Chuong3 Dieu Che

7/30/2019 Chuong3 Dieu Che

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44

CHNG 3

IU CH

3.1. nh ngha

iu ch l qu trnh ghi tin tc vo 1 dao ng cao tn chuyn i xa nhbin imt thng s no (v d : bin , tn s, gc pha, rng xung...)

Tin tc gi l tn hiu iu ch, dao ng cao tn gi l ti tin. Dao ng cao tn mangtin tc gi l dao ng cao tn iu ch.

C 2 loi iu ch; iu bin v iu tn (gm iu tn v iu pha).

3.2. iu bin

iu bin l qu trnh lm cho bin ti tin bin i theo tin tc.Gi s tin tc vs v ti tin vtu l dao ng iu ha:

tVv SSS cos= v tVv ttt cos= vi t>>SDo tn hiu iu bin:

ttVVV tSStdb cos)cos( +=

tcos)tcosV

V(V tS

t

St += 1

tcos)tcosm(V tSt += 1 (1)

Trong : m l h siu ch. tn hiu khng b mo trm trng hoc qu iu ch th1m

t)cos(Vm

t)cos(Vm

tcosVv sttsttttdb +++=22

Vb

t

Hnh 3.1. th thi gian tn hiu iu bin

1/2 mVt 1/2 mVt

t- s t t+ s

Vtvdb

Hnh 3.2 Ph tn hiu iu bin


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45

Ph ca tn hiu iu bin c dng nh hnh 3.2.

Khi tn hiu iu ch c ph bin thin t maxmin SS th ph ca tn hiu iu bin cdng nh hnh 3.3

Vb

t+smax t+smintt-smax t-smin

Vt

Hnh 3.3 Ph ca tn hiu iu bin

Quan h nng lng trong iu bin:Cng sut ti tin l cng sut bnh qun trong 1 chu k ca ti tin:

====T

tt

hdhdt~

R

VdtsinV

TRR

VRIP

0

222

22

2

11

=>2

~2

~t

t

VP

Tng t: ttt

bt PmV

mmV

P ==424

1)

2(

2

1 22

22

~

Cng sut ca tn hiu iu ch bin l cng sut bnh qun trong mt chu k ca tn hiuiu ch:

)m

(PPPP t~bt~t~db~2

122

+=+=

m cng ln th P~b cng ln

Khi m = 12

3 t~db~ PP = v t~bt~ PP 41=

T biu thc (1) suy ra:

Vbmax = Vt(1+m)

Do 22max~ )1(2

1tVmP +

Cc ch tiu cbn ca dao ng iu bin3.2.1 H s mo phi tuyn


3/16

46

)(

...)3()2( 22

st

stst

I

IIK

++=

( )St nI (n 2 ): Bin dng in ng vihi bc cao ca tn hiu iu ch.

( )StI : Bin cc thnh phn bin tn

Trong : It: bin tn hiu ra

VS: gi tr tc thi ca tn hiu vo

A : gi tr cc i

B : ti tin cha iu ch

It

B

A

V

Hnh 3.4. c tnh iu ch tnh

ng c tuyn thc khng thng to ra cc hi bc cao khng mong mun. Trong

ng lu nht l cc hi ( St )2 c th lt vo cc bin tn m khng th lc c.

gim K th phi hn ch phm vi lm vic ca biu ch trong a thng ca c tuyn. Lc

buc phi gim h siu ch m.

3.2.2 H s mo tn sm

Gi : mo : h siu ch ln nht

m0 m

m : H siu ch ti tn sang xt.

H s mo tn sc xc nh theo biu thc :

m

mM o= HocMdB = 20lgM

nh gi mo tn s ny, ngi ta cn

c vo c tuyn bin v tn s: Fsm = f(Fs)

Vs = cteHnh 3.5. c tnh bin tn s

Phng php tnh ton mch iu bin :Hai nguyn tc xy dng mch iu bin :

- Dng phn t phi tuyn cng ti tin v tn hiu iu ch trn c tuyn ca phn t

phi tuyn .

- Dng phn t tuyn tnh c tham siu khin c. Nhn ti tin v tn hiu iuch nhphn t tuyn tnh .


