các loại điều chế số

7/30/2019 cc loi iu ch s

1/40

n mn hc GVHD: Nguyn V Anh Quang

LI NI U

Vo u th k 20 Marconi thnh cng trong vic lin lc v tuyn qua i Ty

Dng, Kenelly v Heaviside pht hin mt yu t l tng in ly hin din tng

pha trn ca kh quyn c th dng lm vt phn x sng in t. Nhng yu t m ra mt k nguyn thng tin v tyun cao tn i quy m.

Chin tranh th gii ln th hai l mt bc ngot trong thng tin v tuyn.

Thng tin tm nhn thng - lnh vc thng tin s dng bng tn s cc cao (VHF)

c nghin cu lin tc sau chin tranh th gii - tr thnh hin thc nh s pht

trin cc linh kin in t dng cho VHF v UHF, ch yu l pht trin ngnh

raa.Vi s gia tng khng ngng ca lu lng truyn thng, tn s ca thng tin v

tuyn vn ti cc bng tn siu cao (SHF) v cc cao (EHF). Vo nhng nm

1960, phng php chuyn tip qua v tinh c thc hin v phng php chuyn

tip bng tn x qua tng i lu ca kh quyn xut hin. Do nhng c tnh u

vit ca mnh nh dung lng ln, phm vi thu rng, hiu qu kinh t cao, thng tin v

tuyn c s dng rt rng ri trong pht thanh truyn hnh qung b, v tuyn o

hng, hng khng, qun s, quan st kh tng, lin lc sng ngn, thng tin v tinh -

v tr v.v... Tuy nhin, can nhiu vi lnh vc thng tin khc l iu khng trnh khi,bi v thng tin v tuyn s dng chung phn khng gian lm mi trng truyn dn.

Chnh v th iu ch tn hiu l mt phn khng th thiu c trong truyn dn

ca thng tin v tuyn. iu ch gip chng ta c th truyn i thng tin hoc tn hiu

mong mun v nhn c nhng tn hiu m mnh mun c.

n c chia lm ba phn :

CHNG 1: TNG QUAN V H THNG THNG TIN

CHNG 2: CC KHI NIM C BN CA IU CHCHNG 3: IU CH MT S LOI TN HIU V V D MINH

HA

Trong khi tm hiu em cn nhiu thiu st mong thy gio Nguyn V Anh

Quang v cc bn gp bi lm hon chnh hn.

Xin chn thnh cm n !

SVTH: L Hoi Giang Trang 1


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MC LC

LI NI U..............................................................................................................1MC LC....................................................................................................................2DANH MC HNH.....................................................................................................4

DANH MC T VIT TT.......................................................................................5CHNG I: TNG QUAN V H THNG THNG TIN....................................6

1.1. S v chc nng c bn..............................................................................61.2. Cc mi trng truyn dn..............................................................................7

1.2.1.Mi trng truyn dn bng cp kim loi...................................................7

1.2.2.Mi trng truyn dn cp quang...............................................................8

1.2.3.Mi trng truyn dn v tuyn..................................................................8

CHNG II: CC KHI NIM C BN CA IU CH...............................10

2.1. Tn hiu iu ch.............................................................................................102.2. Tn hiu sng mang.........................................................................................102.3. iu ch............................................................................................................10

2.3.1.iu ch......................................................................................................10

2.3.2.iu kin iu ch......................................................................................10

2.3.3.S cn thit ca iu ch...........................................................................10

2.4. Phn loi iu ch............................................................................................112.4.1. Phn loi theo tn hiu a vo iu ch.................................................11

2.4.2. Phn loi theo s thay i ca cc tham s ca sng mang....................112.4.2.1. iu ch bin ..................................................................................112.4.2.2. iu ch tn s....................................................................................112.4.2.3. iu ch pha........................................................................................112.4.2.4. iu ch QAM (Quature Amplitude Modulation)................................12

CHNG III : IU CH MT S LOI TN HIU V V D MINH HA.....................................................................................................................................13

3.1. iu ch bin ...............................................................................................133.1.1. iu ch bin tng t (AM)...............................................................13

3.1.2. iu ch bin s (ASK)........................................................................173.1.3. M phng tn hiu AM bng phn mm Matlap......................................18

3.2. iu ch tn s v iu ch pha......................................................................213.2.2. iu ch FSK............................................................................................27

3.2.3. iu ch PSK.............................................................................................28

3.2.3.1. iu ch pha 2 trng thi (2 PSK).......................................................283.2.3.2. iu ch pha 4 trng thi (QPSK).......................................................293.2.3.3. M phng Matlap vi tn hiu iu tn v iu pha.............................31

3.3. iu ch QAM.................................................................................................323.3.1. nh ngha.................................................................................................32

3.3.2. iu ch QAM 16 trng thi (iu ch 16 QAM).....................................32



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3.4. iu ch nh phn............................................................................................343.4.1. iu ch nh phn.....................................................................................34

3.4.2. iu ch kha dch bin nh phn (BASK).........................................34

3.4.3. iu ch kha dch pha nh phn (BPSK)...............................................36

3.4.4. iu ch kha dch tn nh phn (BFSK)................................................37

TNG KT................................................................................................................39TI LIU THAM KHO.........................................................................................40



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DANH MC HNH

Hnh 1.1: Cc phn t ca h thng thng tin...........................................................6Hnh 3.1 : iu ch tn hiu AM mc thp..............................................................13Hnh 3.2 : iu ch tn hiu AM mc cao................................................................13

Hnh 3.3: Dng tn hiu V , V0 v VAM................................................................14Hnh 3.5 : iu ch AM n m...............................................................................15Hnh 3.6: iu ch AM bng mt diot......................................................................16Hnh 3.7 : ...................................................................................................................17Hnh 3.8: iu ch ASK.............................................................................................17Hnh 3.9: Tn hiu iu ch ASK..............................................................................18Hnh 3.10: M phng tn hiu bng phn mm ......................................................21Hnh 3.13: iu ch PM ............................................................................................24Hnh 3.17: Dng sng ca mch iu ch tn s......................................................27Hnh 3.19: Tn hiu iu ch v sng mang sau iu ch ca FSK........................28

Hnh 3.20: ..................................................................................................................28....................................................................................................................32

Hnh 3.26: iu ch on-off: dng sng, b iu ch, ph........................................35Hnh 3.27:...................................................................................................................36Hnh 3.28: iu ch PRK: dng sng, b iu ch, ph..........................................37Hnh 3.29: B thu tng quan PRK.........................................................................37Hnh 3.30: iu ch BFSK: dng sng, b iu ch, ph........................................38



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DANH MC T VIT TT

T vit tt Ngha ca cc t Chc nngBER Bt Error Rate S bt b li

FSK

Frequency Shift Keying

iu ch tn s s

AM Amplitude Modulationiu ch bin

tng t

ASKAmplitude Shift Keying

iu ch bin s

QAM Quature Amplitude Modulationiu bin cu

phngPSK Phase Shift Keying iu ch pha s

BASK Binary Amplitude Shift Keying iu ch kha dchbin nh phn

OOK On off keying iu ch khoa on-off

STR Symbol Timing RecoveryB khi phc ng

b

BPSK Binary Phase Shift Keyingiu ch kha dch

pha nh phn

BFSK Binary Frequency Shift Keyingiu ch kha dch

tn nh phn

QPSK Quadrature Phase Shift Keyingiu ch kha pha

cu phng

FM Frequency Modulationiu ch tn s tng

t

DFM Differential Frequency ModulationPh ca tn hiu iu

tn s



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CHNG I: TNG QUAN V H THNG THNG TIN

1.1. S v chc nng c bn

Hnh 1.1: Cc phn t ca h thng thng tinC ba phn cn bn ca bt c h thng thng tin no cng phi c: My pht,

my thu v knh truyn. Mi phn c mt vai tr nht nh trong vic truyn dn tn

hiu.

