v2 0 Bao Cao Dieu Che Pcm Nhom 10 Chu de 5 Quan 1112

8/11/2019 v2 0 Bao Cao Dieu Che Pcm Nhom 10 Chu de 5 Quan 1112

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Trng i Hc in Lc Khoa in TVin Thng Nhm 10

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TRNG I HC IN LCKhoa in TVin Thng

MN HC I CNG VIN THNG

Ch: Xung M PCM v M Phng Trong MatLab

Ging vin bmn: TS LAnh Ngc

Nhm Sinh Vin Thc Hin: Nhm 10

Sinh vin lp: 5.TVT1


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Li ni u.

Tn hiu bng thng gc c gy ra bi cc ngun thng tin khcnhau,khng phi lc no cng thc hp cho vic chuyn trc tip qua mt knh

cho trc no . Cc tn hiu ny thng c bin di qu trnh chuyn ic ddng. Qa trnh bin i c gi l iu ch. iu ch l mt kthut cho php thng tin c truyn nh s thay i ca tn hiu mang thngtin. iu chc sdng cho c thng tin sv tng t. Trong trng hpthng tin tng t l tc ng lin tc (s bin i mm). Trong trng hpthng tin s, iu ch tc ng tng bc (thay i trng thi). Khi kt hpiu chv gii iu chc gi l modem.Trong truyn dn tng tc th

sdng hai phng php iu chtheo bin v theo tn s.iu chbin

c s dng truyn ting ni tng t (300-3400 Hz). iuch tn tn sthng c sdng cho truyn thng qung b (bngFM), knh m thanh choTV v hthng vin thng khng dy. Ngy nay c rt nhiu cc phng phpiu chkhc nhau nhiu xung m (PCM), iu xung m vi sai (DPCM), iuchDelta (DM), ... Trong thit b ghp knh s thng s dng phng phpghp knh theo thi gian kt hp iu xung m (TDM - PCM). Trong bi bo cony chng ta stm hiu vqu trnh iu chxung m PCM,v ng dng ca ntrong thc t.Trong quas trnh tm hiu v lm bi bo co do knhng ca cc

thnh vin c hn v ti liu cn hn ch nn bi bo co ca chng em cnnhiu thiu st rt mong c thy cho kin nh gi chng em c thhieusu hn vvn iu chxung m PCM ny.

H ni ngy 1 thng 11 nm 2012.


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Mc LcLi ni u2

I: Cc Phng Php M Ha v iu Ch..4

I1: M ha..4I2: iu ch5

II: iu ChXung M PCM.5

II1: Ly mu6

II2: Lng tha.............8

II3: M ha...10

II4: Gii m..13

III: Gi Thiu VMt SPhng Php Mi.15

III1: PCM vi sai...15

III2: PCM thch ng...16

IV: M Phng PCM trong MatLab.18

V: nh Ga Cc Thnh Vin.........26

VI: Ti Liu Tham Kho.27


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I:CC PHNG PHP VM HA V IU CH.

I1.M HA.

Trong cc h thng truyn dn s thng tin c chuyn i thnh mtchui cc thp xung, sau truyn trn ng truyn. Khi ,thng tintng t(nh ting ni ca con ngi) phi c chuyn i vo dng snhcc bbin i A/D. chnh xc ca chuyn i A/D quyt nh chtlng lnh hi ca thu bao. T hp s phi chi tit sao cho ting ni(hoc video) tng tc thc ti to m khng c mo v nhiu lon thit bthu. Hin nay, mong mun ca chng ta l gim khi lng thng tin

ssdng tt hn dung lng mng.Cc bm ho c phn lm 2 loi chnh: m ho dng sng v m hothoi (vocoder). Ngoi ra, cn c cc bm ho lai t hp c tnh ca 2loi trn. Hnh 1.1 minh ho skhc nhau vcht lng thoi v cc yucu tc bit i vi cc loi m ha khc nhau.

M ho dng sng c ngha lcc thay i bin ca tn hiu tng t(ngthoi) c m tbng mt sca gi trc o.Sau cc gi trny c mho xung v gi ti u thu. Dng iu tng tnh tn hiu c ti to trongthit bthu nhcc gi trnhn c. Phngphp ny cho php nhn c mc


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cht lng thoi rt cao, v ng tn hiu nhn c l bn sao nh tht cang tn hiu bn pht.

