ma hoa dieu che

8/7/2019 MA HOA DIEU CHE

1/13

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____________________Chng 2M ha v iu ch II - 1

CHNG 2

M HA V IU CH

PH TN CA TN HIU Ph tn gin on Ph tn lin tc M HA

Cc dng m ph bin K thut ngu nhin ha

IU CH Bin Gc

Xung

Trong truyn thng, tin tc v d liu l tt c nhng g cn trao i, chng c th lting ni, hnh nh, tp hp cc con s, cc k hiu, cc i lng o lng . . . c a vomy pht pht i hay nhn c my thu.

Tn hiu chnh l tin tc c x l c th truyn i trn h thng thng tin.Vic x l bao gm chuyn i, m ha v iu ch.Chuyn i l bin cc tin tc di dng khng in thnh ra tn hiu in.M ha l gn cho tn hiu mt gi tr nh phn v c trng bi cc mc in p c

th c th truyn trn knh truyn v phc hi my thu.iu ch l dng tn hiu cn truyn lm thay i mt thng s no ca mt tn

hiu khc, tn hiu ny thc hin nhim v mang tn hiu cn truyn n ni thu nn c gil sng mang (carrier wave). Mc ch ca siu ch l di ph tn ca tn hiu cn truynn mt vng ph tn khc thch hp vi tnh cht ca ng truyn v nht l c th truynng thi nhiu knh cng mt lc (a hp phn tn s).

Chng ny cp n siu ch v m ha. Nhng trc tin, chng ta cn nhcli mt s tnh cht ca tn hiu qua vic phn tch tn hiu khng sin thnh tng ca cc tnhiu hnh sin v lu n mi quan h tn s-thi gian ca tn hiu.

2 . 1 ph tn ca tn hiu

Trong mt h thng thng tin tn ti 3 dng tn hiu vi ph tn khc nhau:- Loi th nht l cc tn hiu c tnh tun hon c dng hnh sin hoc khng. Mt tn

hiu khng sin l tng hp ca nhiu tn hiu hnh sin c tn s khc nhau. Kt qu ny cc bng cch dng chui Fourier phn tch tn hiu.

- Loi th hai l cc tn hiu khng c tnh tun hon m c tnh nht thi (th d nhcc xung lc), loi tn hiu ny c kho st nhbin i Fourier.

- Loi th ba l tn hiu c tnh ngu nhin, khng c din t bi mt hm ton hcno. Th d nh cc loi nhiu, c kho st nhphng tin xc sut thng k.

Cc loi tn hiu, ni chung, c thc xt n di mt trong hai lnh vc :- Lnh vc thi gian: Trong lnh vc ny tn hiu c din t bi mt hm theo thi

gian, hm ny cho php xc nh bin ca tn hiu ti mi thi im.- Lnh vc tn s : Trong lnh vc ny ngi ta quan tm ti s phn b nng lng

ca tn hiu theo cc thnh phn tn s ca chng v c din t bi ph tn.Trong gii hn ca mn hc, chng ta ch cp n hai loi tn hiu u.

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Nguyn Trung Lp Truyn d liu


2/13

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____________________Chng 2M ha v iu ch II - 2

2.1.1 Ph tn gin on

Tn hiu c tnh tun hon n gin nht l tn hiu hnh sinv(t) = Vm sin (t +) = Vmsin ( 2ft +)

Tn hiu ny c ph tn l mt vch duy nht c bin Vm ti tn s f (H 2.1)

(H 2.1)

Cc dng tn hiu tun hon khc c th phn tch thnh tng cc tn hiu hnh sin, nh

vy ph tn ca chng phc tp hn, gm nhiu vch cc tn s khc nhau.Tn hiu thng gp c dng hnh ch nht m bi php phn tch thnh chui Fourierta thy ph tn bao gm nhiu vch cc tn s cbn f v cc ha tn 3f, 5f, 7f .... (H 2.2).

(a) (b)(H 2.2)

Tn hiu (H 2.2.a) phn tch thnh chui Fourier:

v = .....)cos77

1cos5

5

1cos3

3

1(cos

4Vtttt +

.

Vi = 2 / T = 2fT & f ln lt l chu k v tn s ca tn hiu ch nht.

