非线电(完整版)

1、University of Science and Technology of China非线性电子线路University of Science and Technology of China22250.5()iu uu mA3.5 ,0.8cos( )QQUV u Ut V解:23250.58.38.32.3753.56.3755 21.5QQQQQQQQQu Uu UUUUmAImAmSUVigmsuuuIG 2350.5()iu uumA将代入并计算得:214 1.60.192 1.25.292QQUUmAI 0.8( )QuCOSwt VU5.29216.61510.81IGmsmU

2、 1.2 （2）University of Science and Technology of China3 550.5cos21.5 102cos22 10 ( )utt V 551221.5 10 ,22 10wtt w tt 2350.5iuuu解: 设125(0.5cos2cos)tt23(0.5cos2cos)0.5(0.5cos2cos)1212wtw twtw t计算得直流分量: 2.125ms212:0.9531;:608125;2 :0.125;2:2;1wmAwmA wmA WmA1212:1;3: 0.015625;3: 1;wwmA wmA wmA12122: 0.09

3、375;: 0.75.wwmA wwmA0.5cos2cos12uwtw t1.3 （2）University of Science and Technology of China40.520(1 0.25 )();(1,3)jCupF u0.52020 1 (1 0.25 )jCCu10Lh60.512661110 10201 (1 0.25 )101,8.17 10,3,8.49 10.fHZLCuufHZufHZ 1.4 （2）University of Science and Technology of China（2）静态时:0.984910.022222,(1)CBEBIIII22

4、22(2)0.56CBBEEIIUI解得:22.05KCIImAln778.85CEOBErrBECEOUUUUKESESKCEOrESUUIIeIeIUUmVI 解得:（1）恒流源:2.1University of Science and Technology of China20irUxU00( )BErUUEESKIIeIxI200(20)0.08978Ie0lnln(20)321.52322KBErESCEBEIUUImvIUUmv 解得:查表得:2.12.1（3）由University of Science and Technology of China10irUxU 10102Em

5、rIxIGUxIx所以:查表得:2.22.2（1）由7110LC网络调谐于基波频率50TQRC 101020.18972xIxxIx17.3mGms 7110cos104.74cos10omiLutG U Rtt /1.27%TTHDD xQV VUniversity of Science and Technology of China 330030022EEmriIxIxIIGUxIxUIx所以:2.22.2（2）由713 10LC 网络调谐于三次谐波频率50TQRC 7310cos103.12cos3 10omiLutG U Rtt 32133134.2%9nnTnIxnTHDQIx nV

6、 V223319nnnnnnUniversity of Science and Technology of China2123BBBCCBBRUVVRR12|33.33BBBRRRK2.3BBBEQUUUv1.071EQBEUImARR40.3EQmQIgmSU88.5rUxU150(15)0.1039Ie11502 ( )0.12881( )xI xxIx15irUxU0110ln( ) 2 ( )( )15.94( )mmQIxI xGxgmsxxIx010( )( )2.32cos( )miu tGx Ruw t v查表得:20()22.69LUPmWR电阻分压式偏置电路:2.32.3U

7、niversity of Science and Technology of China2.4 00exp()EBrUEESuIIIx05EEKUIIk0111002 ( )2 ( )( )13.487( )( )EmmQrII xI xGxgmsxIxuxIx010( )9( )cosmLiu tvGxRut 095.26cost V对9V电压源进行诺顿等效10/52KIVkmA 5Rk 570.2EBEUUmV01.886EImA静态时，x =10解得又所以University of Science and Technology of China80110/rad sLC010TQRC2i

8、rUxU100110188188( )(2)cos10104.02cos10 ( )60( )( 4.02cos10 )2.68cos10 ( )60301( )0.25%oCCKLoTu tVI aRtt vu ttt vTHDD xQ 121220C CCPFCC2KImA2.5恒流源偏置差分放大电路:输出端RLC谐振回路:University of Science and Technology of China0oU 2.7120.5iUVVD VD当 时， 截止1220.5iVUVVDVD当 时，导通 截止2112cos333oiUUt122iUVVD VD当 时， 导通13cos22