Mho thoi l bm ho tham s. Thay cho vic truyn tn hiu m t

trc tip dng ca ng tn hiu thoi l truyn mt s tham s m tng cong tn hiu c pht ra nh thno. Cch n gin gii thchs khc nhau gia hai phng php ny l s dng php n dng: nhcang c chi v cc bn nhc th c cc nhc cng sdng. Trong mho dng sng chnh nhng m thanh nhc ang chi c truyn i, cntrong m ho tham sth cc bn nhc c gi ti bn nhn. M ho tham syu cu c mt m hnh xc nh r ng tn hiu thoi c to nh thno. Cht lng s mc trung bnh (m thanh ca thoi nhn c thuc

loi tng hp) nhng mt khc cc tn hiu c thc truyn vi tc bitrt thp.

Bm ho lai gi mt scc tham scngnhmt lng nht nh thngtin dng sng. Kiu m ho thoi ny a ra mt s thohip hp l gia chtlng thoi v hiu qu m ho, v n c sdng trong cc h thng inthoi di ng ngy nay.

I2. IU CH

iu ch l mt k thut cho php thng tin c truyn nh s thay ica tn hiu mang thng tin. iu chc sdng cho cthng tin svtng t. Trong trng hp thng tin tng tl tc ng lin tc (sbini mm). Trong trng hp thng tin s, iu chtc ng tng bc (thayi trng thi). Khi kt hp iu chv gii iu chc gi l modem.Trong truyn dn tng tc thsdng hai phng php iu chtheo

bin v theo tn s.

iu bin c sdng truyn ting ni tng t (300-3400 Hz). iu

tn thng c sdng cho truyn thng qung b (bngFM), knh mthanh cho TV v hthng vin thng khng dy.

II: IU CHXUNG M PCM

Hin nay c nhiu phng php chuyn tn hiu analog thnh tn hiu digital(A/D) nh iu xung m (PCM), iu xung m vi sai (DPCM), iu chDelta


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(DM), ... Trong thit bghp knh s thng sdng phng php ghp knhtheo thi gian kt hp iu xung m (TDM - PCM).

iu chtn hiu analog thnh tn hiu digital cn dng phng php PCMtheo 3 bc nh hnh1.3.

iu chxung m PCM c c trng bi 3 giai on l: Ly mu,

Lng tha v m ha. Bi bo co ny xcp ti ng vn cthtrong ton bqu trnh m ha xung m PCM.

Trc ht phi ly mu tn hiu thoi, tc l chtruyn cc xung tn hiu ticc thi im nht nh. Bc thhai l lng tho bin , nghal chia bin ca xung muthnh cc mc v ly trnbin xung n mc gn nht.

Bc thba m ho xung lng tthnh tm nhphn c m bit.

II1: Ly mu tn hiu analogBin ca tn hiu analog l lin tc theo thi gian. Ly mu l ly bin

ca tn hiu analog tng khong thi gian nht nh. Qu trnh ny ging nhiu chbin , trong cc dy xung c chu kc iu chbin bi tnhiu analog. Do vy cc mu ly c sgin on theo thi gian. Dy mu ny


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gi l tn hiu PAM (iu chbin xung).

thc hin qu trnh ly mu tn hiu bt kphi da vo nh l Nyquist,ni dung ca nh l c pht biu nh sau:

Nu tn hiu gc l hm lin tc theo thi gian c tn ph gii hn t 0

n fmax khi ly mu th tn sly mu phi ln hnhoc bng hai ln tnsln nht trong tn hiu gc, nghal: fm 2fmax. Mt yu t quan trng trong ly mu l pha pht ly mu cho tn hiuanalog theo tn sno cho pha thu ti to li c tn hiu ban u. Theonh l Nyquist, bng cch ly mu tn hiu analog theo tn s cao hn tnht hai ln tn scao nht ca tn hiu th c thto li tn hiu analog banu tcc mu .

i vi tn hiu thoi hot ng bng tn 0,3 3,4 kHz, tn sly mu l8kHz p ng yu cu vcht lng truyn dn: pha thu khi phc tn

hiu analog c mo trong phm vi cho php. Qu trnh ly mu tn hiu

thoi nh hnh 1.4.