My pht x l tn hiu u vo v to ra tn hiu c nhng c tnh thch hp vi

knh truyn dn. Qu trnh x l tn hiu truyn dn ch yu l iu ch v m ha

(modulation and coding).

Knh truyn l mi trng gia im pht v im thu. Knh truyn c th l cp

i, cp ng trc, cp quang, hay mi trng v tuyn. Mi knh truyn u gy ra

suy hao v tn tht truyn dn. V th cng tn hiu b suy gim dn theokhong cch truyn.

My thu ly tn hiu u ra t knh truyn x l v ti to ngc li tn hiu

u pht. Cc hot ng ca my thu bao gm khuch i b vo tn hao truyn

dn, gii iu ch, gii m tn hiu c iu ch v m ha my pht. B lc

cng l mt phn quan trng trong my thu dng chn lc tn hiu mong mun t

knh truyn.

C rt nhiu nh hng khng mong mun xut hin trong qu trnh truyn dn

tn hiu. Suy hao l mt nh hng khng mong mun do gy ra suy gim cng

tn hiu ti my thu. Cc hiu ng khc nh mo (distortion), nhiu (noise), tp m

(interference) lm cho dng tn hiu b thay i do c nh hng nghim trng hn.

Mo l hin tng nh hng n dng sng tn hiu gy ra bi p ng khng l

tng ca h thng nh mong mun. Khng ging nh nhiu v can nhiu, khi khng

c tn hiu th khng c mo. Nu knh truyn l tuyn tnh nhng p ng c mo th



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mo ny c th c sa, hoc c th gim thiu bng b lc c bit gi l b cn

bng.

Can nhiu l nhng tn hiu tc ng t nhng ngun tn hiu khc vo tn hiu

cn truyn nh cc my pht khc, ng dy in. Can nhiu thng xut hin trong

cc h thng v tuyn do nhng anten thng thu nhiu loi tn hiu ng thi. Cc b

lc thng c s dng loi b can nhiu c tn s ngoi di tn ca tn hiu

truyn dn mong mun.

Nhiu hay l cc tn hiu in ngu nhin sinh ra bi cc qu trnh vt l trong h

thng v c t bn ngoi. Khi nhiu tc ng vo tn hiu truyn c th lm gim cht

lng ca tn hiu hay c th lm hng ng truyn. B lc dng gim nhiu mt

phn nhng nhiu khng th loi b hon ton. Nhiu l mt thnh phn c bn to ra

nhng gii hn trong h thng truyn thng.

1.2. Cc mi trng truyn dn

Truyn dn l qu trnh truyn ti thng tin gia cc im kt ni trong mt h

thng hay trong mng vin thng.

Mi trng truyn dn c hai loi l hu tuyn (c dy) v v tuyn (khng dy).

Mi trng truyn dn hu tuyn bao gm cc loi ng dy thng tin nh cp ng

nhiu i, cp ng trc, si quang Mi trng truyn dn v tuyn l khongkhng bao quanh Tri t, chnh l cc tng kh quyn, tng in ly v khong khng

v tr khc (khng phi chn khng).

1.2.1. Mi trng truyn dn bng cp kim loi

Cp kim loi l mt mi trng truyn dn l tng cho vic kt ni cc thu bao

vin thng. C hai loi cp kim loi chnh l cp i ph hp cho truyn tc thp v

cp ng trc c dng cho phn cp tc cao.

Cp i: thng c dng cho d liu tng t. Cht cch in c lm

bng giy. Tuy nhin nu c lm bng nha th tt hn (khng nh cm vi m,

suy hao t ti cc tn s cao) do n c dng nhiu trong cp i hin i. Phn

dn in thng c lm bng ng vi nhiu loi ng knh khc nhau. Cc si li

trong cp c xon vo nhau theo cp 2 hoc 4 dy ty vo ng dng c th. Phn li

chnh ca cp kim loi c hnh thnh nh cc cp dy xon (2 hay 4) theo cc lp

ng tm. Cc cp xon ny c t lin tip nhau v c thay i mt cch ngunhin gim s mt cn bng.



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Cp ng trc: dng cho c hai h thng ghp knh tn s FDM v h thng

ghp knh theo thi gian TDM. Cp ny bao gm li kim loi chnh gia v mt lp

dn khc bao ph bn ngoi c hnh ng. Lp dn hnh ng bn ngoi c lp bo v

chng nh hng ca can nhiu nn c th khng gy nhiu v khng b nh hng

bi cc si cp xung quanh Phn li bn trong bao gm nhiu si dy ng. Phn

dn bn ngoi c cu to t cc l ng, trong mt s loi cp th n c hn cht

thnh ng.

Cp ng trc c th thc hin cho cc tuyn truyn dn dung lng cao (10800

knh thoi trong h thng FDM). Chng thng c lp t theo tng i thc hin

thng tin trn hai hng gia cc tng i ni c lu lng ti tp trung cao.

1.2.2. Mi trng truyn dn cp quang

Truyn sng trn mi trng cp quang. Tuyn thng tin cp quang gm c si

quang v thit b thng tin quang. Thit b thng tin quang gm: chuyn i m nhnh

pht, chuyn i m nhnh thu, chuyn i tn hiu in - quang,

quang in

c im: bng thng rng, c li di khng cn b lp: c tnh mt mt tn hiu

l rt thp khong 0,2 dB/km. So snh vi cc phng tin khc nh i dy trn

khong 30 dB/km. t chu nh hng bi nhiu, an ton, gi thnh bo dng thp,kch thc gn nh, linh hot.

1.2.3. Mi trng truyn dn v tuyn

Thng tin c truyn i xa nh sng in t. Mi trng truyn sng (kh quyn

trn mt t, v tr, nc) chung cho nhiu knh v tuyn.

u im ni bt so vi truyn bng dy cp l khng cn bt k mt ng dy

no, lp t nhanh gn khng cn o bi, chi ph u t t.

Cc tnh cht ca knh truyn thng v tuyn:

Sng v tuyn li nhng tn s ny truyn thng gi l truyn dn tm nhn

thng, d b tc ng bi vt chn.

C suy hao ln (tn s cng cao, suy hao cng ln), thng t ti 140 n

160 dB, khuch i tn hiu kh v s khuch i c tp m.

suy hao ca knh v tuyn thay i trong phm vi rng. Cng trng

in t ti im thu t l nghch vi bnh phng qung ng m sng i qua.



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suy gim ca knh thng tin v tuyn bin i cn do s bin i cc

tham s kh quyn Tri t.

Mo tn hiu pht i do hn ch v ph tn ca n (nng lng tp trung di

tng i hp).

Cn kit v tn s do ngy cng c nhiu h thng v tuyn xut hin.



