Giao Trinh Thong Tin Di Dong 2012

7/30/2019 Giao Trinh Thong Tin Di Dong 2012

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B mn Cng ngh truyn thng Khoa CN T&TT Trng i hc CNTT&TT

Bi ging thng tin di ng ThS. Phm Vn Ngc 1

PH LC

Chng 1:Khi qut chung v thng tin di ng ..................................................... 41.1. Nhng c th ca thng tin di ng ......................................................... 41.2. Lch s pht trin ca thng tin di ng ..................................................... 41.3. Mt s h thng thng tin di ng trn th gii .......................................... 6

1.4. Xu hng pht trin ca thng tin di ng ................................................. 81.5. Mt s kt qu t c mng GSM .......................................................... 8Chng 2:T bo C s thit k h thng .......................................................... 11

2.1. T bo v vic phn b tn s .................................................................. 112.2.1. La chn t bo ................................................................................ 112.2.2. Phn chia knh truyn ....................................................................... 112.2.3. Kch thc nhm N .......................................................................... 12

2.2. Nhiu cng knh v dung lng h thng ................................................ 132.3. Nhiu knh ln cn v k hoch phn chia knh truyn............................ 152.4. Chin lc phn knh v chuyn giao ...................................................... 16

2.5. Trung k v cp dch v ...................................................................... 182.5.1. Knh chung ...................................................................................... 182.5.2. Cp dch v.................................................................................. 182.5.3. Tng i khng nh cuc gi b chn Cng thc Erlang B ............ 192.5.4. Tng i nh cuc gi b chn Cng thc Erlang C ....................... 222.5.5. Hiu sut trung k ............................................................................. 24

2.6. Nng cao dung lng h thng t bo ...................................................... 242.6.1. Chia nh t bo ................................................................................. 242.6.2. S dng ng ten nh hng ............................................................. 25

2.6.3. Phn vng trong t bo ..................................................................... 26Chng 3:M ha ting ni ................................................................................... 283.1. Cc c trng ca ting ni ...................................................................... 29

3.2.1. Hm mt xc sut (pdf) ................................................................ 293.2.2. Hm t tng quan (AFC) ................................................................ 293.2.3. Hm mt ph cng sut (psd)....................................................... 293.2.1. Lng t tuyn tnh .......................................................................... 303.2.2. Lng t phi tuyn ........................................................................... 303.2.3. Lng t thch nghi .......................................................................... 313.2.4. Lng t vct ................................................................................. 32

3.3. Cc b m ha theo dng sng ................................................................. 323.3.1. B iu ch xung m vi phn thch nghi (ADPCM) .......................... 323.3.2. B m ha sng m theo tn s ......................................................... 33

3.4. Cc b m ha theo ngun m (Vocoder) ................................................ 353.4.1. Vocoder knh ................................................................................... 363.4.2. Vocoder hi ...................................................................................... 363.4.3. Vocoder Cepstrum ............................................................................ 363.4.4. Vocoder kch thch bng Voice ......................................................... 36


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3.4.5. Cc b m d on tuyn tnh .......................................................... 363.5. Chn b m ha ting ni trong thng tin di ng ................................... 403.6. nh gi hot ng ca b m ha ting ni ............................................ 41

Chng 4:Cn bng Phn tp M knh ........................................................... 434.1. Gii thiu chung ...................................................................................... 434.2. C s ca k thut cn bng thch nghi .................................................... 44

4.3. C s ca k thut phn tp (t hp t s cc i) ................................... 474.3.1. Cc dng phn tp ............................................................................ 484.3.2. Phn tp pht .................................................................................... 484.3.3. Phn tp thu ...................................................................................... 504.3.4. B thu RAKE ................................................................................... 504.3.5. Ghp xen (interleaving) .................................................................... 51

4.4. M knh................................................................................................... 524.4.1. Cc vn c bn ca m knh ........................................................ 524.4.2. Dung lng knh .............................................................................. 53

4.4.3. M khi ............................................................................................ 544.4.4. M xon ........................................................................................... 65Chng 5:H thng thng tin di ng t ong GSM ............................................... 75

5.1. c im chung ca h thng GSM ......................................................... 755.2. Kin trc h thng GSM .......................................................................... 75

5.2.1. H thng con chuyn mch SS....................................................... 785.2.2. H thng con trm gc BSS ........................................................... 795.2.3. H thng con vn hnh v bo dng OSS ...................................... 81

5.3. Kin trc v tuyn ca GSM .................................................................... 835.4. Cc loi knh trong GSM ........................................................................ 84

5.4.1. Knh lu lng ................................................................................. 855.4.2. Cc knh iu khin ......................................................................... 85

5.5. Cuc gi trong GSM ................................................................................ 875.5.1. Cuc gi t MS ................................................................................ 875.5.2. Cuc gi t mng c nh n MS .................................................... 88

5.6. Cu trc khung ........................................................................................ 895.7. X l tn hiu trong GSM ........................................................................ 935.8. Cu trc mng GPRS da trn nn mng GSM ........................................... 96

5.8.1. Gateway GSN (GGSN) ........................................................................ 975.8.2. Serving GSN (SGSN) .......................................................................... 985.8.3. n v kim tra d liu gi PCU ( Packet Control Unit ) ..................... 995.8.4. HLR, VLR, AUC v EIR ..................................................................... 995.8.5. BSS (Base Station System) ................................................................ 100

5.9. Cc loi knh trong mng GRPS ............................................................... 1015.9.1. Cc knh logic gi ............................................................................. 1015.9.2. Knh lu lng logic gi ................................................................... 1035.9.3. Knh lu lng d liu gi ................................................................ 104

Chng 6:H thng thng tin di ng CDMA ..................................................... 105


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6.1. Gii thiu chung .................................................................................... 1056.2. Kin trc h thng CDMA ..................................................................... 107

6.2.1 Thit b ngi s dng UE (USER EQUIPMENT) ............................. 1096.2.2 Cu trc mng truy cp ....................................................................... 1096.2.3. Mng li CN ..................................................................................... 110

6.3. Dy gi ngu nhin (PN)........................................................................ 111

6.3.1. To dy m....................................................................................... 1116.3.2. Tnh cht ca chui MLSR ............................................................. 1146.3.3. Hm tng quan ca tn hiu m gi ngu nhin............................. 1156.3.4. Dy Gold ........................................................................................ 117

6.4. M trc giao .......................................................................................... 1186.5. Tri ph dy (chui) trc tip (DS SS) ................................................ 1196.6. Hot ng ca tri ph dy trc tip ...................................................... 1216.7. Tri ph nhy tn (FH SS) .................................................................. 1246.8. Hot ng ca tri ph nhy tn ............................................................. 125

6.9. H thng MC DS CDMA.................................................................... 1276.9.1. M hnh h thng MC DS CDMA ............................................... 1286.9.2. My pht ........................................................................................ 1296.9.3. Knh truyn .................................................................................... 1316.9.4. My thu .......................................................................................... 1326.9.5. nh gi c tnh h thng MC DS CDMA ................................. 134

6.10. iu ch OFDM ................................................................................. 1416.10.1. Cu trc khung ............................................................................... 1436.10.2. K thut iu ch ............................................................................ 1446.10.3. Khong bo v v tin t lp ........................................................... 146

6.10.4. To ca s ...................................................................................... 1486.10.5. Ghp xen ........................................................................................ 1496.10.6. Tc dng ca chn CP .................................................................... 150

Chng 7:H thng thng tin di ng th h th 3 .............................................. 1567.1. Yu cu chung i vi h thng thng tin di ngt h h th 3 .............. 1567.2. H thng CDMA .................................................................................. 157

7.2.1. Vng ph sng ca cdma 2000 ....................................................... 1577.2.2. Cu trc ca knh cdma 2000 ......................................................... 158

7.3. W CDMA ........................................................................................... 1617.4. H thng lai ghp TDMA v CDMA (h thng UTRA TDD) .................. 162


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Ni n thng tin di ng l ni n vic lin lc thng qua sng in t bt u

t nm 1897 Gugliemo Marconi thc hin lin lc t t lin vi nhng con tutrn bin bng sng in t. n nm 1980 thng tin di ng mi thc s pht trintrn th gii.

hiu c ta lm php tnh: Mi cuc lin lc gia hai ngi cn mt ngtruyn c lp, mi knh gi s c di thng 3kHz (trn thc t ln hn) th di tn vtuyn l t 0 3GHz ch cho php truyn 3.109/3.103 = 106 cuc lin lc cng mt lc.

phc v hng chc triu ngi c th cng s dng my di ng cng mt lc, ycha k di tn ny cn dnh cho rt nhiu cng vic khc.

Phng php duy nht gii quyt vn nhiu ngi dng c lp trn

mt di tn v tuyn hn ch l: Mt cuc lin lc di ng ny c th s dng ngdi tn ca mt cuc lin lc di ng khc vi iu kin hai cuc lin lc phi xanhau v khong cch sng truyn n nhau nh hn sng truyn gia hai ngi

trong cuc. Do thch hp cho vic qun l ngi ta chia thnh cc phn nh gil t bo (Cellular). Hai cuc lin lc hai t bo xa nhau c th s dng cng mt

di tn s sng in t thng qua vic qun l ca mt trm trung tm t bo. V lthuyt kch thc t bo l rt nh khi c th phc v v s cuc gi cng mt lcch cn mt di tn sng v tuyn hn ch. Phng php ny gi l phng php sdng li tn s.

Tm li, nhng c th ca thng tin di ng l: Phc v a truy cp gn linvi thit k Mng t bo, cc h qa ko theo lin quan n vn ny l: Chuyngiao, chng nhiu, qun l di ng, qun l ti nguyn v tuyn, bo mt, Nhngiu ny khc rt nhiu vi mng v tuyn c nh v lun i hi pht trin nhngcng ngh mi.

! " # $ % &

c bc tranh ton cnh v s pht trin ca thng tin di ng ta im li mt

s mc lch s khi pht trin thng tin di ng trn th gii.

Ta c th la chn lch s pht trin thng tin di ng ca nc M lm inhnh:

Nm 1946: Dch v in thi di ng cng cng c gii thiu ln u 25thnh ph. Mi h thng dng b pht cng sut ln t trn ng ten cao ph sng bn

knh 50km. k thut Push to talk (bn song cng), rng knh truyn l 120kHz(mc d rng bng tn ca ting ni ch l 3khz). y cha phi l h thng t bo,

tn s cha c s dng li, s ngi c phc v rt t.


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Nm 1950: rng knh c thu hp li cn 60kHz, dn n s knh sdng tng gp i

Nm 1960: rng knh ch cn 30kHz.

T nm 1950 1960: Xut hin trung k t ng, dch v IMTS, hiu sut sdng ph km so vi h thng t bo ngy nay.

Trong khong 1950 1960: L thuyt mng t bo ra i tuy nhin cng nghlc cha p ng c.

