【文章內(nèi)容簡介】 9。39。?相39。39。0 39。39。? ?相01 00 100tc?cost? ?mtt??tst??調(diào)制： ?ttmAts cc ?co s)()( ?co sccAt?Polar NRZ ? ?mt2022/3/13 39 ?Demodulator: ??? 用相干用解調(diào)器鑒相器2022/3/13 40 (3).The PSD of the BPSK )()( tmAtg c??ttmAts cc ?co s)()( ?)()( tmjAtg c?22 si n() bg c bbfTP f A TfT????? ????The PSD of s(t) is 1( ) [ ( ) ( ) ]4s g c g cP f P f f P f f? ? ? ? ?The PSD of g(t) is By Eq. (),we get: 2022/3/13 41 ? ?fPgf0 R2RBnull ?RR?fcfRfc ?Rfc ?? ?fPsRB null 2?2022/3/13 42 When we make no ISI filtering (如 作升余弦譜濾波） fcfcfB?cfB?f00f ? ?rfB ?? 10? ?? ?02211TBBfrDr?????? ?ecHf? ?eHf2022/3/13 43 2FSK (FrequencyShift Keying) (1) Discontinuousphase FSK ? Producer of FSK Oscillator f1 Oscillator f2 gate gate Inverse Unipolar NRZ + FSK Discontinuousphase FSK Continuousphase FSK ?Classify: ? ?mt2022/3/13 44 t1co s?t1 00 1? ?tm t? ?tsFSKt00 ???????? ?2c os 2?? tt2022/3/13 45 So FSK signal can be represented as: 可以看成是兩個不同載頻的 ASK 之和。 ? ?? ?1122( ) c o s ( )c o s ( )ccs t m t A tm t A t????? ? ?? ? ?? Frequency field ? ? BffBT 212 ???Where: B bandwidth of m(t) 2022/3/13 46 RB null ?? ?fPmf0 RR?When m(t) is rectangular: ? ? RffB T 212 ???f2fRfc ?Rfc ?? ?fPsRB null 2?1f2022/3/13 47 When we make no ISI filtering (如 作升余弦譜濾波） f2f Bf ?2Bf ?11f? ?eHff00f ? ?rfB ?? 10? ?? ? ? ?? ?212 1 022121TB f f Bf f f rF R r? ? ?? ? ? ?? ? ? ?? ?212 F f f? ? ?峰峰頻偏 ? ?ecHf2022/3/13 48 (2) Continuousphase FSK ? Producer of FSK Frequency Modulator fc ? ?tmN R Zpol ar? ?tsF SK1 00 1??tm t? ?tsFSKt2022/3/13 49 ])(co s [)( ??? dmDtAts tfcc ?????? ??? tf dmDt ??? )()(? The representation ? ?? ?tj cetgtsor ?Re)(: ?? ?tjc eAtg ??)(? ?? ?? ?2112222fdfDdtf t m tdtD ffF????? ? ??? ? ?The continuous phase 2022/3/13 50 BFf??? B bandwidth of baseband signal m(t) ? The frequency field Carson’s rule: ? ?BFBB fT 2212 ????? ?? ? ? ??????????????12,2,22fDffDftmDftffcfcfci???When m(t) is squarewave: 22TB F R? ? ?2022/3/13 51 ?在描述 FSK信號特性時，常用到兩個參數(shù)：載波頻率 fc和數(shù)字調(diào)制指數(shù) h。 ? ?12 /2cf f f?? 12|| 2ff Fh RR? ???2022/3/13 52 (3) Demodulation for FSK ? noncoherent detection a. 適合于頻差較大的場合 ”判發(fā)“，否則”判發(fā)“若判決： 0121 yy ?code BPF(f1) BPF(f2) 整流 整流 LPF LPF 抽樣判決 Envelope detector Envelope detector ? ?tsFSK1y2y2022/3/13 53 b. Frequency detector ? ?tsFSK ? ?tKm2022/3/13 54 ?coherent detection BPF BPF LPF LPF 抽樣判決 ? ?tsFSK ??t1co s?t2co s?1y2ycode頻差較大時適合 頻差較小時，若 正交，則LPF換成積分器可實現(xiàn)最佳接收。 tt 21 c o s,c o s ??”判發(fā)“，否則”判發(fā)“若判決： 0121 yy ?2022/3/13 55 (4)波形相關(guān)系數(shù) ? Discuss: the relation of two carrier of FSK ? ? ? ?? ? ? ?222111c o sc o s????????tAtstAtscc? ? ? ?? ? ? ?? ? ? ?? ? ? ? ? ?? ?12021 1 2 2021 2 1 2 1 2 1 2021 2 1 2 1 212c os c osc os c os2si n si n2bbbTTcTcbcs t s t dtA t t dtAt t t t dtTA?? ? ? ?? ? ? ? ? ? ? ?? ? ? ? ? ????? ? ?? ? ? ? ? ? ? ???????? ? ? ? ?????? ????????波形相關(guān)系數(shù) : 12fRfR????2022/3/13 56 ? ? ? ? ? ?? ?? ?21 2 1 2 1 2120 th a t is :si n si n022bcbbTATT?? ? ? ? ? ??????? ? ? ? ???????????? ? ? ? ? ?1 2 1 222 2 2 2 22d i g i t a l m o d u l a t i n g i n d ew xh e r e :b b bFT f f T F T hRFhR? ? ? ? ? ??? ? ? ? ??????? ?? ? ? ? ? ?302s i n2s i n 2121 ??????? ????hh?????? 時 , 稱 在 [0,Tb]內(nèi)正交 0?? ? ? ? ?tsts21 ,2022/3/13 57 022s in ?hh??(1) For continuousphase FSK 21 ?? ?h?2hh??22s in?2?0?,2,1,2 ???? kkh ??? ? ? ?? ?m in m inm in121 , MS,K224Tkh s tk h F BhR RFst