非线电(完整版)课件

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1、University of Science and Technology of China非线性电子线路第五、六、七、八章习题解析University of Science and Technology of China2课程安排课程安排 第五章习题：5.1 5.3 5.5(c)5.6 第六章习题：6.1 6.2 6.3 6.4 6.6 6.10 6.13 6.14 6.16 6.17 第七章习题：7.1 7.3 7.4 7.6 7.10 7.12(c)7.14 第八章习题：8.1 8.4 8.7 8.6 8.9 8.11 8.16 8.17University of Science and

2、Technology of China35.3University of Science and Technology of China45.3解：(1)高频大幅度的信号 判为开关控制信号tu22当u2的正半周oDuuuu211oDuuuu21211Dgui 22Dgui LoLoRuugRiiu12121221LLogRugRuUniversity of Science and Technology of China55.3当u2的负半周1221LLogRugRu1122ugRKKgRuKuKuuLLoooo综上：University of Science and Technology of

3、 China65.3(2)同(1)KuKuuooo综上：122111LLoRguRgu122212LLoRguRgu1212222111LLLoRgKgRgKguRuUniversity of Science and Technology of China75.3(3)开关控制信号u2u2正半周：oDuuuu211oDuuuu21411Dgui 44Dgui LoLoRuugRiiu14121221LLogRugRuKgRugRKuuLLoo1221University of Science and Technology of China85.5(c)5.5 试分别求出如图所示电路输出电压的表

4、达式。设所有二极管特性均为从原点出发、斜率为gD的直线。且SLUU DLgR/1(c)解：uL为开关信号当uL正半周，VD1导通，VD2截止 oSLDuuuu111DDugi oSLLDLouuuRgRiu1LSLDLDouuRgRgu1University of Science and Technology of China95.5(c)当uL负半周，VD1截止，VD2导通，同理：SLLDLDouuRgRgu1LSLDLDoooouKKuRgRguKuKuu1综上：University of Science and Technology of China105.6University of

5、Science and Technology of China115.6ttttKuuab7732110cos22110cos10cos2.1晶体管足够大，=1kVtKuRVVuiiEEEabEC53.47.021对于输出端RLC网络：sradCRLc/1013解：场效应管开关准模拟乘法器University of Science and Technology of China125.6103分量：kVIC512122.11tZIuCo3110cos410cos2131tRILC410cos216.03t2/101/8LLabLowabHighRRUU%01.02/101/512122.151

6、21212.18LLRRVVUniversity of Science and Technology of China136.6University of Science and Technology of China146.6解：场效应管开关准模拟乘法器ttttKuuCab755110cos221102cos10cos3kVtKuRVVuiiEabEC33.551507.06515021sradLCo/1017sradRCBW/1021215RLC选频网络：University of Science and Technology of China156.6频率成分RLC选频网络：tt7510

7、cos10cos对于kVIC315150231频率成分对于tt7510cos102coskVIC315150212oLjRjH1o4.632arctantt7510cos102cos频率成分的相移University of Science and Technology of China166.6ttRIuoLCo75110cos4510cos212ttRIoLC75210cos4.63102cos5tttuooo75510cos4.63102cos0465.04510cos221.012综合以上分析：University of Science and Technology of China17

8、6.106.10 如图所示的双平衡差分电路中，设所有晶体管均为指数律的。试推导输出电压表达式。若欲用此电路产生DSB信号，u1、u2中哪个信号应该为载波？为什么？解：VT5 VT6组成负反馈的差分放大器：EEECCRuiiii265652VT1 VT2组成的差分放大器：rCrCCCuuiuuiii2tanh2tanh22151521VT4 VT3组成的差分放大器：rCrCCCuuiuuiii2tanh2tanh22161634University of Science and Technology of China186.10输出电压：31424231CCCCCCCCCCCCCCCoiiiiR

9、iiRViiRVurCCCouuiiRu2tanh165rECouuuRRu2tanh212u1为载波，u2为调制信号University of Science and Technology of China196.136.13 证明如图所示的正交复用系统，可以实现用2max的带宽传送两路带宽为max的不同的基带信号。ttsyxcos111 ttsyxsin222解：由图 ttsttsyxyxsincos212211University of Science and Technology of China206.13 ttsttsyxyxsincos212211在接受方：ttsttststts

10、tttsyx2cos2sin2121sincossin21222144 ttsttststttsttsyx2sin2cos2121cossincos21122133分别通过低通滤波器：tsuo11 tsuo22得证！University of Science and Technology of China216.14University of Science and Technology of China226.14解：ui处于正半周，VD1导通，VD2截止，uo=0ui处于负半周，VD2导通，VD1截止ARRRiAUuDTi11211)其中对于理想二极管，UT=0，RD=0 12120001

11、iRRRiuitKuRRtKRRRuutKuuCiCiiCbo12211ttttKuuCio88410cos22110cos10cos5.01University of Science and Technology of China236.14第二级RoCo组成低通滤波器截止频率：oooCRjRjZ1sradCRooC/1014212410cos25.01433tRRuRZuooo410cos354.0118.34tuo第二级输出：University of Science and Technology of China246.142)非理想二级管同理，通过第二级：kiVuARRRiAUuiD