Nm 1976: New York ch c 12 knh phc v 543 khch hng, dch v chtlng km hay b bn

Nm 1981 h thng in thoi di ng Bc u NMT450 tr thnh h thngdch v truyn thng di ng t bo Chu u u tin. H thng ny ra i ch yupht trin cc my iu thoi trn xe hi v xch tay. L h thng k thut Analog,

hot ng trn bng tn 450MHz (453 457.5MHz t MS BTS v 463 467.5MHz

t BTS MS) s dng a truy cp FDMA, iu ch FSK, rng bng tn l 25kHzdo cho php h tr 180 knh

Nm 1986 h thng NMT900 Ty u, h thng ny hot ng trn bng tn900MHz

Nm 1983: Ra i dch v thng tin di ng ci tin (AMPS) bi cng tyAT&T. nh du s ra i in thoi di ng t bo th h 1. FCC phn 40MHz trnph tn 800MHz, Nm 1989; FCC phn thm 10MHz ph cho h thng AMPS (824

849MHz t MS BTS v 869 894MHz t BTS MS) cho dch v ny mi knh c rng bng tn 30kHz, do h thng c 832 knh p (knh song cng mi knh

rng 2*30 = 60kHz). Trong 832 knh c 40 knh ch mang thng tin v h thng. mi thnh ph phn cho 2 nh cung cp dch v. H thng t bo ny hot ng

trong mi trng hn ch giao thoa, s dng li tn s, k thut a truy nhp phnchia theo tn s FDMA, cc i s ngi dng di tn v t chc knh ca h nhsau:

Nm 1991: Ra i h thng t bo s (USDC) theo chun IS 54, h tr 3ngi s dng trn mt knh truyn 30kHz ( DQPSK4/ ). Khi k thut nn ting ni

990 991 1023 1 2 799 990 990 1023 1 2 799

Cc knh pht xuiCc knh pht ngc

824 849 MHzS hiu knh ngc: 7991 N

1023990 N S hiu knh xui: 7991 N

1023990 N

869 894 MHzTn s: 0.030N + 825.0 MHz

0.030(N - 1023) + 825.0 MHz0.030N + 870.0 MHz0.030(N 1023) + 870.0 MHz


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v x l tn hiu pht trin c th tng dung lng ln 6 ln. (kt hp vi TDMA v

tn ti song song vi AMPS trn cng c s h tng) nh du s ra i ca thng tindi ng th h 2.

Cng nm 1991: H thng da trn k thut tri ph c pht trin biQuancom theo chun IS 95 h tr nhiu ngi s dng trn mt di tn 1.25MHz,

phn bit m tri ph trc tip (CDMA). Vi AMPS yu cu SNR >= 18 dB thCDMA yu cu thp hn v cho dung lng cao hn. Ngoi ra b m ha ting ni

tc thay i c th pht hin ting ni khi m thoi s iu khin b pht ch pht

sng khi ni s lm gim mi trng giao thoa v tit kim pin.

Nm 1991 h thng Mng thng tin di ng th h 2 ra i Chu u vi trnph tn 900MHz (890 915MHz uplink v 935 960MHz downlink) s dng kthut TDMA/FDMA

Vn tch hp cc mng trong mt c s h tng cng c t ra t unhng nm 1990.

Nm 1995: Chnh ph M cp giy php trn di tn 1800/2100MHz hahn s pht trin mi cho dch v thng tin c nhn (PCS)

Nm 2000: t chc vin thng quc t (ITU) thng nht mt s hng vchun pht trin cho thng tin di ng a dch v th h 3 theo chun UMTS vCDMA2000.

!

%

H thng thng tin di ng trn th gii phn thnh 3 loi chnh nh sau l: H

nhn tin - in thoi ko di - in thoi t bo trong :

H nhn tin: l loi hnh thng tin di ng bn song cng ngi dng ch nhnc bn tin nhn mt chiu vi mt thit b thu n gin nh mt chic radio vmt m s ring.

in thoi ko di: l thit b cm tay kt ni v tuyn vi mt my ch ttrong nh, my ch c kt ni vi mng in thoi cng cng (PSTN). Tm v

tuyn ko di hp (


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$

%

#

AMPS T bo 1983 FDMA 824 894MHz FM 300KHz

NAMPS T bo 1992 FDMA 824 894MHz FM 10KHz

USDC T bo 1991 TDMA 824 894MHz DQPSK4/ 30KHz

CDPD T bo 1993 FH / gi 824 894MHz GMSK 30KHz

IS 95 T bo

/PCS

1993 CDMA 824 894MHz

1.8 2.0 GHz

QMSK

/ BPSK

12.5MHz

GSC Nhn tin 1970 n cng Vi knh FSK 12.5KHz

POSAG Nhn tin 1970 n cng Vi knh FSK 12.5KHz

FLEX Nhn tin 1993 n cng Vi knh 4 FSK 15KHz

DSC1900(GSM)

PCS 1994 TDMA 1.85 1.99 GHz GMSK 200KHz

PACS Ko di/PCS

1994 TDMA/FDMA

1.85 1.99 GHz DQPSK4/ 300KHz

MIRS SMR.PCS 1994 TDMA Vi knh 16QAM 25KHz

$

%

#

E-TACS T bo 1985 FDMA 900 MHz FM 25KHz

NMT-450 T bo 1981 FDMA 457 470 MHz FM 25KHz

NMT-900 T bo 1986 FDMA 890 - 960 MHz FM 12.5KHz

GSM T bo / PCS 1990 TDMA 890 - 960 MHz GMSK 200KHz

C-450 T bo 1985 FDMA 450 465 MHz FM 20/10KHz

ERMES Nhn tin 1993 FDMA Vi knh 4 - FSK 25KHz

CT2 Ko di 1989 FDMA 864 868 MHz GFSK 100KHz

DECT Ko di 1993 TDMA 1.88 1.9 GHz GFSK 1.728KHz

DCS-1800 Ko di / PCS 1993 TDMA 1.71 1.88 GHz GMSK 200KHz

$

Chun Loi Btu

Truycp

Bng tn iu ch knh

JATC T bo 1988 FDMA 860 925 MHz FM 25KHz

PDC T bo 1993 TDMA 810 1501 MHz DQPSK4/ 25KHz

NTT T bo 1979 FDMA 400/800 MHz FM 25KHz


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NTACS T bo 1993 FDMA 843 925 MHz FM 12.5KHz

NTT Nhn tin 1979 FDMA 280 MHz FSK 12.5KHz

NEC Nhn tin 1979 FDMA Vi knh FSK 10KHz

PHS Ko di 1993 TDMA 1895 1907 MHz DQPSK4/ 300KHz

# $ % &

Hin nay thng tin di ng ang trong giai on pht trin nh v bo, p ng

nhu cu khng ngng tng ca khch hng c v khi lng, cht lng v loi hnhdch v chia lm cc giai on pht trin sau:

T nm 1989 c nhng nghin cu rng ln trn th gii nhm pht trin hthng v tuyn c nhn: Kt hp s thng minh ca mng PSTN, x l tn hiu shin i v cng ngh RF.

Xu hng pht trin mng v tuyn trong nh (indoor) cho php ngi dng ktni my tnh vn phng trong cc ta nh ln (tn s 1.8GHz)

Xu hng chun ho IMT 2000, c quyt nh bi ITU xy dng chun vquy hoch tn s trn ton th gii.

Xu hng pht trin h vin thng v tinh LEO, cng vi s pht trin cngngh v tr, h thng tin v tinh phi hp vi h di ng mt t to nn kt ni toncu thch hp vi mi loi a hnh v loi thng tin.

Hin nay cc quc gia pht trin sau li c c hi i nhanh vo cc ng dng tintin nht v la chn cc m hnh thch hp vi s pht trin ca tng lai.

!

- Cc dch v mng mi v ci thin cc dch v lin quan n truyn s liu nhnn s liu ca ngi s dng, s liu chuyn mch knh tc cao (HSCDS: HighSpeed Circuit Swiched Data), dch v v tuyn gi chung (GPRS: General PacketRadio Sevice) v s liu 14,4 kbit/s.

- Cc cng vic lin quan n dch v thoi nh: Codec ting ton tc ci tin(EFC: Enhanced Full Rate Codec), Codec a tc thch ng v khai thc t do u

cui cc Codec ting.

- Cc dch v b sung nh:chuyn hng cuc gi, hin th tn ch gi, chuyngiao cuc gi v cc dch v chn gi.

- Ci thin lin quan n dch v bn tin ngn (SMS :Short Message Service) nh:mc ni cc SMS, m rng bng ch ci, m rng tng tc gia cc SMS.

- Cc cng vic lin quan n tnh cc nh:cc dch v tr tin trc, tnh ccnng v h tr cho u tin vng gia nh.


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- Tng cng cng ngh SIM.

- Dch v mng thng minh:CAMEL.

- Cc ci thin chung nh:chuyn mng GSM-AMPS, cc dch v ng v, tng

tc vi cc h thng thng tin di ng v tinh v h tr nh tuyn ti u.

Thng tin di ng th h ba s phi l th h thng tin di ng cho cc dch v di

ng truyn thng c nhn a phng tin. Hp th thoi s c thay th bng buthip in t c lng ghp vi hnh nh v cc cuc thoi thng thng trc ys c b sung cc hnh nh tr thnh thoi c hnh

Di y l mt s yu cu chung i vi h thng thng tin di ng th ba ny:

- Mng phi l bng thng rng v c kh nng truyn thng a phng tin. Nghal mng phi m bo c tc bit ca ngi s dng n 2Mbit/s.

- Mng phi c kh nng cung cp rng bng tn (dung lng) theo yu cu.iu ny xut pht t vic thay i tc bit ca cc dch v khc nhau. Ngoi racn m bo ng truyn v tuyn khng i xng chng hn vi:tc bit ng xung v tc thp ng ln hoc ngc li.

- Mng phi cung cp thi gian truyn dn theo yu cu. Ngha l m bo cc ktni chuyn mch cho thoi, cc dch v video v cc kh nng s liu gi cho cc

dch v s liu.

- Cht lng dch v phi khng thua km cht lng dch v mng c nh, nht l

i vi thoi.

- Mng phi c kh nng s dng ton cu,ngha l bao gm c thng tin v tinh.

Bng tn cho IMT-2000 c quy nh ti hi ngh qun l v tuyn th gii

92 (WARC-92) vo nm 1992. Mt di ph 230MHz trong bng tn 2GHz (1885-

2025 MHz, 2110-2200 MHz) c phn chia cho IMT-2000. Tuy nhin s bng

n nhu cu i vi thng tin di ng v cc xu hng a phng tin trong thng tin

di ng khin cho ITU-R d on vo gia nm 1999 v 2000 rng bng tn IMT-

2000 s tr nn khng trong tng lai gn. c bit, ITU-R d bo s thu bao

IMT-2000 s t con s 200 triu thu bao trn ton th gii vo nm 2010, ng thi

ITU-R cng nhn thy cn phi m bo mt bng tn chung ton cu t c gi

thnh thp hn nh vic s dng chung cc thit b u cui IMT-2000 trn phm vi

ton cu v pht trin cc ch tiu k thut chung cho cc thit b u cui. ITU-R c

tnh rng vo nm 2010 s thiu bng thng khong 160MHz cho cc h thng thng

tin mt t v 2 x 67 MHz cho cc h thng thng tin v tinh trn th gii. p ng

d bo ny, hi ngh thng tin v tuyn th gii 2000 (WRC-2000) xut dnh

cc bng tn 800 MHz (806-960 MHz), 1,7GHz (1710-1885 MHz) v 2,5 GHz (2500-

2690 MHz) s dng cho IMT-2000 trn th gii trong tng lai, cn vic phn chia


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thch hp cc tn s trong cc bng tn ny bi mi quc gia s theo nhu cu trong

nc v cc ng dng thng mi khc.

DECTW-CDMA

(TDD)

W-CDMAUplink(FDD)

MSW-CDMA

(TDD)

W-CDMADownlink

(FDD)MS

1885 1900 1920 1980 2010 2025 2110 2170 2200

Tn s (MHz)

MS: ng dng di ng v tinh DECT: H thng vin thng khng dy k thut s

FDD: Song cng phn chia tn s TDD: Song cng phn chia thi gian

Cc tiu chun di ng bng rng mi c xy ng trn c s CDMA hocCDMA kt hp vi TDMA.


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Phng php s dng li tn s dn n vng dch v c chia thnh cc minnh k nhau gi l cc t bo. Mi t bo c mt ng ten trung tm vi cng sut ph

hp qun l cc di ng trong t bo m khng gy nhiu sang cc t bo khc.Vic phn chia ny phi tha mn 2 yu cu:

Din tch cc t bo phi ph kn vng dch v, vng chng ln gia hai tbo k nhau phi cc tiu

Hai t bo s dng cng di tn phi cch nhau xa.

#

!

n gin ta coi a hnh l bng phng l tng, mi t bo nh mt a gic

u. Nu a gic ny lt kn mt phng th cng thc sau y c tha mn

(n 2).1800.n

l= 3600 t y l= 2 +

2

4

n(2 1)

y n l s cnh a gic, ll s a gic c chung mt nh lp kn 3600. Dol, n u l cc s nguyn nn (n 2) phi l c ca 4 do n ch c th nhn cc gi

tr 3, 4, 6 tc a gic u phi l tam gic, t gic hoc lc gic u.

