高频电子线路课后习题及答案

高频电子线路习题集第一章绪论1－1画出无线通信收发信机的原理框图，并说出各部分的功用。答：话筒扬声器上图是一个语音无线电广播通信系统的基本组成框图，它由发射部分、接收部分以及无线信道三大部分组成。发射部分由话筒、音频成。低频音频信号经放大后，首先进行调制后变成一个高频已调波，然后可通过变频，达到所需的发射频率，经高频功率放大后，由天线发射出去。接收设备由接收天线、高频小信号放大器、混频器、中频经放大后，再经过混频器，变成一中频已调波，然后检波，恢复出原来的信息，经低频功放放大后，驱动扬声器。1－2无线通信为什么要用高频信号？“高频”信号指的是什么？答：采用高频信号的原因主要是：（1）频率越高，可利用的频带宽度就越宽，信道容量就越大，而且可以减小或避免频道间的干扰；（2）高频信号更适合电线辐射和接收，因为只有天线尺寸大小可以与信号波长相比拟时，才有较高的辐射效率和接收效率，这样，可以1－3无线通信为什么要进行凋制？如何进行调制？答：因为基带调制信号都是频率比较低的信号，为了达到较高的发射效率和接收效率，减小天线的尺寸，可以通过调制，把调制信号的频谱搬移到高频载波附近；另外，由于调制后的信号是高频信号，所以也提高了信道利用率，实现了信道复用。控制高频载波的某个参数。在调幅方式中，AM普通调幅、抑制载波第二章高频电路基础fB，0QQLO解：1由f得：LC011L（）4651020010262fC4201060.586mH465200422f由B得：00.707QL465103fQ58.12508103BL0.707QC2109Rk02030CCQQ000gggLR100LRLQQR（1R）Rk0LQQQL00L0LCL和Ct求CCCC,t11312min1312max，tttC2601091210tC8t1L（535102601932为L为01QRQ0LL解CC1CLpF,2CC121f)C2010.317)62C接入系数为R1CCL12R2折合到回路两端的负载电阻为Rp20.643.125kLLQ100回路固有谐振阻抗为R2fC6.28108010199k006120Q100有载品质因数Q1.5460LR019911R3.125L答：CCrq0o的Q解：CC060.024总电容C0.024pFCqCC60.024q0qff串联频率f0.998f4.99kHz000.0241Cq01q2C12011109品质因数Q884642602fCr25100.02410153.63120qE42UnS4In1R,HR10R21df0nH220011df1(2fCR)20011125kHz8412输出噪音电压均方值为1U222Rnn0nn41.3710290101251019.865(V)432E-V9.3I=DR020000ID011R),110DDRDD211B=dfdf0nH200D0D111arctan(2fCR)CRD0DDU22n0n020.465101.61044.6410562083(V)319622题9图解解Ts02s13R3EsP2，s130s2s13RFs3R2，23niosno0s1322sFnios31I4kTB2Rno021RR2s3nioRR3s23s123RRs214kTB3s2s133s12RRR24kTBR4kTB3s3R(RR)s23s10s13s132s1322no2s3Fnios3RRsRRRRRRRRR1RR23输出电阻为：RsRR11s2s123RR(RR)(RR)0RRs1s123s1RR23s1输入匹配功率PsmiE2s4Rs2RRE13RRRRssRR1s1RR23输出匹配功率Psmos4R1RR02RR1sPsmiPRRR23N0Rs1FRR1smosRR3s1RRR(RRRR)20s123s1RRR1s3RRRRRR211s2Rs12RRs01U42no02RRU4213RR1RRniosRRs1sRR231s2RR410RR1RRsRs1sRR21sR24R21RRRRRRs01s212sRRRRRRUU122Nno21s2R12sRRsFnio01I42no0RRs2I421RRRRRRnios1s2s122R41s解：SoN0SiNkTBiFS234oN0。