高频电子线路课后习题答案(第五版)张肃文
31解:f01MHz2Δf0.7110699010310(kHz)f01106Q10032Δf0.71010取R10Ω则LCQR0110010159(H)623.14101159(pF)02L(23.14106)2159106
1132解:(1)当01或ω时,产生并联谐振。02L1C1L2C2(2)当01(3)当011或ω02L1C11或ω02L1C11时,产生串联谐振。L2C21时,产生并联谐振。L2C2
1L1)R2jω0LR(12)R2Ljω0CCω0LCCR33证明:Z112RRjω0LR2Rjω0L(12)jω0Cω0LC
34解:1由15C16052450C5352得C40pF(Rjω0L)(R由12C16052100C5352得C-1pF不合理舍去故采用后一个。2L3
L C C’
11180μH02CC23.145351032450401012
35解:Q0112126-120C0R23.141.510100105L011112μH226-120C023.141.51010010Vom110-3Iom0.2mAR5VLomVComQ0VSm212110-3212mV1136解:L2253μH0C23.1410621001012
Q0VC10100VS0.1CCX112100pFCX200pFCCX0L23.14106225310623.1410625310623.14106253106RX47.7ΩQQ02.50.11000L0L1147.7j47.7j796Ω6120CX23.1410200101137解:L220.2μHω0C23.145106501012ZXRXjf05106100Q02Δf0.715010332Δf10025.5510620ξQ06f03510322f0.7,则Q因2Δf0.721kΩ电阻。 00.5Q0,故R0.5R,所以应并上2πf0CωC38证明:4πΔf0.7C0gf02Δf0.7Q
CC0C1520202018.3pF39解:CCi2C2C0C1202020f01141.6MHz6122πLC23.140.81018.310L100C120.810620.9kΩ2020101222RPQ0C2C0C1202020RRiRPR1020.955.88kΩ0C201R5.88103QL28.266ω0L23.1441.6100.8102Δf0.7f041.61061.48MHzQL28.2
312解:1Zf102Zf103Zf1R
313解:1L1L2C1C2ρ101103159μH623.141011159pF226601L123.141015910ηR120M13.18μH60123.14102Zf101M2R223.141063.1810620Ω202L1159106ZP25kΩ12R1Rf1C12020159103Q101L1R1Rf123.1410615910625202042Δf0.75C2f0f01062022228.2kHzQρ1R11031202L2123.149501032159106177pF1Z22R2jL021C022120j23.1410615910620j10061223.141017710Zf101M2Z2223.141063.181060.768j3.84Ω20j10062
315解:RL1591020R1312RPC501015910Rf10M0f0106Q2210032f0.71410
316解:1Rf101M2R21071062052Rab23QRR101L21Rf110710010640k520201M01LR11071062510710010620052f0.711221222210.013f0Q200 I111317解:Q22.522I01.252f1010311Q300103Qf011R11.8312Q0C22.523.14300102000102f1010311Q300103Qf0QQ3022.57.512ω串联联谐L375μHL2C318解:11L2125μω并联联谐L1L2C
II012121Q302
45解:当f1MHz时,β0β0f1fTβ0β0f1fTβ0β0f1fT25050106125010650249当f20MHz时,2当f50MHz时,25020101250106506212.15050106125010625
