Truyn dn s

Truyn dn s

M C LCMC LC.................................................................................................................CC THUT NG VIT TT............................................................................. LI NI U.................................................................................................I. IU CH V GII IU CH S...............................................................1.1. Tng quan v iu ch s...............................................................................1.2. Mt s phng php iu ch s....................................................................II.IU CH KHA DCH PHA PSK............................................................2.1. Gii thiu v kha dch pha PSK...........................................................................2.2. K thut iu ch v gii iu ch BPSK..............................................................2.3. Kha dch pha vi phn DPSK....................................................................................

2.4. Kha dch pha cu phng (QPSK) v Kha dch pha M-ary(MPSK)....................III. IU BIN CU PHNG QAM................................................................

3.1. Gii thiu v iu bin cu phng QAM

3.2. K thut iu ch QAM.

3.3. S iu ch v gii iu ch M-QAM.......................................................................

3.4. ng dng QAM...................................................................................................

IV. KT LUN..............................................................................................................

TI LIU THAM KHO.........................................................................................

CC THUT NG VIT TTAMAmplitude Modulationiu ch bin

ASKAmplitude Shift Keyingiu ch kha dch bin

BPSKBinary Phase Shift Keyingiu ch pha nh phn

PSKPhase Shift Keyingiu ch kha dch pha

QPSKQuadrature Phase Shift Keyingiu ch kha dch pha vung gc

QAMQuadrature Amplitude Modulationiu ch bin cu phng

TLO Transmitter Local OscillatorB dao ng ni pht

RLO Receiver Local OscillatorB dao ng ni thu

I. IU CH V GII IU CH S1.1. Tng quan v iu ch siu ch s la qua trinh x ly ma cac ky hiu s c chuyn i sang dang song tng thich vi cac c tinh cua knh truyn dn. Trong trng hp iu ch bng tn c s (baseband modulation), dang song o thng la cac khung inh dang .Nhng trong trng hp iu ch thng dai (bandpass modulation), cac xung inh dang iu ch m ng hinh sin goi la song mang. Trong truyn n v tuyn thi song mang c bin i thanh trng in t (EM) truyn n ni mong mun. Mt cu hoi t ra rng tai sao cn thit phai s dung song mang cho truyn dn v tuyn cua tin hiu bng tn c s ?.Cu tra li la do s truyn n cua trng in t qua khng gian dc thc hin cung vi vic s dung ng ten. Kich thc cua ng ten phu thuc vao bc song lamda va ng dung cua no. i vi mang in thoai di dng, kich thc ng ten in hinh la /4, vi dai bc song = c/f va c la vn tc anh sang c 3.10 m/s .Xem xt vic truyn mt tin hiu bng tn c s (f=3000Hz) bng cach kt ni no trc tip vi ng ten ma khng co song mang. Khi o kich thc cua ng ten se phai ln n mc nao?.i vi tin hiu bng tn c s 3000Hz th /4=25000 m ( 15 dm) . truyn mt tin hiu co tn s 3000Hz qua khng gian ma khng co iu ch song mang thi yu cu ng ten co rng 25000m( 15 dm ) la khng kha thi. Tuy nhin, nu thng tin bng tn c s c iu ch ln th nht song mang cao hn, vi du 1 song mang 900 MHz, thi ng kinh ng ten tng ng se la 8 cm. Chinh vi li do nay ma diu ch song mang hay iu ch thng dai la mt bc quan trong cho tt ca h thng k ca h thng truyn dn v tuyn .iu ch thng dai con em lai mt s li ich khac trong truyn n tin hiu. Nu co nhiu hn mt tin hiu cung s dung knh truyn n, iu ch co th c s dung tach ring cac tin hiu khac nhau qua vic ghep knh phn chia theo tn s. iu ch co th c s dung giam thiu anh hng cua nhiu qua vic iu ch trai ph, yu cu bng thng h thng ln hn rt nhiu bng thng ti thiu c s dung rong cac ban tin. iu ch co th c s dung t mt tin hiu trong mt bng tn theo yu cu thit k nh vic loc va khuch ai tin hiu. y la trng hp khi tin hiu tn s v tuyn (RF) c chuyn i sang tn s trung tn (IF) may thu. iu ch thng dai (s hoc tng t) la qua trinh trong o tin hiu thng tin c chuyn i sang mt dang song tin hiu hinh sin; i vi iu ch s, chu ky tin hiu hinh sin T tng ng vi b rng cua ky hiu s . Tin hiu hinh sin co 3 c im phn bit tin hiu hinh sin nay vi tin hiu hinh sin khac: bin d, tn s, pha. Do vy , iu ch thng dai co th c inh nghia nh la mt qua trinh ma o bin , tn s hay pha cua song mang v tuyn RF, hoc s kt hp cua ca 3 yu t o c bin di tng ng vi tin hiu thng tin cn c truyn i. Dang tng quat cua tin hiu song mang la:

