Ky thuat truyen so lieu.doc

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K THUT TRUYN S LIU NGUYN QUC NAM

MN: K THUT TRUYN S LIU

N TP

I. CC KHI NIM C BN1. TC BIT V TC BAUD

Tc bit l s bit trong mi giy.

Tc baud l s n v tn hiu trong mi giy.

Tc baud thng b hn hay bng tc bit.Tc baud xc nh bng thng cn thit truyn tn hiu.Tc bit = tc baud . s bit trong mt n v tn hiuRbit= Rbaud.n

n = log2M

Rbit = log2M / Ts = m/ Ts = 1/ Tb

* Ts : chu k tn hiu

* Tb : thi gian ca mi bit

Th d 1:

Mt tn hiu analog mang 4 bit trong mi phn t tn hiu. Nu 1000 phn t tn hiu c gi trong mt giy, xc nh tc baud v tc bit.

Gii:

Tc baud = s n v tn hiu trong 1giy = 1000 baud/giy

Tc bit = tc baud x s bit trong mt n v tn hiu =1000 x 4 = 4000bps.

Th d 2:

Tc bit ca tn hiu l 3000. Nu mi phn t tn hiu mang 6 bit, cho bit tc baud?

Gii:

Tc baud = tc bit/ s bit trong mi phn t tn hiu = 3000/6 =500 baud/giyV d 3: Mt tn hiu s c 8 mc. Cho bit c th truyn bao nhiu bit cho mi mc?

S bit trong mt mc = log2 8 = 3.

Nh th mi mc tn hiu c th truyn c c 3 bit.

V d 4:

Mt tn hiu s c chn mc. Cho bit c th truyn bao nhiu bit cho mi mc?

Gii:

S bit trong mt mc = log29= 3,17.

Mi mc tn hiu truyn i c 3,17 bit. R rng l kt qu ny l khng hin thc. S bit cn truyn i cn l s nguyn l tng ng vi s m 2.

Trong trng hp ny th nn dng 4 bit cho mi mc tn hiu truyn.V d 5:

Mt knh thoi c ri rc ha, c cu to t mt tn hiu tng t c bng thng tn hiu thoi l 4 KHz. ta cn ly mu tn hiu vi hai ln tn s cao nht (hai mu trong mi Hz). Vi gi s mi mu cn 8 bit, hi tc bit l bao nhiu?

Gii:

Tc bit c tnh theo: 2 x 4000 x 8 = 64000 bps = 64 Kbps.

2. TN HAO NG TRUYN (TRANSMISSSION IMPAIRMENT)

C 3 dng tn hao: suy gim, mo dng, nhiu.3.1 Suy gim (Attenuation): L tht thot nng lng.

b suy hao, dng b khuch i tn hiu.

deciBel (dB): c dng o mnh tng i ca hai tn hiu ti hai im khc nhau.

Khi dB m ( tn hiu b suy gim

Khi dB dng ( tn hiu c khuch i.

+ suy gim

Trong : P1 l cng sut pht (im 1)

P2 l cng sut thu (im 2)

+ Khuch i

Trong : P1 l cng sut vo b khuch i (im 2)

P2 l cng sut ra b khuch i (im 3)

V d : Cho mt tn hiu i qua mi trng truyn, cng sut b gim mt na. Hy tnh suy gim theo deciBel (dB).

suy gim =

(-3dB tc l gim i 3 dB, tc l cng sut gim i mt na)V d : Tn hiu c khuch i 10 ln, tc l P2 = 10P1.

khuch i

V d: Mt trong nhng yu t s dng dB l dng php tnh cng trong qu trnh tnh ton tn hao ti nhiu im ni ui nhau.

Tng deciBel ca ng truyn:

tn hiu c khuch i.+ DUNG LNG KNH SHANNON:

Dung lng knh: Xc nh tc truyn d liu cc i theo l thuyt ca mt knh truyn C = B log2(1+S/N) Trong :

C[bps]: Dung lng knh.

B[Hz] : Bng thng ca knh truyn.

S/N : T s cng sut tn hiu trn cng sut nhiu.

S[W]: Cng sut tn hiu; N[W]: Cng sut nhiu.

V d 1: Cho c mt knh truyn rt nhiu nhiu (N = ), t s S/N gn bng 0, nhiu qu mnh lm yu tn hiu. Nh th, dung lng truyn lc ny l: C = B log2(1+S/N)= B log2(1+ 0)= B log2(1)= B . 0= 0

iu ny tc l dung lng knh truyn l zr, bt k bng thng, tc l ta khng th truyn tin qua knh ny.

V d 2: Tnh tc bit cao nht l thuyt ca mt ng cp UTP, vi bng thng 3000Hz, t s S/N l 3162 ln (35 dB).

Nh th, dung lng truyn l thuyt cao nht l:C=Blog2(1+S/N)=3000 log2(1+3162)=3000 log2(3163)= 3000 x11,62= 34.860bps =34,86kbpsNh th, nu mun tng tc truyn d liu trong ng dy UTP, th phi mt l tng bng thng hay ci thin t s S/N.i t dB sang s ln hoc ngc li:

V d: (Sinh vin t lm):

1. o lng hiu nng ca ng dy cp UTP (bng thng 4 KHz), khi tn hiu l 10 volt th nhiu l 5 volt. Tc truyn d liu ti a l bao nhiu ?2. Mt ng dy c t s tn hiu trn nhiu (S/N) l 1000 ln v bng thng l 4000 Hz, tnh tc truyn d liu ti a theo Shannon? 3. Mt tn hiu i t im A n im B. Ti im A, cng sut ca tn hiu l 100 watt, ti im B cng sut cn li 90 watt, tnh suy hao theo dB?

4. Mt knh truyn c suy hao l 10 dB. Khi cho tn hiu 5 watt i qua th cng sut thu bao nhiu?

5. Mt tn hiu i qua ba b khuch i ni ui nhau, mi b c li 4 dB. Hy cho bit li tng? Tn hiu c khuch i bao nhiu ln? 3. TM BNG THNG :Bng thng ca tn hiu hn hp l sai bit gia tn s cao nht v thp nht c trong tn hiu ny.

Th d 1:

Nu phn tch tn hiu tun hon thnh 5 sng hi sin c tn s ln lt l 100, 300, 500, 700 v 900 Hz. Cho bit bng thng ca tn hiu? V ph vi gi s l tt c sng hi u c gi tr ln nht l 10V.

Gii:

Gi fh l tn s cao nht, fl l thp nht, v B l bng thng, th

B = fh - fl = 900 100 = 800 Hz

Ph ch gm 5 gai nhn xut hin ti cc tn s 100, 300, 500, 700 v 900 Hz .

