Do an Tn Phuongdongdoc

8/2/2019 Do an Tn Phuongdongdoc

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n hc k 9===================================================================

TRNG I HC PHNG NGKHOA CNG NGH THNG TIN

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CNG HA X HI CH NGHA VIT NAMc lp - T do - Hnh phc

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N HC K 9

Tn ti: K THUT IU CH TN HIU TRONG CC H THNGTRUYN TIN S HIN I

H v tn : PHM TRN HON

Lp : 504102

Kha : 2004 2009

Ngnh : in t - Vin thng

Ngy thng nm 2008

===================================================================PHM TRN HON


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n hc k 9===================================================================

NHN XT CA NGI HNG DN

im : (Bng ch : )

Ngy thng nm 2008 Gio vin hng dn

PGS . TS. THI HNG NH

===================================================================PHM TRN HON


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n hc k 9===================================================================

LI NI U

Cng vi s pht trin ca t nc, nhng nm gn y, cc ngnh cng

nghip u pht trin mnh m v ngnh cng nghip vin thng cng khngngoi l. Ngy cng c nhiu dch v truyn thng mi v cht lng truyn

thng cng yu cu cao hn dn n s cn thit phi thay i nng cp

ng truyn.

ng trc xu hng nh vy, vic tm hiu v cc vn truyn tin

trong cc h thng vin thng hin i tr nn quan trng i vi sinh vin.

Nhn thc c iu , n tt nghip K thut iu ch tn hiu trong cc

h thng truyn tin s hin i s gii thiu tng quan v cc h thng truyn

tin s, tm hiu v cc vn k thut iu ch tn hiu. B cc ca n bao

gm cc chng:

Chng 1 : Tng quan v h thng truyn tin s

Chng 2 : Cc c im ca truyn dn s v iu ch xung m

Chng 3 : iu ch tn hiu s

iu ch tn hiu s l k thut ngy nay khng cn mi m, song vic

tm hiu cc vn iu ch l cn thit, i hi phi c kin thc su rng, v

lu di. Do vy n khng trnh khi nhng sai st. Rt mong nhn c s

ph bnh, gp ca cc thy c gio v cc bn.

Xin c gi li cm n chn thnh ti PSG.TS. Thi Hng Nh, ngi

tn tnh hng dn em trong sut qu trnh lm n ny.

Xin cm n chn thnh cc thy c gio trong khoa CNTT gip em

trong thi gian qua.

===================================================================PHM TRN HON


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n hc k 9===================================================================

MC LCLi ni u.......................................................................................................

Chng 1. Tng quan v h thng truyn tin s.............................................1.1 Khi qut v h thng truyn tin in t...................................................1.2 Ngun tin v tn hiu ngun......................................................................1.3 iu ch v gii iu ch trong cc h thng truyn tin............................1.4 Phn tch tn hiu.......................................................................................

Chng 2. Cc c im ca truyn tn s v iu ch m xung....................2.1 c im ca truyn tin s........................................................................2.2 Cc dng iu ch xung.............................................................................

2.3 iu ch m xung......................................................................................-Tng quan ......................................................................................................-Ly mu PCM................................................................................................-Tc ly mu,lng t ho v m ho........................................................-Di ng.........................................................................................................-Hiu sut m ho............................................................................................

Chng 3. iu ch tn hiu s.......................................................................3.1 Tng quan v mt s h thng truyn tin s..............................................

3.2 Kho dch bin ASK..................................................................................3.3 Kho dch tnFSK.....................................................................................3.4 Kho dch pha PSK....................................................................................3.5 iu ch bin cu phng QAM..........................................................3.6 Kho dch pha vi phn DPSK....................................................................3.7 Hi phc sng mang..................................................................................3.8 Xc sut bit v t l li bit.........................................................................

===================================================================PHM TRN HON


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n hc k 9===================================================================

Chng 1

TNG QUAN V H THNG TRUYN TIN S

1.1.Khi qut v h thng truyn tin in t:

H thng truyn tin in t l mt h thng s dng cc mch in v cc

thit b in t thc hin cng vic truyn tin t ni ny n ni khc,gi tt

l h thng truyn tin.Thng tin ngun nguyn thu c truyn trong cc h

thng truyn tin c th di dng tng t,v d : ting ni con ngi , m nhc

hoc c th di dng s,ri rc nh nhng ch ci hoc ch s c m ho

di dng nh phn.

Thng tin truyn v x l trong cc h thng truyn tin in t c biu

th di dng cc tn hiu. Tn hiu l i lng vt l mang thng tin v thng

c biu th di hai dng : tn hiu tng t v tn hiu s.

H thng truyn tin truyn cc tn hiu tng t c gi l h thng

truyn tin tng t. H thng truyn tin truyn cc tn hiu s l h thng truyn

tin s.

Trong cc h thng truyn tin c s tham gia ca cc my tnh, tin tc

hoc thng tin c biu th di dng d liu. H thng hoc mng truyn tin

c gi l h thng hoc mng truyn d liu.

Mt h thng truyn tin bt k no cng c th c biu th theo s

khi sau:

===================================================================PHM TRN HON


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n hc k 9===================================================================

Cc k hiu trong s :

m(t) D liu ngun tin a vo thit b pht

s(t) D liu u ra ca my pht sau khi c x l, m ho, iu ch v khuch i a

vo ng truynn(t) Tp nhiu tc ng vo thit b thu. Do c tp nhiu nn r(t) = s(t) + n(t) s(t)

( )m t% Tn hiu u ra thit b thu (nhn tin) ; ( ) ( )m t m t %

Hnh 1.1 M t s khi tng qut mt h thng truyn tin

Bt k mt h thng truyn tin no cng bao gm ba khi chc nng ch

yu : pht, mi trng truyn dn v thu.

Khi pht l mt tp hp gm mt hoc nhiu thit b hoc mch in t

chuyn i thng tin ngun nguyn thu thnh tn hiu thch ng vi mi

trng truyn dn. Khi pht c hai chc nng ch yu l x l tin hiu pht v

to sng mang pht. X l tn hiu pht tc x l tn hiu ngun sao cho thch

ng vi cc yu cu truyn tin. Cc phng php x l c th l: nn, lc, m

ho, s ho, iu ch, truyn tin c th. Mch sng mang pht c nhim v bin

i tn hiu sau x l tn hiu pht sao cho thch ng vi knh truyn dn v

khong cch cn truyn dn.

Knh truyn dn (mi trng truyn dn) thng c nhm vo hai

nhm: nhm knh cng v nhm knh mm. Knh cng v d cc ng dy

in thoi, cp song hnh, cp ng trc,Knh mm v d khng kh, chn

khng, ncNi chung, knh truyn gy suy gim tn hiu v b tc ng ca

tp nhiu lm tn hao v sai lc tn hiu truyn trn knh. Tp nhiu c th docc nhiu t cc ngun nhiu ngoi (nhiu kh quyn, nhiu cng nghip,..) v===================================================================PHM TRN HON

Mchsngmang

X ltn

hiupht

Mitrng

hoc knhtruyn dn

Mch sngmang

X l tnhiu thu

u cuinhn tin

Nguntin


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n hc k 9===================================================================cc tp m bn trong bn thn h thng truyn tin (tp m ni b h thng) gy

nn. Ngoi tc ng ca tp nhiu, knh truyn cn chu tc ng ca cc hin

tng tr, tn hiu vng,

Mch sng mang v x l tn hiu thu l cc qu trnh ngc li ca x ltn hiu pht v mch sng mang pht ti to li ngun tn hiu nguyn thu

c truyn. Do tc ng ca nhiu n(t) trong qu trnh truyn nn b thu cn

c b lc v loi tr nhiu.

1.2. Ngun tin v tn hiu ngun:

Ngun tin trong h thng truyn tin l ni to ra hoc cha cc tin cn

truyn i. Ngun tin c th l s hoc tng t.

Mt ngun tin s to ra mt tp hu hn cc on tin c th c.VD: my

ch.

Mt ngun tin tng t to ra cc on tin oc xc nh trn mt dy

lin tc.VD: microphon.

Mt h thng truyn tin s l mt h thng truyn tin tc t mt ngun s

hoc mt ngun tng t c ri rc ho, s ho ti b thu.

Mt h thng truyn tin tng t l mt h thng truyn tin tc t mt

ngun tng t ti b thu.

Trong cc h thng truyn tin in t, tn hiu l i lng vt l mang

thng tin v thng c biu th di hai dng: tn hiu tng t v tn hiu s.

Thc cht, mt tn hiu s hoc mt dng sng s c nh ngha nh

mt hm thi gian c mt tp ri rc cc gi tr v mt tn hiu tng t hoc

mt dng sng tng t l mt hm thi gian c lin tc cc gi tr.

Gi tr tin tc trong cc h thng truyn tin in t thng c biu th

di dng in p u(t), hoc dng in i(t), lin tc hoc gin on.