Khi s dng ng ten pht trn t ti tm cc a gic ny. Vi t bo lc gic

th cc hnh trn ngoi tip ca 2 a gic k nhau c din tch chng ln nhau

nh nht, Do vy m hnh t bo lc gic c la chn trn thc t.

%

Gi s chng ta c S knh truyn v phn cho mt nh cung cp dch v. Khi

thit k h thng khng th phn tt c S knh ny cho mt t bo v khi lp li iuny t bo bn cnh cc knh cng di tn 2 t bo cnh nhau s gy nhiu lnnhau. Do vy S knh ny phi phn cho mt nhm N t bo (N cn gi l kch thcnhm) nh vy mi nhm c S/N = k knh, ri thit k lp li c nhm t bo ny trn

a bn dch v. iu ny lm cho 2 t bo cng knh xa nhau hn, v hai t bo

cnh nhau ch s dng cc knh truyn khc nhau iu dn n kch thc nhmcng ln, 2 t bo cng knh cng xa nhau.

Nu vng dch v chia lm M t bo th dung lng h thng (s ngi ti ac th s dng cng mt lc) c tnh l T (knh):

T = M.k = M.S/N (2 2)


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Hnh 2 1: Lp li nhm t bo trong vng dich v

T cng thc ny ta thy nu N tng th T gim, nu N gim th T tng. Vy t c dung lng ln nht th N phi tin n 1 (tc l phn tt c knh vo 1 t

bo) song nh ni trn 2 t bo bn cnh nhau s gy nhiu ln nhau. Ngc li m bo chng nhiu tt, N ln s lm cho dung lng h thng gim. La chnkch thc nhm N thch hp l nhim v ca ngi thit k.

Khi la chn t bo hnh lc gic, gi khong cch tm gia hai t bo c knhtruyn ging nhau (cng knh) nm gn nhau nht l D, khong cch ny c tnh

nh sau

D2 = j2(R 3 )2 + i2(R 3 )2 + i.j(R 3 )2 (2 3)

j, i l cc bc dch sang ngang (vung gc vi cnh lc gic) v dchnghing 600 (so vi dch ngang mt gc 600) t bo ny c th trng ln t bo kia,R l bn knh t bo hnh 2 2.

Hnh 2 2: Lin h kch thc nhm v t s D/R

Mt khc so tnh lp li ca lc gic v kch thc nhm cng c tnh:

A

A

3jR

3iR

D i

j

1 2 3

0 1 4 9

1 3 7 13

2 7 12 19

A

C

B

G

F

E

D

A

C

B

G

F

E

D

A

C

B

G

F

E

D


13/164



jijiR

D

R

D

giaclucbaotetichdien

canhDdeugiactamtichdienN .

34/36

2/3

_____

_____.2 222

2

2

2

++==== (2 4)

Kch thc nhm l mt s nguyn N phi tha mn cng thc trn. Thngthng n c la chn gi tr N = 4, 7, 12,

Ph tn 33MHz c phn cho h di ng song cng phn chia theo tn

s c rng knh n l 25kHz. Tnh s knh mi t bo cc trng hp.a. N = 4, b. N = 7, c. N = 12 Nu vng dch v c 50 t bo. Tnh

dung lng h thng trong mi trng hp .

rng knh p l:

25kHz*2 = 50kHz

S knh p c php l:

33000kHz/50kHz = 660 knh p

N = 4 s knh trong mt t bo l:

660/4 = 165 knh p, dung lng knh l C = P*k = 50*165 = 8250


660/7 = 94 knh p C = 50*94 = 4700


660/12 = 55 knh p C = 50*55 = 2450

nh gi nh hng ca nhiu cng tn s do vic s dng li knh truyn

ta c cng thc suy gim sng in t l:

=

0

)()( 0

dd

n

dd pp (2 5)

p(d0) l cng sut sng in t ti khong cch d0, p(d) l cng sut sng

in t ti khong cch d so vi ngun pht, n l s m suy gim sng in t (ch

ph thuc vo mi trng truyn sng).

T s cng sut tn hiu trn cng sut nhiu cng knh gy bi 6 t bo xung

quanh thu ti my di ng c tnh l:

=

=6

1iiI

S

I

S(2 6)


14/164



Hnh 2 3: Nhiu ng knh ln b thu ca my di ng

S l cng sut tn hiu c ch thu ti my di ng cch xa tm t bokhong R (khi ra t bo). I l cng sut khng mong mun cng knh ca trm pht

cc t bo xung quanh gn nht cch my di ng xp x khong D. S dng cngthc tnh D v cng thc suy gim sng in t (2 3) n ( 2 5) ta thu c:

D2 = j2(R 3 )2 + i2(R 3 )2 +j.i(R 3 )2, 22

3R

DN =

( )1 1 1

36 6 6

nn n

n

S R RN

I D D

= = =

(2 7)

Q = D/R cn gi l t s lp li knh, t cng thc (2 2), ( 2 7) trn ta cth suy ra:

Quy tc thit k trn a hnh l tng:

T S/I, n N, D/R

Cc thng s S/I v n b quy nh trc bi mi trng v tnh nng b thu bi

nh ch to thit b, N v D/R c tnh ton thit k bi nh cung cp t bo. xc


15/164



nh tip R phi thng k mt a l ngi s dng (mt d lu lng) v s knh

tng ng trong mi th bo.

My thu di ng hot ng tt i hi S/I > 15dB. Hy tnh h s lp li

knh, kch thc nhm t bo h thng c dung lng ln nht ng vi cc trng

hp suy gim sng in t so mi trng l:

a. n = 4, b. n = 3 (coi rng ch c 6 t bo xung quanh gy nhiu vic ly v cng sut nh nhau)

Vi n = 4 ta chn N = 7, ta c t s lp li knh l

D/R = N3 = 7*3 = 4,583

S/I = 1/6(D/R)n = (1/6)(4,583)4 = 75,3 = 18,66 dB

y l m hnh chp nhn c

Vi n = 3 ta chn N = 7 ta c t s lp li knh l:

D/R = N3 = 7*3 = 4,583

S/I = (1/6)(4,583)3 = 16,4 = 12,05 dB khng tha mn yu cu my thu

Do ta cn tng N, N = 12 (j = 2; i = 2) ta c

D/R = N3 = 12*3 = 6

Khi S/I = (1/6)(6)3 = 36 = 15,56 dB m hnh chp nhn c

#

%

Mt c th ring ca mng thng tin di ng t bo l phi tnh n nhiu

knh ln cn dn n vic phi hoch nh tn s trong nhm t bo khi thit k:

Nhiu knh ln cn gy bi b lc my thu khng l tng v hiu ng xa gnbiu hin r b thu ca trm c s. V b lc khng l tng nn tn hiu khng

mong mun knh ln cn mc d b trit mnh song nu n l tn hiu rt mnh (domt my di ng gn trm c s) vn chui vo b thu gy nhiu tn hiu khng

mong mun ca mt my di ng khc khi my ny xa b thu trm c s. cminh ha trn hnh 2 4:

Gi s my di ng MS1 gn trm c s hn my MS2 20 ln (SMS2/IMS1) =

(20)-n = -52dB (vi n = 4). Nu b lc trung tm ca my thu trm c s c sn dc20dB/octabi th mun chng nhiu knh ln cn di tn pht ca 2 my di ng phicch xa nhau 6 ln rng knh truyn. Do khi lp k hoch phn chia knh

truyn cho cc t bo phi ch n iu ny.


16/164



Hnh 2 4: Nhiu knh ln cn ln b thu trm c s

H AMPS nm 1983 c 666 knh p, nm 1989 c thm 166 knh p

tng cng c 666 + 166 = 832 knh p.

Bng 2 4: Phn knh trong h AMPS cho 2 nh cung cp dch v A v B

Nh cung cp A

1A 2A 3A 4A 5A 6A 7A 1B 2B 3B 4B 5B 6B 7B 1C 2C 3C 4C 5C 6C 7C

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63

64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84

85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 111 112 113 114 115 116

.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333

990 991 .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .

1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 1021 1022 1023

334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354

355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375

376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396

.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

775 776 777 778 779 780 781 782 783 784 785 786 787 788 789 790 791 792 793 794 795

796 797 798 799

Nh cung cp B

Chia cho 2 cng ty dch v mi cng ty 416 knh, tr cc knh ng vai triu khin cn li 399 knh c chia thnh 21 tp con bng cch nh s lin tip,

mi tp con 19 knh. Cc knh trong mt tp cch nhau 21 knh. Trong m hnh lpli 7 t bo (N = 7). Mi t bo c phn chia theo cng thc iA + iB + iCm botrong mt t bo cch nhau t nht 6 knh.

#

&

Vic chia vng dch v thnh cc t bo, tt yu phi gii quyt vn chuyngiao khi ngi dng di chuyn trong khi lin lc. Cc h thng thng tin di ng hini lun phi lm tt hai vic l phn knh v chuyn giao.

Phn knh tnh kt hp phn knh ng: iu ny gn vi mt ngi sdng thay i theo thi gian hoc khi c s t hp bt thng ca nhng

ngi dng my di ng, nn bn cnh mt s knh c phn c nh cn

MS1 S2


17/164



c mt s knh d tr c phn linh hot theo tnh hung c th ca

mng.

Chuyn giao trnh hin tng ping pong v c nguyn tc u tin.

Hnh 2 5: Ngng chuyn giao

Khi my di ng chi chuyn theo tuyn ng nm chnh gia hai t bo c th

xy ra tnh hung chuyn i qua li quyn qun l ca hai trm c s. Hin tng

ny gi l hiu ng ping pong gy qu ti b phn iu khin chuyn giao. kh hiu ng ny cn t mt mc ngng chuyn giao (thng thng l 6dB vi h

thng tng t), tc l khi tn hiu thu c t bo mi phi ln hn tn hiu thuc t t bo c 6dB th mi quyt nh chuyn giao. (hnh 2 5)

iu ny tch mt bin chuyn giao thnh 2 bin khc nhau. Nu t 2 mc

ngng ny khc nhau (theo 2 chiu chuyn ng th c th thay i mm ckch thc qun l ca 2 t bo.

K thut MAHO (my di ng h tr chuyn giao): L k thut gim titnh ton qun l ln tng i hoc cc b iu khin chuyn giao, phnvic o mc v bo co mc tn hiu nhn c t cc trm c s xung

quanh cho chnh my di ng.

Chuyn giao trong h thng ngi dng c tc chuyn ng khc nhau:iu ny lin quan n cu hnh cc t bo chng ph ln nhau c kchthc khc nhau nhm qun l cc di ng c tc rt khc nhau. Ngidng di chuyn tc chm c phn sang h thng cc t bo kch thcnh (p ng ni c mt ngi s dng cao). Ngi dng di chuyn tc

cao c phn sang h t bo kch thc ln (gim tn sut chuyngiao).

A B

6dB6dB


18/164



Hnh 2 6: Phn vng t bo ln kt hp t bo nh (chng cell)

%

#

Knh v tuyn trong thng tin di ng c s dng chung ln lt cho nhiu

ngi, ging trung k tng i hu tuyn. N ch c cp pht khi c yu cu, do ch cn mt s t knh m vn phc v c kh nhiu ngi

Lu lng (A): L i lng o phn thi gian s dng c ch trong 1 gi..

n v o lu lng l Erlang - Erl

Trong 1 gi c 1 cuc gi ko di 30 pht. Ta ni lu lng cuc gi l0.5 Erlang.

Cng lu lng (Au): Din t lu lng trung bnh ca ngi s dng.

Trong 1 gi ngi gi gi cuc gi, mi cuc gi ko di trung bnh

H pht. Cng lu lng s dng l Au =H/60. Nu U ngi s dng trong h

thng ging nhau th khi lu lng tng cng ca c h thng l A = U.A u.

