通信原理概论课后习题答案 马殿光.pdf

1 1Use 1 6 112dh 222dh 1his the height of transmitter antenna measured in feet 2his the height of receiver antenna measured in feet 1dand 2dare measured in miles 12ddd Since 11200hft 60dmiles then 260 6hft 1 2 According to Figure1 3 222 eedrrh Where eris the radius of the Earth 5280ermiles 2222 2eeedrrr hh Sinceehr then 222 2eeedrrr h 2 2edr h 2edr h Since 1mile 5280feet h is measured in feet then 1 25280 5280 mile dhfeetmile feet 2dhmile 1 3Use 1 6 112dh 222dh 12ddd 12dd Since25dmiles then1278 125hhft 1 4Use1 6 112dh 222dh 12ddd Since 1h 60ft 2h 4ft then d 13 8miles 1 5 120 2PP 340 3PP Use 1 8 the average information of source is 4 1 1 log 1 97j j j HPbits P 1 9 00 25P 120 15PP 3456780 07PPPPPP 100 03P Use 1 8 the average information of source is 10 0 1 log 3 08j j j HPbits P 1 15 Use1 10 2log 1 S RB N 10log S SNR dB N a Since 45SNRdB 1200300900BHz then 45 10 2 2 log 1 900 log1 1013 45 S RBKbps N b Since45SNRdB 3200 15001700BHz then 45 10 2 2 log 1 1700 log1 1025 41 S RBKbps N c Since 45SNRdB 32003002900BHz then 45 10 2 2 log 1 2900 log1 1043 35 S RBKbps N 2 2 a Use2 5 for periodic waveform with period 0T 0 2 2 0 0 0 2 1 4 T a dc T a Vv t dt T T Let a 1 then 3 2 1 11 44 dcVv t dtvolts b Use2 11 0 0 2 2 222 0 2 2 0 11 lim 0 8164 TT rms TTT Vv tv t dtv t dtvolts T TT c 2 2 0 816 0 667 1000 rms Pmw R V 2 6 Use2 13 2 14 2 22 2 1 lim T T T Pw tw t dt T 2 2 2 lim T TT EPTw t dt a Use2 52 0 0 2 2 2 0 22 0 2 2 2 111 lim lim lim1lim0 TTTT TTT TTT T Pw t dtt Tdtdt TTTT 0EPTT Since E is nonzero and finite then it is energy signal b 0 0 0 0 2 2 22 2 2 2 0 2 11 lim limcos 111 limcos2lim0 222 TT TT TT o TT T o T Pw t dttdt TT T tdt TT 0EPTT Since E is nonzero and finite then it is energy signal c 00 00 00 00 2 2 2 22 22 2 2 2 00 2 2 2 2 2 00 0 1111 lim cos1cos2 2 111111 12cos2cos 2 12cos2cos4 4422 1311 cos2cos4 828 T TTT oo TTT TT oooo TT oo Pw t dttdttdt TTT tt dttt dt TT tt dt T 0 0 2 2 3 8 T T 3 lim 8 T EPTT infinite Since P is nonzero and finite then it is power signal 2 43 a Use 2 68 2 2 2 2 000 2 00000 2 00 1 lim 1 lim 5 12cos 5 12cos 1 lim2560cos60cos 144cos cos 1 lim2560cos60cos 72cos T T T T w T T T T T T Rw t w tw t w tdt T ttdt T tttt dt T tt T 2 2 00 0 2 72cos 2 2572cos T T tdt Note 1 coscos cos cos 2 b Use2 69 Table2 2 25 36 36 oowwPfF Rfffff 2 71 When 0off xPfhas a max value 0 oxxPfPK a According to the definition of 3 dB bandwidth in Section2 9 1 2o x x P f P f 210 1024 2 B fB In engineering definitions the bandwidth is taken to be the width of a positive frequency band The 3 dB bandwidth is 0 1024B b According to the definition of equivalent noise bandwidth in Section2 9 4 0 2 00 42 4 2 2 4 23 2 2 0 112 22 11 2 1 2 0 125 228 2 2 2 o x eq x B K P f dfK Bdfdf P fKK B ff BB BBB dfB B B f The equivalent noise bandwidth is 0 125B 2 89For RC low pass filter 1 1 c H f j ff where 1 2 cf RC 2 1 1 c H f ff When 0off H fhas a max value 1oH f a According to the definition of equivalent noise bandwidth in Section2 9 2 222 000 2 0 0 11 1 1 11 arctan 124 c eqc cc o ccc f BH fdfdfd ff ffff H f fdxfxf xRC The equivalent noise bandwidth is 1 4RC b According to the definition of zero crossing bandwidth in Section2 9 0H f f In engineering definitions the bandwidth is taken to be the width of a positive frequency band The zero crossing bandwidth