微机原理与接口技术.docx
<计算机技术>习题解第1章1.1 写出下列十进制数的8位二进制补码表示解：(1) 54=(X)IIO1.IOB(2) 37=(X)I(X)IO1.B(3)in=Oiioiiiib(4) 253超过8位补码范围(5)0.1=O.(XX)HO1.(6) 0.63=OJO1.(XX)IB(7) 0.34=0.01011(8)0.21=0.1.10111.2 转换下列二进制数为十进制数(1) 10111101=189(2) I(XX)I(X)I=137(3) O.IOIIIII=95/128=0.7421875(4) O.I1OIO=13/64=0.203125(5) I(X)I1.(X)I.H(X)I1.=153+51/(4=153.796873III(XX)II1.=4551.3 写出卜列带符号数的原码、反码、补码和移码表示(用8位二进制代码表示)解：(1)+112=127-15+112=O1.1.1.O(XK)B+112k=O1.1.I(XXX)B+112“=O1.II(X)(X)B+H2re=I1.I100OOB(2)0.625=OJO100(X)B0.625K=|0.625rt=(0.625h=OJO1.(XX)OB小数无移码(3)-124=127-3=Oiiiiiwb-124%.=1II111O()B1241.=I(XXXX)I1.Br-24,=I(XXX)I(X)B-124w=(X)O(X)100B(4) -0.375=48/128=0.01100B-0.375w=1.()1.1.()(XX)B-0.375Ift=IJ(X)II1.1.BI-0.375*=i.io100oob小数无移码(5) 117=12710=1110101-7Jw=I1.IO1.OIB-17Js=I00010IOB-117U=1OOO1OI1B-117k=00001011B(6) +0.8125=104/128=O.I1.OIO(X)B+0.8125ap+0,81251.4=+0.8125*=0.11010B小数无移码1.4 给出以下机器数.求其真值（用二进制和十进制数）表示解：(1) X=+(32+7)=+39=+0100111B(2) x4=1010II0IBX1.u=I1.O1.(X)IIBX=-IO100IIB=-(64+16+3)=-83(3) X=+10001IOB=+70(4) X=1.()101.IO1.BX=-0101IO1.B=(32+13)=-451.5 已知x“和的值,用补码加减法计算x+y,和xy补,指出结果是否溢出IyJn=O.(XM)IIIyht=1.OOIOIyn=1.10()01Iyh=OJ1.(X)I(1) Xk=OJ1.OI1.,(2) (Xk=OJOIII(3) x=1.O1.O1.O(4) Xk=1.I(M)I1.(1) xU=0.I10II.yH=0.01.1.(X+Y*=xtt+yt1.=0.111.1.0X+Y=+15/16=+().1I1.IBX-Yn=xn+-yk=OJI(XX)X-Y=+12/16=+0.1H1.B(2) xn=0.101.1.1.,(yw=1.OO1.O1.x+>>=b>+yh=1.II100X+Y=-0.00IB=-1/8(X-YJ0=1.1.u+-yh=1.100Io(I,Jfii)(3) Xjtt=1.O1.OIO.yfr=1.10001X+Y"=10.11011X-Yk=(xk+-y,>=11.11001X+Y=0.11011B下溢XY=-(M)O1.1.1.B=-7/32(4) (xh=1.1(X)I1.,Iyh=OJI(X)IX+YH=(X).()II(X)X-Yfr=xk+-yw=10.11010X+Y=(-13+25)/32=12/32=3/8X-Y下浓1.6给出X和y的二进制值，用补码加减法计兑x+y“利X小，并指出结果是否溢出解：(1) X=0.10111Y=OJ1.O1.1.X+Y1.u=01.10010X-Y4=x)u+-y*=11.11100X+Y正源X-Y=-1/8(2) X=OJ1.1.O1.Y=OJOO1.1.X+Y1,F01.1000()IX-Yi=x,H-yk=00.01010X+Y正溢X-Y=1032X=().11011Y=-OJO1.OX+YJt,=(XUX)111X-Y>>=(xtt+-yn=01.011HX+Y=7/32X-Y(上溢)(4) X=-OJ1.1.1.1.Y=OJ1.OI1.IX+YJn=I1.JII(X)X-YU=xJw+-yw.