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Bo co mn hck thut pht thanh truyn hnh

Ging vin hng dn :

TS. NGUYN THU HIN

ti : m ha knh kt ni song song v ni tip

Nhm VI - Lp D11VT4

L c Hng

Trn Vn Hinh

3. L Ngc Hng

HC VIN CNG NGH BU CHNH VIN THNG

THNH VIN TRONG NHM

Nhm 6 - Lp D11VT4

PHNG PHP LUN V NGHIN CU KHOA HC

Nhm 6 - Lp D11VT4

I. Tm tt lch s

II. Cu trc cc b m ha

III. Thnh phn chc nng b m ha

Mc Lc

IV. Gii m

V. So snh


Nhm 6 - Lp D11VT4

CHNG I. TM TT LCH S

Nm 1993, Claude Berrou, Alain Glavieux, Puja Thitimajshima cng vit tc phm Near Shannon limit error correcting coding and decoding: TURBO CODE nh du mt bc tin vt bc trong nghin cu m sa sai. Loi m m h gii thiu thc hin trong khong 0.7dB so vi gii hn ca Shannon cho knh AWGN. Loi m m h gii thiu c gi l m Turbo.

Forney s dng mt b m khi ngn hoc mt b m tch chp vi gii thut gii m Viterbi xc sut ln nht lm b m trong v mt b m Reed-Salomon di khng nh phn tc cao vi thut ton gii m sa li i s lm b m ngoi. C hai kiu kt ni c bn l kt ni ni tipv kt ni song song.


Nhm 6 - Lp D11VT4

CHNG II. CU TRC CC B M HA

2.1. M ha knh kt ni song song-PCCC

S cu trc m ha PCCC (kt ni song song)


Nhm 6- Lp D11VT4


M PCCC l s kt ni song song ca 2 hay nhiu m RSC. Thng thng ngi ta s dng ti thiu 2 b m ho tch chp.

Tc m ho (code rate) ca b m ho PCCC l: r = k/n

Nu c 2 b m RSC th : r = k / (n1+n2)

Mt b m ho RSC c tc m ho tiu biu l r =1/2 , hai b m ho thnh phn c phn cch nhau bi mt b chn ( interleaving)


Nhm 3 - Lp D11VT4

Hnh 1.4 S chi tit m ho PCCC tc 1/3



Nhm 6 - Lp D11VT4

2.2. S cu trc ca b m SCCC (kt ni ni tip).

Hnh 1.1 S n gin m kt ni ni tip

i vi m kt ni ni tip, tc m ho: Rnt=k1k2/n1n2 .


Nhm 6 - Lp D11VT4

2.2. S cu trc ca b m SCCC (kt ni ni tip).

M SCCC l s kt ni ni tip ca 2 hay nhiu m RSC.

Thng thng ngi ta s dng ti thiu 2 b m ho tch chp.


Nhm 6 - Lp D11VT4

Chng III. Thnh phn chc nng b m ha

3.1 B m ha tch chp cc h thng quy( Recursive Systematic Convolutional Code_RSC )

3.2 K thut xa (Puncture)

3.3 Cc b chn (Interleaver)


Nhm 6 - Lp D11VT4

3.1 B m ha tch chp cc h thng quy( Recursive Systematic Convolutional Code_RSC )

3.1.1 M chp tuyn tnh

gl

g1

g0

Sj-m

Sj-1

.....

Sj

Message bits xi

encoded bit yj

B m ha s dng cc thanh ghi dch a thm d vo lung d liu

Cng thc biu din cc bit u ra ng vi cc bit vo l:

yj =Sj-m gl . . . . Sj-1 g1 Sj g0 (mod 2)

m = Sj i gi ; i=0


Nhm 6 - Lp D11VT4

3.1.2 M tch chp h thng quy

M tch chp h thng quy c ly t b m ho tch chp thng thng bng cch hi tip mt trong nhng ng ra m ho thnh ng vo ca n

D

+

+

x

c2

c1

+

D

Hnh 1.7: B m ho RSC c r=1/2 ; K=3

B m ho RSC c biu din l G = [ 1, g2/g1 ] trong ng ra u tin ( biu din bi g1) c hi tip v ng vo, g1 l ng ra h thng, g2 l ng ra feedforward


Nhm 6 - Lp D11VT4

3.1.3 Kt thc Trellis

Hnh 1.8 Cch thc kt thc trellis b m RSC

m ho chui ng vo, kho chuyn bt n v th A, kt thc trellis th kho chuyn bt n v tr B.


