hieu ung on pha trong he thong 256-qam

8/8/2019 Hieu Ung on Pha Trong He Thong 256-QAM

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I HC QUC GIA H NI

TRNGI HC CNG NGH

H Th Thu Cc

HIU NGN PHA TRONG H THNG 256-QAM

KHO LUN TT NGHIPI HC H CHNH QUY

Ngnh: in t- Vin thng

H NI 2005


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I HC QUC GIA H NI

TRNGI HC CNG NGH

H Th Thu Cc

HIU NGN PHA TRONG H THNG 256-QAM

KHO LUN TT NGHIPI HC H CHNH QUY

Ngnh: in t- Vin thng

Cn b hng dn: TS. Trnh Anh V

H NI 2005


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Mc lc

Trang

Li ni u.....................................................................................................................1

Chng 1 CC PHNG PHPIU CH S...................................................3

1.1 Ti sao cn iu ch tn hiu....................................................................31.2 Cc phng php iu ch s cbn.......................................................4

1.2.1 Kho dch chuyn bin -ASK(Amplitude Shift Keying)..........41.2.1.1iu ch tn hiu ASK.......................................................41.2.1.2Gii iu tn hiu ASK......................................................4

1.2.2 Kho dch chuyn tn s-FSK(Frequency Shift Keying).............51.2.2.1

iu ch tn hiu FSK.......................................................5

1.2.2.2Gii iu ch tn hiu FSK................................................61.2.3 Kho dch chuyn pha PSK(Phase Shift Keying).........................7

1.2.3.1iu ch 2PSK (BPSK).....................................................71.2.3.2Gii iu ch tn hiu 2PSK..............................................8

1.2.4 Tn hiu QAM (Quadrature Amplitude Modulation)..................101.2.4.1nh ngha QAM.............................................................101.2.4.2iu ch bin vung gc (QAM)...............................121.2.4.3Gii iu ch v tch tn hiu QAM.................................131.2.4.4c im ca tn hiu QAM............................................141.2.4.5Xc sut xc nh sai tn hiu QAM................................15

Chng 2 N PHA.................................................................................................21

2.1 Mu....................................................................................................212.2 Th no l n pha....................................................................................222.3 Mt s nguyn nhn gy n pha.............................................................23

2.3.1 S dch tn do b to dao ng...................................................232.3.2 nh hng ca hiu ng Doppler...............................................232.3.3 Hiu ng ca ho ba....................................................................24

2.4 Mt ph cng sut ca n pha...........................................................262.5 Hiu ng n pha trong h thng QAM...................................................28


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Chng 3 M PHNG..........................................................................................31

3.1 Mu....................................................................................................313.2 Cu trc, chc nng v hot ng ca cc khi.....................................33

3.2.1 Khi pht s nguyn ngu nhin.................................................333.2.2 iu ch v gii iu ch QAM..................................................333.2.3 AWGN Channel..........................................................................363.2.4 n pha.........................................................................................383.2.5 Khi tnh ton li.........................................................................413.2.6 Gin chm sao........................................................................453.2.7 Khi hin th................................................................................46

3.3 M phng................................................................................................46Kt lun.........................................................................................................................52Ti liu tham kho........................................................................................................53


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Tm tt.

Ngy nay, ng dng cc qu trnh iu ch s trnn ph bin. c bit,qu trnh iu ch s ca tn hiu QAM c dng trong nhiu ng dng thc t nh

truyn hnh s, m thanh s, mng in thoi k thut s, cng ngh viba s,... Mc

d, QAM c s dng rng ri nh vy, nhng cng khng trnh khi cc hiu ng

tn hiu truyn. Mt trong nhng hiu ng l hin tng n pha.

Trong kho lun ny, em trnh by tng quan v cc k thut iu ch s v i

su vo iu ch s tn hiu QAM. Tm hiu vn pha, nguyn nhn gy n pha v

cc hiu ng ca n pha trong h thng QAM. C th, em tm hiu cc hiu ng n

pha trong h thng 256-QAM da vo s khi Phase Noise Effects in 256-QAMtrong MATLAB 7.0 v m phng c t s bt trn nhiu (BER) ca tn hiu. S

dng QAM mc cao s c xc sut gy li cao hn nhng do p ng c tc

truyn cao nn vn c s dng trong cc h thng c nhu cu. Khi thay i t l

BER, mc n pha (phase noise levels) cng lm thay i n pha trong h thng. C

th, BER v mc n pha nh th s sai khc s t hn. Qu trnh m phng s gip

quan st iu ny.


