Thit k Mixer v B lc

Thit k Mixer v B lc

Mch lc

1. Khi nim chung v mch lcB lc l 1 mng li truyn dn tn hiu hon ton vi tn s trong di thng nht nh v c s suy gim trong vng stopband

L thuyt c bn v mch lca. Hm truynHm truyn ca mch lc c nh ngha l t s gia in p tn hiu ra Vo trn in p tn hiu vo Vi.

Hm truyn tng qut theo tham s S (S=jw)

H(s)=

ph thuc vo cu to ca mch. Hm truyn thng gp c dng

, a thc bc 0 vi:

b. Mc ch ng dng - B lc c vai tr quan trng trong cc mch in t. Nhng phn t c s trong mch lc ch gm in tr (R), t in (C), v cun cm (L). - Thng thng gm 2 mch lc RC v RLC. Mch lc RC c dng nhiu v linh kin r v chim t din tch. Cn mch lc RLC t thng dng v c in cm (L) kh tiu chun ha v c gi tr rt ln phm vi tn s thp nn trong thc t kh thc hin v gi thnh t, li cng knh. - Mch lc s lm suy gim nng lng qua n m khng c kh nng khuych i. Kh phi hp tng tr vi cc mch ghp.- b tc cc nhc dim trn ngi ta thm vo cc phn t khuch i nh transistor, vi mch, v.v, c th khuych i tn hiu , phi hp tng tr, iu chnh suy gim.2. Phn loi mch lc Da vo c im cu to, ta phn ra hai loi: mch lc th ng v mch lc tch cc. C hai loi mch lc ny u c cc dng p ng tn s sau:- Mch lc thng cao (High pass Filter)- Mch lc thng thp (Low pass Filter)- Mch lc thng di (Band pass Filter) - Mch lc chn di (Reject Band Filter)Cng sut u ra ca mch lc th ng lun nh hn cng sut u vo trong khi mch lc ch ng cho php ln hn. Nhng i vi sng siu cao tn th mch lc ch ng khng p ng kp nn trong n ny ta ch nghin cu v mch lc th ng. c im ca mch lc th ng l c th m t bng phng php truyn ti v bng phng php chc nng suy gim bin . Trong tn s thp chng ta dng chc nng hm truyn miu t, cn trong tn s cao ta s dng chc nng suy gim bin xt c tnh ca b lc.

a. p ng tn s

Hm truyn:

suy hao:

Attenuation/dB

0c310203040c|H|()|

1Transfer function

p ng tn s ca mch lc thng thp

c|H|()|

1Transfer function

Attenuation/dB0c310203040

p ng tn s ca mch lc thng cao

1|H|()|

1Transfer function2

Attenuation/dB013102030402

p ng tn s ca mch lc thng di

b. hot ng ca mch lc tn s cao, mch lc thng s dng cc yu t nh: in tr , cun cm v t in. V c bn c 2 phng php lc tn s: Phng php tham s nh (image-parameter method (IPM)) Phng php chn-mt (insertion-loss method (ILM))Phng php tham s nh c c im l thit k tng i n gin nhng c 1 bt li rng l p ng tn s bt thng khng th thit lp vo h thng c. Phng php chn-mt chia b lc vo 1 tng mng 2 ca, p ng tn s n nh. V vy trong n ta thit k mch theo phng php chn-mt da trn s suy gim v chn mt ca b lc. Suy hao ca mng 2 ca c tnh bi:

l h s phn x nhn vo b lcThit k b lc kiu chn-mt thng ta bt u vi thit k mch lc thng thp cn bn.Vi in tr ngun v ti l 1 , tn s ct l 1 rad/s.Attenuation/dB

0c = 1310203040

A Filter H()V1()V2()RS =1RL =1

B lc LPP cn bn vi tn s ct rad/sBng cch thay i v tr ca t in v cun cm ta c cc mch khc nhau. B lc bao gm cc phn t phn ng thnh 1 bc thang. Bc ca mch lc chnh l s phn t phn ng. Chuyn i tr khng, tn s thang v tr khng ca ti v ngun c b lc thng cao v b lc thng di.L1=g2L2=g4C1=g1C2=g3RL= gN+11

(a)

L1=g1L2=g3C1=g2C2=g4RL= gN+1g0= 1

(b)

Mch lc thng thp s dng thnh phn LCL2L1C1CNL1=g1L2=g3C1=g2C2=g4RL= gN+1g0= 1L1=g1L2=g3C1=g2C2=g4RL= gN+1g0= 1

Mch lc thng cao s dng LC vi v tr L v C i ch cho nhau mch lc thng thpC2L2L1C1L3C3CNLN

Mch lc thng di, cc t in c thay th bng mng LC song song, cc in dn thay bang mch LC ni tipc.Gii thiu tm tt v thit k b lc thng thp s dng gp cc yu tC 1 s phng php tiu chun thit k mch lc thng thp. 3 phng php trong s c bit n l: Lc Butterworth Lc Chebyshev Lc Bessel tng c bn l tnh xp x p ng bin l tng ca |H()|2 ca b khuch i s dng a thc butterworth, chebyshev v bessel v 1 s chc nng a thc trc giao khc.

(2)K0 v C0 l hng s ph thuc vo loi a thc s dng

PN() l 1 a thc ca bc N.