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II2: Lng tha

Lng tho ngha l chia bin ca tn hiu thnh cc khong u hockhng u, mi khong l mt bc lng t, bin tn hiu ng vi u hoccui mi bc lng tgi l mt mc lng t. Sau khi c cc mc lng tth bin ca cc xung mu c lm trn n mc gn nht.

C hai loi lng tho bin : lng tho u v lng tho khng u

Lng tho uBin tn hiu c chia thnh nhng khong u nhau, sau ly trn ccxung mu n mc lng tgn nht.

Qu trnhlng tho u thhin nh hnh 1.5.


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Bc lng tu bng .nh vy,bin ca tn sgm 7 bc lng tv 8mc (nh st-3 +3 ).mi quan hgia mc lng tnh nhau:

Tng smc lng t= tng sbc lng t+1

Do phi ly trn n mc lng tgn nht,chnh lch gia bin xunglng tv gi trtc thi ca xung ly mu sgy ra nhiu lng tQd (xemhnh 1.6 ).

Bin xung nhiu lng tlun thomn iu kin sau:

Cng sut trung bnh nhiu lng tu c xc nh nh sau:

i vi tn hiu mnh ts:S/N = ( Tn hiu/Nhiu) sln hn tsny ca tn hiu yu. mun san phng tsny gia tn hiu mnh v tn hiu yu phi sdng lng tha khng u.


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Lng tho khng u.

Lng tho khng u da trn nguyn tc: khi bin tn hiu cng ln thbc lng tcng ln (hnh 1.7).


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Trong th d trn hnh 1.7 bin ca tn hiu analog c chia thnh 4 bclng t, k hiu l 1, 2, 3, 4. Nhvy: 1 < 2 < 3 < 4 < ... Cc ngthng song song vi trc honh (t) gi l cc mc lng t, c nh st0 tigc to.

Cc xung ly mu ti cc chu knTm (trong n=0,1,2,...) c ly trn n

mc lngtgn nht.

Mun lng t ho khng u c th s dng mt trong hai phng php:nn - dn analog hoc nn - dn s.

II3:M Ha

Nn - dn analog

Qu trnh nn - dn analog c thc hin bng cch t bnn analog trcbm ho u pha nhnh pht ca thit bghp knh, trong min tn hiuthoi analog v t mt bdn analog trc b gii m u nhnh thu cathit bghp knh, cngtrong min tn hiu thoi analog.

Trong thit bghp knh schto theo tiu chun Chu u sdng bnn -dn theo lut A. Cn theo tiu chun Bc Mv Nht sdng bnn theo lut .

c tuyn ca bnn lut A (sph thuc in p u vo v u ra bnn)biu thbng biu thc


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Trong x= UV/U0 vi Uv l bin in p u vo bnn, cnU0 l in pvo bo ho.

Theo khuyn nghca ITU-T ly A = 87,6. c tuyn ca bnn lut biu thbng biu thc

Theo khuyn nghca ITU-T ly = 255.Tcc biu thc trn c thxy dng c cc ng cong thhin c tuyn

bnn Av . c tuyn bnn phi i xng vi c tuyn bdn khng

gy mo khi khi phc tnhiu. Mun t c tsS/N khong 25dB th slng mc lng tu phi bng 2048. Nh vy mi tm cn 11bit (khng k

bit du).V 211 = 2048 l smc lng tca bin dng hoc m ca tnhiu thoi. Sau khi nn, tn hiu thoi chcn 128 mc. Nu kcbit du chcntm 8 bit. l l doti sao phi thc hin nn tn hiu.

Nn - dn s.Bnn sc t trong min tn hiu s ca nhnh pht v bdn sct trong min tn hiu sca nhnh thu ca thit bghp knh. c tuyn bnn v bdn sda trn c s ca b nn v bdn analog. Bng cch gn

ng ho ng cong c tuyn b nn dn analog theo lut A v thnh


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cc on thng gp khc.

c tuyn ca bnn slut A c tt c13 on thng c dc khc nhau vly tn l bnn sA = 87,6/13 c thhin trong hnh 1.9.