Nhm 6 - Lp D11VT4

3.2 K thut xa (Puncture)

Dng tng tc m ca mt b m ho m khng lm thay i cu trc ca b m ho.

Bn cht ca k thut puncture l lm gim n theo mt qui lut no tc m ho r tng ln.

VD: Hnh 1.9, nu chui h thng c1 vn gi nguyn v cc chui c2 v c3 s c ly xen k. Chui c2 s ly cc bit l v cc bit chn ca chui c3 th b m s c tc 1/2. Khi b gii m nhn c chui bit n th n s thm vo chui ny cc bit 0 ti nhng ch b xo bt. Nh vy c th lm sai lch bit parity nn gim cht lng.


Nhm 6 - Lp D11VT4

3.3. B chn (Interleaver)

c2

M trng s thp

M trng s cao

M h thng

B m ho RSC 1

B m ho RSC 2

B chn

x

c1

c3

Hnh 1.9 B chn lm tng trng s m ca b m ho RSC2 khi so snh vi b m ho RSC1.

B chn c s dng ti b m ho nhm mc ch hon v tt c cc chui ng vo c trng s thp thnh chui ra c t m ng ra trng s cao hay ngc li.


Nhm 6 - Lp D11VT4

B chn ma trn

B chn gi ngu nhin

B chn dch vng

B chn chn-l (Odd-Even)

B chn Smile

3.3. B chn (Interleaver)

Gm cc b chn sau :

B chn khung

B chn ti u

B chn ng dng

B chn S


Nhm 6 - Lp D11VT4

IV. GII M

4.1 Gii thut MAP

4.2 Th tc gii m SISO cho PCCCs

4.3 Th tc gii m SISO cho SCCC


Nhm 6 - Lp D11VT4

4.1 Gii thut MAP

Hnh 2.1 B gii m lp MAP


Nhm 6 - Lp D11VT4

Tch tn hiu nhn thnh 2 chui tng ng cho b gii m 1 v b gii m 2.

vng lp u tin, thng tin A Priori ca b gii m 1 c a v 0. Sau b gii m 1 a ra c thng tin Extrinsic th s c chn v a ti b gii m 2 ng vai tr l thng tin A Priori ca b gii m ny. B gii m 2 sau khi a ra thng tin Extrinsic th vng lp kt thc. Thong tin Extrinsic ca b gii m th 2 s c chn v a v b gii m 1 nh l thng tin A Priori.

Qu trnh gii m lp i lp li nh vy cho n khi thc hin s ln lp quy nh.

Sau vng lp cui cng, gi tr c on c c tnh bng cch gii chn thng tin b gii m th 2 v a ra quyt nh cng.

Gii thut gii m c thc hin qua 4 bc:

4.1 Gii thut MAP


Nhm 6 - Lp D11VT4


Gii iu ch nhn tn hiu t hai dng, mt t DEC1 v mt t DEC2. Tn hiu c nhn thng qua h thng phn chia Equation (3.91) c cho DEC1 v cho DEC2, .

Lp vi . Cho DEC1, khi to .

3. DEC1. Tnh ton ng i ma trn v vi Equations (3.155) v (3.156). Sau s dng Equations (3.164) v (3.167) tnh ton ng ra LLR cho cc bit thng tin.

