lin tc (s tm hiu chng sau) ta c:

2( ) ( )F s

n n

t nT nT

X xs(t) x(t)

(t-nT)

0 T 2T t -T -2T -3T . .

(t-nT)

0 T 2T t -T -2T -3T . .

x(t) x(0)

x(T)

x(2T)

Hnh 1.6 Ly mu

xs(t)

1

-3 -2 -1 0 1 2 3 n

(n) 1

i-3 i-2 i-1 i i+1 i+2 i+3 n

(n-i)

9

vi s=2/T Nh vy:

1 2( ) ( ) * ( )

2s snX X n

T

1( ) ( )s s

n

X X nT

(1.4)

Phng trnh (1.4) cho kt qu ph nh hnh sau y:

Hnh 1.7 c tnh ca x(t) v xs(t) trong min tn s

Ch : Trong min tn s, bin i Fourier ca mt tn hiu mu c tnh tun hon

vi chu k s.

Phn tch v php khi phc li tn hiu lin tc:

Hnh 1.8 Khi phc tn hiu lin tc t tn hiu ri rc bng ni suy

Trong , tn hiu khi phc l tch chp gia tn hiu i vo v p ng xung ca b lc:

xr(t) = xs(t)*h(t)

khi tn hiu i vo l tn hiu mu:

xr(t) = x(nT).(t-nT) v p ng xung ca b lc:

B lc l tng h(t)

vi tn s ct c

s c s - M

xs(t) xr(t)=x(t)

1

X()

-M M 0

1/T

Xs()

-M M 0 s-M s 2s -s -2s

10

sin( )( ) c c

c

T th t

t

cho nn ta s c:

sin[ ( )]( ) ( )

( )c c

rcn

T t nTx t x nT

t nT (1.5)

Trong trng hp c = 12s , th ta c:

sin[ ( )/2]( ) ( )

( )/2s

rsn

t nTx t x nT

t nT (1.6)

Hnh di y minh ha php ni suy do phng trnh (1.6) mang li

Hnh 1.9 Ni suy tn hiu lin tc t nhng tn hiu ri rc thng qua b lc sau.

Ch : Trong tnh hung chng ta khng ly vi tn s s < 2M (hay ni cch khc l ly mu khng nhanh) th chng ta c th gp hin tng ri lon ph hoc ph chng nu khng p dng b lc trc nh minh ha di y:

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Hnh 1.10 Hiu ng ph ri lon v b lc chng chng ph

Th d:

Cho tn hiu f(t) = cos(0t) vi tn s 0 c nh. Tn hiu f(t) c ly mu

vi tn s s v phc hi bng mt b lc thng thp (low-pass) vi tn s s/2. ta c:

Nu s 20 th tn hiu ph hi c ng cos(0t)

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Nu s < 20 th tn hiu ph hi l cos(s - 0t). Ngh l khng phc hi ng tn s ban u.

chng ta c th thy r iu ny qua hnh v minh ha kt qu thc hin trong min tn s v min thi gian nh sau:

Hnh 1.11 Minh ha th d chng ph trong min tn s

13

14

Hnh 1.12 Minh ha th d trong min thi gian V d mt m thanh c tn s 1800Hz c ly mu 3 tn s ly mu khc nhau s cho kt qu nh sau:

Ly mu tn s 8000Hz :

Ly mu tn s 6000Hz :

Ly mu tn s 2666.6667Hz :

Ni chung khi ly mu 1 tn hiu, chng hn nh m thanh pht ra t ging ni ca con ngi c tn s di 2 kHz, song bn cnh cn c nhng tp m c tn s cao hn 2 kHz. Do nu ta ly mu vi tn s 4 kHz (2 ln tn s ging ni) m khng p dng b lc trc, th nhng tp m c tn s cao hn 2 kHz s lm cho ph b ri lon (hay chng ph) nh trong hnh 1.10 trn, kt qu khin cho m thanh khi ti hin li b bp mo tn s, khng cn nghe r c na. Nn vic p dng b lc trc l iu ti quan trng trong vic ly mu. Ngy nay, tn s ly mu m thanh trong cc file m thanh s thng dng t 8 kHz (cht lng thp Low quality) cho n 44.1 kHz (cht lng cao High quality). Ty theo yu cu cng nh iu kin hon cnh m ngi ta chn tn s ly mu cho ph hp.

Trong tn hiu nh, nu chng ta ly mu tn s thp th kt qu nhng tn s cao s khng thu c, m nhng tn s cao tng ng vi nhng ng nt chi tit trn nh, kt qu l nh trng s khng c r nt, hay b mt cc ng nt chi tit. Ngy nay, khi k thut vi x l ngy cng pht trin, iu cho php cc thit b chp nh hot ng tn s cao hn, nn c th ly mu tn s cao hn nhiu so vi trc y, khin cho nhng bc nh thu c rt r nt, vi y cc ng nt chi tit trn bc nh.

1.3 Bin i t tn hiu lin tc ra tn hiu s (A/D converter)

B bin i A/D gm c b ly mu v gi (sample and hold) sau mt b bin i lng t (quantizer) t bin lin tc ra s.

Hnh 1.13 B i tn hiu lin tc ra s (A/D)

Sample & Hold

(B ly mu v gi: bin i t tn hiu

lin tc ra ri rc)

Quantizer

(B lng t: i bin lin tc ra s theo

cc mc lng t)

x(t) xq(nT) x(nT)

15

Hnh 1.13 Minh ha qu trnh lng t.

B bin i bin (Quantizer) bin nhng gi tr lin tc ca x(nT) ra thnh nhng gi tr ri rc theo mt in p quy chiu (reference voltage) vi s mc lng t cho sn. Nh vy mi mc lng t c th biu din bi mt con s, v tin nht cho vic x l l dng s nguyn nh phn. V d, nu chng ta lng t vi tng s 256 mc, th c th dng 8 bt lng t ha, vi mc thp nht s l 00000000 (gi tr 0), mc tip theo l 00000001 (gi tr 1) v mc cao nht l 11111111 (gi tr 255).

Php phn lng thng l chia u, (uniform quantization). V sai s ca phn lng l e(nT) c xem nh l nhiu, v c im ca nhiu ny l c mt ph u (uniform spectral density) v mt xc xut u (uniform probability density).

16

Hnh 1.14 Minh ha php phn lng u vi n=8 mc (3 bt) v khong phn

lng l a volt v bin gii hn i vo l 2 volt.

1.4 B i t tn hiu s ra lin tc (D/A)

B bin i D/A thc tin c xem nh l mt b D/A l tng (i t s ra bin ) theo sau l mt b gi bin bc khng (zero order hold)

17

Hnh 1.15 B i A/D thc tin

B gi bc khng b(t) s to ra tn hiu di dng bc thang trong min thi gian, hay dng (Sinx)/x trong min ph tn s. Do b lc khi phc ngoi vic lm php ni suy, cn phi lm gim (hay b tr) hin tng bc thang ny.

Hnh 1.16 Hin tng bc thang

xt b lc sau:

ta c pT/2(t-T/2)=b(t)

T chng ta c:

xb(t)=x(nT)h(t-nT) = x(nT)pT/2(t-nT-T/2) = pT/2(t)* x(nT). (t-nT-T/2)

xt trong min tn s th:

/2

( ) ( )(a)

sin( /2)( ) . ( )

/2j T

b sb c

TX e TX

T (1.7)

trong :

-T/2 -T/2 0

pT/2(t)

18

+ phn (a) th hin s bin dng trong tn s, hay cn gi l hin tng (sinx)/x ca b gi bc khng + phn (b) th hin tr T/2 ca b gi bc khng b(t) + phn (c) chnh l bin i Fourier ca tn hiu Hnh sau y m t hiu ng (Sinx)/x ca b gi bc khng ca tn hiu trong min tn s:

Hnh 1.17 Hiu ng (sinx)/x ln ph ca tn hiu

b tr li hiu ng ca (sinx)/x, th b lc khi phc Fr() phi tha mn iu kin sau y:

M

M

1, | |sin( /2)( ).

/2 0, | |r

TF

T

hay

19

M

M

/2, | |

sin( /2)( )0, | |

r

TTF (1.8)

Ch : Qua nghin cu ngi ta thy rng: khi tn s ly mu kh cao, ngha l

2s M , th hiu ng bin dng (sinx)/x s t i hay khng ng k. l mt li im nu chng ta c th ly mu nhiu hn tn s ly mu ti thiu m nh l Nyquits i hi.

Ngi ta cn phn tch b lc khi phc Fr() = FC()FA()

trong FA() l mt b lc lin tc thng thp vi nhn (gain) bng 1, tn s

ct l M (ngha l b lc ny ging vi b lc chng chng ph t pha trc

A/D) . Cn FC() s dng ring cho hin tng (sinx)/x. Nh vy ta c:

M

M

1, | |( )

0, | |AF

v,

M

M

/2, | |

sin( /2)( )0, | |

C

TTF (1.9)

Nu c b bin i D/A chng ta s c 2 cch sp xp sau:

Hnh 1.18 Hai cch thc hin b khi phc tn hiu lin tc t tn hiu s

1.5 Tn s chun ha (normalized frequence)

Trong x l tn hiu ri rc, bin i Fourier ca mt tn hiu mu c tnh tun

hon vi chu k s = 2/T. T ngi ta nh ngha tn s chun ha nh sau:

= T = rad/chu k ly mu

20

= /T = rad / giy

Tn s chun ha c chu k 2: hay -< 0 cng xa cng tt.

V nh cu hi a) cho hai hm HC(T) v GC(T) c) So snh gia a) v b) c kt lun gia H v HC b lc no hay hn.

21

Chng 2: H THNG V TN HIU RI RC

2.1 Biu din tn hiu ri rc trong min thi gian

Tn hiu lin tc x(t) c ly mu theo chu k T s c gi tr duy nht ti cc thi im t=nT l x(nT) vi n=0,1,2,...

V mt l thuyt chng ta c th xem qu trnh ly mu c thc hin t

trong qu kh n tng lai v din ra v hn, th khi n=- +. Song trong thc tin thng chng ta s ly mu bt u ti mt thi im t0 no n thi im t1 th kt thc, nn n c th c xem l t 0.. N hay N .. 0 nu chng ta dng php dch.

n gin chng ta s xem chui tn hiu ri rc x(nT) nh l mt chui x(n).

Di y l mt s tn hiu ri rc quan trng thng c s dng trong cc php phn tch v bin i:

Tn hiu xung n v (unit impluse function)

Tn hiu bc n v (unit step function)

Tn hiu sin ri rc

2.1.1 Tn hiu xung n v:

Nh ni phn trc, khi nim xung Delta c nh ngha nh sau:

0n khi 0

0n ,1)(

khin

Hay nu dng php dch chuyn (shift), ta c

in khi 0

in ,1)(

khiin

Hnh 2.1 H thng xung n v

1

-3 -2 -1 0 1 2 3 n

(n) 1

i-3 i-2 i-1 i i+1 i+2 i+3 n

(n-i)

22

mt tn hiu x(n) bt k u c th phn tch thnh tng ca nhiu tn hiu xung vi bin l bin ca x:

( ) ( ) ( )i

x n x i n i

2.1.2 Tn hiu bc n v:

Tn hiu bc n v c nh ngha nh sau:

1, n 0( )

0 khi n 0

khiu n

hay nu dng php dch chuyn ta c: 1, n i

( )0 khi n i

khiu n i

Hnh 2.2 H thng bc n v

2.1.3 Tn hiu sin ri rc:

x(n) = Asin(n+) (2.1) trong , A: bin : tn s chun ha

: Pha

v n+ c tnh theo radian

Mt tn hiu x(n) bt k, c gi l tun hon vi chu k N nu N l gi tr nh nht tha mn iu kin:

x(n)=x(n+N) vi mi gi tr ca n v d:

Sin(n/4)=Sin(n/4+2)=Sin[(n+8) /4], suy ra chu k tun hon

ca Sin(n/4) l 8

Sin(n) khng phi l hm tun hon v khng c gi tr N (nguyn) no tha mn Sin(n)=Sin(n+N). Song tn hiu ny c th xem l mt hm gn tun hon

Hai tn hiu Sin(n/3) v Sin(7n/3) l ging nhau, v:

Sin(7n/3)=Sin[(6n+n)/3]=Sin[2n+n/3]=Sin(n/3)

1

-3 -2 -1 0 1 2 3 .. n

u(n) 1

i-3 i-2 i-1 i i+1 i+2 i+3 .. n

u(n-i)

23

Hnh 2.3 Mt vi tn hiu thng gp

24

2.2 Biu din tn hiu ri rc trong min tn s

Bin i tn hiu sang min tn s gip chng ta thy c nhiu c tnh quan trng ca tn hiu v c th c nhiu phn tch hay tc ng bin i hu hiu. V th php bin i sang khng gian tn s l rt quan trng trong x l tn hiu.