Cc on thng c dc khc nhau, do vy trong cng mt on tn hiu khngbnn. Khi chuyn ton ny sang on khc th tn hiu bnn v khi bin cng ln sbnn cng nhiu.

xy dng c tnh bin ca bnn scn tin hnh ccbc sau y: Trcx c trngcho bin chun ho ca tn hiu u vo bnn (-1 x 1tngng vi 4096 bc lng tu) v trc y c trng cho tn hiu u ra. Trn trc x chia theo khc logarit c s2, na dng u gm cc im 0.1/128, 1/64, 1/32,1/16,1/8,1/4,1/2 v 1. Cn na m c chia ngc li.


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Trn trc y chia thnh cc khong u nhau v na dng gm cc im 0, 1/8,2/8, 3/8, 4/8, 5/8, 6/8, 7/8 v 1. Cn na m c chia ngc li. Tip nh du cc im c bit A, B, C, D, E, F, G v H, trong na dngca ng c tnh, trong im H l im ct nhau ca on thng vung gc

vi trc x ti im c x= 1 v on thng vung gc vi trc y ti im c y= 1.im G l im ct nhau ca on thng vung gc vi trc x ti im c x=1/2v on thng vung gc vi trc y ti im c y=7/8, .... im A l im ctnhau ca on thng vung gc vi trc x ti im c x=1/128 v on thngvung gc vi trc y ti im c y=1/8. Ni hai im k nhau bng mt onthng. Nhvy na dngca ng c tnh bin c tt c8 on thng,mi on c c trng bng t m 3 bit. Trong mi on c chia thnh 16mc, mi mc phn phi tm 4 bit. Na m ca ng c tnh bin c

ly i xng vi na dng qua gc toO. Do 4 on gn gc to 0 c dc nh nhau (trong na dng c hai on OA v OB). Nh vy ton

bng c tnh bin c 13 on thng c dc khc nhau. Na m v na dng ca ng c tnh bin c phn phi tm 1 bt.Bt 0 tng ng vi na m ca ng c tnh bin v bt 1 tng ng vina dngca ng c tnh bin . Tm li, khi cha nn th tn hiu thoi c chia thnh 4096 mc, sau khidng bnn A=87,6/13 th chcn li 256 mc (tc l sbt trong mt tm

gim t12 xung 8).

II.4: Gii MTi pha thu, tn hiu sPCM c chuyn i thnh tn hiu analog qua haibc l: gii m v lc. Tng hp hai qu trnh x l ny gi l qu trnh

chuyn i D/A v c biu din nh hnh 1.10. Gii m l qu trnh ngc li vi m ho. Trong gii m, bt u bng victch cc m nhphn 8 bit ttn hiu PCM (trong hnh 1.10 tng trng tm 3

bit).


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Tip theo, chuyn mi t m nh phn thnh mt xung lng t c bin tng ng vi smc lng tca tm .Hnh 1.11 minh hogii m cctm 3 bt. Tn hiu xung c lng tho u pht c to li uthu bng cch gii m nhvy. Tn hiu xung sau khi gii m c bin chnhlch vi bin xung mu ti pha pht. Hin tng ny gi l mo lng tv

pht sinh do lm trnbin khi lng tho.

Sau,tn hiu xung lng tc aqua blc thng thp. u ra b lcny nhn c tn hiu analog l tn hiu lin tc theo thi gian nhni suy giacc mu ktip nhau nh hnh 1.12.


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III: MT SPHNG PHP M HA MI

PCM tn ti trong 1/4 thp kv cc cng nghmi bt u thu ht sch . Trong thp kcui, m ho thoi tinh vi trln hin thc nhsphttrin ca VLSI (mch tch hp rt ln). PCM ti 64 Kb/s khng cn l cng nghduy nht na. Vic m ho 32 v 16 kbit/s c pht trin, v cc phngphp vocoder cng c pht trin m chyu cu 4.8 Kb/s v t hn. Chng tac thbng mi cch t ti 800bit/s m vn nghe hiu c, nhng ti tc

bit ny khng c khnng nhn dng c li ni ca ngi ni.

