通信考研网www.kyabc.cn(a)picD!8P-Cn6.pig.UP.2(b)图2-20高频等效电路Nas=6p2==%- 0.311gos0g13Q=2元×10.7×10%×4×10-6×100= 37.2 × 10 (s) = 37, 2(μs)ge = 0.2(ms)1.3×1032元×10.7×10 = 19.35 ×10-(F) = 19.35(pF)Cx =ge = 2.86(ms)3.4 × 10-3Cr=2元×10.7×10 =50.60 ×10*"(F) = 50.60(pF)1y/=26.42+36.42=45(ms) 36.4 = - 54°P=arctan26.4g=pig+ga+p2g=0.25×0.2×103+37.2×106+0.3×2.86×103= 307.1 × 10-(s) = 307.1(μs)(2)计算电容C因为C=p,C+C+pCu,且C=（2元/0)L(2x×10.7×10%)2×4×10-6= 55.37 × 1012 (F) = 55.37(pF).24
通信考研网www.kyabc.cn则C=Cg-piCpC=55.37-0.252×19.35-0.3*×50.60=49.61(pF)(3)计算A2△/o7、K,0.1Ato =- pup2ye = - 0.25 ×0.3 × 45 ×10jS4307.1 × 10°68s= 10. 99e s*11Q. =2元×10.7×10°×4×106×307.1×106woLi3gs= 12.114=10.7=0.884(MHz)24f0.7=12.11QK.0.1=V992.考虑y元，且S≥5为稳定，则(ym/=/0.082+0.32=0.31(ms)-0.3=-75°P. = aretan 0.082/ye[A.0, =slyTi+ cos( + )2×45×103/5×0.31×10-3[1+cos(-54°75°]=12.52由于实际放人器的|A|=10.99小于1A。则放大器是稳定的。也可以用|A|代人[A，中可得21 y/STAo'Tyn/[1 + cos(9x + p)]2 ×45×10 3（10.99)×0.31×10-"[1+cos(-54°-75)]=6.48可见|A.。|=10.99的单调谐放大器是稳定的。第三节思考题与习题参考解答2-1已知1C串联谐振回路的C=100pF,J=1.5MHz，谐振时电阻r=5Q，试求：L和Q。1解由fo=2元VLC得L = 1/(2元fo)°C = 1/(2元 × 1.5 × 10°)2 × 100 × 10-12=112.6×10-(H)=112.6(μH)·25*
通信考研网www.kyabc.cnQ, = aol _ 2r ×1.5 × 10° ×112.6× 10*57=212.22-2已知LC并联谐振回路的电感L在f=30MHz时测得L=1μH,Q=100。试求谐振频率f。=30MHz时的C和并联谐振电阻。1解由fo=2元V/LC得C = 1/(2元f)L = 1/(2× 30 ×10°)2×1 ×10-6= 28.14×10-12(F)= 28.14(pF)Ro由Qo=wL得Rg=QwgL=100×2x×30×10°×1×10-6=18.850(kQ)2-3已知LCR并联谐振回路，谐振频率f。=10MHz，电感L在f=10MHz测得L=3pH.Q。=100,并联电阻R=10k2。试求回路谐振时的电容C,谐振电阻R，和回路的有载品质因数。解1.C=1/(2元f起）L=1/2元×10×10)2×3×10*=84.43×1012(F)=84.43(pF)2.电感L的空载损耗电阻以并联表示为R。R=Qagl=100×2元×10×10%×3×10-6=18.850kQ谐振电阻R为R，与R并联值RaR18.850×10= 6.534(ko)Rp=R+R"18.850+103.Q=R/0L=6.534×10/2元×10×10°×3×10*6= 34.662-4某电感线圈L在厂=10MHz时测得L=3uH,Q。=80。试求与L串联的等效电阻r。若等效为并联时，g=？。解1.等效为串联的电阻r。为00l=2元×10×10%×3×10-g=2.360roa80Qa2.等效为并联的g为1Tgo= g LQ。 = 2x × 10 × 10 × 3 × 10 × 80 = 66.3 × 10 (s)= 66.3(μs)2-5电路如图题2-5。给定参数为J。=30MHz，C=20pF，线图L的Q。=60，N2=6,Nz=4,Ns=3.R=10k0,R=2.5k0,R,=8300,Cg=9pF,C=12pF。求LQL。·26
通信考研网www.kyabc.cn图题2-5解1.画高频等效电路根据图题2~5可画图2-21所示等效电路pgpiccpiCsPigr8.1发图2-21高频等效电路其中,P=Nz/N=4/10=0.4;P2=N4s/Nj3=3/10=0.3.gg=1/Ri8=1/R,。2.求Li3由图可知，C=pC+C+pC= 0. 4° × 9 + 20 + 0. 3 × 12 = 22.52(pF)因为。=1/2元C所以L1=12元f)Cs=1/(2元×30×10°)2×22.52×10-12= 1.25(μH)3.求Q11gn=wmLmQ2x×30×10%×1.25×10×60=70.7× 10(s) = 70.7(μs)2 = Piz, + + + pie112.5×10+10×10+70.7×10*+0.33=0.42830= 343.1× 10(s)= 343.1(μs)?27
通信考研网www.kyabc.cnQ=1/gLj3g=1/2元×30×106×1.25×10-6×343.1 ×10-6=12.372-6电路如图题2-6。已知L=0.8μHQ。=100,C,=25pF,C2=15pF,C,=5pF,R=10k0,R=5kn。试求fo、RpQ和2fo.70解由图题26电路可画等效电路如图2-22所示。图题2-6图2-22图题2~6等效电路C111.RRL,P==25/(25+15)=0.625,C,C/(C,+C,)C, + C2R't = (0.625)5 = 12.8(k0)C,C225 × 152.fo=1/2元/LCg，C=C,+=5+14.375(pF)C+C225+15o = 1/2元 /0.8 × 10* ×14. 375 × 10 2= 46.93 × 10(Hz)46.93(MHz)R,R_ 10×12.85.614(k2)3. R =R+R,=10+12.84.Q.=Rp/wnL=5.614×10/2元×46.93×10%×0.8×10623.805.2△fo 7 = fo/Q, = 46.93 × 106/23.80 = 1.79(MHz)2-7晶体管3DG6C的特征频率fT=250MHz，Po=80，求f=1MHz、20MHz和50MHz时对应的|βl。βoIr250解=3.125(MHz)因为Ipl+80BoBo所以1p =1 + (f/3.125)801. f = 1MHz时,Iβ| =76.19V1+(1/3.125)802./=20MHz时，1p|=12.351+(20/3.125)803./= 50MHz时,1βl==4.991+(50/3.125).28