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Kho lun tt nghi p Hiu ng n pha trong h thng 256-QAM

Li ni u

Ngy nay, vi s pht trin ca cng ngh, cc k thut iu ch ngy cng c

ng dng nhiu. X l s l mt loi k thut x l tn hiu bng gc, thng c dngtrong hu ht cc h thng thng tin. c bit, k thut iu ch s QAM c s dng

nhiu trong cng ngh cao, in hnh nh trong v tuyn.

X l tn hiu s QAM c s dng rng ri trong k thut v tuyn. V d, 16-

QAM dng trong mng WLAN, 256-QAM dng trong truyn hnh s, m thanh s, in

thoi di ng s, ... Tu thuc vo yu cu khc nhau ca cc h thng m chng ta s

dng loi tn hiu QAM ph hp. Khi c yu cu v tc truyn dn cao, th chng ta

dng tn hiu QAM mc cao. c bit, trong k thut truyn hnh s, m thanh s hay

in thoi s, do yu cu cao v cht lng m thanh, hnh nh cho nn ngi ta dng

tn hiu 256-QAM. Tn hiu 256-QAM p ng c tc truyn hnh nh cao nhngli b hn ch l xc sut li bt rt ln. Mt trong cc nguyn nhn gy li bt l n pha.

Hin tng n pha xy ra do nhiu nguyn nhn nh: do ni ti trong h thng, do hiu

ng Doppler v khong cch truyn trong thng tin v tuyn l rt ln, hay do cc yu t

ca mi trng,... Khi c hin tng n pha xy ra, tn hiu truyn b sai khc i v khi

ni thu, tn hiu thu c s b li. iu ny xy ra khin cho cht lng tn hiu

truyn gim xung.

h thng truyn hon thin, cn c cng ngh k thut cao c th khc phc c

cc hiu ng ca n pha, nng cao cht lng truyn v phn u tin ti cng ngh sho.

c c bn kho lun hon thin ngy hm nay, em phi dnh nhiu thi

gian, tr tu v cng sc trong sut qu trnh lm kho lun. Mc d trong thi gian ny,

em gp phi khng t kh khn, song nhs quan tm gip , ch bo tn tnh ca

cc Thy gio, C gio v bn b cng nh ngi thn trong gia nh gip em vt

qua.

Trc ht, em xin gi ti Thy gio TS. Trnh Anh V, ngi tn tnh ch bov gip em trong sut thi gian lm kho lun li chc sc kho v lng bit n su

sc. Em xin gi li cm n chn thnh ti tt c cc Thy gio, C gio trong trng

cho em c c nhiu kin thc b ch trong sut thi gian hc tp ti trng.

1

H Th Thu Cc i hc Cng ngh-HQGHN


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Cm n gia nh v bn b dnh nhiu s gip cho em trong thi gian thc

hin kho lun va qua.

H Ni ngy 05 thng 06 nm 2005

Sinh vinH Th Thu Cc

2



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CHNG 1CC PHNG PHP IU CH SChng ny trnh by v cc phng php iu ch s cbn thng s

dng trong cc h thng thng tin. Cc phng php iu ch bin , iu ch

tn s v iu ch pha sc nghin cu. c bit l iu ch M-QAM c nhiu

u im v c s dng kh rng ri trong cc h thng vin thng, nht l

trong h thng truyn thng tc cao. Trc ht, ta xem ti sao li phi iu ch

tn hiu, sau ta s tm hiu chi tit v tng k thut iu ch.

1.1Ti sao cn iu ch tn hiu.Tn hiu bng gc c pht ra bi cc ngun thng tin khc nhau, khng

phi lc no cng thch h p cho vic truyn trc ti p qua mt knh cho trc.