4/16

47

3.2.3 iu bin dng phn tphi tuyn

Phn t phi tuyn c dng iu bin c th l n in t, bn dn, cc n c

kh, cun cm c li st hoc in trc tr s bin i theo in p t vo. Ty thuc vo

im lm vic c chn trn c tuyn phi tuyn, hm sc trng ca phn t phi tuyn c

th biu din gn ng theo chui Taylo khi ch lm vic ca mch l ch A ( =

180o) hoc phn tch theo chui Fourrier khi ch lm vic ca mch c gc ct < 180o (

ch AB, B, C)

Trng hp 1: IU BIN CH A = 180o

Mch lm vic ch A nu tha mn iu kin:

ost EVV


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Thay vD vo biu thc (1) ta nhn c :

iD = a1(E0 + Vtcostt + Vscosst) + a2 (E0 + Vtcostt + Vscosst)2 +

+ a3( E0 + Vtcostt + Vscosst )3 +..... (2)

Khai trin (2) v b qua cc s hng bc cao n 4 s c kt qu m ph ca n c biudin nh hnh 3.8.

Khi a3 = a4 = a5 =.....a2n+1 = 0 (n = 1,2,3) ngha l ng c tnh ca phn t phi tuyn l1 ng cong bc 2 th tn hiu iu bin khng b mo phi tuyn.

tha mn iu kin (*) mch lm vic ch A th m phi nh v hn ch cng sutra. Chnh v vy m ngi ta rt t khi dng iu bin ch A.

Trng hp 2: IU BIN CH AB, B hoc C < 180oKhi < 180o, nu bin in p c vo diode ln th c th coi c tuyn ca n l

mt ng gp khc.

Phng trnh biu din c tuyn ca diode lc :

iD = 0 khi vD 0

= SvD khi vD > 0 S: H dn ca c tuyn

Chn im lm vic ban u trong khu tt ca Diode (ch C).

iD iD

vD

vD

t

t

Vt

Vs

CS

D

+EO

-

Eo

Hnh 3.10. c tuyn ca diode v thca tn hiu vo ra khi lm vic ch C

Rt

Hnh 3.9. Mch iu ch dng Diode


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Dng qua diode l 1 dy xung hnh sine, nn c th biu din iD theo chui Fouriernh sau :

iD = I0 + i1 + i2 + in +

= Io + I1costt + I2cos2tt + I3cos3tt + .......+ Incosntt (3)

I0 : thnh phn dng in mt chiu.

I1: bin thnh phn dng in cbn i vi ti tin

I2, I3.....In : bin thnh phn dng in bc cao i vi ti tin

I0, I1 I3.....In : c tnh ton theo biu thc ca chui Fourrier :

tdtniI

tdtiI

tdiI

ttc

Dn

ttc

D

tc

D

.cos2

.....................

.cos.2

.1

1

0

=

=

=

(4)

Theo biu thc (*) ta c th vit :

iD = S.vD = S( -E0 + VScosst + Vtcostt ) (5)

Khi tt = th iD = 0 :

0 = S.vD = S( -E0 + VScosst + Vtcos) (6)

Ly (5) - (6) =>

tcos)sin(SV

i

)sin(

SV

)sin(

SV

tsin.costsinSV

tdtcos.costcos

.SV

td.tcos).costcos(SVI

)costcos(SVi

tt

tt

ttt

ttt

t

otttt

ttD

=

==

+

=

+

=

=

=

22

1

22

124

1

2

12

4

2

2

12

2

212

2

1

00

0

1

Vy tcos)sin(SV

i tt

= 2

2

11 (7)

y c xc nh t biu thc (6) :t

sso

t

sso

V

tcos.VE

V

tcos.VEcos

=

+= (8)

T biu thc (7) v (8) bin ca thnh phn dng in cbn bin thin theo tn hiu iuch (Vs).