Ccphng php m ho mi gi ra rt nhiu li ch, v chng cho php ccnh khai thc tnggp 2 hay 4 ln dung lng truyn dn thoi trong mngca h m khng cn phi lp t thit b truyn dn mi. Mt trong nhngphng php c thdng l iu chxung m vi sai thch ng, ADPCM. ADPCMcho php truyn thoi vi cht lng gim ti thiu ti 32Kbit/s. Khuyn nghcaITU vADPCM c gi l G.726.

PCM vi sai (DPCM)

Tn hiu c ly mu cho thy mc tng quan cao gia cc mu kcn. Hay ni cch khc, hai mu gn nhau l kh tng tnh nhau. Ngha lsc nhiu li ch nu m ho skhc nhau gia cc mu kcn thay cho mho gi trtuyt i ca mi mu. Trn hnh 1.13 cho thy 4 bit c thc s


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dng thay cho 8 bit. y l tng n trong PCM vi sai (DPCM), y chnh xc vn c gi li mc d khng cn bng tn rng. DPCM utin da trn bn quyn t1952.

PCM vi sai c nhc im l nu tn hiu u vo tng tm thay i qu lngia cc mu, th n khng thc biu din bng 4 bit m sbct.

DPCM thch ng (ADPCM)PCM vi sai thch ng (ADPCM) t hp phng phpDPCM v PCM thch ng. ADPCM c nghal cc mc lngtho c thch ng vi dng ca tn hiu u vo. Kch ccacc bc lng t tng ln khi c lin tip dc ng trong tn


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hiu ko di. Trong hnh 1.14, smu l 6 c thc m tbng 5 bc lng t ln thay cho 10 mu nh. Phng phpny c tn t khnng thch ng y, tc l n to ra khnng

gim cc bc lng t.

Trong m ho ADPCM, sau khi tn hiu vo tng t iqua m ho PCMthng thng, th lung cc mu 8 bit c gi tip ti b m ho ADPCM.Trong bm ho ny, mt thut ton chvi 15 mc lng tc sdng gim di t8 bit xung 4 bit. 4 bit ny khng biu din bin ca mu na,nhng nhc m ho vi sai m 4 bit vn cha thng tin cho php tn hiugc sc ti to bthu.

Mc ca mt mu c d on da trn mc ca mu ng trc. S khcnhau gia mu donv thc tl rt nhv v vy c thm ho bng 4 bit.

Nu c vi mu tip theo thay i ln, th cc bc lng tc thch ng nhm ttrn.

IV: M Phng Phng Php iu ChXung M PCM bng MATLAB.


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1.Gii Thiu C Bn VMATLAB.

Matlab l mt phn mm ton hc ca hng Mathworks tnh ton trn ccsv c tnh trc quan rt cao.L mt mi trng mnh dnh cho cc tnh ton

khoa hcMatlab l vit tt ca Matrix Laboratory. l mt phn mm khoa hc c thit kcung cp vic tnh ton sv hin thha bng ngn nglp trnh cp cao.MATLABcung cp cc tnh nng tng tc tuyt vi cho php ngi sdng thaotc dliu linh hot di dng mng ma trn tnh ton v quan st. Cc dliuvo ca MATLABc thc nhp t"Command line" hoc t"mfiles", trong tp lnh c cho trc bi MATLAB

Hin nay, Matlab c n hng ngn lnh v hm tin ch. Ngoi cc hm ci sn

trong chnh ngn ng, Matlab cn c cc lnh v hm ng dng chuyn bittrong cc Toolbox, mrng mi trng Matlab nhm gii quyt cc bi tonthuc cc phm tr ring. Cc Toolbox kh quan trng v tin ch cho ngidng nhton scp, xl tn hiu s, xl nh, xl m thanh, ma trn tha,logic m,

Ngi dng cng c thto nn cc hm phc vcho chuyn mn ca mnh,lu vo tp M-file dng vsau.

Matlab cn c giao din ha kh p mt v d s dng. Ngi dngc th tnh ton v to nn cc hnh nh ha 2, 3 chiu cho trnh ngdng ca

mnh. Vi cc hnh nh, nu khng chnh v canh trc, phi mu thMatlabthc hin tng mt cch kh ph hp.

V tnh mnh m tr gip gii nhanh cc bi ton k thut, chng ticgng bin son ti liu ny phc vmt t kin thc cbn cho bn c.Tuy nhin, trn cs bn c c thtkhai thc thm cc thnh phn dngring cho minh trong cc Toolbox v Simulink.