Vi 1 knh truyn th lu lng ca 1 knh lun


19/164



%

Cc gi thit b sung xy dng cng thc l:

Khng tnh thi gian thit lp cuc gi

Ngi c cuc gi b chn khi truy cp ln tip theo bnh ng nh nhngngi truy cp khc

S ngi truy cp tun theo phn b Poisson

Thi gian chim knh ca mt cuc gi phn b theo hm m (E.e -t)

C s hu hn knh c s dng

Ta c cng thc:

=

=C

k

k

C

k

AC

Achanbi

0 !!

)Pr( (2 8)

C l s knh trung kA l lu lng tng cng ca h

Prxc sut cuc gi b chn (GOS)

C th xy dng m hnh ngi s dng hu hn kt qu nh hn kt qu tnh

c theo cng thc ny. Ta c th tra theo kt qu theo hnh 2 7.

C bao nhiu ngi dng c th c phc v vi tng i khng nh cuc gi

b chn c cp GOS 5%0, nu s knh trong mt t bo ca h thng l:

a. C = 1, b. C = 5, c. C = 10, d. C = 20, e. C = 100Bit lu lng mi ngi dng l 0.1 Erl

:

T bng trn ta c th a ra c lu lng tng cng.

a. C = 1 Au = 0.1 GOS = 0.005

Ta c A = 0.005 suy ra tng s ngi dng l U = A/Au = 0.005/0.1 = 0.05ngi, ta chn U = 1

b. C = 5 Au = 0.1 GOS = 0.005

Ta c A = 1.13 suy ra tng s ngi dng l U = A/A u = 1.13/0.1 = 11.3 ngi,ta chn U = 11

c. C = 10 Au = 0.1 GOS = 0.005

Ta c A = 3.96 suy ra tng s ngi dng l U = A/A u = 3.96/0.1 = 39.6 ngi,ta chn U = 39


20/164


Bi ging thng tin di ng ThS. Phm Vn Ngc

Hnh 2 7: Gin Erlang B


21/164



d. C = 20 Au = 0.1 GOS = 0.005

Ta c A = 11.1 suy ra tng s ngi dng l U = A/Au = 11.1/0.1 = 111 ngi, ta

chn U = 111

e. C = 100 Au = 0.1 GOS = 0.005

Ta c A = 80.9 suy ra tng s ngi dng l U = A/Au = 80.9/0.1 = 809 ngi, tachn U = 809

Mt thnh ph rng 1300km2 c ph sng bi h thng t bo 7 lp li (N = 7).

Vi mi t bo c bn knh l R = 4km. thnh ph c cp pht bng tn vi ng rngbng tng cng l 40MHz ph v rng bng ca mt knh p l 60KHz. Gi s cp

dch v l GOS = 2% theo cng thc Erlang B vi tng i khng nh cuc gi bchn v lu lng ca mi khch hng trung bnh l 0.03 Erl. Tnh

a. S t bo trong thnh ph

b. S knh trn mt t bo

c. Lu lng cuc gi trn mi t bo

d. Lu lng ca ton thnh ph

e. Tng s ngi c th dng theo GOS = 2%

f. S ngi dng trn mt knh ca ton thnh ph

g. S ngi dng cng mt lc nhiu nht.

a. Din tch mt t bo l: (t bo lc gic) = ( 24

36 R )

2.5981*R2 = 2.5981*(4)2 = 41.57 (km2)

S t bo trong thnh ph l 1300/41.57 = 31 t bo

b. S knh trn mt t bo l:

40000000/(60000*7) = 95 knh (C)

c. C = 95, GOS = 0.02 ta c A = 84 Erlan/ t bo

d. Lu lng ca ton thnh ph

31*84 = 2604 Erlan

e. Mi ngi dng 0.03Erlnn tng s ngi c th c phc v l:

U = A/Au = 2604/0.03 = 86600 ngi


22/164



f. S ngi dng trn mt knh l

86600/(7*95) = 130 ngi/ knh

g. S ngi dng nhiu nht cng mt lc l:

95*31 = 2945 ngi

h thng ny mt cuc gi b chn s c xp hng trong dy nhng cuc gi bchn ch c truy cp.

GOS lc ny s l ch s cho bit kh nng mt cuc gi b chn v phi chi trong

mt thi gian xc nh.

tnh GOS trc ht ta phi tnh xc sut cuc gi b chn theo cng thc Erlang C

=

+=>

1

0 !1!)0Pr(

C

k

kC

C

kA

CACA

A

doicho (2 9)

C th tnh hoc l tra kt qu theo hnh 2 8.

Sau nhn vi xc sut ch t giy trong hng i. Cng thc cui cng l:

)0|Pr(*)0Pr()Pr( >>>=> doichotdoichodoichotdoicho

>=>H

tACdoichotdoicho

)(exp*)0Pr()Pr( (2 10)

Thi gian ch i trung bnh ca h thng s l:

AC

HdoichoD

>= )0Pr( (s)

H t bo lc gic trong c: N = 4 v bn knh t bo R = 1.387 km.

H c phn di tn ng vi 60 knh p. Nu lu lng ngi dng l 0.029Erl,v c 1 cuc gi/gi. Vi GOS l 5% hy xc nh.

a. Bao nhiu ngi dng/km

2

c cung cp bi h ny.b. Xc sut cuc gi (sau khi b chn) phi ch thm t = 10 giy.

c. Xc sut tng cng cuc gi phi ch t = 10 giy.

: Vi bn knh t bo l R = 1.387 km din tch t bo l 5km2

N = 4 v tng s knh p l 60 knh nn s knh / t bo = 60/4 = 15

a. Tra bng Erlan C vi GOS = 5% v C =15 ta c A = 8.8 Erl


23/164


Bi ging thng tin di ng ThS. Phm Vn Ngc

Hnh 2 8: Bng kt qu cng thc Erlan C


24/164



S ngi dng trong mt t bo l 8.8/0.029 = 303 ngi

S ngi dng /km2 l 303/5 = 60 ngi/km2

b. 1= cuc gi/gi H = Au/=0.029 gi = 104.4 giy

Xc sut cuc gi (sau khi b chn) phi ch thm t = 10 giy l

Pr(ch i > t / ch i > 0) = exp( -(C A).t/H)

= exp(-(15 8.8)*10/104.4 = 52.22 %

c. Vi GOS = 5% xc sut tng cng mt cuc gi phi ch 10 giy l:

Pr(ch i > 10s) = Pr(ch i > 0).Pr(ch i > 10s | ch i > 0)

= 0.05*0.5522 = 2.78 %

L lu lng s dng tnh trung bnh trn mt knh trung k. Gi tr ny ph thucvo cch t chc thit k v cp dch v xc nh.

10 knh trung k/t bo vi GOS = 1% (trng hp tng i khng nh cuc gi bchn) khi tra bng Erlan B ta thy chng p ng lu lng cuc gi 4.46 Erlan (tng

ng lu lng mt knh l 0.446 Erl/knh)

Nu t chc knh ny trn 2 t bo th khi tra bng ta thy chng ch p ng c2*1.36 Erl = 2.72 Erln (tng ng 0.272 Erl/knh). Cch t chc u c hiu sut trung

k cao hn cch t chc th 2 vi cng mt GOS song kh nng chng nhiu km hn.

S lng ngi s dng dch v thng tin di ng khng ngng tng, t ra mt

yu cu pht trin nng cao dung lng h thng mt cch c k hot v tnh k tha. Cba k thut chnh nng cao dung lng h thng l: Chia nh t bo, s dng ng ten

nh hng v phn vng trong t bo.

L thay 1 t bo ln b qu ti bng cc t bo nh hn vi cc trm c s t thphn v c cng sut pht nh hn.

Dung lng tng thm l do s dng li knh truyn. (kch thc nhm N khng

i nhng s lng t bo trong vng phc v c tng ln)


25/164



Hnh 2 9: Phn chia t bo.

V d:

Nu dng t bo c Rmi = .Rc khi s c 4 t bo mi thay th t bo c. iu

ny lm s nhm lp li tng ln 4 ln v dung lng tng cng xp x 4 ln m vn ginguyn t s lp li knh (Q) khng lm thay i s phn b knh. Cng sut pht ca

cc t bo mi trong h thng cng phi iu chnh cho ph hp.

Pr (ti bin t bo c) =Pt1R-n = Pr (ti bin t bo mi) = Pt2(R/2)

-n

Vi n = 4 ta c Pt2 = Pt1/16 hay trm pht ca t bo mi phi gim 12dB

Trn thc t, ta duy tr c hai m hnh t bo ln v nh phc v cc i tngchuyn ng vi tc khc nhau v cc knh truyn cng c phn thnh 2 nhm kchc t bo ny trnh giao thoa cng knh, kt hp vi k thut h thp ng ten iu

khin vng ph sng

iu ny s dn n gim giao thoa cng knh cho php cc t bo cng knh gnnhau hn (gim D/R hay gim N, tng s lp li tn s) dn n tng dung lng hthng trong khi kch thc t bo khng thay i

S dng ng ten nh hng c gc 1200 s trm gy nhiu cng knh xung quanhkhng phi l 6 nh s dng ng ten trn m ch l 2. S/I t 17 dB s tng ln 24.2 dB

nn c th gim N t 12 n 7. phng php ny khng thay i s trm c s m chtng thm s ng ten trn mt trm c s iu ny cng s lm gim phn no hiu suttrung k v tng thm s ln chuyn giao.


26/164



Hnh 2 10: S dng ng ten nh hng.

Xt h thng c thi gian trung bnh cuc gi l 2 pht. Xc sut cuc gi b chn l

1%, theo cng thc Erlan B mi t bo c mt cuc gi trong 1 gi v c 399 knh/7 tbo. Khi dng ng ten trn kh nng phc v l 44.2 Erl hay 1326 cuc gi/gi. Nu

dng ng ten nh hng 1200 s ch c 19 knh/1 ng ten vi cng xc sut b chn vtrung bnh cuc gi, mi ng ten phc v 11.2 Erl tc l 336 cuc gi/gi hay 1008 cucgi/gi trong 1 t bo (gim 24% so vi ng ten pht sng trn) trong khi li nng caot s S/I.

V d vi ng ten 600 vi N = 7 s cho S/I = 29 dB cho php dng N = 4 lm tng

dung lng ln 7/4 ln (hiu sut trung k gim 44%)

Phng php s dng ng ten nh hng lm tng s chuyn giao v qu ti ccphn t chuyn mch. Lee a ra gii php l thay trm trung tm ln bng mt s

trm pht cng sut thp hn ph cc vng nh hn trong t bo v cc trm ny ckt ni v iu khin chung mt trm c s ca t bo.


27/164



Tx/Rx

Tx/Rx

Tx/Rx

Hnh 2 -11: Chn vng trong t bo.

Cch b tr ny tt hn nh trn, chuyn giao khng xy ra gia cc ng ten trongcng mt t bo m ch thay i ng ten qun l khi MS di chuyn t vng ny sang

vng khc trong mt t bo. Cc knh c phn chia ng trong khng gian v thi giancho cc vng trong t bo, cn gia cc t bo th lp li nhm knh nh c.

K thut ny thng c dng dc theo cc ng cao tc hay cc hnh lang c

lu lng ln. (vi S/I = 18 dB l i hi thng thng vi h iu ch bng hp FM, viN = 7 th D/R = 4.6, vn dng iu ny cho Dz/Rz = 4.6 chng nhiu cng knh giacc vng th i vi t bo t s ny cho D/R = 3, N = 3. Do vy khi gim N = 7 n N =3 s tng dung lng h thng ln 7/3 = 2.33 ln).