Cc tn hiu ny thng c bin i vic truyn c d dng. Mt trong

nhng qu trnh ny gi l iu ch. Trong qu trnh iu ch ny tn hiu bng

gc dng lm bin i mt vi thng s ca tn hiu sng mang cao tn nh

bin , tn s hoc pha. iu ch l mt thnh phn ca b pht trong h truyn

thng, c lin quan n hiu qu v kh nng ca h.

iu ch gii quyt vn bng truyn. Vi cc tn hiu c cng rng

ph, nu truyn ng thi trn mt knh truyn m khng bin i chng s can

nhiu ln nhau. gii quyt vn ny, ta dng cc ngun tn hiu khc nhau

iu ch cc tn s sng mang khc nhau. Nu tn s sng mang c chn ph

h p, ph ca cc tn hiu dn tri trn mt bng truyn ph hp. Phng php

iu ch ny gi l ghp knh phn chia theo tn s FDM.

iu ch cng gip cho vic bc x d dng. Trong truyn thng v

tuyn, bc x c hiu qu, nng lng sng in tng ten pht phi c kch

thc ti thiu 1/10 chiu di bc sng tn hiu bc x. Vi nhiu tn hiu bng

gc tn s thp nh tn hiu m tn, c bc sng rt ln do vic bc x rt

kh v khng hiu qu. c th bc x, ta cn iu ch tn hiu ny ln min

tn s cao, khi ng ten pht s c kch cph hp hn.

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1.2Cc phng php iu ch s cbn.1.2.1Kho dch chuyn bin - ASK ( Amplitude Shift Keying).1.2.1.1 iu ch tn hiu ASK:

Trong iu ch bin , bin sng mang b thay i t l vi tn hiu

iu ch (tn hiu bng gc). Tn hiu bng gc l tn hiu ng ms(t). Khi

bin ca tn hiu sng mang cosct thay i t l vi tn hiu d liu s(t), kt

qu l ta c sng mang iu ch y(t) = (t/T)acosct. Tn hiu ny vn l tn

hiu ng m, do c gi l kho ng mhay kho dch bin . AVi tn

hiu li vo l phn cc dng NRZ (non-return to zero), li ra bo cc cosct

khi tn hiu xung mc thp 0, v cosct khi tn hiu xung mc cao 1. Tn

hiu iu ch thu c bo pha v c gi l ASKo pha hay kho o pha

(PSK).

Hnh1.1 Siu chASK.

4

1.2.1.2 Gii iu ch tn hiu ASK:Gii iu ch tn hiu ASK c th l kt h p hoc khng kt hp. Vi

phng php gii iu ch thch hp mch phc tp hn nhng chng nh hng

ca nhiu hiu qu hn. Mch l mt b tch sng ca tch gia tn hiu ASK v

sng mang c khi phc ti ch. Trong gii iu ch khng kt hp, hnh bao

ca tn hiu ASKc tch sng bng it. Trong c hai trng h p, b tch

sng km theo mt mch lc thng thp ly i thnh phn sng mang cn d

v mt b to dng tn hiu.

Hnh 1.2a Gii iu chtn hiu ASK khng kt hp.

X ASKD liuy(t)

Sng mangx(t)=cosct

ASKD liu



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Hnh 1.2b Gii iu chtn hiu ASK kt hp.

Cc tnh cht ch yu ca ASK l:

- Dng ch yu trong in tn v tuyn.

- Yu cu mch n gin.

- Kh nhy vi nhiu (xc sut sai s rt ln).

- Nu Fb l tc truyn bt, ph cc tiu Bw ca tn hiu biu

ch ln hn Fb.- Hiu sut truyn c xc nh bi t s gia Fb v Bw b hn 1.

- Baud hay tc Baud c nh ngha nh tc iu ch bng

tc truyn Fb.

1.2.2Kho dch chuyn tn s FSK (Frequency Shift Keying):1.2.2.1 iu ch tn hiu FSK:

Trong dng iu ch ny, sng mang hnh sin nhn 2 gi tr tn s, xcnh bi tn hiu d liu cs 2.

Nguyn tc iu ch FSK:

Gi s c sng mang:

x(t) = a.cos[ct + (t)] = a.cos[(t)] vi (t) = ct + (t)

Ta gi nguyn bin , pha v ch thay i tn s:

i = d(t)/dt = c + d(t)/dt

i l tn s tc thi

d(t)/dt l s thay i tn so vi tn s sng mang.Ta gi l iu tn khi d(t)/dt = kf.s(t)

s(t) l tn hiu sin

kfl h siu tn.

pht lisng mang

XASK

D liu



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Suy ra: (t) = kt

0

f.s().d

Suy ra y(t) = a.cos[ ct + kt

0

f.s().d]

Trong trng hp iu ch s FSK th

=

"0"1

"1"1)(

bit

bitts

Khi

y(t) = a.cos(ct kft) = a.cos(c kf)t.