BesselAmplitude in dBButterworth

Chebyshev

p ng bin bc 4 (N=4) ca b lc Butterworth, chebyshev v bessel

Mi kiu b lc c 1 li th ring ca n. V d nh, b lc chebyshev cho tn s ct nhanh hn, nh hnh () l = 1rad/s, tuy nhin phn di thng s c nhiu gn sng, khng bng phng. Mt khc, B lc Bessel c tn s ct ln nht, suy gim chm nht nhng vng di thng li c tuyn tnh bng phng nht. Cn v b lc Butterworth th mang c im nm gia 2 b lc trn. Da vo c im ca ti nn ta chn b lc Butterworth thit k, v tnh bng phng trong di thng v p ng tn s ct t tiu chun cho php.

d. Mch lc thng thp nguyn bnMt cch tng hp thay th t chc nng hm truyn (2) s dng cc tiu chun k thut c lit k trong ti liu [1] [2], ta c bng thng s ca G1,G2,G3,GN vi N l bc ca b lc thng thp LPP.Ng1g2g3g4g5g6g7g8g9

12.00001.0000

21.41421.41421.0000

31.00002.00001.00001.0000

40.76541.84781.84780.76541.0000

50.61801.61802.00001.61800.61801.0000

60.51761.41421.93181.93181.41420.51761.0000

70.44501.24701.80192.00001.80191.24700.44501.0000

80.39021.11111.66291.96151.96151.66291.11110.39021.0000

Hnh: Bng gi tr ti u cho thit k mch lc thng thp kiu Butterworth (vi g0=1, , N = 110)Ngun: Reprinted from G. L. Matthaei, L. Young, and E. M. T. Jones,Microwave Filters, Impedance-Matching Networks, and Coupling Structures, Artech House, Dedham, Mass., 1980, with permission

Bng trn l gi tr cc yu t bng phng ti a cho b lc thng thp cn bn ( g0=1, , N= 1-10 )g0 l in tr vi (a) l in dn vi hnh (b)gk (k= 1- N) l t cm ca cc cun dy mc ni tip, hay l in dung ca t in mc kiu shunt thit k b lc ta da vo bng s suy gim bin chn bc b lc cho thch hp.

Hnh: S suy gim bin so vi tn s ca b lc Butterworth

V d: Ta s dng bng trn thit k b lc thng thp bc 4 vi .L1 = g1 = 0.7654HL2 = g3 = 1.8478HC1 = g2 = 1.8478FC2 = g4 = 0.7654FL1=0.7654HL2=1.8478HC1=1.8478FC2=0.7654FRL= 1g0= 1

T b lc thit k trn ta c th thit k c b lc c v tr khng ty bng cch phi hp tr khng. Vi cc thng s ban u l: in tr khng ti R0, tn s ct , in tr ban u Rn, in cm Ln,in dung Cn, ta tnh ton c s thay i nh sau:

(1)

Ta khng cn thay i hnh dng ca mch lc m ta ch cn thay i thng s cc phn t ca mch.

Bin i b lc: (a) p ng cng sut b lc nguyn mu rad/s

(b) p ng cng sut b lc bin i thnh tn s ct e. Bin i sang mch lc thng di v chn di

T mch lc thng thp nguyn bn nh ta xy dng trn ta c th bin i sang mch lc thng di v chn di. Vi v l cnh ca di thng th p ng tn s ca di thng c th c s dng bng cch thay th tn s sau y:

(3)Trong

l bng thng phn on ca di thng

Tn s trung tm c chn l trung bnh nhn ca v

Sau khi thay th tn s theo cng thc (3) ta c:Hnh: Bin i tn s thnh thng di v chn di

(a): mch lc thng thp nguyn bn c =1 rad/s(b): p ng mch cng sut lc thng di(c): p ng cng sut mch lc chn di

Cc phn t ca b lc mi sau khi s dung (3) c biu din nh trong bng sau:

Hnh: Biu din gi tr cc phn t ca b lc mi khi thay th trn mch lc thng thp nguyn bn.Qua bng trn ta c th thy trong mch lc thng di cc phn t cm khng c thay th bng chui phn t LC mc ni tip vi:

v Phn t dung khng c thay th bng chui phn t LC mc song song vi:

v

Thit k v tnh ton mch lc thng thpa. La chn yu cu di tn ph hp vi ti

Nh ni trn, trong ti ny ta cn lc b lc thng thp cho tn hiu u vo l 500MHz hay = 3141.6 rad/s.

Trong khun kh n, ta xt trng hp mch lc thng thp c tn s ct rad/s v tn s dng rad/s (fs =900 MHz)b. Tnh ton gi tr cc phn t ca mch lc thng thpTheo phn l thuyt ta c:

T hnhta chn N=4 l thit k mch lc ph hp vi bi ton.T Hnh ta a ra c thng s ca mch:g1 = 0.7654g2 = 1.8478g3 = 1.8478g4 = 0.7654Ta c mch nguyn l:

Vi tn s ct = 3141.6 rad/s v tr khng ca mch l R0=50, theo b cng thc (1) ta c:

Tng t ta tnh c:

Thay vo mch nguyn l ta c:

Khi chy m phng trn ADS ta thu c p ng tn s- bin ca mch lc nh sau: Nhn xt: Nhn vo th ta thy mch cho p ng tn s ph hp vi yu cu bi ton t ra.

c. Tnh ton gi tr ca cc ng Transmission line

6. Mch lc thng diKt hp gia mch lc thng thp v mch lc thng cao ta co mch lc thng di vi di tn i qua l khong nm gia 2 tn s ct.Thit k b lc thng di c di tn t 40MHz-60MHz.

Vi:

ph thuc vo thng s ca vt liu lm mch inVy ta suy ra:

;

;

p ng tn s:

Nguyn Tun AnhPage 15