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3.2.4 iu bin dng phn ttuyn tnh c tham s thay i

y l qu trnh nhn tn hiu dng b nhn tng t

K = 1 vb

~

vS(t)

=~

vt(t)

E0

Hnh 3.11. Mch iu bin dung phn t tuyn tnh

vb = (Eo + VS.cosst) . Vt.costt

vb = EoVt.costt +2

. stVV cos (t+ S) t +2

. stVV cos (t- s) t

Cc mch iu bin c th :a. iu bin cn bng dng diode

i = i1 - i2

i1

i2

EO

D2Cb

Cb

D1

vS vdB

Hnh 3.12. Mch iu ch cn bng dng diodevt

in p t ln D1, D2 :

(1)

+=

+=

tVtVv

tVtVv

ttsS

ttsS

cos.cos

cos.cos

2

1

Dng in qua diode c biu din theo chui Taylo :

(2)

++++=

++++=

...

...

3

23

2

22212

3

13

2

12111

vavavaai

vavavaai

o

o

Dng in ra : i = i1 - i2 (3)

Thay (1), (2) vo (3) v ch ly 4 vu ta nhn c biu thc dng in ra :

i = A cos st + B cos 3st + C [cos (t+ s) t + cos (t- s) t]

+ D [cos (2t+ s) t + cos (2t- s) t] (4)


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51

Trong :

===

++=

23,2,

2

232

32

2

3

2

3

2

31

tStS

S

StS

VVaDVVaC

VaB

VaVaaVA

3ss t+s22t2t-st t+st-sHnh 3.13. Ph tn hiu iu bin cn bng

b. Mch iu bin cn bng dng 2BJT

Hnh 3.14. Mch iu bin cn bng dng 2 BJT

vtVCC

VS vdb

Kt qu cng tng t nh trng hp trn.

c. Mch iu ch vng

Vt

~

D3

D

Cb

Cb D2

D4

D1

vS vdb

Hnh 3.15. Mch iu ch vng

Gi : iI l dng in ra ca mch iu ch cn bng gm D1, D2

iII l dng in ra ca mch iu ch cn bng gm D3, D4

t - s t + s

t


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Theo cng thc (4) mc trn (iu bin cn bng dng diode) ta c c biu thc tnh iI :

iI= A cosst + B cos 3st + C [cos (t+ s) t + cos (t- s) t]

+ D [cos (2t+ s) t + cos (2t- s) (*)

Ta c : iII= iD3 - iD4 (1)

Trong :

(2)...vavavaai

...vavavaai

3

43

2

4241oD4

333

23231oD3

++++=

++++=

Vi v3, v4 l in p t ln D3, D4 v c xc nh nh sau :

tVtVv

tVtVv

sstt

sstt

coscos

coscos

4

3

+=

=(3)

Thay (3) vo (2) v sau thay vo (1), ng thi ly 4 vu ta c kt qu :

iII= - A cosst - B cos 3st + C [cos (t+ s) t + cos (t- s) t]

- D [cos (2t+ s) t + cos (2t- s) t]

idB = iI+ iII= 2C [cos (t+ s) t + cos (t- s) t] (4)

Vy : mch iu ch vng c th khc cc hm bc l ca s v cc bin tn ca2st, do mo phi tuyn rt nh.

3.3. iu chn bin3.3.1. Khi nim

Ph tn hiu iu bin gm ti tn v hai di bin tn, trong ch c cc bin tnmang tin tc. V hai di bin tn mang tin tc nh nhau (v bin v tn s) nn ch cn

truyn i mt bin tn l thng tin v tin tc, cn ti tn th c nn trc khi truyn i.Qu trnh gi l iu chn bin.

u im ca iu ch dn bin so vi iu ch hai bin :

- rng di tn gim i mt na.

- Cng sut pht x yu cu thp hn vi cng mt c ly thng tin.