Th d 2: Tn hiu tun hon c bng thng l 20 Hz. Tn s cao nht l 60 Hz, tm tn s thp nht? V ph ca tn hiu sng hi c bin ging nhau.

Gii:

Gi fh l tn s cao nht, fl l thp nht, v B l kh sng, th

B = fh - fl ( 20 = 60 fl ( fl =60 20 = 40 Hz

Ph cha tt c cc tn s c gi tr nguyn.

Th d 3: Mt tn hiu hn hp khng tun hon c bng thng l 200 kHz, c tn s trung tm l 140 kHz, v bin nh l 20 V. Hai gi tr bin ti hai tn s cc tr l 0. V tn hiu trong min tn s.Gii: Tn s thp nht phi l 40 kHz v tn s cao nht l 240 kHz.

a) Bng thng dng cho ASK(amplitude shift keying)::Tn hiu sng mang (carrier signal):

Trong truyn dn analog th thit b pht to ra tn s sng cao tn lm nn cho tn hiu thng tin. Tn hiu nn ny c gi l sng mang hay tn s sng mang. Thit b thu c chnh nh thu tn s sng mang trong c tn hiu s c iu ch v tn hiu mang thng tin c gi l tn hiu iu ch.

Khi phn tch ph tn hiu iu ch ASK, ta c gi tr ph nh hnh v trong c cc yu t quan trng l sng mang fc gia, cc gi tr fc Nbaud/2 v fc + Nbaud/2 hai bin.

Bng thng cn thit cho ASK c tnh theo:

BW = (1+d) . Nbaud= (1+d) . Rbaud RbaudTrong :

BW: bng thng

Rbaud, Nbaud: tc baud

d: l tha s lin quan n iu kin ng dy (c gi tr b nht l 0)Ta s thy l bng thng ti thiu cn cho qu trnh truyn th bng tc baud.

Th d 4 : Tm bng thng ca tn hiu ASK truyn vi tc bit 2 kbps. Ch truyn bn song cng.

Gii: Trong ASK, tc bit bng tc baud. Tc baud l 2000, nn tn hiu ASK cn c bng thng ti thiu bng tc baud. Nh th, bng thng ti thiu l 2000 Hz.Th d 5: Tn hiu ASK c bng thng l 5000 Hz, tm tc bit v tc baud?

Gii: Trong ASK th tc baud bng bng thng, tc l tc baud l 5000, ng thi do tc bit bng tc baud nn tc bit l 5000 bps.

Th d 10: Cho bng thng h thng truyn ASK 10 kHz (1 kHz n 11 kHz), v ph ASK song cng ca h thng. Tm tn s sng mang v bng thng ca mi hng, gi s khng c khong trng tn s gia hai hng.

Gii: Do h thng ASK song cng nn bng thng trong mi chiu l

BWmi hng = (1/2). BWh thng = 10khz / 2 = 5khz = 5.000 Hz

Tn s sng mang l tn s gia, nh hnh 5.26:

fc thun = 1.000 + 5.000/2 = 3500 Hz

fc nghch = 11.000 - 5.000/2 = 8500 Hz

b) Bng thng ca FSK:

Bng thng cn thit truyn dn FSK chnh l tc baud ca tn hiu cng vi dch tn s (sai bit gia hai tn s sng mang hnh v

BW = /fC0 - fC1/+ Nbaud = (f + NbaudTuy ch c hai tn s sng mang, nhng qu trnh iu ch cng to ra tn hiu hn hp l t hp ca nhiu tn hiu n gin, vi cc tn s khc nhau.Th d 11: Tm bng thng ti thiu ca tn hiu FSK truyn vi tc bit 2kbps. Ch truyn dn bn song cng v cc sng mang cch 3kHz.

Gii: Tn hiu FSK dng hai tn s fC0 v fC1, nn; BW = (fC1 - fC0 )+ Tc baudDo trong trng hp ny th tc bit bng tc baud, nn

BW = (fC1 - fC0 )+ Tc baud = (3.000) + 2.000 = 5.000 HzTh d 12: Tm tc bit ln nht ca tn hiu FSK nu bng thng ca mi trng l 12khz v sai bit gia hai sng mang t nht l 2kHz, ch truyn song cng.

Gii: Vi ch truyn song cng, th ch c 6.000 Hz l c truyn theo mi hng (thu hay pht). i vi FSK, khi c fC1 v fC0 l tn s sng mang.

BW = (fC1 - fC0 )+ Tc baudNn Tc baud = BW - (fC1 - fC0 ) = 6.000 2.000 = 4.000 baud/sng thi, do tc baud bng tc bit nn tc bit cng l 4.000 bpsKhi truyn dn trn di tn c s th bng thng cn thit l t l vi tc bit (bit rate); nu ta mun truyn bit nhanh hn, th cn phi c bng thng rng hn.

Th d 5: Tm bng thng cn c ca knh truyn thng tn s thp nu cn gi vi tc 1 Mbps dng phng php truyn trn di tn c s.

Gii:

Li gii cn ty theo mc chnh xc cn c:

a. Bng thng ti thiu, l B = (tc bit)/2, tc l 500 KHz

b. Tt hn th dng hi bc mt v bc ba, tc l B = 3 x 500KHz = 1,5 MHz

c. Tt hn na l hi bc mt, bc ba v bc nm, B = 5 x 500 KHz = 2,5 MHz

Th d 6: Ta dng hai knh thng tn s thp c bng thng l 100 KHz, cho bit tc truyn bit ti a l bao nhiu?

Gii: Tc truyn bit ti a c th t c nu ta dng sng hi bc mt.

Tc bit l 2 x (bng thng hin c), tc l 200 Kbps.

4. M HA V IU CH4.1 ASK (amplitude shift keying; iu ch s bin ):

+ Khi nim: L qa trnh ly cc bit 1 v 0 lm thay i bin ca tn hiu sng mang (tn s v pha khng thay i).

V d: Gi s c sng mang vc(t)=Vcmsin(2fct + 0);0( vc1(t)=Vcm1 sin(2fct + 1800)(Tn ti trong 1 chu k bit; mt loi n v tn hiu1( vc2(t)=Vcm2 sin(2fct + 1800)(Tn ti trong 1 chu k bit; mt loi n v tn hiuGi s Vcm2 > Vcm1;

V d: Cho mt tn hiu s c dng nh phn nh sau: 01010, bit tc bit l 5 bps. Tn hiu s c iu ch bng phng php ASK. Vi tn s sng mang l fc bng 20Hz, bin i vi bit 1 l 5V, bin i vi bit 0 l 2V v pha ban u ca sng mang l 1800.

a. Hy v tn tn hiu ASK.

b. Tn hiu ASK c phi l tn hiu iu ho hay khng? Gii thch.

c. Tnh tc Baud.

