Điều chế tín hiệu số

7/30/2019 iu ch tn hiu s

1/24

- Chng VII -

7iu ch tn hiu siu ch (modulation) noi chung la lam bin i cac c tnh cua mt tn hiu theo mt tn

hiu khac. Trong h thng thng tin, tn hiu b bin i goi lasong mang (carrier) va tnhiu gy ra s bin i o goi la tn hiu mang tin (information signal). Co th nh nghaiu ch la s bin i cac thng s cua song mang theo tn hiu mang tin.

Chng nay trnh bay viu ch trung tn IF (Intermediate Frequency)hoc iu ch caotn RF (Radio Frequency) (goi chung la iu ch) vi song mang la tn hiu sin co cac thngs la bin , tn s va goc pha.

Muc ch chnh cua iu ch la gn tn hiu mang tin (thng labng gc baseband) vao tnhiu song mang co ph thch hp hn, tao thanh tn hiu thng dai (bandpass signal) :

- Lam cho tn hiu mang tin tng xng vi cac c im cua knh truyn- Kt hp cac tn hiu lai vi nhau (s dung ghep knh phn tn s) ri truyn i qua mt

mi trng vt ly chung.

- Bc xa tn hiu dung cac antenna co kch thc phu hp thc t.

- nh v ph v tuyn nhm gi cho giao thoa gia cac h thng di mc cho phep.

bn thu, qua trnh din ra ngc lai so vi bn phat: tach lai tn hiu mang tin bng gc ttn hiu thng dai. Qua trnh nay c goi lagiai iu ch (demodulation) hay tach song(detection). Phn u cua chng se gii thiu s lc v ly thuyt quyt nh. ng thi gii

thiu c ban v vn thu ti u. y la nhng kin thc cn thit tm hiu v giai iuch trong phn sau.

Tip theo se m ta cac ky thut iu ch nh phn (binary modulation), bao gm: iu chkhoa dch bin ASK (Amplitude Shift Keying), iu ch khoa dch tn s FSK (FrequencyShift Keying), iu ch khoa dch pha PSK (Phase Shift Keying).

so sanh cac s iu ch khac nhau, ta da vao hiu sut ph va hiu sut cng sut.Hiu sut ph (spectral efficiency) la s o tc truyn tin trn bng thng s dung, n vlabit/s/Hz. Mt yu cu t ra i vi ky thut thng tin la truyn tin vi tc ti a trnmt bng thng ti thiu co th c. iu nay c bit ung i vi thng tin v tuyn, vph tn v tuyn rt han hep, va do o, no la mt tai nguyn thng tin quy gia.

Hiu sut cng sut (power efficiency) lin quan n ty sEb/NOi vi mt xac sut libitcho trc. Trong thc t, iu nay ngha la so sanh cng sut tn hiu yu cu bi cac s iu ch khac nhau gi c BER xac nh ng vi mt tc truyn tin xac nh.

Cui chng se gii thiu mt s ky thut iu ch tng hiu sut ph va ky thut iu chtng hiu sut cng sut.

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- Chng VII -

7.1 S lc v ly thuyt quyt nh

Vic quyt nh co anh hng quan trong n cht lng cua h thng thng tin s. V cban, i vi dong tn hiu nh phn co hai kiu quyt nh chnh la:

- Quyt nh mm (nhiu mc)

- Quyt nh cng (hai mc)

Hnh 7.1 so sanh hai loai b thu quyt nh mm va cng. B thu quyt nh mm (hnh7.1a) thc hin lng t hoa tn hiu tc thi cng vi nhiu, s dung cac mc lng t hoacho trc, mi mc c biu din bng mt t vai bit. Hnh 7.1c minh hoa qua trnh quytnh mm 8 mc (3 bit) in hnh. Mi quyt nh mm cha thng tin v kha nng ky t naoc truyn i nhiu nht (000 dn 011 ch nh ky t truyn la0 va 100 n 111 ch nh kyt truyn i la1) va thng tin v tnh hp ly cua quyt nh. Quyt nh mm co th cchuyn thanh quyt nh cui cung (la quyt nh cng) bng cach xem xet day t ma PCMlin tip ri a ra quyt nh v mc ma t o biu din, kt hp vi giai ma iu khin li.

Quyt nh cng (hnh 7.1b) ph bin hn quyt nh mm. Hai chun quyt nh chnh sdung trong trng hp nay la Bayes va Neyman-Pearson. Chun Bayes c dung nhiutrong thng tin s va chun Neyman-Pearson c dung nhiu trong cac ng dung radar.

im khac nhau c ban gia hai chun nay la chun Bayes gia s bit trc s thng k v sxut hin cua s 1 va s 0, con chun Neyman-Pearson th ngc lai. Do vy, chunNeyman-Pearson thch hp vi ng dung radar v thng th s xut hin cua muc tiu lakhng bit trc.