- Tp m u thu gim do di tn ca tn hiu hp hn,

Biu thc ca iu chn bin :Vb (t) = Vt.2

m. cos (t+ s) t

m : h s nn ti tin,t

S

VVm = , m c th nhn gi tr t 0


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3.3.2. Cc phng php iu chn bin

3.3.2.1. iu ch theo phng php lcft2 (ft1 + fS)

CCB1 CCB1LC1 LC2

Dao ng

ft2ft1

Dao ng

ft2 + ft1 + fSvS(t) ft1 + fSft1 fS

Hnh 3.17. S khi mch iu ch theo phng php lc

f1+fmaxf1+fmin

2fmin

f1+fmaf1+fminf1-fminf1-fmax

ft1fsmaxfsmin

0

0

0

0

0

V5

V4

V3

V2

V1

f

f2 + f1+ fmaxf2 +f1+fmin

f

f2 + f1+ fmaxf2 +f1+fminf2 - f1-fmax f2

2 = 2f1 + 2fmin

f2 - f1-fmin

f

f

f

Ph tn hiu theo phng php lc


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54

t : fs = fsmax - fsmin

ft1 : tn s ca ti tn th nht ft2 : tn s ca ti tn th hai

t

SS

t

S

f

ff

f

fx minmax

=

= : h s lc ca b lc.

Trong s khi trn y, trc tin ta dng mt tn s dao ngft1 kh nh so vi ditn yu cuft2 tin hnh iu ch cn bng tn hiu vo vs(t). Lc h s lc tng ln c th lc bc mt bin tn d dng. Trn u ra b lc th nht s nhn c mt tnhiu c di ph bng di ph ca tn hiu vo fs = fsmax -fsmin, nhng dch mt lng bngft1 trn thang tn s, sau a n biu ch cn bng th hai m trn u ra ca n l tnhiu ph gm hai bin tn cch nhau mt khong f = 2(ft1 + fs min) sao cho vic lc ly mtdi bin tn nhb lc th hai thc hin mt cch d dng.

3.3.2.2. iu chn bin theo phng php quay pha

Tn hiu ra ca 2 biu ch cn bng:

VCB1 = VCB cosst costt =2

1VCB [cos (t+ s) t + cos (t- s) t]

VCB2 = VCB sinst sintt =2

1VCB [- cos (t+ s) t + cos (t- s) t]

Cu DiodeCCB1

00

900

00

900

Cu DiodeCCB2

vCB1

MCHINTNGHOCHIU

vCB2

VDB

vS

vt

Hnh 3.18. S mch iu chn bin theo phng php pha

Hiu hai in p ta s c bin tn trn :

VDB = VCB1 - VCB2 = VCB cos (t+ s) t

Tng hai in p ta s c bin tn di :

VDB = VCB1 + VCB2 = VCB cos (t- s) t

3.4 iu tn v iu pha


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3.4.1. Quan h gia iu tn v iu pha

=dt

d(1)

Vi ti tin l dao ng iu ha :

v(t) = Vt . cos (tt + o) = Vt . cos )t( (2)

T (1) rt ra :

(3) +=t

o

)t(dt).t()t(

Thay (3) vo (2), ta c :

v(t) = Vtcos [ ] (4) +t

o

tdtt )().(

Gi thit tn hiu iu ch l tn hiu n m :

vs = Vs cosst (5)

Khi iu tn v iu pha th (t) v (t)c xc nh theo cc biu thc :

(t) = t+ KtVs cosst (6)

(t) = o + KfVs cosst (7)

t : tn s trung tm ca tn hiu iu tn.

Kt.Vs = m : lng di tn cc i

Kf.Vs = m: lng di pha cc i

(t) = t+ m cosst (8)

(t) = o + m cosst (9)

Khi iu tn th gc pha u khng i, do (t) = o.

Thay (8), (9) vo (4) v tch phn ln, ta nhn c :

+

+= osS

mttdt tsintcosV)t(v (10)

Tng t thay (t) trong (9) vo (4) v cho =(t) = cte ta c :

vf(t) = Vt.cos (tt + m cos st+o) (11)

Lng di pha t c khi iu pha : = m cos st

Tng t vi lng di tn :

=dt

d= ms.sin st

Lng di tn cc i t c khi iu pha:

=sm = s.KfVs (12)

Lng di tn cc i t c khi iu tn:


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=KtVs (13)

T (12) v (13) ta thy rng: im khc nhau cbn gia iu tn v iu pha l:

- Lng di tn khi iu pha t l vi Vs v s

- Lng di tn khi iu tn t l vi Vs m thi.