10/40


CHNG II: CC KHI NIM C BN CA IU CH

2.1. Tn hiu iu ch

L cc tn hiu tin tc cn truyn i (tn hiu s v tn hiu tng t bao gm: tn

hiu thoi, truyn hnh, s liu) c tn s thp.2.2. Tn hiu sng mang

L cc tn hiu in cao tn c th ti (mang) c thng tin. Tn hiu cao tn

y mang tnh cht tng i. Mt tn hiu in cao tn c th lm sng mang cho mt

tn hiu iu ch ny nhng li khng th lm sng mang cho mt tn hiu iu ch

khc v chnh n c khi li tr thnh tn hiu iu ch cho mt sng mang c tn s

cao hn.

Mt sng mang tiu biu c cng thc ton hc l:

U(t) = A sin (t + )

Trong : A l bin ca sng mang

= 2f l tn s gc ca sng mang

f l tn s ca sng mang

: pha ca sng mang

Cc tham s A, f v u c th mang thng tin.2.3. iu ch

2.3.1. iu ch

iu ch l qu trnh bin i mt trong cc tham s sng mang cao tn (bin

, hay tn s, hay pha) t l vi tn hiu bng gc BB (Base Band)

2.3.2. iu kin iu ch

1. Tn s sng mang cao tn fc (frequency carry), fc (8 10) Fmax

Trong : Fmax : tn s cc i tn hiu iu ch bng gc

2. Thng s sng mang cao tn bin i t l vi bin tn hiu iu ch

Bandbase m khng ph thuc vo tn s ca n.

3. Bin sng mang cao tn V > Vm (bin tn hiu iu ch bng gc)

2.3.3. S cn thit ca iu ch

Trong thng tin v tuyn: do cc tn hiu tin tc (tn hiu tng t v tn hiu

s) thng c tn s thp nn rt kh trc tip bc x thnh sng in t truyn ixa, nu c th bc x c th nng lng bc x cng rt yu v i hi phn t bc



11/40


x (anten) c kch thc ln. d dng truyn thng tin i xa bng sng in t,

ngi ta phi tin hnh iu ch tn hiu thng tin vo sng mang cao tn, ngha l gi

tin tc cn truyn vo sng mang sau mi cho sng mang iu ch bc x thnh

sng in t truyn i xa. V sng mang c tn s cao nn d bc x thnh sng

in t v i hi phn t bc x c kch thc khng ln.

Trong ghp knh theo tn s: V cc tn hiu thng tin cng loi u c chung

mt bng tn truyn dn (V d: tn hiu thoi c bng tn t 0,3 3,4KHz) nn khi

truyn nhiu tn hiu trn mt ng truyn dn th chng s b ln vo nhau lm cho

pha thu khng thu c tn hiu. truyn c nhiu tn hiu trn cng mt ng

truyn th ngi ta phi iu ch cc tn hiu cn truyn vo cc sng mang khc nhau,

mc ch l chuyn ph ca thng tin cn truyn ln cc vng khc nhau sau mitruyn chung trn mt ng truyn dn. Nh s khc nhau v vng ph ca cc tn

hiu truyn i m pha thu d dng thu c tn hiu.

2.4. Phn loi iu ch

2.4.1. Phn loi theo tn hiu a vo iu ch

- iu ch tng t: tn hiu iu ch l tn hiu tng t

- iu ch s: tn hiu iu ch l tn hiu s.

2.4.2. Phn loi theo s thay i ca cc tham s ca sng mang2.4.2.1. iu ch bin

L tc ng tn hiu iu ch vo sng mang lm cho bin ca sng mang

thay i theo quy lut ca tn hiu iu ch. iu ch bin tng t c gi l AM

(Amplitude Modulation). iu ch bin s c gi l ASK (Amplitude Shift

Keying).

2.4.2.2. iu ch tn s

L tc ng tn hiu iu ch vo sng mang lm cho tn s ca sng mang

thay i theo quy lut ca tn hiu iu ch. iu ch tn s tng t c gi l FM

(Frequency Modulation). iu ch tn s s c gi l FSK (Frequency Shift

Keying).

2.4.2.3. iu ch pha

L tc ng tn hiu iu ch vo sng mang lm cho pha ca sng mang thay

i theo quy lut ca tn hiu iu ch. iu ch pha tng t c gi l PM (PhaseModulation). iu ch pha s c gi l PSK (Phase Shift Keying).



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2.4.2.4. iu ch QAM (Quature Amplitude Modulation)

L phng php iu ch kt hp c iu ch bin ASK v iu ch pha PSK.

Vi iu ch ny th khi tn hiu iu ch tc ng vo sng mang th c bin v

pha ca sng mang u thay i theo quy lut ca tn hiu iu ch.



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CHNG III : IU CH MT S LOI TN HIU V V

D MINH HA

3.1. iu ch bin

3.1.1. iu ch bin tng t (AM)

nh ngha : bin sng mang cao tn t l vi tn hiu iu ch bng gc gi

l iu ch AM.

Hnh 3.1 : iu ch tn hiu AM mc thpng dng : trong truyn hnh truyn thanh

Hnh 3.2 : iu ch tn hiu AM mc caoTn hiu sng mang: Vo = V0.cos0t (1)

Trong : Vo : gi tr bin sng mang

0: tn s gc sng mang

Tn hiu bng gc: Vt = V.cost (2)Trong : V: l tn hiu bin bng gc

: tn s gc tn hiu bng gc

T (1) v (2) ta c tn hiu iu ch bin c dng:

( ) ( )

( ) ( )3.1tcostcosm1V

tcostcosV

V1V

tcostcosVVtV

00

0

0

0

00AM

+=

+=

+=


(3)


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Trong :0V

Vm = l h s iu ch hay cn gi l su iu ch. H s iu

ch m phi tha mn iu kin m 1. Nu m > 1 th mch c hin tng iu ch v

tn hiu mo trm trng (Hnh 3.3).

Trong thc t mmax = 0,7 0,8 m bo thu tn hiu khng b mo. Ta xc nh

m trong thc t bng cch o cc gi tr Vmax, Vmin v p dng cng thc:

( )4.1VV

VV

2

VV2

VV

V

Vm

minmax

minmax

minmax

minmax

0 +

=+

== (4)

Khi m =1 ta c Vmax = 2V0 v Vmin =0.

Bin i cng thc (3) ta c:

( ) ( ) ( )5.1tcos2

mVtcos

2

mVtcosVV

0

0

0

0

00AM+++=

0 5 10 15 20

-3

-2

-1

0

1

2

3

0 5 10 15 20

-4

-3

-2

-1

0

1

2

3

4


(5)

V0

t

0t

V

0

t

VAM

m < 1

t

VAMA

m = 1

0 5 10 15

-6

-4

-2

0

2

4

6

t

VAM

m > 1

Hnh 3.3: Dng tn hiu V, V0 v VAM


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Nh vy khi iu ch n m ph ca tn hiu iu bin AM c ba thnh

phn: ti tin c tn s 0v bin V0, hai dao ng bin c tn s 0 v c bin

2mV

0 nh Hnh 3.4a.

Khi m = 1 th2

VV 0AM = .

Nu ta iu ch mt di m tn

(min max) vo ti tin, ta s c ph

ca tn hiu AM nh Hnh 3.4c.