第三章高频谐振放大器fo0题3－1图解yCCpCpC2000.35180.035202222212固有谐振电导为C2f2465102021037.374Sg0Q8000gpgpgg2212014.6S26236C2f2Q3g0106LfB0QLpp|y|K123g1060Np11pCCC31NNbcbcn11为L答CfTf03QL解B1BB21131B8B11213f为1.60BL1f383o解B1BB121431B8B112314当偏离中心频率10kHz时，根据谐振特性，有：KK21A2)4A1442223KKK1K1328844332(4)2f00444f08833f24222f4(2)404BfB1108831044222411.21033答或CCbLLEEbbbbCIUCLC和UCcc10解根据给定静态特性，得到晶体管的E0.5v,bCdubeuEUcostcciEcmbbb）cm题图(1)当t时晶体管截止，因此但，i=0,uEU242135VccecC位于C点(2)当t时0.5晶体管临界导通i=0,且有UcosE0.5,cos,b3c0.53b0.53因此arccos位于B点。80uEUcos2421。20.5V,ocecc当时(3)t0uEU3V,i=ig(3cos0）2.5g2.5A.cecCccmaxmm位于点。A连接A、B、C三点，就是它的动态特性曲线。i/AcAube1BC0u/Vce39152127333945100.53arccos80oU2121RCI(80)i2.50.472LoC11cmax212.50.472IUC1C1P112.39W22PIE(80)iE0.2862.52417.16Wo0C0C1cmaxCPPP17.1612.394.77WC011DbbSC解因为ICi11CCC1111所以=P222221C1CCCC1CCC1CCSE(SE)SPE2E2U1CC1CC1C1CC12C1C1CUCo1)BU=CECU2PC1R6.7C1A1LZC1LrjLLr110C0LZ0L1LnZ10Ln1nn1n2nn2222h)求βi题315图解hhLiiieffffLhff1ZLhjie2i22ffff11fff2fiefeiC2L2E2123解DvLCCQRLL解01L(2f)C4220RCRCR2fC因,pQ000Bp2QLLL256DRLLQ2R(1Q)122XXXQ(1Q)121Q2s22A21134解DCBAXXXXs3s1s1XXXRRXRpp1p1ssp1pXp2p2ss322p2p2p2p2p2X2222p2p2p2p2p2RX=所以X2p2sRRXX=，2p2RX2s22p22211p2p22p2p2当时，产生谐振，电路简化为图（3），谐振电阻为X将X和RR和XRspps3s2s1XX，Q2s1Rp1sps11spp（）121（）1QR2由2RQX11234RRQ11Q1（）2得3RRXXQ221Q2Xp（1）1，RQR1QXX2s1s212Q22RXQ2XXXR1Q1（2）1Q222Rss2RRXQ2RR1Q1（2）R2s2RR（2）1Q1Qs22RR(1Q)1Q1Q2R2222XXXXXXQQQ(RR12Rs解RRX(1Q)1Qp21QRp1p22s2222(15)152152XLHs2f2s6RXp1p1RR1Q1(15)1p1222p211C544pF2fX230109.766R1005XQ211C265pFIPA2OM2fPPkA21RA2A2A2QkQL21Q1k0LQ1QLf11fL（）fA2答D0312图DRDLR解REcCRcs1LC00ic2iLRuRLcs1(b))输出电压振幅BR1RE22R2E2ULLcLcLccsLcsLcsL2集电极平均电流1U12E2E22IC0cc2R(RR)2csLcsL22直流输入功率2A2PEIc(RR)0C2csL负载上的功率U2REL22PL2Rc123(RR)L22LcsL集电极效率PLPRL(RR)0csL集电极耗散功率2E2RE22PPPcLc(RR)(RR)C0L222csLcsL2ERR2PccscsRR(RR)RR20csLcsLcsL解不平衡转换V37.5T1e575V设功率降低一半的晶体管原来正常输出电压为它贡献合成器U,的电流为，异常输出电压为，该管的负载电阻为R则有：IU0UUU22，UR2R200那么，流过合成器的电流为1I（1+II22合成后的功率为211+IR22与正常相比，有22111+1+I2220.7284I42所以，故障后的合成功率为P=0.72810072.8W3－25起振条件：T(j)，即平衡条件：T(j)，即n0,1,2.....T(j)1TU稳定条件：iiL题图答2（2）LC<LC<LC;2（3）LC=LC>LC;11311311333题图解EcC1C2123题4-5图解EcEc•••vTTVT1123111(2f)10139.438410121620,呈感性，所以15B2fCec1（50,2f31011）(2f310122252(2f)210C1251522fC2fCbc22EcLCCCvCHH1Cb10pFC210HeSizeB12LI解Date:File:2345CK1CCF1CC2C12CC121C1gL60Q011gRLLgggKmmgKgmRrggg2Kg2eRre0LFm0LFeegKmT()KKgFgKg1F20LFmggKgKg,20mFLFm1(1103ggm)K(1K)0LFFIgVeQmT解题4-8图=2211212102.6uceC1CF21K2KFF12igIs回路总电容CCC33.33CC2CC030312C33.3310C43.33C00C33.33260C293.33C00因此得到方程组如下1f201062L(43.33C)代入数值后得,01f101062LC2L(293.33C)0解之得C40pF,L0.76H0C5011503反馈系数K1FCC12g放大器增益Km11Kg2FRR0igg因此，K，即K9mmF1111KKg2F3g2FRRRR00ii111C11K2FK2Fg9g9maxLRRQRRm0i0i1330.3310120.76106111912.8810310009500100IgV12.8826ACQmT235题图解15pF15k47H8.2pF3.3pF5.1k32k50H12k2.2pFKFC1221=fTitleSizeBDate:File:23451=f2.731MHz02max250106810-612ff,f02max1minf1max02min当LC串联支路的电容取68pF时，在回路电抗为0时振荡，即：111501006681011091109114710611整理后得到：15981053.732101.0680412153.7321053.7321041.068159810223196101(53.73247)1031510或2110)319611f122121max1f，f1221max02max当LC串联支路的电容取125pF时,111501010612510121109101911471061将上式整理得：或20610(1510)21121211因此f206102.285MHz,1221211min1或者f，15100.616MHzf12221min102min故在0.616MHz不满足相位条件，也不能振荡。到110119K(10)911047106F191111109471016111111147104710(2f)1853.610f152152152111KK110.9321853.6102.826FminFmin3210.8971853.6102.28532答11C21211112(CC)(CC)10130212C1C751答234300pF200pF3/10pF5MHz4.7H12pF47MHz(b)3解24ER2113R2EC11111234第五章频谱的线性搬移电路iaauauau230123式中：uuuuUcostUcostUcost123112233iω1解12123iaauu)auu)auu)320112321233123aauu)auuuuu)212223011232121323auuuuuuu)2122233121323123aau)auu)uuuuu)2122230112322121323auu)uuu31323333123uuuuuuuuuuu)2121222223233231312qp12，3将，uUtuUtuUt111222333（UU2U2）212aa023222UU2UU2314aUaaat221321113331UU2UU2324aUaaaat12232212333332UU2UU2334aUaat1232231333次频率分量和组合系数为的频率分量为：22U2U2U2acos2tcos2tcos2t1223222123aUUcos()tUUcos()t22212121212aaUUcos()tUUcos()t13131313UUcos()tUUcos()t3232323233U31U324U334a3t3t3t43123UU2UU2cos(2cos(221421431212UU2UU2cos(2cos(21241242121UU2UU2cos(2cos(231431431313UU2UU2cos(2cos(2134132343131UU2UU2cos(2cos(2242432323UU2UU2cos(2cos(2343423232cos(cos(UUU22132133132cos(cos(213213i2i1333i3生UU2UU23U3aUaaa12213214111333UU2132412UUU33122213ua20120u和和12答和2u0guiD0u0E2Ql2Q11g21Qm解U1212由Ucost0得：cost，所以tarccos()22232222232guDt32i234302