S & H

Lngt hoa 8

mcKt hp quyt nh mm/Giai ma iu khin li

(a)

Quyt nh cngQuyt nh mm

Q/ mm 000 010 100 110001 011 101 111

0 1Q/ cng

S & H

Lngt hoa 2

mc

Giai ma iukhin li

Quyt nh cng

(c)

(b)

S 1S 0

Hnh 7.1So sanh b thu quyt nh cng va quyt nh mm

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- Chng VII -

7.2 C ban v qua trnh thu ti u

Trong muc 6.4 va 6.5, ta a xet qua trnh khi phuc tn hiu s. o chnh la qua tnh so sanhgia tr cua tn hiu thu vi mt mc ngng tai thi im ly mu gia bit. Qua trnh naycon c goi laly mu im gia (centre point sampling). T y nay sinh ra mt cu hoi lanu ta ly mu tn hiu tai nhiu thi im khac nhau trong mt bit th co th tng c

tin cy cua quyt nh hay khng? Cu tra li la Co. Hnh 7.2 minh hoa cho giai phap nay.Ta thc hin ly mu tn hiu thu tai n thi im khac nhau, sau o cng cac kt qua lai vinhau va so sanh vi n ln mc ngng, ri a ra quyt nh cui cung. Nu th tngri rac nay se tr thanh tch phn. Ngng quyt nh sau khi ly tch phn tr thanh:

n

)dtvdtv(00 T

0

1

T

0

021 +

yv0 vav1 ln lt la in ap biu din cho s0 va s1. Sau mi bit, u ra cua b tchphn c reset v 0 chun b cho bit tip theo. Ky thut nay c goi la tach I+D

(integrate and dump detection).Tach I+D la mt trng hp ring cua mt ky thut tach tiu goi laloc phi hp (match filtering)

Cc thi im ly mu

Ngng q/nh

1 1 1 1 1 0 0 1 0 0 1 1 1 1 0 1 1 1 1 1

1 0 1 1

1 0 1 1S liu pht

Q/ cui cng

Tn hiuthu

Hnh 7.2Ly mu ti nhiu thi im

7.2.1 Loc phi hp

B loc ng trc mach quyt nh trong b thu c goi la phi hp vi mt xung nao onu no lam cho ty s SNR at gia tr ln nht tai thi im ly mu khi xung o co mt uvao b loc.

ap ng bin va ap ng pha cua b loc phi hp c xac nh nh sau:

Gia s h thng thng tin s truyn i cac xung co dangv(t) nh hnh 7.3a, mt ph nnglng chun hoa ESD nh hnh 7.3b. Nu nhiu trong h thng la nhiu trng th mt phcng sut cua nhiu la NPSD co th c biu din bng ESD khng i trong mt chu kyxung nh hnh 7.3c. Nu chia ph ra cac bng con th t hnh ve ta thy co nhng bng nhbng j co SNR ln va co nhng bng nh bng r co SNR nho hn nhiu. Bt ky bng con naoco cha nng lng tn hiu u co th tham gia vao qua trnh quyt nh. Tuy nhin, d thyrng nhng bng con co SNR cao th se anh hng n qua trnh quyt nh nhiu hn nhngbng con co SNR thp. Do NPSD la hng s nn SNR ty l vi

2

)f(V . Ta cung bit rng

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- Chng VII -

mt cng sut hay nng lng qua mt b loc th ty l vi2

)f(H . iu nay dn n yu

cu v ap ng bin cua b loc phi hp vi gia thit nhiu trng nh sau:

Bnh phng cua ap ng bin cua b loc phi hp phai co dang ging vi dang cua mt ph nng lng cua xung ma no phi hp, ngha la:

222

)f(Vk)f(H=

T0 -1/2T0 1/2T0

NPSD x T0

r j

2

)f(V

(c)

v(t)

(b)(a)

V(f)

Hnh 7.3 (a) Xung truyn (b) Ph in p ca xung tn hiu

(c) Mt ph nng lng ca xung tn hiu v ca nhiu trng

By gi ta xem nh ph cua xung trn hnh 7.3 c tao thanh t nhiu vach ph gn nhau.Theo Fourier, xung truyn i co th c phn tch thanh tng cua v s song cos. Nu co thsp xp tt ca cac song cos ng thi at cc ai tai mt thi im th in ap cua tn hiu(va la cng sut cua tn hiu) se la gia tr cc ai tai thi im o.

