Bai Dieu Che Tuong Tu

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BI THC TP CHUYN :IU CH TNG T1NI DUNGBi 1 iu ch bin Bi 2 Gii iu ch bin Bi 3 iu ch na bng (SSB)Bi 4 iu ch tn sBi 5 Gii iu ch tn sBi 6 iu ch phaBi 7 Chuyn i tn s (FC)Bi 8 Cc b khuch i c la chn (selective amplifiers)Bi 9 H thng truyn thng AMBi10 H thng truyn thng FMPh lc A: Nguyn l hot ng ca my phn tch ph2BI 1 - IU CH BIN (AM)1.1 MC CH:+ Kim tra cc thng s ca mt tn hiu c iu bin+ Kim tra s vn hnh ca mt b iu ch bin (my iu bin)+ Thc hin o c cc thuc tnh trn mt my iu bin+ Phn tch ph ca tn hiu iu bin1.2 C S L THUYT1.2.1 L thuyt chung v k thut iu binXt mt tn hiu sin vm(t) vi tn s f (hnh 1.1): Vm(t) = B.sin(2ft)V mt tn hiu sin khc Vc(t) vi tn s F cao hn: Vc(t) = A.sin(2Ft)Hnh 1.1 a) tn hiu sng mangb) tn hiu iu chc) tn hiu iu chTn hiu vm(t) c gi l tn hiu iu ch (modulating signal), tn hiu vc(t) gi l tn hiu sng mang.Thay i bin ca sng mang vc(t) bng cch cng tn hiu iu ch vm(t) vo A, ta thu c tn hiu iu bin vM(t), c m t bng phng trnh:VM(t) = [A + k.B.sin(2ft)].sin(2Ft) = A[1+m.sin(2ft)].sin(2Ft)Vi k l hng s t l.Phn trm tn hiu iu bin c nh ngha l gi tr:3100 ..AB km i chiu vi hnh 1.1, ch s iu ch m c th c tnh bng cch sau:% 100 .h Hh Hm+1.2.2 Ph ca tn hiu iu binVi cc bin i lng gic n gin, vM tr thnh:] ). ( 2 cos[2] ). ( 2 cos[ .2. ) 2 sin( . ) ( t f FAm t f FAm Ft A t VM+ + T chng ta suy ra rng mt tn hiu ciu ch bin gm ba thnh phn sin:A.sin(2Ft) Sng mang] ). ( 2 cos[2. t f FAm Na bng di] ). ( 2 cos[ .2. t f FAm + Na bng trnHiu qu c bit l s th hin ca tn hiu iu bin trong s bin /tn s. Hnh 1.2 th hin cc thnh phn khc nhau ca tn hiu AM, trong s bin /tn s cng nh s bin /thi gian.4Hnh 1.21.2.3 Cng sut ca tn hiu iu binCng sut tng cng ca mt tn hiu AM l tng ca cc phn lin quan n sng mang v cc na bng di v trn.Xem xt mt tn hiu iu ch dng sin v mt tr ti R, cc thnh phn khc nhau cung cp cc cng sut nh sau: PC = A2/2.R Cng sut kt hp ca sng mang.PL = (m.A)2/8.R Cng sut kt hp ca na bng thpPH = (m.A)2/8.R Cng sut kt hp ca na bng cao5C hai yu t quan trng cn phi ch n l:+ Cng sut kt hp ca sng mang l c nh v khng ph thuc vo s iu ch.+ Cng sut kt hp ca mi na bng ph thuc vo ch s iu ch, v t ti a l 25% cng sut ca sng mang (tng cng 2 na bng l 50%).1.2.4 Tn hiu iu ch khng phi dng sin: PhXem xt mt tn hiu iu ch khng phi l mt sng sin n gin m l mt tn hiu thng thng c ph tn s nm t f1 n f2. Vi iu ch bin ph ny c di chuyn n trn v di sng mang (hnh 1.3)Hin nhin l ph ca tn hiu iu bin rng hn ph ca tn hiu iu ch. Ph ca tn hiu iu bin Bw rng gp i ca tn hiu iu ch:Bw = 2.f2Hnh 1.31.2.4 Cch thc hin k thut iu ch bin Cc mch c s dng to ra iu ch bin phi bin i bin ca mt tn hiu tn s cao (sng mang) vi hm s l bin ca mt tn hiu tn s thp (tn hiu iu ch).Trong mt b pht AM, chng ta ni:6+ iu ch mc cao nu s iu ch c thc hin trc tip trong tng cng sut cui cng. y thng l mt b khuch i trong lp C.