2.M Phng PCM Bng MATLAB.

2.1:Lng THa Tn Hiu.Lng tha l qu trnh ri rc ha tn hiu vbin ,cthl thay thtt c

cc gi trca tn hiu nm trong mt khong xc nh no thnh mt gi trduy nht.Min gi tr ca tn hiu c chia thnh mt shu hn cc khongchia.Nh vy ln ca tn hiu sau khi lng tchc thnhn mt trong shu hn cc gi trcho trc.


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Tp hp cc khong chia gi l sphn hoch ca tn hiu(partition).Tp cc gitrthay thcho mi khong chia gi l bm (codebook).

Matlab biu din phn hoch ca tn hiu bng mt vector m cc phn tca nl cc im ranh gii gia hai khong chia lin tip.V d nu nh tn hiu cmin xch nh R c phn hoch thnh cc khong (-,0],(0,2],(2,4],(4,+].thc thbiu din sphn hoch ny thnh vector:>> partition = [0,2,4];

Tngng vi vector phn hoch tn hiu l vector biu din bm tn hiu. Cc phnt ca n l cc gi tr thay th trong mi khong chia tng ngca phn hoch. Nuta thay thcc gi tr trong khong (-, 0] bng -1, cc gi tr trong khong (0,2]

bng 1, cc gi tr trong khong (2,4] bng 3 v cc gi tr trong khong (4, +]

bng 5 th vector biu din b m s l:>> codebook = [-1,1,3,5];

thc hin qu trnh lng tho, MATLAB cung cp hm quantiz:>> [indx, quant, distor] = quantiz(sig, partition, codebook)

Trong sig l tn hiutr c khi lngt v partition,codebook lnlt l phnhoch v b m lng t.Hm quantiz trv:+ Tn hiu saukhi lng tho quant+vector indxc chiu di bng chiu di tn hiu, mi phn tca n l ch sca gitr tng ng ca tn hiu trong b m lng t. Ni cch khc, vectorindxcho

bit mi phn t ca tn hiu vo nm trong khong chia no. Gia cc vector indx,quant v codebook c mi lin h:

quant = codebook(indx+1);

+Nhiulngt do php lngt ho ny gy ra.( Nhiulngt l sai s bnhphng trung bnh gia tn hiu gc v tn hiu sau khi lng t).

V d:>> parti=[0,2,4];>> codebook=[-1,1,3,5];>> samp=[-3,-1,1,1,5,3,5,4,2,1]>> [index,quantized]=quantiz(samp,parti,codebook);>> index cho biet thu tu muc luong tuans =


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0 0 1 1 3 2 3 2 1 1

>> quantized

quantized =-1 -1 1 1 5 3 5 3 1 1

Trong cc h thng thng tin s ngi ta thng phn hoch min gi tr ca tnhiu thnh q khong bng nhau vi q=2v. Cc gi tr thay th chnh l im giaca mi khong chia.Mi gi tr lng t ny li c biu din bng mt m tnh phn c chiudi v bit.Xt V D:

Bng phng php iu ch PCM.thc hin lng t ha cho tn hiu x(t)=sint,vi

mc lng t l 12, di cc khong chia bng nhau,V tn hiu trc v saulng t.Tn hiu x(t)=sint c min gi tr l [-1.1] nn ta chia thnh cc khong chia c di 0.2.t = [0:.1:2*pi]; % Cc thiim ly mu tn hiu

sinesig = sin(t); % Tn hiu sine cha lng t

partition = [-1:.2:1]; % Phn hoch thnh 12 khong chia

codebook = [-1.2:.2:1]; % B m lng t gm 12 mc

[index,quants] = quantiz(sig,partition,codebook); % Lng tho.

plot(t,sig,'x',t,quants,'.')

legend(Tin hieu goc','Tin hieu sau khi luong tu hoa');

axis([-.2 7 -1.21.2])

Kt qu thu c minh ha trong hnh di:


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Hnh 1

2.2:Ti u ha cc thng s ca qu trnh lng t.