nt2n222022222n233n332221333cos5t......245222所以iK(t)guK(t)g(UUcostUcost)222D2D1122222D222D11时变跨导233costcos2t3222g(t)K(t)gg2DD33cos5t......4522121222210解uuu11LuRguuoLD212D212DL221222D212444t3t5tUt3522211D22uu1212uuuigK(t)(uu)，12D1D212u(uu)igK(t)(uu)12D2D212则负载电流为：iiigK(t)K(tLD1D2D2212D2124435D2221122121213D121211521212215D222212122EcH)H)－＋uouAuRBeEc题图解设:uEuuUcost则：I(t)iBebe3A11RuUcost0e3eB22uEEu1BeeBRREeeeI(t)I(t)uI(t)(1tanhuiC2tanh000AA222V22VTTEUUV1cost1(x)cos(2ntxe2e12RE22n11n1eTRω1L12EUic21cost(x)coste22RE211eeE(x)UuEiRE1costcoste122REoCc2LC21eeEcRRLLiui1o2VVVV1234uuiiiiAB1234VV56ii56I0题图解设:uUcostA11uUcost,B22iiuitanh55A222V2因为Tiiuitanh66A222V4T所以iiiiui=(ii)tanh5656A222Vo24TB而560T560Iui=00BA22oTTVV24IuuEiRER10BA2ocoLcLTTIRE0L2c11(x)cos(2mtcos(2nt12120s1111CCCΩΩΩ1122211ωω1u=u+EUtE,b11bEaIeb00sI(nc0n!nVnnnbTbu23c0123nIIIIIun0000V0V2nTTTTII2t0021VVV22121TTTREb2tsVVTo11TT(2)因为u=u+EuuEUcostUcostEbeU>>Ubcbccb满足线性时变条件，所以ciI(t)g(t)ucC0m显然只有时变静态电流才能产生分量，因此将其展开为级数得I(t)C0cEbVuuIIIIccI(t)aIeIeIucuu......uncV002030nVTTVVVC00s002c3cnTTTT取一次和三次项，包含分量的如cIIIIU133ucuUcostcos3tcost00300VccccVVV443cc3TTTTUUE1U因此，33buRIcostRaIeUcostccVTcccVVVVo00300sc3TTTTuuuEUtUtEbe12b1122bUU21EuEub21b2iI(t)g(t)uaIeaIeuTVTVVcC0m10s0s1TEuI(t)aIeb2VTC00sEu1g(t)aIeVb2VTm0sT将g(t展开为级数得m1Eb1111g(t)aIe1uuu......uVT23nVVVVnVm0s22232n2TTTTTUU1EaIeVb取一次和三次项，包含分量的为3VVcost222T2V0s3TTTUU31Eb而且aIecostUcostV22VVT21V0s31TTTaIUEeVUU3bVVcos()tcos()t0sV122T21213TTT3aIUREUU所以uebcos()t0sV10V2VV2T21o3TTTuiI(1)2VDp11221U2gg；2PPm2PPEuEuDDSSGS2GSV2VVGSPPPPg1m0m02PPV当U|VE|,E时P22PGSGS11V1m02Pg22mQm12显然，在这种条件下，ggm1mQ111222212120111题图解12K(t)cost2222222根据题意，只取分量，则22u2Rg0o0D212101021212R=nRR2f0011g=rRrRfD0212R012121第六章振幅调制、解调及混频cCCCL0cPPP。