B loc, mun at c iu nay, phai co ap ng pha i du vi ph pha cua xung truyn i.Nh vy, xung sau khi loc se co ph pha bng 0 va cac song cos thanh phn se cung at ccai tai cac thi im t = 0, T0, 2T0 ... Trong thc t, co th thc hin c b loc, phaithm vao ap ng pha cua b loc mt dch pha tuyn tnh la ( am bao cho b locnhn qua). iu ny dn n yu cu v p ng pha ca b lc phi hp nh sau:

0Tje

p ng pha ca b lc phi hp i du vi ph pha cua xung ma no phi hp cng thmmt lng dch pha tuyn tnh la

0fT2 , ngha la:

0vfT2)f()f( =

Tom lai, ap ng tn s cua b loc phi hp la:

0fT2j* e)f(kV)f(H=

Co th m rng ap ng bin cua b loc phi hp cho trng hp nhiu bt ky nh sau:

)f(G

)f(Vk)f(H

n

=

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- Chng VII -

vi Gn(f) la PSD cua nhiu

7.2.2 B tach tng quan

Ap dung cng thc bin i Fourier ngc va cng thc v ap ng tn s cua b loc phihp, ta tm c ap ng xung cua b loc phi hp nh sau:

)tT(kv)t(h0

* =

Co th din giai cng thc nay nh sau: ap ng xung cua b loc phi hp la ao ngc xungma no phi hp, b tr i mt khoang thi gian bng rng cua xung. T y ta cung thyrng: co th tm xung ra cua b loc phi hp trc tip t u vao.

Nu xung vao la thc, theo cng thc tch chp, ta tnh c tn hiu ra nh sau:

)(kR)t(vininvvout

=

Vy tn hiu ra cua b loc phi hp la t tng quan cua xung vao.

Hnh 7.4 la b thu trong h thng thng tin s dung c ch tach song tng quan. y, tnhiu vao co khac vi tn hiu tham chiu mt lng, o la nhiu. u ra cua b tng quan seat cc ai tai im cui cua xung vao, tc la sau T0 giy. Do o, nu thi im ly muchnh xac th se dn n SNR ti u.

Hnh 7.4 B thu tng quan

vin(t) = v(t) + n(t)

v(t)

0

0

kT

T)1k(dt

kT0

7.3 iu ch nh phn

iu ch nh phn la kiu iu ch n gian nht va rt ph bin trong thc t. y la trnghp tn hiu mang tin bng gc la tn hiu nh phn. Phn nay se trnh bay chi tit cac kiuiu ch nh phn c ban.

7.3.1 iu ch khoa dch bin nh phn (BASK)

Trong h thng BASK, bin cua song mang tn sfc c chuyn i gia hai gia tr tuy

thuc vao tn hiu bng gc, bin cua song mang gm hai mcA0 vaA1biu din cho haiky t0 va1 tng ng. Trong thc t, dang song BASK gm cac xung "mark" biu din kyt1 va "space" biu din ky t 0. Luc nay BASK con c goi laiu ch khoa on-off OOK(On-Off Keying)va tn hiu BASK c biu din nh sau:

=0,0

1,tf2cosT/tA)t(f c01

scho

scho

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B iu ch OOK co th c thc hin nh la mt khoa chuyn mach n gian, khoa songmang on hay offtuy tn hiu mang tin la1hay 0hoc la mt b iu ch cn bng, nhnsong mang vi vi tn hiu OOK n cc bng gc. Hnh 7.5 trnh bay s iu ch OOKkiu cn bng. Tn hiu iu ch OOK co ph in ap va PSD cn bng quanh tn s songmang .

cf

B tach song OOK co th lakt hp (coherent)hay khng kt hp (noncoherent). Trnghptach song kt hp co th dung mt b loc phi hp (matched filter) n gian (hnh 7.6a), ura cua b loc phi hp at cc ai khi u vao co tn hiu va bng 0 khi u vao khng co tnhiu. Hoc dung b tach song tng quan (hnh 7.6b), yu cu la phai cob khi phuc songmang CR (Carrier Recovery). Tn hiu sau o c ly mu va quyt nh ngng vi ngh ly ra tb khi phuc ng h STR (Symbol Timing Recovery).