Tn sng vi mt bt no :

- i vi bt 0 tn s sng mang l f1, ta c 1 = c -

- i vi bt 1 tn s sng mang l f2, ta c 2 = c + rng bng khi iu ch FSKc tnh l:

Bw = F1 + 2/Tp (F2 - 2/Tp) = F1 F2 +2/Tp = 2(F + 1/Tp)

Trong :

- Bw l rng bng tn.

- Tp l rng xung.

rng bng tn khi iu ch FSK ph thuc vo dch tn F, tc l

khong cch gia hai tn s F1 v F2 v rng bt s liu Tp.

6

.

Hnh 1.3 Siu chFSK

1.2.2.2 Gii iu ch FSK.Mch ph bin nht ca b gii iu ch cc tn hiu FSK l vng kho

pha (PLL). Tn hiu FSKli vo ca vng kho pha ly hai gi tr tn s. in

th lch mt chiu li ra ca b so pha theo di nhng s dch chuyn tn s

ny v cho ta hai mc (mc cao v mc thp) ca tn hiu li vo FSK.

Clock

D liu

:NFSK



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B gii iu ch PLL c km theo mt mch lc thng thp ly i

nhng thnh phn cn d ca sng mang v mt mch to li dng xung to

khi phc dng xung chnh xc nht cho tn hiu iu ch.

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Hnh 1.4 Gii iu chFSK.

Nhng tnh cht ch yu ca FSK:

Ch yu dng trong modem truyn d liu v trong truyn v

tuyn s.

i hi phc tp ca mch mc trung bnh.

t li hn ASK.

Nu Fb l tc truyn bt, ph cc tiu Bw ca tn hiu biu

ch l cao hn Fb. Hiu sut truyn l t s gia Fb v Bw, b hn 1.

Baud hay tc Baud l tc iu ch, bng tc truyn Fb.

1.2.3 Kho dch chuyn pha PSK (Phase Shift Keying).1.2.3.1 iu ch 2PSK (BPSK).

Loi iu ch ny c gi l pha chia 2 hay PSK cs 2 (BPSK) hay

kho ngc pha (PSK). Sng mang hnh sin ly hai gi tr pha c xc nh bi

tn hiu d liu cs 2. K thut iu ch ny dng biu ch vng cn bng.

Dng sng hnh sin li ra ca biu ch l cng pha hay ngc pha (c ngha l

lch pha 1800) vi tn hiu li vo, l hm s ca tn hiu d liu.

VCO

PLLFSK

D liu



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Khi truyn thng tin, cc bt tn hiu cn truyn l 0 v 1, mi bt

ng vi mt trng thi pha ca sng mang v lch pha gia hai bt phi t cc

i. Ngha l:

- i vi bt 0 th tng ng vi gc pha sng mang l 0.

- i vi bt 1 th tng ng vi gc pha sng mang l .Biu thc ton hc ca sng mang by gil:

U0(t) = Um.cos(0t + 0 + 0)

U1(t) = Um.cos(0t + + 0)

8

D liu

Hnh 1.5 B gii iu ch2PSK.

Tn hiu vo dng m RZ n cc, trc khi a ti u vo ca b

trn M th n c a qua b chuyn i sang m lng cc (mc -1 ng vi bt

0 v mc +1 ng vi bt 1).M lng cc c hai mc in p l dng v m s to ra hai trng thi

pha cho dao ng sng mang 00 v 1800.u ra b trn ta c sng mang

iu ch 2PSK.

Nhn vo dng sng mang 2PSK ta thy, iu ch pha 2PSK gc lch pha

gia hai bt l 1800. ng vi thi im chuyn i pha lun c s chuyn i

bin trong mt thi gian ngn hay di. iu bin sinh ra khi thc hin iu ch

pha v iu bin ny gi iu bin k sinh.

1.2.3.2 Gii iu ch 2PSK.B gii iu ch tn hiu 2PSK l gii iu ch kt hp, sng mang c

khi phc t tn hiu iu ch, sau to ra mt s tn hiu sng mang c pha

khc nhau phc v cho qu trnh iu ch. Gii iu ch bng cch nhn cc

b phtsn man

b lcknh

2PSK

B so snh cs 2



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sng mang vi tn hiu iu ch, sau khi qua mch lc thng thp ta thu c tn

hiu d liu.

Qu trnh gii iu ch gm hai bc:

- Khi phc sng mang.

- Gii iu ch.

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Hnh 1.6 B gii iu ch2PSK

Qu trnh gii iu ch 2PSK:

Mt cch ton hc, qu trnh iu ch nh sau:

+sinct l tn hiu tc thi PSKng vi bt d liu 1, trong fc = c/2

l tn s sng mang.