9、oiUUt-1.5-1-0.500.511.522.533.54-1-0.500.511.522.53/ t/oUV/iUV/oUV-1-0.500.51-3-2-10123University of Science and Technology of China132.82.8 增强型MOSFET管的转移特性为：式中 。求下列情况下等效基波跨导Gm1：（1）（2）2/8 . 0VmAnVuVuuiGSGSGSnD202)2(2);(cos3cos1VttUUuQGS).(cos23cos1VttUUuQGSUniversity of Science and Technology of Chi

10、na14解:(1)mSVmAUIGmAItttVVtuDmDGS6 . 116 . 16 . 12cos4 . 0cos6 . 12 . 1)2cos3(8 . 0iMOS)(2)(cos311112D管工作在完全平方律区此时University of Science and Technology of China15(2)mSUIGmAIImAUUIIUUUVVuVtuDmPDnPDSSPQpGSGS728. 12457. 3457. 3)120(2 . 7)120cos1 (224)cos1 (120232coscosMOS21),(cos23111112222212111min管工作在导

11、通角状态。此时University of Science and Technology of China162.10 2.10自生负偏压场效应管放大电路如图E2.8所示。场效应管的参数为：IDSS=6mA，UP=-4V。又知漏极回路的中心频率 ，带宽 。若已知漏极电流的均值分量为1.8mA。求： (1)回路压降幅度及其总谐波失真THD； (2)场效应管的平均功耗和电压增益。srad /1070sradBW/1055University of Science and Technology of China17解:(1)mAIImAIImAIIUUUUUUtUUuPDPDDSSPQPGS799.

12、2)(1103 . 0)(8 . 1)(6)41 (cos)(4coscoscos11000111111111得到由20,8 . 100BWQmAITD%4 . 1)(Q1THD2.981VU2V.11RIUT1D1oDUniversity of Science and Technology of China18(2)76. 3981. 22 .11cos2 .1115u32.11)cos8 . 28 . 1 ( )cos2 .1115(21210020020UiUoAutmWtdtttdiuPdsDds其中University of Science and Technology of Chi

13、na192.12 2.12在如图E2.10所示电路中，场效应管IDSS=10mA，UP=-4V；CG=0.1F,RG=1M；回路中心频率为f0，QT=20。输入信号 计算以下两种情况下的输出电压表达式：(1)(2)(2cos),(2cos222111VtfUuVtfUu;465,1,2 . 0,465. 1,202211kHzfMHzfVUMHzfVU。102211,250,3 . 0,10,38ffkHzfVUMHzfVUUniversity of Science and Technology of China20解:(1)(104652cos625. 020)(2cos2025. 0215

14、 . 2)(,500) 14(5) 1(2)1 (cos22cos31221112cos222cos222111111VttffRIumAUgImSggmSgUuuUuUuUIuigUuIittUUUIFoIFPPPGSGSPGStfuPGSPDSStfuGSDPGSDSSDGSGSGS180224cos1University of Science and Technology of China21(2)mAUgImSggUuuUuuigmAIItUfffffukHzQfBWIFPPGSGSPGStUUuGSDPDGSoTGS40225. 0216815. 2)120(5)(00)41 (58

15、01. 4)120(10)(12038384coscos3838,50020101021111cos111112121160111包含University of Science and Technology of China22)(102cos)4102502cos(11847. 01 1220)4)cos(21)4)cos(21cos20442125007321211121212VtttRItRItRIufffffKHzBWIFIFDo故。和相应的相位差分别偏移，应的幅度为落在通带的边缘，其对，故University of Science and Technology of China232

