Co So Vien Thong_Bao Ve Cac Thanh Phan Chinh

8/8/2019 Co So Vien Thong_Bao Ve Cac Thanh Phan Chinh

1/200

C S VIN THNG

PHM VN TN


2/200

Csvin thng Phm Vn Tn

Trang I.1

Chng I

TIN TC V H THNG THNG TIN

LCH SPHT TRIN CNG NGH VIN THNG IN T. PHN LOI CC NGUN TIN TC V CC H THNG THNG

TIN. S NG XC NH V SNG NGU NHIN. S KHI MT H VIN THNG. S PHN CHIA CC VNG TN S (FREQUENCY

ALLOCATIONS). STRUYN SNG IN T.

SO TIN TC. CC H THNG TIN L TNG. M HA (CODING).


3/200


Trang I.2

LCH SPHT TRIN CNG NGH VIN THNGIN T.

-

T cui th k 18 u th k 19, cng ngh pht thanh v truyn thng bng in cpht trin.-Nm 1820, George Ohm a ra cng thc phng trnh ton hc gii thch cc tn

hiu in chy qua mt dy dn rt thnh cng.-Nm 1830 Michall Faraday tm ra nh lut dn in t trng.-C th coi lch s thng tin d liu c bt u vo nm 1937 vi s pht minh in tn

Samuel F. B.Morse. l h thng truyn cc xung in biu din cho cc du chm v vch(tng ng vi cc s nh phn 1, 0) trn cc ng dy ng nhcc my cin. Cc thp khc nhau ca cc m ny thay cho cc ch, s, du,...c gi l m Morse.

-Nm 1840, Morse ng k sng kin vin tn M.-Nm 1844 ng y in tn u tin c thit lp gia Baltimore v Washington DC.

-Nm 1849, bn tin u tin c in ra nhng vi vn tc rt chm nhng n nm 1860 vntc in t 15 bps.-Nm 1850, i s Boole ca George Boole to ra nn mng cho logic hc v pht trin rle

in. Trong khong thi gian gian ny, cc ng cp u tin xuyn qua i ty dng lpt h thng in tn.

-James Clerk Maxwell a ra hc thuyt in t trng bng cc cng thc ton hc vonm 1980. Cn c vo cc hc thuyt ny Henrich Hertz truyn i v nhn c sng vtuyn thnh cng bng cch dng in trng ln u tin trong lch s.

-Tng i in thoi u tin c thit lp vo nm 1876 (ngay sau khi Alexander GrhmBell pht minh ra in thoi). Nm nm sau Bell bt u dch v gi ng di gia NewYork v Chocago. Cng khong thi gian , Guglieno Marconi ca Italia lp t mt trm

pht sng v tuyn pht cc tn hiu in tn.-Nm 1900, Einstein, mt nh vt l ni ting v hc thuyt tng i vit rt nhiu ti

liu quan trng v vt l cht rn, thng k hc, in t trng v chc lng t. Vo khongthigian ny, phng th nghim Bell ca M pht minh v sng ch ra ng phng in cccho cc knh thin vn xoay c. Tip theo , Le De Forest trthnh ngu khi xng tronglnh vc vi mch in t thng qua pht minh ca ng v mt ng chn khng ba cc. Lc ny,h thng tng i tng t tng c kh nng hot ng khng cn bng chuyn mch.

-Nm 1910, Erwin Schrodinger thit lp nn tng cho chc lng t thng qua cng bca ng v cn bng sng gii thch cu to nguyn t v cc c im ca chng. Vo khngthi gian ny, pht thanh cng cng c bt du bng cch pht sng.

-Nm 1920, Harold .S. Black ca phng th nghim Bell pht minh ra mt my khuch iphn hi m bn m ngy nay vn cn dng trong lnh vc vin thng v cng ng my inm.

-V.K.Zworykin (M) pht minh ra n hnh cho v tuyn truyn hnh v cp ng trc(phng tin truyn dn hiu qu hn cc dy ng bnh thng).

-Cui nhng nm 1940, phng th nghim Bell t ra nn mng cho cho cc cht bn dnc tch hp cao. Howard Aiken ca i hc Harward cng tc vi IBM thnh cng trongvic lp t mt my in ton u tin c kch thc 50 feets v 8 feets. V sau , J.PresperEcker vi Jonh Mauchly ca i chc Pnnylvania pht trin my in ton ln mt bc gi lmy in ton ENIAC. Von Neuman da vo y pht trin my in ton c lu gi chngtrnh.

-

Vo nhng nm 1960, cc loi LSI (Large Scale Interated), cc my in ton mini, cpquang v my phn chia thi gian c pht trin v thng mi ho thnh cng.-Vo nhng nm 1970, truyn hnh nh qua v tinh, cc h thng tng i in t cng ln

lt ra i.


4/200


Trang I.3

Phn loi cc ngun tin tc v cc h thng thng tin.- Mt ngun tin digital ( digital information sourse ) to ra 1 tp hp hu hn cc bn tin (

Message ) c th.V d : My nh ch ; c mt s hu hn cc k t ( bn tin ) c pht ra t ngun ny.

- Mt ngun tin tc analog to ra cc bn tin c xc nh lin tc.V d mt micro: in th ra din t tin tc v m thanh v n c phn b trn mt dy lin

tc nhiu tr gi.- H thng thng tin digital chuyn tin tc t mt ngun digital n thit b thu( Sink ).

- H thng thng tin analog chuyn tin tc t mt ngun analog n Sink.Ni mt cch cht ch, sng digital c nh ngha nh l mt hm theo thi gian v ch c

mt tp hp cc tr gi ri rc. Nu dng sng digital l dng sng nh phn, th ch c hai tr gi.Dng sng analog l mt hm theo thi gian c khong cc tr gi lin tc.

Mt h thng thng tin digital in t thng c cc in th v dng in vi dng sngdigital. Tuy nhin, n vn c th c cc dng sng analog. Th d, tin tc t mt ngun nh phnc th pht n sink bng cch dng mt sng sin 1000Hz din t bit 1 v mt sng sin 500Hz din t bit 0. y ngun tin tc digital c pht n sink bng cch dng cc sng analog,

nhng vn c gi l h thng vin thng digital.Xa hn na, sng analog ny c gi l tn hiu digital v n m t 1 ngun tin digital.

Tng t, mt tn hiu analog m t mt ngun tin analog . T quan im ta thy mt k sVin thng digital cn hiu lm sao phn tch cc mch analog cng nh cc mch digital.

Vin thng digital c nhng li im:

- Cc mch digital tng i r c thc dng.- Khong tc ng ln hn. ( Khong gia cc tr ln nht v nh nht ).- D liu t ting ni, hnh v cc ngun d liu khc c thc trn ln v truyn i trn

cng mt h truyn digital.

- Trong cc h truyn vi khong cch xa, nhiu khng chng cht t repeatern repeater. (Trm pht li ).- Sai s trong d liu c phn tch th nh, d khi c mt lng nhiu ln trn tn hiu thu

c.- Nhiu c thc sa cha ( corrected ) bng cch dng s m ha.

Nhng n cng c nhng bt li:

- Thng thng, n cn mt h rng dy tn ( Band width ) ln hn h analog.- Cn n sng b ha.Vi nhiu u im, cc h digital trnn ngy cng ph bin.

Sng xc nh v sng ngu nhin.Trong cc h Vin thng, ta phn cc dng sng lm hai loi ln: Xc nh v Ngu nhin.- nh ngha: Mt dng sng xc nh c thc m hnh ha nh mt hm hon ton ring

bit ca thi gian.Th d: Nu

w(t) = A cos ( 0t + o )Din t mt dng sng , vi A, 0 , o l cc hng bit. Th dng sng w(t) c ni l

c xc nh.- nh ngha: Mt dng sng ngu nhin khng thc chuyn bit ha hon ton nh l nt

hm theo thi gian v phi m hnh ha 1 cch xc xut. Cc dng sng biu din mt ngunkhng th xc nh c. Th d, trong h vin thng digital, ta c th gi tin tc ng vi bt k

mt mu t no - Mi mu tc biu din bng mt dng sng xc nh. Nhng khi ta xtdng sng c pht t ngun ta thy rng l dng sng ngu nhin, v ta khng bit chnhxc nhng k t sc pht.


5/200


Trang I.4

Do , ta thc s cn thit k h vin thng dng dng sng ngu nhin v tt nhin bt knhiu no c a vo s cng c m t bng mt dng sng ngu nhin. K thut ny cnn nhng khi nim v xc sut v thng k. ( S lm vic phn tch v thit k phc tp hn ).

Nhnng may thay , nu ta trnh by tn hiu bng dng sng tiu biu xc nh, th ta vn cthc hu ht, nhng khng tt c cc kt qu.

S KHI MT H THNG VIN THNG.

Hnh 1.1 S khi ca mt h thng vin thng.Chch mt h Vin thng l truyn mt tin tc t ngun, k hiu l s(t), n Sink. Tin tc

ly ra t Sink k hiu l (t); tin tc c th l digital hay analog, ty vo hc dng. N cth l tin tc v Video, audio hay vi loi khc.

~s

Trong cc h multiplex ( a hp ), c th s c nhiu ngun vo v nhiu Sink. Ph ca s(t) v(t) t p trung quanh f = 0. Chng c gi l nhng tn hiu bng gc ( base

band ).

~s

Khi x l tn hiu: my pht ty iu kin ngun sao cho s truyn c hiu qu. Th d: Trong 1 h digital, n lmt vi x l. Trong h analog, n khng g hn l 1 lc h thng. Trong h lai, n l mch ly

mu tin tc vo ( analog ) v digital - ha c mt bin iu m xung ( Pulse code modulation )PCM.

Tn hiu ra ca khi XLTH my pht cng l tn hiu bng gc v cc tn s tp trung gn f= 0.Khi sng mang: my pht i tn hiu bng gc x l thnh mt bng tn truyn a vo knh truyn.

Th d: Nu knh gm mt cp dy xon ( twisted - pair ) telephone, ph ca sm(t) s nm trongdy m tn ( audio ), t 300 -> 3.700Hz. Nhng nu knh gm cp quang, ph ca sm(t) s l tns nh sng.

- Nu knh truyn i nhng tn hiu bng gc, khng cn dng khi sng mang v sm(t) c thl tn hiu ra ca khi XLTH.

- Khi sng mang th cn khi knh c th ch truyn cc tn s thuc 1 bng xung quanh fc ,vi fc >> 0. Trong trng hp ny sm(t) c gi l tn hiu dy thng ( Band pass Signal ). Vn c thit k c nhng tn s thuc 1 bng quanh fc. Th d, mt i pht bin iu AMvi mt tn s kt hp 850 KHz c sng mang fc = 850 KHz.

S p tn hiu bng gc dng sng s(t) thnh tn hiu dy thng sm(t) c gi l s bin iu( modulation ). ( s(t) l tn hiu audio trong i pht AM ).

Tn hiu dy thng bt k c dng:sm(t0 = s (t) cos [ c(t) + (t) ]

Vi c = 2fc, fc l tn s sng mang.Nu s(t) = 1 v (t) = 0 th sm(t) s l mt tn hiu hnh sin thun ty vi f = fc v bng tn

bng 0.


6/200


Trang I.5

Trong s bin iu bi mch sng mang, sng vo s(t) lm cho R (t) v/hoc (t) thay i nhl mt hm ca s(t). S thay i trong R (t) v (t) lm cho sm(t) c mt kh bng ph thuc vonhng tnh cht ca s(t0 v vo hm p c dng pht ra R (t) v (t).Cc knh truyn:

C th phn chia lm 2 loi: dy mm ( softwire ) v dy cng(hardwire). Vi loi knh dy mm tiu biu nh: Khng kh, chn khng v nc bin. Vi loi

knh truyn dy cng: Cp dy xon telephone, cp ng trc, ng dn sng v cp quang.Mt cch tng qut, knh truyn lm gim tn hiu, nhiu ca knh truyn v / hoc nhiu domy thu khin cho ~s (t) b xu i so vi ngun. Nhiu ca knh c s gia tng t ngun in,dy cao th, snh la hoc nhiu do sng ngt ca mt computer.

Knh c th cha b phn khuch i tc ng, th d: H thng repeater trong telephonehoc nh v tinh tip chuyn trong h thng vin thng trong khng gian. Dnhin, cc b phnny cn thit gi cho tn hiu ln hn nhiu.

Knh cng c th c nhiu ng ( multiple paths ) gia input v output v chng c thigian tr ( time delay ), tnh cht gim bin ( attenuation ) khc nhau. Nhng tnh cht ny c ththay i theo thi gian. S thay i ny lm thay i bt thng ( fading ) tn hiu ng ra caknh. ( Ta c th quan st s fading khi nghe khi nghe 1 i sng ngn xa ).

My thu nhn tn hiu ng ra ca knh v i n thnh tn hiu bng gc.Sphn chia cc vng tN s (Frequency Allocations).

Trong cc h thng tin dng khng kh lm knh truyn, cc iu kin v giao thoa v truynsng th ph thuc cht ch vo tn s truyn.

V mt l thuyt, bt k mt kiu bin iu no (Am, Fm, mt bng cnh - single sideband,phase shift keying, frequency shift keying...) u c thc dng cho bt k tn s truyn no.Tuy nhin, theo nhng qui c quc t, kiu bin iu rng bng, loi tin c truyn cnc xp t cho tng bng tn.

Bng sau y cho danh sch cc bng tn, k hiu, iu kin truyn v cng dng tiu biuca chng.

Bng tn K hiu t tnh truyn Nhng ng dng tiu biu3 - 30KHz VLF

very lowfrequency

Sng t. Suy gim t ngyv m. Nhiu khng khcao

Thng tin di nc

30- 300KHz LFlow frequency

Tng t VLF. t tin cy. Bhp thu vo ban ngy

Hng dn radio cho hihnh

300-3000KHz

MFMediumfrequency

Sng t v sng tri banm. Suy gim t vo ban vnhiu vo ban ngy. Nhiukhng kh

Radio hng hi. Tn s cpcu pht sng Am

3 - 30MHz HFHight frequency

S phn x tn ion cnthay i theo thi gian trongngy, theo ma v theo tns. Nhiu khng kh t ti30Mhz

radio nghip d. Pht thanhquc t. Vin thng qun s.Thng tin ng di chokhng hnh v hi hnh.in thoi, in tn, fax.