- sinct l tn hiu PSKng vi bt d liu 0.

sinct l tn hiu sng mang c pht lp.

Khi tn hiu PSK l +sinct, b gii iu ch cho:(+sinct)( sinct) = sin

2ct =2

1(1- cos2ct) =

2

1-

2

1cos2ct

Cha mt thnh phn mt chiu (+2

1) v mt thnh phn xoay chiu c

tn s gp hai ln tn s sng mang cos2ct. Thnh phn xoay chiu c th lc

bng mch lc thng thp v cn li th dng (+2

1) c trng cho bt 1.

Khi tn hiu PSK l - sinct, b gii iu ch cho:

(+sinct)( sinct) = - sin2ct = -

21 (1- cos2ct) = -

21 +

21 cos2ct.

2PSK ()2 :2PLL

D liuX



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Cha mt thnh phn mt chiu (-2

1) v mt thnh phn xoay chiu c

tn s gp hai ln tn s sng mang cos2ct. Thnh phn xoay chiu b lc bi

mch lc thng thp v cn li th m (-

2

1)c trng cho bt 0.

Nhng tnh cht chnh ca 2PSK:

- Ch yu dng trong pht v tuyn truyn thanh s.

- i hi mch lc phc tp trung bnh.

- Hot ng t li hn FSK.

- Nu Fb l tc truyn bt, ph cc tiu Bw ca tn hiu b

iu ch bng Fb.

- Hiu sut truyn bng 1.

- Baud hay tc Baud bng Fb.

1.2.4 Tn hiu QAM (Quadrature Amplitude Modulation).1.2.4.1 nh ngha QAM:

QAM s dng mt s pha khc nhau c bit n nh l cc trng thi:

16,32,64 v 256. Mi trng thi c nh ngha bi bin v pha xc nh.

Ngha l vic to v xc nh cc symbol kh khn hn mt tn hiu n pha hay

mt n bin. Ti mi thi im s trng thi trn symbol tng s lm ton b d

liu v gii thng tng. Lc iu ch chim bng thng nh vy (sau khi lc)

nhng c hiu qu thay i t nht (theo l thuyt).

Gin chm sao ca QAM:

Gin chm sao miu t bng th cht lng v s mo ca mt tn

hiu s. Trong thc t, iu ny lun c mt t hp li iu ch c th gy kh

khn cho vic tch v nhn bit nu cn nh gi gin chm sao theo phng

php ton hc v thng k. Cc hnh sau s cung cp cc ng dng v gii thch

thng tin ca gin chm sao ca tn hiu iu ch.

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Hnh 1.7 Cc loi gin chm sao ca QAM

Bin m t s khc nhau v h s khuch i ca thnh phn I v Q

ca mt tn hiu. Trong mt gin chm sao, s khng cn bng bin c

th hin bng mt thnh phn tn hiu mrng ra v tn hiu khc b nn li. y

l thc t ci m b thu AGC to nn mt mc tn hiu trung bnh khng i.

Li pha l s khc nhau gia gc pha ca thnh phn I v Q so vi 90 .Mt li pha to ra do nguyn nhn l s dch pha ca iu ch I/Q. Thnh phn I

v Q trong hon cnh ny khng trc giao nhau sau khi gii iu ch.

Nhiu c hiu l tn hiu gi sin c tm thy trong dy tn s truyn

i v thm vo trn tn hiu QAM ti mt vi im trong ng truyn. Sau khi

gii iu ch, nhiu cha trong bng csca tn hiu gi sin tn s thp. Tn s

ca cc tn hiu ny ph hp vi s khc nhau gia tn s ca nhiu sin gc v

tn s sng mang trong bng RF.

Trong gin chm sao, nhiu biu hin trong dng ca mt s xoay

vng con tr chng ln nhau ti mi trng thi tn hiu. iu ny khng p dng

cc iu kin li xy ra cng mt thi im. Gin chm sao biu hin hng

i ca con tr nh l mt vng trn vi mi trng thi tn hiu l tng.

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S trit sng mang hoc d knh l mt loi c bit ca nhiu trong

tn s ca n bng tn s sng mang trong knh RF. D sng mang c thc

thm vo trn tn hiu QAM trong iu ch I/Q. Nhiu Gausse cng c th lm

nhiu tn hiu iu ch s trong sut qu trnh truyn tng t, cho v d trong

knh tng t. Nhiu chng cng thng c mt cng sut xc nh v phnb bin Gauss trn bng thng ca knh. Nu ti cng mt thi gian khng c

nhiu khc, trng thi tn hiu l tng trnh by l hnh m my vng trn.