28/164



Trong thng tin di ng rng di tn c gi tr nh hng ha, do b m hating ni ng vai tr quan trng khng ch vi cht lng ting ni m cn phi m

bo chim dng t bng tn vi tc bit thp. B m ha tc bit thp lm tng dunglng h thng l yu cu quan trng ca cc nh cung cp dch v. Cc b m ha ting

ni c bn c phn thnh 2 loi chnh l: M ha theo dng sng v m ha ngun(vocoder)

Hnh 3 1: Phn loi m ha ting ni

M ha dng sng: L k thut m ha v gii m lin tc bn theo dng sng ca

tn hiu m thanh, v nguyn tc chng c thit k c lp vi ngun m nn cht

lng khng ph thuc vo ngun tn hiu. Kiu m ny chng nhiu tt, khng phctp, gi r song h s nn tn hiu cha cao nh PCM, ADPCM

M ha ngun: Da trn s m phng ngun m c th (ting ni c nghin cutrc cc c tnh khai thc kh nng nn m), k thut ny kh phc tp, c trcao, gi thnh t b li c h s nn ln v cho tc bit m ha thp.

M ho ting ni

M ho dng sng M ho ngun

Min tn sMin thi gian LPC Vocoders

Khng vi phn Vi phn ATCSPC

PCM DELTA ADPCM

APCCVSDM


29/164



Ting ni c mt s tnh cht c th c th a vo thit k b m ha l:

Di tn gii hn (


30/164



NN

kk

N

kk

S

SN

SFM1

1

2

1

21

=

=

= ( 3 3)

Sk l mu tn s th k ca psd. Thng thng tn hiu ting ni trong thi giandi c SFM = 8 v dao ng t 2 n 500 trong khong thi gian ngn.

Lng t l php din t mt di lin tc cc gi tr ca bin tn hiu bng mt

tp hu hn cc gi tr bin ri rc. Lng t tn hiu s lm mo tn hiu trong phmvi m ngi thit k c th nh trc. B lng t n bit c th c L = 2n mc bin rirc.

Trong lng t tuyn tnh cc gi tr lng t phn b u trong ton b tn hiu,

khong cch gia hai gi tr gn nhau nht c gi l bc lng t. mo tn hiulng t t l vi bnh phng bc lng t, bc lng t t l nghch vi s gi trri rc. Mo lng t c tnh l li trung bnh bnh phng nh sau:

[ ] ==T

QQ dttxxfTxfxEMSE

0

22 )]()([1

))(( (3 4)

Trong x(t) l bin tn hiu c ly mu ti thi im t, fQ(x) biu din gi tr

lng t gn x(t) nht. Mo lng t (n lng t) v cht lng ca b lng t cnh gi qua t s cng sut tn hiu / n lng t (SQNR). B iu ch xung m (PCM)dng php lng t 8 bit / mu trong in thoi thng mi. Nu php lng t l tuyntnh th ta c cng thc lin h nh sau:

(SQNR)dB = 6.02n + (3 5)

Trong = 4.77 i vi gi tr nh ca SQNR v = 0 i vi SQNR trungbnh. T phng trnh trn ta thy l c thm mt bit lng t th SQNR c tng thm

6 dB.

Mo lng t c th c gim thiu nu phn b cc gi tr lng t mt cchthch hp (khng phi cch u nh lng t tuyn tnh). Phn b ny da trn hm pdf

nh sau:


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= dxxpxfxxfxE QQ )()]([]))([( 22 (3 6)

T phng trnh trn ta thy mo trung bnh c th c gim thiu bng cch gim

gi tr (x fQ(x))2 ti nhng ni c gi tr p(x) ln. Tc cc ga tr lng t phi nhiu

ti nhng vng bin c xc sut cao (ting ni nh), cn t ti nhng vng bin cxc sut thp (ting ni ln).

Mt phng php thng dng trong in thoi thng mi l b lng t Loga vi

hai k thut l lut dng M v lut A dng chu u. Tn hiu trc ht c i

qua b khuch i nn (dng hm Loga) v sau i vo b lng t tuyn tnh.

Theo lut cng thc ca b lng t loga c dng l:

)1ln(

)(1ln(|)(| 0

+

+=

twtv (3 7)

Trong l hng s dng c gi tr trong di 50 300, gi tr nh ca w(t)

c chun ha bng 1.

Theo lut A cng thc ca b lng t loga c dng l:

+

+

+

=

1)(1

ln1

))(ln(1

1)(0

ln1

)(

)(0

tw

AA

twA

Atw

A

twA

tv (3 8)

Trong A thng c gi tr l 87

fQ(x) f

Q(x)

a. b.

Hnh 3 2: Lng t thch nghi

Nh ni trn l c s khc nhau ca pdf trong thi gian di v thi gian ngn

i vi ting ni. S thay i theo thi gian ca ting ni to nn mt di ng ln (c40 dB). Do vy b lng t cn c iu chnh thch hp bng cch tng bc lng t


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khi cng sut tn hiu li vo tng, v gim khi cng sut tn hiu li vo gim (khi smc lng t khng thay i) ta gi l lng t thch nghi.

Trong php lng t thng thng mi mu bin tng ng vi mt gi tr

lng t hay mt t hp bit biu din. php lng t vc t th mt nhm mu bin ng vi mt nhm mu lng t mi c tng ng vi mt t hp bit c gi lmt vc t lng t trong khng gian vc t lng t (vi lng t thng s mu = 1).

S mu trong nhm lng t c gi l mt vc t mu c lng t. Do tnh chttng quan gia cc mu trong ting ni m php lng t vc t to nn m c nncao vi:

L

n

R2log

= bit / mu (3 9)

Trong n l khng gian lng t.

L l di vc t lng t

R l s bit m ha trn mt mu

Khi n = 256, L = 1, R = 8 bit / mu (ng vi PCM). Khi L = 10 12 c th cho R =

0.5 bit / mu. li lng t vc t c coi l khong cch clit gia vc t lng t vvc t li vo. c s dng cho b m ha ngun m c tc bit thp.

B iu ch PCM khng loi b s d tha thng tin trong tn hiu ting ni (do

tnh tng quan khc 0). B ADPCM khai thc iu ny gim tc bit. Do c tnh ttng quan cao m cc mu bin cnh nhau t khi thay i ln nn di ng trong tnhiu vi phn nh hn nhiu di ng ca bn thn tn hiu. M ha tn hiu vi phn scho tc bit ch l 32Kbps (=1/2 chun PCM) vi cng cht lng tn hiu (khi gii m

dng b tch phn).

Trn thc t ADPCM c kt hp vi thut ton tin on tn hiu, thay cho vicm ha s sai khc gia hai mu cnh nhau l vic m ha s sai khc gia mu tht v

mu d on n (da trn mt nhm cc mu tht trc , sai khc ny cn gi l li don).


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Hnh 3 3: B m ha ADPCM

Mt b m ha PCM thch nghi c tc ly mu l 8kHz, mi mu c biu dinbng nhm 8 bit. Bc lng t c iu chnh li c sau 10ms v ln ca bc

lng t c m ha bng 5 bit. Tnh tc bit truyn, gi tr SQNR nh v trung bnhca b m ha

fz = 8kHz = 8000 Hz

n = 8, s bit thng tin trn mt giy l 8*8000 = 64000 bit/s

Do hiu chnh bc lng t c sau 10ms nn tc bit hiu chnh l:

5*1/(10*10-3) = 500 bit/s

Tc bit truyn ca b m ha l 64000 + 500 = 64500 bit/s = 64.5 kbps

T s SQRN ch ph thuc vo s bit dng m ho mt mu lng t:

SQNRnh = 6.02n + 4.77 = 6.02*8 + 4.77 = 52.93 dB

SQNRTB = 6.02n = 6.02*8 = 48.16 dB

M ha c tin hnh theo cc vng tn s. Ph ca tn hiu ting ni c chiathnh cc di hp hoc bin i thnh cc tn s ri rc v c m ha c lp vi nhau.Ty theo mc quan trng ca cc di tn con hay tn s ri rc m s bit s dng m

ha nhiu hay t

Lng t 4

bit thch nghi

To bcthch nghi

Lng t thchnghi ngcD on

thch n hi

Sr(k)

S(k)

y(k)Se(k)

I(k)

Dq(k)


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3.3.2.1 M ha theo bng con (SBC)

Ph ca tn hiu c chia thnh 4 n 8 di con bng cc b lc. Cc bng con

c di chuyn bin tri v 0 c th dng tc ly mu Nyquist trong min thi

gian v c ly mu vi s bt khc nhau ph thuc bin ph v tiu chun thnh gicca con ngi.

Bng con Tn s (Hz) S bit m ha

1 225 450 4

2 450 900 3

3 1000 1500 2

4 1800 2700 1Tc ly mu c tnh l mu / giy:

Bng 1 2 x (450 225) = 450

Bng 2 2 x (900 450) = 900

Bng 3 2 x (1500 1000) = 1000

Bng 4 2 x ( 2700 1800) = 1800

T dn n tc m ha cha k cc thng tin tiu l:

450x4 + 900x3 + 1000x2 + 1800x1 = 8300 bit/s = 8.3 kbps3.3.2.2 M ha bin i thch nghi

M ha bin i thch nghi l k thut m ha trong min tn s dng phng php

bin i. C tc m ha trong khong 9.6 20 kbps. Cc mu tn hiu trong mt cas c bin i thnh mt tp cc tn s ri rc, bin ca cc tn s ny c lng

t v m ha ring bit truyn i. Ti b thu cc h s lng t c bin i ngc to li tn hiu ban u. Phng php ny dng bin i cosin ri rc (DCT) theo cngthc:

=

+=

1

0 2)12(cos)()()(

N

nx N

knkgnxkX k = 0, 1, 2, , N 1 (3 10)

Trong g(0) = 1 v g(k) = 2 , k = 1, 2, IDCT c thc hin theo cng thc:

=

=1

0 2

)1_2(cos)()(

1)(

N

kc N

knkgkX

Nnx

n = 0, 1, 2, , N 1 (3 11)


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Trn thc t DCT hay IDCT c tnh theo thut ton nhanh p ng thi gianthc t v vic thit k bc lng t khc nhau cc tn s khc nhau cng c thchin theo thi gian c truyn nh thng tin tiu (c 2 kbps) s bit lng t tn s

ri rc t l vi nng lng ca n.

K thut ny dng tch cc thng s ca ngun m v m ha cc thng s nytruyn n ni thu. Ti ni thu cc thng s ny c gii m iu khin mt ngun

m tng t nh ni pht ti to li tn hiu. Cc vocoder c tc bit rt thp nhngphc tp v ph thuc vo ngun m.

C ch chung ca mt ngun m bao gm c ngun kch thch v b lc thngminh v loa.

Ting ni ca chng ta c phn thnh 2 loi:

Voiced l kt qu dao ng gn tun hon ca c quan pht m.

Unvoice c to nn bi s thi kh qua khe hp ca mi v li

Cc thng s ca m hnh ny l cc tn s chnh (tn s pitch


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Voceder knh l h phn tch v tng hp m u tin trong vng tn s. N xcnh cc thng s ca tn hiu trong cc di tn con, sau m ha v hp knh cc di

li. vic ly mu c ng b sau 10 30ms. Cng vi thng tin v nng lng, voice

hay unvoice, tn s pitch c truyn i.

Tng t vi Voceder knh song c th lm tc bit thp hn v dng t tn hiuiu khin hn. Thay v vic gi cc mu ng bao ph cng sut th Vocoder hi ch

gi cc cc i (peak, t nht l 3 nh) ca ng bao ny c th c tc < 1200 bpssong c kh khn trong vic xc nh chnh xc cc cc i nn khng tht s hiu qu.

B Vocoder ny tch ph kch thch v ph c quan pht m bng Fourier ngcca loga ph bin to nn Cepstrum ca tn hiu. Cc h s tn thp th hin trong

Cepstrum l ng bao ca ph c quan pht m. Cc h s tn s cao to nn on xungtun hon l bi s ca chu k ly mu. B lc tuyn tnh dng tch cc h s

Cepstrum ca c quan pht m khi h s kch thch. Trong b thu cc h s Cepstrumca c quan pht m c bin i fourier to nn p ng tn s ca c quan pht mhoc gi nguyn c coi l p ng xung ca c quan pht m. Bng cch nhn chpp ng xung ny vi mt tn hiu kch thch tng hp, ting ni ban u c ti to li.