Gii:Cho: d liu, Rb, ASK, fc, bin , pha ban u.a. V tn tn hiu ASK.0( vc1(t)=2sin(2.20t+1800) V; Tn ti trong 1 chu k bit

1( vc2(t)=5sin(2.20t+1800) V; Tn ti trong 1 chu k bit

Chu k bit Tb=1/ Rb=1/5 = 200msChu k sng mang Tc=1/ fc =1/20 = 50msVy Tb= 4 Tc ( 1 chu k bit cha 4 chu k sng mang

b. Tn hiu ASK khng phi l tn hiu iu ho.V c 2 bin .c. Tc Baud: Nbaud = Rbaud= 5 baud/s(ASK( Rbit = Rbaud)V d: (Sinh vin t lm) Cho mt tn hiu s c dng nh phn nh sau: 110100000000, bit tc bit l 5 kbps. Tn hiu s c iu ch bng phng php ASK. Vi tn s sng mang l fc bng 10 kHz, bin i vi bit 1 l 5V, bin i vi bit 0 l 0V v pha ban u ca sng mang l 00.

a. V tn hiu s trn theo cc dng m: NRZ-L, Manchester v B8ZSb. Hy v tn hiu ASK trong 5 bit u tin. c. Tnh tc Baud.

+ Khuyt im: ASK thng rt nhy cm vi nhiu bin . Nhiu ny thng l cc tn hiu in p xut hin trn ng dy t cc ngun tn hiu khc nh hng c ln bin ca tn hiu ASK.

Phng php ASK thng dng v c gi l OOK (on-off keying). Trong OOK, c mt gi tr bit tng ng vi khng c in p. iu ny cho php tit kim ng k nng lng truyn tin.

+ Bng thng ASK: C v s tn s (khng tun hon). Sng mang fc gia, cc gi tr fc Nbaud/2 v fc + Nbaud/2 hai bin.

Bng thng cn thit truyn tn hiu ASK c tnh theo cng thc sau::

BW = fmax fmin = (fc + Nbaud/2) (fc Nbaud/2) = Nbaud= RbaudBWASK = RbaudTrong : BW: Bng thng [Hz]Rbaud, Nbaud: tc baud [baud/s]

Vy bng thng ti thiu cn cho qu trnh truyn tn hiu ASK bng tc baud (1 hng-trn ng dy).Thc t BW =(1+d)Nbaud; d: l tha s lin quan n iu kin ng dy (c gi tr b nht l 0)V d: Cho mt tn hiu s 01010, tc bit l 5 bps, c iu ch bng phng php ASK. Tn s sng mang fc= 20Hz. Bin i vi bit 1 l 5V, bin i vi bit 0 l 2V. Pha ban u ca sng mang l 1800.

a. Tnh tc Baud.

b. Tnh bng thng ca tn hiu ASK trn.

c. V ph ca tn hiu ASK trn.

Gii:

a. Tnh tc Baud.

Tn hiu ASK, Rbaud= Rbit=5 baud/sb. Tnh bng thng ca tn hiu ASK trn.

ASK, BW = Rbaud=5 (Hz);c. V ph ca tn hiu ASK trn.

V d: (Sinh vin t lm) Cho mt tn hiu s c dng nh phn nh sau: 1101000011110, bit tc bit l 5 kbps. Tn hiu s c iu ch bng phng php ASK. Vi tn s sng mang l fc bng 10 kHz, bin i vi bit 1 l 5V, bin i vi bit 0 l 0V v pha ban u ca sng mang l 900.

a. V tn hiu s trn theo cc dng m: NRZ-I, Manchester vi sai v HDB3.

b. Hy v tn tn hiu ASK trong 5 bit cui.

c. Tnh tc Baud. (5000 Baud/s)d. Tnh bng thng ca tn hiu ASK trn. (5 kHz)e. V ph ca tn hiu ASK trn.

+ Bng thng ca h thng truyn tn hiu thay i theo ch truyn: ng dy c 1 hng truyn (ch n cng): bng thng ca ng dy ti thiu bng bng thng ca tn hiu: BWh thng min = BWng dy min = BWtn hiu.

ng dy c 2 hng truyn nhng khng ng thi (ch bn song cng): bng thng ca ng dy ti thiu bng bng thng ca tn hiu: BWh thng = BWng dy = BWtn hiu= BWmi hng .

ng dy c 2 hng truyn ng thi (ch song cng): bng thng ca ng dy ti thiu: . BWhthng=BWngdymin =BWhng1 +BWhng2+BWbo v=2.BWtn hiu+BWbo vBWbo v: di tn s bo v 2 hng.(l tng bng 0)V d: Tnh bng thng h thng truyn tn hiu ASK vi tc bit l 2 kbps. Ch truyn dn bn song cng.

Gii:

V h thng bn song cng nn: BWh thng = BWmi hng

V iu ch ASK nn Rbit = Rbaud x 1= RbaudSuy ra BWmi hng = BWASK = Rbaud = Rbit = 2000Hz

Bng thng ti thiu ca h thng l BWh thng =2kHz.

V d (Sinh vin t lm): Tnh bng thng h thng truyn tn hiu ASK vi tc bit l 2 kbps. Ch truyn dn song cng.

V d : Cho tn hiu ASK c bng thng 5kHz, tnh tc bit v tc baud.Gii: 5kHz=5000Hz;

V iu ch ASK nn Rbit = Rbaud x n = Rbaud x 1= Rbaud.M BWASK = Rbaud ;

Suy ra tc baud Rbaud = BWASK =5000 baud/s;

Suy ra tc bit Rbit = Rbaud =5000 bps;

(C BWASK (tnh c Rbit , Rbaud )

V d: Cho bng thng ca h thng truyn d liu ASK l 10 kHz (fmin=1 kHz n fmax=11 kHz), ch truyn song cng. Gi s khng c khong trng tn s gia hai hng (BWbo v=0).

a. V ph ca h thng ASK trn.

b. Tnh bng thng ca mi hng.

c. Tnh tn s sng mang mi hng (Hng thun v hng nghch).

Gii:

Cho BWh thng = 10kHz; fmax = 11kHz =11.000 Hz; fmin = 1kHz = 1000Hz ; song cng; BWbo v = 0

a. V ph ASK ca h thng song cng

b. Tnh bng thng ca mi hng.

Do h thng ASK song cng nn BWh thng = 2. BWmi hng + BWbo vSuy ra BWmi hng = BWtn hiu ASK = (1/2). BWh thng = 10kHz / 2 = 5kHz = 5.000 Hz

c. Tnh tn s sng mang mi hng (fc2= Hng thun v fc1= hng nghch).