30- 300MHz VHFVery highfrequency

Gn vi LOS. S tn x gybi nhng thay i nhit .Nhiu khng gian.

Truyn hnh VHF. RadioFM stereo. Trgip khnghnh.

0.3 - 3 GHz

1.0 - 2.0 GHz2.0 - 4.0 GHz

UHFUltra highfrequencyLS

Truyn LOS. Nhiu khnggian.

Truyn hnh VHF. RadioFM Stereo. Tr gip khnghnh.

3 - 30 GHz SHF Truyn LOS. Suy gim do Vin thng v tinh. Radar


7/200


Trang I.6

Bng tn K hiu t tnh truyn Nhng ng dng tiu biu

2 - 4.04.0 - 8.0

8.0 - 12.012.0 - 18.018.0 - 27.027.0-40.0

Supper highfrequency

SC

XKUKKa

Oxi v hi nc trong khngkh. S hp th do hi ncrt cao ti22.2 GHz

microwave links.

30 - 300 GHz

26.5 - 4033.0 - 50.040.0 - 75.075.0 - 110.0110 - 300

EHFExtremely highfrequency

RQV

WMm

Tng t trn. Hi nc hpth rt mnh ti 183GHz.Oxy h p thu ti 60 v 119GHz .

Radar, v tinh, th nghim.

103 - 107 IR (Hng ngoi) nh sng khkin v UV (T ngoi )

Truyn LOS Vin thng quang

Struyn sng in t.Cc c tnh truyn ca sng in tc truyn trong knh truyn dy mm th ph thuc

nhiu vo tn s. iu ny c thy t bng k trn. Phin t c thc chia lm 3 bng

ln: Sng mt t ( Ground ware ), sng tri ( Sky ware ) v sng truyn theo ng tm mt (light of sight ) LOS.

S truyn tn hiu(signal propagation)

a. Truyn sng t

Anten pht

(Transmit antenna) The Earth

Anten thu

(Recieve antenna)

`

b. Truyn sng tri

Anten pht

(Transmit antenna)

Anten thu

(Recieve antenna)The Earth

Ion cu S truyn tn hiu(signal propagation)


8/200


Trang I.7

S truyn tn hiu(signal propagation)

c. Truyn theo ng tm mt

The Earth

Anten thu(Recieve antenna)

Anten pht(Transmit antenna)

Hnh 1.2: s truyn sng in t.

1. Tn s ca sng t nh hn 2 MHz.

y sng in t c khuynh hng truyn theo chu vi tri t. Kiu truyn ny c dngtrong cc i AM. y s ph sng a phng theo ng cong mt t v tn hiu truyn

trn ng chn tri thy c. Cu hi thng c t ra: Tn s th p nht ca sng c thdng l bao nhiu ? Cu tr li l tn s ny ty thuc vo chiu di ca anhten pht. s bc x c hiu qu, antenna cn di hn 1/10 bc sng.V d: Vi sng mang fC = 10KHz, bc sng l:

=C

fC

= ( 3.108m/s )/104Hz = 3.104 mNh vy, mt anten di t nht 3.000m bc x c hiu qu mt sng in t 10KHz!

2. Khong tn s ca sng tri l 2 n 30 Mhz.

S truyn ca sng ny da vo s phn x tng ion ( ion sphere - tng in ly ) v mt t.Nh, c th truyn mt khong rt xa.Tng ion c biu phn b nh sau:

Hnh 1.3: Biu phn b tng ionS ion ha xy ra do s kch thch cc phn t kh bi cc bc x v tr t mt tri. Tng ion

gm cc lp E, F1, F2, D. Lp D ch hnh thnh vo ban ngy v l lp ch yu hp th sng tri.Lp F l lp chnh, lm phn x sng tri v tri t.

Thc t, s khc x tng bc qua cc lp ca tng ion khin tng ny tc dng nh mt vtphn x lm sng tri b phn x trli tri t.


9/200


Trang I.8

Hnh 1.4: S phn x sng tri btng ion.Ch s khc x n thay i theo cao ca tng ion, v mt electron t do thay i.

n = 181

2

n

f

Trong : N: Mt electron t do ( s e-/m3 ).

f: tn s ca sng (Hz).- Di vng ion ha, n = 1- Trong vng ion ha, n < 1 ( V N > 0 ) Sng b khc x theo nh lut Snell:

nsinr = siniTrong : I : Gc n

r: Gc khc x.a. Vi nhng sng c tn s f < 2MHz :

81N > f2 nn n trnn o. Tng ion s lm gim sng n.b. Vi nhng sng c tn s t2 - 30 MHz ( Sng tri ), s truyn sng, gc phn x v

s hao ht tn hiu ti mt im phn x tng ion ty thuc vo f, vo thi gian trong ngy,

theo ma v s tc ng ca vt en mt tri.Ban ngy, N rt ln lm n o. Sng b hp thu, c rt t sng trli tri t.Ban m, N nh nn n < 1. Khi , nu sng truyn t tri t ln tng ion th

r > I. S xy ra hin tng khc x tng bc. Do s phn x nhiu ln gia tng ion v mtt, sng tri truyn i rt xa. V th, c nhng sng tri pht ra t nhng i xa bn kia tri tvn c th thu c trn bng sng ngn.

3. Struyn LOS l phng thc truyn cho cc tn s trn 30 MHz.

, sng in t truyn theo ng thng.Trong trng hp ny f2 >> 81N lm cho n 1 v nh vy c rt t sng b khc x bi tng

ion. Sng s truyn ngang qua tng ny. Tnh cht c dng cho thng tin v tinh.

Cch truyn LOS bt li cho vic truyn thng tin gia 2 trm mt t, khi m ng i tnhiu phi trn ng chn tri. cong mt t s chn ng truyn LOS.


10/200


Trang I.9

Hnh 1.5Anten pht cn phi t trn cao, sao cho anten thu phi thy c n.

d2 + r2 = ( r + h )2

d2 = 2rh + h2 h2


11/200


Trang I.10

nh ngha: So tin tc trung bnh (average information) ca 1 ngun l:

H = P I PP

bitsj jj

m

jjj

m

= = =

12

1

1log

m: S bn tin.PJ : Xc sut ca s gi bn tin th J

Tin tc trung bnh cn gi l entropy.V d: Tm information content (dung lng tin tc ) tin tc ca mt bn tin gm mt word

digitaldi 12digit , trong mi digit c th ly mt trong 4 mc c th. Xc sut ca s gi mtmc bt k trong 4 mc c gi s bng nhau v mc ca mt digit khng ty thuc vo tr gic ly ca digit trc .

Trong mt string gm 12 symbol (digit) m mi symbol gm mt trong 4 mc l4.4.....4 = 412 bits,t hp (word) khc nhau.

V mi mc gm bng nhau tt c cc word khc nhau u bng nhau. Vy:

PJ =1

4

1

412

12

=

hoc

IJ = ( )log log2 12

2

1

1

4

12 4 24

= = bits

Trong v d trn ta thy dung lng tin ( information content ) trong bt k mt bn tin c thno u bng vi dung tin trong bt k bn tin c th khc (24 bits). Vy tin tc trung bnh Hl 24 bits.

Gi s rng ch c 2 mc (nh phn) c cho php cho mi digit v rng tt c cc wordthgn bng nhau Vy tin tc s l IJ = 12 bits cho word nh phn v tin tc trung bnh l H =12bits. tt c word 12 bits s cho 12 bits tin tc v cc word gn bng nhau Nu chng khng

bng nhau mt vi trong cc word 12 bits s cha hn 12 bits tin tc v mt vi s cha t hn.V tin tc trung bnh s cha t hn.

inh ngha:Nhp ca ngun (nate source) c cho bi

R =H

Tbits/sec

H: Tin tc trung bnh.T: Thi gian cn thit gi mt bn tin.nh ngha trn c p dng cho mt ngun digital.

Cc h thng tin l tng.C mt s tiu chun c dng nh gi tn hiu qu ca mt h thng tin . l gi

thnh, rng knh, cng sut truyn, t s s/n ti nhng im khc nhau ca h, thi gian trngang qua h thng. V xc sut bit error ca h digital.

Trong cc h digital, h ti u c thc ngha nh l mt h c xc sut bit error ti thiung ra ca h vi s cng ch v cng sut c pht v rng knh.

iu ny lm ny ra cu hi: liu c th pht minh mt h khng c bit errorng ra d khic nhiu thm nhp vo knh ? Cu hi ny c Claude Shannon tr li l c th, vi vi gi

thit Shannon chng minh rng mt dung lng knh C (bits/sec) sc tnh sao cho nu nhp tin tc R (bits/sec) nh hn C, th xc sut ca bit error tin n zero.Phng trnh ca C l:


12/200


Trang I.11

C = B log2 1+

SN

B: rng knh (Hz) v S/N l tng s cng sut tn hiu trn nhiu ti ng v camy thu digital.

- Trong cc h analog, h ti u c chnh ngha nh l mt h c tng s S/N lnnht ng ra my thu vi s cng ch v cng sut c pht v rng knh.

Ta c tht cu hi: Liu c th thit k mt h thng vi tng s S/N ln v hn ng ra khi nhiu thm nhp vo knh ? Cu tr li l dnhin l khng.m ha (CODING).

Nu d liu ng ra ca mt h thng tin digital c errors, c th gim error bngcch dng mt trong hai k thut :

-Automatic Repeat request (ARQ).-Forward error conection (FEC).

Trong mt h ARQ, khi my thu phn tch c error trong khi d liu, n yu cukhi d liu pht trli.

Trong mt h FEC d liu c pht ra cn c m ha sao my thu c th sa sainh l cc sai s phn tch. Bin php ny cng c xp loi nh s m ha knh, v nc dng sa sai khi knh b nhiu.

S chn la ARQ hay FEC ty vo p dng ring. ARQ thng c dng trong hthng tin computer.

FEC c dng sa sai tr cc knh simplex (1 way).H thng tin vi FEC c vhnh di y. V mt l thuyt dung lng knh

ca Shannon chng t rng mt tr gi v hn ca S/N ch gii hn nhp truyn. l xcsut ca error P(E) c th tin n zero khi nhp tin tc nh hn dung lng knh.

Hnh 1.6

M hov x l Mch sngmang Mch sngmangknh )(

~

tgm g(t) s(t) r(t)

)(~

tm

M hov x l

nhiuTruyn Nhn

B thu tnhiu s

Tn hius


13/200


Trang II.1

Chng II

PHN TCH TN HIU

XEM LI CHUI FOURRIER.PH VCH.BIN I FOURRIER.CC HM K D: ( SINGNLARITY FUNCTIONS ).PHP CHNG (CONVOLUTION).PHP CHNG HNH ( GRAPHICAL CONVOLUTION ).NH L PARSEVAL.

NHNG TNH CHT CA BIN I FOURRIER.NH L V SBIN IU.

CC HM TUN HON.


14/200


Trang II.2

XEM LI CHUI FOURRIER.

1. Mt hm bt k S(t) c thc vit: ( dng lng gic ).

(2.1)

Vi t0 < t < t0 + T ; T1fo

S(t) = a0cos(0) +n=

1

[ an cos 2 nf0t + bn sin 2f0t ]

S hng th nht l a0 v cos (0) = 1.

Vic chn cc hng an v bn theo cc cng thc sau:

- Vi n = 0 ; a0 = 1T

s t dtt

t T

o

o( )+ (2.2)

- Vi n 0 ; an =2

2T

s t nf t dtot

t T

o

o

( ) cos .+

(2.3)

bn =2

2T

s t nf t dtot

t T

o

o

( ) sin .+

(2.4)H thc (2.2) c c bng cch ly tch phn 2 v ca (2.1).

H thc (2.3) v (2.4) c c bng cch nhn c 2 v ca (2.1) cho hm sin v ly tchphn.2. Dng cng thc EULER, c tha dng s(t) trn v dng gn hn ( dng hm m phc ).

EULER ej2nfot = cos 2nfot + j sin 2nfot (2.5)

(2.6)=

Cn e j2nfotS(t) =n

Trn n: S nguyn; dng hoc m. V Cnc nh bi:

Cn =1

T t

t T

o

o +

s(t) e -j2nfot dt (2.7)

iu ny d kim chng, bng cch nhn hai v ca (2.5) cho e -j2nfot v ly tch phn haiv.

Kt qu cn bn m ta nhn c = mt hm bt k theo thi gian c thc din t bngtng cc hm sin v cos hoc l tng ca cc hm m phc trong mt khong.

Nu s(t) l mt hm tun hon, ta ch cn vit chui Fourrier trong mt chu k, chui stng ng vi s(t) trong mi thi im.

V d 1: Hy xc nh chui Fourrier lng gic ca s(t) nh hnh v. Chui ny cn pdng trong khong - /2 < 1< /2 .


15/200


Hnh 2.1 Tn hiu cos(t).-2 2-/2 /2

s(t)

Ta dng chui Fourrier lng gic, vi T = v fo

=1 1

T=

nh vy chui c dng:

n=

1

s(t) = a0 + [ an cos 2nt + bn sin 2nt ]

t

Trong : a0 =1 2

2

2

cos .t dt

+

=

v an =2

22 1

2 1

1

2 12

21

cos .cos .( ) ( )

t nt dtn n

n n

+ +

=

+

+

Ta nh gi bn nh sau:

bn =2 2 2

2T

s t nt dt( ).si n .

