通信电子线路 习题解答

习题(第2章)148.0)/2(11)()(5.055.53.3310150/1052//2)(3.20)(1003.21050)10514.32(1)2(12:1:.120200367.000007.05122620000ffQwMHzfffffQQffBWHHfLfLCv通频带又解习题)(2.21/)/(2/300//150////:)/(||:155.0)(11)(11)(:00000000002002020020KCLQQRRRRRRRRkHzRfQfBWkHzRfQfBWRQRQQRRRRRRRRffffQQwPXXPXPPLLPLppXPXPXpv其中设或习题)(46.2407.0)5.0(5.0cos/)(/)()cos()(53.0419.026.1)(1)(26.1239.0/300/)(iI0.4190.2390.407cos66:75.195.0/833.02/22/%3.83%1002.7/6/)(2.162.7)(2.73.024:''max100max0maxC01011011000VEEUUEEAiIAIiggPPWPPPWIEPbbbbbbcCcccCCCCCC查表解习题。，PPPPPRIIUPIEP。IIRIUR，CCcccfCCcCCCfC故不安全晶体管很可能被毁坏放大器进入欠压区工作等效负载电阻当天线突然短路时答max0/02/2/maxmax,00::.31001211100101习题63.1//////63.1267.0/436.0)70(/)70(/2/)(2/2/)(2/61.0436.0/267.0)70(/)70(/)(//)(//:36372120202201012121212max22221max1111122212max22221max111PPIEPPPIEPPPPPUUiUIUPiUIUPRRUUiUIURiUIURPccccccccffcfcfcfcfccffcfcfccfcfc解题习题。。，P。，，P量的限制感而下限频率受到绕组电上限频率可达无限高传输线变压器的况下在无耗和匹配的理想情题变压器模式起主要作用式传输能量方磁偶合应在低频段主要以电磁感传输能量主要以传输线模式传输线变压器在高频段错题1935)(.1835习题LcLcLinLLLiiinRIUIUZRIUZRRIUIUIURIUIUR2422/8/16/842//2/:21解inRiULRUiIIIU2/U2/ULULII2I2习题32:??P教材什么叫功率分配什么叫功率合成)(31.03.008.0)(452736)(6.1010302102)(2)(102)(2.0)(6.1810302105.3)(5.3)(108.2)(8.2:.1222263046303mSymSypFCmSCSmSgpFCmSCSmSgyyyyCjgyCjgyrefeoeoeoeieieierererefefefeoeoeoeieieie因此而根据解)(68.32216.33107.1016.3347.1014.32102.1121185.27)(102.112100025.020025.02.37)(88.262/::.26060'062222210'00000'210KHzQfBLgQKSgPgPggKLfQLQRgyPPKeeVLoefeV电阻空载谐振谐振时放大倍数解7.401053.14104.2021046514.322)(4.202142035.01835.0200)(53.14104.0035.05535.01080102001046514.32)1(:.336123''02222221'32261232222100222210'gCfQpFCPCPCCSgPgPQCgPgPggeioeioeioe解题)(7.91835.0135.011)(18)3(311053.14108.36035.035.0)2()(4.117.4046511111163'21007.0'pFCPPCNNNNCNNNCpFCCCgyPPKkHzQfBcccnoecbcfeVe6.297.15465)(7.151282)(8)2(:12)2(:3410317.017.037.0117.0BfQkHzfBkHzBfBfLnn又解题iissisRRRuRIP222)(::36号功率信号源输入到网络的信解题iRsRsussRRsiMRuPS，si4|:2即输入信号额定功率信号源最大输出功率RsRsusPMFssiMOMPMssooRRoOMLLoooRRKNRRRSSKRRuRuPSRRRuPLo11)(44|:)(:2222输出信号额定功率输出信号功率ousRRoRoRouLR)(9.4:100)(55.1:10)(49.0:11024100103104005044:397323VENVENVENNNNSKTBNREAFAFAFFFoonFAA解题第四章1.P102题20解:根据“射同余异”或“射同基(集)反”原则:(a),(b),(c)不能振荡;(d)能够振荡(e),(f)振荡条件:CLg312.P106题36解:(a)符合“射基同名”原则,满足相位条件.