Ta thy ngoi ti tin 0 c bin V0 cn

c hai bin tn: bin tn trn c tn s t

(0 - min) n (0 + max) v bin tn

di c tn s t (0 - max) n

(0 + min) i xng qua ti tin.

Thc cht ph ca hai bin khng ng

u nhau m cng xa 0 th bin cng

gim do c tuyn lc ca b cng hng

khng phi l hnh ch nht l tng.

Tn hiu bng gc Tn hiu iu ch AM

Hnh 3.5 : iu ch AM n mNhn xt v iu ch AM:

Cng sut mang khng ti tin th nhiu


0

- 0

+ 0

min

max

0 - max

0+

max0

0

- min

0

+ min

0

0

0

Hnh 3.4: Ph ca tn hiu AM

VAM

V

2

mV0

V0


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Cng sut cao tn ti tin nh hai bin nh nhau v ph thuc vo h s

iu ch m

Bng thng cn truyn ln gp i cn thit gy lng ph v tng nhiu

Xt v tnh hiu qu s dng cng sut cao tn km

D thc hin tn hiu AM v my thu gii iu ch n gin, r

iu ch AM dng trong pht thanh qung b MW-SW

V d: M t mt phng php n gin iu ch AM. Mch gm mt mng trn

tn hiu bng in tr (R1, R2, R3) mt diot nn v mt khung cng hng LC.

Hnh 3.6: iu ch AM bng mt diotTn hiu iu ch c a n u vo R1, tn hiu sng mang c a n u

vo R2. Mng in tr R1, R2 v R3 thc hin trn tuyn tnh hai tn hiu vi nhau theo

nguyn tc cng s hc. Nu tn hiu iu ch l tn hiu hnh sin th tn hiu sau khi

trn (ly trn in tr R3) c dng nh hnh 4.3 (c). Ta thy rng sng mang bin

thin trn nn ca tn hiu iu ch, nhng y cha phi l tn hiu iu ch bin

. y hai tn hiu mi c cng vi nhau, trong khi iu ch l nhn hai tn

hiu vi nhau. Tn hiu sau khi cng c nn qua diode D. Sau khi nn ta thu c

mt dy xung dng l tp hp ca cc na chu k dng ca tn hiu tng, bin

ca cc xung thay i theo quy lut ca tn hiu iu ch hnh 4.3 (d). Cc xung ny

c a n kch thch cho mt mch cng hng song song LC. Khung cng hng

LC ny c tn s cng hng ring fLC

=1

2ng bng tn s ca sng mang.


Ng ra


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Hnh 3.7 :a) Tn hiu iu ch

b) Sng mang

c) Tn hiu trn in tr R3

d) Dng tn hiu qua diode

e) Tn hiu trn khung cng hng

Khi c kch thch th khung LC s dao ng vi tn s ng bng tn s sng

mang, cn bin th ph thuc vo bin ca tn hiu kch thch. Khi xung kch

thch c bin ln th dao ng ly ra trn khung c bin ln, khi bin ca

xung kch thch nh th dao ng ly ra trn khung c bin nh.

Nh ni trn, bin ca cc xung kch thch thay i theo quy lut ca tn

hiu iu ch nn dao ng ly ra trn khung LC cng c bin bin thin ng theo

quy lut ca tn hiu iu ch. y chnh l tn hiu iu ch bin hnh 4.3 (e).

3.1.2. iu ch bin s (ASK)

iu ch ASK l iu ch bin p dng cho tn hiu iu ch l tn hiu s.

Hnh 3.8: iu ch ASK S(t): tn hiu iu ch (tn hiu s)

LO mch dao ng to sng mang (Local Ocsilator)

A.sint: sng mang c to ra t mch dao ng LO

M: B nhn

BPF: b lc thng bng (BPF: Band Pass Filter)



18/40


S(t).Asint: sng mang c iu ch bin

Tn hiu iu ch S(t) l tn hiu s c a n u vo th nht ca b nhn

M. Sng mang A.sint to ra t b dao ng LO c a n u vo th hai ca b

nhn. B nhn M s nhn tn hiu iu ch S(t) vi tn hiu sng mang A.sint. u

ra ca b nhn ta thu c thnh phn c bn l S(t).Asint v mt s thnh phn

khc (v d A.sin2t, S(t).A sin3t ...). Tn hiu ny c a qua b lc thng gii

loi b cc thnh phn khng mong mun. Kt qu u ra ca b lc ta thu c

tn hiu iu ch bin (ASK) l S(t).Asint. Dng sng ra sau iu ch c m

t hnh 4.5.

iu ch ASK c u im l mch in iu ch v gii iu ch u rt n

gin. Tuy nhin c nhc im rt ln l kh nng chng nhiu thp v nhiu d dngtc ng vo vng bin (vng mang tin ca sng mang iu ch ASK).

Hnh 3.9: Tn hiu iu ch ASK3.1.3. M phng tn hiu AM bng phn mm Matlap

Trong tn hiu iu bin, cc bin tn cha tin tc, cn ti tin khng mang

tin tc. Nh vy cng sut ti tin l cng sut tiu hao v ch, cn cng sut bin tn l

cng sut hu ch.

Cng sut ti tin l cng sut bnh qun trong mt chu k ti tin:

Cng sut bin tn:

( ) ( ) ( )

( ) ( ) ( )8.12mPPPP

7.12

mP

R2

1

2

mVPP

2

btbtbt

2

L

2

0btbt

000

000

+

+

=+=

=

==


Po

= (1.6)

V20

2RL

(6)

(7)

(8)


19/40


Khi iu ch su (100%): m = 1 th2

PP 0bt

= (9)

T (3) ta c: VAMmax = V0(1+m)

Do :( )

( ) 2

L

220

maxAM m1P

R2

m1VP

0+=

+= (10)

Khi m = 1 th PAMmax = 4Po (11)

Vy cng sut trung bnh trong mt chu k iu ch:

( )12.12

m1P

2

mPPPPP

2

btAM 0000

+=+=+=

Nu m = 1 th PAM = 3/2 Po (13)

Ptb = 1/3 PAM (14)

H s li dng cng sut:

( )15.1

1m

21

2

m1P

2

mP

P

Pk

2

2

2

AM

bt

0

0

+

=

+

==

Khi iu ch su nht m = 1 th3

1k= tc l cng sut hu ch ch bng mt

phn ba tng cng sut pht i.

Trn thc t tn hiu khng mo m = 0,7 0.8 th3

1k< . y chnh l nhc

im ca tn hiu AM.

H s mo phi tuyn:

( ) ( )

( )

++=

0

00

I

....IIk

23

22

I(t ns) (n 2) l bin cc thnh phn dng in ng vi hai bc cao ca tn

hiu iu ch.

I(t s) l bin cc thnh phn bin tn.


(12)

(15)


20/40


c trng cho mo phi tuyn trong mch iu khin, ngi ta thng dng

c tuyn iu ch tnh hnh trn. Ta c th thy quan h gia bin tn hiu ra v gi

tr tc thi ca tn hiu iu ch u vo.

ng c tuyn iu ch tnh l tng l mt ng thng t C n A. ctuyn iu ch tnh khng thng s lm cho lng bin i bin dao ng cao tn

u ra so vi gi tr ban u (im B) khng t l ng thng vi gi tr tc thi ca

in p iu ch. Do trn u ra thit b iu bin, ngoi cc thnh phn hu ch

(cc bin tn), cn c cc thnh phn bc cao khng mong mun khc. Trong thnh

phn ng lu nht l thnh phn ca tn s t 2s c th lt vo cc bin tn m

khng th lc c.

gim mo phi tuyn, cn hn ch phm vi lm vic ca b iu ch trong

on ng thng ca c tuyn iu ch tnh. Lc buc phi gim su iu ch.