+

V s(t) l mt hm chn theo thi gian, nn s(t) .sin 2nt l mt hm l v tch phn t - /2n /2 l zero. Vy bn = 0 vi mi s(t) l. Chui Fourrierc vit :

s(t) =( )2 2 1

2 1

1

2 12

1

1

+

+

+

=

+

n

n n

n nnt

( )cos (2.8)

Lu : Chui Fourrier cho bi phng trnh trn y c cng khai trin nhca hm tunhon sp(t) nhhnh di y:

Trang II.3

Hnh 2.2 Anh ca s (t) trong bin i Fourier.

sp(t)

-/2 /2-3/2 3/2t

Ph vch

Trong lc tm s biu din chui Fourrier phc ca 1 hm theo thi gian, ta dng mt thas trng lng phc Cn cho mi tr ca n. Tha s Cn c thc v nh l hm ca n. Vy cnn 2 ng biu din. Mt biu din cho sut ca n v mt biu din pha.

ng biu din ny th ri rc. N ch khc zero i vi nhng tr gin on ca trchanh. ( V d: C1/2 th khng c ngha ).

ng biu din Cni vi nf0 gi l ph Fourrier phc. Trong nf0 l lng tng ngvi tn s ca hm m phc m i vi n Cn l mt h s trng lng.

V d 2: Tm ph Fourrier phc ca sng cosin c chnh lu ton sng,s(t) = cos t, nh hnh v di y.


16/200


Trang II.4

|cost|

-/2 /2-3/2 3/2t

Hnh 2.3 Tn hiu |cos(t)|.

Trc ht ta phi tm khai trin chui Fourrier theo dng hm m phc.Vi F0 =

1

, ta tnh tr gi Cn t (2.6) v tm chui Fourrier trc tip.

Tuy nhin v d 1, ta khai trin chui Fourrier di dng lng gic ri, nn c thkhai trin hm cos a v dng hm m phc bng cch dng cng thc Euler:

s(t) =( )2 2 1

2 1

1

2 12

1

1

+

+

+

=

+

n

n n

n nnt

( )cos

Vi cos 2nt = [ ]1

2

2 2

e e

j nt j nt

+

Vy chui Fourrier dng hm m:

s(t) =2

2 2

2

1

21

+ +

=

=

a e a en j ntn

n j nt

n

=2

2 2

2

1

2

1

+ +=

=

a e a en j ntn

n j nt

n

(2.9)

Ta i bin ss hng sau. Vy Cn lin h vi an:Cn =

an2

Vi n > 0

Cn =a n

2Vi n < 0

Cn =2

Trong trng hp ny, Cn l s thc. Nn ch cn v mt hnh.


17/200


Trang II.5

-2/15

2-3

-2-1 1

2/3533

2/3

2/

nf0

Hnh 2.4: Ph vch ca v d 2 .Bin i Fourrier:

Mt tn hiu khng tun hon c xem nh l trng hp gii hn ca mt tn hiu tunhon, trong chu k T ca tn hiu tin n . Nu chu k tin n , tn s cn bn F0 tinn 0. Cc ha tn khp li vi nhau v, trong gii hn, tng chui Fourrier biu din cho s(t) strthnh mt tch phn.

(2.10)

F [.] k hiu cho bin i Fourrier ca [.].

N cn c gi l ph - hai - pha ( Two - Side - Spectrum ) ca s(t), v c hai thnh phntn s dng v m u thu c t (2.10). Gi s s(t) l mt hm thc (vt l).

Mt cch tng qut, S(f) l mt hm phc theo tn s. S(f) c th phn lm hai hm thcX(f) v Y(f) :

S(f) = X(f) + jY(f) (2.11)

Dng trn gi l dng Cartesian, v S(f) c thc biu din trong mt h trc ta

Descartes. Cng c th biu din S(f) trong mt h trc cc. Khi , cp hm thc s trnh bysut v pha.

(2.12)

Vi :

S(f) = X f Y f2 2( ) ( )+ (2.13)

v:

(f) = tan-1 Y fX f( )( ) (2.14)

Dng trn y cn gi l dng cc ( Polar form ).

F [s(t)] = S(f) s t e dt j ft( )

2

S(f) = S(f) ej(f)


18/200


Trang II.6

xc nh nhng tn s no hin hu, ta kho st ph ca xut S(f). ( i khi gi ttl Ph ).

Ph ca mt dng sng ( dng hay th ) c th thu c t nhng php tnh ton hc. Nkhng xut hin mt cch vt l trong cc mch in thc t. Tuy nhin c th dng SpectrumAnalyser quan st mt cch gn ng.

* phc hi li s(t) t bin i Fourrier ca n, ta tnh tch phn sau:

(2.15)s(t) = S f e dt j ft( ) 2

= F -1 [S(f)]

Phng trnh ny thng gi l bin i ngc ca S(f). Hai hm s(t) v S(f) to thnh mtcp bin i Fourrier. Trong , s(t) din t trong phm vi thi gian, cn S(f) din t trong

phm vi tn s.

K hiu cho mt cp bin i Fourrier :

Hoc (2.16)S(f) s(t)s(t) S(f)

Nu tn hiu hoc nhiu c m t trong phm vi ny, th s m t tng ng trong phmvi kia sc bit nhcch dng (2.10) hoc (2.15).

Dng sng s(t) c th bin i Fourrierc nu n tha cc iu kin Dirichelet. Tuynhin, tt c cc dng sng vt l trong k thut u tha cc iu kin .

V d 3: Ph ca mt xung expo.

t s(t) l mt xung expo tt ( Decaying Exponential Pulse ) b ngt ( Switched ) tit = 0.

s(t) = (2.16)e t

t

t >

t + 2

v u(t - - 2) = 0 , > t - 2

Ta c:

u t d d t

t

( ) + = = +

+

2 31 1

2

( V rng t + 2 > -1 hoc t > -3. khong khc, tch phn l zero).

- Nu t - 2 > -1 hoc t > 1,


27/200


Trang II.15

u t d d t

t

( ) = =

2 11 1

2

- Nu t + 2 > +1 hoc t > -1,

u t d d t

t

( ) + = = +

+

2 11 1

2

- Nu t - 2 > 1 hoc t > 3,

u t d d t

t

( ) = =

2 31 1

2

Dng 4 kt qu ta c:

r(t) * s(t) = ( t + 3)u(t + 3) - (t - 1)u(t - 1) - (t + 1)u(t + 1) + (t - 3)u(t - 3)Bn s hng ny v tng ca chng c v nh hnh di y. T v d khim tn ny, tac th thy rng nu r(t) hoc s(t) cha hm nc, th cch tnh php chng trnn rt lng tng.

Hnh 2.12 Php chng ca tn hiu r(t) v tn hiu s(t).

-3

3

3

1-1

-1

4321-4 -3 -2 -1

2

r(t)*s(t)

-(t-1)U(t-1)-(t+1)U(t+1)

(t-3)U(t-3)(t+3)U(t+3)

tt

t t

t

Php chng hnh ( Graphical convolution )

Nu r(t) v s(t) qu phc tp, hoc dng sng khng c bit chnh xc, ta c th dngphp chng hnh. Phng php ny dng nhng quan st v kim tra tng qut m khngphi tnh chi tit cc tch phn. Trong nhiu p dng thng tin, phng php ny th mkhng cn thit phi tnh mt php chng chnh xc.

V d 8: Dng php chng hnh cho 2 hm v d 7.


28/200


29/200


Trang II.17

Hnh trn y trnh by 12 khung ca s di hnh. Vi v dc bit ny, khng bt bucs(t) phi phn x c nh qua gng, v s(t) l mt hm chn.

Nh l din tch ca tch s biu din cho tr gi ca php chng. Din tch ny c vthnh mt chui cc im. C th thy l kt qu ging nh v d 7.

ng ni cc im l ng thng. iu hin nhin, v php chng trthnh tch phn

ca mt hng. Kt qu cho mt hm dc ( Ramp Function ).

r(t)*s(t)

-3 -2 -1 1 32

2

t

Hnh 2.14 Kt qu php chng hnh ca s(t) v r(t).

V d 9: Tnh php chng ( bng hnh ) ca 2 hm sau y: (Sinh vin t gii)

t1-1 31

11

s(t)r(t)

t

Hnh 2.15 Tn hiu s(t) v r(t) .

By gita xem php chng ca mt hm bt k vi xung lc (t).

(T) * s(t) = ( ) ( ) ( ) ( )t s t d s t s t = =

0 (2.42)

Nhvy mt hm bt k chng vi mt xung lc th ginguyn khng thay i.


30/200


Trang II.18

Hnh 2.16 Kt qu php chng hnh ca s(t) v r(t)

Nu ta chng s(t) vi xung lc b di ( Shifted ) (t - t0), ta thy:

(t - t0) * s(t) = ( ) ( ) ( ) (t t s t d s t s t t = =

)

0 00 (2.43)

Tm li, php chng s(t) vi mt xung lc khng lm thay i dng hm ca s(t). C thchgy nn mt sdi thi gian trong s(t) nu xung lc khng xy ra ti t = 0.

Gi ta c khi nim v thut ton gi l php chng . Ta hy tr li php bin iFourrier.nh l vphp chng:

Nu r(t) R(f)

V s(t) S(f)

Th: r(t) * s(t) R(f). S(f) (2.44)

C th chng minh trc tip nh l bng cch tnh bin i Fourrier ca php chng.Ta cng c th chng minh:


31/200


Trang II.19

R(f) * S(f) r(t) . s(t) (2.45)

Bng cch tnh bin i Fourrier ngc.

V d 9: Dng nh l php chng tnh tch phn sau:

sin sin( )3

t

t d

Gii:

Tch phn trn biu din php chng ca 2 hm theo thi gian:

sin3

2

t t

t*

sin

=

x

F

t

tsin

F

t

t3sin

t

t1/2-1/2

1/2-1/2-3/2 3/2

2

t

Bin i Fourrier ca tch phn l tch ca bin i Fourrier ca 2 hm. Hai bin i ny cth xem bng ph lc.

Hnh 2.17 Tch ca hai bin i Fourier t s(t) v r(t).

Ly bin i Fourrier ngc ca tch ny, ta s c kt qu ca php chng. l:

sintt

nh l PaRseval

Dng sng ca mt hm v ca bin i Fourrier ca n th rt t ging nhau. Tuy nhin,mt vi h thc hin hu gia nng lng ca mt hm thi gian v nng lng ca bin i

Fourrier ca n.


32/200


Trang II.20

Dng nng lng ch tch phn ca bnh phng ca hm. T ny c dng v nbiu din tr gi nng lng ( watt - sec ) tiu tn trong in tr1 nu tn hiu l in th hocdng in ngang qua in tr.

Ta c:

r(t) s(t) R(f) * S(f)

F [ r(t) s(t) ] = r t s t e dt j ft( ) ( )

2 (2.46)

= R k S f k dk( ) ( )

V ng thc ny ng vi mi f, ta t f = 0. Khi :

r t s t dt( ) ( )

= R k S k dk( ) ( )

(2.47)

Biu thc (2.47) l mt dng ca cng thc Paseval. N lin quan n nng lng nn taxt trng hp c bit:

s(t) = r * (t)

r*(t) l lin hp ca r(t).

F [ r*(t)] cho bi lin hp ca bin i Fourrier, b phn x qua trc dc. lR*(-f).

Dng kt qu ca (2.47), ta c:

r t dt R f df2 2( ) ( )

= (2.48)

Phng trnh (2.48) chng t rng nng lng ca hm theo t th bng vi nng lng cabin i Fourrier ca n.NHNG tnh cht ca bin i Fourrier1. Thc / o - Chn / l.

Bng sau y tm tt nhng tnh cht ca bin i Fourrier da trn s quan st quan st

hm theo t.

Hm thi gian Bin i Fourrier

A Thc Phn thc chn - Phn o l

B Thc v chn Thc v chn

C Thc v l o v l

D o Phn thc l - Phn o chn

E o v chn o v chn


33/200


Trang II.21

F o v l Thc v l

C th dng cng thc Euler chng minh:

S(f ) = s t e dt j ft( )

2

= s t ft dt j s t ft dt( ) cos ( ) sin2 2

= R + j X

R l mt hm chn ca f v khi fc thay bng -f th hm khng i. Tng t, X l mthm l ca f.

Nu s(t) gi s l thc, R trthnh phn thc ca bin i v X l phn o. Vy tnh cht

A c chng minh.Nu s(t) thc v chn, th X = 0. iu ny ng v X l ( tch ca hm chn v l ) v tch

phn l 0. Vy tnh cht B c chng minh.

Nu s(t) thc v l, R = 0. ( Tnh cht C ).

Nu s(t) o, X trthnh phn o ca bin i v R l phn thc. T quan st n gin ,cc tch cht D, E, F d dng c chng tht.2. Di thi gian ( Time Shift ).

Bin i Fourrier ca mt hm thi gian b di th bng vi bin i ca hm thi giangc nhn bi mt hm expo phc.

(2.49)e-j2fotS(f) s(t - t0)

V d 10: Tm bin i Fourrier ca:

s(t) =1 0 2

0

,

,

<


34/200


Trang II.22

S(f ) = [ ]e dte

j fe e j ft

j ft j f j f

= 2

0

22

2 2

2

= e-j2fsin2

f

f

Kt qu ny c th thu c t vic dng mt hm nc trong v d 4 v tnh cht di thigian. s(t) v d 10 trn y th ging nhv d 4 ( Vi A = = 1), ngoi tr vic dch thigian 1 sec.

4. Di tn s ( Frequency shift ).

Hm theo thi gian tng ng vi mt bin i Fourrier di tn th bng vi hm theo thigian ca bin i khng di tn nhn vi 1 hm expo phc.

(2.50)S(f - f0 ) e

j2fo s(t)

V d 11: Tm bin i Fourrier ca s(t).

s(t) =e t

ph n khac

j t2 1

0

,

,


35/200


Trang II.23

(2.51)as1(t) + bs2(t) aS1(f) + bS2(f)

Trong a, b l nhng hng bt k.

C th chng minh trc tip tnh ngha ca php bin i Fourrier v t tnh cht catuyn tnh ca thut ton tch phn.