47解:gbegmCbeIE10.754mS26β0126501β0500.75410337.7mSrbegm37.710324pF62πfT23.1425010a1rbbgbe1700.7541031bωCberbb23.14107241012700.1yiegbejCbeajb0.754103a2b20.895j1.41mSj23.141072410121j0.1120.12j23.14107310121j0.1yre0.0187j0.187mS2222ab10.1gmajb37.71031j0.1yfe37.327j3.733mSa2b2120.12gjCbcajbjC1rgajbyoegcejCbcrbbgmbcbcbbma2b2a2b21j0.1j23.141073101217037.710320.049j0.68mS210.1gjCbcajbbcAv48解:令AvoAv令Avomm4Q2Δf4Q2Δf41m20.7f0412m得2Δf0.71mf04421Q20.1f04110101211m2mm得2Δf0.12mf044101Q故Kr0.12Δf0.12Δf0.742m41014214
49解:p1gpN2350.25N1320p2N4550.25N13201137.2μS66ω0Q0L2π10.710100410116200102860106228.5μS22442ggpp12goep2gie37.2106Avop1p2yfegΣ0.250.254510312.36228.5102ApoAvo12.32151.3QL2Δf0.71116.3666gΣω0L228.5102π10.710410f010.71060.657MHZQL16.322QL16.3K111.43Q1000fere54o88.5oξtantan2.95222gpp2gie37.21060.252200106g3008.8μSL22p10.25yfeyre26662gsgiegoeg1ξ12286010200103008.81012.95LS451030.31103
1gp410解:110.037mS66Q0ω0L10023.1410.71041012p12goep2gie0.0370.10.320.0820.320.150.158mSR50.30.33824.2221.780.158ggpAvop1p2yfeg22Δf0.7ω0Lgf023.1410.7106241060.158103454.4kHz3Avo4Avo421.784225025.3842Δf0.7452Δf0.714212Δf0.721454.4197.65kHz1044.6kHz14142Δf0.7212Δf0.72Δf0.71044.6454.4590.2kHzAvoAvo2Δf0.721.78454.49.472Δf0.71044.64Avo49.4748042.66Avo4225025.38-8042.66216982.72Avo4Avo411解:CCp12Coe5000.3218501.62pF
L12πf02C123.141.51062501.62101222.5μH
Kr0.11.9不能满足AvoS414解:yfe2.50Cre26.4236.427.742.50.3
417解:L111118μH22312ω0C12π46510100010118118118L36L2L34L566013.5120737373C12C1Co100041004pF13.5C36C2pCi1000401004pF74.5ω0C12π465103100010126g12go201049μSQ0100222ω0C213.52π465103100010123g36pgi49μS0.6210Q010074.5初、次级回路参数相等。若为临界耦合,即1,则13.5314010p1p2yfe74.5Avo74g249106222ω0C122π46510310041012QL606g1249102Δf0.7f04651032210.9kHZQL60
Kr0.13.162420解:vn4kTRΔfn41.381023290100010712.65μV2in4kTGΔfn41.38102329010-310712.65nA
2222421解:vnvn1vn2vn34kT1R1Δfn4kT2R2Δfn4kT3R3Δfn4kT1R1T2R2T3R3Δfn4kTR1R2R3ΔfnTT1R1T2R2T3R3R1R2R32222又inin1in2in34kT1G1Δfn4kT2G2Δfn4kT3G3Δfn4kT1G1T2G2T3G3Δfn4kTG1G2G3ΔfnTT1G1T2G2T3G3R1R2T3R2R3T1R3R1T2G1G2G3R1R2R2R3R3R1