T ta c th lp c hai s khi minh ha qu trnh iu tn v iu pha:

Tch phn

o hm

iu pha

iu tnvS

vS

T/h iu tn

T/h iu pha

Hnh 3.19. S khi qu trnh iu pha v iu tn

3.4.2. Ph ca dao ng iu tn v iu pha

Trong biu thc (10), cho o = 0, t fS

m M=

gi l h siu tn, ta s c biu

thc iu tn : vt= Vtcos [tt+ Mf.sin tt] (14)

Tng t, ta c biu thc ca dao ng iu pha :

vf= Vtcos [tt + M. cos tt] (15)

Trong :M = m

Thng thng tn hiu iu ch l tn hiu bt k gm nhiu thnh phn tn s. Lc tn hiu iu ch tn s v iu ch pha c th biu din tng qut theo biu thc :

Vdt= Vtcos [tt + ]=

+m

1iiSiti )cos(M

Ph ca tn hiu iu tn gm c tt c cc thnh phn tn s t hp : tt + =

m

i

Sii

1

Vi i l mt s nguyn hu t; -i

3.4.3 Mch iu tn v iu pha3.4.3.1 iu tn dng diode bin dung

CV

Rv+

C2

R1RFC

C1

Cv

L V

VV

Hnh 3.20. Mch iu tn dung Diode bin dung v c tuyn ca CV


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L, Cv to thnh khung cng hng dao ng ca mt mch dao ng

C1 : t ngn DC

C2 : t thot cao tn n nh phn cc cho Cv

RFC : cun cn cao tn

R1 : trngn cch gia mch cng hng v ngun cung cp khi Rv thay i ( VPC thay i (

CV thay i theo lm cho tn s cng hng ringVLC

f2

1= ca khung cng hng

LCV thay i, dn n qu trnh iu tn.

3.4.3.2 iu pha theo Amstrongvb1

B1

B2

Tng

Di

pha 900

vS

mVt1

mVt2

Vt1

v

Vb2

vb1

Vt2

Vb2

Hnh 3.21. Mch iu pha theo Amstrong v th vector ca tn hiu

Ti tin t thch anh a n biu bin 1 (B1) v iu bin 2 (B2) lch pha 90o,cn tn hiu iu ch vs a n hai mch iu bin ngc pha. in p ra trn hai biupha :

vb1 = Vt1 (1 + m cos st) cos tt

=2

])cos()[cos(cos 11 ttmVtV ststttt +++

vb1 = Vt1 (1 - m cos st) sin tt

=2

])sin()[sin(sin 22 ttmVtV ststttt +++

th vc tca tn hiu v v vc t tng ca chng = + lmt dao ng c iu ch pha v bin . iu bin y l iu bin k sinh.

1dbV

2dbV

V 1dbV

2dbV

hn chiu bin k sinh chn nh (< 0,35)

3.4.3.3 iu tn dng Transistor in khng

Phn tin khng : dung tch hoc cm tnh c tr s bin thin theo in p iu chttrn n c mc song song vi h dao ng ca b dao ng lm cho tn s dao ng thayi theo tn hiu iu ch. Phn tin khng c thc hin nhmt mch di pha trongmch hi tip ca BJT. C 4 cch mc phn tin khng nh hnh v.


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58

Cch mcmch

S nguyn l th vc t Tr sin khngTham s tng

ng

Mch phnp RC Z = j S

RC

Ltd = S

RC

Mch phnp RL

Z = - jLS

R

Ctd =

R

LS

Mch phnp CR

Z = - jRCS`

1

Ctd = RCS

Mch phnp LR

Z = jRSL Ltd =

RSL

Vi mch phn p RC ta tnh c :

BEVS

V

I

VZ

.== (IC= S.VBE IClun lun cng pha vi VBE)

V

__I VC

_VR

_

V_

_

V_

_

IR

C

V_

_

I

R

L

V_

_

I

I

C

R

R

L

VL

_V

VR

VC

_ VR_

V_

_I

_

_I

_

VL_ V

VR

_

__I


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Z=

Cj

CjR

CjR

CjVS

V

.

1

.

1

.1

.

1

..

+=

+

Nu chn RCj

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