Tn s sng mang l tn s gia:+ fc1 Hng nghch (tn s thp):

fmin = fc1 Rbaud /2 = fc1 BWASK/2 = fc1 BWmi hng /2

Suy ra fchng nghch = fc1 = fmin+ (1/2). BWmi hng = 1.000 + 5.000/2 = 3500 Hz+ fc2 Hng thun (tn s cao):

fmax = fc2 + Rbaud /2 = fc2 + BWASK/2 = fc2 + BWmi hng /2

Suy ra fchng thun = fc2 = fmax - (1/2). BWmi hng = 11.000 - 5.000/2 = 8500 Hz

V d: (Sinh vin t lm) Cho mt tn hiu s c dng nh phn nh sau: 1101000011110. Tn hiu s c iu ch bng phng php ASK. H thng truyn song cng vi bng thng l 10 kHz (fmin=2,5 kHz n fmax=12,5 kHz). Gi s khng c khong trng tn s gia hai hng (BWbo v=0). Bit bin i vi bit 1 l 5V, bin i vi bit 0 l 0V v pha ban u ca sng mang l 900.

a. V tn hiu s trn theo cc dng m: Manchester vi sai v AMI. Cc dng m ho trn dng tn hiu no c kh nng ng b v thnh phn DC bng zr.b. V ph ca h thng ASK trn.

c. Tnh bng thng ca mi hng. (5000Hz)

d. Tnh tn s sng mang mi hng (Hng thun v hng nghch).

e. Hy v tn hiu ASK trong 5 bit cui cho hng thun.( fc2=10kHz)5.3.2 FSK (frequency shift keying):+Khi nim: L phng php m tn s ca tn hiu sng mang thay i biu din cc bit 1 v 0 (bin v gc pha khng thay i).

V d: gi s c sng mang vc(t)=Vcmsin(2fct + 0);Bit 0( ng vi sng mang vc1(t) = Vcm sin(2fc1t+1800); Tn ti trong 1 chu k bit

Bit 1( ng vi sng mang vc2(t) = Vcm sin(2fc2t+1800): Tn ti trong 1 chu k bit

Gi s fc2 > fc1;

V d: Cho mt tn hiu s 01101, tc bit l 5 bps, c iu ch bng phng php FSK. Bin sng mang l 5V, tn s i vi bit 1 l 20Hz, tn s i vi bit 0 l 10Hz v pha ban u ca sng mang l 1800.

a. V tn tn hiu FSK.b. Tn hiu FSK c phi l tn hiu iu ho hay khng? Gii thch.

c. Tnh tc Baud.

Gii:

a. V tn tn hiu FSK

0( vc1(t) = Vcm sin(2fc1t+1800)=5sin(2.10t+1800) V; Tn ti trong 1 chu k bit

1( vc2(t) = Vcm sin(2fc2t+1800)=5sin(2.20t+1800) V; Tn ti trong 1 chu k bit

Chu k bit Tb=1/ Rb=1/5 = 200ms

Chu k sng mang bit 0; Tc1=1/ fc1=1/10 =0,1s= 100ms

Chu k sng mang bit 1; Tc2=1/ fc2=1/20 = =0,05s= 50ms

Vy Tb= 2Tc1 =4Tc2 ( 1 chu k bit cha 2 chu k sng mang fc1 v cha 4 chu k sng mang fc2.

b. Tn hiu FSK khng phi l tn hiu iu ho.V tn s thay i.Tng qut: FSK l tn hiu tng t khng tun honc. Tc Baud: Mt n v tn hiu mang 1 bit nn Rbit = RbaudSuy ra Rbaud= 5 baud/s

Tn hiu FSK : Rbit = RbaudV d: (Sinh vin t lm) Cho mt tn hiu s c dng nh phn nh sau: 110100000000, bit tc bit l 5 kbps. Tn hiu s c iu ch bng phng php FSK. Bin sng mang l 5V, tn s i vi bit 1 l 20kHz, tn s i vi bit 0 l 10kHz v pha ban u ca sng mang l 00.

a. V tn hiu s trn theo cc dng m: NRZ-L, Manchester v B8ZS. (5.1)b. Hy v tn tn hiu FSK trong 5 bit u tin.

c. Tnh tc Baud.( Rbit = Rbaud)

d. Tn hiu FSK c phi l tn hiu tun hon hay khng? Gii thch.

+ Bng thng ca tn hiu FSK:

Ph FSK l t hp ca hai ph ASK tp trung quanh 2 tn s: fC1 (bit 0) v fC2 (bit 1).

Bng thng ca tn hiu FSK: BWFSK = fmax fminBWFSK = fmax fmin = fC2 + (1/2)Rbaud -[ fC1- (1/2)Rbaud ] BWFSK = /fC2 - fC1/+ Rbaud = (f + Nbaud = (f + Rbaud

BWFSK = (f + Rbaud ; ( so snh Bng thng ca tn hiu ASK BWASK =Rbaud ); (f =/fC2 - fC1/ : lch tn s ca 2 sng mang

Nbaud = Rbaud: Tc baud

Rbit = RbaudVy nu bit lch tn s ca 2 sng mang v tc baud th ta c th xc nh bng thng ca tn hiu FSK.V d: Cho mt tn hiu s 01101, tc bit l 5 bps, c iu ch bng phng php FSK. Bin sng mang l 5V, tn s i vi bit 1 l 20Hz, tn s i vi bit 0 l 10Hz v pha ban u ca sng mang l 1800.

a. Tnh tc Baud.b. Tnh bng thng ca tn hiu FSK trn.

c. V ph ca tn hiu FSK trn.Gii:

a. Tnh tc Baud.

FSK, Rbaud = Rbit = 5baud/sb. Tnh bng thng ca tn hiu FSK trn.BWFSK = (f + Rbaud = /20 10/ + 5 = 15Hzc. V ph ca tn hiu FSK trn.

So snh dng ph ca ASK v FSK:

+ u im FSK so vi ASK : FSK trnh c hu ht cc dng nhiu bin .

+ Khuyt im FSK so vi ASK: Nu cng mt tc bit th Bng thng FSK ln hn Bng thng ASK.

V d 11: Tnh bng thng nh nht ca h thng FSK, bit tc bit 2kbps, ch truyn dn bn song cng v cc sng mang cch nhau 3kHz ( lch tn s).