418证明:1Ib1yieVbe1yreVce11Ic1yfeVbe1yoeVce12Ib2yieVbe2yreVce2Ic2yfeVbe2yoeVce2yieVce1yreVcb2Vce1yreVcb2yieyreVce13yfeVce1yoeVcb2Vce1yoeVcb2yfeyoeVce1423得Ic2yfeVbe1yieyreyoeVce1yreVcb2Vce1Ic2yreVcb2yfeVbe1yieyreyoe5Ic2yreVcb2yfeVbe1yieyreyoe5代入4Ic2yoeVcb2yfeyoeIc22yfeyfeyoeyieyoeyreyfeyoeVbe1Vcb26yieyreyfe2yoeyieyreyfe2yoeyfyfeyfeyoeyfeyieyreyfe2yoe由1乘yfeyoe与4乘yre后相加得Ib1yfeyoeIc2yreyieyfeyoeVbe1yreyoeVcb2由6代入消去Ic2得Ib12yieyieyreyieyfe2yieyoeyreyfeyreyoeyreVbe1Vcb2yieyreyfe2yoeyieyreyfe2yoe2yieyoeyreyfeyoeyoyreyieyreyfe2yoe2yieyieyreyieyfe2yieyoeyreyfeyiyieyieyreyfe2yoeyryreyoeyreyyyrereoeyieyreyfe2yoeyfe略同理可证2223422解:vbn4kTr27319702001030.2261012V2bfn41.38102ien2qIEfn21.610191032001030.641016A20f1f20.9510106150010620.952icn2qIC10fn21.6101910310.952001030.321017A2
423证明:fn0A2fdfA2f0120424解:Fn高3dB1.995倍Fn混1FnFn高ff012Qf0Fn中6dB3.981倍dff02Q
Ti6011.207T290F1Fn中11.20713.9811n混1.99510Ap高KpcAp高Ap高0.2Ap高20lg1.8882.76dB2sAp高1.888425解:Fn
PsiPniPPV4RsRRR11nos2s1sPsoPnoPniApApPoPsPoVs4RsRRR PsiPniPsIs24GsGGLrCL1426解:Fn21PsoPnoApPoIs4GsGGLrCLGs427解:A为输入级,B为中间级,C为输出级。APA6dB3.981倍FnFnAApB12dB15.849倍F1FnB12141nC1.72ApAApAApB3.9813.98115.849
428解:不能满足要求。设A前置放大器,B为输入级,C为下一级。PsiPni105F1F11011.9951Fn4FnAnBnCFnAFnA8.1PsoPno10ApAApAApB10100.1
258解:ikv2kV0Vmcos0t1212kV022V0Vmcos0tVmVmcos220t22当VmV0时,ikV022V0Vmcos0t,该非线性元件就能近似当成线性元件来处理,即当V0较大时,静态工作点选在抛物线上段接近线性部分,然后当Vm很小时,根据泰勒级数原则,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iC中的二次谐波振幅达到最大值,C应为60o。cos60oVBZVBB1Vm21VBBVmVBZ2
gVcost513解:im0iIncosntn0当cost0当cost011gVcostdtgVm2m11I1gVmcos2tdtgVm2121gVm2IngVmcostcosntdtn10515解:iiD1iD2I0n为偶数n为奇数
gVcostiD1m024当cost0当cost0gVcostiD2m0当cost0当cost0k11k1,2,3,igVmgVmcos2k0t2k12k1516解:当V01msintsin0t0时,i0;当V01msintsin0t0时,igV01msintsin0t
517解:v0RLiD1iD2RLkv1v2kv1v24kRLv1v2
518解:v0RLi2i3RLi4i1RLi2i4i1i322cos2kω0tsinΩtcos2kω0tigV01msinΩt22m224k14k1k1k1k1,2,3,
2RbRbRbLLLRLb0b1v1v2b2v1v2b3v1v22203b1v1v2b2v1v2b3v1v2b1v1v2b2v1v2b3v1v2223300
b1v1v2b2v1v2b3v1v238RLb2v1v21gm523解:diC23b12b2vBE3b3vBE4b4vBEdvBEdiCdvBEb12b2V0mcos0t3b3V02mcos20t4b4V03mcos30tvBEv0gmtb12b2V0mcos0tgm12b2V0m3b4V03m3b3V02m1cos202b4V03mcos0tb4V03mcos0tcos30t21gm1b2V0m1.5b4V03m2qvBEdiCaISq2gmvBEekTdvBEkTgcdigmtCdvBEVomcos0taIqSVomcos0tekTkTqvBEv023aISqq1q1qVomcos0t1Vomcos0tVomcos0tVomcos0tkT2kT6kTkTqVIqVIqVqVISomcos0tISomcos20tSomcos30tSomcos40tkT2kT6kTkTgm1qV3ISqVomISomkT8kT3234IqV3IS1gcgm1Som22kT16525解:ii1i3i2i4qVomkT23a0a1v0vsa2v0vsa3v0vsa4v0vs34a0a1v0vsa2v0vsa3v0vsa4v0vs234a0a1v0vsa2v0vsa3v0vsa4v0vs234a0a1v0vsa2v0vsa3v0vsa4v0vs23438a2v0vs16a4v0vs16a4vs3v0