+ iu ch tn s thp khi s iu ch c thc hin bi cc tng trc b khuch i cng sut cui cng.Cc thit b bn dn c th c s dng trong trng hp cng sut thp hoc s dng cc n chn khng trong trng hp yu cu cng sut cao.Trong mch cs dng cho cc bi thc tp, iu ch bin cto ra bi mt b khuch i vi sai, trong li ( khuch i) cbin i bi tn hiu iu ch. Mch ny c tch hp vo IC LM1496, cng c th c s dng to ra iu ch bin vi sng mang nn, s nghin cu trong mt bi thc tp khc.1.3 THC TPCc thit b yu cu:+ Modul T10A-T10B+ Ngun mt chiut 12V+ Dao ng k1.3.1. Hot ng ca b iu chHnh 1.4 : S b iu ch71. Thc hin kt ni modul T10A v T10B theo hnh 1.4. Cp ngun 12V cho cc modul v thc hin cc thit lp sau:+ MY PHT CHC NNG: sin (J1), LEVEL khong 0.5V, FREQ khong 1kHz.+ VCO2: LEVEL khong 1V, FREQ khong 450kHz.+ B CN BNG 1: CARRIER NULL2. Ni dao ng k ti li vo ca b iu ch (im 2 v 1), quan st tn hiu iu ch v sng mang (hnh 1.5a/b)3. Chuyn u o im 1 n im 3 (li ra ca b iu ch), quan st tn hiu iu bin (hnh 1.5c). Ch l hnh bao ca tn hiu iu ch tng ng vi dng ca tn hiu iu ch.4. Thay i bin ca tn hiu iu ch v kim tra ln lt 3 iu kin sau: phn trm iu ch nh hn 100% (hnh 1.5c), bng 100% (hnh 1.5d) v vt qu 100% (trn iu ch - overmodulation, hnh 1.5e)5. Thay i tn s v dng sng ca tn hiu iu ch v kim tra s thay i tng ng ca tn hiu iu ch.6. Thay i bin ca tn hiu iu ch v ch l tn hiu iu ch c th b bo ha hay trn iu ch8Hnh 1.5: Cc dng sng ca b iu ch AM91.3.2. Ch s iu ch7. Thit lp cc modul nh trong phn 1.8. Dng dao ng k o (hnh 1.6)+ Bin B ca tn hiu iu ch (im 2 ca T10B)+ Bin H v h ca tn hiu iu ch, v bin C ca hnh bao ca tn hiu iu ch (im 3 ca T10B)9. Tnh gi tr k ca b iu ch, k=C/B. Gi tr ny nh hon 1 mt cht.10. Tnh bin A ca sng mang:2h HA+11. Tnh ch s iu ch m (%):% 100 .h Hh Hm+Hnh 1.6: Tnh ch s iu ch101.3.3. tuyn tnh ca b iu ch12. Thit lp cc modul nh trong phn 113. t dao ng k trong ch X-Y (X=0.2V/div,Y=1V/div). a tnhiu iu ch (im 2 ca T10B) vo trc X, tn hiu iu ch (im 3 ca T10B) vo trc Y.14. Trn dao ng k s xut hin mt hnh thang tng t nh hnh 1.7a; N th hin s bin i ca hnh bao ca tn hiu iu ch nh mt hm ca bin tn hiu iu ch. Phng php hin th ny ch ra s khng tuyn tnh hay mo ca tn hiu iu ch. Tng bin ca tn hiu iu ch v xem s bo ha v trn iu ch ca tn hiu iu ch c th hin nh th no (hnh 1.7b).Hnh 1.7: S tuyn tnh ca b iu ch1.3.4. Ph ca tn hiu AM (cha c my phn tch ph)Ph lc A gii thch nguyn tc hot ng ca my phn tch ph.15. Thc hin kt ni cc modul T10A v T10B theo hnh 1.8. Cp ngun mt chiu t 12V v thc hin cc thit lpsau:11+ MY PHT CHC NNG: sin(J1), LEVEL khong 0.5V, FRQ khong 10kHz+ VCO2: LEVEL khong 1V, FREQ khong 450kHz+ VCO1: LEVEL khong 2V, SHIFFTER 1500kHz,FREQ khaongr 900kHz.+ SWEEP: DEPTH vn ht c theo chiu kim ng h+ RF DETECTOR: LEVEL ht mc theo chiu kim ng h+BALANDCEDMODULATOR1:CARIERNULLvnhtctheochiukim nghhocchiungcli, saochothucmttnhiuAMtilira. OUT LEVEL t ti v tr trung gian.+ BALANCED MODULATOR 2: CARIER NULL v tr trung tm, sao cho mch hot ng nh mt b chuyn i tn s (b iu ch cn bng vi sng mang nn). OUT LEVEL v tr trung gian.