Ngi ta cng dng php bin i Fourier cho tn hiu ri rc v c gi l php bin i DTFT

2.2.1 nh ngha php bin i Fourier cho tn hiu ri rc (DTFT):

Php bin i Fourier cho tn hiu ri rc (Discret Time Fourier Transform hay vit tt l DTFT) l mt php bin i thun nghch, th hin s tng quan gia min thi gian v min tn s, v cho php lin kt vic biu din tn hiu trong min thi gian v trong min tn s.

2.2.1.1 Vi tn s thc:

Php bin i Fourier cho tn hiu ri rc thun c nh ngha nh sau:

( ) ( )j T j nT

n

X e x nT e (2.2)

Php bin i Fourier ca 1 tn hiu ri rc l tng ca cc bin tn

hiu x(nT) c iu ch (modulated) bi e-jnT. Tn hiu bin i X(e jT)

l mt tn hiu lin tc theo tn s v tun hon vi chu k 2/T.

Php bin i ngc c nh ngha bi: /

/

( ) ( ). .2

T

j T j nT

T

Tx nT X e e d

(2.3)

Hai biu thc trn to ra mt cp bin i thun nghch c th vit nh sau:

( ) ( )j Tx nT X e (2.4)

2.2.1.2 Vi tn s chun ha:

Ta c bin i thun

( ) ( )j jn

n

X e x n e (2.5)

Bin i nghch

25

1( ) ( ). .

2j jnx n X e e d

(2.6)

Cp bin i:

( ) ( )jx n X e (2.7)

Mt s im cn lu :

Tn hiu ej l mt tn hiu phc , c th khai trin theo phn thc v phn o theo cng thc Euler di y:

ej = cos()+jsin() (2.8)

suy ra:

e-j

= cos() - jsin() (2.9)

i vi mt tn hiu bin i X(ej), thng l tn hiu phc, chng ta

c th biu din bin A v pha di dng ta cc (pole) hoc phn thc phn o nh sau:

X = A.ej

= A(cos + jsin) = A.cos + jA.sin = Re(X) +j.Im(X)

vi phn thc Re(X) = A.cos

v phn o Im(X) = A.sin

v A = |X| = 2 2 2 2Re( ) Im( ) ( cos ) ( sin )X X A A A Im(X)

Arctag(X)=ArctagRe(X)

mt s sch thng k hiu Arctg = Tan-1 nn:

-1 -1 Im(X)Tan (X)=TanRe(X)

V d: xem xt mt s php bin i di y

Cho x(n) = (n), th ta c:

( ) ( ) ( ) 1j jn jn

n n

X e x n e n e

ej

Re

Im

26

Hnh 2.4 Tng qua gia min thi gian v min tn s ca tn hiu

Cho x(n) = anu(n) vi |a|

27

Hnh 2.5 DTFT ca tn hiu x(n) = an.u(n) vi |a|

28

Hnh 2.5 DTFT ca tn hiu x(n) = a|n| vi |a|

29

30

2.2.1.4 Mt vi cp bin i Fourier ri rc thng dng:

31

32

33

Hnh 2.6 Minh ha cc cp bin i thng dng

2.2.1.5 V d:

34

35

Hnh 2.7 B lc thng di H(ej) c c bng cch ly hiu ca 2 b lc thng

thp H1(ej

) v H2(ej

)

2.3 H thng ri rc tuyn tnh bt bin theo thi gian(LTI system)

H thng ri rc tuyn tnh bt bin theo thi gian ng vai tr quan trng trong nhng h thng thc tin ph bin.

S ca mt h thng biu din nh sau:

Hnh 2.8 S ca mt h thng ni chung

H thng x(n) Tn hiu i vo

y(n)

Tn hiu i ra

2 1 -2 -1

H(ej

)

1

1 -1

H1(ej

)

1

2 -2

H2(ej

)

1

=

_

36

Tn hiu u vo c gi l tn hiu kch thch (excitation), v tn hiu u ra c gi l tn hiu p ng (response). H thng l mt b bin i T (transform) bin i tn hiu kch thch ra thnh tn hiu p ng. C th c vit di dng phng trnh n gin nh sau:

y(n) = T[x(n)] (2.10)

2.3.1 Cc nh ngha c bn:

2.3.1.1 Tnh tuyn tnh

Mt h thng ri rc T c gi l tuyn tnh (Linear) khi v ch khi vi mt t hp tuyn tnh u vo s to ra mt t hp tuyn tnh u ra

nu gi x1(n) v x2(n) l 2 tn hiu u vo c tn hiu u ra tng ng ln lt l y1(n) = T[x1(n)] v y2(n) = T[x2(n)], th t hp tuyn tnh ca 2 tn hiu u vo a1*x1(n) + a2*x2(n) s to ra t hp tuyn tnh 2 tn hiu u ra a1*y1(n) + a2*y2(n).

a1*y1(n) + a2*y2(n) = T[a1*x1(n) + a2*x2(n)] (2.11)

2.3.1.2 Tnh bt bin theo thi gian

Mt h thng T vi u vo x(n) ta thu c y(n)=T[x(n)],

h thng T c gi l bt bin theo thi gian (Time invariant) khi v ch khi: dch chuyn mc thi gian tn hiu u vo s ko theo s dch chuyn mc thi gian tn hiu u ra mt cch tng ng:

y(n-i) = T[x(n-i)] i (2.12)

Mt h thng T c gi l tuyn tnh bt bin (LTI) khi v ch khi tha mn ng thi (2.11)&(2.12)

2.3.1.3 Tnh n nh

Mt h thng ri rc T c gi l n nh (stable) nu vi bt k tn hiu

u vo c c bin hu hn |x(n)|

37

( )n

S h n (2.14)

vi h(n) l p ng xung ca h thng

hiu r c cng thc (2.14) ta cn xem xt mt s phn tch sau y: Cho T l mt h thng tuyn tnh bt bin

tn hiu u vo c phn tch thnh tng ca cc chui xung delta nh sau:

( ) . ( )mm

x n x n m

vi xm l bin

[ ( )] . ( ) [ . ( )]

[ ( )] . [ ( )] . ( )

[ ( )] . ( )

m mm m

m mm m

mm

T x n T x n m T x n m

T x n x T n m x h n m

T x n x h n m

Vy T l h thng n nh khi

| . ( ) | | ( ) | | ( ) |mm m n

x h n m h n m h n

2.3.1.4 Tnh nhn qu

Mt h thng ri rc c gi l c tnh nhn qu nu tn hiu u ra ti thi im n=n0, ch ph thuc vo tn hiu u vo trong qu kh hay hin ti, m khng ph thuc vo cc gi tr ca tn hiu u vo ti cc thi im tng lai. Hay ni cch khc l p ng ca mt h thng nhn qu khng bao gi i trc cc kch thch ca n. H thng LTI l nhn qu khi v ch khi h(n)=0 nu n

38

2.3.2 Biu din h thng bng phng trnh sai bit:

Cc h thng ri rc c biu din bng cc phng trnh sai bit (difference equation) trong khi cc h thng lin tc c biu din bi cc phng trnh sai phn (differential equation)

Phng trnh m t mt h thng tuyn tnh bt bin (LTI) lun c dng

0 1

( ) ( ) ( )N M

i ii i

y n b x n i a y n i (2.15)

Tn hiu u ra ti thi im n l t hp ca M gi tr u ra ca chnh n trong qu kh v (N+1) gi tr ca tn hiu u vo trong qu kh n hin ti.

Trong phng trnh sai bit (2.15) ni trn, nu tt c cc h s ai u bng khng th u ra ca h thng khng ph thuc vo cc gi tr ca chnh n trong qu kh m ch ph thuc vo mt s tn hiu u vo trong hin ti v qu kh m thi. Lc ny h thng s c dng:

0

( ) ( )N

ii

y n b x n i (2.16)

Trong trng hp ny, h thng s c gi l b lc c p ng xung c chiu di hu hn (Finite Impulsional Response filter, hay gi tt l FIR filter).

Trong trng hp ngc li th h thng c gi l b lc c p ng xung c chiu di v hn (Infinite Impulsional Response filter, hay gi tt l IIR filter)

H thng c cho bi phng trnh (2.15) ni trn c th thc hin c bng cch s dng 3 thnh phn: B tr n v (Delay unit), b nhn vi hng s (Gain), v b cng (Add).

2.3.3 Biu din h thng bng p ng xung

Mt h thng LTI c th c biu din bng p ng xung ca chnh n, hay ni cch khc l c biu din bi u ra y(n) = h(n) vi tn hiu u

vo l xung delta x(n) = (n)

p ng ca h thng vi tn hiu u vo x(n) bt k:

Ta phn tch tn hiu x(n) thnh tng ca cc tn hiu xung delta c nhn (gain) vi gi tr xm

39

( ) ( )mm

x n x n m , vi xm ng vai tr bin

th lc ny tn hiu u ra xc nh bi:

( ) [ ( )] [ ( )]mm

y n T x n T x n m

do tnh tuyn tnh nn ta c:

[ ( )] [ ( )]m mm m

T x n m x T n m

do tnh bt bin theo thi gian nn:

[ ( )]= ( )m mm m

x T n m x h n m

vy cui cng ta c:

( ) ( )mm

y n x h n m (2.17)

Biu thc trn (2.17) cn c gi l tch chp ca x(n) v h(n). Vy u ra ca h thng l y(n) c tnh bng tch chp ca x(n) v h(n), v c vit nh sau:

y(n) = x(n)*h(n) (2.18)

p ng xung ca h thng l b lc FIR:

T phng trnh biu din ca h thng b l FIR l

0

( ) ( )N

ii

y n b x n i

ta thay tn hiu u vo x(n) bi xung (n), ta s c p ng xung ca h thng l:

0

( ) [ (n)]= ( )N

ii

h n T b n i

T ta c: khi n=0 ta c h(0)=b0

khi n=1 ta c h(1)=b1

khi n=N ta c h(N)=bN

cc gi tr cn li cho kt qu h(n)=0

40

vy tm li ta c p ng ca b lc FIR l

h(n)=(b0,b1, bN) r rng ng nh tn gi, h thng c p ng xung hu hn gm (N+1) phn t.

p ng xung ca h thng l b lc IIR xc nh tng i phc tp hn, nhng nhng chng sau chng ta s c nhng phng php hiu qu hn

V d cho h thng nh s sau y:

t z(n)=x(n)+y(n)

th phng trnh m t h thng s l:

y(n) = (-3/4).z(n-1)

p ng ca h thng c xc nh qua bng di y:

(bng trn c xy dng da trn phng php tnh hi tip (recursive, vi trng thi ban u ca h thng c gi s l trng thi ngh (bng 0))

Theo kt qu trn bng trn, cui cng ta c:

h(n) = y(n) = (-3/4)n.u(n-1)

p ng xung ca h thng LTI cho php xc nh tnh n nh ca h thng:

nhc li nh ngha:

H thng LTI l n nh khi v ch khi:

41

( )n

S h n

iu kin trn c ngha l tng bin ca p ng xung c gi tr hu hn

H thng LTI nhn qu khi v ch khi: h(n)=0 vi n 0 th s di sang phi, vi n < 0 th di sang tri. 3. Tnh x(m).h(m-n)

4. Tnh tng ca m tch bc 3 c gi tr y(n) tng ng vi n cho trc. 5. Lp li cc bc 2, 3, 4 vi cc gi tr n khc.

Bng cch thay n-m bi m, ta c:

y(n) = x(n)*h(n) =

m

h(m) x(n m)

Nh vy tch chp c tnh giao hon.

D thy rng, nu h(n)= (n-p), ta s c:

y(n) = x(n)*h(n) = x(n)* (n-p) = x(n-p) (2.36)

iu ny c ngha l tch chp ca mt tn hiu vi mt xung n v (unit impulse) l chnh tn hiu , nhng di i p v tr tng ng vi v tr ca xung n v.

V d:

42

Cho h(n) = 6, 5, 4, 3, 2, 1 ln lt ng vi n = 0, 1, 2, 3, 4, 5 v h(n) = 0 vi cc gi tr n khc. Cho x(n) = 1, 2, 3 ln lt ng vi n = 0, 1, 2 v x(n) = 0 vi cc gi tr n khc. Tnh tch chp y(n) = x(n) * h(n).