529解:gc0.5IE262gicgbesIE1r26bbTI0.5E0.55mS26026350.5IE0.50.59.6mS2626gocgce4S2gc9.62Apcmax1047340dB4gicgoc40.550.004QL2fi2465122830.1dBApcApcmax1A1104731pcmaxQQ2f10010000.7IE260.5IE0.50.08530解:gc0.51.54mS22626sIE126rbbTgicgbeIE0.080.1mS2602630222gocgce10S2gc1.542Apcmax592.928dB4gicgoc40.10.01gcGL1.540.1ApcgGg0.010.10.119623dBLicoc
532解:ii1i2i3i422aaa0a1v0vsa2v0vsa3v0vsa4v0vs23423423a0a1v0vsa2v0vsa3v0vsa4v0vs002344a1v0vsa2v0vsa3v0vsa4v0vsa1v0vsa2v0vsa3v0vsa4v0vs38a2v0vs16a4v0vs16a4vs3v0
534解:因存在二次项,能进行混频。只要满足fnfi就会产生中频干扰;当fnf0fi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,可能产生互调干扰;若有强干扰信号,则能产生阻塞干扰。
535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fi465kHz,p、q为本振和信号的谐波次数,不考虑大于3的情况。所以落于535~1605kHz波段内的干扰在fS930kHz和fS1395kHz附近,1kHz的哨叫声在929kHz、931kHz、1394kHz、1396kHz时产生。3.提高前端电路的选择性,合理选择中频等。536解:若满足pf1qf2fs,则会产生互调干扰:p1、q1,f1f277410351.809MHz,不会产生互调干扰;p1、q2,f12f2774210352.844MHz,会产生互调干扰;p2、q1,2f1f2277410352.583MHz,会产生互调干扰;p2、q2,2f1f2277410353.618MHz,会产生互调干扰;p2、q3,2f13f22774310354.653MHz,会产生互调干扰;p3、q2,3f12f23774210354.392MHz,会产生互调干扰;p3、q3,3f1f2377410355.427MHz,会产生互调干扰;p、q大于3谐波较小,可以不考虑。3f2f021S537解:fSf00.8MHz2fS3f02fS2f02fSf00.4MHz2f3f2S0fS0.2MHzf00.6MHz3fs2f0302fsf012MHz2f3f30s0fs2f030fsf020MHz2fsf030fS4MHzf016MHz
539解:若满足pf1qf2fs,则会产生互调干扰。已知f119.6MHz、f219.2MHz、fsf0fi23320MHz,故没有互调信号输出。
64解:PVCCICO240.256WCP0583.3%P622VcmVCC242Rp57.62P02P025Icm12P02P0250.417AVcmVCC24Icm10.421.67Ic00.25gcc查表得c77o
2ηVCC20.71266解:gcθc1.56查表得θc91oVcm10.8P0Ik2R2214W11PCPP01P141.7Wη00.7cIc09067解:icmax282mAoα0900.319
Ic1mα190oicmax0.5282141mA22222VcmIkmR20LIkmR20LRCIkmR20LRIkmR68证:P022L2RP2L220LRC
i2.269解:VcmVCCvcminVCCcmax2421.25Vgcr0.822211RpIc21m2000.14122W22P02ηc74%VCCIc0300.09 P0Ic0icmax070o2.20.2530.5566AIcm1icmax170o2.20.4360.9592APVCCIc0240.556613.36W11P0VcmIcm121.250.959210.19W22PCPP013.3610.193.17WCP010.1976.3%P13.36
2Vcm21.252Rp22.162P0210.1922VcmVCC242610解:R1Rp1442P02P022XC1C1R114414.4QL1011221pF62fXC123.14501014.4R2R22QL11R12002001001114416.95XL1L1XL116.950.054H2f23.145010610144QLR1R2200XC22112.57QL1QLXL110011016.9511C21239pF62fXC223.1450102.57