[ ]as t bs t e dt a s t e dt b s t e dt j ft j ft j ft1 22

12

22( ) ( ) ( ) ( )+ = +

= aS1(f) + bS2(f)

V d 12: Tm bin i Fourrier ca s(t).

s(t) =

1 1 0

2 0 1

1 1 20

,

,

,,

< 2fm

Hnh 4.4: Bin i F ca 2 sng AM.Nu cc tn sca 2 sng bin iu khng cch nhau xa lm, c 2 c thdng 1 anten,

mc d chiu di ti u ca anten khng nhnhau cho c 2 knh [trong thc t, mt anten c

dng cho c 1 khong tn s.

Ta nhn mnh li rng, cc tn hiu c thc tch ra nu chng khng b ph ln nhau (hoc v thi gian, hoc v tn s ). Nu chng khng ph nhau v thi gian, c th dng cccng hay cc Switchs tch. Nu chng khng ph v tn s, cc tn hiu c th tch ra bi cclc dy thng. Vy, mt h thng nh hnh 4.5 c th dng tch sng mang b bin iu.

fc2-fc2

H2(f)

fc1-fc1

1

1

H1(f)

BPF

s1(t).cos2fc1t+

s2(t).cos2fc2t

H1(f)

H2(f)

s1(t). Cos2fC1t

s2(t). Cos2fC2t

Hnh 4.5: S tch 2 knh.


65/200

CSVin Thng Phm Vn Tn

Trang IV.5

Nu nhiu tn hiu c truyn trn cng mt knh, ch c thc tch ra ti my thubng cc lc dy thng. Cc lc ny ch tip nhn, mt trong cc tn hiu hin din trong tn hiubin iu mong mun.

TD: Mt tn hiu cha thng tin c dng:

s(t) =sin2 t

t

Tn hiu ny bin iu bin mt sng mang c tn s 10Hz. Hy v dng sng AM vbin i F ca n.

Gii: Sng AM c cho bi phng trnh:

sm(t) =sin2 t

t

cos 20t

Hm ny c v nh hnh 4.6:Hnh 4.6: Dng sng AM

cos 20t l sng mang.

- Khi sng mang bng 1 ( t =10

k), sm (t) = s(t).

- Khi sng mang bng -1, t =k

10

1

20+ , sm(t) = -s(t).

v dng sng AM. Ta bt u v s(t) v nh qua gng ca n -s(t). Sng AM chmmt cch tun hon vo mi ng cong ny v thay i bin gia nhng im tun hon .

Trong hu ht trng hp thc t, tn s sng mang cao hn rt nhiu so vi th d trn.Bin i F ca s(t) c vhnh 4.7 ( Xem ph lc chng II )

Hnh 4.7: nh Fourier ca s(t)Bin i F ca sng bin iu c tnh nhnh l bin iu.

Sm(f) =S(f -10) S(f 10)

2

+ +(4.7)


66/200


Trang IV.6

Hnh 4.8: Tn ph ca sng bin iuV Sm (f) c suy t S(f) bng cch di tt c cc thnh phn tn s ca s(t) mt khong l

fC, ta s c th hi phc li s(t) t sm(t) bng cch di cc tn s bi cng mt tr theo chiungc li.

nh l bin iu chng t rng php nhn mt hm thi gian vi mt hm Sinusoide sdi nh F ca hm thi gian i ( c chiu ln v xung ) trong min tn s. Vy nu ta li nhnSm(t) vi mt hm sin ( tn s sng mang ), th nh F s di lui xung n tn s thp ca n.Php nhn ny cng di nh F ln n 1 v tr gia khong 2fC, nhng thnh phn ny d dng

b loi bi mt lc h thng. Tin trnh ny vhnh 4.9.

S hi phc ca s(t) c m t bi phng trnh (4.8)sm(t). cos 2fCt = [ s(t) cos 2fCt ] cos 2fCt= s(t) cos2 2fCt

=s t s t ( ) ( )+ cos 4 f tC

2(4.8)

Ng ra lc h thng l /2s t( )

sm(f)

Hnh 4.9: S hi phc tn hiu t sng bin iu.Tin trnh ny gi l hon iu ( Demodulation ).

BIN IU BIN SNG MANG C TRUYN 2BNG CNH

( Double - Side Band Transmitted Carrier AM ). DSBTCAM.By gi ta ci bin thm s bin iu AM, bng cch cng vo sng bin iu mt phn

ca sng mang.


67/200


Trang IV.7

s(t)

Hnh 4.10.Hnh 4.10 ch s cng mt sng mang hnh sin thun ty vo sng bin iu DSBSCAM.

Kt qu cho bi phng trnh (4.8)sm(t) = s(t) cos 2fCt + A cos 2fCt (4.9)

y l kiu bin iu AM sng mang c truyn 2 bng cnh. ( DSBTC AM). Khc vikiu AM sng mang b nn 2 kiu AM sng mang c truyn c cha mt thnh phn r rng

ca sng mang ( A cos 2fCt ).nh F ca TCAM l tng ca bin i F ca SCAM v bin i F sng mang thun ty.

Bin i sng mang l mt cp xung lc fC.

Hnh 4.11: Bin i F ca TCAMDng sng c th vit li ( T phng trnh 4.9 )sm(t) [A+s(t)] cos 2fCt (4.10)

Hm ny c th v theo cch v dng sng SCAM. Trc ht, ta vng bin [A+s(t)] vnh qua gng -[ A + s(t)]. Sng AM chm tun hon vo 2 ung bin v thay i bin iugia nhng im tun hon . Hnh v 4.12, cho mt s(t) hnh sin ( th d ting hut so vomt microphone ).

- Hnh 4.12a Tn hiu s(t) hnh sin- Hnh 4.12b Dng sng DSBTCAM vi gi tr ca A nh hn bin a ca s(t); Aa; A0.

- Hnh 4.12d Dng sng DSBTCAM khi A=0.


68/200


Trang IV.8

Hnh 4.12


69/200


Trang IV.9

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Hnh 4.12

HIU SUT

S cng thm sng mang vo sng bin iu s lm cho s hon iu d dng hn. Ci gim ta phi tr l hiu sut. Mt phn ca nng lng c truyn dng gi sng mang v nhvy khng mang mt thng tin hu ch no.

Ta thy t phng trnh (4.9) : Cng sut sng mang l cng sut ca A cos2fCt, hayA

22

watts. Cng sut ca tn hiu l cng sut ca s(t) cos2fCt, l tr trung bnh ca s2(t) chia

2. Cng sut trung bnh ca s2(t) th n gin l ca s(t), hay PS. Vy cng sut ca tn hiu l

P 2S .

Cng sut truyn ton phn l tng ca 2 s hng ny.Ta nh ngha hiu sut l t s ca cng sut tn hiu cng sut ton phn:

=P

A P

S2

S+(4.10)

TD: Gi s ta xem dng sng hnh 12c, v t A bng vi bin ca hnh sin. Vy hiusut l 33%.


70/200


Trang IV.10

CC KHI BIN IU:

Hnh 4.13 S ca cc khi bin iu AM.- Hnh 4.13a: H thng to nn DSBSC AM.- Hnh 4.13b,c: H thng to nn DSBTC AM.

Hnh 4.13: Khi bin iu AM

Ti sao sbin iu th khng tuyn tnh ?Ta bit, bt k mt h tuyn tnh v khng i theo thi gian no iu c mt output m

bin i F ca n l tch ca nh F ca input vi H(f). Nu bin i ca tn hiu vo bng zero

trong mt khong tn s no , th nh F ca output phi cng bng zero trong khong y.Ngha l, tnh cht tng qut ca h tuyn tnh khng i theo thi gian l n khng th cho rabt k mt output no nu khng c input ng vo.

Vy c mt h tuyn tnh khng theo t no c th cho sm(t) ng ra khi nhn s(t) ng vo? Ni cc khc, ta c th tm c hay khng mt H(f) no cho:

Sm(f) = S(f) . H(f)

Hnh 4.14R rng, cu tr li l khng.


71/200


Trang IV.11

S bin iu l mt tin trnh di tn. V khng c mt h tuyn tinh no thc hin ciu .

Mt h phi tuyn v thay i theo t, ni chung, l rt phc tp. Tuy nhin, trong trngh p bin iu, ngi ta c th thc hin c bng 2 kiu gin ti p: Bin iu cng (Gatedmudolator) v bin iu theo lut bnh phng (Square - Law Mudolator ).Bin iu Cng:

Da vo s kin: Php nhn s(t) vi mt hm tun hon bt k s to ra mt chui sngAM vi nhng sng mang l bi s ca tn s cbn ca hm tun hon. Hnh_4.15

Hnh 4.15: Tch ca s(t) v hm cng tun hon

Output ca mch nhn (hnh 4.15)

s(t)P(t) = s(t) (4.11)

=+ )tf2cos(

1a c0 n

nn

a

fc L tn s cbn ca hm tun hon. an , cc h s chui F. Gi s P(t) l hm chn ( trnh phi vit cc s hng sin trong chui )

Lc BPF s chn tt c, ch tr thnh phn no trong chui m ta s chn. Kt qu l ng ra c mt sng AM. Mch lc iu hp vi tn s cbn, nhng n s c thiu hp vimt trong nhng ha tn ca sng AM, c tn s sng mang cao hn. Trong thc t, ta chnnhng ha tn thp (V cc h s F lm gim bin tn hiu khi n tng).

P(t) l mt hm cng gm mt on xung tun hon. (Hnh 4.16)


72/200


Trang IV.12

Hnh 4.16: Hm cng

* V P(t) lun bng 0 hay bng 1, mch nhn c th xem nh c cch hot ngon/off ( hoc switch ).

Output ca BPF tm c bng cch khai trin P(t) thnh chui F v tm a1.

21 =a

)().()( tPtstsm =

sm(t) =2

s(t) cos2fCt (4.12)

Phng trnh (4.12) c vit cho hm cng c na thi gian cao v na thi gian zero.Nhng sng AM vn c to ra vi bt k tr gi no ca chu k thao tc ca xung.

B phn to hm cng c th l thng hoc tc ng hnh 4.17 ch b phn bin iugm 2 thnh phn thng.

B phn to hm cng

Hnh 4.17a: Mch to xung cng thng dng Switch.

-

R

- +

1

4

3

2

s(t)

+

-

c2(t)

+cos2fct

Hnh 4.17b: Mch to xung cng thng dng diode.- Hnh 4.17a, SW ng ngt tun hon. Khi SW h, tn hiu ra bng tn hiu vo. KhiSW ng, tn hiu ra bng zero. R l in trngun. Bt li ca SW chc l ng ngt chm.Tn sng ngt ca SW phi bng tn s sng mang ( hoc c s, nu ta chn 1 ha tn ). Vitn s sng mang cMHz, SW chc khng thp ng kp.

- Hnh 4.17b: Sng ngt thc hin nhcu diode. Khi cos2fCt dng ( im Bc in th dng hn im A ), c 4 doide b kha: Mch tng t nh hnh 4.17a khi SW h,tn hiu ra l s(t). Ngc li khi cos2fCt m ( im B c in th m hn im A ). C 4 diodedn: mch ging nh hnh 4.17a khi SW ng. Gii hn duy nht cho mch ng ngt ny l tnsng ngt ca loi Diode c dng. ( Tnh khng l tng ca cc diode, thng l thi gianhi phc ( recovery time ) ca in dung mi ni kh ln so vi chu k sng mang ).


73/200


Trang IV.13

- Hm cng cn c th to c bng cch dng cc linh kin tc ng, nhtransistor hot ng gia vng kha v vng bo ha. Mt transistor kha, tng ng vi mtSW h. Mt transistor bo ha, xem nh mt SW ng.

- Hnh 4.18, trnh by mt kiu mch bin iu di l bin iu vng (ringmodulator). Sng mang l mt sng vung, c a vo mi gia ca 2 bin th. Output l mt

phin bn b cng ha ca input, ch cn lc l c c sng AM .

Bin iu Theo Lut Bnh Phng.Loi ny da vo nh lut: Bnh phng ca mt tng 2 hm c cha mt s hng l

tch ca 2 hm :[s1(t)+s2(t)]

2= s12 (t) + s2

2 (t)+2 s1(t).s2(t)Nu s1(t) l tn hiu cha tin v s2(t) l sng mang, ta c:

[ s(t) + cos2fCt ]2 = s2(t) + cos2 2fCt + 2s(t) cos2fCt (4.13)S hng th 2 chnh l sng AM mong mun. Ta phi tm cch tch n ra khi 2 thnh

phn kia. Ta bit, s tch sn gin, khi chng khng ph nhau ( trong phm vi thi gianhoc phm vi tn s ). R rng, chng ph nhau v thi gian. Vy, ta hy xem phm vi tn s.

Cc xung lc ti gc v 2fC kt qu ca s khai trin lng gic

Cos2 = 2cos21 + ng cong lin tc gia ( tn s thp ) ch bin i F ca s2(t). Ta khng bit dng

chnh xc ca s(t). Nhng ch bit rng nh F ca n b gii hn nhng tn s nh hn fm. Bini F ca s2(t) b gii hn nhng tn s di 2fm. Mt cch thy iu l xem bin i Fca s2(t) l php chng ca S(f) ln chnh n. Php chng hnh cho thy bin i ny i tzero n 2fm. Cch khc, l xem s(t) nh l tng ca cc hnh sin c tn s (ring) di fm. Khi

bnh phng tng ny, ta c kt qu l tt c cc tch ca cc s hng. iu ny sa n tngv hiu ca cc tn s khc nhau ( dng lng gic). Khng c tng hay hiu no vt qu 2fmnn tn s gc khng vt qu fm.


74/200


Trang IV.14

Hnh 4.18: Bin iu vng


75/200


Trang IV.15

Hnh 4.19: Bin i F ca (4.13)Hnh 4.19 cho thy khi fC >> 3fm th cc s hng khng ph nhau ( v tn s ). Vy c th

tch chng bng mt lc BPF c sng AM. Trong hu ht cc trng hp thc t, fC>>fm, nniu kin ny d tha.