Hnh 7.5 iu ch on-off: dang song, b iu ch, ph

(e)

(d)(c)

(b)

- fc fc-1/T0 1/T0

T0

1

0

(a)

0

(a) Tn hiu bng gc (b) Ph in ap cua tn hiu bng gc (c) B iu ch OOK

(d) Tn hiu OOK thng dai (e) Ph in ap cua tn hiu OOK thng dai

Kiu tach song khng kt hp c s dung ph bin hn, v du dung b tach song ngbao (envelope detector) khi phuc tn hiu bang gc, sau o tach song I+D. Trc b tachsong la b loc thng dai tng ty s C/N (hnh 7.6c). Mt cach khac, b tach khng kt hpco th c cu truc thanh hai knh tng quan tach thanh phn ng pha (I) va vungpha (Q) cua tn hiu, sau o bnh phng thanh phn I va Q ri cng lai (hnh 7.6d). S spxp nay khc phuc c yu cu v ng b pha song mang. Kiu nay co ve phc tap nhngvi s phat trin cua cng ngh VLSI, chung tr nn nho, nhe va re hn so vi b loc va tachng bao trong cac thit k truyn thng.

Thi im quyt nh b quyt nh bn thu laf(nT0) va in ap quyt nh la:

=0,0

1cho,kE)nT(f

1

0 scho

s

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- Chng VII -

trong oE1 la nng lng chun hoa cha trong ky t1 vakla hng s, co th t bng 1.

Trng hp s dung b loc phi hp, xac sut li c tnh nh sau:

=

2/1

0

1

e N

E

2

1erf1

2

1P

Trong o N0la PSD cua nhiu, erf(x) la ham li (cho sn trong bang hoc tnh).

t )EE(2

1E

01+= la nng lng trung bnh theo thi gian trn mt ky t, lu y vi

OOK th E0 = 0, ta c:

=

2/1

0

e N

E

2

1erf1

2

1P

(d)

(c)

(b)

OOKtsin

c

tcosc

OOK

OOK

OOK Loc phihp

STR

CR

0T

0dt

STR

STR

BPF

0T

0dt

0T

0 dt

( )2

( )

2

STR

(a)

Hnh 7.6 B thu OOK kt hp va khng kt hp

(a) B thu dung b loc phi hp (b) B thu tng quan (c) B thu tach song ng bao (d) B thu 2 knh

S dung cac quan h:

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- Chng VII -

N/BCTN/E

)V(BNN

)V(T/EC

00

2

0

2

0

=

=

=

y C la cng sut song mang thu ly trung bnh trn tt ca cac ky t va N la cng sutnhiu chun hoa trong mt bng thng B (Hz). T cac quan h trn ta c:

=

2/12/1

0

e N

C

2

)BT(erf1

2

1P

Tach khng kt hp nhay vi s thay i pha trong ky t thu. iu nay se lam giam CNR thu,nhng khng ang k. Do o trong thc t, hu ht s dung tach OOK kiu khng kt hp

7.3.2 iu ch khoa dch pha nh phn BPSK (khoa ao pha)

Trong h thng iu ch BPSK, tn hiu bng gc c gn vao song mang bng cach thayi pha cua song mang tuy thuc vao tn hiu bng gc, ngha la:

( )

( )

+

=

0),tf2cos(T/tA

1,tf2cosT/tA)t(f

c0

c0

scho

scho

V nguyn tc, co th chon bt ky nhng thng chon trang thai ngc pha, tc la

. Kiu iu ch nay con c goi la iu ch khoa ao pha PRK (Phase ReversalKeying). Hnh 7.7 trnh bay b phat PRK, dang song tn hiu va ph.

0180=

(e)

(d)(c)

(b)

- fc fc-1/T0 1/T0

T0

1

-1

(a)

0

Hnh 7.7 iu chPRK: dang song, b iu ch, ph

(b) Tn hiu bng gc (b) Ph in ap cua tn hiu bng gc (c) B iu chPRK

(d) Tn hiu PRK thng dai (e) Ph in ap cua tn hiu PRK thng dai

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10/24

- Chng VII -

0dt)tf2cos()tf2cos(2

T

0

1

0

=

th khi u ra cua mt knh trong b thu kt hp la ti a th u ra cua knh kia se ti thiu.Sau khi tr cac tn hiu cho nhau th in ap quyt nh trong b thu BFSK trc giao se gingnh trng hp PSK

f2f2

0

1

f2 f

f

f

(f1+ f2)/2

(e)

(d)

(c)

f f2-1/T0 1/T0

T0

1

0

0

(a)

f

(b)

Hnh 7.9 iu chBFSK: dang song, b iu ch, ph

(a)Tn hiu bng gc (b)Ph in ap cua tn hiu bng gc (c) B iu chBFSK(d) Tn hiu BFSK (e) Ph in ap cua tn hiu BFSK

