高频电子线路答案(张肃文版)

3解：fMHz02Δf11099010(kHz)63071106fQ10002Δf1010307取R10ΩQR100102314106L159(H)0.11CL((pF)2141015910)262601132解：(1)01或ω02并联谐振。LC1LC21211(2)或ω振。并联谐振。01LC1102LC2112(3)或ω01LC102LC2121L1Lω)Rω1)2R2ωCCω1002C3R0012RRωLR2Rω1)0jωC0ω200341由15C450C5354022100由125351CC-22。111802LC2145351045022C3120LCC’113521223141510100105..0CR0612011L061122C2314151021001012..00110-3VomR02Iom5VVQV212110-2123LomCom0Sm11253μH36解：LC214102100102612010011VCVQ1000SCC1100pFC200pFXCCL2314102253106X2.6X021410253102141025310LL6666R477Ω00QQ2501100X011ZRj477j477j796Ω2141020010C612XX0X1137解：L2.ωC21451050102061251061003fQ021501030072100255102036ξQ351060f0因2Δf22f，则Q0.5QR0.5R，所以应并上电阻。0.70.7002πfCωC3明：4πΔfCg002ΔfQ07f007CCC202020202020139解：CC513pF20CCCi2011f0416MHz2πLC2314081018310...612L209kΩRQ100202010P0C121222202020CCCRRRR.102095588Ω.k20120iPC0188103RQ2ωL214416100810L660f416102648MHz2Δf007QL312解：1Z0f12Z0f13ZRf1ρ1033解：1LL159μH1214101260111159pFCC26212L011231410159106.ηR120M18μH121410620123141031810..M220Ω662Z01f1R202L159106Z25kΩ251RRC20201591012P1f112314101591066.L0113QRR202011f11020ff2kHz642Δf22200ρR10307Q1111177pF5C22L22314950102159106.3021ZRjL122202C022120j21410159102010066214101771061222141031810.20100159106M266Z0768j84Ω01Zf122L3解：R20RRC5010159101231PR0M0f1f1014106Q2210002f30.710102M276316解：1R2001f1R5221010010L276Rab40k01RR5201f121010M76201R51101001076L3Q20001R512f11122210.0130.722fQ2000I111317Q22.5I1.252210102f301Q1Q300103f011R11.822.523.1430010QC3120I111Q2I2210102f301Q1Q300103f0QQ3022.57.512ωL375318解：LC211125μLω2LLC12β504解：当fMHz时当f20MHz时490β22250102f6110f250106Tβ5012.150β5020102f6110f250106Tβ50当f50MHz时0β5050102f6110250106fTI10754mS47解：gE26β126501be0βg50075410377mS03rbem377103g24pFCm2πf214250106beTa1rg17007541031bbbebωCr2141024101270017bebbgjCajb0.754102314102410j.10.1j3712yiebebea2b212012089541mS1jj2141031012gjCajb7yre0.0187j0.187mSbcbcab1012222gajb37.7101j0.13y37.327j3.733mSmab101fe2222gjCajbajb1ygjCrgjCrgbcbcbca2b2bcab22oecebbmbbm1j0.11010.049j0.68mSj2141031017037.710712322m21fmAvAvo144得2Δf4210m42Q2Δff07.Q4070m21fmA2得2Δf4101vm0424A1001QΔf.Qf4vo01024101m242Δf2Δf101m故K01074r011121421mm455NNNN49解：p025p2025234513202011311372μSgωQL2π10710100410p66001137210200102860102285μSggpg21pg22..666442poeie2ppy10250254510..3Avo1232fe5106gΣAA123322povo11Q163gωL5102π10710410L666Σ0107106f2Δf0657MH016307QZL2163Q.2K11.143L100Q054885.ooξtantan.295fereie22gpg..372100252001022626g30088μS.pL20252p1ggg1ξ22860102001030088101295g..26662S1sieoeyyL45100311033fere114解：1g0037mSQωL10021410710410p66001ggpgpg003701030082030150158mS22222pR1oeie5ppy10303384222Avo21782feg01582622ωLgf214107104100158104544kHz6307003AA21783844vo4vo114221221454419765kHz4407407210446kHz52f070712142Δf21044645445902kHz07072..21784544Avo07.947vo2Δf1044607947804266AvoA4.4.4vo.