V d: Cho tn hiu iu bin vi h s iu ch m = 2, tn s iu ch = 10

KHz. Tn hiu ti tin c bin V0 = 5mV v tn s 0 = 1 MHz. Vit phng trnh tn

hiu iu ch v tn hiu iu ch? V dng tn hiu iu ch?

Gii:

Ta c V0(t) = 0.005 cos (2*106

) t

Theo bi ra:0V

Vm = V = mV0 = 2*0.005 =0.01

Tn hiu iu ch l:

V = 0.01 cos (2*104) t

M phng tn hiu iu ch:

fc=10^6;fm=10^4;

T=1/fc;

t=0:T/200:100*T;



21/40


VAM(t)=0.005*cos(2*pi*fc*t).*[1+2*cos(2*pi*fm*t)];

plot(t,VAM(t))

Title('DC-AM,m>1')

Hnh 3.10: M phng tn hiu bng phn mm3.2. iu ch tn s v iu ch pha

3.2.1. iu ch tn s v pha (FM, PM)

PM l dng iu ch pha trong tn s sng mang cao tn khng i, ch c pha

sng mang cao tn bin i t l vi iu ch.

Nu nh tn hiu bng gc lm thay i tn s sng mang ta c iu ch tns FM.

Nu nh tn hiu bng gc lm thay i pha ban u ta c iu ch pha PM.

Ta c mi quan h gia pha v tn s:

Ti tin dao ng iu ha:

V0(t) = V0cos(0t + 0) = V0cos (t) (17)

T cng thc (16) ta c:

( ) ( ) ( ) ( )3.3tdtttt

0

+=

Thay (18) vo (17) ta c biu thc:

[ ] )4.3()t(dt)t(cosV)t(V 00 +=

Gi thit tn hiu iu ch l tn hiu n m:V(t)= Vcost (20)


0 0.2 0.4 0.6 0.8 1

x10-4

-0.015

-0.01

-0.005

0

0.005

0.01

0.015DC-AM,m>1

= (3.1)d

dt

(16)

(18)

(19)


22/40


Khi iu ch tn s th tn s ca ti tin s bin thin t l vi tn hiu iu ch

v chng c xc nh theo cc biu thc sau:

(t) = 0 + cost (21)

Trong : : lng di tn cc i

Khi ta c ch s iu tn:

k: h s t l

(t) = 0 + cost (23)

: lng di pha cc i. Khi ch s iu pha:

mp= kV= (24)T (21) ta c:

= kV (25)

Nn khi V= const th = const nhng khi thay i th mf cng thay i.

T (16) n (25) ta nhn thy khi V= const th mp = const, nhng di tn khi

iu pha th tng t l vi tn s iu ch theo biu thc:

( )

( )11.3tsin.dt

d=

=

Nh vy iu khc nhau c bn gia iu ch tn s v iu pha l lng di tn

khi iu pha t l vi bin in p iu ch v tn s iu ch, cn lng di tn khi

iu ch tn s ch t l vi bin in p iu ch m thi.

Thay (21) v (23) vo (3) ta c:

VFM(t) = V0cos(0t +

sint + 0)

VPM(t) = V0cos(0t + cost + 0)

Ph tn hiu ca iu ch tn s v iu ch pha:

Cng thc (27) v (28) c vit li nh sau vi 0 = 0:

VFM(t) = V0cos(0t + mfsint) (29)

VPM(t) = V0cos(0t + mpsint) (30)

Khi iu ch n m, ph ca tn hiu iu tn v iu pha ch cha thnh phn

0 v nhiu thnh phn tn s bin (0 n) vi n = 1, 2, 3. Bin ca cc thnh

phn tn s bin t l vi hm s Bessel loi 1 bc n


mf

= k = (3.7)V

(22)

(26)


23/40


Khi ch tnh n cc thnh phn c Jn(mf) 0,01 th b rng di tn iu tn ca

tn hiu iu tn chim l:

DFM = 2(mf + fm +1)max (31)

Khi mf > 1 ta c biu thc gn ng

DFM 2mfmax = 2 (32)

Nh vy rng di tn ca tn hiu iu tn khng ph thuc tn s iu ch. i vi tn hiu iu pha, rng di tn ca n c xc nh gn ng:

DPM 2mpmax = 2max. (33)

Vy rng di tn ca tn hiu iu pha ph thuc iu ch tn s.

Khi mf 1 th ch c mt cp bin tn c bin ln hn 5% bin di tn.

Do : DFM 2max (34)

Trong trng hp rng di tn ca tn hiu iu tn bng rng di tn ca

tn hiu iu bin, ta gi l iu tn di hp. Ngc li khi m f >1 ta gi l iu tn di

rng.


J1 J

2

J0

Hnh 3.11: Gi tr h s Bessel i vi J0, J

1, J

2

ph thuc mf

mf0,20,40,6

0,8

1

J

2J

3

J1

J0 J

1 J2

J3

Jn(m)

0

Hnh 3.12: Ph ca hm Bessel


24/40


Ta c th thy ph tn bng hp ca FM ging vi ph ca AM. c ng dng

dng trong thng tin thoi FM vi di tn (5-15 KHz)

ng dng: ph tn bng rng c tnh chng nhiu cao dng trong pht thanh FM

stereo, ting TV, Viba, truyn hnh v tinh. Theo chun FCC, di tn cc i FM

pht thanh v ting TV 75 KHz.

Hnh 3.13: iu ch PM(a): Tn hiu ban u

(b): Tn hiu sng mang

(c): Tn hiu sau iu ch

Nhn xt v iu ch FM :

Cng sut AM thay i theo iu ch. Cng sut FM khng i.

Bng thng: v l thuyt bng thng FM v cng ln nhng trn thc ti khi bng thng FM c coi l khng i vi cng thc:

Tnh chng nhiu: thng tin FM c tnh chng nhiu cao hn AM, chnh

v vy iu ch FM thng c chn h thng thng tin iu ch tng t.


(b)

(a)

(c)


25/40

mV01

V01

VAM1

VAM2

mV02

V02

V

V

V

0

Hnh 3.15: th vecto ca tnhiu iu pha Amstrong


V d: iu ch pha theo Amstrong

Tn hiu ti tin c to ra t b dao ng thch anh ( c tc n nh tn s

cao), c a ti hai b iu bin (B1 v B2) lch pha nhau 900. Cn tn hiu iu

ch m tn Vc a n hai mch iu bin ngc pha. in p ra trn hai b iu

bin l:

VAM1 = V01(1 + mcost) cos0t

= V01cos0t + (m/2) V01 (cos(0t + ) + cos(0t - )t)

VAM2 = V02(1 + mcost) cos0t

= V02cos0t + (m/2) V01 (sin(0t + ) + sin(0t - )t)


Dao ngthch anh

B1

B2

Di pha900

Tng

Hnh 3.14: iu ch pha theo Amstrong

V


26/40

L

CD

+Vcc

R1

R2

CoC2

C1

Tn hiuiu ch

Ti mchdao ng

Hnh 3.16: Mch iu ch tng t FM


T th ta thy tng cc dao ng iu bin V = VAM1 + VAM2 l mt dao

ng iu ch v pha v bin . iu bin y l iu bin k sinh. Mch c nhc

im l di pha nh.