Vocoder kch thch bng Voice khng cn tch pitch v cc thao tc tch voice khi

ting ni. N thc hin vic t hp v truyn dn PCM bng tn thp vi Vocoder knh bng tn cao hn. Tn hiu pitch c to nn ti b tng hp bng cch tch sng, lc

bng v ct tn hiu bng c s to nn tn hiu ph bng phng vi nng lng ti tica pitch. Hot ng ti tc 7200 9600 bps. Cht lng ca chng tt hn vocoderkch ng bng pitch.

3.4.5.1. Vocoder LPCy l lp cc vocoder trong min thi gian, chng tch cc c im quan trng t

dng sng thi gian. Mc d kh phc tp song chng rt ph bin trong cc vocoder tc thp v c th cho cht lng ting ni tt tc 4.8kbps

H LPC m hnh c quan pht m nh b lc ton im cc vi hm truyn cm t:


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=

+=

M

k

kkzb

GzH

1

1)(

(3 12)

y G l h s khuch i, z-1 biu din ton t tr n v. Kch thch b lc hoc

l xung ti tn s pitch l n trng ngu nhin ty thuc on ting ni l voice hayunvoice. Cc h s ca b lc c xc nh dng k thut d on tuyn tnh ging nh

trong b ADPCM. Song thay cho vic truyn cc gi tr lng t ca tn hiu li giamu tin on v my thc (ADPCM), h thng LPC truyn ch nhng c tnh chn lcca tn hiu li l cc thng s nh: h s khuch i, tn s pitch, quyt nh voicehay unvoice, cho php xp x tn hiu li chnh xc. Ti b thu cc thng tin trn c

dng t i to li tn hiu li kch thch b lc tng hp. Cn b lc tng hp ti bthu c iu khin bi cc h s on nhn c. Trn thc t nhiu b LPC phtcc h s b lc trong biu din tn hiu li v c th tng hp trc tip ti b thu.

Xc nh h s d on: tn hiu d on l mt tng c trng s ca p mu trc (p di t 10 15).

Hnh 3 5: S chung ca b m ha v gii m

n

p

k

knkn esas += =

1 (3 13) en l li d on. Cc h s d on c xc nh qua vic lm ti thiu nng

lng trung bnh E ca tn hiu li:

= =

=

==

N

n

p

kknk

N

nn saeE

1

2

01

2

(3 14)

B m

Lc LPC

Phn tchPitch

Quyt nhVoice/unvoice

M ha Gii m

M ha

M ha

Gii m

Gii m

K

n

h

Tng

H

p

Kch thch

Livo


38/164



Vi a0 = -1. Thng li c tnh trong ca s 10ms tng ng vi N = 80. Cho viphn ca E theo am bng 0 ta c:

=

=

== p

kknk

N

nmn

m

sasa

E

01

02

= =

=p

k

N

nkknmn ass

0 1 0

0

== =

p

kkmkaC

(3 15)

Vi Cm l h s tng quan gia cc mu r, m. Sau khi xc nh C m gii phng

trnh trn c th xc nh c cc h s d on. Cc h s d on thng khng c

m ha truyn i trc tip (v cn n 8 -10 bit/h s) m truyn cc h s phn nh(chng c di ng nh hn ch cn 6 bit / h s). Nh vy mt b d on bc 10 tng

s bit dng cho cc thng s m hnh trn 1 frame l 72 (bao gm 5 cho khuch i v 6cho chu k pitch). Nu cc thng s c c lng li c sau 15 30ms th tc bit sl 2400 4800 bps. Vic m ha cc h s phn nh c th c ci tin khi dng thmcc b bin i phi tuyn (t trc b m ha) s lm gim nhy ca cc h s phn

nh i vi li lng t. Chng thng dng l bin i t s loga (LAR).

+=

)(1

)(1log))((tanh)( 10

1

kR

kRkRkLAR

n

nnn

(3 16)

Cc LPC khc nhau trong cch to ra tn hiu li ti b thu. C 3 cch c bn: cch

th nht rt ph bin dng 2 ngun kch thch ti b thu, mt ngun l n trng v ngunkia l on xung c tc pitch, vic chn ngun ny hay ngun kia kch thch phthuc vo quyt nh voice/unvoice ca b pht. K thut ny gp kh khn trong victch thng tin v pitch b pht (ngoi ra s ng pha gia cc thnh phn hi ca xungkch thch to nn ting trong ting ni tng hp). Vn ny c loi b trong 2 cch

sau l kch thch bng a xungv kch thch bng sch m

3.4.5.2. LPC kch thch bng a xungVic kch thch bng xung n/chu k pitch lun lm mo tn hiu. Atal ngh

dng nhiu xung (8 xung/chu k) v iu chnh v tr v bin cc xung ny ti thiuli trung bnh bnh phng c trng s. K thut ny c gi l MPE LPC lm cht

lng ting ni tt hn khng ch do li d on c xp x tt hn m cn do nkhng i hi tch pitch. S xung c th gim vi voive c pitch cao bng cch kt hp

mt b lc tuyn tnh vi vng pitch trong b tng hp.

3.4.5.3. LPC kch thch bng m (CELP)


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phng php ny b m ha v gii m c chung mt sch m nhng tn hiukch thch ngu nhin (n trng Gauss). Vi mi tn hiu ting ni b pht s tm trongsch m tn hiu kch thch ln b lc LPC cho tn hiu gn ging nht, sau ch vic

truyn ch s ca tn hiu kch thch ny n b thu. Da vo ch s ny b thu s tm ra

tn hiu kch thch thch hp. B m ha theo k thut ny i hi hn 500 triu php tnhnhn v cng / giy. Chng c th cho cht lng ting ni cao vi vic kch thch cm ha 0.25 bit / mu v c tc 4.8 kbps

Hnh 3 6: B m ha CELP

Xt vic m ha cm 5 ms ca ting ni. Ti tn s ly mu 8 KHz mi cm c 40mu, vi vic m h bit / mu s cho chng ta 10 bit / cm. V vy s c 2 10 = 1024cc dy di 40 mu c th. Mi dy 40 mu ca tn hiu kch thch c ly chun li v

cho i qua b lc quy to ra chu k voice v hiu chnh ng bao ph.Nhng mu ting ni c pht li ti li ra b lc th 2 c so snh vi tn hiu

ting ni nguyn bn to nn tn hiu vi phn (li). Li ny c x l qua b lc

tuyn tnh vi h s khuch i cao tn s th cm mnh, v khuch i nh tn sth cm thp. Mc d i hi tnh ton nhanh song tin b ca DSP v cng ngh VLSI to ra ng dng thi gian thc ca CELP. (h CDMA theo chun IS 95 s dngCELP ti tc 1.2 14.4 kbps. Nm 1995 Qual.com ra QCELP13 l b m ha

CELP 13.4 kbps hot ng trn knh 14.4 kbps.)

3.4.5.4.

LPC kch thch bng li (RELP)C s ca RELP lin quan n k thut DPCM trong m ha dng sng. Trong loi

PLC ny, sau khi c lng cc thng s m hnh (h s LP hay cc thng s lin h),cc thng s kch thch (quyt nh voice/unvoice, pitch, h s khuch i) t mt khungting ni. Ting ni c tng hp li ti b pht v tr i tn hiu gc c tn hiu li.Tn hiu li c lng t, m ha v pht n b thu cng vi cc thng s LPC. Ti b

B d on B d on

Lc thcm

Mchbnh phng

Mchtrung bng

i thcm

Ngunkch thch

Li

Ting ninguyn bn


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thu tn hiu li c cng vi tn hiu tng hp. Cht lng ting ni c nng cao nhvic cng thm tn hiu li.

Hnh 3 7: B m ha RELP

Cc thng s cn c xem xt l:

- Cht lng cm th ting ni khi dng m nn

- Gi thnh v dung lng h thng

- Ngoi ra cn phi tnh n tr ca b m ho, phc tp ca thut ton

- Yu cu cng sut ngun, s cnh tranh ca cc tiu chun khc.

- Kh nng chng ni trn ng truyn.

Ni chung khi ting ni c m ho thnh t bt, lng thng tin trn mi bit stng ln dn n cp bo v n trn ng truyn li phi tng ln. Vic bo v cng

cn c phn bit tu theo vai tr quan trng ca tng bit trong khung ting ni.

La chn b m ho cng ph thuc vo kch c t bo. Khi kch thc t bo nhc th s dng cc b m ho tc cao (v hiu sut s dng ph cao v khng cn

nhiu bit bo v trn ng truyn). CT2 v DECT dng cho t bo rt nh (Microcell)c th s dng b m ho ADPCM 32kbps m khng cn m knh v b cn bng. Cct bo ln th iu kin knh truyn km nn cn dng thm cc m sa li khi cn

dng cc b m ho c tc thp.

K thut a truy cp trong h thng cng l nhn t nh hng n hiu sut s

dng ph v do vy cng c tnh n khi la chn b m ho ting ni. H t bo sUS TDMA lm tng dung lng ca h tng t cng tn ti (AMPS) ln 3 ln dng

b m ho VSELP 8kbps. H CDMA do kh nng giao thoa tt v bng truyn rng cho

B m/ca s +

Mha

Phn tchST - LP

Tng hp LPC

Tham s LP

Li

Li ramha

Quyt nhvoice/unvoice,

khuch ipitch

-


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php dng b m ho tc thp m vn khng b nh hng li ng truyn khi dngthm m hiu chnh li.

K thut iu ch cng c xem xt khi la chn b m ho ting. Khi dng iu

ch c hiu sut bng cao c th gim yu cu tc thp i vi b m ho ting.

H di ng c di knh pht thun t 810 - 826 MHz v di knh pht ngc l 940 956 MHz, gi s rng c 90% rng bng dnh cho knh lu lng, h yu cu h tr1150 cuc gi cng mt lc dng k thut FDMA, s iu ch c hiu sut ph l

1.68bps/Hz. Knh truyn i hi dng m sa li FEC . Tm gii hn trn v tc cab m ting.

rng bng thng dnh cho lu lng l: 0.9*(826 810) = 14.4 MHz

S ngi dng ng thi cc i l 1150 ngi nn rng cc i l 14.4/1150 =12.5 kHz.

Hiu sut ph l 1.68b/Hz nn tc knh cc i l:

16.8*12500 = 21kbps

FEC nn tc d liu thc s l 21 * 1/2 = 10.5 kbps

Vy tc m ha ting phi chn thp hn tc ny

B m ha c s lng bit phn b theo vai tr quan trng ng gp vo cht lngtn hiu. M ha tin hnh trn on 20 ms (260 bit li ra b m ha). 50 bit u (loi 1)l quan trng nht c bo v CRC 10 bit v FEC 1/2. 132 bit tip theo (loi 2) bo v

CRC 5 bit v 78 bit cui khng c bo v li. Tnh tc d liu knh tng cng tc.

Gii:

S bit knh cho loi 1 l (50 + 10)*2 = 120 bit

S bit knh cho loi 2 l: 132 + 5 = 137 bit

S bit knh cho loi 3 l: 78 bit

Tng s bit c truyn trong 20 ms l: 120 + 137 + 78 = 335 bit

C 2 cch nh gi mt b m ho ting ni trong vic cung cp cht lng ting

ni.


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- Cc php o khch quan cung cp mt gi tr nh lng v mc ti to ting

ni v xp x tn hiu gc. MSE theo bin cho mo bin , MSE theo tn s

c trng s. SNR tng on cc ch s r rng.

- Cm nhn ch quan ca ngi nghe. Trong khi cc ch s khch quan rt c ch

cho thit k ban u v m phng th s nh gi tng hp cui cng li cn n

s cm nhn ch quan ca ngi nghe.

B m ho ph thuc nhiu vo ngi ni, cc php th ch quan c tin hnhtrong cc mi trng khc nhau m phng ging cuc sng thc nh ting n, nhiungi ni Php th DRT l php o d hiu ca nhiu ngi dng c s rng rng

ri. Trong php th nhng t cng vn c s dng nh l those dose c s dng th xem kh nng phn bit ca ngi nghe. chnh xc trong php th ny l 75

90%.

