模电答案-童诗白第七章

1、8第七章自测题(1)(1)(1)VC带阻(2 )X (3)V(2) F ( 3)(2)带通(4)E (4) A(3)低通(5) C(6) D(4)有源四、U|1UO1 = rU|2RfR1/R4L R3 + R4U|3(b)1Uo = fuo1dtRC _ R2Uo2 = - U|R1"r4R3'R3Uo -R4kuO习题UOR2 R4 kRR U|7.1(1)反相同相(2)同相7.2(1)同相比例(2)反相比例(5 )反相求和(6)乘方7.3UO1 = ( Rf/R) UI=10 u|1反相 (3)(3)微分同相反相(4)(4 )同相求和同相反相7.47.57.67.77.

2、87.9UI/V0.10.511.5UO1151014UO21.15.51114P7.3(a)所示。R= 20k Q ,UI = 11 U|可采用反相比例运算电路，电路形式如图由图可知 Ri = 50k Q , Um = 2U|o.,.日仃 UMUM , Um - Uo|R2 TR4 +lR3即 一=+-R2R4R3Uo =52um = T04U|Rf = 2M Q o(1) uo= 2 U|= 4V ( 2) Uo= 2 U|= 4V ( 3)电路无反馈，(4) uo= 4 U|= 8V(1) 10.4(2) 10(a) UO= 2 U|1 2 U|2 + 5 U|3( b) UO= 10

3、U|1 + 10 U|2 + U|3(c) uo= 8 ( U|2 U|1)(d) Uo= 20 Un 20 U|2 + 40 U|3 + U|4因为均有共模输入信号，所以均要求用具有高共模抑制比的集成运放。10 . 1(b) U|C = U|2 中U|311 117.10(a) uic = U|3Uo= 14V o357.117.127.13(a)(b)Uo7.1440 1(d) Uic =rU|3 +二 U|441 41(c) U|C =-U|29|LUZ / r2 = 0.6mA(1 ) UO2 = Up2= 10 ( U|2 U|1 )UO = 10 ( 1 + R2/R1)( U|2

4、 U|1 )或 Uo= 10 ( Rw /R1)( U|2 U|1 ) uo= 100mVUO= 10 ( 10 /R1min ) ( UI2max UI1min ) V=14V , R1mine 71k QR2max= Rw R1min &( 10 0.071 ) k Q = 9.93 k QUmi R4UoUoiR (U|1 +U|2) 一R3(石+R；)Ui1U|2Um= 035 =二十二 一二-R1R2R5U=UM -iR4 R4 = -(R3 + R4 +)(Z时+ UJ2) r'RT瓦丿R,"RUI1RUo1 +(1 + R)U|-R(1R4R4R4R3R

5、5R5+ )U|1 +(1 + h)U|2 =(1 + 害)(U|2 -U|1) R|R4R4uo= 10 ( U|1 + U|2 + U|3 )1uo =RCJ, u|dt +uo(t1)t1当U|为常量时1Uo =-Ui (t2 -t1 ) +UO(t1) RC=100U|(t2 -t1) + UO(t1)1-5 U| (t2 -t1)+uO(t1)10 X10o若 t= 0 时 uO = 0,贝U t= 5ms 时 uO= 100 X 5X 5X 10 V = 2.5V。当 t= 15mS 时，UO= -100 X (-5) X 10X 10 3+( 2.5) V = 2.5V。因此输出

6、波形为7.15输出电压与输入电压的运算关系为UO= 100U| (t211)+ U| Uc (t1)，波形如图下所示。7.16(a)Uo =(c)(d)uoUoUo-RC1空UIRi1賦2 00側dUI C1Ui 10此-2uidtdtfUIdt RC RiC2=10 fuIdt+ 业)dt = -100 J(U|1 + OUjdtR2UiUo1 Uo7.17(1) UO1 = Uo Ui , Uc= Uo, ic =1 1Uo = Jicdt =-fuId-10 fuIcltCRCUo = 10uiti= 10X( 1)X tiV = 6V，故 ti = 0.6S。即经 0.6 秒输出电压达

7、到6V。7.18U02 = _1Jui dt = 2 JU| dt = 2 uoRiCUo =-JuIdt7.19(1) Ua = 7V, Ub = 4V, uo= 2 UD uo3Uc = 1V, Ud = 2V , Uo = 2 Ud = 4V。UO3 = _1'UaRiC50"03 x10t 止 28.6mS7.20=c4? f ? ? ? ?ku /Ml? ? ?本？?址？ ？本？/ ui2UI2_I7.21(1)上为“-”，下为 “ + ”37(2) uo =kuoui2,UI1uoR + Rf uO10(R + Rf)U|1U|2，(-0.1UoU|2)，所以Uo

8、R3 . ui1 ui2 ,=(中)ku I3R1R27.22r ,ui1 UI2 ,(a) uo = R3(+)=kuoui3R1R2R42R4 2 3 R4(b) uo =丄如k uI - uIR2R3R17.237.247.257.267.27方框图如图7.3.9所示，N = 5时为5次方电路；N = 0.2时为5次幕电路。 (1)带阻滤波器(2)(a)高通滤波器(b)将两个滤波器串联，图略。带通滤波器 高通滤波器(3)低通滤波器(C)带通滤波器(4)低通滤波器(d)带阻滤波器Au (s)SR2C,为高通滤波器。1+SR1CAu (s)1，为高通滤波器。R1 1+SR2C7.287.297.