SQR

Hnh 4.20: Mch bin iu bnh phng.Hnh 4.20 ch ton th mt khi bin iu theo lut bnh phng. Cc b phn tng c th

l tc ng, thng hay op.amp.- B phn bnh phng th khng n gin. Bt k mt linh kin phi tuyn no cng u

cho mt tn hiu ra tng ng vi mt tn hiu vo bi mt h thc m ta c th khai trin thnhchui ly tha. Gi s khng c s tch tr nng lng, ngha l output ti bt k thi im noch ph thuc vo input ti cng thi im , ch khng kn nhng tr gi trc .

Vi y(t) l output v x(t) l input:

y(t) = a0 + a1x(t) + a2x2(t) + a3x

3(t) + .... (4.14)

S hng m ta lu l a2x2(t). V ta tm cch ta tm cch tch n khi cc thnh phn khc.

Linh kin phi tuyn c chn dng phi cbn l mt linh kin c c tnh bnh phng. Thd diode

an trong phng trnh (4.14) phi c tnh cht:

an 2

C vi iu cn ni thm v s phi tuyn.Nu cc shngng vi n = 1 v n = 2trongchui chim u th(bin ln) th kt qu l sng TCAM. Hn na, Nu an nh qu ( vi n > 2), sng AM vn c nu lm cho s(t) tht nh. Vy sn(t) 1, v TCAM vn cnchim u th. y l mt trng hp khng mong mun, v bin ca sng qu nh.

* Cc diode bn dn c c tuyn rt ging vi lut bnh phng ( trong vng hot ngca n ).

S khi ca mt mch bin iu cn bng (balance modulator) vhnh 4.21. H nycng sng mang cos2fCt vi tn hiu cha tin s(t), sau a chng vo linh kin phi tuyn (

bnh phng ). S vn hnh cng c lp li vi -s(t) . Mch tng s ly hiu sca 2 tn hiu

ra, lm loi bs hng ca ly tha l trong khai trin (4.14). V d, xem s hng ly tha 3.


76/200


Trang IV.16

Khi khai trin [s(t)+cos2fCt]3, S hngphln bng tn ca sng AM ls2(t)cos2fCt. S hngny khng i du khi -s(t) c thay vo s(t). Nh vy ti mch tng (thc ra l tr ) chng strit nhau. S hng m ta mun ly, s(t).cos2fCt , si du khi -s(t) c thay cho s(t). Vymch s lm tngibin tn hiu.

Ta cng nhrng, khishng bc mt b trit, nn tn hiu ra ca khi bin iu cn bng

l SC AM. ( Bin iu AM sng mang b nn ).

Mch in thc t ca bin iu bnh phng vhnh 4.22. y l mch transistor kiu Echung. Mch dng s phi tuyn ca transistor to nn tch ca tn hu vi sng mang. Mchc iu hp chn C, lc b nhng ha tn khng mong mun.

Hnh 4.21: Khi bin iu AM cn bng

Hnh 4.22: Mch bin iu bnh phng

Cc mch bin iu bnh phng thc t d thit kn ngc nhin! Thc vy, Chngthng hin hu ngoi mun. Cc sn phm ca s bin iu xut hin trong mch in mtkhi cc linh kin in t ba vo vng hot ng phi tuyn. V vy, ngi ta thng c ngnnga mt mch hot ng nh mt mch bin iu khng mong mun.

Hnh 4.23 l mch ca mt my pht AM bin iu chn C. Ch cn thay i in th tcthi t vo chn B ca Transistor do s bin i bin ca tn hiu trong tin s(t). Sng xut

hin ti nh ca mch iu hp chn C l tng ca VCC v tn hiu s(t). Nh vy, cbn ta lm thay i in th tc thi do bin ca s(t) thay i.

s(t)

SQR

SQR

SQR

SQR


77/200


Trang IV.17

Ng ra ca mch l mt lc BPF, nhm gim thiu cc ha tn sinh ra do s hat ng phituyn ca transistor.

Hnh 4.23: Mch pht AM bin iu chn C

CC KHI HON IU ( Demodulators)

hon iu cho sm(t) v sau

C(t) c ng b ha v c tn s v phavi sng mang c thu.

Ta ni t trc rng s(t) sc hi phc t sm(t), bng cch cho tn hiu qua mt lc LPF. ( loai sng mang ).

Hnh 4.24 l s khi ca mt mch hon in ng b(Synchronous Demodulator) hayhon iu kt hp. Gi nh vy v mch dao ng to s


78/200


f

1/21/4 1/4

fm-fm

f

-2fc 2fc

G(f)

1

-fm fm

S(f)

Hnh 4.24: Hon iu AMV mch nhn ca hnh v nhn khng khc vi mch nhn dng trong mch bin iu, ta

c th tin on nhng ci bin ca mch bin iu cng v bnh phng c th p dng c y.

C hai loi hon iu ng bHon iu Cng:

Trc ht, hy kho st s dng mch bin iu cng hon iu mt sng DSBSCAM:

Trang IV.18

Hnh 4.25: Hon iu cng

P(t) l mt hm cng gmmt chui xung tun hon bin n v.

P(t) = a0 + a

=1nn cos2nfCt


79/200


Trang IV.19

Vy tn hiu vo ca LPF l:

sm(t) P(t) = s(t) cos2fCt

+

=1nCn0 tfcos2 naa

= a0 s(t) cos2fCt + 2s(t) [ ]tf1)2+cos(ntf2)1cos( CC1nn +

=na (4.15)

Quan tm n thnh phn bc 1:

sm(t).P(t) = a0.s(t).cos2fct + a1.s(t).cos22fct

2

tf4cos)t(sa

2

)t(satfs2a0.s(t).co c11c

++=

Vy output ca LPF cho bi:

V s hon iu c hon tt.

so(t) =21 a1s(t)

Ta ni v hot ng ca hon iu cng cho mt sng AM SC. By gi,nu ta thay A + s(t) cho s(t) trong phng trnh (4.15) ( trng hp TCAM).Ta s thy rng hon iu cng s to ra mt tn hiu ra.

Biu thc trnh by tn hiu cha tin gc b di bi mt hng. Nu h cha linh kin linlc ac, hng s khng sut hin output. Nu tt c mch khuch i trong h lin lc dc, ta cth loi bng cch dng mt t ni tip tng i ln, n np n tr trung bnh ca tn hiu.

so(t) =21

a1[A + s(t)]

Ta gi s tr trung bnh ca tin s(t) l zero. Nu n khng ng, s loi b hng cng sloi vi tn hiu khc. May mn, hu ht s(t) u c tr dc l zero.Hon iu Bnh Phng:

Ta kho st hiu qu ca vic cng sng AM vo mt sng mang thun ty, ri sau bnhphng tng:

[sm(t) + A cos2fCt ]2 (4.16)

Trc ht, hy xem trng hp sng mang b nn SCAM. Phng trnh (4.16) trnn:

sm(t) = s(t). Cos2fCt

( )[{ 2C Atstfcos2 + ]} = cos22fCt + [s(t) + A]2

=2

Cos4A]+[s(t)A]+[s(t) 22 tfC+ (4.17)

- S hng th nh l mt sng AM xung quanh mt sng mang tn s 2fC. Vy c th tch

n ra d dng bng mt lc LPF.


80/200


Trang IV.20

- S hng th nht c th khai trin:

s2(t) + A2 + 2A s(t).

Nhng tn s cha s2(t) ph vi s(t), v chng khng th tch ra. Tuy nhin, gi s rng ta

dng mt lc LPF tch tt c s hng

2

2

A+s(t)

ra khi thnh phn c tn s 2fc .

Nhl lc ny phi cho qua nhng tn s ln n 2fm. Vy ta hi phc bnh phngcatng ca A v s(t). Ta s ly cn bc 2 ca n c:

A)t(s2

1As(t)0,707 .+=+

A* S ly sut ca mt tn hiu sa n mt dng mo. Th d, tn hiu l mthnh sin thun, sut ca n c dng sng sin chnh lu 2 bn k vi tn s cbn gp i tn s

gc. Tn hiu chnh lu khng ch cha mt tn sn, m bao gm nhiu ha tn. [ nu ta nghen loa, sng sin gc s cho mt tng thun, trong lc sng sin chnh lu 2 bn k s cho mttng s - Thnh phn ha tn - cao hn mt bt ]. Nu tn hiu gc l mt hn hp nhiu tns, s mo s nghim trng hn.

B* Nhng gi s A ln sao cho s(t) + A khng bao gic tr m, th A+s(t) s

bng s(t) + A. Khi , ta hon iu c. Ngha lsng mangc thm vo my thu hon iuphi c bin ln hn hay bng tr m ti a ca s(t).

By gita xem vic hon iu sng TCAM. Trong vic hon iu, cn thit phi to limt bn sao hon chnh ca sng mang. iu ny kh thc hin, tr khi sng AMcha mt shng tun hon c tn s bng tn s sng mang. iu ny t nhin a ta n vic phi dngTCAM. Thc vy, phng trnh (4.16) l kt qu t vic bnh phng sng TCAM thu c mkhng cn cng thm mtsng manga phng(ni local) (ti my thu ).

s(t)

Hnh 4.26: Khi hon iu bnh phng cho TCAM.

Hnh 4.26 l khi hon iu cho TCAM. Bin sng mang A ln lm cho A +s(t) khng m.

C* i vi sng SCAM, cn phi thm mch to (bn sao ca) sng mang ti my thu.Bn sao ny cn c ng b ha vi sng mang thu c ( ph hp v tn s v pha). Thngmy thu c mt mch dao ng ni thc hin vic ny.

Ta hy xem hu qu ca s khng ph hp v tn s v pha. Gi s mch dao ng nihnh 4.24 b lch tn bi fv lch pha bi . Khi , output ca mch nhn l:


81/200


Trang IV.21

sm(t) cos [ 2 (fC+f)t + A]

= s(t) cos2fCt cos [ 2 (fC+f)t + A]

= s(t)

++++

2])t2(2[cos

2]t2[cos ff Cf (4.18)

y cng l input ca LPF ca khi tch sng ng b, output ca n l:

s0(t) = s(t)2

]t2[cos f + (4.19)

( S hng th nh ca (4.18) c thnh phn tn s 2fC + f nn b loi )

Biu thc (4.19) cho thy mt tn hiu l s(t) nhn vi mt hm Sinusoide ti tn sfHertz. Ta gi sfnh, v ta c lm cho n 0. nh l bin iu ch rng so(t) c mt bin i

F vi cc tn s trong khong nfm + f. D LPF c thit k ch cho qua cc tn s ln fm , nhng n vn cho qua ton b fm + f,v f


82/200


fC -2

BW< f < FC +

2

BW

Ly F -1:

so(t) = A cos2fCt + +

2BWf

2BWf

C

c

C

)fS(f cos2fCt + df (4.21)

Tch phn ca phng trnh (4.21) gii hn bi:

t21

Smax (f)BW.

Vy:

Mt mch lc vi kh bng tht hp s ch cho qua s hng th nht, ( thnhphn sng mang thun ty ).

Hnh 4.27: S hi phc sng mang dng BPF trong TCAM.

Mt cch khc hi phc sng mang l dng vng kha pha (phase - lock loop). Vngkha pha s kha thnh phn tun hon input to nn mt sinusoide c tn s sng mang.

Hnh 4.28: Vng kha pha

Hnh 4.29: Hi phc sng mang trong TCAM bng PLL

Tch Sng Khng Kt Hp ( Incoherent Detection ):

Cc khi hon iu ni trn cn phi to li sng mang my thu. V tn s sngmang phi chnh xc v pha phi ng phi hp ( matched ) ng ti b phn tch sng, nnsng mang ti pht xem nh l mt thng tin chnh xc v thi gian (timing information) cn

phi c truyn ( n my thu ). V l do , cc khi hon iu trn gi l tch sng kt hp (

Trang IV.22

So pha

VCO

v0(t)Input

Hi tipTn hiu chun


83/200


Trang IV.23

Incoherent Detection ).

Nhng nu thnh phn ( s hng ) sng mang ln trong TCAM, ta c th dng kiu tchsng khng kt hp. Trong , khng cn phi to li sng mang.

Gi s sng mang ln sao cho A + s(t) > 0. Hnh 4.30. Ta bit, hon iu bnh

phng th hiu qu cho trng hp ny.

Hnh 4.30: TCAM vi A + s(t) > 0

Ta nhc li, nh hnh 4.26, output ca khi bnh phng:

[A + s(t)]2 cos22fCt =21 [ ] [ ] tfcos4s(t)As(t)A C

22+++

Output ca LPF ( cho qua nhng tn s ln n 2fm) l:

s(t) =[ ]

2

s(t)A 2+

Nu by gita gi s rng A ln sao cho A + s(t) khng bao gim, th output ca khicn hai l:

so(t) = 0,707[ A + s(t) ]V s hon iu c hon tt

s(t)

Hnh 4.31: Tch sng bnh phng


84/200


Trang IV.24

Tch sng chnh lu:

Khi bnh phng c thc thay bng mt dng phi tyn khc. Trng hp cbit, xem mch tch sng chnh lu ( Rectifier Detection ) nh hnh 4.31.

Chnh lu LPFs1(t)sm(t)

Hnh 4.31: B tch sng chnh lu.

)f(H

fm-fm

Xem mt sng DSBTCAM:

[ ] ft2cos.)t(sA)t(s m +=

Mch chnh lu c th l na sng hoc ton sng.Ta xem loi mch chnh lu ton sng ( Full - Wave Rect ) Chnh lu ton sng th tng

ng vi thut ton ly tr tuyt i. Vy tn hiu ra ca khi chnh lu l:s1(t) = A + s(t)cos2fCt

V gi s A + s(t) khng m, ta c th vit:s1(t) = [ cos2f]s(t)+A Ct

Tr tuyt i ca cosine l mt sng tun hon, nh hnh 4.32.

)tcos(

Hnh 4.32

Tn s cn bn ca n l 2fC. Ta vit li s1(t) bng cch khai trin F :s1(t) = [ A + s(t) ] [ ao + a1 cos4fCt + a2 cos8fCt + a3 cos12fCt +.... ]

Vy output ca LPF l:

so(t) = ao [ A + s(t) ]V s hon iu hon tt.