Gii:

Tm tt:

Rbit = 2kbps= 2000bps ;

(f = 3kHz = 3000Hz;

Ch truyn bn song cng; Tnh BWh thng min= ?V h thng truyn bn song cng nn:

BWh thng min= BWmi hng= BWFSK =(f + Rbaud Trong FSK, Rbit =Rbaud ; suy ra Rbaud = 2000 baud/sSuy ra BWh thng min = (f + Rbaud = 3.000 + 2.000 = 5.000 Hz = 5 kHz;

Vy bng thng nh nht ca h thng FSK trn l 5 kHz.

V d (SV t lm): Tnh bng thng nh nht ca h thng FSK, bit tc bit 2kbps, ch truyn dn song cng v cc sng mang cch nhau 3kHz ( lch tn s).V d (SV t lm): Tnh bng thng ln nht ca h thng FSK, bit tc bit 2kbps, ch truyn dn song cng v cc sng mang cch nhau 3kHz ( lch tn s).V d (SV t lm): Tnh bng thng nh nht ca h thng FSK, bit tc bit 2kbps, ch truyn dn song cng v cc sng mang cch nhau 3kHz ( lch tn s), bng thng bo v l 1kHz.V d 12: Tnh tc bit cc i ca tn hiu FSK nu bng thng ca h thng l 12kHz v lch tn s ca gia hai sng mang t nht l 2kHz, ch truyn song cng.

+ Tm tt: H thng FSK; Ch truyn song cng; (fmin = 2kHz= 2000Hz; BWh thng = 12khz=12.000Hz; Tnh Rbit max ?

+ Gii: Xc nh BWFSK= BWmi hngV h thng truyn song cng nn BWh thng= 2.BWmi hng + BWbo v (BWbo v=0)Suy ra: BWmi hng= BWFSK =(1/2)BWh thng = 12kHz/2 = 6kHz= 6.000Hz

Xc nh RbitTrong FSK: BWFSK = BWmi hng = (f + Rbaud ; Rbit = Rbaud.Suy ra Rbit= BWmi hng - (f (1)

Theo (1) Khi (fmin( Rbit Max Suy ra Rbit Max = BWmi hng - (fmin= 6.000 2.000 = 4.000 bps = 4 kbps

Vy tc bit cc i ca tn hiu FSK l 4 kbps.

V d (Sinh vin t lm): Tnh tc bit nh nht ca tn hiu FSK nu bng thng ca h thng l 12kHz v lch tn s ca gia hai sng mang ln nht l 2kHz, ch truyn song cng.

V d (Sinh vin t lm): Tnh tc bit nh nht ca tn hiu FSK nu bng thng ca h thng l 12kHz v lch tn s ca gia hai sng mang ln nht l 2kHz, ch truyn bn song cng.V d: (Sinh vin t lm) Cho mt tn hiu s c dng nh phn nh sau: 1101000011110. Tn hiu s c iu ch bng phng php FSK, lch tn s l 5kHz. H thng truyn song cng vi bng thng l 22,5 kHz (fmin=2,5 kHz n fmax=25 kHz). Gi s khong trng tn s gia hai hng l 2,5 kHz (BWbo v=2,5 kHz). Bin sng mang l 5V, pha ban u ca sng mang l 900.

a. V tn hiu s trn theo cc dng m: RZ, Manchester vi sai v AMI. Cc dng m ho trn dng tn hiu no c kh nng ng b v thnh phn DC bng zr.

b. V ph ca h thng FSK trn.

c. Tnh bng thng ca mi hng. (10kHz)

d. Tnh tn s sng mang mi hng (Hng thun v hng nghch).

e. Hy v tn tn hiu FSK trong 5 bit cui cho hng tn s thp, tn s i vi bit 1 l 10kHz, tn s i vi bit 0 l 5kHz.f. V ph y h thng FSK trn.5.3.3 PSK (phase shift keying):

+Khi nim: Pha ca sng mang thay i biu din cc bit 1 v 0 (bin v tn s khng i).

V d: gi s c sng mang vc(t)=Vcmsin(2fct + 0);0( vc1(t)=Vcm sin(2fct+ 01) ; Tn ti trong 1 chu k bit

1( vc2(t)=Vcm sin(2fct+ 02) ; Tn ti trong 1 chu k bitThng thng ta chn 02 , 01 sao cho lch pha ca 2 tn hiu l ln nht

max = /02 - 01/= 1800V d: Cho mt tn hiu s 01101, tc bit l 5 bps, c iu ch bng phng php PSK. Bin 5V. Tn s sng mang 20Hz. Pha i vi bit 1 l 1800, pha i vi bit 0 l 00.

a. V tn tn hiu PSK.

b. Tn hiu PSK c phi l tn hiu iu ho hay khng? Gii thch.

c. Tnh tc Baud.

Gii:

a. V tn tn hiu PSK

0( vc1(t) = Vcm sin(2fct + 01) =5 sin(2.20t + 00) V ; Tn ti trong 1 chu k bit

1( vc2(t) = Vcm sin(2fct + 02) = 5 sin(2.20t + 1800) V; Tn ti trong 1 chu k bit

Chu k bit Tb=1/ Rb=1/5 = 200ms

Chu k sng mang Tc=1/ fc=1/20 = 50ms

Vy Tb= 4Tc ( 1 chu k bit cha 4 chu k sng mang fc.

b. Tn hiu PSK khng phi l tn hiu iu ha.V c 2 pha.c. Tc Baud: Nbaud = Rbaud= Rbit =5 baud/sVy trong PSK hoc 2-PSK tc bit bng tc baud; Rbaud= Rbit2-PSK (BPSK; Binary): Tn hiu PSK c 2 pha. ( c 2 loi tn hiu, mi loi cha 1 bit)

2n-PSK : Tn hiu PSK c 2n pha. ( c 2n loi tn hiu, mi loi cha n bit)V d:

Tn hiu 4 PSK (22 PSK) : Tn hiu PSK c 22 =4 pha. ( c 4 loi tn hiu, mi loi cha 2 bit)

Tn hiu 8 PSK (23 PSK) : Tn hiu PSK c 23 =8 pha. ( c 8 loi tn hiu, mi loi cha 3 bit)

+ Bng thng ca PSK:

Dng ph ca PSK ging ASK nn Bng thng ca PSK ging bng thng ASK

BW2-PSK = Rbaud ;Nbaud=Rbaud: Tc baudTng qut: BW2n -PSK = Rbaud

+ u im PSK (2-PSK, BPSK): Khng b nh hng nhiu bin

Bng thng hp (nh hn bng thng ca FSK)

+ So snh bng thng v nh hng ca nhiu bin ln cc tn hiu PSK, ASK, FSK: BWASK = Rbaud ; nhiu bin ; BWFSK = (f + Rbaud ; khng b nh hng nhiu bin ; BWPSK = Rbaud ; khng b nh hng nhiu bin Vy tn hiu PSK khng b nh hng ca nhiu bin v c bng thng nh nhtTn hiu PSK kh iu ch hn ASK v FSK+ Gin trng thi pha: 0( vc1(t)= 5 sin(2.20t+00) V ; 1( vc2(t)= 5 sin(2.20t+1800)

Vy tn hiu PSK khng b nh hng ca cc dng nhiu tc ng nh ASK, ng thi cng khng b nh hng ca yu t bng thng rng nh FSK. iu ny c ngha l mt thay i nh ca tn hiu cng c th c my thu pht hin, nh th thay v ch dng hai thay i ca tn hiu t mt bit, ta c th dng vi bn s thay i thng qua dch pha ca hai bit.