cC22CC）uuuuX++XuCuuuXX滤波器uuCCCCccΩΩCDiii012iiigK(t)(uu)gK(t)(uu)Lb12DccDccDcc4gDcccccos()tcos()tcc13cc21Dc332212DccDccDccDcccDcc4gDccDcccos()tcos()tccgUt1ttcos(3)cos(3)Dcc3ccLd12DccDccDcc22gcos3t.....UcostDccccccccbCCcΩΩU及Uuu求22C012ΩoCo)2uuucc2221012D1,u2uuac2c22012uu2uuccc222120120112c12Ccu(t)iiR2RaUcost2RaUUcostcosto12LL1L2Cccc）uc2uuD1u,uD2uc2在忽略负载的反作用时，ui1gK(t)ugK(t)uc2DcD1DcuigK(t)ugK(t)uc2DcD2Dc2uiiR2RgK(t)uo12LLDc12222RgUcostcos3tcos5t.....cost352LDccc因此，与(1)相比，输出信号也中包含了的基频分量和、c频率分量,但多了的奇次谐波与的组合频率cc（2n+1)分量cu11i1K(t)K(t)ucrRRrRR2cD1cDLDLu11iK(t)K(t)ucrRRrRR22cD2cDLDLu(t)iiRo12L2RUL1222costcos3tcos5t.....costrRR235cccDLR设T,TT。CCΩΩgRD12cD12cu:c0uuucu0coCoc22Ut3cccUocgD2LDDD12o2当uRLo20L22210L121212ΩΩΩdCCCC题图uuuC1D1ccCuuC2DcD2DcCD3CDcD3DcCD4C4DcD2DcCiiiii4gccCDccCccDccDc4gcos5t.....UcostDDDcccoLL4gcos5t.....UcostDDDLcccD1037CCLE,eceuωtmut。4CCouECe2EcTeeU44EccceeUtt1tte7REc4ee47oCoL生37CoC2E2VC时变静态电流为：Eu22GSV4D0DSSP61030.50.25cos10t时变跨导为：cos10t22IE3DSSGSVV4PP31030.50.25cos10t3DmC6t2376t237uC6100.50.25cos10t230.31010.5cos10tcos10t4oD337600.50.25cos10t2310.5cos10tcos10t(V)37ADAMC22CCCcos(11CCCukuK(t)CMC2)t)t1MCC22t3t5t2CCC12212M12M2SgRgsCCDidR2iD33Rod因为361m2m5100.51570030601D6sso36s求u.oR1022ipi026oD33R0.9du(t)iR4100.5cos10t103636sspu(t)KU0.921.8(V)ods）3363spV2t2t()36u(t)Kt)2t)V3odsCRΩ1212C211fH23622HCuC1uuCF时211fH3622sCrrrrsdU令m=,则：sUruuuUcos()tUcostD1rsscrrUcostcostUsintsintUcostscscrrrcUcostUcostUsintsintsrcscU(t)cost(t)m1c12式中U(t)UUcostUsint22m1rssUUUcostU22rrssUUU2U1costU12mcostssUr2rrr2m3nU1mcost(1)mcostm2n1nnn!rn2Ucost(t)arccossUUcost1rsuuuU(t)cost(t)D2rsm2c22U(t)UUtUt22m2rssUUUtU22rrssUUU2U1tU12mtssUr2rrr2m3nU1mtmtm2nnn!rn2Ut(t)2sUUtrsuuuoo1o22m3n，mcost(1)12KUmcostKUm2nnn1n!drdrn2u2KUmtodr1234C解已知TT,即CC44CCCC44541C5555525544tcos39t...32C22222ututuutuuttia+au+au23C023ωωωωωωSSLLLSLSLS解aUcos2t12g(t)auauaUt232Lm2L3L2LLLaU3aU322cos2taUtLL222LLLggtgcos2tm0m1Lm2Lgm12gC1aU2Lgm22aU32gC2L4U。L0g。C000SSSLSQL0Eu0QL当时Q0由tarccos(1)QL00LL所以g=0CQLSE当时02QEEcost0得：02QL0LLaut2nL或n0,1,2,3....i=K(t)auL02K(t)t2n33LL11233costdtsint3LLL22333C当时Q由Eu0Ecost0tarccos(0)2QL0LLg2m1LLL当时E0Q1由EuEt0t)023QL0LL32DPP0L123LCVP2uEuuLSdiIEu2DLVVPPIII22VVV22LLPpp231.51033LV322pggm12CC1gggn2R0L0L1122363SUUgn1.88IUg737.93106SSULgg；CsLIE2EI1GSVGSDSSSP1122VV2GSGSPSP1010064E10160,E5322GSGSV=4E2(V),UE2(V)PGSGSIEu31DSSGSLLVV44PPLg0.0011(mS)mQ3）iI(t)g(t)uDD0mS2EuI(t)I10.5t2LVD0LP20.001ttLLg(t)u0.001tUtmSLSS0.001ttLSt0.45)t)t66SLSLSAA,I0.00181.8mA,I0.45100.45I10166ωω、LSSLω+ωLSIiiiu1VIuu1Vtanhtanh025602Re22ITT12Iu21IeT21C3CeU频率分量n=0,1,2...)