y, A(t) la bin bin i theo thi gian va la goc pha bin di theo thi gian. co th vit cu th hn:

Khi o:]Vi la tn s goc cua song mang va la pha. F va u c s dung biu thi cho tn s. Khi f c s dung thi n vi la Hert z (HZ) con khi c s dung thi n vi la radian/s, gia chung co mi quan h : .1.2 Mt s phng php iu ch sCac loai iu ch/giai diu ch thng dai c ban c minh hoa hnh ve 1. Khi may thu bit c pha cua song mang tach cac tin hiu thi qua trinh x li o goi la tach song kt hp ( coherent detection); Khi may tu khng s dung thng tin tham khao v pha, thi qua trinh x li c goi la tch song khng kt hp ( noncohernt detection). Trong truyn thng s, cc thut ng gii iu ch v tch song thng c dng thay th cho nhau, mc d gii iu ch nhn mnh s ti to, khi phc dng sng, cn tch song lin quan ti qu trnh a ra quyt nh v k hiu thu c .Trong tch sng kt hp l tng, c sn my thu mt nguyn mu ca mi tn hiu n. Nhng dng song nguyn mu c gng sao chp nguyn tn hiu mi kha cnh , thm ch c pha ca song v tuyn . My thus au c kha pha (phase locked) vi tn hiu n. Trong qu trnh gii iu ch , my thu nhn v kt hp ( tng qua ) tn hiu n vi mi bn sao nguyn mu ca n.iu ch/gii iu ch kt hp c phn loi l: Kha dch pha (PSK), Kha dch bin (ASK), Kha dch tn s (FSK), iu ch pha lin tc (CPM) v lai ghp gia cc phng php. V d v iu ch pha lin tc CPM l Kha dch pha cu phng b (OQPSK), kha dch nh nht/ ti thiu (MSK), v iu ch lai ghp l iu bin cu phng (QAM).Gii iu ch khng kt hp c cc h thng m b gii iu ch c thit k hot ng khng cn bit v gi r tuyt i ca pha tn hiu vo; do vy khng yu cu vic c lng pha. Cho nn u im ca h thng khoogn kt hp so vi kt hp l gim c phc tp, nhng xc sut li ( li tng ln. hnh 1 nhng loi iu ch/gii iu ch c lit k trong ct khng kt hp l DPSK,FSK,ASK,CPM, v iu ch lai.

Kt hp (Coherent)Khng kt hp (NonCoherent)

Kha dch pha - PSKKha dch pha vi phn-DPSK

Kha dch tn s-FSKKha dch tn s-FSK

Kha dch bin -ASKKha dch bin -ASK

iu ch pha lin tc-CPMiu ch pha lin tc-CPM

iu ch lai-Hybridiu ch lai-Hybrid

II. IU CH KHA DCH PHA PSK (Phase Shift Keying)2.1. Gii thiu v kha dch pha PSKKha dch pha l mt dng iu ch gc, bin khng i. Kha dch pha cng tng t nh iu ch pha thng thng, ch c khc l PSK c tn hiu u vo l tn hiu nh phn v pha u ra c s lng gii hn.iu ch PSK s dng b iu ch ngoi nh b iu ch pha LiNbO3. Ti my thu, tn hiu PSK c gii iu ch bng h thng tch sng ng tn hoc i tn, tn hiu trung tn IF c gii iu ch ng b hoc khng ng b.Trong phng php iu ch PSK, tn s v bin ca sng mang c gi khng i trong khi pha ca n dch theo mi bit dng truyn d liu.