===================================================================PHM TRN HON


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n hc k 9===================================================================

Hnh 1.2. M t dng tn hiu tng t v s

Khi mt ng truyn tin c thit lp truyn tin t ngun tin n ni

nhn, mt dy cc phn t c s ca ngun tin s c truyn i vi mt phn

b xc sut no , dy ny c gi l on tin.

maxS Ngun tin c th l ngun tin nguyn thu hoc c s b x l. Cc

ngun tin nguyn thu phn ln l nhng hm lin tc theo thi gian f(t) hoc lhm bin i theo thi gian cng cc thng s khc. Phn ln cc tin nguyn

thu mang tnh lin tc theo thi gian v mc, ngha l c th biu din mt

thng tin no di dng mt hm s s(t) tn ti trong khong thi gian 1 2( , )t t

vi cc gi tr bt k trong phm vi (Smin,Smax) nh hnh 1.3.

===================================================================PHM TRN HON


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n hc k 9===================================================================

Cc ngun tin nguyn thu c th c a trc tip vo knh truyn

i hoc c th qua cc php bin i x l trc khi a vo knh truyn tin s

phi c s ho hoc m ho. Php bin i tn hiu ngun tng thch vi

knh truyn c gi l php iu ch.

1.3.iu ch v gii iu ch trong cc h thng truyn tin:

Trong thc t, cc tn hiu thng tin nguyn thu khng th truyn c

xa trn cc ng truyn dn cp kim loi, si cp quang hoc trong tng khng

gian kh quyn, do cn phi iu ch tn hiu thng tin nguyn thu vi

mt tn hiu tng t c tn s cao hn gi l sng mang. Tn hiu sng mang

c nhim v mang thng tin trong h thng truyn tin. Tn hiu thng tin c th

iu ch vi sng mang hoc theo bin , theo tn s hoc theo gc pha. Vic

iu ch c hiu n gin l qu trnh bin i mt hoc nhiu c tnh ca

sng mang theo s bin i ca tn hiu thng tin.

Trong cc h thng truyn tin c hai dng iu ch c bn, l iu ch

tng t v iu ch s.

Mt h thng truyn tin trong nng lng c truyn v thu di

dng sng tng t (tn hiu bin i lin tc theo thi gian) c gi l h

thng truyn tin tng t.

Thut ng truyn tin s (digital communication) trong thc t k thut

bao gm c truyn dn s v radio s.

Truyn dn s (digital transmission) l h thng truyn dn trong cc

xung s c truyn gia hai hoc nhiu im trong h thng truyn tin. Vi

===================================================================PHM TRN HON

s(t)

m a xS

minSt2

t1

t

Hnh 1.3. Hm s(t) ca ngun tin nguyn thu lin tc


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n hc k 9===================================================================truyn dn s th khng cn c sng mang v cc thng tin ngun c th l dng

s hoc tng t. Nu thng tin l dng tng t th cn phi chuyn i thnh

dng s trc khi truyn v c chuyn i tr li dng tng t pha u

thu. Cc h thng truyn tin s c ng truyn vt l gia pht v thu l dykim loi hoc si cp quang.

Radio s l vic truyn cc sng mang tng t c iu ch s gia hai

hoc nhiu im trong h thng truyn tin. radio s, tn hiu iu ch v tn

hiu c iu ch l cc xung s. Cc xung s c th t mt h thng truyn

dn s, t mt ngun s VD: t mt my tnh hoc t mt tn hiu c m ho

nh phn. cc h thng radio s th mi trng truyn dn c th l cc

phng tin vt l hoc khng gian t do (tng kh quyn). Cc h thng truyn

tin tng t c pht trin trc v ngy nay ang c thay th bi cc h

thng truyn tin s do nhng im li th ca n.

Biu thc (1.1) l biu thc tng qut ca mt sng in p hnh sin bin

i theo thi gian ca mt tn hiu sng mang cao tn :

( ) sin(2 )u t U ft = +

(1.1)

Trong :

u(t) l sng in p hnh sin bin i theo thi gian;

U l bin (V);

f l tn s (Hz);

l gc lch pha (rad).

Cc k thut iu ch c th c m t nh sau:

===================================================================PHM TRN HON


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n hc k 9===================================================================

Nu tn hiu thng tin l tng t v bin (U) ca sng mang c

bin i t l vi tn hiu thng tin th iu ch bin AM (amplitude

modulation) c to ra; Nu nh tan s (f) ca sng mang bin i t l vi tn

hiu thng tin th iu ch tn s FM (frequency modulation) c to ra v nu

gc pha ( ) ca sng mang bin i t l vi tin hiu thng tin th iu ch pha

PM (phase modulation) c to ra.

Nu tn hiu thng tin l s v bin (U) ca sng mang bin i t l

vi tn hiu thng tin th tn hiu c iu ch s c gi l kha dch bin

ASK (amplitude shift keying); Nu tn s (f) ca sng mang bin i t l vi

tn hiu thng tin th iu ch c gi l kho dch tn FSK (frequency shift

keying) v nu gc pha ( ) ca sng mang bin i t l vi tn hiu thng tin

th iu ch c gi l kha dch pha PSK (phase shift keying). Nu nh c

bin v gc pha cng bin i t l vi tn hiu thng tin th iu ch l bin cu phng QAM (quadrature amplitude modulation).

Vic iu ch c thc hin pha pht bi mch c gi l b iu

ch. Sng mang c tc ng bi mt tn hiu thng tin c gi l sng c

iu ch hoc tn hiu c iu ch. Vic gii iu ch l qu trnh ngc li

ca iu ch chuyn i sng mang c iu ch thnh thng tin ban u.

Gii iu ch c thc hin pha thu bi mch gii iu ch.

===================================================================PHM TRN HON

Tn hiu iu ch Dng iu ch cto thnh

Tng t AM FM PM

S ASK FSK PSK

QAM


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n hc k 9===================================================================

C hai nguyn nhn cn phi thc hin vic iu ch trong cc h thng

truyn tin in t l:

1.Cc tn s thp kh bc x t anten di dng sng in t.

2.Cc tn hiu thng tin thng c di tn ging nhau v nu nh cc tnhiu t hai hoc nhiu ngun pht cng thi gian th chng s can nhiu ln

nhau. VD: cc i pht thanh thoi v pht thanh m nhc c gii tn audio

trong khong t 300Hz n 15000Hz, chng khng can nhiu ln nhau th

phi chuyn i thng tin ca chng thnh cc bng tn khc nhau hoc knh

khc nhau.

Hnh 1.4 m t s d khi n gin ho quan h ca tn hiu iu ch,

sng mang tn s cao v sng c iu ch trong mt h thng truyn tin in

t.

Hnh 1.4 S khi n gin ho mt h truyn tin c sng mang c iu ch

===================================================================PHM TRN HON

B iu

ch(chuyni ln)

Thng tintn hiu

iuch(tn sthp)

Khuch

i(chuyni xung)

Gii iuch

Tn hiuc giiiuch,thngtin(tn sthp)

To sngmang(tns cao)

To sngni(tn scao)

Mi trngtruyn dn


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n hc k 9===================================================================1.4. Phn tch tn hiu:

Khi thit k cc mch ca h thng truyn tin in t thng cn phi

phn tch d on hiu nng ca mch da trn c s phn b cng sut v hn

hp tn s ca tn hiu thng tin. Vic lm thun li nht l nh vo vicphn tch tn hiu qua cc biu thc ton hc.

1.4.1. Cc tn hiu hnh sin

Phn tch tn hiu l vic phn tch ton hc cc tham s nh tn s,

rng di tn v mc in p ca tn hiu. Cc tn hiu in l cc in p hoc

dng in bin i theo thi gian v n c th c biu th bi mt di cc

sng sin hoc cosin.

( ) sin(2 )u t U ft = + hoc ( ) cos(2 )u t U ft = + (1.2)

( ) sin(2 )i t I ft = + hoc ( ) cos(2 )i t I ft = + (1.3)

trong :

u(t) l in p hnh sin bin i theo thi gian;

i(t) l dng in hnh sin bin i theo thi gian;U l in p nh (V);

f l tn s (Hz);

l gc lch pha (rad);

I l dng in nh (A);

2 ft = l tc gc (rad/s).

Hm sin v cosin c s dng tu ph thuc vo im tham chiu ca

tn hiu.

Cc biu thc trn l i vi dng sng tn s n, c chu k lp li v

thng c gi l sng tun hon. Cc sng tun hon c th c phn tch

trong min thi gian hoc min tn s.

===================================================================PHM TRN HON


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n hc k 9===================================================================

1.4.2. Cc sng tun hon khng sin

Mt sng tun hon no m n bao gm nhiu hn mt sng sin hoc

cosin l dng sng khng sin hoc gi l sng phc hp. phn tch cc sngphc hp thng s dng chui ton hc Fuorier.