=10.00110+=-0.00100=-1/8X-Y(下溢)(5) X=-OJIO1.1.Y=OJO100X+YU=1I.IIOIIX-Y*="+1.y产10.10001+=-0,00111.=-732(6) X=-O-HO1.OY=-OJ1.OO1.X+YJ,F10.01101X-Ytt=1.xJo+-y¼=I1.-I1.1.1.1.X+Y(下溢)X-Y=-O-OOOO1.=-1/32(7)X=-IOinoiY=+1101101X+Yh=(XXX)I(XXX)X-Yw三xw+-yw=oooo+=6X-Y=54(下溢)(8)X=+11IO11OY=-X+Ym.=000101001(X-Yw=x0+-yH=011001.1.X+Y=41.X-Y=-61(上溢)(9)X=+1101110Y=+1010101X+Y»=O1.IO(XX)H(X-Yb>=(h+-y11=ooooooX+Y(上溢)X-Y=25(10)X=-IIIIIIiY=-(X+Y<=I(XX)IO1.(X)x-Y)n=xb>+-yk=111101110X+Y(下溢)X-Y=-OO100IO=-181.7写出下列数据的浮点数表示，基数为2,设阶码为5位(含1位阶符)，尾数为I1.位(含I位里符)，要求尾数用补码，阶码用移码。(1)125(1(2)IO1.OI2(3) -0.00138o(4)237%(5)-1101012(6)1.1.1.1.112解(1) 125.0=O1I1I1O12=0.1111IOI×27表示为00111,1111101000(2) 101012=0.10101×25表示为001.01j010100000(3)0.()013810=1447.03488/28=-144722°="0.00000,00001,01101,001118(1024+256+12S+32+7=1447)×2,=1,10111.(-9b=11001)=1,10111.(4) 237o=I1.ionoi2=0.11101101×28=0,01000.1110110111(5) -1IO1O12=0.110101×21,00110.0010110000(6) 2=0.1011111×27=0.00111.10111111.8用32位二进制浮点数表示，阶码9位(其中1位为阶符)，尾数23位(其中1位为尾符)，要求阶码为移码表示，尾数为补码表示.请问：(1)最大正数是多少？(2)最小正数是多少？(3)肯定值最大的负数是多少？解：(1)最大正数X.xxxxxxxxx.XXXXXXXXXXXXXXXXXXXXXX9位22位+2?”×22位=+225s×(222-1.)222=2w×(222-1)(2)最小正数+2256×0.=2-256|/222=2-278(3)肯定值最大的负数(最小数)1.011111111.=22552.1用真值表验证下列公式:(1)A+BC=(A+B)(A+C)解：ABC+BC(+B)(+C)OOOOOOO1OOOIOOOOIIIIIOO1IIOI11IIOIIII11I(2)A+AB=A+B解：ABA+ABA+BOOO0OI1IIOII1II1-W(3)(A+B)(A+B)=AB+AB解：AB(A+B)<A+B)/KB+ABOOOOO1II1OII(4)B+AB+B+B=1解：ABAB+B+B+AB100I101I1I0I1I1I1(5)ABC=OBC解:ABCABCAOBOC0000O00111010110110O100111010O1100O11111(6)C+AB=C(A+B)解：，ABCC+ABGA+B)000I1001000I0I10I100I000010100II0001I1002.2 写出下列表达式的对偶式：(1) F=(A+B)(A+C)(C+DE)+G解：F'=(B)+(AO+(C(D+E)G(2) F=A+B÷C+D+D+A+B解：F=a11cj5Da7b(3) F=B(AC)+B(AC)解：F=B(AC)+B(C)F=反AC)+B(AC)(4) F=(AB)(CD)解：F=(B)(COD)+(B)(CD)=<AB)(CD)=<ab)(cd)=(AB)<CD)=(A0B)(CD)2.3 写出下列函数的“反”函数：(1) F=AB+AB解：F=(A+B)(A+B)(2) F=ABC+ABC+ABC+ABC解：F=(A+B+C)-(A+B+C)-(A+B+C)-(A+B+C)(3) F=(A+BC)(A+DE)解:F=(A(B+C)+(A(D+E)(4) F=XYI解：F=XY=XOY=XY+XY=XY=XY=XY2.4 应用公式化简法化简下列各函数式：(1) F=ABC+ABC+ABC+ABC解：F=ABC+ABC+ABC+ABC=AB(C+C)+