+ 4-PSK (QPSK- Quadrature- cu phng):

Tn hiu 4 PSK (22 PSK) : Tn hiu 4-PSK c 22 =4 pha. ( c 4 loi tn hiu, mi loi cha 2 bit)

V d: Cho mt tn hiu s 0110101100, tc bit l 10 bps, c iu ch bng phng php 4-PSK(QPSK). Bit sng mang c bin 5V, tn s 20Hz v pha c biu din nh sau: 00 pha ( 00 ; 01 pha (900 ; 10 pha ( 1800 ; 11 pha (2700 (-900).a. V tn tn hiu QPSK.

b. Tn hiu QPSK c phi l tn hiu iu ho hay khng? Gii thch.

c. Tnh tc Baud.d. V gin trng thi pha ca tn hiu QPSK.Gii:

a. V tn tn hiu QPSKTn hiu QPSK ( 4-PSK) c 4 pha. ( c 4 loi tn hiu), mi pha mang thng tin 2 bit.Theo bi ta c cc loi tn hiu trong QPSK nh sau:00( vc1(t)= Vcm sin(2fct + 01) = 5 sin(2.20t+00) V ; Tn ti trong 2 chu k bit

01( vc2(t)= Vcm sin(2fct + 02) = 5 sin(2.20t+900)V; Tn ti trong 2 chu k bit

10( vc3(t)= Vcm sin(2fct + 03) = 5 sin(2.20t+1800)V ; Tn ti trong 2 chu k bit

11( vc4(t)= Vcm sin(2fct + 04) = 5 sin(2.20t -900); V Tn ti trong 2 chu k bit

Chu k bit Tb=1/Rb=1/10 =0,1s=100ms

Chu k sng mang Tc=1/ fc=1/20 =0,05s= 50ms

Vy Tb= 2Tc ( 1 chu bit cha 2 chu k sng mang fc.

Vy 2Tb= 4Tc ( 2 chu bit cha 4 chu k sng mang fc.

Vy tn hiu QPSK ca tn hiu s 0110101100 nh sau:

b. Tn hiu QPSK khng phi l tn hiu iu ha. V c pha thay i.c. Tc Baud: Nbaud = Rbaud= (1/2)Rbit =5 baud/sVy trong 4-PSK (QPSK) Rbit= 2.RbaudTng qut h thng 2n-PSK, Rbit= n.Rbaudd. Gin trng thi pha QPSK:

Tng qut h thng 2n-PSK, cc pha cch nhau 3600/2n+ Bng thng ca QPSK: Ging bng thng ASK

BW = Rbaud Nbaud=Rbaud: Tc baud+ u im QPSK: khng b nh hng nhiu bin , nu cng 1 bng thng cho trc th tc ca d liu ln hn tc ca cc phng php iu ch khc (2-PSK, ASK, FSK).

+ Tng t, ta cng c cc phng php iu ch pha khc 2n - PSK, c n bit biu din 1 pha, khong cch gia cc pha l 3600/2n.

T , c th pht trin ln 8PSK. Thay v dng gc 900, ta thay i tn hiu t cc gc pha 450. Vi 8 gc pha khc nhau, dng ba bit (mt tribit), theo quan h gia s bit to thay i vi gc pha l ly tha ca hai. ng thi 8PSK cng cho php truyn nhanh gp 3 ln so vi 2 PSK, nh minh ha hnh 33.

+ Bng thng dng cho 2n -PSK: Bng thng ti thiu dng cho truyn dn 2n -PSK th tng t nh ca ASK (Bng tc Baud).Bng thng ti thiu dng cho truyn dn 2n -PSK th tng t nh ca ASK, tuy nhin tc bit ti a th ln hn nhiu ln. Tc l tuy c cng tc baud ti a gia ASK v PSK, nhung tc bit ca PSK dng cng bng thng ny c th ln hn hai hay nhiu ln nh minh ha hnh 5.34.

V d 13: Tm bng thng ca h thng QPSK(4 PSK), vi tc 2kbps theo ch bn song cng.

Gii:

V h thng bn song cng nn BWh thng= BWmi hng= BWQPSKPhng php iu ch 4 PSK, 1 pha (n v tn hiu) cha 2 bit,

Rbit = 2x Rbaud ; Suy ra Rbaud = (1/2). Rbit=1000 baud/s;

M BWPSK = Rbaud ; Suy ra BWQPSK = 1000Hz.

BWh thng= BWmi hng= BWQPSK = 1000Hz.V d 14: Cho tn hiu 8PSK c bng thng 5.000 Hz, tm tc bit v tc baud?

Gii:

Phng php iu ch 8 PSK, 1 pha (n v tn hiu) cha 3 bit,

Rbit = 3x Rbaud ; M BW8-PSK = Rbaud ; Suy ra Rbaud =5000 baud/s ;

Suy ra Rbit = 3x Rbaud =15.0000 bps=15kbps ;5.3.4 QAM (quadrature Amplitude Modulation - iu ch bin cu phng)

PSK b gii hn t kh nng phn bit cc thay i gc pha nh ca thit b, iu ny lm gim tc bit.

+ Khi nim: QAM l phng thc kt hp gia ASK v PSK sao cho ta khai thc c ti a s khc bit gia cc n v tn hiu.

(QAM l qu trnh ly d liu s lm thay i bin v pha ca sng mang, tn s khng thay i)

V d: Cho mt tn hiu s 101100001000010011110111, tc bit l 24 bps, c iu ch bng phng php 8-QAM (8 loi n v tn hiu), tn s sng mang 16Hz, gin pha nh hnh v.

a. V tn tn hiu 8-QAM.

b. Tn hiu 8-QAM c phi l tn hiu iu ho hay khng? Gii thch.

c. Tnh tc Baud.

d. Tnh bng thng 8-QAM.Gii:

a.V tn tn hiu 8-QAM.