* By gi, ta hy xem cch m khi tch sng trn hi phc li sng mang. Hnh4.33 ch rng s chnh lu ton sng th tng ng vi php nhn sng vi mt sng vung.(ti tn s fC ). l tin trnh ly tr tuyt i ca phn m ca sng mang. N tng ng vi

s nhn cho -1. Vy, mch chnh lu khng cn bit tn s sng mang chnh xc, m ch thc


85/200


86/200


Trang IV.26

Hnh 4.35: Tch sng nh* By ginu ta u thm mt in trx in cho t. Mch hnh 4.36 l mch

tch sng bao hnh. Output s c dng expo gia cc nh. Nu chn la thi hng RCthch hp,th output s xp xvi bao hnh. V mch tc ng nh mt mch tch sng. Output c cha

sng d ( tn s fC) nhng iu khng h g, v ta ch quan tm n nhng tn s di tn sfm.

Hnh 4.36: Tch sng bao hnh

Thi hng RC phi ngn sao cho bao hnh c th vch nhng thay i trnh ca sng

AM . Cc nh cch nhau ti nhng khong bng vi tn s sng mang, trong lc chiu cao ththeo bin i ca bin ca s(t).

Ta xem trng hp s(t) l mt hm sin thun ( tn s fC). N s c kh nng thay i tr

nh nhanh nht. Ti tn s ny, cc nh thay i t mt tr max n min trong21

fm sec. Mch

cn 5 ln thi hng t 0,7% tr cui cng ca n. Vy nu ta t thi hng RC n 10% ca

mf1 , Th mch tch sng bao hnh c th hot ng tn s cao nht. V d, vi fm = 5kHz,

thi hng s chn l50

1m sec. ( hoc 20s).

Bin iu v Hon iu bng IC


87/200


Trang IV.27

Cc mch bin iu v hon iu c th dng IC.Cc ICny c cha nhng mch khuchi Visai a vo vng bo ha hoc m phng mt giao hon in t. (Electronnic Commulator ).

- Hnh 4.37, IC MC1496 c s dng nh mt bin iu TCAM. Mch tng t c thdng pht ra SCAM, bng cch chn li tr s cc in trtrong mch hiu chnh sng mang.

- Hnh 4.38, cng dng chip ny hon iu cho TCAM. Sng mang trong mch cthc bng cch thc tn khuch i cao tn vo vng bo ha. Nh vy, output ca tn ny gingnh mt sng vung ti tn s fC. Sng mang ny c a vo mt trong nhng ng v ca MC1496. Ng ra phi l LPF, hi phc tn hiu cha thng tin.

Carrier

1K

51

+Sm(t)

0.1uF

+12V

1K

6,8K

U3

MC14961

4

6

5

2 3

10

8

7

8

9

51

s(t)

1K

+8V

1 3

2

51

3,9K 3,9K

-Sm(t)

Hnh 4.37: Bin iu AM


88/200


Trang IV.28

6,8K

+12V

51

U3

MC14961

4

6

5

2 3

10

7

9

8

+12V

3,9K

10K

+8V

1K

3,9K

+12V

10K

U3

Amplifier/Limiter

600

s(t)50K

1K

+12V

51

Sm(t)

Hnh 4.38: Hon iu cho TCAM

TRUYN MT BNG CNH (single sideband) SSB:

Trong cc h thng AM m ta ni trn, khong tn s cn thit truyn tn hiu lbng gia fC - fm v fC + fm kh bng tng cng l 2fm

Trong vic khai thc cc i pht AM, ngi ta xem tng ph nh l ti nguyn thinnhin . Vic bo qun cho n l mt ch tiu quan trng. Nu kh bng cn thit cho mi knhrng qu, Th si pht sng cng mt lc s t i. Ta tm mt phng php c th gi thng

tin m kh bng th nh hn 2fm.Truyn mt bng cnh l k thut cho php truyn phn na kh bng cn thit cho AMhai bng cnh.

Hnh 4.39: nh ngha cc cnh bng

Hnh 4.39 nh ngha cc bng cnh. Phn ca sm(t) nm trong bng trn sng mang gi lbng cnh trn ( upper - sideband ). V phn di sng mang gi l bng cnh di (lower -sideband). Mt sng AM 2 bng cnh th bao gm c bng cnh trn v bng cnh di.

Ta c th dng cc tn cht ca bin i F chng t rng 2 bng cnh ny ph thuc lnnhau. Bin i F ca sng AM c to nn bng cch di ( shifting ) S(f) ln v xung, nh

bit. Bng cnh di to nn do phn f m ca S(f); v bng cnh trn do phn f dng ca S(f).Ta gi s rng tn tc s(t) l mt hm thc. Vy sut ca S(f) th chn v pha th l. Phn f m cth suy t f dng bng cch ly phc lin hp.

Tng t, bng cnh di ca sm(t) c th suy t bng cnh trn. V cc bng cnh khng

c lp, ta c th truyn tt c cc thng tin cbn bng cch gi i ch mt bng cnh.


89/200


Trang IV.29

Hnh 4.40: Bin i F ca cc bng cnhHnh 4.40 ch bin i F ca bng cnh trn v bng cnh di ca sng AM, ln lt k

hiu l susb(t) l slsb(t). Sng AM 2 bng cnh l tng ca 2 bng cnh.

sm(t) = susb(t) + sLsb(t) (4.22)V sng SSB ch chim mt phn ca bng tn b chim bi sng DSB, n tha 2 yu cu

ca mt h bin iu. l, bng cnh chn tn s sng mang ring, ta c th chuyn sng biniu thnh mt khon tn s, m truyn i mt cch hiu qa. Ta cng c th dng nhng

bng khc nhau cho nhng tn hiu khc nhau (tc fc khc nhau). Nn, cng lc c th truyn inhiu tn hiu (a hp).

Ch cn mt vn cn chng t. l, thng tin gc c thc hi phc t sng cbin iu SSB. V sng bin iu c thc to ra bi cc mch tng i n gin. Vy ta xtn cc khi bin iu v hon iu.Kh

i Bi

n

iu Cho SSB:

V bng cnh trn v bng cnh di tch bit v tn s, cc mch lc c th dng chnbng cnh mong mun. Hnh 4.41, ch khi bin iu cho bng cnh di (LSB). C cc cch to bng cnh trn (USB). Ta c th hoc thay i dy thng ca lc BPF ch nhn USB, hocc th ly h s gia DSB v LSB.

Hnh 4.41: Khi bin iu cho LSB, SSB


90/200


Trang IV.30

Hnh 4.42: Khi bin iu cho USB, SSB

Cc mch lc 2 hnh bn phi tht chnh xc, v khng c dy tn bo v no gia bng

cnh trn v bng cnh di.* Mt phng php khc to ra SSB. S khi vhnh 43 ( dng LSB - SSB ).

Gi s s(t) l mt Sinusoide thun ty. Vi trng hp n gin ny, s phn tch ch cn nlng gic.

S(t) = cos2fCtSng DSB Amc dng:

sm(t) = cos2fCt + cos2fCt

=2

)tf(fcos2)tf(fcos2 mCmC ++ (4.23)

S nhn dng cc bng cnh trong trng hp n gin ny tht r rng: S hng th nhtl bng cnh di, s hng th nh l bng cnh trn.

Hnh 4.43: Bin iu cho LSB, SSB

By gita khai trin bng cnh di:

sLsb(t) =2

fm)t(fcos2 C

=2

tft.sin2fsin2+tft.cos2fcos2 mCmC (4.24)


91/200


Trang IV.31

SSB

Vy ta c th thy ti sao s khi hnh 4.43 c th to ra LSB. S hng th nht caphng trnh (4.24) l sng DSB AM. S hng th nh c c l do s di pha 900 cho misng Cosine.

S trn y c th ci bin to ra bng cnh trn (USB). Ch cn thay b phn tngbng mt b phn ly hiu s hai outputs ca 2 mch nhn.Khi Hon iu Cho SSB:

Khi hon iu ng b hnh 4.44 c th dng hon iu SSB

Hnh 4.44: Hon iu ng b

* V phng din tn s, ta bit s nhn cho mt Sinusoide s lm di tn bin

i F c chiu ln v chiu xung.- Hnh 4.45, ch bin i F ca susb(t) khi nhn n vi mt Sinusoide ti tn s fC.

- Hnh 4.46, ch kt qu tng ti vi tn hiu sLsb(t). Trong c 2 trng hp, mtlc LPF s hi phc li bn sao ca tn hiu cha thng tin gc.

SUSB(f)

Hnh 4.45: Bin i F ca hon iu USB v SSB


92/200


Trang IV.32

SLSB(f)

tf2cos).t(sF cLSB

-fc fc

-2fc 2fc

Hnh 4.46: Bin i F ca hon iu LSB v SSB* V phng din thi gian ta thy:

2

tfcos2+tfs(t)sin2tf2s(t)cos mCC2

fSSB(t) cos2fCt = (4.25)

Du + cho LSB v du - cho USB. Khai trin lng gic

= 4

tfs(t)sin4tf2s(t)cos+s(t) CC2

(4.26)

Output ca LPF (vi mt input nh vy ) s ls(t)/4

V ta hon tt c s hon iu.

* Ghi ch: (t) l bin iHilbertca s(t)

(t) =

.d

t

s(t)1V

=

dt

)(s1-s(t) s(t)

Bin i Hilbert ca mt hm thi gian c c bng cch quay tt cthnh phn tn si mt gc 900.

V d: s(t)= cos(2fCt+) (t)= sin(2fCt+)

BIN IU M TRC PHA:

Ta chng t rng nhng tn hiu khng ph nhau v tn s v thi gian th c th tch rakhi nhau. DSBAM gia s tch bit v tn s v thi gian th c th tch bit tn s cc knhkhng b giao thoa vi nhau. Nhng n phi cn dng kh bng rng gp i SSBAM.

Tuy nhin, trong trng hp 2 tn hiu DSBAM c gi i ng thi m c tn s v thigian ph nhau, chng vn c th tch ra ti my thu. Thc vy, bin iu bin trc pha sthc hin c vic y. ( Quadrature Amplitude Modulation QAM ) .


93/200


Trang IV.33

Hnh 4.47: My thu QAM

Gi s, c 2 tn hiu s1(t) v s2(t) c tn s gii hn nh hn fm. Hai tn hiu ny bin iu2 sng mang c tn s bng nhau.

s1m(t) = s1(t).cos2fCt

s2m(t) = s2(t).sin2fCtTng ca 2 sng:

AM = sm1(t) + sm2(t) = s1(t). cos2fCt + s2(t).sin2fCtMc d hai sng ph ln nhau, nhng chng c th tch ra bi my thu nh hnh v trn.

- Tn hiu ng vo LPF1:

sa(t) = [s1(t) cos2fCt + s2(t) sin2fCt].cos2fCt

= s1(t).cos22fCt + s2(t).sin2fCt.cos2fCt

=2

1[s1(t)+ s1(t) cos4fCt + s2(t) sin4fCt]

Mch lc LPF1 s ch cho qua s hng th nht, l s1(t)/2

- Tn hiu ng vo LPF2:

sb(t) = s1(t) cos2fCt.sin2fCt + s2(t) sin22fCt

=2

1[s1(t) sin4fCt + s2(t) - s2(t) cos4fCt]

Ng ra ca LPF1 l s hng th hai, s2(t)/2

BIN IU BNG CNH ST ( vestigial sideband ) VSB.

Bin iu SSB c li hn DSB v mt s dng tn s. l SSB ch dng phn na khbng cn thit tng ng ca DSB. Nhng SSB c bt li l kh thit k mt my pht v mtmy thu c hiu qu. Mt vn ni bt ca SSB l vic thit k mch lc loi b mt bngcnh - Tnh cht pha ca mch lc s to nn sng d. Vic ny s gy hu qu xu. V d, trongtruyn hnh, kh bng rng hn trong truyn thanh (ting ni). S mo pha tn hiu video gynn hiu ng offset ln hnh nh c qut, ( to ra bng ma )- mt ngi rt nhy vi dng monh vy (hn l s mo tng t ca ting ni).

Vy ta c l do ni n mt kiu bin iu nm gia SSB v DSB. l kiu bngcnh st (VSB). [ Mt bng cnh b loi tr khng hon ton bi mch lc trnh mo ].


94/200


Trang IV.34

VSB c xp xcng kh bng tn vi SSB v khng kh thit k mch hon iu. Nh tngi, VSB c cha phn st li ca bng cnh th nh (khng loi b hon ton nh SSB).

Hnh 4.48: Bin iu VSB

Mch lc c dng cho VSB khng ging nh trong SSB - n khng cht ch.

sm(f)

+fc-fc

H(f)

-fc +fc

-fc +fc

Sm(f).H(f)

Hnh 4.48 ch bin i ca DSB, c tnh mch lc v bin i ca output.

Nu SV(f) l bin i F ca tn hiu VSB, th:SV(f) = Sm (f)H(f) = [ s(f + fC) + s(f - fC)]H(f) (4.27)

Output ca b hon iu ng b c bin i:

S0(f) = f,2

)ff(S)ff(SCVCV

++< fm (4.28)

Thay (4.27) vo (4.28), ta tm c:

S0(f) =4

)ff(H)ff(H)[f(S CC ++ (4.29)

Phng trnh (4.29) c dng t cc iu kin cho mch lc.

Tng nm trong [ ] c vhnh 4.49. Vi mt H(f) tin biu.


95/200


Trang IV.35

Hnh 4.49: Lc BPF cho VBS

Gi s rng mt s hng sng mang c cng vo (TCAM). Sng mang c truynVSB c dng

sv(t) + A cos2fCt

S hng sng mang ny c rt ra ti my thu bng cch dng hoc mt lc bng rt hphoc mt vng kha pha. Nu s hng sng mang ln, c th dng tch sng bao hnh [ ta thy iu SSB. , sng mang ln hn nhiu so vi tn hiu. Cn DSB, sng mang chcn ln cng cvi tn hiu. i vi VBS, Bin sng mang th nm gia 2 kiu y ].