16、.14 2.14在如图E2.12(a)所示电路模型中，耦合电容足够大，VD为指数律二极管，其IS=2exp(-30)mA，ui(t)为图(b)所示的周期波形。 (1)画出输出电压波形并注明电平值； (2)求输入电压低电平为-520mV时电路的自生负偏压系数。University of Science and Technology of China24解:(1)由图有： rirCrirCriCroUUUUSDUUUUSUUUSUUSDDCDCeeIieeIeIeIiiiii所以因为即：的均值电流为流经耦合电容*020,C2mVUIUUieeeIimVUeedtdteeiSrCDUUUUUUUSD

17、iUUUUUUUUriCrirCriririri58026084020ln:,(*)101101)260(101109101130130130330有式代入将故：此处University of Science and Technology of China25所以mVumVUmVuUmVuUuuuuoioiiCiCio840,260580,0)(580时当时当University of Science and Technology of China26(2)同(1)中)9ln(20ln2)109101(1091011301301230330ririrCririririrCUUSrCUUUUSU

18、UUUUUUUUUSDeIUUeeIedtdteemAeeIi故有)(19520参考方向相反和iCmVUiCUUUUiCUUdUdUeedUdUiririUniversity of Science and Technology of China272.16 2.16在如图E2.14所示电路中，C=100F，r=50，非线性电阻特性为 。计算当输入信号为图E2.14(a)所示时，电阻r所消耗的功率。)(232mAuuioooUniversity of Science and Technology of China28解：由图有6 . 12322)238(2322232300C22222CciiC

19、CCoooooororCiCoUUuuUUUuuuuruiiiiiuUu而即：的均值电流为稳态时，流经电容mWdtudtudtruUUVuiVuioCcii1 .1605. 0)07. 0(20105. 0)07. 0(201201P-0.07VU061.23221210222022002rC2得到所以University of Science and Technology of China292.18 2.18求如图E2.16所示电路在不同输入和不同负载条件下回路的有载QT及输出电压uo表达式。 (1) ,负载如图(a)所示，VD为指数律二极管。t(mA)4cos10(t)i7iUniver

20、sity of Science and Technology of China30解：xUxIxIUUxxIxImAxIxIIIUGUIIoroooooio00325. 01125. 01)()()3()2()()(22)()(2) 1 (5 . 040101010111联立上面三个方程有：基波非线性负载方程为基波线性负载方程为：University of Science and Technology of China31)(10cos3286. 0)(3.84mAI)2(3286V. 0U57.1295775. 000325. 01338V. 0U96072. 0)()(13961. 000

21、325. 01312V. 0U95736. 0)()(127o1oo01o01VttuxxxIxIxxxIxIxo故：式有：，代入，最后求得，时，当，时，当因为University of Science and Technology of China3233.1675.115 . 05 . 040040075.113268. 084. 301NSNSNSTooNSGGGGGGQGGCQmSVmAUIGUniversity of Science and Technology of China2.18（1）VD为指数律二极管 74cos10ii tt2,0.5,0.02RkLHCF7110/rad

22、 sLC0400QCRTQ假设足够高 1cosooutUt流过VD直流分量2DKiImA基波线性负载方程：11ooiUIIR 1102oKIxIIIx126oUxmV由指数律：10.3268oUV11/11.74NSooGIUms016.34TNSGQQGGUniversity of Science and Technology of China2.18（2） 74cos10ii ttTQ假设足够高 1cosooutUt非线性器件导通角13arccosoU基波线性负载方程：11ooiUIIR基波电流：13.73oUV11/0.304NRooGIUms0249TNSGQQGG1211693cos

23、1ooImAdmAU 11.135oImA 3.73cosouttV3TVV开启电压University of Science and Technology of China2.18（3） 4iii t导通角/ ii t激励电流源基波分量 1cosooutUt1arccosToUU基波线性负载方程：111ooiUIIR11.152oUV11/0.284NRooGIUms0256TNSGQQGG1142cossin0.903iiiiImAdmA 10.327oImA 1.152cosouttVTQ 足够高折线二极管导通角：1PoTIG UU 11oPII 29.77oUniversity of