000( vc1(t)= Vcm1 sin(2fct + 01) = 2 sin(2.16t+00) V ; Tn ti trong 3 chu k bit

001( vc2(t)= Vcm2 sin(2fct + 01) = 5 sin(2.16t+00)V; Tn ti trong 3 chu k bit

010( vc3(t)= Vcm1 sin(2fct + 02) = 2 sin(2.16t +900)V ; Tn ti trong 3 chu k bit

011( vc4(t)= Vcm2 sin(2fct + 02) = 5 sin(2.16t +900); V Tn ti trong 3 chu k bit

100(vc5(t)= Vcm1 sin(2fct + 03) = 2sin(2.16t+1800) V ; Tn ti trong 3 chu k bit

101( vc6(t)= Vcm2 sin(2fct + 03) = 5 sin(2.16t+1800)V; Tn ti trong 3 chu k bit

110( vc7(t)= Vcm1 sin(2fct + 04) = 2 sin(2.16t-900)V ; Tn ti trong 3 chu k bit

111( vc8(t)= Vcm2 sin(2fct + 04) = 5 sin(2.16t -900); V Tn ti trong 3 chu k bitChu k bit Tb=1/ Rb=1/24 (s);Chu k sng mang Tc=1/ fc=1/16 (s);Ta c 3Tb = 2Tc, suy ra 3 chu bit s tn ti 2 chu k sng mangTn hiu s 101100001000010011110111b. Tn hiu 8-QAM khng phi l tn hiu iu ho, v c nhiu bin v nhiu pha.

c. Tnh tc Baud.

Rbaud = (1/3)Rbit = 8 baud/s

d. Tnh bng thng 8-QAM.

Bng thng ca tn hiu QAM bng bng thng ASK v bng tc baud

BWQAM = BWASK = Rbaud;

Suy ra BWQAM = 8Hz.+ Tng t ta cng c cc dng iu ch 2n-QAM. Vi n l s bit cha trong mt n v tn hiu, 2n : l s loi n v tn hiu.

Quan h hnh hc ca QAM c th th hin di nhiu dng khc nhau nh trong hnh sau, trong minh ha 3 cu hnh thng gp ca 16-QAM.

Trng hp u dng 3 bin v 12 pha, gim thiu tt nhiu do c t s gia gc pha v bin ln nh ITU - ngh. Trng hp th hai, 4bin v 8 pha, theo yu cu ca m hnh OSI, khi quan st k, ta s thy l cu hnh theo dng ng trc, khng xut hin yu t giao nhau gia cc bin v pha. Thc ra, vi 3 x 8 ta c n 32 kh nng. Tuy nhin khi mi s dng phn na kh nng ny, th sai bit gc pha o lng c gia tng cho php c tn hiu tt hn ri. Thng thng th QAM cho thy t b nh hng ca nhiu hn so vi ASK (do c yu t pha)

+ Bng thng ca QAM:

Bng thng ti thiu cn cho truyn dn QAM th ging nh ca ASK v PSK, ng thi QAM cng tha hng u im ca PSK so vi ASK.

+ So snh tc bit/tc baud:

Gi s tn hiu FSK c dng truyn tn hiu qua ng thoi c th gi 1200 bit trong mt giy (1200 bps). Mi tn s thay i biu din mt bit; nh th th cn c 1200 phn t tn hiu truyn trong 1s. Trong tc baud, cng l 1200 bps. Mi thay i ca tn hiu trong h thng 8QAM, c biu din dng ba bit, nh th vi tc bit l 1200 bps, th tc baud ch l 400. Trong hnh 38, cho thy h thng dibit c tc baud ch bng phn na tc bit, v trong h tribit th tc baud ch cn mt phn ba tc bit, v trng hp quabit th tc baud ch cn mt phn t tc bit.

Bng B.1 nhm so snh tc bit v tc baud trong nhiu phng php iu ch s - tng t.

Dng iu chS bit trong mt n v tn hiuBits/BaudTc Baud Tc Bit

ASK, FSK, 2-PSK

4-PSK, 4-QAM

8-PSK, 8-QAM

16-QAM

32-QAM

64-QAM

128-QAM

256-QAM1 Bit

2 Bit

3 Bit4 Bit5 Bit6 Bit7 Bit8 Bit1

2

3

4

5

6

7

8N

N

N

N

N

N

N

NN

2N

3N

4N

5N

6N

7N

8N

V d 15: Gin trng thi pha gm 8 im cch u nhau trn mt vng trn. Bit tc bit l 4800 bps, hy tnh tc baud .Gii:

y l dng 8PSK (23 =8)

Cc pha cch nhau 3600/8 = 450 Mt n v tn hiu cha 3 bit.

Nh th tc baud l Rbaud = (1/3)Rbit = 4.800/3 = 1600 baud/sV d 16: Tnh tc bit ca tn hiu 16QAM, bit tc baud l 1000.

Gii:

y l dng 16 QAM (24 =16)

Mt n v tn hiu cha 4 bit.

Nh th tc bit l Rbit= 4 Rbaud = 1.000 x 4 = 4.000 bps.V d 17: Tm tc baud ca tn hiu 64QAM bit c tc bit 72.000 bps.

Gii:

y l dng 64 QAM (26 =64)

Mt n v tn hiu cha 6 bit.

Nh th tc baud l Rbaud = (1/6)Rbit = 72.000/6 = 12.000 baud.5.4 CHUYN I TNG T - TNG T (iu ch tng t)

+ Khi nim: L qu trnh thay i mt trong cc thng s ca sng mang cao tn (Dng iu ho) bi tn hiu tin tc (d liu tng t).

+ S khi:

+ Sng mang cao tn (Dng iu ha) c 3 thng s : Bin , tn s v pha nn ta c 3 phng php iu ch tng t l:

AM (Amplitude Modulation): iu bin (iu ch bin ) FM (Frequency Modulation) ): iu tn (iu ch tn s) PM (Phase Modulation) ): iu pha (iu ch pha)

5.4.1 iu bin (AM):

+ Khi nim: L phng php m bin sng mang c thay i theo tn hiu iu ch (tin tc), tn s v gc pha sng mang khng i.

- Tn hiu iu ch (tin tc) tr thnh hnh bao ca sng mang.

Tn hiu AM c bin thay i theo tn hiu tin tc.

+ Bng thng ca tn hiu AM: BWAM = 2 Fi max= 2 BWi

Vi Fi max: l tn s cc i ca tin tc.

BWi (BWm): l bng thng ca tin tc.

+ Chun AM trong truyn thanh:

Bng thng ca tn hiu thoi thng l 5 KHz. Bng thng cc i ca 1 knh AM l 10 KHz. FCC (Federal Communication Commission- Tiu ban thng tin lin bang-chun M) mi i AM c bng thng l 10 KHz.