( ch : gin ny c th thu c cc kiu khc ca nhiu v vy n l

s khc bit m vic to dng pha khng th lm c ).

Rung pha hoc n pha trong tn hiu QAM do h thng nhn v pht tn

hiu li trong hng truyn hoc bi biu ch I/Q. N c th xut hin khi

khi phc hoc loi b sng mang ti y. Khc vi s miu t li pha, rung pha

l mt lng c th thng k c l hiu ng ngang nhau ca I v Q. Tronggin chm sao, rung pha th hin bi cc trng thi tn hiu b dch i so vi

tn hiu gc.

1.2.4.2 iu ch bin vung gc (QAM).Mt tn hiu iu ch bin vung gc QAM (Quadrture-Amplitude-

Modulated signal) s dng hai sng mang vung gc l cos2ct v sin2ct,

mi sng mang c iu ch bi mt chui c lp cc bt thng tin. Cc sng

tn hiu c truyn i c dng:

um(t) = AmcgT (t) cos2ct + AmsgT (t) sin2ct m=1,2,...,M (1.1)

Trong {Amc} v {Ams} l cc tp cc mc bin nhn c bng

cch nh x cc chui k bt thnh cc bin tn hiu. V d, mt gin chm

sao tn hiu 16-QAM nhn c bng cch iu ch bin tng sng mang

bng 4-QAM. Ni chung, cc gin hnh sao tn hiu hnh vung c sinh ra

khi tng sng mang trong hai sng mang c iu ch bi PAM.Tng qut hn, QAM c thc xem nh mt dng hn hp ca iu

ch bin s v iu ch pha s. Nh th, cc dng sng tn hiu QAM c

truyn c th biu din theo:

12

umn(t) = AmcgT (t) cos(2ct+n) m=1,2,...,M1, n=1,2,...,M1. (1.2)



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Nu M1 = 2k1 cn M2 = 2k2 th phng php iu ch bin v pha

kt hp dn n vic truyn dn ng thi k1+k2=log2M1M2 digit nh phn xy

ra vi mt tc symbol l Rb/(k1+k2). Vic biu din hnh hc cc tn hiu cho

bi (III.2.1) v (III.2.2) l biu din bng cc vcttn hiu hai chiu c dng:

( )mssmcsm AEAEs = m=1,2,...,M (1.3)

13

b lcht t

iu chcn bn

bin i nitip_songsong

b dao ngd liu

QAMnh phn Quay pha

900

b lcht t

iu chcn bn

+

Hnh 1.8 S khi chc nng ca mt biu chQAM

1.2.4.3 Gii iu ch v tch tn hiu QAM.Gi s rng mt lng dch pha sng mang c a vo trong qu trnh

truyn dn tn hiu qua knh. Thm vo , tn hiu thu c b nhiu lon bitp m cng Gauss. V vy, r(t) c thc biu din theo:

R(t) = AmcgT (t) cos(2ct + ) + AmsgT (t) sin(2ct + ) +n(t) (1.4)

Trong l lng dch pha ca sng mang v n(t) = nc(t) cos2ct ns

sin2ct.

Tn hiu thu c c tnh tng quan vi hai hm cstrc giao c

dch pha

1(t) = gT (t) cos(2ct + )

2(t) = gT (t) sin(2ct + ) (1.5)

Nhc minh ho trn hnh 1.9, cn cc b tng quan c ly mu

ri c a ti b tch tn hiu. Mch vng kho pha (PLL) trn hnh 1.10 c



20/59


lng lng dch pha sng mang ca tn hiu thu c v b lng dch pha

ny bng cch dch pha 1(t) v 2(t) nh ch ra trong (1.5). ng h trn hnh

1.9 c gi thit l ng b vi tn hiu thu c sao cho cc li ra ca cc b

tng quan c ly mu ti cc thi im ly mu chnh xc. Vi cc iu kin

ny, cc li ra t hai b tng quan l:rc = Amc + nccos - nssin

rs = Amc + nc sin - nscos (1.6)

Trong

nc =2

1T

0

nc(t)gT(t)dt

ns =2

1T

0

ns(t)gT(t)dt (1.7)

Cc thnh phn n l cc bin ngu nhin Gauss khng tng quan, trung

bnh 0 vi varian N0/2.

