《组成习题答案》PPT课件.ppt

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1、Computer Organization,exercise,2-1Give the sign mag、1s compl、2s compl representation of the follow number(word length is 8),where MSB is the highest bit(sign bit)，LSB is the lowest bit。If the number is decimal,the decimal point is behind the MSB;if the number is integer,then the decimal point is beh

2、ind the LSB.（1）35/64（2）23/128（3）127（4）1(decimal representation)（5）1(integer representation),2-1 Solute：（1）-35/64 binary-100011/1000000=-0.100011=-0.1000110 sign mag：1.1000110 1s compl：1.0111001 2s compl：1.0111010（2）23/128 binary10111/10000000=0.0010111 sign mag：0.0010111 1s compl：0.0010111 2s compl：

3、0.0010111,2-1 Solute：（3）-127 binarysign mag：11111111(1,1111111)1s compl：10000000(1,0000000)2s compl：10000001(1,0000001)（4）-1(decimal)There are no representation of-1 in sign mag and 1s compl 2s compl：1.0000000（5）-1(integer)sign mag：10000001(1,0000001)1s compl：11111110(1,1111110)2s compl：11111111(1,1

4、111111),2-2 Rank these numbers：(small to large)16，1010.11B，25.3Q，X1 2s compl=10001101，1CH，X2 1s compl=01001101，0110.1001BCD，X3 sign mag=10101011，-X4 2s compl=10111111，-X5 2s compl=10100101 Key：（1）unify all the forms,generally decimal is selected（2）transformation between B,Q,H and BCD；,2-2 Solute：16，

5、1010.11B，25.3Q，X1 2s compl=10001101，0110.1001BCD，1CH，X2 1s compl=01001101，X3 sign mag=10101011，-X4 2s compl=10111111，-X5 2s compl=10100101 AN:1010.11B=10.75,25.3Q=21.375,X1=-01110011B=-1150110.1001BCD=6.9,1CH=28,X2=01001101B=77X3=-0101011B=-43,X4=01000001B=65,X5=01011011B=91 small to large:X1，X3，011

6、0.1001BCD,1010.11B，16，25.3Q，1CH，X4，X2，X5,3-1 设计一个A、B、C三人表决电路，当表决某个提案时，多数人同意提案通过，同时A具有否决权，用与非门实现。设计步骤列出真值表 设A、B、C三个人，表决同意用1表示，不同意时用0表示；Y为表决结果，提案通过用1表示，提案未通过用0表示，同时还要考虑A的否决权。写出逻辑表达式画出逻辑图,3-1 Solute：,写出逻辑表达式：Y=AB+AC,3-1 Solute：,A,C,Y,B,6-1 Given a 32 bit floating-point number，1 bit for sign,8 bit for E

7、,represented by biased code,23 bit for M,represented by 2s Complement；radix is 2:（1）Maximal binary；（2）Minimal binary；（3）range；（4）positive and negative normalized representation float-point that closest to 0,6-1 Solute：规格化浮点数的最大正数值是由尾数的最大正数值与阶码的最大正数值组合而成的;最小正数值是由尾数的最小正数值与阶码的最小负数值组合而成的。在负数区间;最大负数值是由尾数

8、的最大负数值与阶码的最小负数值组合而成的;最小负数值是由尾数的最小负数值与阶码的最大正数值组合而成的。,6-1 Solute：f-p format is X=2EM，8 bit E(biased code)，with a range of-128+127；23 bit M(2s C)，with it max positive value of Mmax=1-2-23,min positive is Mmin=2-23,max negative value of Mmax=-2-23,min positive value of Mmin=-1（1）binary format of Maximum

9、：Positive Xmax=2127（1-2-23）=2127 0.111111=11111100000（23 bits 1，104 bits 0）Negative Xmax=2-128（-2-23）=-0.0000001（151 bits 0）（2）binary format of Minimum：positive Xmin=2-1282-23=0.0000001（151 bits 0）Negative Xmin=2127（-1）=-10000000（127 bits 0）,6-2 Transform the follow decimal to IEEE754 32 bit float-p

10、oint（1）27/64（2）-27/64,6-2 Solute:Transform the follow decimal to IEEE754 32 bit float-point(1)27/64=11011X2-6=1.1011X2-2S=0；E=2127125=01111101B；M=1011 0000 0000 0000 0000 000result：0011 1110 1101 1000 0000 0000 0000 00003ED80000H,6-2 Solute:Transform the follow decimal to IEEE754 32 bit float-point(