+ CERAMIC FILTER vn ht c theo chiu kim ng h16. t dao ng k trong ch X-Y (X=0.2V/div, Y=50mV/div). Ni my pht SWEEP (im 1 ca T10A) n trc X, v tn hiu tch (im 3 ca T10A) n trc Y.17. Thay i tn s ca sng mang (VCO2) cho n khi ao ng k hin th ging nh hnh 1.9. nhn c dng sng tt nht hy iu chnh: d lch ca my pht SWEEP (DEPTH); bng thng ca b lc t gm (variable capacity); CARRIER NULL ca b iu ch cn bng 2.18. Dng sng quan st c chnh l ph ca tn hiu AM, bao gm sng mang v hai na bng.19. Thay i tn s, bin v dng sng ca tn hiu iu ch v kim tra s thay i ca ph.12Hnh 1.9: Ph ca tn hiu AM13BI 2 - GII IU CH BIN (AM)2.1 MC CH:+ Hot ng ca b tch sng hnh bao+ S mo ca tn hiu tch sng: s gn sng v mo do ct cho+ Hiu sut tch sng+ Hot ng ca b tch sng AM ng b2.2 C S L THUYT2.2.1 Tch sng hnh baoVic tch tn hiu iu ch t mt tn hiu AM c th thc hin bng cch s dng mt b tch sng hnh bao. C th thy trong thc t tn hiu iu ch cu thnh hnh bao ca dng sng tn hiu AM (hnh 2.1).B tch sng hnh bao n gin nht bao gm mt b lc RC ni vo sau mt diode (hnh 2.2). Hot ng ca n tng t nh mt b chnh lu na sng, in p li ra bm theo gi tr ln nht ca sng mang. Bi v bin ca sng mang bin i, bng cch chn cc gi tr thch hp cho R v C, li ra ca b tch sng c th ti to mt cch chnh xc cc bin i ny.Hnh 2.1: Tn hiu AMHnh 2.2: B tch sng hnh bao2.2.2 S mo ca tn hiu tch sngTn hiu gii iu ch c hai loi mo tn hiu:14+ Nu hng s thi gian RC qu nh tng ng so vi chu k sng mang, hnh bao xp x dng sng b nh hng bi gn sng (ripple), s nh hng cng r rng hn vi gi tr RC nh (hnh 2.3).+ Nu hng s thi gian RC qu cao tng ng vi chu k tn hiu iu ch, tn hiu tch sng s bm theo hnh bao, nhng thnh thong s b nh hng bi lut gim theo hm m (mo do ct cho, hnh 2.4) Hnh 2.3: Ripple Hnh 2.4: Mo do ct cho2.2.3 Gi tr RC nh nht: ti thiu ha rippleMc ch ca vic chn gi tr RC l lm gim bin ca ripple n mc cc tiu. t c kt qu ny, hng s thi gian RC phi ln hn rt nhiu so vi chu k T ca sng mang, do trnh s phng in qu nhanh ca C trong khong gia hai nh tn hiu AM ta phi c:RC>>THnh 2.5 ch ra hai v d ca tn hiu tch sng, thu c vi hai gi tr RC khc nhau. Nh rng gi tr ca RC cng khng c ln qu, v nu qu ln n cng s gy mo.15Hnh 2.5: Gim ripple2.2.4 Gi tr RC ln nht: mo do ct choMt iu kin khc ca vic la chn gi tr RC thch hp c th hin trong hnh 2.6Hnh 2.6: Mo do ct choNu hng s thi gian qu ln, c th xy ra trng hp nh sau: t thi im t1, li ra ca b tch sng khng bm theo hnh bao ca tn hiu iu bin AM na m tip tc h thp theo lut phng in ca C, ngha l khng ng theo mong i ca chng ta na.Do diode ca b tch sng s b kha cho n tn t2, khi li ra ca b tch sng t n mc ca hnh bao. Trong khong thi gian t2-t1, li ra cao hn gi tr ln nht ca tn hiu iu bin, v diode b phn cc ngc.Kiu mo tn hiu gii iu bin ny c gi l mo do ct cho (diagonal cutting).16 Gi tr cc i ca RC c tnh sao cho t C phng in trn in tr R ln hn hoc bng hnh bao ca tn hiu AM h thp xung.mmfC R) 1 (.1.2maxvi :m l ch s iu chfmax l tn s ln nht ca tn hiu iu chCh rng trong trng hp m=1, iu kin ny khng th c tha mn, v tn hiu tch sng tt nhin s b mo.2.2.5 Hiu sut tch sngHiu sut ca b tch sng hnh bao dng diode c nh ngha l t s gia bin ca tn hiu li ra b tch sng vi bin ca hnh bao ca tn hiu AM li vo. Gi thit rng ti tn s ca sng mang, dung khng ca t C [1/(2FC)] nh hn tr khng R rt nhiu (gi thit c kim tra nu RC>>1/F gim ripple), hiu sut thc t ph thuc vo t l gia R v tr khng vi sai (differential resistance) rdca diode (hnh 2.7)Hnh 2.7: Hiu sut tch sng2.2.6 S cn nhc cui cng trong la chn RC trong b tch sng hnh bao+ t c hiu sut tch sng cao, in tr R phi ln hn rt nhiu so vi tr khng vi sai ca diode (R>>rd), ti thiu ripple trong tn hiu tch sng, tch RC phi ln hn nhiu so vi chu k sng mang T (T