Phn loiC 2 loi PSK thng dng: Loi th nht dng hai tn hiu sng mang i din cho bit 1 v bit 0, hai sng mang ny khc pha nhau 180. V tn hiu ny ch l nghch o ca tn hiu kia nn loi ny c gi l PSK pha phi hp. iu bt tin ca loi ny l ti my thu i hi phi c sng mang tham chiu so pha vi tn hiu thu, do cn phi thc hin ng b pha gia my thu v my pht. Kt qu dn n mch gii iu ch phc tp hn.Loi PSK th 2 gi l PSK vi sai. Vi loi ny s dch chuyn pha xy ra ti mi bit hay mi symbol, khng cn quan tm ti chui bit 0 hay 1 ang c truyn. Gi s vi iu ch 2-PSK vi sai th mt s dch pha 90 tng ng vi tn hiu hin hnh ch nh 0 l bit k tip, trong khi s dch pha 270 ch bit 1 l bit k tip. Nh vy mch gii iu ch ch cn xc nh ln ca s dch pha thay v phi xc nh gi tr tuyt i ca tng pha. mch iu ch ch khi no thay i trng thi ca d liu mi i pha ca sng mangV mt ton hc ta c th xc nh bng thng ca PSK. y chng ta trnh by tn hiu s nh phn di dng lng cc v mc m ca tn hiu s l kt qu i pha 180 ca sng mang. Tn hiu d liu biu din di dng chui Fourier nh sau:S(t)=4/[cost-1/3 cos3t +1/5cos5t-... ]T suy ra:SPSK=4/[cosct.cost-1/3 cosct cos3t +...]

m(t)

s(t)

SPSK (t)

Hnh 2.1: iu ch pha tn hiu nh phn 1011001Nng lng tn hiu:

f1 -3f0 f1 -f0 f0 f1+ f0 f1+3f0 f0 -thnh phn tn s c bn =1/2 tc bitHnh 2.2: Bng thng tn hiu PSKYu cu v rng bng i vi ASK v PSK l ging nhau th hin hm mt ph cng sutPPSK = (A/4) Ph ca PSK khng cha cc hm Delta Dirac hay xung tn s mang, v do l dng iu ch nn sng mang.2.2. K thut iu ch v gii iu ch PSK hai trng thi nht qun, BPSK nht qun.

Bn tin: m(t)Kha dch pha nh phn (BPSK) lin quan n s dch pha ca sng mang hnh sin 0 hoc 180 tng ng vi tn hiu nh phn n cc u vo. S to tn hiu iu ch BPSK c miu t hnh...

Sng mang cos (2fct)Tn hiu BPSK

Accos(2fct+Dpm(t)