11、2)-27/64=-11011X2-6=-1.1011X2-2S=1；E=2127125=01111101B；M=1011 0000 0000 0000 0000 000result：1011 1110 1101 1000 0000 0000 0000 0000BED80000H,6-3 Transform decimal-0.75 to IEEE754 32 bit single precision float-point number,6-3 Solute:Transform decimal-0.75 to IEEE754 32 bit single precision float-poi

12、nt number-0.75=-3/4=-11X2-2=-1.1X2-1S=1；E=1127126=01111110B；M=1000 0000 0000 0000 0000 000result：1011 1111 0100 0000 0000 0000 0000 0000BF400000H,6-4 Transform 0C0B00000H IEEE754 32 bit float-point number to Decimal,6-4 Solute:Transform 0C0B00000H IEEE754 32 bit float-point number to Decimal0C0B0000

13、0H=1 10000001 0100 0000 0000 0000 0000 000S=1；e=E-127=(10000001 01111111)B(2)10；1.M=(1.01)B=(1.25)10result：（-1）1 X（1.25）X 22=-1 X 1.25 X 4=-5.0,6-5 Transform(0.15)10 to Normalized floating-point numbers representation阶码：Sign of m:1-bitExponent:8-bit,integer,biased code,radix=2mantissa:23-bit,decimal

14、,sign-mag,radix=2,implied leading 1,Sign of mantissa,exponent,mantissa,6-5 Transform(0.15)10 to Normalized floating-point numbers representation阶码：Sign of m:1-bitExponent:8-bit,integer,biased code,radix=2mantissa:23-bit,decimal,sign-mag,radix=2,implied leading 1Solute:0.15D 0.001001100110011001 22S=

15、0；M001 1001 1001 1001 1001 1001=199999HE=-2+128=126D 01111110=7EHN 0 01111110 001 1001 1001 1001 1001 1001 3F199999H,Sign of mantissa,exponent,mantissa,6-6 Transform(0.1)10 to Normalized floating-point numbers representation阶码：(1)Sign of m:1-bitExponent:8-bit,integer,biased code,radix=2mantissa:23-b

16、it,decimal,sign-mag,radix=2,implied leading 1(2)Sign of m:1-bitExponent:7-bit,integer,biased code,radix=2mantissa:6-bit,decimal,sign-mag,radix=16,Sign of mantissa,exponent,mantissa,6-6 Transform(0.1)10 to Normalized floating-point numbers representation阶码：(1)Sign of m:1-bitExponent:8-bit,integer,bia

17、sed code,radix=2mantissa:23-bit,decimal,sign-mag,radix=2,implied leading 1Solute:0.1D 0.00011001100110011001 23S=0；M 100 1100 1100 1100 1100 11004CCCCCH；E-3+128=125D 01111101=7DH；N 0 01111101 1001 1001 1001 1001 1001 100 3ECCCCCCH,Sign of mantissa,exponent,mantissa,6-6 Transform(0.1)10 to Normalized

18、 floating-point numbers representation阶码：(2)Sign of m:1-bitExponent:7-bit,integer,biased code,radix=2mantissa:6-bit,decimal,sign-mag,radix=16Solute:0.1D 0.00011001100110011001 0.199999999H 20S=0；M0001 1001 1001 1001 1001 1001 B199999H；E=064=64D 1000000B=40H；N 0 1000000 0001 1001 1001 1001 1001 1001

19、40199999H,Sign of mantissa,exponent,mantissa,6-7 Let X=2-0100.10010010,Y=2-011(-0.11110111)，e=6 bits，m=10 bits（all with 2 sign bits,2s c）X+Y?,6-7 Let X=2-0100.10010010,Y=2-011(-0.11110111)，exponent:6 bits，with 2 sign bits,2s compl representationmantissa:0 bits，with 2 sign bits,2s compl representatio

20、n X+Y?An：Floating-point number:X 2s compl=11 1110；00.10010010Y 2s compl=11 1101；1 1.00001001Exponent equalization111110-111101=000001ExEyY 2s compl=11 1110；1 1.10000100(1)(2)Mantissa add/sub00.10010010+1 1.10000100(1)=00.00010110(1),6-7 Let X=2-0100.10010010,Y=2-011(-0.11110111)，exponent:6 bits，with

21、 2 sign bits,2s compl representationmantissa:0 bits，with 2 sign bits,2s compl representation X+Y?An：Step(2)result is：X+Y2s c compl=11 1110；00.00010110(1)(3)normalization:After left normalization：X+Y2s c=11 1011；00.10110100(4)Rounding has no effect(5)Check the result:not overflow,not zero result：X+Y=2-101 X 0.10110100,