Bi gii: C th gii bng phng php trc tip hoc gin tip

Phng php trc tip:

y(n) = x(n)h(n) =

m

h(m) x(n m)

V x(-m) = (3, 2, 1) ln lt cho m = (-2, -1, 0), ta c bng sau :

Ct kt qu y(n) c tnh t cng thc trn. V d vi n = 3 : y(3) = 6x0 + 5x3 + 4x2 + 3x1 + 2x0 + 1x0 = 26

Phng php gin tip (tnh t cc xung (impulses)):

Phn tch x(n) thnh tng ca cc xung: x(n) = 1.(n) + 2.(n-1) + 3.(n-2)

V vy : y(n) = h(n) + 2h(n-1) + 3h(n-2)

v ta c bng sau :

43

V d :

Cho h thng IIR bc 1 c m t nh sau : y(n) = x(n) + a.y(n-1) Tnh p ng xung ca h thng Tnh p ng ca h thng vi tn hiu u vo l x(n) = bnu(n).

Bi gii

Vi x(n) = (n), phng php tnh trc tip v lp li (iteration) cho ta cc gi tr trong bng sau :

N 0 1 2 3 n

x(n) 1 0 0 0 0

y(n-1) 0 1 a a2 an-1

y(n) 1 a a2

a3 an

V vy y(n) = anu(n) = h(n).

xc nh p ng ca h thng trong trng hp x(n) = bnu(n), ta s dng tch chp:

m m

mnm nuabmnhmxny )()()()(

V y l tng theo m, c lp vi an v bng cch tnh tng trn theo m t 0 n n, ta c :

)()()/(1

)/(1)(

111

nuba

banu

ab

abany

nnnn

Vi a = b, gi s ta t a = b(1 + ) v cho tin v 0, ta c :

y(n) = (n+1) an u(n).

44

Kt qu trn c th c kim chng bng phng php tnh lp li (iteration) qua vic xy dng mt bng tng t ri tnh theo tng ct tm ra p ng xung ca h thng.

Tch chp c th c tnh gin tip, hoc i khi c th tnh n gin hn bng cch s dng bin i Fourier hoc bin i z, s c trnh by cui chng ny.

Trong chng sau, mt phng php khc tnh tch chp s c trnh by : phng php tch chp trn (circular convolution).

Tch chp c tnh giao hon

2.5 Biu din h thng LTI bng p ng tn s

p ng ca mt h thng LTI vi mt tn hiu u vo bt k c th c tnh n gin hn vi s gip ca bin i DTFT bng cch s dng cc tnh cht sau :

Tnh di (shift) x(n-m) e jm X(ej ) (2.37)

Tch chp x(n)*h(n) X(ej) H(ej ) (2.38)

V cc cp bin i thng dng.

V d: (ging v d trc)

Cho h thng IIR bc 1 c m t nh sau :

y(n) = x(n) + a.y(n-1)

Tnh p ng xung ca h thng.

Tnh p ng ca h thng vi tn hiu u vo l x(n) = bnu(n).

Bi gii :

Bin i DTFT ca phng trnh m t nu trn c th c vit nh

sau :

Y(ej ) = X(e

j) + a.e

-jY(e

j).

45

Vi x(n) = (n), ta c :

X(ej) = 1 v Y(e

j) = H(e

j).

Phng trnh trn tr thnh:

H(ej) = 1/(1-a.e

-j) h(n) = anu(n).

Vi tn hiu u vo x(n) = bnu(n), ta c X(ej) = 1/(1-b.e-j) v:

).1(

1

)().1(

1

)().1)(.1(

1)(

jjjj

j

ebba

b

eaba

a

ebeaeY

Kt qu bin i ngc:

)()(11

nuba

bany

nn

Kt qu bin i DTFT H(ej) ca p ng xung h(n) cng biu din mt c tnh quan trng: chnh l p ng tn s ca h thng. Thc t, nu x(n) = e

jn l mt tn hiu sin phc (complex), p ng ca h thng s l:

y(n) =

m

h(m)x(n-m) =

m

h(m)ej(n-m)

= ejn

m

e-jm

h(m)

y(n) = ejn

H(ej

)

Tn hiu u ra ca h thng cng l mt tn hiu sin phc nhng c nhn thm vi mt thnh phn c trng ca h thng H(ej). Thnh phn c trng ca h thng H(ej) c gi l p ng tn s ca h thng, tnh theo tn s ca tn hiu u vo. Tn hiu ejn c th xem nh l vect c trng ca h thng H(ej).

Hn na,nu biu din H(ej) = A ej dng module-phase, ta c tn hiu u ra :

y(n) = Aejn

ej

.

Nu x(n) = ejn tng ng vi tn s -, v nu h(n) l tn hiu thc, ngha l:

H(ej

) = H*(ej

),

Ta c :

y(n) = Ae-jn

e-

.

Bng cch kt hp hai tn s v , v d nh nu x(n) = cos(n), ta c:

46

y(n) = Acos(n+)

Tn hiu u ra ca mt h thng LTI lun lun l tn hiu sin vi cng tn s, nhng bin v pha ph thuc vo p ng tn s H(ej) c tnh ti tn s .

2.6 Tng quan (Correlation)

Nu tch chp l tm p ng ca h thng vi tn hiu u vo bt k, th php tng quan li tng ng vi cc ng dng nng lng ca tn hiu. Tng quan c th c s dng tm hoc nh v mt tn hiu bit trong mt tn hiu b nhiu.

Tng quan cho ca hai tn hiu x(n) v y(n) c nh ngha nh sau :

rxy() =

m

x(m+ ) y*(m) , = 0, 1, 2, ..

(2.39)

vi y* l lin hp phc (complex conjugation) ca y v biu din ch s (theo thi gian) ca di (shift) hay tr (delay) ca x(n) so vi y(n).

Mt nh ngha khc tng ng ca rxy() :

rxy() = x(m) y*(m ) , = 0, 1, 2, ..

(2.40)

Nu ta i vai tr ca x(n) v y(n) trong biu thc trn, ta c s tng quan:

ryx() =

m

y(m) x*(m ) , = 0, 1, 2, ..

So snh vi nh ngha ban u ca rxy(), ta c s lin h:

rxy() = ryx*(-) hay ryx() = rxy

*(-). (2.41)

Nh vy ta c th ni ryx() l lin hp phc v o ngc thi gian ca rxy() v ngc li.

S tng t trong nh ngha ca php tng quan v tch chp l kh ln. Xt hai biu thc nh ngha:

47

rxy() =

m

x(m)y*(m )

v x(n)* y(n) =

m

x(m) y(n m)

ta c th kt lun l php tng quan c th c tnh thng qua tch chp.

rxy() = x()*y*(-) (2.42a)

hay ryx() = y()*x*(-) (2.42b)

Hm t tng quan rx() hay n gin l s tng quan c nh ngha nh sau:

rx() = rxx() =

m

x(m + ) x*(m)

=

m

x(m) x*(m ) , = 0, 1, 2, ..

Trong trng hp ny, t tnh cht ca php tng quan cho, ta c:

rx() = rx*(-)

(2.43)

Ni cch khc, nu tn hiu x(n) l thc, rx() = rx(-), hay hm t tng quan ca x(n) l mt hm chn.

Hn na, ti gi tr c bit = 0, ta c:

m

xx Emxr2

)()0( (2.44)

Gi tr gc ca hm t tng quan ca mt tn hiu chnh l nng lng ca tn hiu .

Mt tnh cht quan trng ca hm t tng quan:

rx(0) |rx()| (2.45)

C ngha l, hm t tng quan t gi tr cc i ti gc = 0.

48

Ni cch khc, nu y(n) = x(n ) l tn hiu di ca x(n) i v tr cho trc v nu ta s dng y(n) vi php tng quan nh v tnh hiu x(n), kt qu nh v s t cc i nu = 0 hay khi hai tn hiu x(n) v y(n ) tht trng nhau.

Bt ng thc ni trn c th c gii thch nh sau:

t y(n) = x(n - ), xt mt chui tn hiu t hp ax(n) + y(n) vi a l mt hng s cho trc. Nng lng ca chui tn hiu ni trn c tnh nh sau :

=

n

2 =

n

[ax(n)+y(n)][ax(n)+y(n)]* 0

Tng ni trn lun lun 0, c lp vi gi tr a. Bng cch khai trin tng ni trn, ta c :

= |a|2rx(0) + arx() + a*rx(-) + ry(0)

= |a|2rx(0) + arx() + a

*rx

*() + rx(0) = (|a|

2 +1)rx(0) + 2Re{arx()}

Ta c rx() = | rx()|ej()

vi () l pha ca rx(). Gi s chn a = -e-j(). Tng

ni trn tr thnh:

= 2rx(0) - 2|rx()| 0.

y chnh l iu phi chng minh.

T bt ng thc ni trn, ta c th nh ngha hm t tng quan chun ha (normalized autocorrelation function) trnh cc vn v li (gain)

x() = rx()/ rx(0) 1 (2.46)

Vi hai tn hiu x(n) v y(n) c nng lng tng ng Ex v Ey, ta c th chng minh mt cch tng t (xem sch tham kho ca tc gi Proakis) c :

| rxy()|2 ExEy

Tng t, ta c hm tng quan cho chun ha (normalized cross- correlation fucntion)

xy() = rxy()/[rx(0) ry(0)] (2.47)

V d :

49

Xt cc tn hiu x(n), y(n) v t(n) c di l 5, c minh ha theo hnh sau :

Vi cc gi tr n ngoi khong (0..4), c ba tn hiu u bng 0. a. Xc nh hm t tng quan ca x(n). b. Xc nh hm tng quan cho ca x(n) v y(n). c. Xc nh hm tng quan cho ca x(n) v t(n).

Bi gii:

a. Ta c th tnh theo 2 cch: trc tip v theo p ng xung.

Phng php trc tip:

Ta c : rxx() =

m

x(m + ) x*(m)

tnh rxx(), ta c th s dng bng tnh sau y :

50

Hng x(m) th hin tn hiu cho vi 5 xung n v lin tip tnh t v tr m = 0. Cc hng x(m+) th hin 11 tn hiu b di (shift) ca x(m), vi thay i t -5 n 5.Vi mi gi tr , rx() l tng cc tch ca 2 hng x(m) v x(m+). V d, vi = -5, tch ca x(m) vi x(m+) l 11 gi tr 0, nn tng ni trn hay rx(-5) = 0. Vi = -4, ta ch c 1 tch duy nht c gi tr khc 0 (m = 4), nn rx(-4) = 1. Cc gi tr khc c tnh tng t.

Nu ta v rx() theo , ta c th thy dng ca n l mt tam gic cn vi gi tr cc i ti gc. tnh hm chun ha x(), ta ch cn chia cc gi tr rx() trn vi nng lng ca tn hiu Ex = 5. Gi tr cc i ca x() l 1, t c ti =0. S ging nhau gia x(m) v x(m+) l tuyt i 100% khi chng trng nhau (=0).

Phng php dng p ng xung:

51

Trong phng php ny, ta s s dng tch chp tnh : rxx() = x()* x

*(-). Do x() c 5 gi tr khc 0, hay ni cch khc l 5 xung, t v tr 0 n 4,

nn x(-) l 5 xung t 0 n -4. Ta c th biu din x(-) nh sau :

x(-) = () + (+1) + (+2) + (+3) + (+4). Tch chp = x()x*(-) c tnh da theo bng sau :

b. tnh tng quan cho, ta s dng cng thc :

rxy() = x()*y*(-)

vi y(-) = () + (+1) + (3/2)(+2) + (+3) + (+4)

v

Ey = 19/4 = 4.75 & (ExEy)

= 4.8734

-5 -4 -3 -2 -1 0 1 2 3 4 5

x()*(1/2)() 0 0 0 0 0 0.5 0.5 0.5 0.5 0.5 0

x()*(+1) 0 0 0 0 1 1 1 1 1 0 0

x()*(3/2)(+2) 0 0 0 1.5 1.5 1.5 1.5 1.5 0 0 0

x()* (+3) 0 0 1 1 1 1 1 0 0 0 0

x()*(1/2)(+4) 0 0.5 0.5 0.5 0.5 0.5 0 0 0 0 0

rxy() = x()*y(-) 0 0.5 1.5 3 4 4.5 4 3 1.5 0.5 0

xy() 0 .10 .31 .62 .82 .92 .82 .62 .31 .10 0

Tam gic y(n) ging vi hnh ch nht x(n) khong 92% nng lng.

c. Ta cng s dng cng thc tch chp tnh tng quan cho ca x(n) v t(n)

rxt() = x()*t*(-)

vi t(-) = (9/8)[() + (+1) - (+3) - (+4)]

52

v

Et = 4.(9/8)2 = 81/16 = 5.0625 & (ExEt) = 5.03115.