-2250253880426621698272..vo4vo4411解：CCpC500031850162pF221oe11225μH.L2πfC2231415102.5016210..6120K不能满足r0.1y26.436.422414A7.74fe2.5C2.50.3S011118μH417解：LωC122π46510321000101201118118118LLLL6051207373733623456CCC100041004pF121o52CCpC1000401004pF22745362iωC2π4651010001049μS312gg201001610012oQ0ωC135.2π46510100010249μS312gp36gi0621022023Q7451000初、次级回路参数相等。若为临界耦合，即，则5140103ppy1745Avo742fe249106gωC2π46510100410312Q6001249106gL12465103f2Δf22109kH06007QZLK16r014解:4kTRf4138102901000101265v2..237nn1265nAi24kTGf438102329010-3107nn421解：vvv2v24Δf4Δf4kTRΔfkTRkTR222nn1kTRTRTRn2n311n2n33n4Δf112233n4ΔfkTRRR123nTRTRTRT112233RRR123又iiiikTG4Δf4Δf4kTGΔfkTG22222nn1n2n311n2n33n4kTGTGTGΔf112233n4kTGGGΔf123nTGTGTGRRTRRTRRTT112233123231312GGGRRRRRR12312233114证明：1IyVyV11be1iereIyVyV2c1be11feoeIyVyVyVyVVyVyyV3b22ce21cb21cb21beiereierereiereIyVyVyVyVVyVyyV4c22ce21cb21cb21befeoefeoeoefeoe23得IyVyyyVyVc2be11cb2feiereoereIyVyV5V221ccbrefe1yyyiereoeIyVyV5代入4IyVyyc2cb21refec22yyycboefeoeiereoeyyyyyyyy2oeIVbeV6cb2fefeoeieoerefec2yy2y1yyy2yyierefeoeierefeoeyyyyyfefeoefyyy2yfeierefeoeyyyyy2yoyieoerefeoeyyy2yrerefeoeie由1乘yy与4乘y后相加得feoereIyyIyyyyVyyV1c2be1cb2feoereiefeoereoe由6代入消去I得c2yyyyy2yyyyyyy2ieIVbeViereiefeieoerefereyoereb1y2y1y2cb2yyyyierefeoeierefeoeyyyyy2yyyy2ieyyrefeiereiefeieoeyy2yiyieierefeoeyyyyyyyrreoerereoeyfereyyy2yie同理可证2refeoe略422解：v4kTrf41.3810273192000.226V2bn233122bni2en2qIf21.6101910320030.641016A2En0.950.950210102f611500106ff1910321.61010.952000.32i2cn2qI1317A2C0nAf21f4证明：f00A2f2Qn0ff012Q0f424解：FdB倍0F6dB981倍n高n中60TiTF11207290n混F1F11207139811..FF99510n混n中KA02Ann高AA高pcp高20lg1888276dB高p高A888p高114RRRPPPnoPAni1Ps2R4解：F1siPPnissssAPPPV4RR2RRnsonopposossPPPsI2GGGrCL4解：F1siPPnissLGAPI4GGGrCL2nsonopossLs427解：A为输入级，为中间级，为输出级。A6dB3.981倍A12dB15.849倍PApBF1F12141FF1.72nBnCAAA3.9813.98115.849nnApApApB4解：不能满足要求。设A前置放大器，为输入级，为下一级。PP105F1F11011.9951FnFFF8.1sininBnCPP104AAA10100.1nAnAnAsonopApApB25解：ikvkVVcost20m011kVVVcostVVcost2022222000mmm当VV时，ikVVVcost，该非线性元件就能近似当成线性元件来处理，2m000m0即当V较大时，静态工作点选在抛物线上段接近线性部分，然后当V很小时，根据泰0m勒级数原则，可认为信号电压在特性的线性范围内变化，不会进入曲线弯曲部分，故可只取其级数的前两项得到近似线性特性。512解：为了使i中的二次谐波振幅达到最大值，应为60。