Cc bc gii m:


Nhm 6 - Lp D11VT4

4. Chn cung cp u vo LLR cho DEC2: .

5. DEC2. Trc ht s dng Equations t c ng ma trn v . Sau :

Nu , s dng Equation trong conjuction vi (3.167) tnh ton ng ra ca LLR cho chn thm bit thng tin. Gii chn (deinterleaver) n ca ln lp tip: . Tng , tr v bc 3 v bt u lp tip.

Nu . S dng Equations (3.164) nhn c y thng tin cho cc bit chn. Gii chn v cc quyt nh da trn cc bt truyn nh sau:

i n bc 1 gii m khung tip theo.



Nhm 6 - Lp D11VT4


Hnh 2.3: Gii m SCCC vi b iu ch chung SISO.


Nhm 6 - Lp D11VT4


Cc bc gii m:

Tn hiu nhn s dng Equations (3.91) u vo LLRs cho t m bit ca m trong.

. i vi gii m trong, khi to , .

Gii m trong. Tnh ton ma trn ng v vi Equations (3.155) v (3.156). Sau s dng Equations (3.164) v (3.167) tnh ton u ra LLRs cho bit thng tin ca m trong.

B chn dng ca u vo LLR cho bit t m ca m ra.


Nhm 6 - Lp D11VT4

Gii m u ra. Trc ht tnh ton ng ma trn v s dng Equations (3.155) v (3.156). t , v th s khc bit n t s c tnh. Sau :

Nu , s dng Equations (3.163) v (3.165) tnh ton u ra LLR cho bt t m ca m u ra. Chn dng t u vo LLR ca m trong. Tng , tr v bc 3 v bt u chu k mi.

Nu , s dng Equations (3.164) (vi ) trn vn thng tin ca m ra. Quyt nh da trn bit truyn nh sau:

Tr li bc 1 gii m khung tip theo.



Nhm 6 - Lp D11VT4

V. SO SNH

5.1 Turbo Codes lai:

Mt PCCC c th to ra t mt SCCC bi ta xn tt c cc bit chn l i. Khi nim ny a n s thc t l nhiu thit k ang hybrid mi c th xy dng t mt m SCCC gc bng cch thay i s bit chn l i ta xn

M lai c th c thit k kt hp li th ca PCCC v SCCC, do , thu hp khong cch gia hai m. m lai vi mt s lng ln cc bit chn l i thng s c nhng tnh cht ging nh PCCC v thc hin tt hn so vi SCCC trong vng SNR thp trong khi nhng m c chn l i bit l thng t nhiu s c nhiu c tnh ging nh SCCC v c mt tng thp hn PCCC ti SNR cao. Hiu sut ca cc m c hin th trong Hnh 3.2 v Hnh 3..

25


Nhm 6 - Lp D11VT4

Hnh 3.1: So snh hiu nng ca mt thng thng r = 1/3 PCCC v PCCC c to ra bi ta xn tt c cc bit chn l i ca mt r = 1/3 SCCC gii m.



Nhm 3 - Lp D11VT4

Hnh 3.2 So snh hiu sut BER ca r = 1/3 PCCC, SCCC v m lai vi kch thc khung hnh = 512 bit so vi Es/N0dB.



Nhm 6 - Lp D11VT4

5.2. S khc nhau gia m SCCC v m PCCC


Nhm 6 - Lp D11VT4

KT LUN

Qua bi tiu lun ny chng ta bit r hn v m ha knh kt ni ni tip (SCCC), v song song (PCCC), cc cu trc b m ha b chn, cc th tc ca b gii m cng nh nhng u nhc im ca n so vi cc loi m khc. Cui cng, mt ci nhn su sc vo cc k thut c s dng tin hnh thit k v phn tch nh thit k M lai c cung cp. Cc m c xut l cc k linh hot. Thit k h thng c th c hng li t nghin cu ny t by gi h s c mt lot cc m ha ty chn la chn Vy nn n c ng dng vo cc h thng thng tin yu cu cht lng, tc cao. V hin ang c nghin cu rt nhiu.

Thank You !

ng vo Ng ra

B m ha 1 r=k1/n1 B m ha2 r=k2/n2

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