Mt php o khc cng c chp nhn l DAM. Cc php th ni chung l kh

xp hng v i hi phi c h tham chiu. H thng xp hng ph bin hin nay l MOS.

Mt vn kh khn nht cho b m ho khi truyn t MS n trm c s BTS sau

gii iu ch thnh tn hiu tng t. Li ra c m ho truyn i nh tn hiu s qua

ng dy dn hay v tuyn. Tnh hung ny gi l tn hiu Tandem, c c xu hngthi phng li bit nhn c ti trm c s. Tn hiu Tandem gy kh khn cho bo v

song l mt tiu chun nh gi quan trng cht lng m h. Khi mng di ng tngtrng, s c nhiu yu cu gi t di ng ti di ng, kiu lin lc ny s c 2 ngun

nh hng c lp n cht lng ting ni, n v Tandem.


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Lin lc bng sng in t ngoi u im thch hp cho ngi dng di ng, khngphi chi ph nhiu cho vic xy dng c s h tng th li c kh khn l tn hiu c tc li bit cao i vi bt k k thut iu ch no. Bn cnh tnh suy gim sng theo

ng truyn, tp nhiu AWGN, cn c mo tn hiu, cn cha k n ngun gynhiu bt thng khc. Do xut hin nhiu k thut s l tn hiu m bo victruyn dn thng tin c chnh xc, c cc k thut di y.

Cn bng, phn tp, m knh l 3 k thut c th dng c lp hay phi hp cithin cht lng tn hiu nhn c khi truyn tn hiu qua knh v tuyn.

Cn bngs kh giao thoa gia cc tn hiu (ISI) gy ra do hiu ng a ng v rng ph ca knh truyn hp hn rng bng cn thit do tn hiu b tri di rav chng so le ln nhau, l tr ngi chnh hn ch tc knh truyn. B cn bng

trong my thu c nhim v sa li dng tn hiu thu c cho ging tn hiu trc khitruyn ci thin tc ng truyn, l yu cu rt cao trong cc b thng tin hini. Do knh truyn thay i theo thi gian mt cch ngu nhin nn tnh cht quan trngca b cn bng l phi thch nghi c theo knh truyn.

Phn tp l k thut nhm khai thc tnh a ng c lp ca knh truyn nngcao t s tn / tp m khng i hi tng cng sut ca my pht. Nu k thut cn bngcoi hiu ng a ng l i th th phn tp li coi n l ng minh. Phn tp c ththeo khng gian, tn s v cc tnh ca tn hiu. Khi phn tp theo khng gian k thut

ny thng i hi dng nhiu anten thch hp cho b thu trm c s, khi anten ny thutn hiu yu th anten kia c th thu c tn hiu mnh ci thin c tn hiu thu tng

hp.

M knh l k thut chuyn i trc dng d liu ti b pht (khc vi cc k

thut khc l tin hnh b thu) nhm to nn mt quy lut no . Ti b thu quy lutny c kim tra pht hin li gy nn bi ng truyn v ty theo cp m n cth hiu chnh c li ny. Gi phi tr cho k thut ny l gim tc thng tin v phi

tng thm nhiu bit truyn, m knh c bc cng cao th tc thng tin cng gim. Chai k thut m knh chnh l M khi v M xon:

M khi tin hnh trn tng khi xc nh ca d liu


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M xon tin hnh lin tc theo dng d liu

Ty theo yu cu ca h thng tin c th m 3 k thut trn c s dng vi mc khc nhau.

B cn bng thch nghi s dng mt dy th nghim chiu di c nh c bit

trc (thng thng l mt dy gi ngu nhin) ti b thu da vo hiu chnh li tnhiu sau ng truyn. Dy th nghim c pht xen k vi d liu ni pht.

Ti ni thu dy th nghim c bit trc ban u. Mt thut ton c s dng

hiu chnh li dy thu c nhm lm cho s khc bit vi dy bit trc l ti thiu.Khi hiu chnh xong mo knh coi nh c b tr v ngay sau c th truyn mton d liu qua vi s gi nguyn hiu chnh ny. Do knh lin tc bin i nn ngaysau on d liu li n dy th nghim lin tc hiu chnh mo knh.

Hnh 4 1:M hnh cn bng knh.

Thng thng b cn bng l mt b lc ngang ri rc hu hn, tn hiu li vo ti

thi im k k hiu l yk, cc tn hiu trc lu gi trong cc thanh ghi l yk-1, yk-2, yk-N. Tn hiu li ra ph thuc trng thi tn hiu ny v ph thuc vo cc h s hiuchnh (trng s) tng ng ti thi im k.

iu ch B ht Knh Radio

B thu RFTn IFB tch

Quyt nhB cnbngTn hiuc to li

Tn hiubng cs x(t)

nb(t)

y(t)

d(t)

e t

n tngng

x(t)


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Hnh 4 2: B cn bng thch nghi

K hiu vc t li vo (dy tp dt nhn c ti b thu sau khi qua ng truyn)

ti thi im k l:

= [yk, yk-1, yk-2, , yk-N]T

Vc t trng s hiu chnh l:

= [w0k, w1k, w2k, , wNk]

Li ra ca b cn bng ti thi im k khi c biu din:

=

=N

nnknkk ywd

0 (4 1a)

Dng k hiu i s vc t c th vit:

kd =

=

(4 1b)

So snh vi tn hiu tp dt bit trc vi tn hiu nhn c c hiuchnh li, c mt sai s (li) l:

ek= xk-

kd = xk-

= xk-

(4 2)

Bnh phng li l:

ek2 = xk

2 +

- 2xk.

(4 3)

Ly trung bnh bnh phng li theo thi gian ta c:

E[ek2] = E[xk

2] +

.E[

]. 2E[xk.

].

yk yk-1 yk-2 yk-NZ-1 Z-1 Z-1

W0k W1k W2k WNk

-+xk bit

Li ra kd ca b cn

bng

Li ekThut ton thch nghi cp nht trng s


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Ch rng cc trng sWk c th rt ra ngoi trung bnh theo thi gian (v khng

thay i trong lc truyn d liu qua knh).

K hiu:

P = E[xk

] = E[xkyk, xkyk-1, xkyk-2, , xkyk-N]T (4 4)

L vc t tng quan cho gia tn hiu mong mun (dy tp dt nguyn bn) vtn hiu nhn c khi cha hiu chnh.

K hiu:

R = E[

] =

221

1212

11

212

...

................

...

...

NkkNkkNkkNk

Nkkkkkkk

Nkkkkkkk

yyyyyyy

yyyyyyy

yyyyyyy

(4 5)

L ma trn tng quan li vo, phng trnh trung bnh bnh phng li c vitl:

E[ek2] = E[xk

2] +

.R. 2PT (4 6)

Ly o hm trung bnh bnh phng li theo vc t trng s v cho bng 0 (zero),

ta c phng trnh.

02.2,...,,10

==

=

PWR

W

E

W

E

W

E

W

E

N

Hay vc t trng s ti u l PRW . 1= gi l phng trnh Wiener

y chnh l c s cho thut ton tnh trng s ca b lc nhm hiu chnh tn hiu

nhn c gn st vi tn hiu th knh bit trc

Li ti thiu ca b cn bng c vit li l:

Emin = MMSE = E[xk2

] PT

.R-1

.P = E[xk2

] PT

.W (4 7)Trn thc t vic ti thiu ha trung bnh bnh phng li (MMSE) c thc hin

quy bng thut ton o hm ngu nhin. Thut ton cn c gi l ti thiu trung

bnh bnh phng (LMS), y l thut ton cn bng n gin nht, n ch cn 2N + 1php ton trong mt bc tnh quy. Cc h s ca b lc c cp nht theo cc

phng trnh di y vi bin n l k hiu bc tnh quy:


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)().(.)()1(

)()()(

)().()(

* nynenwnw

ndnxne

nynWnd

NkNN

kkk

NT

Nk

=+

=

=

Vi N l s tng lm tr trong b cn bng v l c bc iu khin tc hi tv s n nh ca thut ton.

B cn bng LMS lm cc i t s tn hiu trn mo vi mt di b lc hn

ch. Nu tn hiu thu c c c tnh phn tn thi gian ln hn thi gian tr trong blc th b lc cn bng khng th lm gim mo. Tc hi t trong LMS l lu v ch c

mt tham s iu khin tc thch nghi . m bo s thch nghi khng tr thnh

s mt n nh, gi t phi chn sao cho

=


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Dng bt ng thc Chebysep, M cc i khi Gi = ri/N dn n:

=

==M

ii

M

i

i

i

iiM N

rG

rG

1

2

2

2

2

1)(

2

1(4 10b)

Ni cch khc trng s t hp phi t l vi ln tn hiu li vo ca knh ts tn hiu / n ca knh tng ng ta thu c tn hiu t hp cc i.

C th thc hin phn tp theo nhiu cch: thi gian, tn s, khng gian, a ng v

phn cc. tn dng c ton b li ch ca phn tp, cn phi thc hin kt hp

pha thu. Cc b kt hp phi c thit k sao cho sau khi hiu chnh tr v pha cho

cc ng truyn khc nhau, cc mc tn hiu vo phi c cng theo vect cn tp m

cng ngu nhin. Nh vy khi ly trung bnh t s tn hiu trn tp m SNR u ra s ln

hn u vo mt my thu.

1) Phn tp v m

2) Phn tp vi m

Phn tp khng gian

Phn tp tn s

Phn tp phn cc

Phn tp thi gian

Cng ging nh phn tp thu, phn tp pht da trn nim l pha pht s pht i

cc mu tn hiu khng tng quan cc mu ny s b pha inh khng tng quan

pha thu. Phn tp pht cho php gim t s Eb/It yu cu hay cng sut pht yu cu trn

knh v nh vy tng dung lng h thng. C th thc hin phn tp pht theo cc cch

sau.

1) Phn tp pht a sng mang (Multicarrier Transmit Diversity)

2) Phn tp pht trc giao

3) Phn tp pht khng gian - thi gian (STTD)

S khi ca h thng phn tp pht khng gian v thi gian dc cho hnh 4 3


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Hnh 4 - 3. S khi h thng phn tp pht khng gian v thi gian (STTD)

4) Phn tp pht chuyn mch thi gian (TSTD)Phn tp pht phn chia thi gian

S khi ca phn tp pht chuyn mch phn chia thi gian c th hin trong

hnh 4 4. Trong s phn tp ny,chuyn mch thc hin chn anten pht c

mt trong hai ng truyn.Tn hiu c chuyn mch hoc theo mt mu c quyt

nh BTS hoc theo cht lng tn hiu thu c t MS.

Hnh 4 - 4. S h thng phn tp pht chuyn mch phn chia thi gian

S phn tp pht chn la ging nh s hnh 4 - 3, tuy nhin y chuyn

mch c iu khin bi tn hiu la chn anten pht nhanh (AS) c pht i t MS

B mSTTD

MUXan

xen

Mho

knh

phihptc

M

UX

M nh knh v m ngu nhin di

Hoa tiu phn tp

Tx anten 1

Hoa tiu

TPCTx anten 2

MUX

MUX

an

xen

M

hoknh

ph ihptc M nh knh v m ngu nhin di

Hoa tiu

TPC

Chuynmch

Tx Anten1

Tx Anten2


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trong thi gian chuyn giao. Gi tr ca bit AS c xc nh trn c s o knh vt l

iu khin chung(CCPCH) pht i t anten c xt.

Cc k thut phn tp thu c s dng gim nh hng ca pha inh v ci thin tin cy ca thng tin m khng cn tng cng sut pht hoc rng bng tn.

nim cn bn ca thu phn tp l nu thu hai mu tn hiu c lp, th cc mu ny

s b pha inh khng tng quan. iu ny c ngha l xc sut tt c cc mu ny ng

thi thp hn mt mc nht nh thp hn nhiu so vi xc sut ca mt mu ring l.