V d: Gii thiu mt phng php n gin thc hin iu ch tn s.

Trong mch in gm:

CD l mt diode bin dung

R1, R2: to thnh mt mch phn p nh im lm vic cho CD.

Co: t dn tn hiu iu ch vo mch

C2: mt t dn tn hiu c tr khng gn nh bng 0 i vi bng tn cng tc

nhm m bo cho CD dng nh c mc song song vi khung LC 1.

Khi cha c tn hiu iu ch a vo mch th trn diode bin dung c mt inp tnh c xc nh bi mch phn p R1 v R2 do diode ny c mt gi tr in

dung CD khng i. Lc ny mch dao ng s hot ng vi tn s khng i:

fL C CD

=+

1

2 1 ( )

Khi c tn hiu iu ch a vo mch, nu bin ca tn hiu iu ch tng th

in p trn diode bin dung tng, lm cho tr s in dung CD ca diode gim, dn

n tn s dao ng ca mch dao ng tng. Nu bin ca tn hiu iu ch gim

th in p trn diode bin dung gim lm cho in dung CD tng dn n tn s cng

tc ca mch dao ng gim.

Nh vy, tn s dao ng ca mch dao ng b khng ch bi tn hiu iu ch

u vo v bin thin ng theo quy lut ca tn hiu iu ch. ng ra ca mch

dao ng ta thu c mt tn hiu c iu ch tn s.



27/40


3.2.2. iu ch FSK

Khi tn hiu iu ch l tn hiu s

th iu ch tn s c gi l FSK

(Frequency Shift Keying: kha dch tn

s).

Hin nay c nhiu kiu iu ch

FSK: 2 FSK, 4 FSK, 8 FSK ... iu ch2 FSK l iu ch tn s m sau khi iu

ch sng mang ch c hai tn s. Tng

t vi 4FSK th sau khi iu ch sng

mang c 4 tn s ... . n gin trc

ht ta xt iu ch 2FSK.

Mch gm:

LO1 v LO0: hai b dao ng c lp: LO1 to ra sng mang c tn s f1, LOo

to ra sng mang c tn s fo.

Mo v M1: cc cng truyn dn c iu khin bi tn hiu iu ch.

: b cng c nhim v kt hp cc sng mang fo v f1 to thnh tn hiu 2

FSK.

Nguyn l hot ng ca mch hnh 3.13nh sau: Mch dao ng LO0 to ra

tn hiu sng mang c tn s f0 a n ng vo ca cng truyn dn Mo. Mch daong LO1 to ra sng mang c tn s f1 a n ng vo ca cng truyn dn M1. Tn


Hnh 3.17: Dng sng ca mch iu ch tn s.a) Tn hiu iu ch b) Sng mang

c) Tn hiu sau khi iu ch

LO0

LO1

Tn hiuiu ch

M0

M1

Tn hiu iu ch

Hnh 3.18: S khi mch iu ch2FSK

t

t

U

(a)

(b)

(c)

t


28/40


hiu iu ch c a n iu khin cc cng truyn dn Mo v M1. Khi tn hiu

iu ch l bit 0 th cng truyn dn M1 khng c iu khin nn khng cho tn hiu

sng mang f1 i qua, cng truyn dn Mo c iu khin nn cho tn hiu sng mang

fo i qua. Khi tn hiu iu ch l bit 1 th cng truyn dn Mo khng c iu khin

nn khng cho sng mang fo i qua, cng truyn dn M1 c iu khin cho sng

mang f1 i qua. Mch cng kt hp cc tn hiu ng ra ca cc cng truyn dn Mo

v M1. Kt qu l ng ra ca mch cng ta thu c sng mang iu ch 2 FSK

gm hai tn s f0 v f1 ng vi cc bit s liu a vo iu ch.

Hnh 3.19: Tn hiu iu ch v sng mang sau iu ch ca FSK3.2.3. iu ch PSK

3.2.3.1. iu ch pha 2 trng thi (2 PSK)

(a)

(c)

Hnh 3.20:

(a): S khi ca mch iu ch 2 PSK(b): Dng sng ca iu ch 2 PSK

(c): Biu pha ca iu ch pha 2

PSK

(b)

iu ch pha hai trng thi l iu ch m sau khi iu ch sng mang iu

ch c hai trng thi pha so vi pha ca sng mang cha iu ch.

S liu nh phn cn iu ch trc ht c a vo mch chuyn m bin

i t m NRZ n cc sang NRZ lung cc. Sau , s liu c a n b nhn M



29/40

Bin ini tip

sang songsong

Chuynm

Chuynm

90o

VCO

Z1

Z2

M1

M2

S2(t)

S1(t)

Tnhiu

QPSK

Tnhiuiuch

Hnh 3.21: S khi ca iu ch pha 4 trng thi


nhn vi sng mang 2. cos( )A t o + to ra t mch to dao ng sng mang.

ng ra ca b nhn ta thu c mt tn hiu m thnh phn chnh l d(t)

2. cos( )A t to + + v mt s thnh phn pht sinh khng mong mun. Cc tn hiu

ny c a qua mch lc bng loi b cc thnh phn khng cn thit. Kt qu, ng ra ta thu c mt tn hiu iu ch 2 PSK:

U t d t A t o( ) ( ) . cos( )= +2 . Trong : d(t) l s liu s cn iu ch, o: pha

ban u ca sng mang, A 2 : bin ca sng mang.

V b nhn M nhn tn hiu iu ch d(t) vi sng mang 2. cosA t theo min

pha nn c th vit li U(t) nh sau: U t A t t o( ) .cos( )= + +2

Trong t l gc lch pha do s liu iu ch gy ra ti thi im iu ch ( t =0 hoc t = ). D dng nhn thy rng khi s liu iu ch l bit 0 th d(t) = -1 v

sng mang sau iu ch l:U t A t o( ) .cos( )= + +2 0 .

Khi s liu iu ch l bit 1 th d(t) = 1 v sng mang sau iu ch l:

U t A t o( ) .cos( )= + +2

Nh vy sng mang sau iu ch pha c hai trng thi l ng pha vi sng mang

cha iu ch (khi iu ch bit 0) v ngc pha vi sng mang cha iu ch (khi iuch bit 1). Pha thu s da vo s chuyn pha ca sng mang thc hin gii iu

ch.

3.2.3.2. iu ch pha 4 trng thi (QPSK)

iu ch pha 4 trng thi cn c gi l iu ch 4 PSK hay iu ch cu

phng pha hay iu ch QPSK, l phng thc iu ch pha m sng mang sau iu

ch c 4 trng thi pha, mi trng thi pha mang hai bit thng tin.



30/40

00(45o)

01(135o)

11(225o)

10(315o)

Y

X

0 0 0 1 1 11 0 1 1

0 0 1 1 1

0 1 1 0 1

U

t

t

t

t

t

S liuiu ch

Lungs S1

Lung

s S2Sngmangcha

iu chSngmang

iuch

(a)

(b)Hnh 3.22: Dng sng (a) vbiu pha ca iu ch

QPSK(b)


Mch gm:

S/P: mch chuyn i lung s t ni tip sang song song c nhim v chia

lung s iu ch thnh hai lung c tc bng nhau v bng 1/2 lung s vo.