Qu trnh lng t lm pht sinh sai s gia tn hiu lng t vi tn hiu thc.Sai s ny cgi l nhiu lng t .Ta c th nh gi nhiu lng t bng hm quantiz cp

phn trc .Vn t ra y l lm th no ti thiu ha loi nhiu ny.Nhiu lng t ph

thuc vo cch phn hoch v tp cc gi tr lng t c chn.Nu s khong chia cng nhiuth nhiu lng t cng nh.Tuy nhin gii php ny khng phi lc no cng thc hin c donhng hn ch khch quan trong h thng.Mt phng php n gin hn ti thiu hanhiu lng t mc chp nhn c m khng cn phn hoch tn hiu thnh qu nhiukhong l phng php ti u ha theo quy lut Lloyd.Gii thut ny da trn nguyn tchun luyn,ngha l cc thng s ca qu trnh lng t s c chn v cp nht mt cch thchnghi vi tng loi tn hiu a vo lng t.

Qa trnh ti u ha trn c thc hin bng hm lloyd ca MATLAB communicationsToolbox.

>> [partition, codebook] = lloyds(training_set, ini_codebook, tol)hoc:>> [partition, codebook, distortion, rel_distortion] = lloyds(...)

training_set l tp d liu dng hun luyn. l mt tn hiu tiu biu ca nguntin tc m ta cn lng tho;


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ini_codebook l b m khi u do ta chn;

tol l sai s chophp ca qu trnh hun luyn (gi tr mc nh nu khng nhp thng sny l 10-7.Cc ktqu xut ra gm phn hoch v b m ti u ho (partition vcodebook ),nhiu lng t tuyt i (distortion) va tngi (rel-distortion) caqu trnh lng t.

V d sau y minh ha s ti u ha ca cc thng s ca qu trnh lng tdng hm lloyd trong MATLAB.Nhiu lng t pht sinh t hai qu trnh lng tha cha ti u v ti u s c so snh vi nhau.Xt v dThc hin li qu trnh lng t ha tn hiu x(t)=sint.vi cc thng s lng t c ti u ha.So snh nhiu lng t trc v sau ti u ha.% Btu vi cc thng slng t nh v d 14-2

t = [0:.1:2*pi];

sig = sin(t);

partition = [-1:.2:1];

codebook = [-1.2:.2:1];

% Thc hin ti u ho vi b m khiu l codebook.

[partition2,codebook2] = lloyds(sig,codebook);

[index,quants,distor] = quantiz(sig,partition,codebook);

[index2,quant2,distor2] = quantiz(sig,partition2,codebook2);

%So snh nhiu lng ttrc v sau khi tu ho

[distor, distor2]

Kt qu xut ra ca s lnh ca MATLAB:

ans =

0.0148 0.0024


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Sau khi ti u ha nhiu lng t gim i l: 0.0148/0.0024= 6.2 ln.2.3:Nn v dn tn hiu

Trong cc ng dng x l tn hiu thoi (speech processing), trc khi lng tho,ngi tathng thc hin nn (compress) tn hiu theo hm logarithm, mc chl tn hiu mcbin nh s thayi nhiu mc hn so vicc gi tr bin ln. do sai s lng ttng i cc mc bin nh v ln s khng chnhlch nhau nhiu nh i vi trng hp khng nn.

khi phcli ng tn hiu ban u th sau khi gii m, ta phi a qua mtbgin tn hiu (expander) c c tuyn truyn t l nghch o ca c tuyn ca bnn (compressor).S kt hp ca b nn v b gin tn hiu gi chung l b nn gintn hiu (compander).

Hai lut nn gin thng c s dng trong x l tn hiu thoi l lut dngBc M v lut A dng chu u:

MATLAB cung cp hm Compand thc hin nn gin d liu.Hm ny h trhai lut nn gin A v ni trn.

>> out = compand(in, param, v, method)

In l tn hiu vo cn out l tn hiu raV l bin nh ca tn hiu voParam l thng s ca lut nn gin


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Method c th nhn mt trong cc gi tr sau:

Vd .Thc hin qu trnh lng tho tn hiu hm m theo hai phngphp:lng tho u v l ng tho c nn gin theo lut. So snh saisbnh phng trung bnh ca hai phng php.

Vi phng php u tin ta s dng trc tip hm quantizi vi tn hiu lytha. Tn hiu c phn hoch thnh cc khongu nhau c chiu di bng 1.