Nh vy nu ta kt hp cc mu ny mt cch thch hp th ta s c tn hiu tng hp

t b nh ca pha inh hn nhiu so vi mt tn hiu n l. Hin c nhiu kiu phn tp

khc nhau: thi gian, tn s, khng gian, gc, nhiu tia v phn cc.

Trong thc t cc k thut phn tp thu c th c p dng hoc BTS hoc MS

mc d mi kiu ng dng u c cc c trng ring. Thng thng my thu phn tp

ch c s dng trm gc. V gi thnh ca b kt hp phn tp cao, nht l khi s

dng nhiu my thu. Ngoi ra cng sut my pht ca MS b gii hn bi tui th ca

acqui. BTS c th tng cng sut pht hoc cao anten ci thin din ph sng cho

MS. Thng thng cc h thng phn tp c thc hin my thu ch khng my

pht v khng cn thm nng lng cho h thng phn tp thu. Do tnh o ln gia

ng truyn ln v xung, nn cc h thng phn tp thc hin MS cng ging nh

cc h thng phn tp BS.

Trong my thu RAKE nhn c cc phin bn dch thi ca chui ngu nhin,

tn hiu thu phi i qua ng tr trc khi c ly tng quan cho v c kt hp.

Trong h thng tri ph CDMA tn hiu c tri ph bng dy gi ngu nhin. Dy

gi ngu nhin ny c tng quan thp khi lch nhau hn 1 chp. Do vy nu cc thnhphn a ng c thi gian tr lch nhau ln hn 1 chp th khi b tng quan ti ni thu

bt ng b vi 1 ng truyn, tn hiu ng truyn kia do tng quan thp s biuhin nh l n m khng cng vo lm mo tn hiu nh k thut bng hp thng thng,do c th khng cn dng n b cn bng. Do tnh cht ny m b thu RAKE dng 3

b tng quan bt ng b vi 3 ng truyn tr khc nhau ri sau khai thc tnhphn tp v mt thi gian ca 3 tn hiu thu c bng cch hiu chnh pha v t hp li.


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Li ra ca mi b tng quan c trng s theo mnh ca tn hiu cc i tn hiutng hp.

Hnh 4 5: B thu RAKE

Ghp xen l mt cch ch ng to ra s phn tp v thi gian ngay b pht mkhng mt thm tng ph no. K thut ny pht trin mnh in thoi di ng th h 2

khi b m ha ting ni c s ha. Li ra ca b m ha ting c mt dy cc bit quantrng cn phi phn tn chng hoc xp xp li v mt thi gian nu c can nhiu hocsuy gim mnh chng khng b nh hng cng mt lc. C 2 cch ghp xen l ghp xenkhi v ghp xen xon.

Ghp xen khi l vic a d liu (tng khi n x m bit) vo mt bng ch nht (m

hng v n ct) theo cch in y tng ct sau li c ra ln lt theo hng (hnh v),cc bit quan trng khi s c tch bit nhau bng m bit. Ti b thu gii ghp xen s

lm tc ng ngc li v in vo theo hng v c ra theo ct.

r(t) haytn hiuCDMAa ng

B tng quan 1

B tng quan 2

B tng quan M

Z1

Z2

ZM

Z Z m(t)

> c = 0

c = 0 thuc C(n,k)

c1 = u1G, c2 = u2G => c1 + c2 = cj

M khi tuyn tnh nu t hp tuyn tnh (modul 2) ca hai t m l mt t m.

M khi trong cc bit bn tin c gi nguyn, ch b xung thm cc bit d gil m h thng. M h thng cng c tnh cht tuyn tnh v c u im thc hin ngin.

Cu trc ca m khi h thng l:u1, u2, , uk,p1,p2, , pn-k

Trong k bit bn tri (u1, u2, , uk) l cc bit bn tin

n-k bit bn phi (p1,p2., pn-k) l cc bit kim tra. Bit kim tra l t hp tuyn tnhca k bit bn tin li vo

pi = p1iu1 + p2iu2 + + pk,iuk i = 1, , n-k (4 15)

vi

=ii

iiij uthuocphukhongpneu

uthuocphupneup

_____0

____1

Phng trnh trn xc nh mt cu trc ton hc ca m khi tuyn tnh (n, k) ta c

th vit li di dang vector ma trn nh sau:

(vi l ma trn c kch thc kx (n-k))

Cc t m = [ ] to nn mt khng gian con k chiu ca khng gian n chiu:

Do ta c th vit li l:


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= [ ] trong Ik l ma trn n v hay =

Ma trn G l ma trn nh phn c kch thc k x n c dng

=

knkkk

kn

kn

ppp

pppppp

G

,2,1,

,22,21,2

,12,11,1

..1..00

..............................

..0..10..0..01

Ma trn c gi l ma trn sinh, c k hng phi l c lp tuyn tnh (tc biu

din mt hng phi l t hp tuyn tnh ca cc hng cn li) t ta c

Tnh cht tuyn tnh c suy ra t h thng

Cch biu din khc mi lin h gia bit kim tra v bit bn tin l:

khi

= [

]

Tk

P

I=

S dng tnh cht cng modul 2 ta c HGT = 0 tc v phi l ma trn zero hay ta c

(4 - 16)

Khi ma trn H c gi l ma trn kim tra chn l. Khi tn hiu truyn

khng li th phng trnh (4 - 16) c tho mn

V d: Xt m lp li n = 5, k = 1 ta c:

= [1, 1 1 1 1] v =

1000

0100

0010

0001

,1

,1

,1

,1

Khong cch hamming ti thiudmin l c nh ngha l khong cch Hamming (s cc bit khc nhau trong 2 t

m) nh nht gia cc t m. T tnh cht ng ca t m dmin bng trng lng (s phnt 1) nh nht ca mt t m

dmin = min dH( ) vi ( 4 17)

M khi tuyn tnh (n, k) c th tch v hiu chnh c ti a

li nu v ch nu


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t0 [(dmin-1)/2] ly phn nguyn

Vy ta c kh nng c th sa li l t1


59/164



[ ]

==

1011000

1110100

1100010

0110001

4PIG

Ta c dmin = 3 do c th hiu chnh ti a c 1 li

Khi ta c ma trn kim tra H c dng

=

100

010

001

1

0

1

110

111

101

H

V cc mu li vi c trng li nh sau

Mu li c trng (s) S th tbit li

0000000 000 0 b li

1000000 110 1

0100000 011 2

0010000 111 3

0001000 101 4

0000100 100 5

0000010 010 6

0000001 001 7

Tp cc vc t li ng vi mt c trng to nn mt tp con trong tp m v mkhi tuyn tnh c 2n-k tp con nh vy khng chng ln nhau v ph ht khng gian tm. (4 17) l mt h n-k phng trnh tuyn tnh c th cho thng tin pht hin li.

Song n khng xc nh vc t li (n phn t) hay chnh xc hn xc nh duy nhtvector li: s c 2k mu li tho mn h phng trnh ny. Tuy nhin thng tin c trnggip cho php d tm li t 2n kh nng xung cn 2kkh nng.

4.4.3.2. M Hamming


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M Hamming l loi m khi tuyn tnh dng (2m 1, 2m -1 m) c khong cch tithiu bng 3 v ma trn kim tra chn l n gin c kch thc mx(2m 1) trong ttc cc dy nh phn di m tr dy ton 0 nhn theo ct.

Vi m = 3, ta c m (7 , 3) ma trn sinh l

[ ]

==

1011000

1110100

1100010

0110001

4PIG

Tng ng ta ta c ma trn kim tra chn l l :

=

100

010

001

1

0

1

110

111

101

H

T m to ra t ma trn sinh ny c dmin = 3, nn c kh nng pht hin v hiu

chnh 1 bit li. Cc mu li 1 bit v cc c trng gii m nh sau:

Mu li c trng(s)

0000000 000

1000000 110

0100000 011

0010000 111

0001000 101

0000100 100

0000010 010

0000001 001

T m nhn c sau ng truyn c nhn vi ma trn kim tra chn l H

tnh c trng s, t gi tr s nhn c so vi bng trn ta nhn c v tr bit li.

4.4.3.3. M d tha vng (CRC)M vng l mt trng hp ring ca m khi tuyn tnh:


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u im ca m vng l c th biu din nh biu thc ton v c s thc hin

n gin. M vng yu cu dch vng ca n cng l mt t m.

biu din ta c t m = c0c1cn-1 l mt a thc m

c(X) = c0X

0

+ c1X

1

+ +cn-1X

n-1

(4 20)Trong X l mt bin thc, cc h s ca a thc sinh l bin nh phn ca t m

Lu tha ca c(X) biu din dch bit theo thi gian

Khi Xc(X) = c0X + c1X2 + cn-1X

n (4 20a)

Thm cn-1 + cn-1 vo a thc trn do cn-1 + cn-1 = 0 (tnh cht cng modul 2) do a

thc trn tr thnh

Xc(X) = cn-1(Xn + 1) + cn-1 + c0X + cn-2X

n- 1

= cn-1(Xn + 1) + c(1)(X) (4 20b)

Vi c(1)(X) l c(X) dch vng i 1 bit

Tng t ta c:

Xic(X) = q(X)(Xn + 1) + c(i)(X) (4 20c)

Vi c(i)(X) l c(X) dch vng i i bit ng thi q(X) l phn d ca php chia Xic(X)

cho Xn + 1

Hay c(i)(X) = Xic(X) mod (Xn + 1) (4 21)

a thc Xn+ 1 v cc nhn t ca n ng vi tr chnh trong vic to m vng.

chn g(X) l a thc bc ti gin n-k v l tha s ca Xn + 1 c dng

knkn

i

ii XXgXg

=

++= 1

1

1)( (4 22)

gi = 0 hoc 1, g(X) gi l a thc sinh m vng theo cch sau:

c(X) = a(X)g(X) trong a(X) l a thc c bc k-1 v c(X) c to theo cch nys tho mn phng trnh (4 22) v g(X) l nhn t ca Xn + 1.

Gi s cho a thc sinh g(X) v cn m ho cho bn tin (k bit) thnh m vng

h thng (n, k) theo cu trc (n k) bit d thm vo v (k bit) bn tin, khi t m

phi c dng:


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c(X) = b(X) + Xn-km(X) ( 4 23)

Hay a(X)g(X) = b(X) + Xn-km(X)

Chia c 2 v cho g(X) ta c:

)()()(

)()(

XgXbXa

XgXmX kn += (4 24)

Phng trnh trn cho thy phn d ca php chia Xn-km(X) cho g(X) do qu

trnh m ho nh sau:

Bc 1: Nhn a thc bn tin m(X) vi Xn-k

Bc 2: Chia Xn-km(X) cho a thc sinh g(X) tm phn d

Bc 3: Cng phn d vi Xn-km(X) nhn c t m c(X)

V d: Tm t m ca vc t d liu 1010 theo loi m d tha vng (7, 4) ta c n =7, n k = 3 v:

x7 +1 = (x + 1)(x3 + x + 1)(x3 + x2 + 1).

Chn g(x) = x3 + x2 + 1

d(x) = x3 + x

T ta c: xn kd(x) = x6 + x4

V

Ta c th vit: c(x) = x3d(x) + r(x) = q(x)g(x)

= x3(x3 + x) + 1

=x6 + x4 + 1

V t m tm c s l: c = 1010001

3 bit cui c thm vo 4 bit d liu tho mn tnh chia ht, s pht m tun

hon nh sau:

x6 + x4

x6 + x5 + x3

x5 + x4 + x3

x5 + x4 x2

x3 + x2

x3 + x2 + 1

1

x3 + x2+ 1x3 + x2 + 1 = q(x)


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Hnh 4 8: S to khi Reed - Solomon

4.4.3.4. Hot ng ca m khi tuyn tnhM khi tuyn tnh c th thc hin gii m quyt nh cng hoc mn. Trong s

Giao Trinh Thong Tin Di Dong 2012

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Transcript of Giao Trinh Thong Tin Di Dong 2012