Lu , nu di tn hiu (H 2.2.a) ln mt khong V theo trc tung th ph tn cthm thnh phn mt chiu (H 2.3)

(a) (H 2.3) (b)

v = V + .....)t77

1t5

5

1t3

3

1t(

4+

coscoscoscos

V

Xt trng hp chui xung ch nht vi rng


3/13

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____________________Chng 2M ha v iu ch II - 3

v= .....)cos33x

sin3xcos2

2x

sin2xcos

x

sinx(

T

2V

T

Vttt

+++

vi x = / T

(a) (H 2.4) (b) Ph tn trong trng hp = 0,1T

Nhn thy bin ca ha tn th n xc nh bi Vn =

nx

sinnx

T

2V

(H 2.4.b) l ph tn ca tn hiu (H 2.4.a) cho trng hp = 0,1 T. Trong trng hpny tn su tin ca tn hiu c bin t tr 0 l 10f.

Nu xem bng thng BW ca tn hiu l khong tn s m bin tn hiu t gi tr0 u tin (v nng lng tn hiu tp trung trong khong tn s ny) ta c:

BW xc nh bi:sin(nx) = 0

nx = n/ T = n/ T = 1/hay BW = nf = n/T = 1/

2.1.2 Ph tn lin tci vi chui xung trn khi T cng ln khong cch ph vch cng thu hp li v

khi T , chui xung tr thnh mt xung duy nht v ph vch tr thnh mt ng conglin tc c dng bao hnh ca bin ph trc y (H 2.5).

ng cong xc nh bi:

V(f) = Vf

fsin

(a) (b)

(H 2.5)

2.2. M ha

Vic to m c tn hiu trn cc h thng s c th thc hin mt cch n gin lgn mt gi trin th cho mt trng thi logic v mt tr khc cho mc logic cn li. Tuynhin s dng m mt cch c hiu qu, vic to m phi da vo mt s tnh cht sau:

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4/13

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____________________Chng 2M ha v iu ch II - 4

- Ph tn ca tn hiu:Nu tn hiu c cha tn s cao th bng thng ca tn hiu v ca h thng phi rngNu tn hiu c thnh phn DC c th gy kh khn trong ghp ni, th d khng th

ghp tn hiu c thnh phn DC qua bin th v kt qu l khng cch ly in c.Trong thc t, s truyn thng xu nht cc cnh ca bng thng.

V cc l do trn, mt tn hiu tt phi c ph tn tp trung gia mt bng thngkhng qu rng v khng nn cha thnh phn DC.

- Sng bThng my thu phi c kh nng nhn ra im bt u v kt thc ca mt bit

thc hin sng b vi my pht. Nn nhl trong ch truyn ng b, my pht v thukhng to ra xung ng h ring r m my thu phi phc hi xung ny t chui d liu pht s dng. Nh vy tn hiu truyn phi to iu kin cho my thu phc hi xung ng hntrong chui d liu, c th l phi thng xuyn c s bin i gia cc mc ca tn hiu.

- Kh nng d sai tin cy trong mt h thng thng tin s l rt cn thit do my thu phi c kh

nng d sai sa cha m vic ny c thc hin d dng hay khng cng ty vo dng m.- Tnh min nhiu v giao thoaCc dng m khc nhau cho kh nng min nhiu khc nhau. Th d m Bipolar-AMI

l loi m c kh nng pht hin c nhiu.

- Mc phc tp v gi thnh ca h thngCc c tnh ny ca h thng cng ty thuc vo dng m rt nhiu

2.2.1 Cc dng m ph bin

Di y gii thiu mt s dng m thng dng v c s dng cho cc mc chkhc nhau ty vo cc yu cu c th v cc tnh cht ni trn (H 2.6)

- Nonreturn - to - zero - Level (NRZ - L)0 = mc cao1 = mc thpy l dng m n gin nht, hai trin th cng du (n cc) biu din hai trng

thi logic. Loi m ny thng c dng trong vic ghi d liu ln bng t, a t . . . .

- Nonreturn - to - zero inverted (NRZI)0 = chuyn mc in th u bit1 = khng chuyn mc in thu bit

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5/13

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____________________Chng 2M ha v iu ch II - 5

(H 2.6)

NRZI l mt th d ca m vi phn: s m ha ty vo s thay i trng thi ca ccbit lin tip ch khng ty thuc vo bn thn bit . Loi m ny c li im l khi gii mmy thu ch cn d s thay i trng thi ca tn hiu th c th phc hi d liu thay v phiso snh tn hiu vi mt tr ngng xc nh trng thi logic ca tn hiu . Kt qu l ccloi m vi phn cho tin cy cao hn.