-90Dch pha

Hnh 2.3: S to tn hiu iu ch BPSKTn hiu BPSK c biu din bi:s(t) = Accos [ct+Dcm(t) ]Trong m(t) l tn hiu d liu bng gc lng cc. thun tin, m(t) c gi tr nh l 1 v l dng xung ch nht.Gi ta s ch ra rng BPSK l mt dng tn hiu AM khai trin biu thc, ta c:s(t) = Accos(Dpm(t))cosct - Acsin(DPm(t))sinctNhc li, m(t) c gi tr 1 v cos(x) v sin(x) l hm chn v l ca x, minh ha cho tn hiu rt gn BPSK l:=(AccosDp)cosct - (AcsinDp)m(t)sinctMc ca sng mang hoa tiu c thit lp bng chnh lch gi tr nh, =Dp.i vi tn hiu s c iu ch gc, ch s iu ch s, h, c nh ngha l:h = trong 2 l lch pha ln nht t nh n ch (rad) trong khong cch thi gian yu cu gi 1 k t, Ts. i vi tn hiu nh phn, thi gian k t c tnh bng thi gian bit Ts = Tb Mc ca sng mang hoa tiu c thit lp bng lch gi tr nh, = D, vi m(t) = 1. N u Dp nh, sng mang hoa tiu s c bin tng i ln so snh vi d liu; do ; c rt t cng sut trong dng d liu ( gm thng tin ngun). ti a hiu su t tn hiu ( kh nng li thp), cng sut ca dng tn hiu cn c ti a ha. iu ny t c bng cch = DP=90=/2 rad, tng ng ch s iu ch s ca h=1. Trong trng hp h=1, tn hiu BPSK tr thnh: s(t) = -Acm(t)sinctChm sao tn hiu BPSK c minh ha c th hnh v di y:Bit2 1Phase

010180

BitsHnh 2.4: Chm sao tn hiu iu ch BPSKTrong sut phn ny, gi s = 90, h=1 c s dng cho tn hiu BPSK ( loi tr cc giai on khc). Biu thc 6.9 ch ra rng BPSK tng ng vi tn hiu DSB-SC vi dng sng d liu bng gc lng cc. ng bao dng phc cho tn hiu BPSK l: g(t)=jAcm(t) cho BPSKChng ta c c mt ph cng sut PSD cho ng bao phc l:Pg(f) = Tb(cho BPSKPh ca tn hiu BPSK c minh ha hnh..... Bng thng t 0 n 0 cho BPSK l 2R, ging vi ASK.

Hnh 2.5: PSD ca tn hiu thng s di BPSK tch BPSK, b tch kt hp c s dng nh minh ha trong hnh v di y:

Tn hiu ra nh phnTn hiu vo BPSKB lc thng thp

cosct Hnh 2.6: B tch tn hiu BPSK (tch kt hp)

Qu trnh iu ch:

Hnh 2.7: Qu trnh iu ch BPSKLung nh phn n cc u vo b(t) i vo. B chuyn i mc chuyn i cc k hiu 0 v 1 vo dng lng cc vi + v -. Thy r, mi khong thi gian ca tn hiu iu ch, a ln b nhn l si. Lung ny c s dng iu ch sng mang 1(t)=-cos(2fct) t b dao ng ni pht TLO. u ra ca b iu ch ta nhn c sng BPSK mong mun.Qu trnh gii iu ch:

Hnh 2.8: Qu trnh gii iu ch BPSKB gii iu ch BPSK bao gm mt b tng quan v c cp ti ch mt cp tn hiu chun 1(t). Cc sng chun ny c to tao ra t b khi phc sng mang. Thi im khi u tch phn cho mt bit c ng b bi mch khi phc xung nh thi. Cc u ra ca b tng quan c so snh vi 1 ngng 0V. Nu y1>0 th quyt nh c thc hin thin v k hiu 0, v nu y10 th quyt nh c thc hin thin v k hiu 0 i vi u ra ca knh ng pha I pha trn, nhng nu y10 th quyt nh thin v k hiu 0 i vi u ra ca knh vung gc pha di, nhng nu y2 =p dng nh l cosin cho chm sao tn hiu 8-PSK, ta c: + - 2cos(45) = => =T , ta c cng sut pht trung bnh ca 4-PSK v 8-PSK ln lt l: = = Cng sut pht b sung i vi 8-PSK at c xc sut li nh nhau l:P = 10 = 5,3392dBnh gi hiu nng ca cc k thut iu ch:Xc sut li bit ca iu ch BPSK kt hp c xc nh theo biu thc di y:Pe = erfcvi Eb l nng lng xung v N0 i din cho mt ph cng sut nhiu.Hnh.... biu din xc sut li bit (BEP) ca cc dng iu ch M-PSK vi M=2,4,8,16,64.i vi h thng M- ary, xc sut li bit l:Pe=Q y hm Q(u) c xc nh theo biu thc:Q(u)=dz