12、Ti0012.10004.01121tKuuCbotKtKuuCCio000799.09976.021000799.0410cos354.0118.39976.04tuo410cos354.0117.34tuotKRRkVuuCii210012.10004.0University of Science and Technology of China256.16解：放电时间常数oooooCkRCRRR1/11252/1/1oooRRRRk交流负载与直流负载比：当=0.4时，k=0.80862.08.016.018.08.012maxmk会发生失随失真University of Science a

13、nd Technology of China266.16改变以满足要求：得到关于k的方程：关于的方程：012maxmk05.018.02kk87287.01252k025426.063565.0522978.00University of Science and Technology of China276.16当=0.2978时，等效的交流阻抗：输出电压：截止频率：kRRRRooo3645.41/1CoooojRCRjRjZ11sradCRooC/1025.215oC2425.21arctanarctan 1/8.02oooRRRZtsu125/25.2112410cos8.02ooooRR

14、Rtuootu2410cos286.05University of Science and Technology of China287.107.10 设 (V)，求下列两种情况下差分混频器的输出电压表达式：(1)(mv)(2)(V)解：差分对为受控“恒流源”偏置 RLC选频网络：ttuRF76108cos10cos16tuLO810cos26tuLO810cos6.2kVukVVuIRFRFK5.153.9500157.010sradLCo/10217sradCRBWL/1021216University of Science and Technology of China297.10(1)

15、(mv)tuLO810cos26 txIIKC81110cosZIuCo110 kttRxuLo5.15102cos410cos28.0162110761 tkVttx77611010cos5.153.9108cos10cos8.0161rLOUUx VkRxL456.05.156211ttuo76102cos410cos5657.01456.010其中：University of Science and Technology of China307.10(2)(v)差分对表现出动态限幅特性，电流开关特性:tuLO810cos6.2其中：tKIiKC810tkVtt87610cos2215.1

16、53.9108cos10cos8.016ZiuCo10 kttRuLo5.15102cos410cos28.0162211076VkRL232.15.156221ttuo76102cos410cos5657.01232.110University of Science and Technology of China317.147.14 求如图示电路的混频损耗LC。设所有的二极管均为理想二极管，且其通断受控于uLO(t)解：二极管无源混频器当uLO的正半周：oLORFDuuuu1oLORFDuuuu211Dgui 22Dgui LoRFLoRuugRiiu221122LRFLogRugRugRF

17、ouu 依题意，理想二极管University of Science and Technology of China327.14混频的中频分量：当u2的负半周，两个二极管同时截止，故：tKuuLORFottULORFRFcos221cosRFRFIFUUU1221LIFIFRUP22RFinRFRFRUP,22dBPPLIFRFC93.622中频输出功率：射频输入功率：LRFinRR,正半周：负半周：RFinR,LRFRFRUP2212University of Science and Technology of China338.98.9 在图所示电路中，设VT1、VT2的=1。(1)若 (

18、v)，求输出电压及调幅指数(2)若 (V)，求保证寄生调幅度1%，最大频偏的最大允许值。解：(1)差分对为恒流源偏置 RLC选频网络：ttui7410cos105.2cos5.0152.0tidtu04710cos10cos5.02026520riUUx ttxIiKC741110cos105.2cos5.01sradLCo/1017sradRCBW/105.221214University of Science and Technology of China348.9输出电压：ZiuCo19 ttRxIuKo74110cos4105.2cos25.019ttuo7410cos4105.2co

19、s354.01634.09 ttxIiKC741110cos105.2cos5.01University of Science and Technology of China358.9(2)：宽带调频信号通过RLC网络，不同频率表现的阻抗不同，产生寄生调幅效应，调幅度：mmax正比于中心频率表现的阻抗 mmin正比于最大频偏表现的阻抗%1minmaxminmaxmmmmojRjZ1%11122 RRRRsrad/10505.04University of Science and Technology of China368.118.11假定如图所示的变容管直接调频电路中，变容管的特性为 基带信

20、号为：pFuCj5.121540)(10cos1354mVt 其他参数见图。计算产生的调频波的中心频率fo、最大频偏、调频指数mf、信号带宽BWCR University of Science and Technology of China378.11解：Cj的直流偏置：VVUQ440604010pFUUCCQjjQ2015.10pFC57.34MHzLCfo56.6021University of Science and Technology of China388.11pFCopen51.9678.0openjQjQCCCApFCclose15571.0closejQjQCCCB03.0U

21、UUMQ00241.0211BAMDkHzDffo97.145166.91/maxfmkHzmBWfCR10.2952/12maxUniversity of Science and Technology of China398.168.16 如图所示为简化的相位鉴频器，设传递函数(1)写出VD1，VD2上高频电压的表达式；(2)用矢量图法讨论鉴频特性的形状；(3)若 ，求输出电压表达式。解：ojUUjHarctan2exp12)(tso21D1UUU21D2UUUo arctanUniversity of Science and Technology of China408.161U1U2U2U1DU2DUsin121UUDsin121UUD 12DDoUUtu ooUtuarctan21sin22 tsUtuo2sin1sin12U)(tsoUniversity of Science and Technology of China41谢谢大家！谢谢大家！