- Bipolar - AMI0 = khng tn hiu (hiu th = 0)1 = hiu th m hoc dng, lun phin thay i vi chui bit 1 lin tip

- Pseudoternary0 = hiu th m hoc dng, lun phin thay i vi chui bit 0 lin tip1 = khng tn hiu (hiu th = 0)Hai loi m c cng tnh cht l s dng nhiu mc in th to m (Multilevel

Binary), c th l 3 mc: m, dng v khng. Li im ca loi m ny l:- D to ng bmy thu do c s thay i trng thi ca tn hiu in mc d cc

trng thi logic khng i (tuy nhin iu ny ch thc hin i vi mt loi bit, cn loi bitth hai sc khc phc bi k thut ngu nhin ha)

- C iu kin tt d sai do s thay i mc in th ca cc bit lin tip gingnhau nn khi c nhiu xm nhp s to ra mt s vi phm m my thu c th pht hin ddng.

Mt khuyt im ca loi m ny l hiu sut truyn tin km do phi s dng 3 mcin th .

- Manchester0 = Chuyn t cao xung thp gia bit1 = Chuyn t thp ln cao gia bit

- Differential ManchesterLun c chuyn mc gia bit0 = chuyn mc u bit

1 = khng chuyn mc u bitHai m Manchester v Differential Manchester c cng tnh cht : mi bit c ctrng bi hai pha in th (Biphase) nn lun c s thay i mc in thtng bit do to

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6/13

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____________________Chng 2M ha v iu ch II - 6

iu kin cho my thu phc hi xung ng h to ng b. Do c kh nng t thc hinng b nn loi m ny c tn Self Clocking Codes. Do mi bit c m bi 2 pha in thnn vn tc iu ch (Modulation rate) ca loi m ny tng gp i so vi cc loi m khc,c th , gi s thi gian ca 1 bit l T th vn tc iu ch ti a (ng vi chui xung 1 hoc 0lin tip) l 2/T

2.2.2 K thut ngu nhin ha (Scrambling techniques)

khc phc khuyt im ca loi m AMI l cho mt mc in th khng i khi cmt chui nhiu bit 0 lin tip, ngi ta dng k thut ngu nhin ha. Nguyn tc ca kthut ny l to ra mt s thay i in th gi bng cch thay th mt chui bit 0 bi mtchui tn hiu c mc in th thay i, dnhin s thay th ny sa n cc vi phm lutbin i ca bit 1, nhng chnh nhcc bit vi phm ny m my thu nhn ra c binphp gii m thch hp. Di y gii thiu hai dng m c ngu nhin ha v cdng rt nhiu trong cc h thng tin vi khong cch rt xa v vn tc bit kh ln:

- B8ZS : l m AMI c thm tnh cht: chui 8 bit 0 lin tc c thay bi mt chui8 bit c c bit 0 v 1 vi 2 m vi phm lut o bit 1- Nu trc chui 8 bit 0 l xung dng, cc bit 0 ny c thay th bi 000 + - 0 - +- Nu trc chui 8 bit 0 l xung m, cc bit 0 ny c thay th bi 000 - + 0 + -Nhn xt bng m thay th ta thy c s vi phm lut o bit 2 v tr th 4 v th 7

ca chui 8 bit.- HDB3 : l m AMI c thm tnh cht: chui 4 bit 0 lin tc c thay bi mt chui

4 bit c c bit 0 v 1 vi 1 m vi phm lut o bit 1S thay th chui 4 bt ca m HDB3 cn theo qui tc sau:

Cc tnh ca xung trc S bt 1 t ln thay th cui cng

L chn-+

000- +00+

000+ -00-

S vi phm lut o bit xy ra bit th 4 trong chui 4 bit.Ngoi ra h thng Telco cn c hai loi m l B6ZS v B3ZS da theo qui lut sau:

- B6ZS: Thay chui 6 bit 0 bi 0 - + 0 + - hay 0 + - 0 - + sao cho s vi phm xy ra bit th 2 v th 5

- B3ZS: Thay chui 3 bit 0 bi mt trong cc chui: 00 +, 00 -, - 0 - hay + 0 +, tytheo cc tnh v s bit 1 trc (tong t nh HDB3).