B tch tn hiu ti u tnh cc metric khong cch

D(r,sm) = |r sm|2, m = 1,2,..., M (1.8)

Trong r = (rc,rs) v sm cho bi (1.3).

T

dt0

(.) ly

mu

14

Hnh1.9 Gii iu chv tch tn hiu QAM.

PLL

T

dt0

(.) lymu

gr(t)

Tnh metricD(sm)

ngh

X

X

X

X

1(t)

t/h thuc QAM

Quay pha 900

2(t)



21/59


1.2.4.4 c im ca tn hiu QAM.Tn hiu QAM l s kt hp ca iu ch bin ASK v iu ch pha

PSK, do n mang cc c im ca ASK v PSK. Ngoi ra n cn mang mt

sc im khc do s kt hp ny.

Khi tn hiu sng mang c cc gi tr bin v pha l hng s bt k thph tn s ca sng mang cng khng thay i. Nh vy, c th truyn d liu

c tc bt cao hn qua mt knh cho trc, ta c th s dng cc loi iu ch

ASK hoc PSK. y l u im ca iu ch ASK v PSK so vi FSK v trong

FSK mun truyn d liu c tc bt cao hn th cn tng rng ph ca

knh truyn. Hiu sut s dng ph ca iu ch QAM l cao hn iu ch FSK.

S mc bin hoc pha ca sng mang trong iu ch ASK hay PSK

cng ln th cho php mang nhiu thng tin hn, nhng s lng ny b gii hn

do nhiu knh truyn. S mc cng tng ko theo phc tp trong mch iu

ch v gii iu ch cng tng.

Vi iu ch n-PSK sng mang truyn ng thi n bt thng tin. S lng

pha cn c l 2n, n tng lm cho lch gia hai pha k tip l = 2/2n gim

rt nhanh, do rt d b nhiu tc ng lm li bt.

iu ch 8PSK cng p ng kh nng truyn bng iu ch QAM,

nhng tn hiu QAM c xc sut li bt t hn tn hiu 8PSK, do trong tn hiu

QAM ch s dng iu ch 4PSK cn 4 gi tr pha so vi iu ch 8PSK cn s

dng 8 mc pha khc nhau. V vy, xc sut li ca 4PSK ch bng 50% xc sut

li ca tn hiu 8PSK. Bin ca sng mang trong iu ch QAM c 2 mc, do

c tht chnh lch cc gi tr bin ln c th khng nhiu.

1.2.4.5 Xc sut xc nh sai tn hiu QAM.Tn hiu QAM c thc biu din nh sau:

um(t) = AmcgT (t) cos2ct + AmsgT (t) sin2ct 0 t T (1.9)

Vi Amc v Ams l bin ca cc thnh phn vung gc (chng mang

thng tin v g(t) l tn hiu xung. Vctbiu din tn hiu ny l:

um = [ Amc g21

Ams g21 ] (1.10)

xc nh xc sut xc nh sai tn hiu QAM, ta phi xc nh cc

im tn hiu. Ta bt u vi tn hiu QAM c M = 4im. Hnh 1.10 m t hai

15



22/59


tp hp bn im tn hiu. Tp hp th nht l tn hiu iu ch pha bn mc v

tp hp th hai l tn hiu QAM hai mc bin , k hiu l A1 v A2 vi bn gi

tr pha. Do xc sut xc nh li gn vi khong cch nh nht gia hai im tn

hiu v ta c d(e)min = 2A vi c hai loi tn hiu. Cng sut trung bnh ca tn hiu

pht i (trn cstt c cc tn hiu l ng xc sut) vi tn hiu bn mc l:

Pav =4

1.4.2A2 = 2A2 (1.11)

Vi tn hiu hai mc bin , bn mc pha, cc im tn hiu nm trn

hai ng trn bn knh A, 3 A v d(e)min = 2A, ta c:

Pav =4

1[ 2.3.A2 + 2.A2 ] = 2A2 (1.12)

Nh vy vi cc ng dng trong thc t, t l sai s ca hai tn hiu ny

l nh nhau. Ni cch khc, khng c s khc bit gia hai loi tn hiu ny khis dng trong thc t.

Hnh 1.10 Hai tp hp bn im tn hiu.

A2

.

.

.

.

A.

