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n hc k 9===================================================================
TRNG I HC PHNG NGKHOA CNG NGH THNG TIN
---------------o0o--------------
CNG HA X HI CH NGHA VIT NAMc lp - T do - Hnh phc
--------------o0o--------------
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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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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