Chui Fuorier c s dng trong phn tch tn hiu biu th cc thnh

phn sin ca mt dng sng khng sin (c ngha l bin i tn hiu theo min

thi gian thnh tn hiu theo min tn s). Ni chung, cc chui Fuorier cho mt

hm tun hon no nh mt dy s cc s hng cc hm s lng gic theo

biu thc sau:

0 1 2 1 2( ) cos cos 2 ... cos sin sin 2 ... sinn nf t A A A A n B B B n = + + + + + + + +

trong = (1.4)

Biu thc (1.4) biu th dng sngf(t) bao gm mt gi tr mt chiu (dc)

l A0, mt chui cc hm cosin v sin trong cc hng sau c tn s l bi s

ca tn s s hng th nht. Biu thc cng ni ln rng, mt dng sng tun

hon no bao gm mt thnh phn mt chiu v mt chui cc sng hi dngsin v cosin. Sng hi c tn s l bi s nguyn ca tn s c bn. Tn s c

bn l sng hi bc nht v bng tn s ca dng sng. Bi s bc hai ca tn s

c bn c gi l hi bc hai, bi s bc ba c gi l hi bc ba, Nh vy,

biu thc (1.4) c th vit:

f(t) = dc + sng c bn + hi bc 2 + hi bc 3 + + hi bc

1.4.3. Chui Fuorier ca dng sng ch nht

Khi phn tch cc mch truyn tin in t thng gp cc dng sung ch

nht. Hnh 1.5 m t mt dng sng xung ch nht vi rng xung v chu

k xung T. Chui Fuorier i vi mt xung ch nht i xng chn c dng nh

sau:

( ) ( ) ( )2 sin sin 2 sin

( ) cos cos 2 ... cos2t t

U U x x nxu t t t n t T T x x nx

= + + + + (1.5)

===================================================================PHM TRN HON


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n hc k 9===================================================================

trong : u(t) l sng in p bin i theo thi gian:

l rng ca xung ch nht (s);T l chu k ca song xung ch nht;

x l ( / T );

n l hi bc n;

U l bin nh ca xung.

Xung c rng xung cng hp th thnh phn mt chiu dc trong chui

Fuorier cng b, do bin ca hi bc n trong biu thc (1.5) s l:

2 sinn

U nxU

T nx

= (1.6a)

hoc:

( )

( )

sin2n

n TUU

T n T

= (1.6b)

trong :Un l bin nh ca hi bc n;

n l hi bc n;

U l bin nh ca sng ch nht;

T l chu k xung ch nht.

Hm (sinx)/x c s dng biu th cc dng sng xung tun hon.

Hm (sinx)/x l mt sng hnh sin tt dn trong nh k tip sau c gi tr bhn nh trc n. Hnh 1.6 m t hm (sinx)/x.

===================================================================PHM TRN HON

t

U

T

Hinh 1.5. Dng sng xung ch nhtH


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n hc k 9===================================================================

1.4.4. Cng sut v ph nng lng

Mc ch ca knh truyn tin l truyn nng lng in t, t ngun n

ni nhn. Nh vy mi quan h gia nng lng c truyn i v nng lng

thu c l yu t quan trong. iu dn n vic xem xt mi quan h gia

nng lng v cng sut theo tn s.

Cng sut in l tc m nng lng c pht tn, cung cphoc c s dng v n l hm m bc hai ca in p hoc ca dng in

(P=E2/R hoc P=I2R). Theo quan h v cng sut th vi biu thc Fuorierf(t)

s c biu th l [f(t)] 2.

Hnh 1.7 m t ph cng sut ca mt dng sng ch nht vi t s T

l 0,25. Ging nh ph in p theo tn s, n cn nhiu mi con v mt mi

chnh rng hn.

T hnh 1.7 thy rng, cng sut trong mt xung c phn tn trong mt

ph tn tng i rng v phn ln cng sut nm trong mi chnh. Do , nu

rng di tn ca knh truyn tin rng c th cho qua cc tn s nm trong

mi chnh th n cng truyn c phn ln nng lng cha trong xung n

pha thu.

===================================================================PHM TRN HON

(sinx)/x

x

Hnh 1.6. M t hm (sinx)/x.H


17/57

n hc k 9===================================================================

Chng 2

CC C IM TRUYN DN S V IU CH XUNG M

2.1. c im ca truyn dn s

Truyn dn s so vi truyn dn tng t c nhng u im sau:

1. Truyn dn s c tnh khng nhiu tt hn nhiu so vi truyn dn tng

t. Cc xung s t b tc ng ca nhiu lm thay ir hoc bin dng so vi tn

hiu tng t. ng truyn dn s th cc c tnh v bin , tn s v gc

pha khng cn phi nh ra mt cch chnh xc nh knh truyn tng t. Cc

xung truyn dn s s c nh ra theo khong thi gian mu hoc mc trn,

mc di ca xung theo mt ngng no . chnh xc v bin , tn s v

gc pha truyn dn s khng quan trng lm.

2. Tn hiu s thun li v d dng hn nhiu trong cc qu trnh x l v

ghp knh so vi tn hiu tng t. Vic x l tn hiu s y c hiu l x

l cc tn hiu tng t theo cc phng php s. X l tn hiu bao gm lc,

cn bng v chuyn dch pha.Cc xung s c th c nh d dng hn tn hiu

tng t. Tc truyn ca cc h thng s c th thay i mt cch d dng

===================================================================PHM TRN HON

-4f -3f -2f -f f 2f 3f 4f

Tn sT

= 0,25

Hnh 1.7. Ph cng sut ca mt xung ch nht c t s = 0,25


18/57

n hc k 9===================================================================thch ng vi cc mi trng khc nhau v thch nghi vi cc dng thit b khc

nhau.

3. cc h thng truyn dn s dng cc b ti to tn hiu trong khi

truyn dn tng t dng cc b khuch i tn hiu. Tp m trong cc mchkhuch i l tp m cng, do t s tn hiu trn tp m u ra b khuch

i s b xu hn v nu ng truyn dn tng t dng nhiu b khuch i

th t s tn hiu/tp m (S/N) s cng xu. Trong khi truyn dn s s dng

cc b ti to tn hiu c t s tn hiu/tp m u ra bng t s tn hiu/tp m

u vo b ti to. Cng v l do m khong cch truyn dn s c th ln

hn nhiu so vi truyn dn tng t.

4. Vic o lng v lng gi cc tn hiu s n gin hn nhiu so vi tn

hiu tng t c bit l khi cn so snh hiu nng cc h thng.

5. Cc h thng s thch hp hn nhiu trong vic nh gi hiu nng li.

Li truyn trong cc tn hiu s c th c pht hin v sa li mt cch d

dng, c kh nng chnh xc hn nhiu so vi cc h thng tng t. Tuy vy,

truyn dn s cng c nhng nhc im sau:

a. Vic truyn dn cc tn hiu tng t c s ho cn phi c rng

di tn ln hn nhiu so vi vic truyn tn hiu tng t khng s ho.

b. Cc tn hiu tng t mun truyn dn s th trc khi truyn phi c

chuyn i thnh tn hiu s v ti pha thu phi chuyn i ngc tr li, c

ngha l phi tn thm mch m ho v gii m.

c. Truyn dn s yu cu phi c s ng b thi gian chnh xc gia ng

h pht v thu. Nh vy, cc h thng s cn phi c cc mch hi phc ng h

trong tt c cc my thu, gy thm tn km.

d. Cc h thng truyn dn s l khng tng thch vi cc phng tin

truyn dn tng t c in.

Hnh 2.1 m t so snh cht lng truyn tin cu mt tn hiu tng t v

mt tn hiu s c truyn qua mt knh truyn co hm truyn t H(f), khng

tuyn tnh.

===================================================================PHM TRN HON


19/57

n hc k 9===================================================================

===================================================================PHM TRN HON

u

H(f )

f

t

Uvo

Ura

H(f )

H(f )

Uvo

H(f )

Ura

t

f

t

u

a) Tn hiu ra tng t b mo dng

b) Tn hiu ra s khng b mo dng

Hnh 2.1. Tn hiu tng t v s qua knh c hm truyn t khng tuyn tnh.


20/57

n hc k 9===================================================================

2.2. Cc dng iu ch xung

iu ch xung l bin i cc thng tin tn hiu t dng tng t ca

ngun tin nguyn thu thnh dng cc xung ri rc truyn trn knh truyn

tin s n ni nhn. Trong cng ngh k thut s c bn dng iu ch xung

thng c s dng l: iu ch rng xung PWM (pulse width

modulation), iu ch v tr xung PPM (pulse position modulation), iu ch

bin xung PAM (pulse amplitude modulation) v iu ch m xung PCM

===================================================================PHM TRN HON


21/57

n hc k 9===================================================================(pulse code modulation). Hnh 2.2 m t bn dng iu ch xung .

===================================================================PHM TRN HON

a)

b)

c)

d)

e)

f)

Hnh 2.2. M t cc dng iu ch xung:a) Tn hiu tng t d) iu ch v tr xung PPM

b) Xung mu e) iu ch bin xung PAM

c) iu ch rng xung PWM f) iu ch m xung PCM


22/57

n hc k 9===================================================================

1. PWM. iu ch rng xung PWM cn gi l iu ch khong thi

gian tn ti xung PDM (pulse duration modulation) hoc iu ch di xung

PLM (pulse length modulation). y, rng ca xung t l vi bin ca

tn hiu tng t.