24、Science and Technology of China363.13.1如图所示的甲类功率放大器电路中，若不计变压器损耗，且晶体管UCES=0，ICEO=0，发射极电阻RE上滞留压降可以忽略不计，放大器输入信号充分激励。（1）若输出变压器Tr2的匝数比n=4，放大器实现最佳匹配。求RL所获不失真最大功率PO及相应的放大器效率c；（2）在维持ICQ不变的条件下，不用变压器，直接将扬声器接到集电极，再求不失真最大输出功率PO和c；（3）在维持ICQ不变的条件下，将变压器外接负载增大到16，求此时的不失真最大输出功率PO和c；University of Science and Technolo

25、gy of China373.1VVVcco12max1282LLRnRmARVILCCCQ75.93解：（1）mVRVPLOO5 .5622122maxmax%50maxCQCCOCIVP（2）mAICQ75.938LRmVRIPLCQO16.35212max%125. 3maxCQCCOCIVP（3）2562LLRnRmVRVPLCCO25.281212max%25maxCQCCOCIVPUniversity of Science and Technology of China380.45LIA3.3如图所示为互补对称功放电路，设在正弦激励的情况下，负载电流幅度为 ，求：（1）负载获得的功

26、率（2）电源提供的功率PDC和放大器效率C解：0.45LIA213.542LLLPIRW电源的平均电流为 的均值分量LICCII25.012CCCCDCCCVIPIVW70.7%LCDPPUniversity of Science and Technology of China39如图所示电路中，设输入信号足够大且不考虑晶体管安全（1）说明图中各二极管和电阻的作用（2）设晶体管饱和压降UCES=0.3V，RL=8，求负载最大不失真功率150.314.7OUV213.512OOUPWR3.4解:VT1、VT2组成互补乙类 推挽电路；VT3推动级 VT4、VT5两个集电极短接的三极管及二极管；提供

27、偏置减少交越失真R1 R2 提供偏置University of Science and Technology of China403.53.5 如图所示的有输入缓冲级的互补对称乙类放大器，设R1=R2=2k，RL=100，所有晶体管=60，UBE(npn)=UEB(pnp)=0.6V，计算电流增益。（提示：VT1、VT2、VT3、VT4分别组成射极跟随器。因此有uo=ui）University of Science and Technology of China413.5解：由于乙类放大器，故VT1，VT2始终导通，VT3，VT4轮流导通。 当输入信号正半周时，VT3导通，对交流信号有：06

28、. 01/1131RUiiiEb12biii LiEoRUii338.459ioiiiA联立以上方程可以解得：University of Science and Technology of China42cos090oTBbUEU CCCUVcos1CCCCCCcrCmbVUVUG Ug U(1)0.48mbcrCCg UG UA(1 cos )0.48CPmbmbIg Ug UA3.9谐振功率放大器工作在临界状态，已知VCC=30V，-EB=UT=0.6V，=0.96，临界动态线斜率Gcr=0.4S。求 ，Po，RT，PC的值CM M点横点横标标University of Science a

29、nd Technology of China430001.128CDCCCCPPPI UPW01111( )3.45622CCCPCCPI UIVW 2012120CTURP00( )CCPII 101001( )275.4%2( )CCCDCCCCI UPPI V 3.911( )CCPII University of Science and Technology of China443.103.10 一谐振功率放大器工作在欠压状态。若已知VCC=12V，RT=110，PO=500mW，c=68.5%。为了进一步提高效率，在维持VCC，RT，PO不变得前提下，将电流的通角减小到60o，并使放

30、大器不工作在饱和区。（1）这时，c转换效率多大？（2）如何调整放大器，才能使放大器的电流通角减小到60o？ CCccVU0121mWRUPTcO500212解：（1）VCC RT PO不变 ，得Uc Ic不变VUc49.10%4 .7860602101CCcoocVUUniversity of Science and Technology of China453.10（2）bTBUVE arccosEB增大或Ub减小？)cos1 (bmCPUgI)(1CPcII 如果Ub减小，则减小，与题设矛盾 University of Science and Technology of China463.