Lu l k thut ngu nhin ha khng lm gia tng lng tn hiu v chui thay th

c cng s bit vi chui c thay th.(H 2.7) l mt th d ca m B8ZS v HBD3.

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7/13

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____________________Chng 2M ha v iu ch II - 7

B = Valid bipolar signal; V = Bipolar violation

(H 2.7)

2.3 iu ch

Bin iu hay iu ch l qu trnh chuyn i ph tn ca tn hiu cn truyn n mtvng ph tn khc bng cch dng mt sng mang chuyn chtn hiu cn truyn i; mcch ca vic lm ny l chn mt ph tn thch hp cho vic truyn thng tin, vi cc tn ssng mang khc nhau ngi ta c th truyn nhiu tn hiu c cng ph tn trn cc knhtruyn khc nhau ca cng mt ng truyn.

Mt cch tng qut, phng php iu ch l dng tn hiu cn truyn lm thay imt thng s no ca sng mang (bin , tn s, pha....). Ty theo thng sc lachn m ta c cc phng php iu ch khc nhau: iu ch bin (AM), iu ch tn s

(FM), iu ch pha M, iu ch xung PM . . . ..

2.3.1 iu ch bin ( Amplitude Modulation, AM )

Xt tn hiu cao tne(t) = Ac cos(ct + ) (1)

Tn hiu AM c c bng cch dng tn hiu g(t) lm bin i bin ca e(t).Biu thc ca tn hiu AM l:

eAM(t) = [(Ac +g(t)]cosct (2)n gin, ta b qua l lng khng i trong AM.Nhng tnh cht cbn ca AM d dng c xc nh nu ta bit tn hiu g(t).Xt g(t) l tn hiu h tn:

g(t) = Em cosmt (3)Nh vy:

eAM(t) = (Ac +Em cosmt)cosct = Ac[ 1 + (Em/Ac) cosmt]cosct= Ac[ 1 + macosmt] cosct (4)

Trong ma = Em/Ac gi l ch s bin iu(H 2.8) v dng sng v ph tn ca tn hiu AM.

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8/13

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____________________Chng 2M ha v iu ch II - 8

(a) (H 2.8) (b)

thy c ph tn ta trin khai h thc (4)eAM(t) = Ac cosct + (maAc/2)cos(c + m)t + (maAc/2)cos(c - m)t (5)

T (H 2.8b) ta thy bng thng ca tn hiu iu ch bng hai ln tn s ca tnhiu h tn v c chia ra lm hai bng cnh. iu ch bin l mt qu trnh tuyn tnhnn mi tn s ca tn hiu h tn to ra mt bng thng v trong trng hp tn hiu h tn

gm nhiu tn s khc nhau th bng thng ca tn hiu bin iu l:BW = 2fm(max)

fm (max) l tn s h tn cao nht.D liu s c thc truyn bng phng php iu ch AM, trong trng hp ny

gi l k thut di bin (ASK, Amplitude- Shift Keying). Bit 1 c truyn i bi sngmang c bin E1 v bit 0 bi sng mang bin E2. (H 2.9) minh ha tn hiu ASK

(H 2.9)

2.3.2 iu ch gc (Angle modulation)

Ta cng bt u vi sng mang cha iu ch:e(t) = Ac cos(ct + ) = Ac cos(t) (6)

Nu c thay i tng ng vi ngun thng tin, ta c tn hiu iu ch tn s (FM) v

nu (t) thay i ta c tn hiu iu ch pha (M).Hai k thut iu ch ny cbn ging nhau v c gi chung l iu ch gc.