2. PPM. iu ch v tr xung l v tr ca cc xung vi rng hng s

trong khe thi gian bt buc c bin i ph hp vi bin ca tn hiu

tng t.

3. PAM. iu ch bin xung l bin ca xung c rng hng s,

v tr hng s, c bin i ph hp vi bin ca tn hiu tng t.

4. PCM. iu ch m xung th tn hiu tng t c ly mu v c

chuyn i thnh dy s nh phn ni tip, c chiu di c nh truyn. S

cc nh phn ph hp vi bin ca tn hiu tng t.

iu ch bin xung PAM thng l mt dng iu ch trung gian ca

cc dng iu ch khc, VD: PSK, QAM v PCM. iu ch PAM t khi c

dng ring r. iu ch rng xung PWM v iu ch v tr xung PPM c

dng trong cc h thng truyn tin c bit (thng trong qun i) v t c

dng trong cc h thng dn dng. iu ch m xung PCM l mt dng iu ch

c dng ph bin trong truyn tin s.

2.3. iu ch m xung

2.3.1. Tng quan

iu ch m xung PCM ch l mt k thut iu ch c m ho theo s

ho v c m t hnh 2.2. Thut ng iu ch thc ra y dng khng

===================================================================PHM TRN HON


23/57

n hc k 9===================================================================c chnh xc lm, bi v thc cht y khng phi l iu ch tn hiu nh cc

dng iu ch khc m n l mt dng m ha ngun tin. PCM th cc xung

c bin v chiu di c nh. PCM ch l mt h thng nh phn trong c

hoc khng c xung trong khe thi gan c xc nh trc m iu tng ng vi trng thi logic 1 hoc trng thi logic 0. Cc tn hiu PWM,

PPM v PAM nhiu khi chng khng l nh phn v cng khng c trng cho

mt n v nh phn (bit). Hnh 2.3 m t s khi n gin ho ca mt h

thng truyn dn PCM n cng, n knh.

Cc b lc di thng c nhim v gii hn tn s ca tn hiu tng t

u vo trong bng tn chun l 300 3000 Hz. B ly mu v gi c nhim v

ly mu tn hiu tng t u vo theo chu k v bin i n thnh tn hiuPAM nhiu mc. B chuyn i tng t/s A/D hoc ADC (analog to

digital converter) c nhim v chuyn i cc mu PAM thnh cc m PCM

song song v sau tip tc chuyn i d liu t song song thnh ni tip

a ra u ra ca ng truyn. Nu ng truyn c khong cch ln c th c

nhng b lp ti to li cc m PCM.

B chuyn i s/tng t D/A hoc DAC (digital to analog converter)c nhim v chuyn i m PCM song song thnh cc tn hiu PAM. Mch gi

===================================================================PHM TRN HON

HNH 2.3 S KHI N GIN HO MT H THNGTRUYN DN PCM N CNG N KNH


24/57


25/57

n hc k 9===================================================================ly mu t nhin u ra l mt dy cc xung c khong cch ging nhau v c

nh khng phng nh hnh 2.4b.

phng php ly mu t nhin th ph tn ca tn hiu u ra c ly

mu khc vi ph tn ca mt mu l tng. Bin ca cc thnh phn tn sc to ra bi cc xung hp, c rng hu hn s gim so vi cc sng hi

cao hn theo quan h (sinx)/x. Cng v vy m trong mch cn phi dng b lc

thng thp trc khi hi phc.

Phng php thng dng nht ly mu cc tn hiu thoi trong cc h

thng PCM l ly mu nh phng trong kt hp c vic ly mu v gi. Mc

ch ca mch ly mu v gi y l ly mu tn hiu tng t u vo theo

chu k mu v chuyn i cc mu thnh dy cc mc PAM c bin

khng i. Ly mu nh phng lm thay i ph tn v n s gy nn mt sai

s c gi l sai s khu v sai s s lm cho mch hi phc b thu

PCM khng th hi phc c chnh xc tn hiu tng t nguyn thu ban u.

Hnh 2.5a m t s nguyn l mch ly mu.

y FET lm vic nh mt chuyn mch tng t n gin (trn s

c k hiu l Q1). Khi FET trng thi tr khng thp, Q1 ng v in pmu tng t qua Q1 n t C1. T C1 c np v thi gian Q1 ng mch c

===================================================================PHM TRN HON

Hinh 2.5


26/57

n hc k 9===================================================================gi l thi gian khe h hoc thi gian thu nhn. C 1 lm nhim v nh l mch

gi. Khi FET trng thi tr khng cao, Q1 h (ngt mch) v t C1 b ngn

khng c ng phng, do in p ly mu c lu gi. Thi gian lu

gi ca t in c gi l thi gian chuyn i A/D, bi v trong khong thigian b chuyn i ADC s chuyn i in p mu thnh m PCM. Thi

gian thu nhn (khe h) cn phi rt ngn m bo s chuyn i t b mo

dng khe h nh hn so vi ly mu t nhin v yu cu b chuyn i tng

t/s chm hn.

Hnh 2.5b m t tn hiu tg t u vo, xung ly mu v dng sng qua

C1. y c mt iu cn ch l tr khng u ra ca Z1 l tr khng ca C1

cn phi c gi tr nh m bo cho thi gian np RC l hng s, np v

phng nhanh.

2.3.3. Tc ly mu, lng t ho v m ho

Tc ly mu ti thiu (fs) c xc nh theo nh l Nyquist c th

c s dng cho h thng PCM. Tc ly mu Nyquist c th biu th vi

biu thc:

fs 2fa (2.1)

trong : f s l tc ly mu ti thiu (Hz);

fa l tn s ln nht dng ly mu (Hz);

===================================================================PHM TRN HON


27/57

n hc k 9===================================================================

Thc cht, mch ly mu v gi l mt b iu ch bin . Chuyn mch

l mt phn t khng tuyn tnh c hai u vo: xung ly mu v tn hiu tng

t u vo. Nh vy s c s trn phi tuyn xut hin gia hai tn hiu . Hnh

2.6 m t c trng vng tn s ca ph u ra ca mch ly mu v gi.

Nu nh tn s fs nh hn hai ln tn s fa th hin tng mo xut hin.Hin tng c gi l mo do chng ln. Hnh 2.6a m t vng tn s ca

ph tn u ra ca mch ly mu v gi. u ra gmf c hai u vo nguyn

thu (m tn v tn s c bn ca xung ly mu), tng ca cc tn s (fs fa)

tt c cc sng hi ca fs v fa (2fs fa, 3fs fa,)

Bi v xung ly mu l dng sng lp li, cho nn n s to ra mt dy cc

sng hi dng sin. Mi mt sng dng sin s li c iu ch bin v tonn s mo chng ln (h.2.6b).

V d 2.1

Cho h thng PCM vi tn s am cc i u vo l 4 kHz. Hy xc

nh tc ly mu cc tiu v tn s chng ln to ra nu nh c tn s audio

l 5 kHz k tip c a vo mch ly mu v gi.

Bi giiTheo nh l Nyquist, ta c:

===================================================================PHM TRN HON

Hnh 2.6


28/57

n hc k 9===================================================================

fs 2fa do f s 8 kHz

Nu nh tn s audio 5 kHz c vo b ly mu th ph tn u ra

c m t hnh 2.7 sau y. T hnh v, c th thy rng, tn hiu 5 kHz gy

nn tn s chng ln 3 kHz so vi ph tn audio nguyn thu.

B lc thng di u vo nh trong hnh 2.3 cn c gi l b lc chngchng ln. Tn s ct pha trn ca n c chn sao cho tn s khng ln

hn mt na tn s ly mu k tip c a vo b ly mu v gi, nh vy

gii hn mo chng ln c th xy ra.

i vi iu ch PCM, tn hiu tng t u vo c ly mu, sau

c chuyn i thnh dy ni tip m nh phn. M nh phn c truyn n

pha thu sau c chuyn i tr li tn hiu tng t. Cc m nh phn sdng cho PCM l cc m n bit, trong n l s nguyn dng ln hn 1. Cc

m thng c s dng cho PCM l cc m nh du theo bin trong

mt bit quan trng nht l bit nh du v cc bit cn li c dng cho cc mc

bin .

Bng 2.1 m t mt m PCM vi n bit trong n = 3. Bit nh du mu

(logic 1 = dng v logic 0 = m). Hai bit cn li c trng cho mc bin .

===================================================================PHM TRN HON

Hnh 2.7


29/57

n hc k 9===================================================================

Nh vy, vi hai bit bin th c kh nng cho 4 mc dng v 4 mc m.