Khi cng mt sng mang vo, hiu sut s gim. S d dng trong vic thit k mt mchtch sng bao hnh khin h ny c chn dng trong truyn hnh.

AM STEREO.

Ta ch gii thiu nhng im ch yu v AM stereo. S phn gii su hn cn n nhng

hiu bit v bin iu pha, m ta s ni chng 5.Nguyn l AM Stereo l gi 2 tn hiu audio c lp trong kh bng 10kHz nm trong mi

i pht thanh thng mi. Nhng hiu chnh cn thit c th tng thch vi cc my thumono ang hin hu (nu 2 tn hiu biu din cho 2 knh tri v phi, th mt my thu mono phihi phc tng ca 2 tn hiu ny).

Nu 2 tn hiu k hiu l sL(t) v sR(t), tn hiu tng hp c th vit :q(t) = sL(t) cos2fCt + sR(t) sin2fCt (4.30)

Nu c 2 tn hiu sL(t) v sR(t) l tn hiu aodio vi tn s ti a l 5kHz, q(t) chim dytn gia fC - 5kHz n fC+5KHz. ( kh bng tng cng l 10kHz ).

Tn hiu tng hp c th vit li nh l mt Sinusoide duy nht:q(t) = A(t) cos[2fCt+(t)] (4.31)

Trong : A(t) = (t)s(t)s 2R2L +

(t) = -tan-1

(t)s

(t)s

L

R

Mch tch sng bao hnh trong mt my thu mono s to A(t). l mt phin bn b moca tng ca 2 knh v khng cn cho yu cu tng thch.

Hnh 4.50 Ch s ca khi bin iu v hon iu. Khi v chm chm l mt vng

kha pha, c dng hi phc sng mang. Output ca vng kha pha l cos(2fCt-450)


96/200


Cc hm thi gian khc c ghi trong hnh l:

s1(t) = (2fCt - 450)

s2(t)= cos2fCt

s3(t)= sin2fCt

s4(t) = sL(t) cos

2

2fCt + sR(t) sin2fCt + cos2fCt

s5(t)= sL(t) sin2fCt cos2fCt + sR(t) sin22fCt

s6(t)=2

(t)sL

s7(t)=2

(t)sR

Hnh 4.50: H thng AM STEREO

=1/2sL(t)

Trang IV.36


97/200


Trang IV.37


98/200


Trang V.1

Chng V:BIN IU GC

TN S TC THI. BIN IU TN S (FREQUENCY MODULATION). BIN IU PHA. FM BNG HP (NARROW BAND FM). PM BNG HP. FM BNG RNG (WIDE BAND FM). HM BESSEL. KHI BIN IU. KHI HON IU. FM STEREO.

SO SNH CC H.


99/200


Trang V.2

TN S TC THI.

Xem mt sng mang cha b bin iu

sC(t) = A cos(2fCt + ) (5.1)Nu fC b thay i ty theo thng tin m ta mun truyn, sng mang c ni l c biniu tn s. Cn nu b lm thay i, sng mang b bin iu pha. Nhng nu khi fC hay bthay i theo thi gian, th sC(t) khng cn l Sinusoide na. Vy nh ngha v tn s m tadng trc y cn c ci bin cho ph hp.Xem 3 hm thi gian:

s1(t) = A cos 6t (5.2a)s2(t) = A cos (6t +5) (5.2b)s3(t) = A cos (2t e-t ) (5.2c)

Tn s ca s1(t) v s2(t) r rng l 3Hz. Tn s ca s3(t) hin ti cha xc nh. nh nghatruyn thng ca ta v tn s khng p dng c cho loi sng ny. Vy cn mrng khi nim

v tn s p dng cho nhng trng hp m tn s khng l hng.Ta nh ngha tn stc thi theo cch c th p dng c cho cc sng tng qut. Tn s tcthi c nh ngha nh l nhp thay i ca pha.

t s(t) = A cos (t) dtd

)t(f2 i

= (5.3)

fi : tn s tc thi, Hz. Nhl c 2 v ca phng trnh (5.3) c n v l rad/sec.Nh vy trong th d trn, tn s tc thi ca cc tn hiu cho ln lt l 3Hz; 3Hz v e-t

(1 - t) Hz.Th d 1:Tm tn s tc thi ca cc sng sau:


100/200


Trang V.3

Hnh 5.2Th d 2. Tm tn s tc thi ca hm sau y:

s(t) = 10 cos2[1000t + sin 10t ]Gii:Ap dng nh ngha tm:

t10cos101000dt

d

2

1)t(fi +=

=

fic vhnh 5.3.

Hnh 5.3

BIN IU TN S (FREQUENCY MODULATION).

Bin iu FM c pht minh bi Edwin Armstrong nm 1933 [cng l ngi pht minhmy thu kiu i tn (superheterodyne - siu phch)]. Trong bin iu FM, ta bin iu tn stc thi fi (t) bi tn hiu s(t). V cng v c th tch bit cc i vi nhau, ta phi di tn s(t)ln n tn s sng mang fC.

Ta nh ngha bin iu FM nh l mt sng vi tn s tc thi nh sau:fi (t) = fC + Kfs(t) (5.5)

Trong : fC l tn s sng mang (hng s) v Kf l hng s t l, thay i theo bin cas(t). Nu s(t) tnh bng volt, Kfc n v l Hz/v hoc 1/v.sec .

V tn s l o hm ca pha, nn

(t) = 2 fo

t i ()d = 2 [fCt + Kf o

t s()d] (5.6)

Gi siu kin u bng zero, sng bin iu c dng:fm(t) = A cos (t).

+=

t

0

fcf d)(sKtf2cosA)t(m (5.7)

Nhl, nu t s(t) = 0, phng (5.7) s thnh mt sng mang thun ty.Td . V sng AMSC v FM cho cc tn hiu thng tin nh hnh 5.4.Gii:


101/200


Trang V.4

Hnh 5.4

m1(t)

sm1(t)

s1(t)

t

sm2(t)

s2(t)

t

m2(t)

Hnh 5.4Tn s ca fm(t) thay i t fC + Kf[min . s(t)] n fC + Kf[max . s(t)].


102/200


Trang V.5

Bng cch lm cho Kf nh mt cch ty , th tn s ca fm(t) c thc gi mt cchty xung quanh fC. iu lm tit gim c kh bng.

Nh l s bin iu th khng tuyn tnh cho s(t). Nu thay s(t) trong phng trnh (5.7)bng mt tng gm nhiu tn hiu th sng FM kt qu khng l tng ca cc sng FM thnhphn. iu ng, v:

Cos (A + B) cosA + cosB.Ta chia bin iu FM lm 2 nhm; ty thuc vo cca Kf. Vi Kf rt nh ta c FM bng

hp; v Kf ln ta c FM bng rng.

BIN IU PHA.

Khng c s khc bit cbn gia bin iu pha v bin iu tn s. Hai ty thngc dng thay i cho nhau. Bin iu mt pha bng mt sng th cng nh bin iu o hmca n (tn s) vi sng y.

Sng bin iu pha cng c dng:pm(t) = A cos (t).

Trong (t) c bin iu bi s(t). Vy:(t) =2 [fCt + Kp s(t)] (5.8)Hng s t l Kp c n v V-1. Sng PM c dng:

(5.9)pm(t) = A cos 2 [fCt + Kp s(t)]

Khi s(t) = 0, sng PM trthnh sng mang thun ty.Ta c th lin h PM vi FM bng cch dng nh ngha ca tn s tc thi:

fi (t) = fC + Kpds

dt

(5.10)

Trng rt ging vi (5.5), trng hp ca FM.Thc vy, khng c s khc bit gia vic bin iu tn s mt sng mang bng s(t) v

vic bin iu pha ca cng sng mang bng tch phn ca s(t). Ngc li khng c g khcnhau gia vic bin iu pha ca mt sng mang bng s(t) v bin iu tn s cng sng mangy bng o hm ca s(t).

V vy, tt c cc kt qu sau y th chuyn d dng gia 2 loi bin iu.

FM BNG HP (NARROW BAND FM).

Nu Kfrt b, ta c th dng php tnh xp xn gin phng trnh ca sng FM.

=

t

0fcf)ds(K+tf2cosA)t(

m(5.11)

trnh vic lp li nhiu ln, ta t g(t) l tch phn ca tn hiu cha tin.

= t

0

d)(s)t(g (5.12)

Phng trnh (5.11) trnn:

fm(t) = A cos 2[ ]f t + K g(t)c f (5.13)

Dng lng gic, khai trin hm cosine:


103/200


Trang V.6

fm(t) = Acos2fCt . cos2Kfg(t) - A sin2fCt . sin2Kfg(t) (5.14)

Cosine ca mt gc b 1. Trong khi sin ca n gn bng chnh n.

Vy, nu Kf nh sao cho 2Kf g(t) biu din cho mt gc rt nh, ta c th tnh xp x

phng trnh (5.14):

fm(t) Acos2fCt - 2Ag(t) Kfsin2fCt (5.15)

Php tnh ny tuyn tnh vi g(t) v nh vy tuyn tnh vi s(t). Ta c th tnh bin i F

ca n (vi mt t kh khn) nh sau:

Bin i F ca g(t) lin h vi s(t) bi:

G(f) =S(f)

j2 f

Ly bin i F ca (5.15):

( ) ( )[ ] ( ) ( )

++

+++=

fcfffS

fcfffS

j4AK2ffff

2Afm(f) ccfcc (5.16)

Hnh 5.5: Bin i F ca sng FM.FM bng hp c 3 vn :- Tn s c th tng cao n mc cn thit truyn i c hiu qa, bng cch

iu chnh fCn tr mong mun.- Nu tn s sng mang ca ngun tin ln cn cch n t nht 2fm, th cc tn hiu

cha nhng ngun tin khc nhau c th truyn cng lc trn cng mt knh.- s(t) c th hi phc t sng bin iu. V phn sau ta s thy, cng mt khi

hon iu c th tch sng cho FM trong c 2 trng hp Kfnh v Kfln.Kh bng ca sng FM l 2fm, ng nh trng hp AM hai cnh. Th d dng ting hut

so (ti a 5000Hz) bin iu mt sng mang. Gi s s di tn ti a l 1Hz. Nh vy, tns tc thi thay i t (fC - 1)Hz n (fC + 1)Hz. Bin i F ca sng FM chim mt bng gia

(fC - 5000)Hz v (fC + 5000)Hz.R rng, tn s tc thi v cch thc m n thay i gp phn (c 2) vo kh bng caFM.

Gi l Bng hpkhi Kf nh, l v khi Kf tng, kh bng s tng t tr ti thiu 2fm.

PM BNG HP.

Bin iu pha bng s(t) th ging nh bin iu tn s bng o hm ca s(t). V o hmca s(t) cha cng khong tn s nh s(t), nn kh bng ca PM bng hp cng chim vng tns t gia fC - fm v fC + fm. Tc l kh bng rng 2fm.

Vi FM bng hp, tr max ca 2kf g(t) l mt gc rt nh (Trong g(t) l tch phn cas(t)).


104/200


Trang V.7

Vi PM bng hp, 2Kp s(t) phi l mt gc rt nh. iu ny cho php tnh xp x cosinev sine (s hng th nht trong chui khai trin).

FM BNG RNG (WIDE BAND FM).

Nu Kfnh khng cho php tnh xp x nhphn trn, ta c FM bng rng. Tn

hiu c truynfm(t) = A cos 2[ ]fct + Kfg(t) (5.17)

Trong g(t) l tch phn ca tn hiu cha tin s(t). Nu g(t) l mt hm bit, bin i Fca sng FM s tnh c. Nhng trong nhng trng hp tng qut, khng th tm bin i Fcho sng FM, v s lin h phi tuyn gia s(t) v sng bin iu. Nhng phn gii thc hintrong phm vi thi gian.

Ta gii hn trong mt trng hp ring, dng tn hiu mang tin l mt Sinusoide thun ty.iu ny cho php dng lng gic trong phn gii.

S(t) = a cos 2fmta: hng s bin .

Tn s tc thi ca sng FM c cho bi:fi (t) = fC + aKfcos 2fmt (5.18)

Sng FM c dng:

fm(t) = A cos 2 fctaKff

sin2 f m tm

+

(5.19)

Ta nh ngha ch s bin iu :

aK

ffm

, : khng n v (5.20)

fm(t) = A cos (2fCt + sin2fmt)fm(t) = Re {A exp (j2fCt +j sin 2fmt)} (5.21)

Hm expo trong (5.21) phn thnh mt tch, trong tha s th 2 c cha tin. l: expo(j sin 2fmt).

l mt hm tun hon, chu k 1/fm.Khai trin chui F phc, tn s fm.

+

=

=n

tf2jnn

tf2sinj mm eCe (5.22)

H sF cho bi:

=

m

m

mm

f1

f1

tf2jntf2sinjmn dteefC (5.23)

Tch phn ca (5.23) khng tnh c, n hi t ti mt tr gi thc. Tr gi thc l mthm ca n v . N khng phi l mt hm ca fm. Tch phn c gi l hm Bessel loi mt,k hiu Jn().

HM BESSEL.

Hm Bessel loi 1 l gii p ca phng trnh vi phn:

x2

d y

dxx

dy

dx

2

2

++ ( x

2

- n2

) y( x ) = 0


105/200


106/200


Trang V.9

hm Bessel gim rt nhanh khi n tng. l mt tnh cht chnh tc tim kh bng ca sngFM.

Hnh 5.7: Jn (10) l mt hm ca n.

Trli phng trnh (5.23), ta thy cc h sFourierc cho bi: Cn = Jn ().V sng FM trnn:

fm (t) = Re (5.25)Aj fc t Jn

jn f m te en

2

( )

=

2

nf t

V ej2fct khngl mt hm ca n, ta em vo du tng:

fm (t) = Re A Jn j t f c nfmen

( ) ( ) 2 +

=

V ly phn thc:

fm (t) = A J fn Cn

m( ) cos ( ) 2 +=

(5.26)Ta rt gn sng FM thnh tng ca cc Sinusoids. Bin i F ca tng ny l mt chui

xung lc.