2Vcm/2RP0.5P0;1RP增加一倍,放大器工作611解:于过压状态,Vcm变化不大,P02Icm2RP减小一半,放大器工作于欠压状态,Icm变化不大,P0RP/22P0。612解:krr1r1
22VCCVCEsatVcm120.5613解:RP662P02P021211157.4%L1L211112Q1Q2k2100150.032Q1Q2M
设QL10C1则XC1RP666.6QL1011241pF2fXC123.141086.6RLR1QR2LLXC2150P1105016625.5C211290pF2fXC223.141085.5QLRPRL1066501112.522QL1QLXC2101105.5
XL1L1XL112.519.9nH2f23.14108C1和Ra将R1C1和R2C2串联电路改为R1614证:XC12C2并联电路,并设2XCR1212R1R1XC12XC222R2R2R2XC2R1QL2R222XCXC22R2XC2R1212XCXC12R1XC122XCXC11,即R12匹配时R1R2R1R1222R222R1XC11QLR2XC2XC2R21QRR2L211XL12R12R21XC22XCXC12XC222R1XC1R2XC22XCR12R1R2R1QL1R1R2R1QL12XC122222R1XCR12XCXC21QL1QLXC21QL11R21QXLC2C1和Rb将R1C1和R2L1串联电路改为R1XC12L1并联电路,并设2XCR1212R1R1XC12XL12R2R22R2XL1R1QL2R22XL1XL12R2XL1R1212XCXC12R1XC122XC1XL11,即R12匹配时R1R2R1R12R2222R1XC11QLR2XL1XL1R21QRR2L211XC22R2R121XC12XLXL12XC122R2XL1R1XC12XCRRRQRQR2RQRQ212121L21L21L21L2R1XC1XL11QL1QLQLXL11QL1QLR21QXLL11天线断开,工作于过压618解:状态,集电极直流电表读数减小,天线电流表读数为0;2天线接地,工作于欠压状态,集电极直流电表读数略增,天线电流表读数增加;3中介回路失谐,工作于欠压状态,集电极直流电表读数略增,天线电流表读数减小。1PAPPCPk10316W619解:2k3cPA685.7%PPC103PPC10370%P10PA660%P10
620解:当kkc时,k11若k625解:
r150%则rr1r1r1Q1Q2kc1Q01QQ0933kcQ
r190%则r9r1故kr1r
627解:260o160o,P0减小,工作于欠压状态。
VV628解:Ri939RLI3I
VVBB0.61.45629解:VbmBZ6Vocoscos70VBVbmVBB61.454.55ViCmaxIcm1P0VBVBZ4.550.61.98A22iCmax170o1.980.4360.86AVcmVCCgcriCmax241.9822.02VIcm1Vcm0.8622.019.47W22QL10PA1P19.478.52WQ01000
630解:VbmVBZVBB0.61.56.14Vcoscos70oVBVbmVBB6.141.54.64ViCmaxIcm1P0VBVBZ4.640.62.02A22iCmax170o2.020.4360.88AVcmVCC0.92421.6VIcm1Vcm0.8821.69.5W22QL10PA1P1Q01009.58.8W0
a电路可能振荡,属于电75解: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;b、c、d电路不可能振荡;f电路在L2C2L3C3时有可能振荡,属于电容反馈式振荡电路g电路计及Cbe可能振荡,属于电容反馈式振荡电路。 1有可能振荡,属于电容76解:反馈式振荡电路,f1f2f0f3;2有可能振荡,属于电感反馈式振荡电路,f1f2f0f3;4有可能振荡;属于电容反馈式振荡电路,f1f2f0f3;356不可能。
77解:
1f0721解:11100MHz7122πLCCd23.141020510C1L362gd11RPQ30.06~0.08V1fq726解:620510121075.27mS
11.5~1.5001MHz727解:1.657101.657104.14MHzd0.4S200Cq21.110521.11050.105pFd0.4d30.43Lq43.543.514mHS200d0.4rd42500B425000.2521.2ΩS200S200C03.961023.9610219.8pFd0.41.051.05Qq104d1040.416800B0.251.6571061.6571062d0.11mmfq15106