Quan h gia hm li v hm Q(u) c cho nh sau:Erfc(u) = 2Q(u; Q(u)= erfc(

Hnh 2.14: Xc sut li bit ca iu ch M-PSK

Bi tp v d: Cho h thng BPSK truyn qua knh AWGN vi mt ph cng sut No/2 = W/Hz, Eb = T/2, T l chu k ca bt v A l bin tn hiu. Xc nh gi tr ca A xc sut li bit t c , nu tc d liu l:a.10kbpsb.100kbpsc.1MbpsGii:Xc sut li bit ca iu ch BPSK kt hp c xc nh theo biu thc: Pe = erfcTa c:Pe = erfc = erfc = QVi Pe=, chng ta tra bng : = 4,74 => = 44,9352.a. Nu tc d liu Rb=10kbps. Ta c : T = = s.Khi bin tn hiu: A = = 6,7033.b. Tng t vi tc d liu Rb = 100kbps A = 2,12.c. Tng t vi tc d liu Rb = 1Mbps A = 6,703.

III. IU BIN CU PHNG QAM.3.1. Gii thiu v iu bin cu phng QAM. Phng php iu ch M-QAM l phng php nng cao hiu qu ca mt knh truyn m khng cn tng cng sut pht hay tng rng bng thng. Vic iu ch hai thnh phn ng pha v pha vung gc l mt cch c lp vi nhau cho ta mt s iu ch mi gi l iu ch bin vung gc (hay cu phng) M trng thi (QAM). Nh vy, trong s iu ch ny sng mang b iu ch c v bin ln pha.3.2. K thut iu ch QAM:Trong h thng PSK, cc thnh phn ng pha v vung pha c kt hp vi nhau to thnh mt tn hiu ng bao khng i. Tuy nhin, nu loi b loi ny v cho cc thnh phn ng pha v vung pha c th c lp vi nhau th ta c mt s iu mi gi l iu bin cu phng iu ch bin sng mang QAM (iu ch bin gc) . s iu ch ny, sng mang b iu ch c bin ln pha. iu ch QAM l c u im l tng dung lng truyn dn s.Dng tng qut ca iu ch QAM, 14 mc (m-QAM) c xc nh nh sau:

Trong ,E0 : nng lng ca tn hiu c bin thp nht ai , bi : cp s nguyn c lp c chn ty theo v tr bn tin. i=1,2L.Dng c s ca chm tn hiu M-QAM l dng ca hai tn hiu ASK c L trng thi. Nh vy, tn hiu Si(t) gm hai thnh phn sng mang c pha vung gc c iu ch bi mt tp tn hiu ri rc nn c tn l iu ch bin vung gc.C th phn tch Si(t) thnh cp hm c s:

Ta cc im bn tin l v vi :

i vi 16-QAM ta c L=4

Thnh phn ng pha v vung pha trong 16-QAM

Chm tn hiu ca 16-QAM

Hnh 3.1: Chm tn hiu M-QAM3.3 S iu ch v gii iu ch M-QAMB iu ch

Hnh 3.2: Qu trnh iu ch M-QAMHot ng ca b iu ch B phn lung (demux) chuyn i lung nh phn b(t) tc bit Rb=1/Tb u vo thnh bn lung c lp, trong hai bit l c a n b chuyn i mc nhnh trn cn hai bit chn c a n b chuyn i mc nhnh di. Tc k hiu trong trng hp ny s bng Rs=Rb/4.Cc b bin i mc chuyn i 2 mc vo L mc () to ra cc tn hiu L mc tng ng vi cc u vo ng pha v pha vung gc. Sau khi nhn hai tn hiu L mc vi hai sng mang c pha vung gc c to t b dao ng ni pht TLO (Transmitter Local Oscillator) ri cng li ta c tn hiu M-QAM.B gii iu ch