oCCVVBZ121cos60BBVVVoV2BBmBZmgVcost当cost0当cost05解：im0iIcosntnn0111IIgVtdtcosgV101mmgVcostdtgV22mm211n为偶数为奇数gVgVcostcosIntdt01n2mnm5解：iiiD1D2gVcost当cost0当cost0gVcost当cost0iimm00当cost0D1D2241k1kigVgVt2k1mm20k15解：当V1msintsint时，i；00当V1msintsint时，igV1msintsint00002cos2kωtsinΩtcos2kωti1sinΩt22m00gVm4k14k1022k1k1k225解：vRiiRkvvkvv4kRvv0LL1212L125解：vRiiRiiRiiii0L23L41L2413Rbbvvbvvbvv23L01122123123Rbbvvbvvbvv2L0112212312Rbbvvbvvbvv23L01122123123Rbbvvbvvbvv2L01122123128RbvvL212di5解：1gb2bvbv4bv23Cdv123BE4BEmBEBEdigtb2bVcostbVtbVcos4cost2233Cdv120030m040m0mmBEv0vBE3b2bVcostbV1cos2bV2costbVcoscos3tt2332120m030m040m040m00g2bVbV3m120m40m1ggbVbV32c120m40mdiaIqSkTq2gveBEvkTBECdvmBEdiaIqSkTqcosgtVcosteVkTomtC0dv0momBEv0vBEaIqSkTq1q21q3VcostV1costVcostVcost2kT6kTom0kTom0om0om0IqV6kTqVomkTqVomkT2IqV34ItIcoscoscoscosttt234SomSom2kTS0S0003IqVqVomkT3gISom8kTm1S1IqVIqV3ggSomSom22kT16c1kT5iiiii1324aavvavvavvavv234010s20s30s40saavvavvavvavv234010s20s30s40saavvavvavvavv234010s20s30s40s4aavvavvavvavv23010s20s30s40savv16avv16avv340320ss4s0I26E0.5I0.50.5529解：g0.5E9.6mS2626c2I1rEsbb26TI0.526350.55mSggicE26be0gg4Socce9.6240.550.004g21047340dBAc4ggpcmaxicoc22224652dB122830.1QQf1104731AA1ALiQ2f10010pcpcmaxpcmax000.7I26E0.5I0.50.081.545解：g0.5E2626c2I1rsEbb26TI0.080.1ggicE262630be0gg10Socceg21.5440.10.012A592.9dBc4ggpcmaxicoc21.540.1gGg219623dBApcLcgGoc0.010.10.1Lic532iiiii1234aavvavvavvavv234010s20s30s40saavvavvavvavv234010s20s30s40s4aavvavvavvavv23010s20s30s40s4aavvavvavvavv23010s20s30s40savv16avv16avv340320ss4s0534解：因存在二次项，能进行混频。只要满足ff就会产生中频干扰；当fffnin0时产生镜像干扰。由于不存在三次项，不会产生交调干扰；有二次项，可能产生i互调干扰；若有强干扰信号，则能产生阻塞干扰。5解：1.此现象属于组合频率干扰。这是由于混频器的输出电流中，除需要的中频电流外，还存在一些谐波频率和组合频率，如果这些组合频率接近于中频放大的通带内，它就能和有用中频一道进入中频放大器，并被放大后加到检波器上，通过检波器的非线性效应，与中频差拍检波，产生音频，最终出现哨叫声。2.因f465kHz，p、q为本振和信号的谐波次数，不考虑大于535i~1605kHz波段内的干扰在f930kHz和f1395kHz929kHz、SS931kHz1394kHz1396kHz3.提高前端电路的选择性，合理选择中频等。5解：若满足pfqff，则会产生互调干扰：12spqff77410351.809MHz，不会产生互调干扰；12pq，f2f774210352.844MHz，会产生互调干扰；12pq，2ff277410352.583MHz，会产生互调干扰；12pq，2ff277410353.618MHz，会产生互调干扰；12pq，2f3f2774310354.653MHz，会产生互调干扰；12p、q，3f2f3774210354.392MHz，会产生互调干扰；12p、q，3ff377410355.427MHz，会产生互调干扰；12pq谐波较小，可以不考虑。3f2f2537解：1ff0.8MHzS02f3f2S0S02f2fff0.4MHz0S2f3f2S0S0f0.2MHzf0.6MHz0S3f2f302ff12MHz0s2f3f30s0s02f30fff20MHz0s2ff30s0s0f4MHzf16MHz0S539pfqff，则会产生互调干扰。已知f19.6MHz、12s1f19.2MHz、fff23320，故没有互调信号输出。2s0i64解：PVI240.256WCCCO5P83.3%0P6C242V2V257.6RcmCC2P2P25p002P2P250.417AIcm100VV24cmICC0.421.67gcm1I0.25ccc0查表得77oc2ηV20712.1086解：θ156θ91.gCCoVccccmPIR214W220k11PCPP1P147Wη07.0900cI67解：i282mAc0α900319cmaxo0Iα90i05282141mAoc1m1cmax11PRI20001412W22220pc1mP2ηc74%0VI30009CCc0V2I2R2L2I2R2L2I2R2L2RI2R6证：P00022L2R2L20L2P02.2i69解：VVvVCC2421.25Vcmaxg0.8cmCCcmincr0.2530.556670IiAoc0cmax0A702.20.4360.9592Iicm1ocmax1PVI240.556613.36WCCc011PVI21.250.959210.19W220cmcm1PPP13.3610.193.17WCP0010.1976.3%P13.36C21.252V222.16Rcm2P210.19p024V2V21442610解：RRcm2PCC2P221p00R14414.4X110C1QL11C221pFfX23.14501014.416C1R200X16.952200144L1RRQ111001122L1Xf16.950.054HL1L123.145010610144200QRLRX112.5712Q1QX10011016.95C22LLL111C1239pFfX23.1450102.5726C2611解：1R增加一倍，放大器工作于过压状态，V变化不大，PV/2R0.5P；2cmPcm0P02R减小一半，放大器工作于欠压状态，I变化不大，PIR/22P。