非線電(完整版)

1、University of Science and Technology of China非線性電子線路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將代入并計算得: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: 設(shè)125(0.5cos2cos)tt23(0.5cos2cos)0.5(0.5cos2cos)1212wtw twtw t計算得直流分量: 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）靜態(tài)時: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網(wǎng)絡(luò)調(diào)諧于基波頻率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 網(wǎng)絡(luò)調(diào)諧于三次諧波頻率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電阻分壓式偏置電路: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對9V電壓源進行諾頓等效10/52KIVkmA 5Rk 570.2EBEUUmV01.886EImA靜態(tài)時，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電路:輸出端RLC諧振回路:University of Science and Technology of China0oU 2.7120.5iUVVD VD當(dāng) 時， 截止1220.5iVUVVDVD當(dāng) 時，導(dǎo)通 截止2112cos333oiUUt122iUVVD VD當(dāng) 時， 導(dǎo)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 增強型MOSFET管的轉(zhuǎn)移特性為：式中 。求下列情況下等效基波跨導(dǎo)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區(qū)此時University of Science and Technology of China15(2)mSUIGmAIImAUUIIUUUVVuVtuDmPDnPDSSPQpGSGS728. 12457. 3457. 3)120(2 . 7)120cos1 (224)cos1 (120232coscosMOS21),(cos23111112222212111min管工作在導(dǎo)

11、通角狀態(tài)。此時University of Science and Technology of China162.10 2.10自生負偏壓場效應(yīng)管放大電路如圖E2.8所示。場效應(yīng)管的參數(shù)為：IDSS=6mA，UP=-4V。又知漏極回路的中心頻率 ，帶寬 。若已知漏極電流的均值分量為1.8mA。求： (1)回路壓降幅度及其總諧波失真THD； (2)場效應(yīng)管的平均功耗和電壓增益。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在如圖E2.10所示電路中，場效應(yīng)管IDSS=10mA，UP=-4V；CG=0.1F,RG=1M；回路中心頻率為f0，QT=20。輸入信號 計算以下兩種情況下的輸出電壓表達式：(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。和相應(yīng)的相位差分別偏移，應(yīng)的幅度為落在通帶的邊緣，其對，故University of Science and Technology of China232

16、.14 2.14在如圖E2.12(a)所示電路模型中，耦合電容足夠大，VD為指數(shù)律二極管，其IS=2exp(-30)mA，ui(t)為圖(b)所示的周期波形。 (1)畫出輸出電壓波形并注明電平值； (2)求輸入電壓低電平為-520mV時電路的自生負偏壓系數(shù)。University of Science and Technology of China24解:(1)由圖有： rirCrirCriCroUUUUSDUUUUSUUUSUUSDDCDCeeIieeIeIeIiiiii所以因為即：的均值電流為流經(jīng)耦合電容*020,C2mVUIUUieeeIimVUeedtdteeiSrCDUUUUUUUSD

17、iUUUUUUUUriCrirCriririri58026084020ln:,(*)101101)260(101109101130130130330有式代入將故：此處University of Science and Technology of China25所以mVumVUmVuUmVuUuuuuoioiiCiCio840,260580,0)(580時當(dāng)時當(dāng)University of Science and Technology of China26(2)同(1)中)9ln(20ln2)109101(1091011301301230330ririrCririririrCUUSrCUUUUSU

18、UUUUUUUUUSDeIUUeeIedtdteemAeeIi故有)(19520參考方向相反和iCmVUiCUUUUiCUUdUdUeedUdUiririUniversity of Science and Technology of China272.16 2.16在如圖E2.14所示電路中，C=100F，r=50，非線性電阻特性為 。計算當(dāng)輸入信號為圖E2.14(a)所示時，電阻r所消耗的功率。)(232mAuuioooUniversity of Science and Technology of China28解：由圖有6 . 12322)238(2322232300C22222CciiC