Xac sut li trong b thu kt hp i vi BFSK trc giao c tnh la:

=

2/1

0

eN

E

2

1erf1

2

1P

Biu din xac sut li theo ty s CNR se co dang nh sau:

=

2/12/1

0

eN

C

2

)BT(erf1

2

1P

Trng hp BFSK khng tng quan th gia hai ky t BFSK se co mt s tng quan naoo. Ky hiu cp ky t BFSK laf1(t) va f2(t), h s tng quan chun hoa c nh ngha nhsau:

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- Chng VII -

2

2

2

1

21

)t(f)t(f

)t(f)t(f=

BFSK

(b)

(a)BFSK

LPF

Mach quytnh

BPF

VCO@ f1

LPF

VCO@ f1

(c)

BFSK

tsin1

tcos1

0T

0 dt

0T

0dt

( )2

( )2

tsin2

tcos2

0T

0dt

0T

0dt

( )2

( )2

Mach quytnh

BPF

Mach quytnh

Hnh 7.10B thu BFSK

(a) B thu kt hp dung PLL (b) B thu khng kt hp (c) B thu khng kt hp hai knh

tnh toan thc t, co th vit lai h s nh sau:

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==00 T

0

21

21

T

0

21

0RMS2RMS1

dt)t(f)t(fEE

1dt)t(f)t(f

T

1

ff

1

Trong h thng BFSK, ta co E1 = E2 = E, do o:

012

012012

T

0

21

0 T)ff(

T)ff(cos[]T)ff(sin[dt)tf2cos()tf2cos(

T

2 0

==

Hnh 7.11 la th cua h s theo 2(f2 - f1)T0. Cac im 0 biu din cho h thng BFSKtrc giao va im 1.43 biu din cho h thng BFSK ti u. Ti u y theo ngha la titkim c 0.8dB ty s CNR so vi BFSK trc giao.

1

2(f2-f1)T0

1.43

3

2

Hnh 7.11 th cua

7.3.4 Khi phuc song mang

Tach song kt hp trong cac b thu va xet yu cu phai co mt tn hiu tham chiu trungpha vi pha cua song mang trong tn hiu. V du vi b thu PRK, ta co th thc hin khiphuc song mang theo mach vong bnh phng nh hnh 7.12.

PRK

2PLL

LPF

fx2 VCO

2/

LPF

Giai /ch

( )2

Hnh 7.12 Vong binh phng khi phuc song mang

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14/24

- Chng VII -

V nguyn tc, nu dang tn hiu la xung vung th hiu sut ph la0 bit/s/Hz. Bi v brng ph cua xung vung la v han. Tuy nhin, ta co th nh ngha hiu sut ph danh nhda vao bng thng danh nh. Bng thng danh nh la b rng cua bup song chnh cua phcua tn hiu. Bang 7.2 tom tt cac hiu sut ph danh nh cua ba h thng iu ch trn.

BASK BFSK trc giao BPSK

Tc d liu (bit/s) 1/T0 1/T0 1/T0

Bng thng danh nh (Hz) 2/T0 (n+4)/2T0 2/T0

Hiu sut ph danh nh (bit/s/Hz) 1/2 2/(n+4) 1/2

Bang 7.2Quan h hiu sut ph cua cac kiu iu ch khoa nh phn

(n la s im 0 trn th hnh 7.11)

T y ta nhn thy: trong cac phng phap iu ch nh phn trn khng co phng phapnao s dung bng thng c bit hiu qua., noi cach khac la hiu sut ph thp. Phn sau setrnh bay v cac phng phap iu ch lam tng hiu sut ph.

7.4 iu ch tng hiu sut ph

Nh a gii thiu t u chng, hiu sut ph, ky hius, c nh ngha la tc truyn

tin trn mt n v bng thng chim dung. Ngha la co th tnh c hiu sut ph nh sau:

)Hz/s/bit(B

HRs

s=

y Rsla tc ky hiu, H la entropy, tc la lng tin trung bnh cha trong mt ky hiuvaB la bng thng chim dung.

Tc ky hiu la:

0sT/1R =

Entropy trong h thng co M ky hiu c lp va ng xac sut la:

MlogH2

=

Do vy, hiu sut ph la:

)Hz/s/bit(BT

Mlog

0

2

s=

T y ta thy, mun tng hiu sut ph, ta tng s lng ky hiu M ln va giam tch T0Bxung. y chnh la cach c s dung trong cac ky thut iu ch tng hiu sut ph.

Ta bit bng thng ti thiu la:0

T/1B = . Do vy tch T0B ti thiu la bng 1.

