30第七章信号的运算和处理自测题一、(1)√(2)×(3)√(4)×二、(1)C(2)F(3)E(4)A(5)C(6)D三、(1)带阻(2)带通(3)低通(4)有源四、I3142O2O43'O43I12O2O1OI343421f2I21I1fO1)b(d1)1()()a(uRkRRRukuRRuRRuRRutuRCuuRRRRRRRuRuRu∥习题7.1(1)反相同相(2)同相反相(3)同相反相(4)同相反相7.2(1)同相比例(2)反相比例(3)微分(4)同相求和(5)反相求和(6)乘方7.3uO1=(-Rf/R)uI=-10uIuO2=(1+Rf/R)uI=11uIuI/V0.10.511.5uO1-1-5-10-14uO21.15.511147.4可采用反相比例运算电路,电路形式如图P7.3(a)所示。R=20kΩ,Rf=2MΩ。7.5由图可知Ri=50kΩ,uM=-2uI。IMO3OM4M2M34210452uuuRuuRuRuiiiRRR即7.6(1)uO=-2uI=-4V(2)uO=-2uI=-4V(3)电路无反馈,uO=-14V。(4)uO=-4uI=-8V7.7(1)10.4(2)107.8(a)uO=-2uI1-2uI2+5uI3(b)uO=-10uI1+10uI2+uI3(c)uO=8(uI2-uI1)(d)uO=-20uI1-20uI2+40uI3+uI47.9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。7.10(a)uIC=uI3(b)I3I2IC1111110uuu31(c)I2IC98uu(d)I4I3IC4114140uuu7.11IL≈UZ/R2=0.6mA7.12(1)uO2=uP2=10(uI2-uI1)uO=10(1+R2/R1)(uI2-uI1)或uO=10(RW/R1)(uI2-uI1)(2)uO=100mV(3)uO=10(10/R1min)(uI2max-uI1min)V=14V,R1min≈71kΩR2max=RW-R1min≈(10-0.071)kΩ≈9.93kΩ7.13))(1()1()1()1()1()b())(()()a(I1I245I245I11345I245O145OI113O12I21I15434344MO5M2I21I15342I21I13MuuRRuRRuRRRRuRRuRRuuRRuRuRuRRRRRRiuuRuRuRuiiiRuRuRuRRRR(c)uO=10(uI1+uI2+uI3)7.14)(d11OIO21tutuRCutt当uI为常量时)()(100)()(10101)()(11O12I1O12I75112IOtuttututtututtuRCuO-若t=0时uO=0,则t=5ms时uO=-100×5×5×10-3V=-2.5V。当t=15mS时,uO=[-100×(-5)×10×10-3+(-2.5)]V=2.5V。因此输出波形为7.15输出电压与输入电压的运算关系为uO=100uI(t2-t1)+uI-uC(t1),波形如图下所示。ut/V/mS0-2.552.555152535327.16(a)tuutuCRuRRud100d1III1I12O(b)II3I21I1O2dd10ddutuuCCtuRCu(c)tutuRCud10d1I3IO(d)tuutRuRuCud)5.0(100d)(1I2I12I21I1O7.17(1)uO1=uO-uI,uC=uO,RuRuuiIOO1CtutuRCtiCud10d1d1IICO(2)uO=-10uIt1=[-10×(-1)×t1]V=6V,故t1=0.6S。即经0.6秒输出电压达到6V。7.18tuuututuCRud2d2d1IOOII1O27.19(1)UA=7V,UB=4V,UC=1V,UD=-2V,uO=2UD=-4V。(2)uO=2UD-uO3mS6.28471010501163A1O3tttuCRu7.20ttuu/mS/mS/V/VIO012.557.5-2.55152535对数运算电路对数运算电路减法运算电路指数运算电路uuI1I2uI1uI2k337.21(1)上为“-”,下为“+”(2))1.0(I2Of'OfI1I2O'OuuRRRuRRRuukuu,,所以I21IfO)(10uuRRRu7.22I143I2342I24O2I21I1I33OI3O2I21I13'O)b()(-)()a(uRRukRRkuRRuRuRukuRuukuRuRuRu-7.23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次幂电路。7.24(1)带阻滤波器(2)带通滤波器(3)低通滤波器(4)低通滤波器7.25(a)高通滤波器(b)高通滤波器(c)带通滤波器(d)带阻滤波器7.26将两个滤波器串联,图略。7.27为高通滤波器。,为高通滤波器。,11)()a(1)()a(21212CsRRRsACsRCsRsAuu7.28uO1:高通。uO2:带通。uO2:低通。7.29参阅P362~P363。7.30Ωk6404kΩ160π2121221p0uppRRRRRCffAQAffu。,代入数据,得出因为。,=,所以因为7.31参阅7.5节。7.32略。