-5 -4 -3 -2 -1 0 1 2 3 4 5

x()* () 0 0 0 0 0 1 1 1 1 1 0

x()*(+1) 0 0 0 0 1 1 1 1 1 0 0

x()*0.(+2) 0 0 0 0 0 0 0 0 0 0 0

x()* (-1).(+3) 0 0 -1 -1 -1 -1 -1 0 0 0 0

x()*(-1).(+4) 0 -1 -1 -1 -1 -1 0 0 0 0 0

rxt() = x()*t(-) 0 -1.13 -2.25 -2.25 -1.13 0 1.13 2.25 2.25 1.13 0

xt() 0 -.22 -.45 -.45 -.22 0 .22 .45 .45 .22 0

Cc gi tr tng quan xt() cho thy gia hai tn hiu x(n) v t(n) khng c s tng quan hay s ging nhau.

2.7 Bin i Z

Vai tr ca php bin i z (Z - Transform) trong h thng x l tn hiu ri rc cng tng t nh vai tr ca php bin i Laplace trong h thng x l tn hiu lin tc. Nu bin i DTFT tng ng vi cc tn hiu dng sin, th bin i z s ph hp vi nhng tn hiu vi tn s phc z.

Bin i z cho php ta biu din tn hiu hay h thng theo mt cch khc, thng qua cc gi tr cc (pole) v zero.

nh ngha :

Bin i z ca mt tn hiu ri rc c nh ngha nh sau :

n

nznxzX )()( (2.48)

V d:

Nu x(n) = (n) th X(z) = 1. Nu x(n) = (n-m) th X(z) = z-m. Nu x(n) = anu(n) th X(z) = 1 + az-1 + a2z-2 + a3z-3 +....

Min hi t (Region of convergence, ROC)

53

Trong v d cui cng trn, X(z) c biu din bng mt tng v hn. Cu hi t ra y l: c tn ti mt cch biu din X(z) ngn gn hn? Cu hi ny dn ta n vic nghin cu min hi t ca bin i z.

Min hi t ca bin i z nm trong mt phng z, nh ra phm vi hnh hc trong cc tng v hn c kh nng hi t.

Min hi t thay i ty theo bn cht ca tn hiu v c tm tt trong bng sau:

54

55

V d:

Cho chui tn hiu m c nh ngha nh sau: x(n) = anu(n). Bin i z ca tn hiu trn c vit nh sau:

X(z) =

0n

(az1

)n

S dng tng mt phn (partial sum) ca chui (geometric series), ta c:

SN = 1 + c + c +.. +cN-1

= c

c N

1

1

Vi c l mt hng s bt k. Nu |c| < 1, tng v hn (khi N tin n ) ni trn s tin n:

S = S = 1/(1-c).

Trong chui tn hiu cho, c = az1, v theo biu thc tnh tng v hn trn, ta c bin i z rt gn :

X(z) = 1/(1-az1

)

Mt khc, t iu kin |c| = |az-1| < 1, ta c min hi t

|z| > |a| (ROC)

Ni cch khc, bin i z s lun ng khi tn s phc z nm ngoi ng trn c bn knh ln hn |a|.

Trong vng pha trong ng trn tn ti im cc (pole) z = a ca X(z) lm cho X(z) khng xc nh. Tm li, ROC ca X(z) khng th cha cc im cc ca X(z).

By gi ta xt chui x(n) = -anu(-n-1) l tn hiu c dng anti-causal.

Bin i z ca tn hiu c tnh bi cng thc :

Y(z) = -

1

n

(az1

)n = -

1n

(z/a)n = 1-

0n

(z/a)n .

S dng tng chui nh trong v d trc, ta c:

56

Y(z) = 1/(1-az1), vi |z| < |a|

Kt qu ny ging hon ton vi kt qu v d trc. im khc nhau duy nht chnh l min hi t, y l |z| < |a|. Min hi t trong v d ny l pha trong ng trn bn knh |a|. Cng tng t nh trong v d trc, min hi t khng cha cc im cc (pole).

Xt chui x(n) = anu(n) + bnu(-n-1) l tn hiu khng c tnh nhn qu (non-causal). Bin i z ca tn hiu c th c suy ra trc tip t 2 v d trc :

X(z) = [1/(1-az1

)] [1/(1-bz-1)]

Ta hy xt min hi t ca tn hiu : min hi t tng ng vi phn tn hiu a

nu(n) l |z| > |a|, cn min hi t ng vi phn bnu(-n-1) l |z| < |b|.

V vy, nu |b| < |a|, min hi t s khng tn ti. Ngc li, nu |b| > |a|, min hi t ca bin i z ni trn l hnh vnh khn nm gia hai ng trn c bn knh ln lt l |a| v |b|, tng t nh trong hnh 2.9 minh ha v min hi t.

Xt chui x(n) = anu(n) - bnu(n) l mt tn hiu c tnh nhn qu (causal).

Bin i z ca tn hiu c th d dng tnh c t cc v d trc :

X(z) = [1/(1-az1

)] [1/(1-bz-1)]

Bin i z c dng ging nh v d trc, tuy nhin min hi t c xc nh bi |z| > |a| v |z| > |b|, hay ni cch khc l |z| > max(|a|, |b|).

Ghi ch:

T nhng v d trn, lu l vi mt bin i z cho trc, c th tn ti nhiu hm biu din trong min thi gian khc nhau. Hm biu din trong min thi gian l duy nht khi min hi t c ch r.

Min hi t ca mt tn hiu nhn qu (causal) lun lun l pha ngoi ca mt ng trn c bn knh l mt hng s thch hp. V vy ta c th ngm hiu l cc tn hiu nhn qu (causal) v bin i z ca n trong min hi t to nn mt cp bin i.

57

Bin i Z ngc (Inverse Z-Transform)

Bin i z ngc dng tm li hm biu din theo thi gian. Bin i ny c nh ngha bi mt tch phn ng khp kn (contour integral) trong min hi t ca mt phng z. Tuy nhin, trong thc t cch tnh theo tch phn ny rt t dng.Thng thng, i vi cc bin i z c dng phn thc, ta c th phn tch thnh cc thnh phn n gin (partial fraction expansion) hay cc bin i thng dng (xem phn ph lc v cch phn tch phn thc thnh tng nhiu phn thc n gin). i khi, xc nh bin i ngc, ta c th s dng php chia n gin hay cc bin i thch hp.

Bin i z ca mt s tn hiu causal thng dng :

x(n) causal X(z)

(n) 1

(n-m) z-m

u(n) z/(z-1)

anu(n) z/(z-a)

n.u(n) z/(z-1)

nu(n) z(z+1)/(z-1)

ancos(n)u(n) [z-a.z.cos()]/[z-2azcos()+a]

ansin(n)u(n) [azsin()]/[z-2azcos()+a]

Aansin(n+)u(n) A[zsin+azsin(-)]/[z-2azcos()+a]

n-1

u(n-1) ln[z/(z-1)]

(n!)an-m

u(n)/[(m!)(n-m)!] z/(z-a)m+1

(n+m)!anu(n)/(n!m!) zm+1/(z-a)m+1

Mt s tnh cht ca bin i z

58

ax(n)+by(n) aX(z)+bY(z) x(n-m) z-mX(z) x(n)*h(n) X(z)H(z) n.x(n) -z.d[X(z)]/dz a

nx(n) X(z/a)

59

2.8 Biu din h thng LTI trong min Z

BIU DIN H THNG BNG HM TRUYN (TRANSFER FUNCTION)

Gi x(n), y(n) v h(n) ln lt l l tn hiu u vo, tn hiu u ra v p ng xung ca h thng. T tnh cht ca tch chp :

y(n) = x(n)*h(n) Y(z) = X(z)H(z) (2.49)

ta c th xem H(z) nh l hm truyn ca h thng.

P NG TN S CA H THNG :

jez

j zHeH

)()( (2.50)

Ni cch khc, p ng tn s ca mt h thng c th c kho st khi cho z di chuyn trn ng trn n v (unit circle) ej trong mt phng z, ng ngha vi vic thay i t - n . Hnh minh ha 2.10 sau th hin mi quan h gia X(z) v X(e

j).

Hnh 2.10 Mi lin h gia X(z) v X(ej)

60

BIU DIN H THNG BNG POLE V ZERO

Cho h thng LTI c m t bi phng trnh sai phn (difference equation) sau :

N

i

M

i

ii inyainxbny0 1

)()()( (2.51)

Hm truyn H(z) tng ng ca h thng s l :

M

i

i

i

N

i

i

i

za

zb

zXzYzH

1

0

1

)(/)()( (2.52)

Ta c th biu din hm truyn ni trn di hai dng khc nh sau :

NM

M

N zpzpzpz

zzzzzzbzH

))...()((

))...()(()(

21

210 (2.53a)

Hay

NM

M

i

i

N

i

i

z

pz

zz

bzH

1

10

)(

)(

)( (2.53b)

Cc gi tr phc zi v pi ln lt c gi l cc zero v cc pole ca hm truyn H(z) hay ca h thng. Cc gi tr zi, nm t thc ca H(z), s lm cho H(z) = 0 khi z = zi. Cc gi tr pi, nm phn mu thc ca H(z), s lm cho H(z) tr nn v nh khi z = pi.

H(z) c th hon ton c xc nh (tr gi tr li b0) bi cc pole v zero

ca n.

Nu h thng l thc, cc h s ai v bi cng l nhng gi tr thc. Cc gi tr zi c th l nhng s thc hay l nhng cp lin hp phc. Cc gi

tr pi cng c tnh cht ging nh vy.

Trong mt h thng n nh, tt c cc gi tr pole s nm trong ng trn n v |z| = 1. Cc gi tr zero, ngc li, c th nm bt k

v tr no trong mt phng z.

Nu tt c cc gi tr zero ca mt h thng LTI u nm trong vng trn

61

n v, h thng c gi l c pha ti thiu (minimum-phase).

Nu tt c cc gi tr zero ca mt h thng LTI u nm ngoi vng trn n v, h thng c gi l c pha ti a (maximum-phase).

Nu thay i tt c hay mt vi zk bng zk-1 ca mt h thng c pha ti thiu, cc tnh cht v pha ca h thng s b thay i, nhng cc tnh

cht v bin c gi nguyn (hnh 2.11).

Nu z nm trn ng trn n v ti tn s , ngh l z=ej, v nu nh tt c cc gi tr trong biu thc H(z) dng pole-zero u biu din di

dng bin -pha, ngha l:

kj

kk eBzz )( v kjkk eApz

)(

Th p ng tn s H(ej) c th c xc nh bi:

jjbA

B

eHk k

kk

k

k

k

kj

)exp()( 0

Mt h thng dng FIR s c xc nh bng cc zero ca n. Tt c cc pole ca h thng u nm gc ta ca mt phng z.

Mt khc, nu mt h thng c mt cp zero i xng hnh hc zk v zk

-1, th h thng c pha tuyn tnh. Kiu h thng ny thng v rt

hay c s dng cho b lc c pha tuyn tnh v khng hi quy (non-

recursive, Hnh 2.12c).

Mt h thng c gi l all-pass hay thng sut nu bin ca p ng tn s l 1 (hng s). Cc gi tr zero v pole ca h thng ny u

nm ngoi ng trn n v. Cc gi tr pole v zero s c dng tng

cp i xng hnh hc qua ng trn n v, ngha l zk = pk-1

. (hnh

12.a)

Mt h thng c gi l ton cc (all-pole) nu, tr im gc, h thng ch c cc m thi. (Hnh 2.12 b)

i vi h thng LTI c tnh nhn qu (causal), s pole s ln hn hay bng s zero, hay M N.

63

64

Chng 3: BIN I FOURIER RI RC

Biu i Fourier ri rc (DFT, Discret Fourier Transform) l mt cng c cho php tnh gn ng (xp x) bin i Fourier bng my tnh.

3.1 DFT cho tn hiu ri rc tun hon

Ta xt li cp bin i Fourier cho tn hiu ri rc DTFT v IDTFT

n

jnj enxeX )()( (DTFT)

deeXnx jnj )(2

1)( (IDTFT)

Trong trng hp tn hiu l ri rc v tun hon vi chu k N xp(n) = xp(n +mN) vi m = 0, 1, 2,..

Nu ta thay xp(n) trc tip vo nh ngha ca DTFT ta s gp vn v s hi t khi n thay i t - n .

Ta nh ngha mt cp bin i thun nghch vi N im cho tn hiu tun hon xp(n) vi chu k N nh sau :

1

0

)/2()()(N

n

knNj

pp enxkX (DFT)

1

0

)/2()(1

)(N

k

knNj

pp ekXN

nx (IDFT)

nh ngha ny c im tng t nh chui Fourier cho tn hiu lin tc.