19、CCoooooororCiCoUUuuUUUuuuuruiiiiiuUu而即：的均值電流為穩(wěn)態(tài)時，流經(jīng)電容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求如圖E2.16所示電路在不同輸入和不同負載條件下回路的有載QT及輸出電壓uo表達式。 (1) ,負載如圖(a)所示，VD為指數(shù)律二極管。t(mA)4cos10(t)i7iUniver

20、sity of Science and Technology of China30解：xUxIxIUUxxIxImAxIxIIIUGUIIoroooooio00325. 01125. 01)()()3()2()()(22)()(2) 1 (5 . 040101010111聯(lián)立上面三個方程有：基波非線性負載方程為基波線性負載方程為：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，時，當(dāng)，時，當(dāng)因為University of Science and Technology of China3233.1675.115 . 05 . 040040075.113268. 084. 301NSNSNSTooNSGGGGGGQGGCQmSVmAUIGUniversity of Science and Technology of China2.18（1）VD為指數(shù)律二極管 74cos10ii tt2,0.5,0.02RkLHCF7110/rad

22、 sLC0400QCRTQ假設(shè)足夠高 1cosooutUt流過VD直流分量2DKiImA基波線性負載方程：11ooiUIIR 1102oKIxIIIx126oUxmV由指數(shù)律：10.3268oUV11/11.74NSooGIUms016.34TNSGQQGGUniversity of Science and Technology of China2.18（2） 74cos10ii ttTQ假設(shè)足夠高 1cosooutUt非線性器件導(dǎo)通角13arccosoU基波線性負載方程：11ooiUIIR基波電流：13.73oUV11/0.304NRooGIUms0249TNSGQQGG1211693cos

23、1ooImAdmAU 11.135oImA 3.73cosouttV3TVV開啟電壓University of Science and Technology of China2.18（3） 4iii t導(dǎo)通角/ ii t激勵電流源基波分量 1cosooutUt1arccosToUU基波線性負載方程：111ooiUIIR11.152oUV11/0.284NRooGIUms0256TNSGQQGG1142cossin0.903iiiiImAdmA 10.327oImA 1.152cosouttVTQ 足夠高折線二極管導(dǎo)通角：1PoTIG UU 11oPII 29.77oUniversity of

24、Science and Technology of China363.13.1如圖所示的甲類功率放大器電路中，若不計變壓器損耗，且晶體管UCES=0，ICEO=0，發(fā)射極電阻RE上滯留壓降可以忽略不計，放大器輸入信號充分激勵。（1）若輸出變壓器Tr2的匝數(shù)比n=4，放大器實現(xiàn)最佳匹配。求RL所獲不失真最大功率PO及相應(yīng)的放大器效率c；（2）在維持ICQ不變的條件下，不用變壓器，直接將揚聲器接到集電極，再求不失真最大輸出功率PO和c；（3）在維持ICQ不變的條件下，將變壓器外接負載增大到16，求此時的不失真最大輸出功率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如圖所示為互補對稱功放電路，設(shè)在正弦激勵的情況下，負載電流幅度為 ，求：（1）負載獲得的功

26、率（2）電源提供的功率PDC和放大器效率C解：0.45LIA213.542LLLPIRW電源的平均電流為 的均值分量LICCII25.012CCCCDCCCVIPIVW70.7%LCDPPUniversity of Science and Technology of China39如圖所示電路中，設(shè)輸入信號足夠大且不考慮晶體管安全（1）說明圖中各二極管和電阻的作用（2）設(shè)晶體管飽和壓降UCES=0.3V，RL=8，求負載最大不失真功率150.314.7OUV213.512OOUPWR3.4解:VT1、VT2組成互補乙類 推挽電路；VT3推動級 VT4、VT5兩個集電極短接的三極管及二極管；提供

27、偏置減少交越失真R1 R2 提供偏置University of Science and Technology of China403.53.5 如圖所示的有輸入緩沖級的互補對稱乙類放大器，設(shè)R1=R2=2k，RL=100，所有晶體管=60，UBE(npn)=UEB(pnp)=0.6V，計算電流增益。（提示：VT1、VT2、VT3、VT4分別組成射極跟隨器。因此有uo=ui）University of Science and Technology of China413.5解：由于乙類放大器，故VT1，VT2始終導(dǎo)通，VT3，VT4輪流導(dǎo)通。 當(dāng)輸入信號正半周時，VT3導(dǎo)通，對交流信號有：06