Thc t hin nay a co cac h thng iu ch vi M ln n 64, 128, . . ., 1024

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7.4.1 iu ch khoa dch pha M mc

iu ch khoa dch pha M mc, goi tt la MPSK, la trng hp m rng cua PSK 2 mc,tng s trang thai pha t2 ln 4, 8, 16, . . ., 2i. Gian pha (con goi la chom sao) cho trng

hp M = 16 nh trn hnh 7.14, y co 16 pha phn bit.

Hnh 7.14 Gian pha cua PSK 16 mc

Xac sut li trong h thng MPSK vi c xp x bi Stein va Jones nh sau:4M

=

2/1

0

eN

E

Msinerf1P

Co th vit lai theo ty s CNR, vi quan h C = E/T0 va N = N0B nh sau:

=

2/1

2/1

0eN

C

Msin)BT(erf1P

iu ch nhiu mc y co th xem nh la qua trnh sp xp n bit nh phn vao trong mtky hiu M mc, mi ky hiu la mt xung IF. Do o, li tach song trong mt ky hiu chnh lamt vai li bit trong day bit ma hoa tng ng. Vy xac sut li bit Pb khng ch phu thucvao xac sut li ky hiu Peva entropy H ma con phu thuc vao s sp xp bit va kiu li xuthin. V du nh, trong h thng PSK 16 mc nh hnh 7.14, kiu li hay xay ra nht la nhmln gia hai trang thai pha ln cn. Nu dung ma Gray sp xp cac ky hiu nh phn th chco mt li bit n khi co li ky hiu. Trong trng hp nay, xac sut li bit la:

Mlog

PP

2

e

b=

Ta co quan h gia nng lng ky hiu va nng lng bit la:

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Mlog

EE

2

b=

T y co th biu din Pbtheo Eb/N0 nh sau:

=

2/1

0

b2

2

bNEMlog

Msinerf1

Mlog1P

hoc theo ty s CNR nh sau:

=

2/1

2/1

0

2

bN

C

Msin)BT(erf1

Mlog

1P

V cac ky hiu trong tn hiu PSK nhiu mc u co ph bin nh nhau nn b rng ph

cua tn hiu ch phu thuc vao tc ky hiu (goi la tc baud) va dang xung ch khngphu thuc vao M. i vi dang xung vung, bng thng danh nh cua tn hiu PSK M mcla2/T0 Hz. Trong trng hp nay, hiu sut ph la:

Mlog5.02s

= (bit/s/Hz)

Hiu sut ph ti a co th c la trong trng hp bng thng c tnh cho mt bn 1/T0,ngha la:

Mlog2s

= (bit/s/Hz)

Bang 7.3 so sanh mt vai thng s cua cac h thng PSK.

Eb/N0 yu cu choPb=10

-6

Bng thng tithiu

Hiu sut phti a

CNR yu cu cho bngthng ti thiu

PRK 10.6 dB Rb 1 10.6 dB

QPSK 10.6 dB 0.5Rb 2 13.6 dB

8-PSK 14.0 dB 0.33Rb 3 18.8 dB

16PSK 18.3 dB 0.25Rb 4 24.3 dBBang 7.3So sanh cac ky thut iu chPSK

7.4.2 iu bin cu phng QAM (Quadrature Amplitude Modulation)

i vi b phat, ta co th kt hp iu bin vi iu pha cai thin s phn b trang thaipha trong chom sao. Phng phap nay c gii thiu ln u tin vao nm 1960 vi 2 mcbin (tng ng 2 vong) va8 mc pha trn mi vong (hnh 7.15a). V y la s kt hpgia iu bin va iu pha nn no c goi la iu ch khoa bin /pha APK

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(Amplitude/Phase Keying). Sau o ngi ta nhn thy rng nu s im vong trong giamcon 4, s im vong ngoai tng ln 12 (hnh 7.15b) th cac im trn chom sao cach unhau hn. n nm 1962, ngi ta a ra mt cach chon bin va pha khac (hnh 7.15c).Ngi ta nhn thy kiu nay d thc hin hn vaPe co c cai thin i chut. V tn hiuAPK nay co th xem la cp ASK nhiu mc c iu ch trn cac song mang vung gocnn no thng c goi la iu bin cu phng QAM. Nh vy QAM la mt trng hpring cua APK. Tuy nhin, thc t i khi ngi ta dung thay i gia hai thut ng APK vaQAM.