Nu ta so snh cc biu thc nh ngha ca bin i ngc, ta c th thy l php

tch phn (integral) c thay bng php tng v = (2/N)k. Nh vy tn s ch

c kho st ti nhng gi tr ri rc (2/N)k vi k = 0, 1,, N-1. Ni cch khc, DFT c th xem nh mt phin bn c ly mu ca DTFT.

Cn lu l tn hiu bin i Xp(k) cng tun hon vi chu k N.

Xp(k+mN) = Xp(k).

V d

Cho chui xung tun hon vi chu k N = 4 c nh ngha nh sau :

x(n) = 1 vi n = 4m (vi m = .., -1, 0, 1,..) x(n) = 0 vi cc gi tr n khc.

Bin i DFT ca chui trn c tnh nh sau :

3

0

)4/2( 1)()(n

knj

pp enxkX

65

Ngc li, nu ta thay Xp(k) = 1 vo biu thc tnh bin i ngc IDFT, ta c:

3

0

2/32/22/)4/2( )1(4

1)(

4

1)(

k

njnjjnknj

pp eeeekXnx

hay : xp(0) = () (1+1+1+1) = 1; xp(1) = () (1+j-1-j) = 0;

xp(2) = () (1-1+1-1) = 0; xp(3) = () (1-j-1+j) = 0;

v.v

Hnh 3.1 V d 1 DFT

Tm bin i ngc IDFT 16 im ca Xp(k) c nh ngha bi:

Xp(k) = 1 vi k = 2+m16 hay k = 14+m16

Xp(k) = 0 vi cc gi tr k khc.

Bng cch thay Xp(k) trong nh ngha IDFT, ta c:

)()(16

1)( 16/28

15

0

16/4)16/2( nj

k

njknj

pp eeekXnx

)4/cos(8

1)(

16

1)( 216/416/4 neeenx njnjnjp

Hnh 3.2 V d 2

Tm bin i DFT ca )sin()( 0nnx vi N/320

0 2 4 8 6 14

1

k

x(n)

0 1 2 4 3 8

1/8

k

X(k)

5 6 7 -2

0 1 2 4 3 8

1

k

x(n)

0 1 2 4 3 8

1

k

X(k)

5 6 7

66

Ta c tn hiu x(n) tun hon vi chu k N: x(n)=x(n+N). Bng cch biu din:

NnjNnj ej

ej

nx /32/32

2

1

2

1)(

V bng cch xc nh x(n) vi nh ngha ca IDFT ta c:

,2

1)3(

jX

jNXX

2

1)3()3(

Cc gi tr khc ca X(k) u bng 0.

3.2 DFT CHO TN HIU RI RC C CHIU DI HU HN

Xt tn hiu ri rc x(n) c nh ngha vi chiu di hu hn N. p dng nh ngha ca php bin i DFT v IDFT, ta c :

1

0

)/2()()(N

n

knNjenxkX (DFT)

1

0

)/2()(1

)(N

k

knNj

p ekXN

nx (IDFT)

Trong min thi gian, xp(n) l phin bn tun hon nhiu chu k ca tn hiu ri rc x(n) c chiu di hu hn.

Trong min tn s, X(k) l phin bn ly mu ri rc N im ca tn hiu bin i DTFT, nu ta so snh vi:

n

jnj enxeX )()( (DTFT)

Vi = 2(/N)k

Tm li, X(k) c th i din cho X(ej) nu N ln.

Ghi ch :

Cn lu n mi quan h trc giao sau y :

1

0

))(/2(N

n

nmkNj Ne vi k = m hay k = m+ pN, p nguyn

= 0 vi cc gi tr k khc.

67

3.3 Cc tnh cht ca bin i DFT

Tnh cht Tn hiu theo thi gian DFT

1. Tuyn tnh a.x(n)+b.y(n) aX(k)+bY(k)

2. i xng + Nu x(n) chn th X(k) l thc v chn

x(-n)=x(n)

+ Nu x(n) l th X(k) l o v l x(-n)=-x(n)

X(k)=XRe(k)=XRe(-k)

X(k)=jXIm(k)=-jXIm(-k)

3. Di (Shift) x(n-n0)

02

( )j k nNx n e

02

( )j knNX k e

X(k-k0)

4. Tch chp vng (cicular

convolution)

x(n)*h(n)

vi x(n) v h(n) tun hon vi cng chu k

X(k)*H(k)

5. Nhn tn hiu x(n).h(n) (1/N)X(k)*H(k)

6. Tn hiu thc x(n) thc X(k)=X*(N-k)

7. nh l Parseval

1 12 2

0 0

1( ) ( )

N N

n k

x n X kN

8. Mi lin h vi DTFT

22( ) ( ) ( )j kN

jz e k

NX k X z X e

9. IDFT thng

qua DFT

n(n)=IDFT{X(k)} = (1/N){DFT(X*(k))}

*

Lu : K hiu p (trong chui xp(n) ) khng c vit m c hiu ngm nh trong bng trn.

3.4 Cch chn N trong tnh ton DFT

Cch chn s mu N tnh ton ph thuc mt phn vo kiu ca tn hiu ang xt: tun hon, khng tun hon, chiu di v hn hay hu hn, ... Cch chn cng ph thuc vo cng c x l s s dng, v d nh bin i Fourier nhanh (Fast Fourier Transform, FFT) i hi N c dng 2n. Bin i FFT s c trnh by phn sau.

Mc ch ca chng ta l xem xt v hiu cc kt qu ca vic chn N cng nh cch b li cc sai s nu c c th c c kt qu tt hn.

Tn hiu tun hon Mt cch l tng, N phi l bi s ca chu k tun hon ca tn hiu, nu gi tr ny c bit trc. Trong trng hp ngc li, N c th chn bt k, nh N = 2n. Nu cn thit, ta c th dng mt hm ca s (window function) gim hin tng nhe ph (spectral leakage).

V d:

68

Cho tn hiu x(n) = cos (2n/6). Chu k tun hon ca tn hiu l 6 (xem hnh minh ha di y).

Hnh 3.3 Tn hiu x(n)=cos(2n/6)

Bng cch s dng N ln lt bng 6, 12 v 16, ta c kt qu bin i DFT ln lt trong cc hnh (a), (b) v (c) trang sau. Trong hnh (a), ta c hai xung ti v tr 6/6 = 1 v 6 (6 / 6) = 5. Trong hnh (b) ta c nhiu im hn ( phn gii cao hn) hnh (a) v cc xung nm cc v tr 2 v 12 2 = 10. hnh (c), cc xung chnh nm xung quanh cc v tr 16/6 = 2.67 v 13.33. Hn na, nhiu v tr c gi tr khc 0, khng ging vi l thuyt (theo l thuyt th cc v tr ny bng 0). Hin tng ny c gi l nhe ph (spectral leakage).

Trong min thi gian, hin tng nhe ph (spectral leakage) c minh ha trong hnh sau y vi N ln lt bng 6,12 v 16:

-20 -15 -10 -5 0 5 10 15 20-1.5

-1

-0.5

0

0.5

1

1.5

69

Hnh 3.4- DFT ca tn hiu x(n)=cos(2n/6) vi (a) N=6; (b) N=12; (c) N=16

Hnh 3.5 (a) Tn hiu x(n) = cos (2n/6), (b) N = 6, (c) Lp li nhiu ln on tn

hiu x(n) vi N = 6 c xp(n).

1 2 3 4 5

3

k

X1[k]

0

2 4 6 8 10

6

k

X2[k]

0

3 5 9 12 14

8

k

X3[k]

0 1 2 4 6 7 8 10 11 13

(a)

(b)

(c)

70

Khi N bng vi chu k tun hon hay l bi s ca chu k tun hon, vic lp li nhiu ln on tn hiu x(n) vi N gi tr s cho ta tn hiu ging vi tn hiu tun hon ban u. Tuy nhin, vi cc gi tr N khc th kt qu s khc:

Hnh 3.6 (a) Tn hiu x(n) = cos (2n/6), (b) N = 16,

(c) Lp li nhiu ln on tn hiu x(n) vi N = 16 c xp(n).

Lu hin tng khng lin tc ti bin gia cc on trn hnh minh ha.

gim hin tng nhe ph (spectral leakage), ta s nhn tn hiu vi mt hm ca s (window function). Ta c th chn mt trong rt nhiu hm ca s hin c sn, tuy c th cc hm ny khng ti u nhng li hu dng. Cc hm ca s l mt ch nghin cu quan trng trong cc nghin cu v thit k, thi cng cc b lc s. Thng thng ta c th s dng cc hm ca s thng dng sau y:

o Hanning : w(n) = ()(1 cos(2n/(N-1)).

o Hamming : w(n) = 0.54 0.46cos(2n/(N-1)).

Di y l kt qu ca v d trc s dng ca s Hanning c nh ngha nh

sau: w(n) = 0.50 0.50cos(2n/N).

Bng cch nhn tn hiu vi hm ca s, ta c th gim c s khng lin tc ca tn hiu ti bin gia cc on tn hiu v hin tng nhe ph (spectral leakage) cng gim i ng k.

71

Hnh 3.8 Dng ca s gim nhe ph

Tn hiu c chiu di hu hn:

Trong trng hp tn hiu c chiu di hu hn, vic chn N ln hn chiu di ca tn hiu bng cch thm vo cc gi tr 0, s tng phn gii ca tn hiu bin i, nh hnh minh ha di y:

72

Tn hiu khng tun hon c chiu di v hn:

Trong trng hp ny, ch c bin i DTFT l c th biu din chnh xc tn hiu trong min tn s. i vi bin i DFT, ta phi chn N ln.

3.5 TCH CHP VNG (CIRCULAR CONVOLUTION)

Tch chp vng c s dng cho tn hiu tun hon. Tuy nhin, phm vi s dng ca tch chp vng khng ch dng li cc tn hiu tun hon. Vi bin i DFT v phin bn tnh nhanh ca DFT l FFT, ta c th tnh tch chp thng thng thng qua tch chp vng. Trong phn ny, ta s nghin cu v tch chp vng, sau s m rng vi vic p dng tch chp vng tnh tch chp thng thng.

Tch chp vng vi tn hiu tun hon:

Tch chp vng ca hai tn hiu tun hon chu k N l mt tn hiu cng tun hon vi chu k N. Tch chp vng c nh ngha nh sau:

73

1

0

1

0

)()()()()(*)()(N

i

N

i

ppppppp inxihinhixnhnxny

Trong biu thc trn, tng theo i c tnh ch trong mt chu k vi N gi tr. Trong biu thc tng u tin, ta cn c 2 tn hiu xp(i) v hp(- i). Ch s (n-i) tng ng vi vic di tn hiu tun hon hp(-i) i n v tr v bn phi nu n > 0 lun tn ti N gi tr ca hp(n-i). y l nhng lu c bn quan trng tnh tch chp vng.

Hnh minh ha 3.9 th hin theo ct tng bc tnh tch chp vng.

Vi vic s dng DFT v IDFT, tch chp vng ca hai tn hiu cng c th c tnh trong min tn s.

xp(n)hp(n) (tch chp vng) Xp(k)Hp(k)

hay xp(n)hp(n) (tch chp vng) = IDFT[Xp(k)Hp(k)].

74

Hnh 3.9 M t cch tnh php tch chp vng

75

Tch chp (thng) ca hai tn hiu c chiu di hu hn :

Cho hai tn hiu c chiu di hu hn ln lt l N1 v N2. Ta c th d dng thy l tch chp ca hai tn hiu cho l mt tn hiu c chiu di l N1 + N2 1. tnh tch chp thng thng bng tch chp vng, ta chn chiu di N chung cho c hai tn hiu ang xt tha iu kin N N1+N2-1 bng cch thm cc gi tr 0 vo cc tn hiu ban u.

Hnh 3.10 Tnh tch chp thng 2 tn hiu di hu hn bng tch chp vng

Tch chp ca mt tn hiu c chiu di hu hn vi mt tn hiu c chiu di v hn :

Cho tn hiu h(n) c chiu di hu hn N1 v tn hiu x(n) c chiu di v hn. Ta c th xem x(n) l chui lin tip ca cc tn hiu x1(n), x2(n) ... vi mi tn hiu c chiu di hu hn N2. Tch chp ban u c th c bin i nh sau :

...)()(

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

21

21

nyny

inhixinhixinhixnhnxnyiii

76

Trong v d di y, gi tr ca N1, N2 v N ln lt l 3, 5 v 7.