28、. 01/1131RUiiiEb12biii LiEoRUii338.459ioiiiA聯(lián)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諧振功率放大器工作在臨界狀態(tài)，已知VCC=30V，-EB=UT=0.6V，=0.96，臨界動態(tài)線斜率Gcr=0.4S。求 ，Po，RT，PC的值CM M點橫點橫標標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 一諧振功率放大器工作在欠壓狀態(tài)。若已知VCC=12V，RT=110，PO=500mW，c=68.5%。為了進一步提高效率，在維持VCC，RT，PO不變得前提下，將電流的通角減小到60o，并使放

30、大器不工作在飽和區(qū)。（1）這時，c轉(zhuǎn)換效率多大？（2）如何調(diào)整放大器，才能使放大器的電流通角減小到60o？ CCccVU0121mWRUPTcO500212解：（1）VCC RT PO不變 ，得Uc Ic不變VUc49.10%4 .7860602101CCcoocVUUniversity of Science and Technology of China453.10（2）bTBUVE arccosEB增大或Ub減小？)cos1 (bmCPUgI)(1CPcII 如果Ub減小，則減小，與題設(shè)矛盾 University of Science and Technology of China463.

31、11解：放大器依然工作在臨界狀態(tài)：A 放大器正常工作，晶體管沒有損壞，飽和區(qū)特性沒有改變，又UBEMAX沒有改變，故知放大器依然臨界B Uc沒有變化，ICP沒有變化，M點位置沒有變，依然臨界。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：輸出的交流功率：University of Sci

32、ence and Technology of China483.11性能變化的原因：TcoRUP221減小，故RT增大University of Science and Technology of China49(1) 放大器工作在欠壓狀態(tài)(2) 由圖知: 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：由圖知 : 3.13假設(shè)諧振功放的輸出特性如圖中的折線MNK所示，試根據(jù)動態(tài)特性計算功放的輸出功率和轉(zhuǎn)換效率University of Science and Technology of China513.14University of Science and Technology of China523.16University of Science and Technology of China533.16解：（1）由串并聯(lián)轉(zhuǎn)換：LLeRXQ11TCeR

34、XQ2211111LeLXQX22211CeCXQXLeLRQR211TeTRQR221LTRR 11212eTLeQRRQ0111211CCLXXX匹配條件：12111111CLCXXCjX12112CCL諧振條件：故：University of Science and Technology of China543.16（2）將數(shù)據(jù)代入（1）中計算：1700121LeLTRQRR40011eLLQRXHL27. 11408. 211212eTLeQRRQpFC29. 52pFC98. 21University of Science and Technology of China554.1(a

35、)- - - - + + + + 不符合起振相位條件不符合起振相位條件University of Science and Technology of China564.1(b)+ + - - - -+ + 符合起振相位條件符合起振相位條件University of Science and Technology of China574.1(c)+ + + + + + + + + + 符合起振相位條件符合起振相位條件University of Science and Technology of China584.1(d)符合起振相位條件符合起振相位條件當(dāng)當(dāng)L3L3、C C串聯(lián)的阻抗串聯(lián)的阻抗呈容性

36、時呈容性時University of Science and Technology of China594.1(e)不符合起振相位條件不符合起振相位條件University of Science and Technology of China604.1(f)當(dāng)當(dāng)L1L1、C2C2并聯(lián)阻抗并聯(lián)阻抗呈感性呈感性且且L2L2、C1C1并聯(lián)阻抗并聯(lián)阻抗呈容性時呈容性時符合起振相位條件符合起振相位條件University of Science and Technology of China614.1(g)不符合起振相位條件不符合起振相位條件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)能夠起振能夠起振Un

38、iversity of Science and Technology of China674.3(c)能夠起振能夠起振University of Science and Technology of China684.3(d)能夠起振能夠起振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，畫出實際的振蕩器電路。(1)基極調(diào)諧、集電極耦合的變壓器耦合振蕩器，采用PNP三極管，正電源工作，發(fā)射極交