Tng cng kh nng ca m l 23 = 8.

bng 2.1

M PCM nh trong bng 2.1 mt s ti liu cn gi l m nh phn

gp, bi v nu ly ng nt t lm chun v gp li th na phn trn v na

phn di c cc gi tr nh phn bin trng nhau. Trong trng hp ny c

hai gi tr m c n nh cho mc 0 vn, l 100 cho (+0) v 000 cho (-0).

Vi v d trn, bin cho mi kch c bc l 1 vn. Do in p cc i c

th c m ho trong mch l 3 V (111) hoc -3 V (011). Nu nh bin ca

mu vt qu khong lng t ln nht th s xut hin mo v c gi l mo

qu ti.

Vic phn nh cc m PCM theo cc gi tr tuyt i bin nh vy c

gi l lng t ho. Bin ca mi bc ti thiu c gi l bc chuyn

trng thi v n c gi tr bng in p ca bit c trng s b nht LSB (least

significant bit)

Hnh 2.8 m t mt tn hiu tng t u vo, xung ly mu, tn hiu

PAM tng ng v m PCM. Tn hiu tng t c ly mu ba thi im.

Mu th nht xut hin ti t1 khi in p tng t c gi tr chnh xc l +2 V.

M PCM tng ng vi xung mu 1 l 110. Mu th hai xut hin ti thi im

t2, khi in p ca tn hiu tng t l -1 V. M PCM tng ng ca n l

001.

===================================================================PHM TRN HON


30/57

n hc k 9===================================================================

xc nh m PCM i vi mt mu c th, c th thc hin n

gin php chia in th ca mu cho bc chuyn trng thi, chuyn chng

thnh m nh phn n bit v sau cng thm bit du. V d vi mu 1, bit du l

1 ch in p dng, m bin l 10 tng ng vi s 2 nh phn. Mu 3 xut

hin thi im t3. in p tng t thi im l 2,6 V. Khng c m chomc bin . Trong cc b ADC ly mc gn ng, v d ly 111 hoc +3 V.

Nh vy s xut hin sai s v c gi l sai s lng t Qe (quantization

error). Sai s n cng tng ng nh tp m trng cng. Cng v vy m sai

s lng t cn c gi l tp m lng t Qn (quantization noise) v bin

cc i ca n bng mt na in p ca bc ti thiu (VLSB/2). Trong v d

trn, Qe = 1V/2 = 0,5 V.

===================================================================PHM TRN HON

Hnh 2.8


31/57

n hc k 9===================================================================

2.3.4 Di ng

S cc bit PCM c truyn trn mu c xc nh bi mt vi thng

s, trong bao gm bin cc i c th chp nhn c in p u vo,

bc chuyn trng thi v di ng. Di ng DR (dynamic range) l t s gia

bin cho php ln nht trn bin cho php nh nht c th c gi m

bi b DAC. Di ng DR c biu th bi biu thc:

ax

min

mUDRU

= (2.2)

Trong Umin bng bc chuyn trng thi v Umax l bin in p cc

i m b DAC c th thc hin gii m.

Nh vy c th biu th:

DR =Umax

Bc chuyn trng thiVi h thng m t trong bng 1.1 th:

===================================================================PHM TRN HON

Hnh 2.9


32/57

n hc k 9===================================================================

33

1

VDR

V= =

Thng thng, di ng c biu th bng dB, do :

ax

min

320 lg 20 lg 9, 54

1m

UDR dB

U= = =

Gi tr ca di ng bng 3 ni ln rng, t s gia in p ln nht v

in p nht c gii m bng 3.

Nu nh c yu cu bc chuyn trng thi nh hn, v d 0,5 V m vn

gi gi tr ca gii ng l 3 th lc cn phi gim in p cc i u vo

cng vi h s gim l mt na, tc l:

1, 53

0,5DR = =

Nh vy, Umax c gim i 2 ln v di ng l c lp vi bc chuyn

trng thi. Nu nh bc chuyn trng thi gim i 2 ln (0,25V) v gi nguyn

gi tr bin in p u vo, th lc di ng cn phi tng gp i tc l:

1, 56

0,25DR = =

S cc bit s dng trong m PCM ph thuc vo di ng. Vi m PCM

hai bit th bin cc tiu c th gii m c m nh phn l 01 v bin cc

i c m nh phn l 11. Nh vy, t s ca m nh phn cc i trn m nh

phn cc tiu l 3, c cng di ng. Do m nh phn cc tiu lun lun c gi

tr 1, cho nn DR s l s nh phn cc i ca h thng. Do , xc nh s

bit cn thit i vi mt m PCM th biu thc quan h ton hc sau y cs dng:

2n 1 > DR

V vi gi tr cc tiu ca n th:

2n 1 = DR (2.3a)

Trong : n l s bit ca m PCM, bao gm c bit du;

DR l gi tr tuyt i ca di ng.

===================================================================PHM TRN HON


33/57

n hc k 9===================================================================

Ti sao li 2n 1 ? Mt m PCM c s dng cho OV, nh vy n

khng quan tm n di ng. Do

2n = DR + 1 ( 2.3b)

tnh gi tr ca n, y ta s chuyn sang logaritLg2n = lg(DR + 1) nlg2 = lg(DR + 1)

lg(3 1) 0,6022

lg 2 0, 301n

+= = =

Nh vy, vi di ng l 3 th m PCM cn c 2 bit.

Di ng cng c th biu th di dng dB

ax( )

min

20lg mdBU

DR U= (2.4a)

hoc 2 lg(2 1)n= (2.4b)

trong n l s bit ca PCM. Vi cc g tr n ln th gi tr di ng c th tnh

theo biu thc tng ng:

( ) 20 lg(2 ) 20 lg 2 6n

dBDR n n (2.5)

Biu thc (2.5) c ngha l khong 6 dB trn di ng bit vi h thngPCM c gii m tuyn tnh.

Hnh 2.9 m t hm truyn t u vo/u ra i vi mt b chuyn i

tng t/s (cn c gi l b lng t tuyn tnh). Trn hnh m t tn hiu

tng t vo l ng thng (tc mt ng dc) v tn hiu c lng t cc

i ca chng l ging nhau i vi mi bin u vo v c m t hnh

2.9c.Hnh 2.10 m t cng tn hiu tng t u vo nh hnh 2.8 nhng

c ly mu tc nhanh hn. Trn nhn thy rng, vic gim thi gian

gia cc mu (ngha l tng tc ly mu) s to ra tn hiu PAM gn ging

vi tn hiu tng t u vo. V y cng cn lu rng, vic tng tc ly

mu khng lm gim sai s lng t ca mu.

Bng 2.2 sau y lit k gi tr di ng cho m PCM c n bit vi cc gi

tr n n 16.

===================================================================PHM TRN HON


34/57

n hc k 9===================================================================Bng 2.2. Gi tr di ng cho m PCM c n bit

S cc bit (n) trong m PCM S cc mc M = 2n Gi tr di ng (dB)0 1 01 2 6,02

2 4 123 8 18,14 16 24,15 32 30,16 64 36,17 128 42,18 256 48,29 512 54,210 1024 60,2

11 2048 66,212 4096 72,213 8192 78,314 16384 84,315 32768 90,316 65536 96,3

V d 2.2

Cho h thng PCM vi cc thng s sau:

- Tn s ln nht ca tn hiu tng t u vo l 4 kHz

- in p cc i c gii m ti u thu l +2,25 V===================================================================PHM TRN HON

Hnh 2.10


35/57

n hc k 9===================================================================- Gi tr di ng l 46 dB

Hy xc nh:

- Tc ly mu ti thiu;

- S bit ti thiu c s dng trong m PCM;- Bc chuyn trng thi;

- Sai s lng t.

Bi gii

p dng nh lut Nyquist biu thc 2.1, tc ly mu ti thiu l:

fs = fa = 2(4 kHz) = 8 kHz

xc nh gi tr ca di ng, theo biu thc 2.4a.

ax ax

min min

46 20lg 2, 3 lgm mU U

dBU U

= =

2,3 ax

min

10 199,5mU

DR DRU

= = =

Thay th vo (2.3b) c:

lg(199,5 1)7,63

lg 2n

+= =

S nguyn gn nht ln hn 7,63 l 8; Do 8 bit c s dng cho bin

. Do bin in p u vo l 2,25 V cho nn cn cng thm vo mt bit

du. Nh vy tng cng s bit ca m PCM trong trng hp ny l 9 v tng

cng s m ca PCM l 29 = 512 (trong s c 255 m dng, 255 m m v

hai m 0).

xc nh di ng thc, thay th vo biu thc (2.4b) c:DR = 20lg255 = 48,13 dB

xc nh bc chuyn tip trng thi, em chia bin cc i + hoc

cho s m khc 0, dng hoc m s c:

Bc chuyn trng thi = ax 82, 55 2, 55

0,012 1 2 1 256 1

m

n

U= = =

V

Sai s lng t cc i l:

===================================================================PHM TRN HON


36/57

n hc k 9===================================================================

Qe =Bc chuyn trng thi

=0,01

= 0,005 V2 2

2.3.5. Hiu sut m ho

Hiu sut m ho l ch s bng s ni ln hiu qu m PCM c s

dng. Hiu sut m ho l t s gia s bit ti thiu yu cu t c mt di

ng no cho s bit thc t ca m PCm c s dng.