Hnh 3.11 Tnh tch chp thng bng cch phn on mt tn hiu di

3.6 ng dng ca DFT trn tn hiu lin tc

Do s ph bin ca phin bn tnh nhanh FFT, DFT tr thnh mt cng c hu dng thc tin cho vic tnh ph ca cc tn hiu lin tc. thc hin vic tnh ton ny mt cch tt nht c th (v c sai s), ta s gii hn tn hiu lic tc ang xt trong mt on thi gian nht nh ri tin hnh ly mu tn hiu c c N mu cn thit cho vic tnh DFT.

77

Hnh 3.12 Khi nim tnh FT mt tn hiu lin tc bng DFT

Hnh minh ha trn th hin khi nim ca mt h thng dng tnh gn ng tn hiu bin i Fourier H() ca mt tn hiu lin tc h(t) thng qua bin i DFT.

bin i h(t) sang hp(t), ta phi chn khong cch ly mu T v s im ly mu N. Tch s NT tng ng nh vy vi chiu di ca tn hiu dng lm phn tch. T nh

ngha ca DFT, ta cng thy khong cch gia 2 mu trong min tn s l =

2/(NT). Ngoi ra, bin ca H() v Hp(k) cn khc nhau bi tha s 1/T. Nh vy

nu N kh di v T kh nh, ta c th c s xp x gia H() v Hp(k). Mi lin h gia h(t), hp(n) v H(), Hp(k) c minh ha trong Hnh 3.13 di y.

Hnh 3.13 S gn ng ca DFT tnh ph mt tn hiu

u tin, c tn hiu c chiu di hu hn, ta nhn tn hiu h(t) vi mt hm ca s (window function) w(t). Ph ca tn hiu

)()()( twthth c tnh bi )(*)(2

1)(

WHH , theo tnh cht ca bin i

Fourier. Ta c th tnh c :

78

2/

2/sin)(

NT

NTNTW

Do c dng hm (sinx/x), tn hiu W() gm nhiu thy (lobe), vi thy (lobe)

chnh c mt na rng l 2/(NT). Hm ca s W() cho ta s gii thch v nguyn nhn gy ra sai s : s gii hn chiu di. Hnh 3.14 s cho thy thm iu ny. Hm W() c th gy ra s che lp cc gi tr ca hai tn hiu sin c tn s gn nhau. V vy, vi gi tr N cho trc, sai bit tn s ti thiu 0 l

2/(NT), ngha l 0 > 2/(NT).

Hn na, hm W() c nhng thy ph (secondary lobe) c th gy ra s che lp cc thnh phn tn s c gi tr nh ng gn mt thnh phn tn s c gi tr ln (cc thy ph ca thnh phn tn s mnh che lp thy chnh ca cc thnh phn tn s yu hai bn). Trong trng hp ny, ta c th s dng hm ca s kiu Hamming hay Hanning, v.v...

79

Hnh 3.14 Hiu qu ca s gii hn Thi gian & chnh xc ca ph vi DFT

Sau ta i tn hiu )( th thnh tn hiu ri rc )()( nThnhp bng php ly mu

vi tn s 2/T. Bin i Fourier ca phin bn lin tc hS(t) ca tn hiu hp(t) c tnh bi :

p

j

S

Tj

ST

pH

TeHeH )

2(

1)()(

80

S lin h ny cho thy kh nng xy ra s chng ph (overlapping spectral). y cng l ngun gc sai s th hai trong qu trnh x l. Hnh 1.15 tng kt c hai ngun gc ca s sai s cp.

Hnh 3.15 Ngun sai trong cch xp x bi DFT cho ph H() ca tn hiu lin tc

h(t)

kt thc, t N mu ca hp(n), bin i DFT cho ta N mu ca hm HS(ej

) ti N

tn s ri rc : = k2/N (rad/mu) hay = k2/(NT) (rad/s).

Hp(k) = HS(e j2k/N) cc tn s = k 2/NT, vi k = 0, 1, 2, 3, N-1

81

Trong Hnh 3.16, ta c mt tn hiu lin tc hnh tam gic i xng vi chiu di hu

hn 2. Trong Hnh 3.16, ta chn 3 trng hp:

1) N1 = 8 v T1 = /4 cho 1 = 2/(N1T1) = /, hay NT = 2;

2) N2 = 16 v T1 = /4 cho 2 = 2/(N2T2) = /(2), hay NT = 4;

3) N3 = 16 v T1 = /8 cho 3 = 2/(N3T3) = /, hay NT = 2;

V tn hiu lin tc c chiu di hu hn v v ta chn kh tt NT 2, ta c th loi ra ngun sai s th nht m ch cn suy nghim v ngun sai s th hai: s ly mu trong min thi gian v hin tng chng ph. Hnh 3.16 minh ha nh hng ca vic chn T v N.

Hnh 3.16 - nh hng ca s chn T v N trn s xp x gia DFT v ph H() ca

tn hiu lin tc h(t)

82

Hnh 3.16 (tip theo) - nh hng ca s chn T v N trn s xp x gia DFT v ph

H() ca tn hiu lin tc h(t)

83

3.7 Bin i Fourier nhanh (Fast Fourier Transform, FFT)

Bin i FFT l mt thut ton tnh DFT vi nhng gi tr N c bit. Mc ch ca bin i FFT l gim s tnh ton khi tnh DFT. Hin nay c nhiu mch tch hp (IC) c th tnh FFT theo thi gian thc (real-time). Thut ton ph bin v thng dng nht l cc thut ton chn N theo tiu chun N = 2m.

Cc thut ton FFT da trn hai k thut l gim nhp (decimation) trong min thi gian (Cooley-Tukey) v gim nhp trong min tn s (Sande-Tukey). Ti liu ny ch trnh by k thut th nht v xem k thut th hai l tng t.

Gim nhp (decimation) trong min thi gian (Cooley-Tukey)

Xt X(k) l bin i DFT ca x(n), c nh ngha bi:

1

0

)/2()()(N

n

knNjenxkX

n gin trong vic k hiu, ta c th nh ngha h s c bn ca bin

i FFT nh sau: NjN eW/2 .

H s ny c nhiu tnh cht kh th v:

*)(

2/

2

)()(

)( knNnNk

N

kn

N

kn

N

nNk

N

Nnk

N

kn

N

WW

WW

WWW

vi k hiu cho lin hp phc.

Trong biu thc trn ca bin i FFT X(k), nu x(n) l tn hiu phc, th

tch x(n) vi e-j(2/N)kn cn 4 php nhn cho mi gi tr n. Vi N gi tr n, cn thit tnh tng, ta cn 4N php nhn cho mi gi tr k ca X(k). Vi N gi tr k, s php nhn cn thit s l 4N2. phc tp s t l vi N2.

Thut ton gim nhp trong min thi gian gm vic nhm cc s hng chn v s hng l ca tng trong biu thc nh ngha ca X(k) thnh 2 nhm ring bit :

122

)()()(pn

kn

N

pn

kn

N WnxWnxkX

Sau thay n bi 2p hay 2p+1 ta c:

1)2/(

0

21)2/(

0

2 )12()2()(N

p

k

N

pk

N

N

p

pk

N WWpxWpxkX

1)2/(

0

2/

1)2/(

0

2/ )12()2()(N

p

kp

N

k

N

N

p

kp

N WpxWWpxkX

84

Trong biu thc cui cng, tng u tin tng ng vi nh ngha DFT ca XE(k) vi N/2 im ca chui x(2p) vi cc ch s chn; tng th hai XO(k) tng ng vi N/2 im ca chui x(2p+1) vi cc ch s l. V vy ta c th vit:

)()()( kXWkXkX Ok

NE vi k = 0, 1, 2, , (N/2)-1 (a)

Vi nhng gi tr k khc, k = N/2, N/2 + 1, , bng cch thay k bng (k + 1/2N) trong biu thc cui cng, ta c:

)2

()2

()2

( 2/N

kXWWN

kXN

kX ON

N

k

NE

Do XE(k) v XO(k) tun hon vi chu k N/2, v do 12/ NNW , biu thc cui

cng c th c vit li nh sau:

)()()2

( kXWkXN

kX Ok

NE vi k = 0, 1, , (N/2)-1 (b)

V vy, nu XE(k) l x(0), x(2), x(4), x(6), .. v XO(k) l x(1), x(3), x(5), x(7), ... th trong trng hp vi N = 8, ta c cu trc (configuration) th hai c biu din hnh minh ha pha sau (cng c th xem thm m t chi tit li mt co (lattice) cui chng).

Mt cch tng t, ta c th phn tch bin i XE(k) thnh:

)()()( 2/ kXWkXkX EOk

NEEE vi k = 0, 1, , (N/4)-1 (c)

)()()4

( 2/ kXWkXN

kX EOk

NEEE vi k = 0, 1, , (N/4)-1 (d)

vi XEE(k) l bin i DFT ca x(0), x(4), x(8),.. v XEO(k) ) l bin i DFT ca x(2), x(6), x(10),.. i vi XO(k), ta c:

)()()( 2/ kXWkXkX OOk

NOEO vi k = 0, 1, , (N/4)-1 (e)

)()()4

( 2/ kXWkXN

kX OOk

NOEO vi k = 0, 1, , (N/4)-1 (f)

vi XOE(k) l bin i DFT ca x(1), x(5), x(9),.. v XOO(k) l bin i DFT ca x(3), x(7), x(11),..

Cc biu thc trn gii thch cho cu trc th ba trong hnh minh ha (Hnh 3.17), minh ha cho nguyn tc tnh FFT vi N = 8 theo thut ton gim nhp thi gian.

Mt khc, cc biu thc (a) v (b) to nn mt n v tnh ton bng li mt co (lattice) cho bin i FFT nh sau:

85

BWAY kN v BWAZk

N

Tng t, cc biu thc (c) v (d) hay (e) v (f) cng a n cch biu din tng t (xem thm m t chi tit li mt co cui chng).

Mt php bin i FFT N im gm log2N tng tnh ton. Mi tng tnh ton bao gm 1/2N php nhn phc hay 2N php nhn thc. V vy ta c ti a 2Nlog2N php nhn cho bin i FFT N im. phc tp ca FFT gn nh thay i tuyn tnh theo N. Trong khi th phc tp ca DFT l 4N2. Vi N = 256, DFT cn 262144 php nhn trong khi FFT ch cn 4096 php nhn.

(Bin i FFT vi thut ton gim nhp thi gian v gim nhp tn s, di dng li mt co (lattice) c minh ha nhng trang cui)

86

Hnh 3.17 Minh ha nguyn tc FFT vi N=8.

(a) DFT 8 im (b) DFT 8 im da trn 2 DFT 4 im (c) DFT 8 im da trn 4DFT 2im

87

M T CHI TIT LI MT CO (LATTICE) Li mt co ca biu thc (a) v (b)

Gim nhp thi gian (Cooley-Tukey)

Lu : Ch s ca u vo l i ngc nh phn (reverse-binary) ca u ra. u ra theo th t, u vo khng. Cho nn trc khi lm FFT, ta phi sp xp li tn hiu i vo.

u vo u ra (0) = 000

(4) = 100

(2) = 010

(6) = 110

(1) = 001

(5) = 101

(3) = 011

(7) = 111

(0) = 000

(1) = 001

(2) = 010

(3) = 011

(4) = 100

(5) = 101

(6) = 110

(7) = 111

88

Gim nhp tn s (Sande-Tukey) :

Lu : Ch s ca u vo l i ngc nh phn (reverse-binary) ca u ra. Ngc li vi trng hp trc, sau khi lm FFT ta phi sp xp li tn hiu u ra theo th t.

89

Chng 4: CU TRC B LC

Vic nghin cu cu trc b lc l iu cn thit v quan trng nhm nhng mc ch sau:

Thc hin b lc mc logic hay mc vt l nhng vn gi c nhng tnh cht quan trng ca b lc.

Nghin cu v pht trin l thuyt v thit k b lc.

Chng ny s trnh by ch yu nhng cu trc thng dng cho b lc FIR v IIR. Mt vi b lc c bit s c cp n, nh:

B lc hnh lc (comb filter)

B lc ly mu tn s (frequency sampling)

B lc li mt co (lattice)

B lc n gin iu khin c (tunable filter)

4.1 Cu trc b lc FIR

B lc ri rc khng quy thng c c tnh l p ng xung ca n c chiu di hu hn (finite impulse response).