2cmPcm0P0r111612解：57.4%LL11rrk11112QQM21100150.032QQk212122VV120.521266V26RCCCEsatcm2P2PP00R66设Q10则X6.6PL10LC1Q11241pFCfX23.14106.618C1R505.5XLRR501101C21Q122L66LP11290pFCfX23.14105.528C2106650QRLRLQXX1112.5P11021105.5L1Q2LL12.5C2X19.9nHL11f23.14108RQ614证：将RC和RC串联电路改为RC和RC并联电路，并设Xa111221122C1LX2X2C2RXRR2122R2C1C2RR1RXXXXC1X1R212C1222C22RX212C1C1RX222C2C2X21X2C2RX匹配时RR，即RR1R1RC1X121R212C11Q2222C22LRX2C2R1Q122R1LR21R22XXXXXRXRX1C1C2212C1C1222C2C2R21XX2RR1RQ11RR1RQ1RX1C1X2L22L22QXR212C1C1R212C1X1Q1Q2X1QC2LLC2LLC2Rb将RC和RL串联电路改为RC和RL并联电路，并设X1Q11211121C1LX2X2L1R21R22R2C1L1RR1RXXXXC1X1R212C1RX222L12RX212C1C1RX222L1L1X21X2L1匹配时RR，即RR1R1RC1X121R212C11Q2RX222L12LRX2L1R1Q122RL1R22XR21XXXXXC2C1R222L1L1RX212C1C1X2RR1RQ1RQ1RRQ1RQR12L2L22QXL21L22QXR21X2C1X1Q1Q1Q1QL1LLLL1LLLL16解：1；236191PPPP316WACkPA6285.7%70%PP3kCPP3P6360%ACPPcr16解：当kk时，1则rrk1rrQck1c1r1131则r9r故k3k1rrQQQQk1cQ1120906解：6276060P。oo210VV33I6解：R99RIiLVV0.61.456V629VbmBZBBcoscos70oVVV61.454.55VBbmBBVV4.550.6iBZ1.98AB22Cmax701.980.4360.86Iicm1VVgiAoCmax1241.9822.02VcmCCIVcrCmax0.8622.01P09.47Wcm1cm2210100QPA11P9.478.52WLQ00VV1.56解：VbmBB6.14VBZcoscos70oVVV6.141.54.64VBbmBBVV4.640.6iBZ2.02AB22CmaxA702.020.4360.88Ii1VVoCmax1V0.92421.60.8821.6cmCCIVP09.5Wcm1cm2210QPA11P9.58.8WL100Q007解：a电路可能振荡，属于电感反馈式振荡电路；e电路可能振荡，属于电容反馈式振荡电路；h电路可能振荡，属于电容反馈式振荡电路；bcd电路不可能振荡；f电路在LCLC时有可能振荡，属于电容反馈式振荡电路2233g电路计及C可能振荡，属于电容反馈式振荡电路。be7解：1有可能振荡，属于电容反馈式振荡电路，ffff；12032有可能振荡，属于电感反馈式振荡电路，ffff；12034有可能振荡；属于电容反馈式振荡电路，ffff；1203356不可能。77解：117解：1f2πLCC0.723145d11C1QL352g5.27R7dP3006~008V6571065710667解：1f414MHzd04211105qSd200C211100105pF504qd304314mHL55qS200d04Ωr42500B42500025d212S200S200198pFC9610296102040d0505025Q10d10041680044Bq65710657102d011mm66f15106q7271~23728解：恒温槽、稳压电源、高稳定度克拉泼振荡电路、共集电极缓冲级等。729解：并联cb型（皮尔斯）晶体振荡电路。9解：iI1mcosΩtcosωta0IIIcosωtmcosωΩtmcosωΩt220a0a0222IIImm22222有效值aaIm21a2294解：1v25107cos2πt03cos2tsin210t625sin210875sin21005000sin2π995000t.6375sin21010000sin2π990000.2包络25107cos2πt03cos210000tVV251070325峰值调幅度m04max025上V0VV252510703谷值调幅度m10min25下V0119解：1m时PPmP10025W244aωΩωΩa0T00112m0时PPmP03100225W2244aωΩωΩa0T0096ibvbv3131197解：1PPmP075000612.5W2244ωΩωΩa0T00P2P1225WωΩωΩ005000PP0Tη10kW2P0avη05.07m2.21a50001P0T22P051245kW3P0avηη.9解：1m时a11PPm2P1000250W44ωΩωΩa0T00PPPP10002502501500W00ωΩωΩT002m0时a11PPm2P07100012