2.3.2.1 iu ch tn s (FM)

Tn s(t) l gi tr bin i theo thi gian ca (t), ngha l:

(t) =dt

td )((7)

Vy tn s ca tn hiu cha iu ch l:

(t) = cc

dt

td

=+ )(

(8)

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9/13

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____________________Chng 2M ha v iu ch II - 9

Gi s tn hiu iu ch l g(t), theo nh ngha ca php iu ch tn s, tn s tcthi ca sng mang l:

(t) = c[1 + g(t) ] (9)Thay (9) vo (7):

(t) = +=+ g(t).dttg(t)].dt[1 cc (10)

Thay vo pt (6):

eFM(t) = }g(t).dttcos{A ccc + (11)Biu thc (11) cho thy tn hiu g(t) c ly tch phn trc khi c iu ch.Xt trng hp g(t) l tn hiu h tn c dng hnh sin:

g(t) =c

cosm(t) (12)

l di tn v ml tn s ca tn hiu h tn

(t) = t.dtcost mccc

+

= ct + mfsinmtvi mf = / m l ch siu ch. l t s ca di tn v tn s ca tn hiu

iu ch (h tn).e

FM(t) = Ac cos{ ct + mfsinmt} (13)

thy ph tn ca sng FM ta trin khai biu thc (13):

eFM

(t) = AcJ0(mf) cosct + AcJ2n(mf) [ cos(ct + 2ncosmt) + cos(ct - 2ncosmt)]AcJ2n+1(mf) { cos[c t + (2n+1)cosmt] - cos[ct - (2n+1)cosmt]} (14)

J l hm Bessel theo mfv n c mi tr nguyn t 0 n .T (14) ta thy sng FM gm thnh phn cbn c tn s ca sng mang v bin

cho bi s hng th I , J0(mf) , v cc bng cnh cho bi cc s hng cn li.V n ly mi gi tr t 0 n nn ph tn ca sng FM rng v hn, tuy nhin do

nng lng tn hiu gim rt nhanh vi tn s cao nn ngi ta xem bng thng trong FM xpx bng:

BW = 2(mf.m + m) = 2( + m) rad/s(H 2.10) cho dng sng v ph tn ca sng FM

(H 2.10)

Cng nh trong trng hp AM, tn hiu d liu s cng c truyn bng phngphp FM. K thut ny c gi l k thut di tn (FSK: Frequency- Shift Keying).

FSKc dng rng ri trong truyn s liu. Trong FSK bit 1 c truyn i bi tns fm v bit 0 bi tn s fs v d, trong h thng truyn s dng tiu chun ca hng Bell bit 1

c truyn bi tn s 1070 Hz (fm) v bit 0 bi tn s 1270 Hz (fs).(H 2.11) minh ha tn hiu iu ch FSK

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10/13

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____________________Chng 2M ha v iu ch II - 10

(H 2.11)

2.3.2.2 iu ch pha ( M )

T phng trnh (6) nu gc pha (t) thay i theo tn hiu thng tin ta c iu chpha. Vy:

ePM

(t) = Ac cos[ ct + mp g(t)] (15)Trong mp l di pha cc i

Tn s tc thi cho bi:

i(t) = d(t)/dt

= c + mpdg t

dt

( )

Nu g(t) c dng cosmt th:i(t) = c - mpmsinmt (16)ePM (t) = Ac cos[ ct - mpmsinmt ] (17)

So snh (17) v (13), xem mpl ch siu ch pha, tng ng vi mftrong FM,ta c th xc nh c bng thng ca tn hiu M

BW = 2(m + mpm) rad/s (18)mpm = ep

l di tn tng ng ca M

So snh (11) v (15) ta thy k thut ca FM v M c cng cs. im khc bit ltrong FM ta ly tch phn ca tn hiu h tn trc khi iu ch cn trong M th khng.

iu ch pha l k thut rt tt truyn s liu. Trong k thut di pha, PSK(Phase-Shift Keying), cc bit 1 v 0 c biu din bi cc tn hiu c cng tn s nhng cpha tri ngc nhau.

(H 2.12) m t mt tn hiu PSK.

(H 2.12)

2.3.3 iu ch xung ( Pulse modulation)

y l phng php dng tn hiu h tn iu ch sng mang l tn hiu xung (c tns cao hn), cn gi l phng php ly mu tn hiu h tn. Mc d cc tn hiu tng tc ly mu bi cc gi tr ri rc, nhng cc mu ny c th c bt c gi tr no trongkhong bin i ca tn hiu h tn nn h thng truyn tn hiu ny l h thng truyn tng

t ch khng phi h thng truyn s.