(c)(b)(a)

Hnh 7.15 Chom sao cua cac tn hiu iu ch

(a) APK 16 (8,8) (b) APK 16 (4,12) (c) QAM 16 (4 x 4)

Gia s ca M trang thai u c lp thng k va ng xac sut, Carlson a tnh xp x xac sutli trong h thng QAM vi nhiu Gauss nh sau:

> la nng lng trung bnh trn mt ky hiu QAM. Nu dang xung la hnh chnht th:

0

2

T)1M(2

v

3

1E

>=<

y la sai khac in ap gia hai mc bin canh nhau vaTv 0 la dai mt ky hiuVit theo ty s CNR, ta c:

=

2/1

0

2/1

2/1

eN

C

)1M(2

BT3erf1

M

1M2P

Nu dung ma Gray sp xp cac bit doc theo cac truc ng pha va vung pha trn chom saoQAM, xac sut li bit se la:

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- Chng VII -

B phat QPSK co th thc hin t hai b phat PRK c sp xp trn hai knh la knh I vaknh Q, mi knh hoat ng tc bit bng mt na tc cua toan b h thng QPSK.

tf2sin2

1c

tf2cos2

1c

tf2cosc

Knh IRs= Rb/2

Knh Q

Rs= Rb/2

Ni tip/song song

2/

D liunh phnRb=1/Tb

QPSKRs= Rb/2

Hnh 7.17S b phat QPSK

Hnh 7.18S b thu QPSK

)4/tf2cos(c

Q/ ngng

QPSK

Rs= Rb/2

)4/tf2cos(c

+ D liu

nh phnRb=1/Tb

Song songNi tip/

Tch phn Q/ ngng

Tch phn

Hiu sut ph cua QPSK cao gp i so vi PRK. Bi v cac ky hiu trong mi knh chimcung dai tn s va co bng thng bng mt na so vi PRK cung tc .

Xac sut li Pe trong h thng QPSK ro rang se xu hn PRK, bi v min quyt nh trngian chom sao giam t mt na mt phng xung con mt phn t. Tuy nhin xac sut liPbth lai ging nhau, v mi knh I hay Q trong h thng QPSK u c lp vi nhau (trcgiao nhau). V nguyn tc, co th truyn tn hiu knh I trc knh Q, do o ban tin tng semt thi gian truyn gp i. Tuy nhin trong thc t, mi ky hiu trn knh I hoc knh Qchim thi gian gp i va nng lng giam i mt na so vi ky hiu trong PRK nn nnglng cua ban tin tng (va do o la nng lng trn mt bit Eb) trong QPSK va PRK la nhnhau (hnh 7.19). Xac sut li bit Pbla:

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=

2/1

0

b

bN

Eerf1

2

1P

Vit theo ty s CNR la:

=

2/1

2/1

bbN

C

)BT(erf12

1

P

y Tbla mt na dai ky hiu QPSK, T0. y la ly do noi rng BER cua QPSK vaPRK la nh nhau.

Xac sut li ky tPe c tnh t xac sut bit trn knh I b li, hay bit trn knh Q b li, hayca hai bit trn hai knh b li, ngha la:

2

bbbbbePP)P1()P1(PP ++=

A

0

-A

2 4 6 8

Es = Eb

1 3 5 7

1 2 3 4 5 6 7 8

Es = Eb

Eb

2/A 0

2/A

2/A 0

2/A

T/h nh phn

T/h knh I

T/h knh Q

Hnh 7.19 Dong bit vao, dong bit knh I va Q tng ng va s phn bnng lng bit

7.5 iu ch tng hiu sut cng sut

Co mt s h thng thng tin lam vic trong mi trng sn co bng thng rng nhng lai bhan ch v cng sut tn hiu. Nhng h thng nh vy cn phai da vao cac ky thut iuch tng hiu sut cng sut at c mt ty l li va mt tc d liu nao o. Noichung, co th cai thin tc d liu bng cach tng s lng ky hiu b phat (tng M). vic tng s lng ky hiu khng lam giam khoang cach gia cac im trn chom sao vakhng lam tng cng sut phat, co mt phng phap la s dung tn hiu nhiu hng- athm cac truc trc giao nhau vao gian chom sao.

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iu ch khoa dch nhiu tn s MFSK (Multi-frequency Shift Keying) la mt trng hp cuaiu ch nhiu hng nhm tng hiu sut cng sut. Cac ky hiu c thit k sao cho trcgiao nhau tng i mt. Hnh 7.20 ch ra rng ph in ap cua tn hiu MFSK trc giao laxp chng cua ph tn hiu OOK.