40、流接地。University of Science and Technology of China764.5 (2)負電源工作，基極交流接地，交流等效電路如圖所示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：不變：不變，則應(yīng)有University of Science and Technology of China814.6(3) (4) (6) (7) (3) 導(dǎo)線不可以，會將C1交流短路 高頻扼流圈可以，起到隔交流通直流的作用 (4) 只能去掉一個，起到對反饋回路隔直流的作用 (6) 2端測量比較合理，頻率計探頭具有接入電容，在1端會影響電路諧振頻率 (7) UBEUBEO 電路振蕩以后BE結(jié)處于大電流狀態(tài)，壓降有所降低晶體管工作特性 若相等說明電路沒有起振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、得反饋系數(shù)University of Science and Technology of China854.10University of Science and Technology of China864.10LLn221LLL/2mELCgGnGGg1gngAFmmsggmm044.11min電阻分壓式：msRRUUVUIgBErBEQBBrEQmQ457.561教材P13頁University of Science and Technology of China874.10諧振頻率：kHzCLLfOSC35.6432121振蕩穩(wěn)定時： 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電阻分壓式：msRRUUVUIgBErBEQBBrEQmQ66.101 334. 066.10556. 3min1msmsgggxGmQmmQmVFxUUro67. 785. 8xMHzCCLfOSC5 .612141232754123gCCQosco %2 . 01xDQTHDo輸出電壓：諧振頻率：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(差分對右側(cè)差分對右側(cè))871.19 0.1278 423. 9204. 1)( 1: min1min2xmsmsgggxggggngLMnmdQmdmdQmdmdmdmd起振條件University of Science and Technology of

47、China4.14(差分對左側(cè)差分對左側(cè))94 v0.203)(LC3958. 721322LooscoRxaufMHzCLf輸出電壓為：的三次諧波諧振回路提取振蕩信號University of Science and Technology of China954.15University of Science and Technology of China964.15解：（1）tUuoocos非線性電阻特性特性曲線： ooufi 01coscos2dUfIoooU1arccosoooUUI21sin11211arccos111121ooonUUUG振蕩平衡條件：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思考的一個問題：思考的一個問題：非線性電阻的特性方程： ufi tUUuoQocos電壓激勵： 求：Gn1方法二： 方法一： tUUfufioQoocostItIIooo2coscos21oonUIG11 tUUuoQuftgcostgtgg2coscos2

49、10?11gGnUniversity of Science and Technology of China99tgtggtg2coscos)(210tUUtgtggioQocos2coscos210方法二應(yīng)該這樣計算： 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）直流負載線方程QQQIRU

50、RUE21工作在負阻區(qū)：VUVQ6 . 02 . 0VUQ4 . 0?。罕ＷC直流負載線與負阻特性只有一個交點： 5 . 12121RRRR22121RRRRkR31. 21kR638. 02?。嚎山獾茫篣niversity of Science and Technology of China1024.16（2）振蕩頻率：MHzCCLfdOSC45.4821msRGGn2 . 011平衡條件：University of Science and Technology of China1035.3University of Science and Technology of China1045.3解

51、：(1)高頻大幅度的信號 判為開關(guān)控制信號tu22當(dāng)u2的正半周oDuuuu211oDuuuu21211Dgui 22Dgui LoLoRuugRiiu12121221LLogRugRuUniversity of Science and Technology of China1055.3當(dāng)u2的負半周1221LLogRugRu1122ugRKKgRuKuKuuLLoooo綜上：University of Science and Technology of China1065.3(2) 同(1)KuKuuooo綜上：122111LLoRguRgu122212LLoRguRgu1212222111

52、LLLoRgKgRgKguRuUniversity of Science and Technology of China1075.3(3)開關(guān)控制信號u2u2正半周：oDuuuu211oDuuuu21411Dgui 44Dgui LoLoRuugRiiu14121221LLogRugRuKgRugRKuuLLoo1221University of Science and Technology of China1085.5 (c)5.5 試分別求出如圖所示電路輸出電壓的表達式。設(shè)所有二極管特性均為從原點出發(fā)、斜率為gD的直線。且SLUU DLgR/1(c)解：uL為開關(guān)信號當(dāng)uL正半周，VD1導(dǎo)