31、11解：放大器依然工作在临界状态：A 放大器正常工作，晶体管没有损坏，饱和区特性没有改变，又UBEMAX没有改变，故知放大器依然临界B Uc没有变化，ICP没有变化，M点位置没有变，依然临界。University of Science and Technology of China473.11 CCccVU0121 %60%687258. 170700101010101oocc 9559. 101o89.26 112121CPccCoIUUIP 4356. 01940. 07089.261111ooooPPWPo3375. 1由放大器效率公式：输出的交流功率：University of Sci

32、ence and Technology of China483.11性能变化的原因：TcoRUP221减小，故RT增大University of Science and Technology of China49(1) 放大器工作在欠压状态(2) 由图知: 34 1414242CCVV34 14102CUV0.80.165crAGSV0.10.11bmmbccUgg UUVU 0.8CPIA078.46cos10.2CPmbIg U 3.12University of Science and Technology of China500.8120.811.2cCCUVV0.4cos110.50

33、.8CPmbIg U 06011( )0.1564CCPIIA 011875.842CcPI UmW10( )183.7%2( )C 解：由图知 : 3.13假设谐振功放的输出特性如图中的折线MNK所示，试根据动态特性计算功放的输出功率和转换效率University of Science and Technology of China513.14University of Science and Technology of China523.16University of Science and Technology of China533.16解：（1）由串并联转换：LLeRXQ11TCeR

34、XQ2211111LeLXQX22211CeCXQXLeLRQR211TeTRQR221LTRR 11212eTLeQRRQ0111211CCLXXX匹配条件：12111111CLCXXCjX12112CCL谐振条件：故：University of Science and Technology of China543.16（2）将数据代入（1）中计算：1700121LeLTRQRR40011eLLQRXHL27. 11408. 211212eTLeQRRQpFC29. 52pFC98. 21University of Science and Technology of China554.1(a

35、)- - - - + + + + 不符合起振相位条件不符合起振相位条件University of Science and Technology of China564.1(b)+ + - - - -+ + 符合起振相位条件符合起振相位条件University of Science and Technology of China574.1(c)+ + + + + + + + + + 符合起振相位条件符合起振相位条件University of Science and Technology of China584.1(d)符合起振相位条件符合起振相位条件当当L3L3、C C串联的阻抗串联的阻抗呈容性

36、时呈容性时University of Science and Technology of China594.1(e)不符合起振相位条件不符合起振相位条件University of Science and Technology of China604.1(f)当当L1L1、C2C2并联阻抗并联阻抗呈感性呈感性且且L2L2、C1C1并联阻抗并联阻抗呈容性时呈容性时符合起振相位条件符合起振相位条件University of Science and Technology of China614.1(g)不符合起振相位条件不符合起振相位条件University of Science and Techno

37、logy of China624.2(a)+ + + + + + - - -* *University of Science and Technology of China634.2(b)+ + - -+ + * *University of Science and Technology of China644.2(c)+ + + + + + * *University of Science and Technology of China654.3(a)不能起振不能起振University of Science and Technology of China664.3(b)能够起振能够起振Un

38、iversity of Science and Technology of China674.3(c)能够起振能够起振University of Science and Technology of China684.3(d)能够起振能够起振University of Science and Technology of China694.4(a)X X X X University of Science and Technology of China704.4(a)改正后改正后University of Science and Technology of China714.4(b)X X X X

39、 X X X X University of Science and Technology of China724.4(b)改正后改正后University of Science and Technology of China734.4(c)X X X X University of Science and Technology of China744.4(c)改正后改正后University of Science and Technology of China754.54.5按下列要求，画出实际的振荡器电路。(1)基极调谐、集电极耦合的变压器耦合振荡器，采用PNP三极管，正电源工作，发射极交