1

A2d=2A

Xt trng hp QAM vi M = 8. C nhiu tp hp cc im tn hiu, v

ta xt bn tp hp cc im tn hiu nh trn hnh 1.11, tt c cc loi tn hiu u

c hai mc bin v khong cch nh nht gia hai im tn hiu l 2A. Cc gi

tr (Amc,Ams)c chun ha bi A. Gi s cc tn hiu ng xc sut, cng sut

trung bnh ca tn hiu truyn i l:

Pav = =

M

mM 1

1(A2mc + A

2ms ) =

M

A2

=

M

m 1

(a2mc + a2

ms ) (1.13)

vi (amc, ams) l to cc im tn hiu c chun ho bi A.

16



23/59


17

Hnh 1.11Bn tp hp im tn hiu QAM tm im (M=8)

Hai tp hp tn hiu (a) v (c) c cc im tn hiu trong mt hnh ch

nht v Pav = 6A2. Tn hiu trong hnh (b) c cng sut trung bnh Pav = 6,83A

2 v

hnh (d) l 4,73A2. Nh vy tn hiu (d) yu cu cng sut thp hn 1dB so vi

tn hiu th nht v 1,6dB so vi tn hiu th hai vi cng mt xc sut li. Loi

tn hiu ny l loi tn hiu QAM vi M=8 tt nht do yu cu v cng sut nh

nht vi khong cch cc tiu gia hai im tn hiu cho.

Vi M 16, c nhiu kh nng la chn tn hiu QAM trong khng gian

hai chiu. V d, ta c th chn tn hiu nhiu mc bin . Loi tn hiu QAM

vi M=16 ny l mrng ca tn hiu QAM vi M=8 ti u. Tuy nhin tn hiu

loi ny khng phi l tt nht trong knh AWGN.

Tp hp tn hiu QAM ch nht c u im l d dng to ra t hai tn

hiu PAM iu ch vo cc tn hiu pha vung gc. Hn na, chng d dng

trong gii iu ch. Mc d chng khng phi l tn hiu QAM vi M 16 tt

nht, cng sut trung bnh yu cu ch ln hn mt cht so vi tn hiu ti u

cho mt xc sut xc nh sai (vi cng mt khong cch cc tiu). V nhng l

do , tn hiu QAM M mc hnh ch nht thng c s dng trong thc t.

(c) (d)

(a) (b)

2 (3 )1.

( )1..

..

.

. .

. .

( )CC.1.. .

....

.

.

.

.

.

..

...

.

.

.

.

.

. .2

2

( )1.1

( )1.1

( )1.1

( )1.3 ( )1.3

( )2.2( )2.2

0.31+

( )2.0



24/59


Tn hiu QAM hnh ch nht vi M=2k vi k chn tng ng vi hai

tn hiu PAM trong tn hiu sng mang vung gc, mi tn hiu c M = 2k/2

im tn hiu. Do cc tn hiu trong cc thnh phn pha vung gc c th phn

tch mt cch r rng ti b gii iu ch, xc sut xc nh sai ca tn hiu QAM

c th xc nh d dng t xc sut xc nh sai ca tn hiu PAM. Xc sut xcnh ng ca tn hiu QAM M mc l:

Pc = (1- M )2 (1.14)

ViM

P l xc sut xc nh sai ca tn hiu PAM M mc vi mt na

cng sut trung bnh trong mi tn hiu l vung gc ca tn hiu QAM tng

ng. Sa i xc sut xc nh sai ca tn hiu PAM M mc, ta c:

))1(

3()11(20NM

QM

P avM

= (1.15)

Vi0N

av l SNR trung bnh ca mi k hiu. Xc sut xc nh sai k

hiu tn hiu QAM M mc l:

( )211MM

PP = (1.16)

Ch rng kt qu ny ng vi k chn. Vi k l th khng c h thng

PAM M mc tng ng. Tuy nhin c th d dng xc nh tc xc nhsai cho tp hp cc im tn hiu hnh ch nht. Nu s dng b xc nh ti u

da trn o khong cch th xc sut xc nh sai k hiu b chn trn bi:

MP

( )

2

01

3211

NMQ av

( )

01

34

NM

kQ av

(1.17)

Vi k 1 v0N

av

l SNR trung bnh ca tng bt. Xc sut sai k hiu

c v trn hnh 1.12 theo SNR trung bnh tng bt.

Vi tn hiu QAM khng ch nht, ta c th xc nh gii hn trn ca

xc sut sai bng cch s dng gii hn hp:

( )( )

hieu ung on pha trong he thong 256-qam

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