53、通，VD2截止 oSLDuuuu111DDugi oSLLDLouuuRgRiu1LSLDLDouuRgRgu1University of Science and Technology of China1095.5 (c)當(dāng)uL負半周，VD1截止，VD2導(dǎo)通，同理：SLLDLDouuRgRgu1LSLDLDoooouKKuRgRguKuKuu1綜上：University of Science and Technology of China1105.6University of Science and Technology of China1115.6ttttKuuab7732110cos221

54、10cos10cos2 . 1晶體管足夠大，=1kVtKuRVVuiiEEEabEC53 . 47 . 021對于輸出端RLC網(wǎng)絡(luò)：sradCRLc/1013解：場效應(yīng)管開關(guān)準模擬乘法器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：場效應(yīng)管開關(guān)準模擬乘法器ttttKuuCab755110cos221102cos10cos3kVtKuRVVuiiEabEC33 . 551507 . 06515021sradLCo/1017sradRCBW/1021215RLC選頻網(wǎng)絡(luò)：University of Science and Technology of China1156.6頻率成分RLC選頻網(wǎng)絡(luò)：tt7510cos10cos對于kVIC315150231頻率成分對于tt7510c

56、os102coskVIC315150212oLjRjH1o4 .632arctantt7510cos102cos頻率成分的相移University of Science and Technology of China1166.6ttRIuoLCo75110cos4510cos212ttRIoLC75210cos4 .63102cos5tttuooo75510cos4 .63102cos0465. 04510cos221. 012綜合以上分析：University of Science and Technology of China1176.106.10 如圖所示的雙平衡差分電路中，設(shè)所有晶體管

57、均為指數(shù)律的。試推導(dǎo)輸出電壓表達式。若欲用此電路產(chǎn)生DSB信號，u1、u2中哪個信號應(yīng)該為載波？為什么？解：VT5 VT6組成負反饋的差分放大器：EEECCRuiiii265652VT1 VT2組成的差分放大器 ：rCrCCCuuiuuiii2tanh2tanh22151521VT4 VT3組成的差分放大器 ：rCrCCCuuiuuiii2tanh2tanh22161634University of Science and Technology of China1186.10輸出電壓：31424231CCCCCCCCCCCCCCCoiiiiRiiRViiRVurCCCouuiiRu2tanh1

58、65rECouuuRRu2tanh212u1為載波，u2為調(diào)制信號University of Science and Technology of China1196.136.13 證明如圖所示的正交復(fù)用系統(tǒng)，可以實現(xiàn)用2max的帶寬傳送兩路帶寬為max的不同的基帶信號。 ttsyxcos111 ttsyxsin222解：由圖 ttsttsyxyxsincos212211University of Science and Technology of China1206.13 ttsttsyxyxsincos212211在接受方： ttsttststtstttsyx2cos2sin2121sinc

59、ossin21222144 ttsttststttsttsyx2sin2cos2121cossincos21122133分別通過低通濾波器： tsuo11 tsuo22得證！University of Science and Technology of China1216.14University of Science and Technology of China1226.14解：ui處于正半周，VD1導(dǎo)通，VD2截止，uo=0ui處于負半周，VD2導(dǎo)通，VD1截止ARRRiAUuDTi11211) 其中對于理想二極管，UT=0，RD=0 12120001iRRRiuitKuRRtKRRRu

60、utKuuCiCiiCbo12211ttttKuuCio88410cos22110cos10cos5 . 01University of Science and Technology of China1236.14第二級RoCo組成低通濾波器截止頻率：oooCRjRjZ1sradCRooC/1014212410cos25 . 01433tRRuRZuooo410cos354. 0118. 34tuo第二級輸出：University of Science and Technology of China1246.142) 非理想二級管同理，通過第二級 ：kiVuARRRiAUuiDTi0012.