is from 0 to c According to the definition of absolute bandwidth in Section2 9 Outside the absolute bandwidth 0H f f In engineering definitions the bandwidth is taken to be the width of a positive frequency band The absolute bandwidth is from 0 to 2 92Use Table2 2 2 0 2 000 0 sin T f S fF s tT SaT fT T f When 0off S fhas a max value 00 S fT a According to the definition of absolute bandwidth in Section2 9 Outside the absolute bandwidth 0S f f In engineering definitions the bandwidth is taken to be the width of a positive frequency band The absolute bandwidth is from 0 to b According to the definition of 3dB bandwidth in Section2 9 2 2 0 1 2 S f S f 4 0 1 2 SaT f 0 4 1 0 8408 2 SaT f 01T f Use A 9 Page 555 1 0 8415Sa 0 1 f T In engineering definitions the bandwidth is taken to be the width of a positive frequency band The 3 dB bandwidth is 0 1 T from 0 to 0 1 T c According to the definition of equivalent noise bandwidth in Section2 9 4 2 24 00 22 00 0 0 4 00 00 000 11 111 33 eq o BS fdfTSaT f df T S f T SaT f dT fSax dx TTT The equivalent noise bandwidth is 0 1 3T d According to the definition of zero crossing bandwidth in Section2 9 0S f 0 21 T fk 0 1 21 fk T First zero point blow above 00f Let k 1 k 0 0 1 f T In engineering definitions the bandwidth is taken to be the width of a positive frequency band The zero crossing bandwidth is 0 1 T from 0 to 0 1 T 3 4For natural sampling sw tw t s t where s k tkT s t s t is a periodic function with period Ts Use2 110 0 jnt sn nn s th tnTc e Where t h t H fF h tSaf Use 2 112 0000 ncf H nffSa nf Use2 103 00 1 cos nn n s tDDnt Use2 106 2 107 000Dcf 0022 nnDcfSa nf 0nnc 00000 11 2 cos 2 cos nn s tffSa nfntdd Sa n dnt 0df 0 1 0 1 2 0 1 cos cos 2 cos cos cos 2 cos cos 2 cos cos 2 cos sss s k ss k k n s s k k n v tw tntw t s tnt w t ddSa k dktnt w t dntdSa k dktnt d Sa n dnt w t dntdSa k dkt cos cos2 s s ntd Sa n d d Sa n dnt Where n is a positive const After v t pass a low pass filter outvtd Sa n dw t b Cd Sa n d 3 7The quantizing noise is half of step size 1 2100 q P nR 2n RR M 11 22 2100 n RRP R M 50 2n P 50 log2log n P log 50 50 3 32log log2 P n P Where R is Range M is quantizing levels is step size 3 8 a According to sampling theorem log22 100200sanafBHz b Use the solution to 3 7 5050 3 32log3 32 log8 96 0 1 nbits P Let 9nbits c Use 3 14 200 91 8sRf nKbps d For binary PCM 221 l Ll Use 3 52 1 8 11 8DR lKKbaud Use 3 74 2 01 1 channelB Dr r 1 2 channel D Br For minimum ofchannelB let 0r min 1 8 900 22 channel DK BHz 3 9Use3 18 6 0230 averagedB S ndB N Let n 5bits PCM word Use 3 14 8540sRf nKKbps 8508 170sec47 22 40 sec MMbytebit byte TKhours RKbit 3 14 a Use 3 16a 2 2 3 14 1 e pkout SM NMP 2 2 3 10log10log30 14 1 e pkoutdBpkout SSM dB NNMP Where 4 10eP bit error rate 2nM M is quantizing steps Let n 5bits PCM word 5 2232 n M b According sampling theorem log22 2 75 4sanafBKKHz Use 3 15b 5 4527PCMsBf nKKHz 3 19 a 2164 l Llbits level b Use 3 28 1 1250 sec1250 0 8 Nsymbol Dsymbolbaud Tms c Use 3 52 4 12505RDlKbps 3 29Use 3 47 1nnnede e0 is reference digit has two possible value 0 or 1 0 1 2 3 4 5 6 7 8 9 10 11 di 0 1 1 0 1 0 0 0 1 0 1 ei 1 1 0 1 1 0 0 0 0 1 1 0 ei 0 0 1 0 0 1 1 1 1 0 0 1 3 34 a Use the solution to 3 7 5050 3 32log3 32 log5 64 1 nbits P Let n 6bits PCM word According to sampling theorem log22 2 75 