40、流接地。University of Science and Technology of China764.5 (2)负电源工作，基极交流接地，交流等效电路如图所示University of Science and Technology of China774.5University of Science and Technology of China784.6University of Science and Technology of China794.6(1)University of Science and Technology of China804.6(2)pFpFFFpF1307C

41、606C:)2)(1 ()2(464. 0CC2CC,2) 1 (414CCCCCCCCCC212121212121210式得由即：不变：不变，则应有University of Science and Technology of China814.6(3) (4) (6) (7) (3) 导线不可以，会将C1交流短路 高频扼流圈可以，起到隔交流通直流的作用 (4) 只能去掉一个，起到对反馈回路隔直流的作用 (6) 2端测量比较合理，频率计探头具有接入电容，在1端会影响电路谐振频率 (7) UBEUBEO 电路振荡以后BE结处于大电流状态，压降有所降低晶体管工作特性 若相等说明电路没有起振Uni

42、versity of Science and Technology of China824.6(5)EminibinmbbeEbembbeiRgguiggiuRugiuu)1 (/1:/)(得到iniciinbcguigugiiUniversity of Science and Technology of China8321122121CCCnCCCnUniversity of Science and Technology of China84gugnuino/1)( 22212inBCLgGnGnGng377 1/ F21212211EinfinofRggnnuuTgnugnFuunCCC求

43、得反馈系数University of Science and Technology of China854.10University of Science and Technology of China864.10LLn221LLL/2mELCgGnGGg1gngAFmmsggmm044.11min电阻分压式：msRRUUVUIgBErBEQBBrEQmQ457.561教材P13页University of Science and Technology of China874.10谐振频率：kHzCLLfOSC35.6432121振荡稳定时： min1mmgxG 1965. 0457.5604

44、4.11min1msmsgggxGmQmmQmmVnUUUxorf26223.11VUo75. 1University of Science and Technology of China884.12University of Science and Technology of China894.12333. 02111CCCn9255. 067.668 . 68 . 612332CCCn/2122mECgGnGnggUgnUebmo/2ggnUUAmebo2031. 020017. 62123CCUUFof1AFmsggmm556. 3minUniversity of Science and

45、 Technology of China904.12电阻分压式：msRRUUVUIgBErBEQBBrEQmQ66.101 334. 066.10556. 3min1msmsgggxGmQmmQmVFxUUro67. 785. 8xMHzCCLfOSC5 .612141232754123gCCQosco %2 . 01xDQTHDo输出电压：谐振频率：University of Science and Technology of China914.13解：（1）VUB3 . 2mAIK22104CiUmVU（2）4riUUx10/1/11222122CCCCUUiRVUR04. 12Unive

46、rsity of Science and Technology of China924.13kILMURCR65. 41122 mAxIIKC118. 111（3）VRIUCo47. 415 . 612LRQo %98. 01xDQTHDUniversity of Science and Technology of China934.14(差分对右侧差分对右侧)871.19 0.1278 423. 9204. 1)( 1: min1min2xmsmsgggxggggngLMnmdQmdmdQmdmdmdmd起振条件University of Science and Technology of

47、China4.14(差分对左侧差分对左侧)94 v0.203)(LC3958. 721322LooscoRxaufMHzCLf输出电压为：的三次谐波谐振回路提取振荡信号University of Science and Technology of China954.15University of Science and Technology of China964.15解：（1）tUuoocos非线性电阻特性特性曲线： ooufi 01coscos2dUfIoooU1arccosoooUUI21sin11211arccos111121ooonUUUG振荡平衡条件：msRGGn5 . 011Un

48、iversity of Science and Technology of China974.15（2）：同（1）理VUo9093. 2mWGUPnoo116. 22112解得：msUGon5 . 043621VUo708. 2mWPo833. 1University of Science and Technology of China98思考的一个问题：思考的一个问题：非线性电阻的特性方程： ufi tUUuoQocos电压激励： 求：Gn1方法二： 方法一： tUUfufioQoocostItIIooo2coscos21oonUIG11 tUUuoQuftgcostgtgg2coscos2

