北邮2010-2011学年《通信原理》期末考试试题(B卷)答案.pdf
20102011BDIFIFRBHFRB2A6D2BIFRAYDJFDBGQGJDJDTCHBLF1EPC1C7E8 BP30EPBQ(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11)A B D C D A E D B D A(12) (13) (14) (15) (16) (17) (18) (19) (20) (21) (22)A F A C A A E B C C A(23) (24) (25) (26) (27) (28) (29) (30)C C D A AAACBBACCABADBBAAJ BPA114EPBQ(1) BP6EPBQ1 1 1 1 1 11 0 1 0 1 0 11 +1 1 +1 1 +1 1 0 0 0 1 1 1 1 1 +1 1+1 1 +1 1 +1 +1 +10 1 0 1 0 0 01 1 1 1 0 0(2) BP4EPBQD9DCF1GBDOGB1(a)BP4EPBQAQFEF1GBDOGB1(b)EJDWEXF0DNCIB5C6+ABEPFKGGEK(a) DPSKD9DCF1GB(b) DPSKAQFEF1GBGB1:1D2 BPA112EPBQ(1) BP3EPBQs=8kHzEN = 64kbpsENb= 96kbps(2) BP3EPBQE1EW = 30kHzENs(1+)=ENblog2(1+)=96log2(1+)=30ENlog2 =3.2(1+)AYlog23.2ENAIlog2 =4EN =16ENDNBT =1/4AY(3) BP3EPBQD9DCF1GBDOGB2(a)(a) MQAMD9DCF1GB(b) MQAMAQFEF1GBGB2:BP3EPBQAQFEF1GBDOGB2(b)D3 BPA112EPBQ(1) BP2EPBQD9DC2()EEEN =b0w()cos(2c)dENFRw()F0AYH2DUARENAPARH2DU2=0AYBP2EPBQcos(2c),0 bARCDH2F0b2ENAPARE0AB022=0b4AY(2) BP2EPBQD9DC1()EEENARH2DU1=b01()cos(2c)d =2AY(3) BP2EPBQFR1,2AVCEENEKEXDLCIDKDLF01+22=4(4) BP4EPBQ(e2)=Pr(4)=12erfc(2b80)2FM BPA110EPBQBP4EPBQ1=s(1,0)EN2=s(cos2,sin2)12 =s(1cos2)2+sin22=s2(1cos2)=2ssinBP6EPBQB4DWGBDOGB3AY(a) 2PSK (b) 4PSK (c) 8PSKGB3:BK BPA112EPBQ(1) BP4EPBQAR4ERDWCDAIDUF01= 3,2= 1,3=1,4=3ENFTDDCEFJH214AYDTF9 = 4=3EEENDWCDARAIDUDTDWF02 4AY(2) BP4EPBQDTF9 = 4=3EEEN =3AYFRCSEEBFERD12 4CBH2AXEPG1ENFRCSBFERD11,+1CBH2AXEPG1AYAPCS()=12 10 G1EM(3) BP4EPBQE( )2 = 4=E2 = 4=210122d =133FW BPA110EPBQ(1) BP6EPBQ(),s(),()AREQFODOGB4(a)AXGB4(b)AXGB4(c)(2) BP4EPBQ()F0DSBC6EBEN()F0SSBC6EBEN(a) ()(b) s()(c) ()GB4:4