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11/13

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____________________Chng 2M ha v iu ch II - 11

Ty theo thng s no ca xung thay i theo tn hiu h tn, ta c : iu ch bin xung (pulse amplitude modulation, PAM), iu ch v tr xung (pulse position modulation,PPM), iu ch rng xung (pulse width modulation, PWM)

2.3.3.1 iu ch bin xung ( PAM)

Khi mt chui xung hp vi tn s lp li cao p(t) c iu ch bin bi tn hiusin tn s thp m(t), ta c siu ch bin xung. Tn hiu sau khi iu ch l tch ca haitn hiu m(t).p(t) c dng sng l cc xung vi bin thay i theo dng sng h tn m(t) (H2.13).

(H 2.13)

a-/ Mu PAM tnhin (Natural PAM sampling)

Khi bin xung iu ch c nh theo dng ca tn hiu m(t), ta c mu PAM tnhin (H 2.13).

Kt qu ca phn 2.1.1 cho thy tn hiu p(t) c th phn tch thnh cc thnh phn:

Vo + Vn.cos(nst)vi Vo = V/Tsl thnh phn DC v s = 2/Ts l tn s ca p(t).Nh vy, m(t).p(t) bao gm:

m(t).Vo = m(t).V/Ts v m(t).Vn.cos(nst)Tm li, tch m(t).p(t) c cha dng sng ca tn hiu iu ch (tn hiu cn truyn)

trong thnh phn tn s thp m(t).Vo v c th phc hi bng cch cho sng mang iu chqua mt mch lc h thng.

Thnh phn ha tn c dng Vnm(t)cos(nst) tng t nh tn hiu iu ch 2 bngcnh trit sng mang (Double Sideband Suppressed Carrier, DSBSC).

Ph tn ca tn hiu PAM vi h tn l m(t) = sinmt c dng nh (H 2.14)

(H 2.14)

Trong (H 2.14) M(f) l ph tn ca tn hiu di nn v fm l tn s cao nht ca tnhiu ny. T (H 2.14) ta cng thy ti sao tn s xung ly mu fs phi t nht hai ln ln hnfm . Nu M(f) c phc hi t mch lc h thng, phn cch t M(f) ti di tn k cnphi ln hn 0, ngha l W > 0

W = fs - fm - fm > 0 hay fs > 2 fm

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12/13

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____________________Chng 2M ha v iu ch II - 12

b-/ Mu PAM nh phng (Flat-top PAM)

y l mu PAM c dng rng ri do d to ra sng iu ch. Dng sng cho (H2.15) cc xung sau khi iu ch c nh phng ch khng theo dng ca h tn.

(H 2.15)

Mc d khi phc hi tn hiu t mch lc h thng s c bin dng do on nhphng nhng v b rng xung thng rt nh so vi chu k Ts nn bin dng khng ng k.Nu s bin dng l ng k th cng c th loi b bng cch cho tn hiu i qua mt mchb tr.

Tn hiu PAM t c dng pht trc tip do lng thng tin cn truyn cha trongbin ca xung nn d bnh hng ca nhiu. PAM thng c dng nh l mt bctrung gian trong mt phng php iu ch khc, gi l iu m xung (pulse codemodulation, PCM) v c dng trong a hp thi gian truyn (TDM).

2.3.3.2 iu ch thi gian xung (Pulse -time Modulation, PTM)

iu ch thi gian xung bao gm bn phng php (H 2.16). Ba phng php u tptrung trong mt nhm gi l iu ch rng xung (Pulse-width modulation, PWM) (H2.16d, e, f), phng php th t l iu ch v tr xung (Pulse-position modulation, PPM) (H2.16g).

Ba phng php iu ch rng xung khc nhau im cnh ln, cnh xung hayim gia xung c gi cnh trong khi rng xung thay i theo tn hiu iu ch.

Phng php th t, PPM l thay i v tr xung theo tn hiu iu ch trong khi brng xung khng i. (H 2.16) minh ha cho cc cch iu ch ny.

Lu l k thut PTM tong t vi iu ch FM v M, tn hiu c bin khngi nn t bnh hng bi nhiu.

Ph tn ca tn hiu iu ch bng phng php PWM, PPM ging nh ph tn catn hiu iu ch FM (H 2.16h), ngha l c nhiu ha tn nn khi s dng PWM v PPMngi ta phi gia tng tn s xung ly mu hoc gim di tn ( gii hn bng thng catn hiu v tng s knh truyn).

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13/13

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____________________Chng 2M ha v iu ch II - 13

(H 2.16)

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Nguyn Trung Lp T d li

ma hoa dieu che

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