Trong phn ln trng hp, cu trc ca mt b lc FIR khng c vng hi tip (feedback). Trong mt vi trng hp c bit, mt filter FIR c th c thc hin mt cch hiu qu vi mt vng hi tip

B lc FIR c din t bi mt phng trnh vi phn (difference equation) nh sau:

y(n) = b0x(n) + b1x(n-1) + .. + bN(x-N). (4.1)

Nh vy, hm truyn t hay chuyn i H(z) ca b lc c th vit di dng:

H(z) = Y(z)/X(z) = b0 + b1z-1

+ b2z2

+.. +bNz-N

(4.2)

NN

i

iN

i

iN

i

i zzbzbzH

00

)( (4.3)

Trong phng trnh ny, N l bc (order) ca h thng v N+1 l s h s bi. Cn ghi ch ny khi thao tc vi b lc trong Matlab.

H(z) gm N cc ti gc z = 0 v N zero. p ng xung (impulse response) ca h thng c to bi N+1 gi tr lin tip bi.

90

Dng trc tip :

S dng b tr, b nhn v b cng, b lc FIR c th xy dng bi cu trc sau :

Hnh 4.1 Cu trc trc tip cho FIR.

Cu trc ny hu ch cho c s thc hin bng phn mm mc logic.

Tuy nhin, trong khi thc hin phn cng, ta cn phi n thi gian truyn (propagation time) d liu x(n) qua nhng b nhn v b cng. Nu thi gian truyn gn 1 chu k mu, chng ta c th s dng k thut pipelining/interleaving, ngha l song song chn tng khc li vi nhng b tr (delay) ph sau vi php tnh. Nhng b tr ph thm vo s gi gi tr i vo cho n thi im sau. Th d:

Hnh 4.2 Cu trc trc tip phn cng vi b tr ph

Cu trc chuyn v (transposition)

Khng mt tnh tng qut, ngi ta xt b lc FIR bc 3 sau y:

H(z) = b0 + b1z-1

+ b2z-2

+ b3z-3

.

91

Bng cch o ngc th t ca php cng v nhm li cc ton hng theo tha s z-1, ta c th vit phng trnh tng ng:

H(z) = (((b3z-1

+ b2)z-1

+ b1)z-1

+ b0. (4.4)

Mt cch thc hin khc l da trn l thuyt chuyn v nh sau:

a - Nghch o hng ca tt c cc nhnh, ng vo, ng ra

b - i cc nt (node) hoc im r nhnh thnh cc b cng v cc b cng thnh cc im r nhnh.

Li ch ca cu trc chuyn v l c th khng cn nhng b lc ph.

Th d: Thc hin b lc FIR bc 3 c a ra trn.

Hnh 4.3 Cu trc chuyn v & th d cch chuyn v

B lc FIR c pha tuyn tnh

Mt b lc FIR c gi l i xng nu cc h s ca n tho iu kin :

bn = bN-n vi n = 0, 1, 2, (4.5)

Mt b lc FIR c gi l phn i xng (antisymmetry) nu cc h s ca n tho iu kin :

92

bn = - bN-n vi n = 0, 1, 2, (4.6)

Khi mt b lc FIR l i xng hay phn i xng th pha ca n

() l tuyn tnh (linear) hay tr hon (delay) nhm () l hng s.

Delay nhm () c nh ngha bi :

() = -d()/d . (4.7)

Trong lnh vc video hay audio vi chnh xc cao, mt b lc c pha tuyn tnh s rt l cn thit v quan trng.

Cn ch rng, cc zero ca mt b lc FIR c pha tuyn tnh khng ch l nhng cp lin hp phc (complexes conjugate) m cn l nhng cp i xng nhn (geometrical symmetry) qua chu vi vng trn n v trong mt phng z.

Thng v tit kim, sau mt php nhn, ngi ta lm trn s bit ca kt qu. Php thc hnh ny c th ph hy s tuyn tnh ca pha. V d, nu Q[.] l php lm trn th nhng tnh ton sau y, s khng a ra cng mt kt qu.

Q[b0x(n) + bNx(n-N)] Q[b0x(n)] + Q[bNx(n-N)]

trnh vn ny, nu th d b0 = bN, th ngi ta gom : Q[b0x(n) + bNx(n-N)] = Q[b0(x(n) x(n-N))] .

Nh vy, tuyn tnh ca pha s c gi nguyn.

Cu trc ny c gi l cu trc vng nh hnh sau y: Ngoi ra, s b nhn s gim mt na. iu ny cho ta mt li ch khc ca cu trc vng trong vic thit k b lc FIR c pha tuyn tnh.

93

Hnh 4.4 Cu trc vng cho b lc tuyn tnh

Cu trc dng tng (cascade) hay dng ni tip (series)

Mt b lc FIR c th thc hin bi mt dy nhng b lc n gin mc ni tip thnh tng. Thng thng, mt b lc c th phn thnh nhng phn on (section) bc 2 v mt phn on bc 1 nu bc ca b lc l s l. Mt b lc FIR c pha tuyn tnh c th c phn tch thnh nhng section bc 4.

k

k zHzH )()( vi k = 1, 2, 3,

V Hk(z) = bk0 + bk1z-1

+ bk2z-2

vi RIF tng qut,

Hk(z) = bk0+bk1z-1

+bk2z-2

+bk3z-3

+bk4z-4

vi FIR c pha tuyn tnh.

Mt thun li c th c ca cu trc ny l chnh xc ca h s ca b lc.

Th d: Cho mt b lc FIR c nh ngha bi

H(z) = 1 (3/8)z-1 + (1/32)z-2.

Nu chng ta phn tch n thnh 2 section bc 1, th chng ta c:

H(z) = (1 ()z-1)(1 (1/8)z-1).

Nu chnh xc bin cn 5 bit cho cc h s (1/32 v 3/8 coefficient) ca b lc ban u, th nhng section phn tch ra ch cn 3 bit nh chnh xc cho bin .

Cu trc quy (recursive)

94

Trong vi trng hp, mt b lc FIR c th thc hin mt cch tit kim di dng quy.

Sau y l mt th d c bit nhng c ch trong thc hnh: Cho mt b lc FIR c N+1 h s, c dng :

H(z) = 1 + az-1

+ az-2

+ .. +aNz

-N. (4.9)

(Khi a = 1, ngi Anh gi n l b lc box car)

Cu trc ri rc hoc chuyn v s cn N b nhn. Nhng nu chng ta s dng tnh cng ring phn (partial sum), th b lc a ra s tr thnh:

H(z) = (1 a N+1z-N-1)/(1 az -1) (4.10)

B lc trn ch cn 2 b nhn nh hnh sau:

Hnh 4.5 Cu trc quy cho b lc FIR (4.9)

Ta s tr li cu trc ny phn b lc ly mu tn s.

4.2 CU TRC B LC IIR

B lc IIR l mt lp ln khc ca b lc s.

V mt khi nim, n tng t nh nhng b lc lin tc (analog) nh nhng b lc c in ni ting: Butterworth, Chebyshev, Elliptique, v.v

V mt cu trc, c tnh ca b lc l s c mt ca vng hi tip (feedback). Trong on ny.

Phng trnh sai phn biu din b lc IIR l:

M

i

i

N

i

i inyainxbny10

)()()( (4.11)

Hm truyn t H(z) ca b lc:

95

M

i

i

i

N

i

i

i

za

zb

zX

zYzH

1

0

1)(

)()( (4.12)

Dng trc tip 1 & 2:

Dng 1 tng ng vi s thc hin trc tip t phng trnh sai phn. Cu trc ny khng ng dng qui tc v s b tr (delay) khng phi l ti thiu. Dng 2 gim s b nh mc ti a ca N v M. Cu trc ny nhn c bng cch thc hin H(z) theo th t mu s trc v t s sau v bng cch hp nht nhng b tr hon (delay). Dng ny l 1 dng qui tc (canonic).

Hnh 4.6 Dng trc tip, b lc IIR

Cu trc chuyn v :

96

Hnh 4.7 Cu trc trc tip v chuyn v ca mt section bc 2.

Cu trc dng tng (cascade) hay dng ni tip (series)

Trong trng hp ca b lc IIR, chng ta thng phn tch thnh section bc 2 ni lin tip (dng chui) hay ni song song (Hnh 4.8). Nhng cu trc ny khng ti u. Tuy nhin n c ngha trong thc hnh v l do gin d l section bc 1, bc 2 l d hiu v quen thuc nht.

Trong trng hp phn tch thnh tng ni tip (cascade/series), ta c :

H(z) = H1(z)H2(z)H3(z)..Hk(z).. (4.13)

Mi tng c dng chung nh :

2

2

1

1

2

2

1

10

1)(

zaza

zbzbbzH

kk

kkkk (4.14)

97

Hnh 4.8 Cu trc ni tip & Cu trc song song

Vi mt h thng thc, nhng h s ca mi section cng thc. Nu bk2=ak2=0, ta c mt section c bit c bc 1. Mi section bc 2 c th c thc hin hoc vi cu trc trc tip dng 2 hoc vi cu trc chuyn v.

Th d: (sinh vin t gi)

V m hnh dng chui ca b lc sau y, bit rng b lc c mt zero ti z = -1 v mt cc (pole) ti z = - :

H(z) = (23+40z-1

+36z-2

+19z-3

)/(10+9z-1

+8z-2

+3z-3

)

Cu trc song song :

Trong trng hp ny, H(z) c vit :

H(z) = H0 + H1(z) + H2(z) +..+ Hk(z) +.. (4.15)

Nu bc (order) ca mu s ln hn bc ca t s, th section tng qut Hk(z) c dng:

2

2

1

1

1

10

1)(

zaza

zbbzH

kk

kkk

Th d: (sinh vin t gii)

V m hnh dng song song ca b lc sau, bit rng b lc c mt zero to z=-1 v mt cc (ple) ti z=- :

H(z) = (23+40z-1

+36z-2

+19z-3

)/(10+9z-1

+8z-2

+3z-3

)

98

4.3 NHNG CU TRC B LC C BIT

B lc hnh lc (Comb Filter)

Mt b lc s c gi l hnh rng lc hay hnh lc nu p ng tn s ca n l

tun hon trong khong n .

Mt b lc hnh lc G(z) c th thc hin c khi ngi ta thay i t mt b lc ri rc bt k H(z) mi b tr n v (unitary delay) z-1 bi mt b tr bc N, z-N, ngha l N b tr n v. Nh vy ta c :

G(z) = H(zN) hay G(e

j) = H(e

jN) (4.17)

Th d:

a)

Hnh 4.9 Bin thnh b rng lc t b lc thp

(b) Cho b lc H(z) = 1 z-1 c bin tn s |H(ej)| = 2(1-cos).

H(z) c mt zero ti z = 1 hay = 0 trong khong t - n .

Cho G(z) th hin b lc hnh lc tng ng:

G(z) = 1- z-N

v |G(ej

)| = 2(1-cosN).

99

B lc ny c N zero ti (2k/N). Hnh 4.10 m t b lc hnh lc vi N=20.

Hnh 4.10 Mt b lc rng lc n gin bc N

B lc FIR t ly mu min tn s (Sampled Frequency)

Cho mt b lc FIR c chiu di M (tc c bc N = M-1) nh sau:

1

0

)(M

n

n

n zhzH (4.18)

Hay trong min tn s:

1

0

)(M

n

jn

n

j eheH (4.19)

Gi s p ng H(k) ca M tn s ri rc ca b lc l c bit ti:

= k = k(2/M), vi k = 0, 1,.. , M-1, (4.20)

1

0

/2)/2( )()(M

n

Mjnk

n

Mjkj eheHeH (4.21)

Ngha l b lc c ch nh bi nhng p ng cho sn ti mi tn s ri rc.

100

By gi ngi ta mong mun thc hin b lc ny bng cch s dng IDFT trn (4.21) tm hn v sau thay th nhng kt qu nhn c trong biu thc (4.18) H(z), ta c:

1

0

211/2)().(

)1(

)(1)(

M

kMkj

M

zHzHze

kH

M

zzH

(4.22)

Ni mt cch khc, b lc c th c thc hin bng mt b lc hnh lc H1(z)=(1-z

-M)/M theo sau mt b lc H2(z) gm M b lc ph song song c dng H(k)/(1-

ej2k/M

z-1) vi k = 0, 1, .., M-1.

Ghi nhn: trong mi b lc ph, d rng H(k) l s thc, h s ej2k/M thng vn l s thc (complex number).

Hnh 4.11 B lc FIR s tp ly mu t tn s

trnh im , ngi ta dng lin h sau y. Nu p ng xung hn ca b lc gc FIR l s thc, ta c:

H(k) = H*(M-k) =H

*(-k), k = 1, 2, ..

Bng cch kt hp nhng gi tr ca nhng b lc ph c cc cc (poles) lin hp phc (complexes conjugate), chng ta to c nhng b lc ph mi ch c nhng h s thc.