，1VCT要求输出滤波器为带通滤波器，中心频率为，带宽为。C如果调制信号是多频信号，则应该保证滤波器的高频截止频率等于2倍调制信号的最高频率分量，即2.max12（1mcost)cost(x)cost2IC1C3Ce频率分量n1,2,3...)，要求输出滤波器2n2nC频截止频率大于调制信号的最高频率分量.max122IS1L3LeUx=2n11VLSTILS120021LD21120C0.7120C0.70.7fFF。0C12H212f20ILC0.7ACDFGBEABCD0000555/FGE000ILsLsI3SI3LSSLLSSL3和。解信号，它与本振信号混频，产生了接近中频的干扰信号。此时本振频率为f=fLLJILJf4If当fLLJI3fSJ1J1J2件，433fffILs6解p1fp1p1~fSfISIqpfqpqpf0.55~25MHzSp+q6,6p1qpp1当时2,f20.4550.91MHz，(1)p1,q2,qpSfff1.635MHz,组合干扰f2f0.455MHzfLSILSIp1当时(2)p2,q3,3,f30.4551.365MHz,qpS组合干扰-3f2f0.455MHzffff1.82MHz,LSISLIp1当时0.6825MHz,(3)p2,q4,1.5,f1.50.455qpSfff1.1375MHz,组合干扰4f-2f0.455MHzfLSISLIMHzMHzMHZff.ILs解和时此时，f，本振频率f干扰信号频率f，且SLSIJf=fJLI时fffSLSIJf4LJIffSJ1ffJ2I解f-ff-ff-f=f-fSJ1J1J2SJ1J1J23第七章频率调制与解调6解根据给定条件，可以看出，/F4dt/,10Hz44mmmmP3SmU2C(4)因为调角波的功率就等于载波功率，所以P2RL33Ωutf7c()tku310tt333ft1810cos310t333t03333tttdt1810cos31006sin10t3u(t)5cosFMtt7tt6sin10V35cos10t73t(t的积分限不定，所以0(a)FM和PM分别称为2FSK和2PSK。PM信号的波形与DSB信号的波形相同，故在数字调制中，可用产生信号。(c)5UUΩUΩΩ解FfCm(1)B2(fF)2(505)sm当(2)UffUmmmfB2(f)2(1005)F当s不变时，加倍时，最大频偏不变，但调频指数减小一倍，Fmm(3)Uf所以带宽为B2(fF)2(5010)120kHzsm当、都加倍时，最大频偏加倍，但调频指数不变，所以带宽为(4)UFmfB2(f)2(10010)220kHzFsmffFcm解)nB2(fF)2(5015)130(kHz)smmaxff83.750.0583.7(MHz)cmff83.750.0583.8(MHz)cm所以，伴音信号瞬时频率的变化范围为100kHz，从到83.7MHz83.8MHz和。3ΩkfυV3Ω解。fBfmfSm3Ω。fmfBfSm信CCj=j0EVQKf2解Cj012uCCC(t)j0j0j12Eu12E1Q12EQQCjQ1mtCCC其中，CjQj012Ej0124j03QU22m12E1249Q111f1mcost41f(t)1mcost4LCLCCjjQ13f1mcostmcost......22432C3f3ffmCmcostfmcos2tC2C264464C31因此：=，fmf2f64432CCC111f,fCfmff418432mC2mCf1fkfmUCf31Kf2m2mfm33Kf2即mm11107f443mCL:C，j0，E=4QφΩ解CCCC(t)j0j0j0ju(t)6tuE4111Q0.6uuCjQ1mcost121t4可以看出，0.5,0.5,C68.7mpFjQ11f(t)f1mcost4LCCj13f1mcostmcost......22432C313f1mmcostmcos2t......2264464C因此得到：11（）1f（MHz）13.6LC21068.710C6123jQ3（）（）2fmf221CC1（）（）3f44mCf（）4kmU3f33（）（）5fmf