Z1 v Z2: cc mch chuyn i m c nhim v bin cc lung s S 1 v S2 t m

NRZ n cc thnh m NRZ lng cc.

VCO: b to dao ng sng mang c tn s iu khin bng in p.

90o: mch dch pha 90o bin sng mang A.cos (t + ) thnh

A.sin (t + ).

M1 v M2: hai b nhn thc hin nhn cc tn hiu s S1(t) v S2(t) vi cc sng

mang A.cos(t + ) v A.sin (t + ) tng ng. : mch cng c nhim v cng cc tn hiu sau cc b nhn M1 v M2 c

tn hiu QPSK

Tn hiu c biu thc ton hc l:

S1(t).Acos(t + ) + S2(t).Asin(t + ) = A.cos (t + + t)

t: gc dch pha do iu ch pha sinh ra


Nguyn l hot ng ca mch iuch pha nh sau: S liu iu ch a nmch bin i t ni tip sang song songc bin thnh hai lung s c tc

bng nhau v bng 1/2 lung s vo. Cc

lung s S1 v S2 sau c chuyn mt NRZ n cc sang NRZ lng ccnh cc mch chuyn m Z1 v Z2. Saukhi chuyn m ta thu c cc lung stng ng l S1(t) v S2(t). B nhn M1nhn tn hiu S1(t) vi sng mangA.cos(t + ) do b dao ng VCO tora, b nhn M2 nhn tn hiu S2(t) visng mang A.sin(t + ) do b daong VCO to ra v qua dch pha 90o.Sau cc b nhn M1, M2 ta thu c cctn hiu 2 PSK l S1(t).Acos(t + ) vS2(t).Asin(t + ). Hai tn hiu nyc cng vi nhau qua mch cng v ng ra ca mch cng ta thu c tn hiu4 PSK (QPSK).


31/40


3.2.3.3. M phng Matlap vi tn hiu iu tn v iu pha

V d 1: Cho tn hiu c di tin c bin 5V, tn s 90 MHz v mt tn hiu

iu ch c bin 5V, tn s 15 KHz. H s t l k = 1 KHz/V. Tm biu thc ca tn

hiu iu ch v tn hiu iu ch? V dng sng ca tn hiu iu ch?

Gii:

Ta c: mf = kV

= 3.010*15

5*10003

=

Biu thc ca tn hiu iu ch:

V = 5 cos (2 15*103) t

Biu thc ca tn hiu iu tn:

VFM (t) = 5 cos [ (2*90*106 ) t + 0.3 sin (2*15*103) t]

M phng dng tn hiu iu tn:

fc=90*10^6;fm=15*10^3;

T=1/fc;

t=0:T/20:10*T;

VFM(t)=5*cos(2*pi*fc*t+.3*sin(2*pi*fm*t));

plot(t,VFM(t))title('DC-FM')

:

V d 2: Cho tn hiu ti tin c bin 5V, tn s 25 KHz v mt tn hiu iu

ch c bin 5V, tn s 1000 Hz. H s t l k = 2.5 Hz/V. Hy tm biu thc ca tn

hiu iu ch v tn hiu iu ch (PM)? V dng sng ca tn hiu iu ch

(PM)?

Gii:

Ta c: mp = k*V = 2.5*5 = 12.5



0 0.2 0.4 0.6 0.8 1 1.2 1.4

x10

-7

-5

-4

-3

-2

-1

0

1

2

3

4

5

DC-FM


32/40

Chia 4

BiniA/D

BiniA/D

VCO

90o

A1

A2M2

M1S(1)

S(2)

Tn hiu 16QAM

Tn hiuiu ch

Hnh 3.23: S khi iu ch 16 QAM


V (t) = 5 cos (2*103) t


VPM (t) = 5 cos [(2*25*103) t+12.5 cos (2*103) t]

M phng dng tn hiu iu ch (PM):

fc=25*10^3;fm=10^3;

T=1/fc;

t=0:T/20:10*T;

V=5*cos(2*pi*fc*t+12.5*sin(2*pi*fm*t));

plot(t,V)

title('DC-PM')

3.3. iu ch QAM

3.3.1. nh ngha

iu ch QAM (Quature Amplitude Modulation: iu ch bin vung gc) l

phng php iu ch kt hp gia iu bin v iu pha. phng php ny, sng

mang sau iu ch c bin v pha thay i theo quy lut ca tn hiu iu ch.

3.3.2. iu ch QAM 16 trng thi (iu ch 16 QAM)

p ng yu cu thng tin tc cao, ngi ta phi tng s trng thi pha v

bin ca sng mang c c s trng thi pha nhiu v khong cch gia cc

trng thi pha ln. Xt v d sau:


0 0.5 1 1.5 2 2.5 3 3.5 4

x10-4

-5

-4

-3

-2

-1

0

1

2

3

4

5

DC-PM


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Mch iu ch 16 QAM gm c:

- Mch bin i ni tip sang song song c nhim v chia lung s u vo thnh 4 lung

c tc bng nhau v bng 1/4 lung s vo.

- Mch dao ng VCO: l b dao ng to sng mang c tn s iu khin bng in p.

- Mch dch pha 90o: bin i sng mang A.cos (t + ) thnh sng mang

A.sin (t + ).

- B bin i D/A c nhim v nhn hai lung s c hai mc bin u vo bin

i thnh mt tn hiu c 4 mc bin (tn hiu tng t) ng ra.

- M1, M2: cc b nhn c nhim v nhn cc sng mang vi tn hiu sau cc b bin i

D/A.- B cng c nhim v cng cc tn hiu sau khi nhn M1 v M2 to thnh sng mang

16 QAM.

Biu pha ca sng mang sau iu ch 16 QAM:

Nguyn l hot ng:

S liu cn iu ch c a n mch chuyn i ni tip sang song song, c chia

thnh 4 lung D1, D2, D3, D4 c tc bng nhau v bng 1/4 lung s vo. Cc lung D1,

D2 c a n b bin i D/A 1 chuyn thnh tn hiu tng t 4 mc A 1. Cc

lung D3, D4 c a n b bin i D/A 2 chuyn thnh tn hiu tng t 4 mc A 2.

Sau tn hiu tng t A1 c nhn vi sng mang A.cos(t + ) do b dao ng

VCO to ra ti M1, u ra ca M1 ta thu c tn hiu:

S1 = A1.Acos . (t + ). Tn hiu tng t A2 c nhn vi sng mang A.sin(t + )do b dao ng VCO to ra v qua dch pha 90o ti M2, u ra ca M2 ta thu c tn


0000 0001

0010 0011

1100

1110

0100

1101

0101

0111 0101

1000 1010

101110011111

Y

X

Hnh 3.24: Biu pha ca sng mang 16 QAM


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hiu: S2 = A2Asin(t + ). Cc tn hiu S1 v S2 c cng vi nhau to thnh tn

hiu 16 QAM. Hnh 3.23 m t dng sng ca iu ch 16 QAM:

3.4. iu ch nh phn

3.4.1. iu ch nh phn

iu ch nh phn l kiu iu ch n gin nht v ph bin trong thc t. yl trng hp tn hiu mang tin bng gc l tn hiu nh phn.