49、10?11gGnUniversity of Science and Technology of China99tgtggtg2coscos)(210tUUtgtggioQocos2coscos210方法二应该这样计算： 再解4.15（1）： tttg2cos22sincossin22)(42sin2211oonUIGmWPVUoo137. 2924. 2University of Science and Technology of China1004.16University of Science and Technology of China1014.16解：（1）直流负载线方程QQQIRU

50、RUE21工作在负阻区：VUVQ6 . 02 . 0VUQ4 . 0取：保证直流负载线与负阻特性只有一个交点： 5 . 12121RRRR22121RRRRkR31. 21kR638. 02取：可解得：University of Science and Technology of China1024.16（2）振荡频率：MHzCCLfdOSC45.4821msRGGn2 . 011平衡条件：University of Science and Technology of China1035.3University of Science and Technology of China1045.3解

51、：(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 China1055.3当u2的负半周1221LLogRugRu1122ugRKKgRuKuKuuLLoooo综上：University of Science and Technology of China1065.3(2) 同(1)KuKuuooo综上：122111LLoRguRgu122212LLoRguRgu1212222111

52、LLLoRgKgRgKguRuUniversity of Science and Technology of China1075.3(3)开关控制信号u2u2正半周：oDuuuu211oDuuuu21411Dgui 44Dgui LoLoRuugRiiu14121221LLogRugRuKgRugRKuuLLoo1221University of Science and Technology of China1085.5 (c)5.5 试分别求出如图所示电路输出电压的表达式。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

53、通，VD2截止 oSLDuuuu111DDugi oSLLDLouuuRgRiu1LSLDLDouuRgRgu1University of Science and Technology of China1095.5 (c)当uL负半周，VD1截止，VD2导通，同理：SLLDLDouuRgRgu1LSLDLDoooouKKuRgRguKuKuu1综上：University of Science and Technology of China1105.6University of Science and Technology of China1115.6ttttKuuab7732110cos221

54、10cos10cos2 . 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 China1125.6103分量：kVIC512122 . 11tZIuCo3110cos410cos2131tRILC410cos216. 03t2/101/8LLabLowabHighRRUU%01. 02/101/512122 . 15121212 . 18LLRRVVUniversity of Science and Tec

55、hnology of China1136.6University of Science and Technology of China1146.6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 China1156.6频率成分RLC选频网络：tt7510cos10cos对于kVIC315150231频率成分对于tt7510c

56、os102coskVIC315150212oLjRjH1o4 .632arctantt7510cos102cos频率成分的相移University of Science and Technology of China1166.6ttRIuoLCo75110cos4510cos212ttRIoLC75210cos4 .63102cos5tttuooo75510cos4 .63102cos0465. 04510cos221. 012综合以上分析：University of Science and Technology of China1176.106.10 如图所示的双平衡差分电路中，设所有晶体管

57、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 China1186.10输出电压：31424231CCCCCCCCCCCCCCCoiiiiRiiRViiRVurCCCouuiiRu2tanh1

58、65rECouuuRRu2tanh212u1为载波，u2为调制信号University of Science and Technology of China1196.136.13 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 China1206.13 ttsttsyxyxsincos212211在接受方： ttsttststtstttsyx2cos2sin2121sinc

59、ossin21222144 ttsttststttsttsyx2sin2cos2121cossincos21122133分别通过低通滤波器： tsuo11 tsuo22得证！University of Science and Technology of China1216.14University of Science and Technology of China1226.14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 12120001iRRRiuitKuRRtKRRRu

60、utKuuCiCiiCbo12211ttttKuuCio88410cos22110cos10cos5 . 01University of Science and Technology of China1236.14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 China1246.142) 非理想二级管同理，通过第二级 ：kiVuARRRiAUuiDTi0012.