Hiu sut m ho c biu th bi biu thc ton hc:

Hiu sut m ho =s bit ti thiu (bao gm c bit du)

100%s bit thc t (bao gm c bit du)

Nh vy, hiu sut m ho nh cc s liu trong v d 2.2 s l:

Hiu sut m ho =8,63

.100 95,89%9

=

Chng 3

IU CH TN HIU S

3.1. Tng quan v mt s h thng truyn tin s

Nh phn tch trong chng mt, truyn tin s bao gm c truyn

dn s v radio s.

c im phn bit mt h thng radio s vi mt h thng radio s

AM, FM hoc PM truyn thng trc nay dng quen thuc l cn c vo bn

cht ca tn hiu iu ch. C hai h thng radio s v radio tng t u s

dng cc sng mang tng t; tuy vy, iu ch tng t th tn hiu iu ch

l tn hiu tng t cn iu ch s thi tn hiu iu ch l tn hiu s. C haidng iu ch s v iu ch tng t th thng tin ngun c th hoc l tn hiu

s hoc l tn hiu tng t.

Hnh 3.1 m t s khi n gin ho mt h thng radio s.

===================================================================PHM TRN HON


37/57

n hc k 9===================================================================

B tin m ho pha pht lm nhim v chuyn i mc sau m ho

hoc nhm d liu n vo trong t iu khin iu ch vi sng mang tng

t. Sng mang iu ch c o gt (c lc), c khuch i sau

a ra mi trng truyn dn truyn n pha thu. Cc tn hiu nhn c

u thu s c lc, khuch i v sau chuyn n mch gii iu ch tito li thng tin ngun nguyn thu. Mch ng h v mch hi phc sng

mang c nhim v ly ra thng tin ng h v sng mang t tn hiu n

c iu ch.

3.2.Kho dch bin, ASK

K thut iu ch s n gin nht l iu ch bin s hoc cn gi l

iu ch kho dch bin (ASK amplitude shift keying) hoc iu ch ng -ngt (OOK on off keying). y l dng iu ch bin sng mang s dng

c hai n bin vi tn hiu iu ch a vo c dng nh phn. Biu thc ton

hc ca tn hiu iu ch s vi tn hiu nh phn c th c biu th bi biu

thc:

( ) ( ) ( )1 os2

am m c

Au t u t c t

= +

(3.1)

Trong :

===================================================================PHM TRN HON

Hnh 3.1


38/57

n hc k 9===================================================================

uam(t) l sng c iu ch bin s;

um(t) l tn hiu nh phn c iu ch (V);

A/2 l bin ca sng mang cha iu ch (V);

c l tn s gc ca sng mang (rad/s).Trong biu thc (3.1), tn hiu iu ch [um(t)] l dng sng nh phn

khng chun ho, trong +1V = logic 1 v 1V = logic 0.

Do , vi logic 1 u vo th um(t) = +1 biu thc (3.1) s l:

( ) [ ] ( ) ( )1 1 os os2

am c c

Au t c t Ac t

= + =

V vi logic 0 u vo th um(t) = - 1 v biu thc (3.1) s l:

( ) [ ] ( )1 1 os 02

am c

Au t c t

= =

Nh vy, vi iu ch 100% th sng c iu ch nhn cc gi tr hoc

Acos(ct) hoc 0 tng ng vi logic 1 v 0 u vo. Sng mang trong trng

hp ny l dng hoc ng mch hoc ngt mch, cng v vy m iu ch

kha ng - ngt (OOK on off keying). Trong mt s ti liu, iu ch bin

s cn c gi l sng lin tc bi v sng mang ca chng (trng hpng mch) c bin , tn s v gc pha l hng s.

===================================================================PHM TRN HON

Hnh 3.2


39/57

n hc k 9===================================================================

Hnh 3.2 m t cc dng sng u vo v u ra ca mt my pht iu

ch bin s. Mt dng sng OOK c th c iu ch hoc c s lin kt

vi nhau hoc ri rc khng c s lin kt nhau cng vi s khc nhau v h s

iu ch.Vic s dng cc sng mang tng t iu ch theo kiu iu ch bin

s truyn thng tin s l c cht lng tng i thp nhng d thc hin v

mch in n gin. V vy dng iu ch ny t c s dng cc h truyn

tin s c dung lng v hiu nng cao.

Tn hiu ASK c th c tch sng bng mt b tch sng ng bao

(tch sng khng kt hp) hoc mt b tch sng tch (tch sng kt hp) v n

l mt dng tn hiu AM. Cc b tch sng ny c m t hnh 3.3a v hinh

3.3b.

c tch sng ti u ASK ngha l t c xc sut li bit thp

khi tn hiu ASK u vo b tp m Gauss trng k sinh lm sai lch - cn phi

s dng tch sng tch vi vic s dng lc thch ng. Trng hp ny c

minh ho trong hnh 3.3c, y cc dng sng ti cc thi im khc nhau ca

mch in c minh ho trong trng hp thu mt tn hiu ASK tng ng vichui d liu 1101.

===================================================================PHM TRN HON

Hnh 3.3


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n hc k 9===================================================================3.3. Kho dch tn, FSK

3.3.1. iu ch FSK

Kho dch tn (FSK frequency shift keying) l mt dng iu ch n

gin khc, c hiu nng tng i thp ca iu ch s, FSK nh phn l mtdng iu ch gc c bin khng i tng t nh iu tn truyn thng FM,

vi tn hiu iu ch y l tn hiu nh phn. Cc tn hiu nh phn s lm

thay i tn s sng mang theo hai mc tng ng. Biu thc tng qut ca

FSK nh phn l:

ufsk = Uccos{2[fc + um(t)f]t} (3.2)

trong : ufsk l dng sng FSK nh phn;

Uc l bin cc i sng mang (V);

fc l tn s sng mang (Hz);

f l di tn cc i (Hz);

um(t) l tn hiu iu ch nh phn u vo (1).

Trong biu thc (3.2), di tn cc i ca sng mang f, l t l vi

bin v cc ca tn hiu vo nh phn. Tn hiu iu ch [Um(t)] l dng sng

nh phn chun ho trong logic 1 = +1 v logic 0 = - 1. Nh vy, vi logic 1

u vo U ch s (t) =1 th biu thc (3.2) s l:

ufsk(1) = Uccos[2(fc + f)t]

vi logic 0 u vo, Um(t) = - 1 th biu thc (3.2) s l:

ufsk(t) = Uccos[2(fc - f)t]

tn hiu FSK nh phn, tn s sng mang s b dch chuyn (b di tn)

ph thuc vo tn hiu nh phn u vo. Bi tn hiu nh phn c hai mc l 0

v 1, cho nn tn hiu u ra cng s b dch tn ng vi hai tn s: tn s vt

hoc tn s logic 1, fm v tn s trng hoc tn s logic 0, fs. Khong tn s cch

bit nhau gia tn s vt fm v tn s trng fs c gi l lch tn hoc

di tn f (tc fm = fc + f v fs = fc - f). Cc tn s vt v tn s trng c gi tr

tu ph thuc vo thit k h thng.

===================================================================PHM TRN HON


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n hc k 9===================================================================

Hnh 3.4 m t dng sng ca tn hiu nh phn u vo v tn hiu u ra

ca mt b iu ch FSK. T hnh v nhn thy rng, tn hiu nh phn u

vo bin i t logic 1 n logic 0 v ngc li, cn tn hiu u ra c tn s

bin i t tn s vt fm n tn s trng fc v ngc li, tng ng vi s bini ca tn hiu nh phn u vo. Tn s ln nht l fc + f v tn s nh nht

l fc - f.

3.3.2. Tc bit v baud cu FSK

tn hiu FSK nh phn th khong thi gian chuyn i tn s u ra

tng ng vi s chuyn i tn hiu nh phn u vo. Do tc chuyn

i u ra l bng tc chuyn i u vo.

iu ch s, tc chuyn i u vo b iu ch c gi l tc

bit fb v c th nguyn l bit/s, k hiu l bps. Tc bin i ti u ra ca b

iu ch c gi l tc baud.

Baud thng hay b nhm ln vi tc bit. Baud l tc chuyn i c

gi tr bng nghch o thi gian ca mt phn t tn hiu ti u ra. Vi FSK

th thi gian ca mt phn t tn hiu u ra l thi gian ti thiu m tn s vt

hoc tn s trng c to ra v n bng thi gian ca mt bit n c tb. Nh

hnh 3.4 s chuyn i t tn s vt sang tn s trng v ngc li l cng vi

tc chuyn i t logic 1 sang logic 0 v ngc li. V vy, FSK th thi

gian ca mt phn t tn hiu u ra v thi gian ca bit l bng nhau, tc

bit v baud l bng nhau.