3.4.2. iu ch kha dch bin nh phn (BASK)

Trong h thng BASK, bin ca sng mang tn s fc c chuyn i gia hai

gi tr ty thuc vo tn hiu bng gc, bin ca sng mang bao gm hai mc A0 v

A1 biu din cho hai k t 0 v 1 tng ng. Trong thc t, dng sng BASK gm cc

xung mark biu din k t 1 v space biu din k t 0. Lc ny BASK cn c

gi l biu din kha on-off OOK v k hiu BASK c biu din nh sau :

+

=

tfT

tA

tfT

tA

tf

00

00

2cos

2cos

)(

B iu ch OOK c th c thc hin nh l mt kha chuyn mch n gin,

kha sng mang on hay off ty tn hiu mang tin l 1 hay 0 hoc l mt b iu ch

cn bng, nhn sng mang vi tn hiu OOK n cc bng gc. B tch sng OOK c


0 0 0 0 0 0 1 1 1 1 0 0

U

t

t

t

Tn hiuiu ch

Sng mangcha iu

ch

Sng mang iu ch

Hnh 3.25: Dng sng mch iu ch 16 QAM

Cho s 1

Cho s 0


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th l kt hp hay khng kt hp. Trng hp ny tch sng kt hp c th dng mt

b lc phi hp, u ra ca b lc phi hp t cc i khi u vo c tn hiu v bng

0 khi u vo khng c tn hiu. Hoc dng b tch sng tng quan, yu cu l phi

c b khi phc sng mang CR. Tn hiu sau c ly mu v quyt nh ngng

vi ng h ly ra t b khi phc ng h STR.

Kiu tch sng khng kt hp c s dng ph bin hn, c cu thnh hai

knh tng quan tch thnh phn ng pha v vung pha ca tn hiu, sau bnh

phng thnh phn ng pha v vung pha ri sau cng li. S sp xp ny khc

phc c yu cu v ng b pha sng mang. Kiu ny phc tp nhng vi s pht

trin ca cng ngh VLSI, chng tr nn nh, nh v r hn so vi b lc v tch

ng bao trong cc thit k truyn thng.

Hnh 3.26: iu ch on-off: dng sng, b iu ch, ph(a): Tn hiu bng gc

(b): Ph in p ca tn hiu bng gc

(c): B iu ch OOK

(d): Tn hiu OOK thng di

(e): Ph in p ca tn hiu OOK thng di

Thi im quyt nh b quyt nh bn thu l f(nT 0) v in p quyt nh l:

=0

)( 10kE

nTf

Trong E l nng lng chun ha cha trong k t 1 v k l hng s, c th t

bng 1.

Trng hp s dng b lc phi hp, xc xut li c tnh nh sau:


Chn s 1Chn s 0


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=

2

1

1

1

2

11

2

1

N

EerfPe

Trong N l PSD ca nhiu, erf(x) l hm li

t ( )2121 EEE += l nng lng trung bnh theo thi gian trn mt k t, vi OOK

th E0=0, ta c:

=

2

1

02

11

2

1

N

EerfPe

Hnh 3.27:(a): B thu dng b lc phi hp

(b): B thu tng quan

(c): B thu tch sng ng bao

(d): B thu 2 knh

S dng cc quan h:

N

BCT

N

E

VBNN

VT

EC

0

0

20

2

0

)(

)(

=

=

=

y C l cng sut sng mang thu ly trung bnh trn tt c cc k t v N l cng

sut nhiu chun ha trong mt bng thng B(Hz).3.4.3. iu ch kha dch pha nh phn (BPSK)



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Trong h thng BPSK, tn hiu bng gc c gn vo sng mang bng cch

thay i pha ca sng mang ty thuc vo tn hiu bng gc:

+

=

tfTtA

tfT

tA

tf

00

00

2cos

2cos

)(

V nguyn tc, c th chn bt k nhng thng chn trng thi ngc pha,

tc l =1800. Kiu iu ch ny gi l iu ch kha o pha PRK.

Hnh 3.28: iu ch PRK: dng sng, b iu ch, ph(a): Tn hiu bng gc

(b): Ph in p ca tn hiu bng gc

(c): B iu ch PRK

(d): Tind hiu PRK thng di

(e): Ph in p ca tn hiu PRK thng di

Tch sng PRK l kiu tch sng kt hp, c th thc hin bng b lc phi hp hoc

tng quan. V k t 1 v 0 ngc pha nn ch cn b thu mt knh.

Mc in p quyt nh sau khi lc:

= kE

kEnTf )( 0

Hnh 3.29: B thu tng quan PRK3.4.4. iu ch kha dch tn nh phn (BFSK)

BFSK biu din s 1 v s 0 bng cc xung sng mang vi hai tn s l f1 v f2:


Chn s 1

Chn s 0

Chn s 1

Chn s 0


38/40


=

tfT

tA

tfT

tA

tf

20

10

2cos

2cos

)(

B iu ch BFSK thng l mt b dao ng iu khin s. C th thy tn hiuBFSK l xp chng ca hai tn hiu OOK, do vy ph in p ca tn hiu BFSK l

xp chng ca hai ph in p OOK, mt biu din tn hiu cho f1 v mt OOK biu

din cho f2. Ph tn hiu b chng ln, nu khong cch gia f1 v f2 ln th s chng

ln coi nh khng ng k.

B tch sng BFSK c th kt hp hay khng kt hp. Tch sng khng kt hp

c t s CNR khng tt bng tch sng kt hp ging nh trong h thng OOK.

Nu cc k t nh phn trong h thng BFSK trc giao c ngha l :

=0

0

21 0)2cos()2cos(T

dttftf

Khi u ra ca mt knh trong b thu kt hp l ti a th u ra ca knh kia s ti

thiu


Chn s 1

Chn s 0

Hnh 3.30: iu ch BFSK: dngsng, b iu ch, ph

(a): Tn hiu bng gc

(b): Ph in p ca tn hiu bnggc(c): B iu ch BFSK(d): Tn hiu BFSK(e): Ph in p ca tn hiu BFSK


39/40


TNG KT

iu ch l qu trnh khng th thiu trong h thng thng tin. Bn trong h

thng thng tin, khi iu ch thc hin qu trnh lc, khuch i tn hiu v cho ra tn

hiu t b li nht c th.Tn hiu ta thu c c tt hay khng ph thuc ln vo qu trnh iu ch v

khong cch ng truyn.

Vi vic m phng cc loi iu ch thng qua phn mm Matlap cho ta thy r

hn v cc tn hiu c ph tn nh th no.

Trong qu trnh lm n khng trnh khi thiu st rt mong nhn c s

ng gp t pha gio vin hng dn Nguyn V Anh Quang v cc bn.

Em xin chn thnh cm n!



40/40


TI LIU THAM KHO

[1]. Ti liu: Thng tin v tuyn _Ging vin: Nguyn V Anh Quang

[2].www.tailieu.vn

[3].www.vntelecom.org[4].www.Wikipedia.org
http://www.tailieu.vn/http://www.vntelecom.org/http://www.wikipedia.org/http://www.tailieu.vn/http://www.vntelecom.org/http://www.wikipedia.org/

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