(d)根据“射同余异”原则,满足相位条件要求:LCfg21),max(2121030233032202fffCLfCLfg3.P104题27解:(1)(2)L2C1CiC)(25.0:10500102000510200051021213122121mHLLCCCCLfg解得255.0200051021CC。，ZZNNZNN。Z。，。R，C。，C。C。pFCpFCCCCCCCiie对原边无影响测量仪表可视为开路从原边看端准确停振交流负反馈不满足三端振荡器要求开路基直流电位只需一个隔断集隔直通交作用是可以加仅使放大器直流电流增可以解得降低一半,)(22)6(:)5(”“,)4(,)3()(3593)(458:1025.02110500255.021:21221’5432121213321P106题34解:2CJbReReCcE1CL补充题:解:(1)H1H9pF250串联型晶体振荡器MHz5)2(。短路线的作用晶体起选频)3(。很高频率稳定度)4(补充题:解:)(800225132222)/(2513)(62107.3726HzFBsradMHzfccP168题18解:)(09.0)(01.008.025.02)(08.01002422222WPPPWPmPWRUPccLccP169题34解:Ic1c1c2c3c4c2c3c4cP170题39解:cucu1Duu2Du1i2iLiLRou2121221121)2()(iiiiiTtKugitKugiuuuuuuLLLcDDDDcDcDtRUgtRUgtRUgRitutttUgtUgtKugugTtKtKugTtKtKugicLDcLDcLcDLLoccDccDDcDcDccDL)3cos(32)cos(2cos)()3cos34cos4(coscos)(1)2()()2()('P170题40解:(C)能检波,(a),(b),(d)不能.P170题42解:81.01.36coscos1.36)(63.0107.41253333333OdOdDKradRrRg)(58.281.059.063.012514.3cossin)cossin()(9.281.027.42)(35.227.42KrgRKKRRKRRdDidii或或。RRmRRRRR。mmRCsmmmsRCg无底部切割失真满足无惰性失真满足直交直交68.07.41010||1)(105.55.010525.0115.0)(107.41001.0107.425322563P172题49]})sin()sin(2sin])cos()cos(2cos])sin()sin(2sin])cos()cos(2cos2)cos(coscos)1(:121212121212121221221121ttttttttUmUtUttmUukui解。ttRUmURiuttUmUiLLo存在振幅失真经低通滤波cos])cos[(2cos])cos[(2:12211221。tRUmURiutUmUittUmUtUtmUukuiLLo存在相位失真时当存在频率及相位失真时当经低通滤波,;,])cos[(4])cos[(4:]})cos[(]){cos[(4)cos()cos(2)2(212121212121212121221121P172题50)(5012102)4()(12)1(2)3()(5)2()(515)1(5)(121022::223WRUPkHzFmBradmkHzFmfmkHzFLffmfmf调频信号解)(5012102)4()(12)1(2)3()(5)2()(515)1(5)(121022:223WRUPkHzFmBradmkHzFmfmkHzFLppmpmp调相信号P172题51)(2105)120(2205100)(100250:,)(110)1(210550:kHzBFfmkHzfFVVkfkHzFmBFfmmfmfmfmf不变加倍解)(12010)15(251050)(1025)(50:,)(22010)110(21010100)(1025)(100250:kHzBFfmkHzFkHzfFVkHzBFfmkHzFkHzfFVmfmmfm加倍不变加倍及P168题22答:反向偏置状态.P168题24答:直接调频是指用调制信号直接控制载波的瞬时频率,以产生调频信号.间接调频是指利用调相来实现调频.P173题54)(102cos502)()()(102cos10210)()(102sin10)(:3333mVttStfStutdttdtttDDo