3.3.3. B pht FSK

Hnh 3.4, m t b pht FSK nh phn n gin ho v n cng tng t

nh mt b iu ch FM truyn thng trong thng s dng b to sng c

in p c iu khin (VCO voltage controlled oscillator). Tn s sng

mang c chn chnh gia tn s vt v tn s trng.

===================================================================PHM TRN HON


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n hc k 9===================================================================

Logic 1 u vo s lm dch chuyn tn s VCO v tn s vt v logic 0

u vo s lm dch chuyn tn s VCO ngc li v tn s trng. Tn hiu

u vo nh phn bin i qua li gia logic 1 v logic 0 cho nn tn hiu u ra

cng s c bin i qua li gia tn s vt v tn s trng.

b iu ch FSK, f l di tn cc i ca tn s mang v c gi tr

bng hiu s gia tn s mang v tn s vt hoc tn s trng (hoc bng mt

na hiu s ca tn s vt v tn s trng).

B iu ch VCO FSK cng c th c thc hin theo phng php

qut trong di tn cc i l tch ca in p u vo nh phn v nhy

ca b VCO. Theo phng php qut, di tn c th c biu th bi biu

thc:

f = um(t)kt (3.3)

Trong : f l di tn cc i (Hz);

um(t) l in p tn hiu iu ch nh phn cc i (V);

===================================================================PHM TRN HON

Hnh 3.4


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n hc k 9===================================================================

kt l nhy di tn (Hz/V).

Vi FSK nh phn th bin ca tn hiu u vo ch c mt trong hai

gi tr, mt ng vi logic 1 v mt ng vi logic 0. V vy, di tn cc i l

mt hng s. Do in p nh ca logic 1 v logic 0 l ging nhau cho nn bin

ca di tn i vi logic 1 v logic 0 cng ging nhau. mt s dng iu ch

khc , gi tr ca tc bit v baud l khc nhau.

Cn lu rng, c mt s ti liu c cch gi ngc li, trc iu ch

gi l baud v sau iu ch gi l tc bit.

3.3.4. rng di tn ca FSK

u ra ca mt b iu ch FSK c quan h vi u vo nh phn nh m

t hnh 3.5 trong logic 0 tng ng vi tn s trng fs, v logic 1 tng ng

vi tn s vt fm cn fc l tn s mang.

di tn cc i c biu th bi:

2

m sf ff

= (3.4)

Trong : f l di tn cc i (Hz);

fm l tn s vt (Hz);

fs l tn s trng (Hz).

T hnh 3.5 cng thy rng FSK c hai sng hnh sin ca tn s fm v fs.

Cc sng hnh sin dng xung c ph tn theo hm sinx/x.

V vy, c th m t ph tn u ra ca tn hiu FSK nh hnh 3.6. Gi

thit rng, cc nh ca ph tn cng sut cha phn ln nng lng, th lc

rng di tn ti thiu cn thit cho tn hiu FSK c th i qua c th biuth theo biu thc gn ng:===================================================================PHM TRN HON


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n hc k 9===================================================================

B = | (fs + fb) (fm fb)| = (|fs fm|) + 2fb (3.5)

V do fs fm c gi tr 2f, cho nn rng di tn c th theo biu thc

gn ng:

B = 2f + 2fb = 2(f + fb) (3.6)

Trong : B l rng di tn ti thiu (Hz);

fm l tn s vt;

fs l tn s trng;

f l di tn nh ti thiu (Hz).

V d 3.1

Mt tn hiu FSK c tn s vt l 49 kHz, tn s trng l 51 kHz v tc

bit u vo l 2 kbps. Hy xc nh:

a) di tn cc i

===================================================================PHM TRN HON

Hnh 3.5Hnh 3.6


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n hc k 9===================================================================

b) rng di tn ti thiu

c) Baud u ra

Bi gii

a) di tn cc i c tnh theo biu thc (3.4)

49 511

2 2

m sf f kHz kHzf kHz

= = =

b) rng di tn ti thiu, tnh theo biu thc (3.6)

B = 2(f + fb) = 2(1000 + 2000) =6 kHz

c) Vi FSK, baud c gi tr bng tc bit v bng 2000 bps.

y c th nhn xt rng, biu thc gn ng (3.6) tnh rng di

tn cng ging nh quy tc Carson xc nh di tn trong sng FM c ch s

trung bnh. Ch s c s khc nhau gia hai biu thc l i vi FSK th tc

bit fb c thay th cho tn s iu ch fm tn hiu FM.

Cc hm Bessel cng c th s dng xc nh gi tr gn ng ca

rng di tn ti thiu ca tn hiu FSK. Nh m t hnh 3.5, tc chuyn i

nhanh nht tn hiu nh phn NRZ (khng tr v khng) xut hin ti cc

chuyn tip gia 0 v 1 ( c ngha l sng ch nht), bi v ti to ra chu k.

Do tn s cao nht trong sng xung ch nht c gi tr bng tc lp li casng ch nht v i vi tn hiu nh phn th n c gi tr bng mt na tc

bit. C ngha l:

2

ba

ff = (3.7)

Trong : f a l tn s c bn ln nht ca tn hiu iu ch nh phn (Hz);

fb/2 l tc bit (bps).

Biu thc i vi ch s iu ch FM cng c gi tr i vi FSK, l:===================================================================PHM TRN HON


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n hc k 9===================================================================

a

fh

f

= (3.8a)

Trong : h l ch s iu ch FM, cng cn c gi l h s h trong FSK;

Fa l tn s c bn ca tn hiu iu ch nh phn (Hz);f l di tn cc i (Hz).

Ch s iu ch (t s di tn) trong trng hp xu nht to ra rng

bng tn ln nht, ch s iu ch xu nht hoc rng bng tn ln nht xut

hin ti cc gi tr cc i ca di tn v tn s tn hiu iu ch. di tn cc

i FSK l khng i v c quan h vi ch s iu ch, tn s c bn, tc

bit u vo theo biu thc:

` 2

2

m s

b

f f

hf

= (3.8b)

hocm s

b

f fh

f

= (3.9)

trong : h l h s;

fs l tn s trng;

fm l tn s vt;

fb l tc bit (bps).

V d 3.2

S dng hm Bessel, hy xc nh rng di tn ti thiu ca tn hiuFSK ; vi tn s vt l 49 kHz, tn s trng l 51 kHz v tc bit l 2 kbps.

Bi gii

Ch s iu ch tnh theo (3.9) s l:

49 51 21

2 2

kHz kHz kHzh

kpbs kbps

= = =

T bng hm Bessel c ba di bin ng ch i vi ch s iu ch. Vvy rng di tn c th l:

===================================================================PHM TRN HON


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n hc k 9===================================================================

B = 2(3.1000) = 6000 Hz

rng di tn c xc nh y ph hp vi rng di tn c

tnh trong v d 3.1.

3.3.5. B thu FSK

Hnh 3.7 m t mt b thu tn hiu FSK n gin ho. Tn hiu FSK u

vo c ng thi a n cc b lc di thng BPF (bandpass filter) sau khi

qua b tch cng sut. B lc di thng ch cho qua tn s vt hoc tn s trng

sau a n b tch sng ng bao. Cc b tch sng ng bao a ra

cng sut cho mi di thng a n vo b so snh to tn hiu d liu u

ra. B tch tn hiu FSK nh trn l loi b tch khng lin kt.

Hnh 3.8 m t s khi ca b thu FSK lin kt. Tn hiu FSK vo

c nhn (trn) vi tn hiu sng mang c hi phc c cng tn s v pha

vi tn s sng mang pha pht. Tuy nhin hai tn s pht (tn s vt v tn s

trng) vn l khng lin tc. Vic to li tham chiu ti ch lin kt cho c hai

tn s trong thc t l rt kh khn, do b tch FSK lin kt t c s dng.

===================================================================PHM TRN HON

Hnh 3.8


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n hc k 9===================================================================

Mch in thng dng nht c dng gii iu ch cc tn hiu nh phn

iu ch FSK l vng kha pha PLL (phase locked loop), c m t hnh

3.13 (mc 3.3.6).

Nguyn l hot ng ca b gii iu ch FSK dng vng kha pha PLL

(PLL FSK), cng tng t nh nguyn l hot ng ca b gii iu ch PLL FM iu tn. u vo vng kho pha c s lch nhau gia hai tn s vt v

tn s trng , in p chnh lch mt chiu dc u ra ca b so pha tng ng

vi lch tn s trong mt tng ng vi logic 1 v mt tng ng vi

logic 0. Nh vy, u ra ca b gii iu ch s c hai mc (nh phn) c trng

cho tn hiu FSK u vo. Tn s hot ng ca b PLL bng tn s gia ca

b iu ch. S chnh lch v in p ca b so pha tng ng vi s bin itn s u vo v n s dao ng xung quanh gi tr OV.

iu ch FSK nh phn c hiu nng sai s km hn so vi iu ch FSK

hoc QAM v chng cng t c s dng trong cc h thng v tuyn s c

hiu nng cao. N ch c s dng mt cch hn ch cc modem d liu c

hiu nng thp, gi thnh thp, khng ng b hoc truyn d liu trn cc

ng tn hiu tng t, tn hiu thoi.

