模拟电子电路基础课后习题答案第二章

习题2.12.000ovV3.000vV若放大器理想，则3.000vV若放大器非理想，则()ovAvv故2.000100/3.020(3.000)ovAVVvv习题2.2(1)(2)(4)21/vARR1/vAVV则210Rk令110Rk2/vAVV则220Rk令110Rk0.5/vAVV则120Rk令210Rk(3)100/vAVV则21000Rk令110Rk习题2.4要求设计1234234ovvvvvA+-A+-v1v2v3v4R1R2R5R3R4R6R7vo由图可得77776677134123453451234()ooRRRRRRRRvvvvvvvvRRRRRRRR其中一种结构如图所示取340Rk则7120Rk430Rk取5120Rk则620Rk120Rk210Rk习题2.6因为该运放线性工作，故虚短和虚断同时成立434iRvvvRR又12ovvvRR24134(1)oiRRvvRRR24134(1)vRRARRR习题2.9A+-R4R3R5R2R1vi1vi2vi3vi4voR6由图可得546RR532RR410Rk取5360,30RkRk则52634126//(1)1////RRRRRRRR又51634216//(1)3////RRRRRRRR5126341////(1)1//RRRRRRR即5216342////(1)3//RRRRRRR126////PRRRR取可得19pRR23pRR(*)代回(*)式可得12//2.25pRRR61.8pRR10PRk取61118,90,30RkRkRk则习题2.11i1221122124344330.04sin(),30.04sin()i0.08sin()234.04sin(),34.04sin()1.52.02sin(),2()8.08sin()iiiABiiBcDBoBAvtVvtVvvtmARvviRtVviRvtVRvvvvtVRRRvvvtVR令。。。习题2.1301()()toivtvtdtRC(1)100.5()10.51itmsvtmstms000.5()100011000()0.5(0.5)0.5tootmsvtdttVtmsvmsV当时，则当时，111000RC0.50.51()(0.5)1000(1)0.51000(0.5)()1(1)0tooomstmsvtvmsdttVtmsvmsV当时，则当时，vo(V)-0.5t(ms)0.51(2)习题2.13000.5()100022000()0.5(0.5)1tootmsvtdttVtmsvmsV当时，则当时，0.50.51()(0.5)1000(2)12000(0.5)()1(1)0tooomstmsvtvmsdttVtmsvmsV当时，则当时，200.5()20.51itmsvtmstms0vo(V)-1t(ms)0.51111000RC(3)习题2.13000.5()50021000()0.5(0.5)0.5tootmsvtdttVtmsvmsV当时，则当时，0.50.51()(0.5)500(2)0.51000(0.5)()1(1)0tooomstmsvtvmsdttVtmsvmsV当时，则当时，200.5()20.51itmsvtmstms11500RCvo(V)-0.5t(ms)0.51习题2.15maxomax3(1)0.5,5,2VSR31.8220,7.95,0.79521050,318.322102000~200kHz105ioopoooiiMoitdBSRvVvVVffkHzVSRvfkHzvVvVfSRSRvmVfkHzVvffkHzfkHz则，为输出不失真的最大频率。（2）则不失真的。（3）若则。而。有用工作频率范围为。（4）'maxmaxmaxmax,31.810,210,110-1V1VoooiSRvVVfvvVvV则则对应的。有用的输入电压范围为（，）。