+ Cc i AM pht cc tn s sng mang t 530 kHz n 1700 KHz (540 kHz n 1600 KHz). Cc tn s pht ny phi cch vi t nht l 10 KHz (mt bng thng AM) nhm trnh giao thoa.

+V d: Nu mt i pht dng tn s 1100 KHz th tn s sng mang k khng c php nh hn 1110 KHz.

V d 18: Cho tn hiu tin tc vi bng thng 4KHz, hy tm bng thng ca tn hiu AM. Gi s khng xt theo qui nh ca FCC.

Gii:

BWi= 4KHzTn hiu AM c bng thng BWAM = 2 BWi= 2 x 4KHz = 8 KHz5.4.2 FM (iu tn):

+ Khi nim: L qu trnh m tn s sng mang bin thin theo bin tn hiu tin tc, bin v pha ca sng mang khng i.

+ Bng thng tn hiu FM: BWt = BWFM = 2 (Fi max+ fm ) = 10. BWiVi :Fi max l tn s cc i ca tin tc.

fm : l di tn cc i.

BWi = BWm : l Bng thng ca tin tc.

Bng thng ca tn hiu m thanh khi pht theo ch stereo thng l 15 KHz. Mi i pht FM cn mt bng thng ti thiu l 150 KHz. + Chun FM trong truyn thanh: C quan FCC cho php bng thng mi knh FM 200 KHz (0,2 MHz)

Di tn t 88 MHz n 108 MHz, Cc i phi c phn cch t nht 200 KHz trnh trng sng mang. Tm t 88 MHz n 108 MHz c khong 100 knh FM, c th pht cng lc 50 knh.

V d 19: Cho tn hiu tin tc c bng thng 4 MHz, c iu ch FM. Hy tm bng thng ca tn hiu FM , khng xt n qui nh ca FCC.

Gii:

Theo cng thc tnh bng thng ca FM:

BWt = BWFM = 2 (Fi max+ fm ) = 10. BWiM BWi = 4MHz

Nn BWt = BWFM = 10 x 4 MHz = 40 MHz.

EMBED Visio.Drawing.11

EMBED Visio.Drawing.11

_1346304382.vsdBin

Frequency

fmin fc1 fc2 fmax

fC1 hng nghch

fC2 hng thun

_1346312883.vsdConstellation diagramGin pha

1

0

Bit Phase

0 0

1 180

Bits

_1347350370.vsd

_1349710990.unknown

_1405142466.unknown

_1347360075.vsd

_1346314005.vsdGin trng thi pha

10

00

2 bit Pha

00 0 01 90 10 180 11 270

Dibit(2 bits)

01

11

_1346327332.vsd

_1346328373.vsdt

t

t

_1346328583.vsd

_1346325506.vsd

011

110

100

111

8-QAM2 bin , 4 pha

010

101

000

001

_1346326229.vsdAmplitude

1 second

3 bits 101

3 bits 100

Bit rate : 24

Baud rate : 8

Time

1 baud

1 baud

1 baud

1 baud

1 baud

1 baud

1 baud

1 baud

3 bits 001

3 bits 000

3 bits 010

3 bits 011

3 bits 110

3 bits 111

3Tb=2Tc

_1346325446.vsd

011

110

100

111

8-QAM2 amplitude, 4 phases

010

101

000

001


1 baud

1 second

1 baud

1 baud

1 baud

1 baud

2 bits 01

2 bits 10

2 bits 10

2 bits 11

2 bits00

Bit rate : 10

Baud rate : 5

Time

2Tb


Bit rate : 5

Baud rate : 5

1 baud

1 second

1 baud

1 baud

1 baud

1 baud

1 bit 0

1 bit 1

1 bit 1

1 bit 0

1 bit 1

Time

Tb=0,2s

Tb=0,2s

Tc1=0,1s

Tc2=0,05s

_1346308314.vsd


1 baud

1 second

1 baud

1 baud

1 baud

1 baud

1 bit 0

1 bit 1

1 bit 1

1 bit 0

1 bit 1

Bit rate : 5

Baud rate : 5

Time

Tb

Tc

_1346307642.vsdf

_1346305034.vsd


Frequency

1000 3500 6000 8500 11.000

fC hng nghch

fC hng thun

_1346304945.vsd

_1322823837.unknown

_1324744980.vsdImpairment

Attenuation

Noise

Distorsion

_1343751781.vsd

_1322824055.unknown


1 baud

1 second

1 baud

1 baud

1 baud

1 baud

1 bit 0

1 bit 1

1 bit 0

1 bit 1

1 bit 0

Bit rate : 5

Baud rate : 5

Time


Frequency

1000 3500 6000 8500 11.000

fC(backward)

fC(forward)

_1231135113.vsdAnalog/Analogconversion

_1231135212.vsdAnalog/Analogmodulation

AM

FM

PM

Analog/Analogmodulation

_1231135302.vsdfC

fC

fC

fC

fC

fC

fC

fC = Carrier frequency of the station

530 KHz

10 KHz

1700 KHz

_1231135298.vsdNo station here

No station here

No station here

fC

fC

fC

fC

fC

fC = Carrier frequency of the station

88 MHz

200 KHz

108 MHz

_1231135204.vsd

_1231131194.vsd

_1231131223.vsd

16-QAM

3 amplitudes, 12 phases


16-QAM


16-QAM

_1231131131.vsdConstellation diagram

100

000

Tribit Phase

000 0 001 45 010 90 011 135 100 180 101 225 110 270 111 315

Tribits(3 bits)

010

110

001

111

011

101

_1230291379.vsdPoint 1

Point 2

Transmission medium

Point 3

Amplifier

Hnh 39


Frequency

Nbaud/2

Nbaud/2

fC1 fC0

BW = fC1 fC0 + Nbaud

fC0

fC1


1 baud

1 second

1 baud

1 baud

1 baud

1 baud

1 bit 0

1 bit 1

1 bit 0

1 bit 1

1 bit 0

Bit rate : 5

Baud rate : 5

Time


Bit rate : 5

Baud rate : 5

1 baud

1 second

1 baud

1 baud

1 baud

1 baud

1 bit 0

1 bit 1

1 bit 1

1 bit 0

1 bit 1

Time

_1230291414.vsdPoint 1

Point 2

Transmission medium

Point 3

Point 4

Transmission medium

Amplifier

1 dB

7 dB

- 3 dB

- 3 dB

Hnh 40

_1231130844.vsd

_1200211271.unknown

Documents

Ky thuat truyen so lieu.doc