Vi phng php th hai , u tin ta s dng hm compand nn tn hiu theolut , sau dng hm quantiz lng t ho tn hiu thu c v cui cng lidng hmcompand phc hi tn hiu ban u.

Kt qu xut ra gm hai sai s bnh phng trung bnh ca hai phng php v thbiu din tn hiu gc v tn hiu c nn .C th thy rng sai s lng t phng php th hai nh hn so vi phng php th nht.

Mu = 255; % Thng scho bnn gin theo lut

sig = -4:.1:4;

sig = exp(sig); % Tn hiu hm mcn lng t

V = max(sig);

% 1. Lng thou (khng nngin) .

[index,quants,distor] = quantiz(sig,0:floor(V),0:ceil(V));

% 2. Sdng cng phnhoch v bm lng tnhng c nn trc khi lngt% ho v gii nn sau

compsig = compand(sig,Mu,V,'mu/compressor');


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[index,quants] =quantiz(compsig,0:floor(V),0:ceil(V));

newsig =compand(quants,Mu,max(quants),'mu/expander');

distor2 = sum((newsigsig). 2)/length(sig);

[distor, distor2] % Hin thsai slng ta hai phngphp.

plot(sig); % Vtn hiu gc .

hold on;

plot(compsig,'r--'); % Vtn hiu nn gin.

legend('Tin hieu goc','Tin hieu nen gian','Location','NorthWest')

Sai s lng t:

ans =0.5348 0.0397

th dng tn hiu:

BNG NH GI CC THNH VIN TRONG NHM


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I: Danh sch cc thnh vin nhm 5.

1: Nguyn Quc QunNhm Trng

2: Nguyn Vn Hng

3: Nguyn Vn Khoi

4: Hong Vn Lm

5: Nguyn Vn Tin Lm

II: Ni dung ng gp ca cc thnh vin trong nhm.

Phn 1: Cc Phng Php M Ha v iu Ch.Nguyn Quc Qun vi Nguyn Vn HngPhn 2: iu ChXung M PCMC Nhm

Phn 3: Gii Thiu Mt SPhng Php Mi

C Nhm

Phn 4: M Phng PCM Trong MathLab

Cnhm

Chu trch nhim tng hp v Thit kni dung

Nhm Trng v Nguyn Vn Hng

III: nh Ga Ca Trng Nhm

Qua qu trnh lm vic th em thy cc thnh vin trong nhm cng nhit

tnh v tt ccc thnh vin cng nlc ht mnh v cng vic ca cnhm. Nhng i lc cng vn cn tnh trng da dm ni vo cc thnhvin khc trong nhm, nhng cc bn cng rt hng hi v nhit tnh lmvic khi c nhm trng triu tp v giao nhim v. Mi thnh vin cng hon thnh tt nhim vca mnh khi trng nhm giao v khon cngvic cho mi thnh vin. Chnh v vy m em xin c nh gi cc thnh


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vin nh sau.Theo s nh gi ca em th chmang tnh c nhn, tc l snhn xt cc thnh vin l hon ton nm trong snhn nhn ca c nhn em.Cha phi l ca cnhm, vy mong thy chly bi nh gi ca em tham kho thi . Di y l bng nhn xt v nh gi ca em.

H ni ngy 15 thng 11 nm 2012

Tn Thnh Vin Mc HonThnh CngVic cGiao

Thi LmVic

im

Nguyn Quc Qun 10/10 10/10 10/10Nguyn Vn Hng 10/10 10/10 10/10

Nguyn Vn Khoi 10/10 10/10 10/10Hong Vn Lm 10/10 10/10 10/10Nguyn Vn TinLm

9/10 10/10 9,5/10

VI: Ti Liu Tham Kho

[1].Gio trnh Matlab v ng dng trong vin thng-TS.Phm Hng Lin

[2].Thc Hnh XL STn Hiu Vi Matlab-TS.HVn Sung.[3].Lun Vn Tt Nghip Kho St Tn Hiu iu ChBng Matlab-DDHQGTP.HCM

[4].Webside:http://www.mathworks.com/

v2 0 Bao Cao Dieu Che Pcm Nhom 10 Chu de 5 Quan 1112

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