Nh vy, vi M l l, ta c:

2/)1(

11/21/212 1

)(

1

)(

1

)0()(

M

kMkjMkj ze

kH

ze

kH

z

HzH

(1-z-M

)/M H(0)/(1-z-1

)

H1(z)

B lc hnh

rng lc

H(1)/(1-ej2/M

z-1

)

H(2)/(1-ej22/M

z-1

)

.

x(n) y(n)

H2(z)

H(M-1)/(1-ej2(M-1)/M

z-1

)

101

2/)1(

121

1

12 )/2cos(21

)()(

1

)0()(

M

k zzMk

zkBkA

z

HzH

(4.23)

v vi M l chn:

1)2/(

121

1

112 )/2cos(21

)()(

1

)2/(

1

)0()(

M

k zzMk

zkBkA

z

MH

z

HzH

(4.24)

Trong : A(k) = H(k) + H(-k)

B(k) = H(k)e-j2k/M

+ H(-k)ej2k/M

Hnh 4.12 B lc FIR s thc ly mu t tn s

Cu trc b lc bng ly mu min tn s c u im l tit kim trong vic thc hin nhng b lc c bng tn hp (narrow band), N ln v ch c vi gi tr khng bng zero ca H(k).

Th d:

y(n) (1-z-M

)/M H(0)/(1-z-1

)

H1(z)

B lc hnh

rng lc

H(M/2)/(1+ z-1

) ch cho M chn

1

1 2(1) (1)

(1 2cos(2 1/ )A B z

M z z

1

1 2( ) ( )

(1 2cos(2 / )A k B k z

k M z z

x(n)

H2(z)

1

1 2( /2 1) ( /2 1)

(1 2cos(2 ( /2 1)/ )AM B M z

M M z z

102

Hnh 4.13 Th d b lc FIR s thc ly mu

Hnh 4.13 m t mt b lc ly mu tn s vi M = 5, H(1) v H(-1) l bng 3.

Hm truyn t ca b lc di dng pole-zero hay di dng s thc c vit:

)1)(1(

))5/2cos(1)(5/6()1()(

15/215/2

15

zeze

zzzH

jj

21

15

)5/2cos(21

))5/2cos(1)(5/6()1()(

zz

zzzH

(4.25)

103

)1)5/2cos(2(

))5/2cos(()1()5/6()(

24

5

zz

z

z

zzH

H thng ny c 6 zero zi v 6 cc pi :

z1=1, z2,3=ej2/5

, z4,5=ej4/5

, v z6=cos(2/5)

p1,2,3,4=0, v p5,6=ej2/5

.

Hai zero v cc chng kht nhau ti ej2/5, h thng hiu qu l bc 4.

Thc vy, ly IDFT ca H(k), ta c:

1

0

5/25/2/2 )5/2cos(5

6)33(

5

1)(

1)(

M

k

njnjMknj neeekHM

nh , ngha l:

h(0)= 6/5, h(1)=h(4)=(6/5)cos(2/5), h(2)=h(3)=(6/5)cos(4/5).

Cng ch rng h(4)=h(-1) v h(3)=h(-2), ta c b lc FIR tng ng dng cong (un):

H(z) = (6/5) [1 + (z + z-1

)cos(2/5) + (z2 + z-2)cos(4/5)]. (4.26)

(4.25) v (4.26) u cn 3 php tnh nhn thc hin H(z).

Cu trc li mt co (lattice) cho FIR:

Cu trc ny c t ngnh vt l a cht phng theo nhng tn cu trc ca v qu t. Trong XLTHS cu trc ny rt hay cho nhng b lc thch nghi hay ng hp (adaptive filtering) hoc trong vic XLTH ting ni. Cu trc ny c m t theo s sau y:

Hnh 4.14 Cu trc li mt co (lattice) cho FIR

104

Cu trc ny c th dng thit lp hm truyn t:

M

k

k

M zkzH1

)(1)( (4.27)

Cu trc gm nhiu tn hay t bo (cell) ni lin tc. Mi t bo c hai u vo v hai u ra. Vi t bo p, hai u ra, ta c:

fp(n) = fp-1(n) + Kpgp-1(n-1) (4.28a)

gp(n) = Kpfp-1(n) + gp-1(n-1) (4.28b)

Ngoi ra, t bo 1: x(n) = f0(n) = g0(n) (4.29) v t bo cui cng M: y(n) = fM(n). (4.30).

Nhng nhn Kp trong (4.27) c gi l h s phn chiu.

Trong min z, (4.28) tr thnh:

Fp(z) = Fp-1(z) + Kp.z-1

.Gp-1(z) (4.31a)

Gp(z) = Kp.Fp-1(z) + z-1

.Gp-1(z) (4.31b)

Di dng ma trn:

)(

)(1

)(

)(

1

1

1

1

zG

zF

zK

zK

zG

zF

P

P

P

P

P

P (4.31c)

Nu ta nh ngha nhng hm truyn t TH x(n) t u vo: Ap(z) = Fp(z)/X(z) (4.32a)

Bp(z) = Gp(z)/X(z), (4.32b)

phng trnh (4.31c) tr thnh

)(

)(1

)(

)(

1

1

1

1

zB

zA

zK

zK

zB

zA

P

P

P

P

P

P (4.33)

Dng suy lun quy np, t (4.33), ta c th chng minh l:

Ap(z) = Ap-1(z) + Kpz-p

Ap-1(z-1

) (4.34)

Bp(z) = z-p

Ap(z-1

). (4.35)

(Chng minh dnh nh bi tp).

By gi nu ta nh ngha hm truyn t ca h thng l H(z) (4.27), ta c:

H(z) = Y(z)/X(z) = AM(z) (4.36)

105

Hn na, nu ta nh ngha Ap(z) v Bp(z) nh:

p

k

k

pp zkzA1

)(1)( (4.37)

p

k

k

pp zkzB0

)()( (4.38)

Cu trc li mt co nh cu trc tin on hai chiu:

Thay th (4.37) vo (4.35), ta c:

p

k

p

k

k

p

k

pp zkpzkzB0 0

)()()( (4.3.9)

Ngha l, nu p ng xung ca Ap(z) theo th t l:

p(0) = 1, p(1), p(2), .., p(p-1), p(p)

p ng xung ca Bp(z) s l o ngc:

p(0) = p(p), p(p-1), p(p-2), .. p(1), p(0) = 1 = p(0).

Ap(z) trong (4.37) tng ng nh vy vi h thng sau y (Hnh 4.15, p = 3) m t mt b tin on xui hay trc (forward predictor).

Hnh 4.15 B tin on xui

Tht vy, nu TH u ra fp(n) c xem nh sai s thi im n, th b lc ny tng ng vi mt s tin on tr gi hin ti ca x(n) da vo p qu kh ca n. l x(n-1), x(n-2), ..,x(n-p).

Trong khi , Bp(z) m t bi Hnh 4.16 vi p = 3 tng ng vi mt b tin on ngc (backward predictor):

106

Hnh 4.16 B tin on ngc

Gi tin on ngc theo ngha l TH i vo thi im (n-p) c nh gi tr bi p mu tng lai ca n. l x(n-p+1), x(n-p+2), .., x(n).

Ni mt cch khc, cu trc li mt co thc hin cng mt lc hai s tin on ngc v xui (backward & forward predictions). Hai tin on ny u l tuyn tnh, bi v gi tr on l kt qu ca mt tuyn hp (linear combination) vi nhng h s .

Phng trnh: Bp(z) = z-p

Ap(z-1

) (4.35)

cng chng t nu z0 l mt zero ca Ap(z) th 1/z0 l mt zero ca Bp(z).

Thut ton Levinson-Durbin:

Tr s t K: tnh hm truyn t H(z) ca h thng nh ngha nh:

M

k

k

MM zkzAzXzYzH1

)(1)()(/)()( (4.27)

ta phi tm cch tnh nhng h s ca t K1, K2, ..KM cho sn trong s Hnh 4.14 ca cu trc li mt co.

Khai trin phng trnh (4.34), ta vit:

Ap(z) = Ap-1(z) + Kpz-p

Ap-1(z-1

)

1

0 1

11

1

0

1

0

1

0

1

)()(

)()()(

p

k

p

k

k

pp

k

p

p

k

k

p

p

k

p

k

p

p

k

p

k

p

zkpKzk

zkzKzkzk

Cn bng hai v ca phng trnh trn theo lu tha k, ta c thut ton quy (recursive) Levinson-Durbin sau y:

K:

Tnh p = 1, 2, . ., M;

107

k = 0, p(0) = 1 (4.40a)

k = p p(p) = Kp (4.40b)

k = 1, .., p-1, p(k) = p-1(k) + Kpp-1(p-k). (4.40c)

Tr s K t :

tnh ngc li t ra K, ta tch ring z-1Gp-1(z) trong phng trnh (4.31b) v th vo (4.31a). Ta c:

(1-Kp2)Ap-1(z) = Ap(z) KpBp(z)

Phng trnh ny cho:

K:

Tnh p = M, M-1,.., 2, 1:

k = p Kp = p(p) (4.41a)

k = 0, p(0) = 1 (4.41b)

k=1, 2, , p-1 21 1

)()()(

p

ppp

pK

kpKkk

(4.41c)

(Nu c mt Kp no = 1, AM(z) c nghim trn vng trn n v. Ta phi t tha s loi nghim ny i, v dng thut ton trn cho h thng gim bc).

4.4 HIN TNG PHN LNG T CC H S:

4.4.1 NHY (SENSIBILITY):

Trong thc t, ta khng th c nhng h s vi chnh xc v hn. Cc h s s c phn lng t ho. Sau khi phn lng, v tr ca nhng cc hay zero s thay i. Nhng cc c th i ra ngoi vng trn n v trong min z v to ra s bt nh (instability) cho h thng.

Ngi ta vn khng c phng php my mc no tm c cu trc tht hay. Thng thng, ta phi th vi ln vi nhng tn hiu i vo tng t. Tuy vy, ta vn c th on c mt vi hiu ng ca s bin i.

Cho H(z) = N(z)/D(z), (4.67)

trong :

108

N

k

N

k

k

kk zazpzD1 1

1 1)1()( (4.68)

Khi phn lng mt h s ak, ta s c kkk aaa v cc cc tr thnh

kkk ppp . Khi phn lng mi h s ak, s bin i ca cc pi bng:

N

k

k

k

ii a

a

pp

1

(4.69)

Bi v

k

i

pzpzka

p

z

zD

a

zD

ii

.)()(

Ta c:

iipzpzkk

i

z

zD

a

zD

a

p

)(/

)( (4.70)

T s ca phng trnh (4.70) bng:

k

ipz

k

pzk

pza

zD

i

i

)(

Mu s ca phng trnh (4.70) bng:

N

i

N

im

mi

pz p

pp

z

zD

i

)(

)(

Thay th 2 phng trnh sau cng vo (4.70), sau vo (4.69), ta c:

N

kN

im

ki

kN

ii

pp

pp

1 )(

(4.71)

Phng trnh ny din t s bin i ca cc pi vi bin i ca mi h s ak. Ta nhn thy pi s to nu:

|pi| 1, nhng cc gn vng trn n v

N to, bc cao

(pi pm) nh, cc t hp gn nhau.

109

iu kin s mt ch ni ta phi thn trng khi cc gn vng trn n v. iu kin s hai ta khng lm g c ch tr thay i, nu c, nhng t im (specification) ca b lc. Tuy nhin iu kin ny, khng quan trng lm. iu kin s ba, tri li, c th cho ta cch gim s bin i ca cc pi. Tht vy, ta c th tng (pi pk) bng cch phn tch b lc bc N ra nhiu bc lc nh bc 1 hay 2 v iu chnh nhng b lc ny cho chnh xc. cng l l do cho cu trc dng tng (cascade) hay dng song song. Tuy nhin, theo mt vi tc gi, cch ny khng hn l ti u.

L lun tng t cho phn nhng zero.

4.4.2 Cu trc kt hp (coupled structure) :

Cu trc ny ch dng cho nhng b lc bc 2 vi nhng cc bng s phc (complex number). Cho hm truyn t sau y:

H(z) = 1/(1 (2rcos)z-1 + r2z-2).

Nhng cc ca ng truyn t l p = rej. Hai h s tng ng l a1 = -2rcos v a2 = r

2.

Phn lng h s a2 bng nhng hng s ci cho sn, s ch cho cc nm trn nhng vng trn r = (ci)

1/2. Nhng vng trn ny s khng cch khong u nhau trong vng trn n v trong min z. Tri li, s c t vng trn gn trung tm, m s c nhiu vng trn nm st vn