Hnh 3.3: S gii iu ch M-QAMHot ng ca b gii iu ch Tn hiu thu c a ln 2 nhnh ng pha v vung pha, sau c nhn vi 2 hm trc giao ging pha pht c to ra t b dao ng ni thu RLO (Receiver Local Oscillator). Nh tnh cht trc giao m ta tch c 2 thnh phn tn hiu. Tn hiu sau c a qua b tng quan ly mu, nh gi ngng (so snh vi L-1 ngng) thu c k hiu. Sau cng hai chui s nh phn c tch ra ni trn s kt hp vi nhau b bin i song song vo n tip khi phc li chui nh phn pha pht (c tnh chui pht ).3.4. ng dng QAM:Cc loi iu ch QAM.

STTLOI IU CHS BIT/I(Q)S BIT/SYSMBOLS TRNG THI

14QAM124

2(QPSK)2416

316QAM3664

425648256

Nh vi nhiu n iu ch k thut s, biu chm sao l mt i din hu ch. Trong QAM, cc im chm sao ny thng c b tr trong mt vung vi khong cch dc v ngang bng nhau, mc d cu hnh khc l c th (v d nh Cross-QAM). T trong vin thng k thut s d liu thng nh phn, s lng cc im trong li in thng l mt sc mnh ca 2 (2, 4, 8 ...). K t khi QAM thng vung, mt s trong s ny l him cc hnh thc ph bin nht l 16-QAM, 64-QAM, 128-QAM v 256-QAM. Bng cch di chuyn n mt chm sao bc cao, c th truyn ti cc bit trn mi k hiu. Tuy nhin, nu nng lng trung bnh ca chm sao ny l vn nh c (bng cch lm mt so snh cng bng), cc im phi c xch li gn nhau v do d b nhiu v tham nhng khc, iu ny dn n mt t l li bit cao hn v do bc cao QAM c th cung cp nhiu d liu hn t ng tin cy hn so vi di QAM, cho nng lng lin tc c ngha l chm sao. Nu d liu, t gi vt ra ngoi nhng ngi c cung cp bi PSK-8 c yu cu, n l nhiu hn bnh thng di chuyn n QAM k t khi n t c mt khong cch ln gia cc im ln cn trong mt phng IQ do phn phi cc im ng u hn. Cc yu t phc tp l nhng im khng cn tt c cc bin nh nhau v do , b gii iu phi by gi chnh xc pht hin c hai pha v bin , hn l ch giai on. 64-QAM v 256 QAM-thng c s dng trong truyn hnh cp k thut s v cc ng dng modem cp. Ti Hoa K, 64-QAM v 256-QAM l cc n iu ch u quyn cho cp k thut s (xem QAM tuner) theo tiu chun ca SCTE trong tiu chun ANSI / SCTE 07 nm 2000. Lu rng nhng ngi tip th nhiu ngi s tham kho nhng lc QAM-64 v QAM-256. Ti Anh, 16-QAM v 64-QAM hin ang c s dng cho truyn hnh mt t k thut s (Freeview v Top Up TV) v 256-QAM c ln k hoch cho Freeview-HD. H thng truyn thng c thit k t c mc rt cao hiu qu quang ph thng thng s dng cc chm sao QAM rt dy c. Mt v d l G.hn ITU-T tiu chun cho ni mng thng qua h thng dy in hin ti nh (cp ng trc, ng dy in thoi v ng dy in), trong s dng cc chm sao ln ti 4096-QAM (12 bit / biu tng). Mt v d khc l cng ngh VDSL2 cho cp ng xon, c chm sao kch thc ln ti 32768 im.

KT LUN M-QAM l mt trong nhng s iu ch M trnh thi thng c dng hn so vi s 2 trng thi truyn s liu trong knh bng tn hn ch. Vic s dng M-QAM s gim c rng bng tn n= so vi BPSK

TI LIU THAM KHO1. Gio trnh Truyn Dn S 2. Gio trnh C s K Thut Thng Tin V Tuyn

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