4sanafBKKHz Use 3 14 5 4632 4sRf nKKbps b For multilevel PCM 283 l Llbits level Use 3 52 32 4 310 8DR lKKbaud c Use 3 74 2 01 1 channelB Dr r 1 2 channel D Br For minimum ofchannelB let 0r min 10 8 5 4 22 channel DK BKHz 3 45 a 216 n M 4 nbit PCMword For binary PCM 221 l Ll Use 3 74 22 4 5 33 110 5 channelBK DKbaud r Use 3 52 DR l 5 3315 33RDlKKbps c Use 3 14 sRf n 5 33 41 33sfR nKHz According to sampling theorem log2sannfB log 1 33 667 22 s ann fK BHz 3 48 a Use 3 25 3 26b 6 024 7720log ln 16 024 7720log ln 12556 0210 140 dB S nnndB N 8 32n Let n 9bits PCM word According to sampling theorem log22 3 46 8sanafBKKHz Use 3 14 6 8961 2sRf nKKbps c Use 3 80a 3 80b For DPCM 6 0240 315 dB S ndB N For best case 15 let n 5bits PCM word For worst case 3 let n 8bits PCM word According to sampling theorem log22 3 46 8sanafBKKHz Use 3 14 sRf n For best case 6 8534sRf nKKbps For worst case 6 8854 4sRf nKKbps 3 57 Use 3 8 1122 11 22 sssss kk w tw kT h tkTwkThtkT Where 1 t h t 2 t h t For TDM system w1 is sampled at Ts 2Ts 3Ts 4Ts w2 is sampled at 1 2 sT 3 2 sT 5 2 sT Use 3 10 12 111 2 sss ss kk WfHfW fkfHfWfkf TT Where 11 HfF h tSaf 22 HfF htSaf 4 2 2 10002 1000 2sin 2 10002 2 jj ee m tt j Note sin 2 jj ee j Use Table 2 2 1000 1000 M fF m tjff Use Table 4 1 for DSB SC cg tA m t 50 1000 1000 cG fA M fjff Use 4 15 1 2 251000 1000 1000 1000 cc cccc V fG ffGff jffffffff Note ff 4 3 a Use 4 17 22222 0 1 2 1000 1 0 1000 1111 4sin 2 1000 222 1 cos 400011 5042500 22 T vccPgtA mtAt dt T t dtwatts 2500 50 50 v s L P Pwatts R b Use 4 18 2 2 2 maxmax 50 2sin2000 111100 100 2250250 PEP L g tt Pwatts R 4 6For RC low pass filter 1 1 c Hf j ff 2 1 1 c Hf ff arctan cHfff Where 5 11 15 9 22 3 1416 10 cfKHz RC a Use 2 151 The phase delay is 6 1115 arctan8 02 10sec 22 3 1416 1515 9 d K THf fKK b Use 4 27b The group delay is 2 6 2 1 1 11 221 111 5 29 10sec 2 3 141615 91 15 15 9 g cc df T dffff KKK c 2 1 1 11 221 g cc df T dffff Where 15 9cfK 4 9 a cossin j ej cosRe j e sinRe j je sin 2 jj ee j cos 2 jj ee 100sin 500cos100sin Re 100500100 Re500 100 Re500 100 2 2 Re 500200sin ca cac caa aa c c cacca jt jtjt jtjtjt jtjt jt jt a s tttt jeeje ej ee ee ej j j et Use 4 9 Re cjt s teg t The complex envelope for the modulated signal 500200sinag tt AM modulation is involved Use Table 4 1 2 500200sin500 1sin 1 5 aacg tttAm t The modulating signal 2 1sin 5 am tt b Use 4 3a 4 3b 2 Re1sin 5 ax tg tt Im0y tg t c Use4 4a 4 4b 500200sin aR tg tt 0otg t d Use 4 17 2 2 2 2 2 5542 2 2 554 2 11 1 500200sin 22 50 1 1 2 5 102 10 sin4 10 sin 100 1 11 cos2 2 5 102 10 sin4 102 7 1002 T a vT LL T aa T T a a T g ttdt P T P RR ttdt T t tdtKwatts T 4 10 a S fF s t 100sin 500cos100sin caccas tttt 50 25050 5050250 2505050 505050 502502 caca cccaca cacac ccaca cacaca cac S fjffffff ffffjffffff jfffjfffff ffjfffjfff jfffjfffjfff jfffff 50cff Note sin 22 c c c c j j j j c eej tee j sin 2 ccc j Ftffff 1 cos 22 c c c c jj jj c ee tee 1 cos 2 cccFtffff b 500200sinag tt 500100100 aaG fF g tfjffjff 500100100aaGffjffjff Use 4 12 1 2 1 500100100 2 500100100 250250 50 505050 cc ccaca ccaca ccca cacaca S fG ffGff ffjfffjfff ffjfffjfff ffffjfff jfffjfffjfff Note ff 4 16 1 n nnVV e 1VV 1 2110 367VVeV 2 3210 0498VVeV Use 4 47 3 2 2 1 37 03 n n V THD V 4 20Use 4 46 0 0 cosn n V tVnt Use 2 95 2 96 2 97 0nb 00 0 00 0 0 0 0 2 2 0 2 2 00 4 4