===================================================================PHM TRN HON

Hnh 3.9


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n hc k 9===================================================================

Modem FSK. Hnh 3.9 m t v d mt modem dung iu ch FSK c

s dng trong ng truyn d liu tc thp thng qua ng dy in thoi.

3.3.6. iu ch FSK c pha lin tc

iu ch FSK c pha lin tc (CP FSK) l FSk nh phn m cc tn svt v tn s trng c ng b vi tc bit u vo. S ng b y ng

rng, gia chng c quan h thi gian vi nhau ch khng nht thit phi chnh

xc bng nhau.

iu ch CP FSK th tn s vt v tn s trng c chn sao cho s

cch bit ca cc tn s so vi tn s gia ng bng mt bi s l ca mt

na tc bit, tc l (fm v fs = n(fb/2), trong n l s nguyn l.Nh vy s

m bo tn hiu tng t u ra c pha lin tc, khng b gin on khichuyn tip t tn s vt (logic 1) sang tn s trng (logic 0) hoc ngc li.

Hnh 3.10 m t mt dng sng FSK c pha khng lin tc. y khi

u vo c s chuyn tip t logic 1 sang logic 0, hoc ngc li, th tn hiu

u ra c s bin i gin on v pha. Khi xut hin s gin on v pha nh

vy th b gii iu ch s lm vic trong trng thi khng n nh, dn n sai

s thu.

===================================================================PHM TRN HON

Hnh 3.10


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n hc k 9===================================================================

Hnh 3.11 m t dng sng ca mt b iu ch FSK c pha lin tc.

y, khi u vo c s chuyn tip t logic 0 sang logic 1 hoc ngc li, th

tn hiu u ra khng c s gin on v pha v gc ca tn hiu vn c

bin i lin tc.Tn hiu loi iu ch CP FSK ny c hiu nng sai s bit tt hn so vi

cc b iu ch FSK thng thng, v t s tn hiu/tp m cng tt hn. Nhc

im ca b iu ch CP FSK l kt cu mch in phc tp hn.

y, nu nh hiu s gia tn s vt v tn s trng l bng mt na tc

bit u vo, tc fs fm = 0,5 fb th ch s iu ch h = 0,5. y l trng

hp hiu s cc tiu gia tn s vt v tn s trng v loi CP FSK ny c

gi l kho dch cc tiu (MSK minimum shift keying).

3.4. Kho dch pha, PSK

Kho dich pha (PSK phase shift keying) l mt dng iu ch gc, bin

khng i. Kho dich pha cng tng t nh iu ch pha thng thng ch

c khc l PSK th tn hiu co l tn hiu nh phn v pha u ra l c s

lng gii hn.

3.4.1. Kho dch pha nh phn, BPSK

===================================================================PHM TRN HON

Hnh 3.11


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n hc k 9===================================================================

kha dch pha nh phn (BPSK binary phase shift keying) th hai pha

u ra c th l vi mt tn s sng mang n (nh phn c ngha l 2).

Trong hai pha th mt pha tng ng vi logic 1 v mt pha tng ng

vi logic 0. Nu nh trng thi ca tn hiu nh phn u vo thay th hai gcpha u ra cng bin i lch pha nhau 1800. Cng v vy m tn hiu BPSK

cn c tn gi khc l kho do pha (PRK phase reversal keying) hoc iu

ch nh pha.

1. B pht BPSK

Hnh 3.12 m t s khi n gin ca mt b pht BPSK. y, b

iu ch cn bng lm vic nh mt chuyn mch o pha. Ph thuc vo logic

ca tn hiu nh phn u vo m sng mang c a n u ra c s lch

pha nhau 1800 so vi sng mang tham chiu ca b to sng.

Hnh 3.13a m t s khi b iu ch vng cn bng. B iu ch cn

bng c hai u vo. Sng mang ng pha vi b to sng tham chiu v d

liu s nh phn. cho b iu ch cn bng lm vic mt cch chnh xc th

in p nh phn u vo cn phi ln hn in p nh ca sng mang. iu

m bo cho cc in p u vo iu khin ng - m trng thi lm viccc dit D1-D4. Nu tn hiu nh phn u vo l logic 1 (in p dng) th

===================================================================PHM TRN HON

Hnh 3.12


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n hc k 9===================================================================cc dit D1 v D2 c phn cc thun v m, cn cc dit D3 v D4 phn cc

ngc v ng (hnh 3.13b). Vi phn cc nh trn th in p u ra ca bin

p T2 s ng pha vi in p u vo ca bin p T1.

Nu nh tn hiu nh phn u vo l logic 0 (in p m), th lc ccdit D1 v D2 phn cc ngc v ng, cn cc dit D3 v D4 th phn cc thun

v m (hnh 3.13c). Kt qu l in p u ra bin p T 2 s lch pha 1800 so

vi in p u vo T1.

Hnh 3.14 m t bng chn l, th pha v th khng gian trang thi

ca mt b iu ch BPSK.

2. rng di tn ca tn hiu BPSK

B iu ch cn bng l mt b iu ch tch v tn hiu u ra l tch cahai tn hiu u vo. b iu ch BPSK th tn hiu sng mang u vo c

nhn vi tn hiu nh phn. Nu nh +1 V c trng cho logic 1 v 1 c trng

cho logic 0 v sng mang u vo (sinct) c nhn vi nhau c ngha l gi tr

ca sinct s c nhn vi +1 v 1. Gi tr th nht c trng cho tn hiu

ng pha vi sng ca b to sng tham chiu v gi tr th hai lch pha 1800

vi sng ca b to sng tham chiu.

===================================================================PHM TRN HON

Hnh 3.14


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n hc k 9===================================================================

Ti thi im m u vo c s chuyn i gi tr logic 0 v 1 th tn

hiu u ra c s chuyn i pha 0 v 180 0. Nh vy vi tn hiu BPSK, tc

baud ( u ra) bng tc chuyn i bit (bps) u vo, v rng bng

tn ln nht u ra xut hin khi d liu nh phn u vo l mt dy bini 1/0. Tn s c bn (fs) ca dy bin i 1/0 bng mt na tc bit (fb/2).

V ton hc, u ra ca BPSK t l vi gi tr:

u ra BPSK = [sin(2fat)] x [sin(2fct)] (3.10)

trong : f a l tn s c bn cc i ca tn hiu nh phn u vo (Hz);

fc l tn s sng mang tham chiu (Hz).

V phi ca (3.10) c th vit:

( ) ( )1 1cos 2 cos 22 2

c a c af f t f f t +

Nh vy, rng di tn Nyquist ti thiu ly c hai bn ca tn hiu s l:

fc + fa (fc fa) = 2fa

v do fa = fb/2 trong fb l tc bit u vo cho nn:

2

2

bb

fB f= = (3.11)

Trong B l rng bng tn Nyquist c hai bn.

===================================================================PHM TRN HON

Hnh 3.15


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n hc k 9===================================================================

Hnh 3.15 m t quan h pha theo thi gian i vi dng sng BPSK. Ph

tn u ra ca b iu ch BPSK l dng hai n bin c sng mang b trit

trong cc tn s bin cao nht v bin thp nht cc sng mang mt gi tr l

mt na tc bit. Nh vy, rng bng tn ti thiu (fN) cn m bo chotn hiu BPSK xu nht i qua l bng tc bit u vo.

V d 3.4

Mt b iu ch BPSK c tn s sng mang l 70 MHz v tc bit

u vo l 10 Mbps. Hy xc nh cc tn s cc i v cc tiu ca bin trn v

bin di, ph tn u ra, rng di tn v gi tr baud.

Bi gii

S dng biu thc (3.10) ta c:

Tn hiu u ra = (sinat)(sinct)

= [sin2(5MHz)t][sin2(70MHz)t]

= ( ) ( )1 1

cos 2 70 5 cos 2 70 52 2

MHz MHz t MHz MHz t +

Tn s bin di Tn s bin trn

Tn s cc tiu bin di (LSF lower side frequency)

LSK = 70MHz 5MHz = 65MHz

Tn s cc i bin trn (USF upper side frequency)

USF = 70MHz + 5MHz = 75MHz

Ph tn u ra trong iu kin tn hiu nh phn u vo xu nht c

biu th nh sau:

===================================================================PHM TRN HON


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n hc k 9===================================================================

rng bng tn Nyquist ti thiu (fN)

fN = 75MHz 65MHz = 10MHz

v gi tr baud s l:

baud = fb hoc 10 Mbaud

3. M ho M mc

===================================================================PHM TRN HON

B = 10 MHz

65MHz 70MHz 75MHz

Sng mang b trit


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n hc k 9===================================================================

===================================================================PHM TRN HON


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n hc k 9===================================================================

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