2不能3不能,普通三极管没有负阻特性。
728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。
93解:iI1macosΩtcosω0tIcosω0tI有效值IImacosω0Ωtmacosω0Ωt22222IIImama222222maI122
1v2510.7cos2π5000t0.3cos2π10000tsin2π106t94解:25sin2π106t8.75sin2π1005000sin2π9950003.75sin2π1010000sin2π9900002包络2510.7cos2π5000t0.3cos2π10000tVV2510.70.325峰值调幅度mmax00.4上V025V0Vmin252510.70.31V025
1211ma1时95解:Pω0ΩPω0ΩmaP0T10025W441212ma0.3时Pω0ΩPω0ΩmaP0T0.321002.25W44
396解:ib1vb3v不包含平方项,不能产生调幅作用。
1Pω0ΩPω0Ω1ma2P0T10.725000612.5W97解:44Pω0Ω2Pω0Ω1225W谷值调幅度m下2PP0avηP0T500010kWη0.520.72maP0T15000122P0av12.45kW3Pηη0.5
1ma1时98解:121maP0T1000250W44P0P0TPω0ΩPω0Ω10002502501500WPω0ΩPω0Ω2ma0.7时121maP0T0.721000122.5W44P0P0TPω0ΩPω0Ω1000122.5122.51245W Pω0ΩPω0Ω99解:ff0f1f2f3f45202001780800010005kHz
910解:i1b0b1vvΩb2vvΩb3vvΩ23i2b0b1vvΩb2vvΩb3vvΩ233v0i1i2RR2b1vΩ4b2vvΩ6b3v2vΩ2b3vΩ32b1RVΩcosΩt3b3RV02VΩcosΩt1.5b3RVΩcosΩt2b2RV0VΩcosω0Ωt2b2RV0VΩcosω0Ωt1.5b3RV02VΩcos2ω0Ωt1.5b3RV02VΩcos2ω0Ωt30.5b3RVΩcos3Ωt
输出端的频率分量:、3、0、20
P010.125912解:m121210.5P90T121P0P0m2P0T10.1250.42910.845kW22
9131vAtvtVcostvBtvt2若D1D2开路,则vAtvBtvtvAB03若D1D2短路,则vAtvtVcostvB0vABtVcost918解:RR13R2ri2470010005101335R2ri2470010003Rd333.141000.57R5104700Kdcos0.87VKdmaVim0.870.30.50.132V0.132P6.33W2R2133522VimVimKd0.520.87P41.7W2RidR5104700APP6.330.152P41.7RRR1R22R22ri2235010005102350R22ri2235010000.55R1R251047001中间位置919解:2最高端RRR1R2ri247001000510R2ri2470010000.26R1R25104700
R2的触点在中间位置会产生负峰切割失真,而在最高端不会。920解:由RR1R25~10k11R1~R2510取R26kR11.5kR2ri262RR11.53kR2ri262R31maR93取ma0.31-ma210.32C0.0187FmaRmax0.3900023.143000Ce1minri210.26F23.143002000取C1C20.01F取Ce20F33Rd333.141000.5R60001500Kdcos0.9R9000Rid5k能满足要求2Kd20.9
0C23.144651032001012921解:GP5.84SQ0100f0465QL23.52f0.720Q0100622GPp24goeGP5.840.31005.8410QL23.5p340.1532gid4700
ikmV1costcos1tV0cos0t1kmV1V0cos10tcos10t4cos10tcos10t11当01时,vSkmRLV1V0costcostkmRLV1V0coscost42无失真,只影响输出幅度。1当01时,vSkmRLV1V0cos10tcost2有失真。12v1mV1cos1t21ikmV1cos1tV0cos0t21kmV1V0cos10tcos10t41vSkmRLV1V0cos10t4当01时,只产生相移;当01时,有失真。
1v1mV1costcos1t924解:
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