=

Hnh 7.20 Ph cua tn hiu MFSK trc giao

f1 f2 f3

Tc d liu tng ln hoan toan la do bng thng tng ln. V mi ky hiu cha lng tin la(bit) nn hiu sut ph danh nh cua MFSK trc giao la:MlogH

2

2)1M)(2/n(

Mlog2

s+

= (bit/s/Hz)

y n la s im 0 trn th hnh 7.11. Vi n = 2, hiu sut ph la:

1M

Mlog2

s+

= (bit/s/Hz)

Tn hiu nhiu hng cung co th at c bng cach dung cac mu bit ma hoa trc giao(hnh 7.21). Ly tng th mi ky hiu ma hoa mang lng tin la MlogH

2= (bit), nhng

at c s trc giao tng i mt, cac ky hiu phai co chiu dai bng M s nh phn (xemhnh 7.21). Bng thng danh nh cua cac ky hiu nay laM/T0 (Hz) i vi tn hiu bng cban va la2M/T0 Hz i vi tn hiu thng dai, y T0/M la khoang thi gian cua mt s nhphn. Nu cac ky hiu ng xac sut th hiu sut ph danh nh cua tn hiu thng dai ma

hoa trc giao la:

M2

Mlog2

s= (bit/s/Hz)

Va hiu sut ph ti a (vi T0B = M) la:

M

Mlog2

s= (bit/s/Hz)

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- Chng VII -

Trong trng hp tach song kt hp ti u co nhiu Gauss, cac ky hiu ng xac sut va conng lng bng nhau, xac sut li ky hiu c gii han bi:

0

eN2

Eerf1)1M(

2

1P

y E la nng lng ky hiu. Nng lng trong mt bit la: Mlog/EE2b

= T hnh 7.21 ta thy i vi mt bit bt ky trong mt ky t cho sn, ch coM/2 li ngoai tngsM-1 li ky hiu se dn n bit o b li. Do o xac sut li bit trong tn hiu MFSK trcgiao la:

ebP

1M

)2/M(P

=

000

001

Tp ky hiu trc giao Cac bit tin tng ng

111

110

101

100

011

010

Hnh 7.21 Tp ma trc giao gm 8 ky t

ng dung chnh cua h thng MFSKla trong h thng in bao Piccolo, co 32 tn s va tc ky hiu la 10 ky hiu trong 1 giy, mi ky hiu la 100 ms.

Mt ng dung khac cua MFSK hin ang c trin khai rt rng rai la FDM trc giao(OFDM) dung nhiu trong truyn hnh v tinh, trong LAN khng dy. . .

TOM TT CHNG

1. iu ch IF hoc RFc s dung trc ht la dch ph cua tn hiu s ln mt dai tns phu hp. iu nay giup cho bng tn ma tn hiu chim phu hp vi bng thng cuang truyn, ghep knh theo tn s hoc la cho phep tn hiu c bc xa bng cacanten co kch thc thc t.

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- Chng VII -

2. so sanh cac kiu iu ch khac nhau, dung hai thng s lahiu sut phvahiu sutcng sut. Hiu sut ph co n v la bit/s/Hz, va hiu sut cng sut lin quan n gia trEb/N0 yu cu i vi mt xac sut li bit cho trc.

3. Co ba ky thut iu ch chnh cho tn hiu nh phn. o la ASK, FSK va PSK. Trong cacphng phap nay, bin , tn s va pha cua tn hiu song mang b thay i theo tn hiu

nh phn vao.4. ASK va FSK u s dung tn hiu trc giao. B thu ASK va FSK co th dung loai khng

kt hp, vi u im la s n gian trong thit k. PSK yu cu phai tach song kt hp,phai khi phuc song mang, do o kho thc hin hn va t hn. Tuy nhin co th tranhvic nay bng cach ma hoa vi sai trc khi iu ch- goi la DPSK H thng DPSK khngat c xac sut li tt nh PSK nhng bu lai kinh t hn.

5. PSK yu cu cng sut tn hiu thp hn so vi ASK va FSK vi cung mt xac sut li.

6. Trong truyn thng, ngi ta s dung ph bin ba phng phap iu ch la PSK, DPSK

va FSK khng kt hp.7. M-PSK, APK, QAM la cac ky thut iu ch tng hiu sut ph. c im chung cua cac

kiu iu ch nay la s lng ky hiu iu ch nhiu hn 2.

8. APK yu cu cng sut tn hiu thp hn MPSK vi cung mt xac sut li. Hiu sut phcua APK va MPSK nh nhau. QAM la mt trng hp ring cua APK, d thc hin hnva co xac sut li c cai thin i chut.

9. MFSKla kiu iu ch tng hiu sut cng sut. Cng sut phat tit kim la v tng brng ph vi mt xac sut li cho trc.

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