107/200


Trang V.10

Hnh 5.8: Bin i F ca FM, i vi tin tc l Sinusoids.Ta ang gp phi mt rc ri ln ! Bin i ny mrng theo c 2 chiu t tn ssng

mang. N c mt kh bng rng v hn. D Jn() tin n zero ti vi tr gi, nhng kh bngrng th khng b gii hn. Nh vy, ta khng th truyn c hiu qu v cng khng th phi hp

nhiu ngun tin ring l vo chung mt knh ( Multiplexing ) ( v trng f ).Vi khng i, cc hmJn() tin n zero khi n tng. Vi s chn la ,shng J0()tin n zero v sng mang b loi. Trong trng hp AM, s loi b sng mang lm tng hiusut. Nhng i vi FM, s loi sng mang khng c li g c v cng sut ton phn gikhng i.

a * tnh xp x kh bng ca sng FM, ta xem cc xung hnh 5.8. Trc ht, ta chnmt trnh. T hnh 5.6, ta thy rng, nu < 0,5 th J2() < 0,03. Cc hm Bessel bc caohn (n > 2) th nh hn. Ti =0,5, J1 l 0,24. Vi nhng tr nh ny ca , bin i Fhnh 5.8chbao gm 5 xung lc gn sng mang. l, thnh phn ti sng mang v 2 thnh phn cch fm k t sng mang. iu , cho mt kh bng l 2 fm. Ta bit iu v nhng tr rt nhca (aKf/fm) tng ng vi iu kin bng hp.

b * By gi, gi skhng nh, th d = 10. Nhng tnh cht m ta ni trn ch rngJn(10) s gim nhanh chng, khi n > 10. Xem hnh 5.8, ta thy nhng thnh phn c ngha lsng mang v 10 ha tn mi bn ca sng mang. Mt cch tng qut: Vi ln,s s hng(thnh phn) mi bn ca sng mang l (c lm trn snguyn ). iu cho mt kh

bng l 2fm.Gn y, Jonh Carson a ra nh lut: Kh bng ca sng FM th xp x bng hm ca tn

s tn hiu cha tin v ch s bin iu:BW 2(fm + fm) (5.27)

iu tha nhn 2 trng hp gii hn. Vi rt nh, kh bng 2fm v ngc li vi ln, kh bng 2fm.

Thay = aKf/fm vo (5.27): BW 2(aKf+fm) (5.28)* Ta nhli tn s tc thi c cho bi phng trnh (5.18):

fi (t)=fC + aKfcos2fmtTa thy rng fm l nhp thay i ca fi (t) ,trong lc aKf l tr ti a m n di tn t sng

mang - c 2 i lng y iu tham gia vo kh bng ca sng FM.Th d: Tm bng xp x ca cc tn s b chim bi sng FM vi sng mang c tn s

5khz, Kf= 10Hz/V v:

a) s(t) = 10 cos10t.b) s(t) = 5 cos20t.

c) s(t) = 100 cos2000t.Gii:

a) BW 2(aKf+fm) = 2[10(10)+5] = 210Hz.b) BW 2(aKf+fm) = 2[5(10)+10] = 120Hz.c) BW 2(aKf+fm) = 2[100(10)+1.000] = 4khz.

Bng ca nhng tn s b chim:

a) 4895 n 5105 Hz.b) 4940 n 5060 Hz.c) 3 n 7 Khz.


108/200


Trang V.11

Phng trnh (5.28) c khai trin cho trng hp c bit ca mt tn hiu cha tin hnhSinusoide. Nu s bin iu l tuyn tnh, th ta c th p dng cng thc ny cho thnh phn tns cao nht ca s(t) tm kh bng. Nhng, FM th khng tuyn tnh nn cch y khng ng.

Ta s tm mt cng thc tng t cho trng hp tng qut. Hnh 5.9, ch tn s tc thica trng hp c bit m tn hiu cha tin Sinusoide v trng hp tng qut.

Hnh 5.9: Tn s tc thi

akf

Trong trng hp s(t) hnh sin, aKf l di tn ti a ca tn s so vi fc. V trongtrng hp tng qut di tn ti a tng t k hiu l f. Cng thc tng qut cho (5.28) l:

(5.29)

Nu f rt ln so vi fm, ta c FM bng rng, v tn s ca sng mang thay imt khong rng, nhng vi nhp chm. Tn s tc thi ca sng mang thay i chm t fC-fn fC+f. Nh vy sng FM xp x vi mt Sinusoide thun trong mt thi gian di. Ta c

th nghl n l tng ca nhiu Sinusoide vi cc tn s nm gia 2 gii hn. Nn bin i F thgn bng vi s chng ( Superposition ) cc bin i F ca nhng sinusoide y tt c nm tronggii hn tn s. Vy thc hp l gi s rng kh bng th xp xvi brng ca khong tn

s

ny, hoc 2f.

BW 2( f + fm )

Nu f rt nh, ta c mt sng mang thay i trong mt khong rt nh ca tns, nhng vi nhp nhanh. Ta c th tnh gn ng bng 2 mch giao ng ti nhng gii hntn s. Mi giao ng c Cng ha trong na thi gian ton th. Bng ca cc tn s bchim bi output ca H 5.10 l t fC - f - fm n fC + f + fm.

Vi f nh, kh bng l 2fm .Ta thy kh bng ca sng FM tng vi s tng tr gi ca Kf. Vim ny, s dng FM

bng hp ( vi kh bng ti thiu 2fm ) l hp l. Nhng, FM bng rng li c u im v tritnhiu hn c FM bng hp v AM.

Hnh 5.10: Xp x ca FM bng hp


109/200


Trang V.12

V d: Mt sng mang 10MHz c bin iu FM bi mt tn hiu Sinusoide c tn s5KHz, sao cho di tn ti a ca sng FM l 500KHz - Tm bng xp x ca cc tn s bchim bi sng FM.

Gii:Kh bng xp x

BW 2(f + fm).BW 2(500KHz + 5KHz) = 1.010 KHz .

Vy bng ca tn s b chim th tp trung quanh tn s sng mang, v trong khong t9.495 n 10.505KHz. Tn hiu FM th d ny l bng rng. Nu n l bng hp, kh bng sch l 10KHz.

Th d: Mt sng mang 100KHz b bin iu FM bi mt tn hiu sinusoide c bin 1V. Kfc tr 100Hz/V.

Tm kh bng xp x ca sng FM nu tn hiu bin iu c mt tn s 10KHz.Gii:

Ta li dng php tnh xp x ca Carson:BW 2(f + fm)

V tn hiu cha tin s(t) c bin n v, di tn ti a fc cho bi kf, hoc100Hz .fm l 10 Khz, tn s ca tn hiu bin iu. Vy :

BW 2(100Hz + 10 Khz) = 20.200Hz .V fm rt ln so vi f , y l tn hiu FM bng hp. Kh bng cn thit truyn

cng tin tc khi dng DSB AM s l 20KHz, xp x vi kh bng ca sng FM ny.V du: Mt sng bin iu gc c m t bi:

(t) = 10 cos[2 x 107t + 20cos1000t]Tm kh bng xp x ca sng ny.

Gii:fm l 500Hz. tnh f, trc ht ta tm tn s tc thi:

fi (t) = 12

ddt

( 2 x 107t + 20cos1000t ).

= 107-10.000 sin 1000t . di tn ti a ca 10.000 sin1000t, hoc 10KHz. Vy kh bng xp xc cho bi:

BW 2( 10.000 + 500 ) = 21khz .R rng y l mt sng FM bng rng v f rt ln so vi fm. Nhl ta khng bit y l

bin iu tn s hoc pha khi tm kh bng.

KHI BIN IU.

Ta thy sng FM c kh bng gii hn chung quanh sng mang fC. Nh vy tiu chunth nht ca mt h thng bin iu c tha. Ta c th truyn tin mt cch hiu qu bngcnh chn fC trong mt khong ring. V ta cng c th Multiplexing nhiu tn hiu ng trong

cng mt knh bng cnh lm cc tn s sng mang ln cn cch bit nhau sao cho bin i Fca ca cc sng FM khng ph nhau v tn s.

Tiu chun th 2, l chng tc s(t) c thc hi phc t sng bin iu gc. Vcc khi bin iu, hon iu c th thc hin c trong thc t.

Ta bt u xem li FM bng hp - dng sng c din t bi phng trnh (5.15).fm(t) = A cos2[fct - Kfg(t)]

fm(t) = A cos2f

ct - 2A g(t) K

fsin 2f

ct (5.30)

Phng trnh ny tc khc a n s khi nh hnh 5.11.- Biu thc tng ng cho PM bng hp:


110/200


111/200


Trang V.14

Hnh 5.14: Mch pht FMMt mch dao ng cao tn to sng mang, c tn s quyt nh bi mch iu hp ( hoc

thch anh ) u song song vi mt doide bin dung (Varicap). in dung ca varicap c th thayi bng cnh lm thay i dng chy ngang qua n (nu phn cc thun) hoc in tht ln 2u n (nu phn cc ngc). S thay i in dung ca varicap s lm thay i tn s ca mchgiao ng. Nu dng hay thi ngang qua varicap thay i t l vi tn hiu cha tin th tn sca mch giao ng thay i t l vi tn hiu ny. V sng FM sc to ra.

Trong hnh 5.14. Bn phi D l mch giao ng m tn sc lm thay i. Bn tri D lmch phn cc v ghp tn hiu s(t) vo doide D. T C2 c tr rt ln so vi tr ca in dungVaricap, nn ch c tc dng cch ly DC. RFC, cun chn cao tn, nga tn hiu dao ng ghpngc li ngun phn cc. C1: t phn dng.

KHI HON IU.

Xem dng sng bin iu FM nh sau:fm(t) = A cos2( fct + Kf s()d ) .

0

t

S hon iu hi phc li s(t) gm 2 loi:- Tch sng phn bit ( Discriminator ), tch mt thnh phn tn s ra khi cc thnh phn

khc v chuyn sthay i tn sthnh thay i bin ri tch sng ging nh AM.- Vng kha pha ( Phase - Lockloop ) phi hp mt dao ng ni vi sng mang c

bin iu.

1. Tch sng phn bit. (discriminator)

A. Ly o hm mt Sinusoide l tin trnh nhn Sinusoide vi tn s tc thica n:

d

dt

= -2A [ fc + Kfs(t) ] sin2(fct + Kf0

t

s()d ) .


112/200


Trang V.15

Hnh 5.15: o hm ca sng FMGi s tn s tc thi th ln hn nhiu so vi fm (hp l vi thc t). Thnh phn sng

mang lp y vng gia bin v nh qua gng ca n. Thc t, vng din tch gia ngbin trn v ng bin di b che kn do tn s qu cao ca sng mang. Nh vy, ngay c khitn s sng mang khng l hng, bao hnh ca sng vn c nh ngha:

2A[fC + Kfs(t)] (5.34)S thay i cht t ca tn s sng mang s khng ng k bi mt tch sng bao hnh.Trong cc h thng tin thc t, fC >> Kf s(t). Vy lng nm trong ngoc ca (5.34) th

dng, v ta c th bu tr tuyt i.Tm li: Mt mch vi phn v sau l mt tch sng bao hnh s c th dng hi

phc li s(t) t sng FM.

Hnh 5.16: Hon iu FM.

Nu s bin iu l PM, th output ca h hnh 5.16 l o hm ca s(t). Khi cn thmmt mch tch phn ng ra ca h.

Hm h thng ca mch vi phn:H(f) = 2 jf

Hnh 5.17: c tuyn Sut ca mch vi phn.c tuyn Sut c vhnh 5.17. Sut ca output ca mch vi phn th t l tuyn tnh

vi tn s ca input. Nh vy mch vi phn i FM thnh AM. Khi mt mch vi phn dng nhth, ta gi n l mt discriminator.

b. C mt loi Discriminator khc. Bt k h thng no c mt sut hm hthng gn - Tuyn tnh vi tn s trong khong dy tn ca sng FM siui FM thnh AM.

Th d: Mt BPF s lm vic nh mt Discriminator nu cho n hot ng trn mt

khong gii hn ca kh bng, nh hnh 5.18.

)f(H

f

Gn tuyn tnh

Hnh 5.18


113/200


Trang V.16

Ta c th chng minh s tuyn tnh ca BPF Discriminator theo cch thc tng t nhkhi bin iu cn bng.

Xem mch in hnh 5.19. Na trn ca my bin th L1 v C1iu hp ti fa . Na dimy bin th v C2iu hp ti fb.

1D

2D

Hnh 5.19: Tch sng dc

Hnh 5.20: DiscriminatorMch in trn y gi l tch sng dc ( Slope Detector ) v n dng on dc ca c

tuyn mch lc tch sng.

C. By gita trli khi vi phn gc. Ta s thy mt cch tip cn khc. Ta cth tinch o hm mt cch gn ng bng vi tn hiu ca hai tr mu casng:

(t) - ( t - to ) tod

dt

.

iu ny dn n khi hon iu nh hnh 5.21.V mt s di thi gian th tng ng vi mt s di pha, nn khi ny gi l hon iu

di pha ( Phase Shif Demodulator ).

)t(mf )t(s

Hnh 5.21: Hon iu di pha.

2. Vng kha pha (phase - lockloop).

Vng kha pha PLL l mt mch hi tip, c thc dng hon iu sng bin iugc. Mch hi tip thng c dng gim thiu error (v zero). Trong trng hp PLL, errorl mt hiu pha gi tn hiu ng vo sng FM v mt tn hiu chun hnh sin. (VCO) .


114/200


Trang V.17

)cos(2

121

Error

PLL tch sng FM: Trc ht, xem mch so pha; gm 1 mch nhn v mt lc LPF.

Cho hai tn hiu vo

Documents

Co So Vien Thong_Bao Ve Cac Thanh Phan Chinh