数字逻辑第二版毛法尧课后题答案

第一章数制与码制1.1把下列各进制数写成按全展开的形式14517.23910=4×103+5×102+1×101+7×100+2×10-1+3×10-2+9×10-3210110.01012=1×24+0×23+1×22+1×21+0×20+0×2-1+1×2-2+0×2-3+1×2-43325.7448=3×82+2×81+5×80+7×8-1+4×8-2+4×8-34785.4AF16=7×162+8×161+5×160+4×16-1+A×16-2+F×16-31.2完成下列二进制表达式的运算110111+101.10121100-111.011310.01×1.0141001.0001÷11.10110111+101.101=11100.10111000.000-00111.011=10000.10110.01×1.0110010000100110.110110.0111101100100011.5如何判断一个二进制正整数B=b6b5b4b3b2b1b0能否被410整除解：b1b0同为0时能整除,否则不能,1.6写出下列各数的原码、反码和补码,10.101120.00003-10110解：0.1011原=0.1011反=0.1011补=0.10110.0000原=0.0000反=0.0000反=0.0000-10110原=110110-10110反=101001-10110反=1010101.7已知N补=1.0110,求N原、N反和N解：N原=1.1010N反和=1.1001N=-0.10101.8用原码、反码和补码完成如下运算10000101-0011010解10000101-0011010原=10010101∴0000101-0011010=-00101010000101-0011010反=0000101反+-0011010反=00000101+11100101=11101010∴0000101-0011010=-00101010000101-0011010反=0000101补+-0011010补=00000101+11100110=11101011∴0000101-0011010=-00101011.8用原码、反码和补码完成如下运算20.010110-0.100110解20.010110-0.100110原=1.010000∴0.010110-0.100110=-0.0100000.010110-0.100110反=0.010110反+-0.100110反=0.010110+1.011001=1.101111∴0.010110-0.100110=-0.0100000.010110-0.100110补=0.010110补+-0.100110补=0.010110+1.011010=1.110000∴0.010110-0.100110=-0.0100001.9分别用“对9的补数“和”对10的补数完成下列十进制数的运算12550-123解：12550-1239补=2550-01239补=25509补+-01239补=02550+99876=02427∴2550-123=+24272550-12310补=2550-012310补=255010补+-012310补=02550+99877=02427∴2550-123=+24271.9分别用“对9的补数“和”对10的补数完成下列十进制数的运算2537-846解：2537-8469补=5379补+-8469补=0537+9153=9690∴537-846=-309537-84610补=53710补+-84610补=0537+9154=9691∴537-846=-3091.10将下列8421BCD码转换成十进制数和二进制数1011010000011201000101.1001解：10110100000118421BCD=683D=10101010112201000101.10018421BCD=45.9D=101101.111021.11试用8421BCD码、余3码和格雷码分别表示下列各数157810211001102解：57810=0101011110008421BCD=100010101011余3=10010000102=1101100011G解：11001102=1010101G=10210=0001000000108421BCD=010000110101余31.12将下列一组数按从小到大顺序排序110110012,135.68,2710,3AF16,001110008421BCD110110012=21710135.68=93.75103AF16=43110001110008421BCD=3810∴按从小到大顺序排序为：2710,001110008421BCD,135.68,1101100123AF16,2.1分别指出变量A,B,C,D在何种取值时,下列函数的值为1第二章逻辑代数基础2.2用逻辑代数的公理、定理和规则证明下列表达式2.4求下列函数的反函数和对偶函数2.5答：1正确2不正确,A=0时,B可以不等于C3不正确,A=1时,B可以不等于C4正确2.6用代数法化简成最简“与或”表达式ABCD00011110000111101110111111111000ABCD000111100001111011101111111110001111000000001111ABCD00011110000111100000111111110000ABCD00011110000111101111ABCD000111100001111011111111ABCD0001111000011110G是F的子集ABCD00011110000111101d0ddd10d1101d0d11111111ABCD000111100001111011111111ABCD00011110000111101111ABCD00011110000111103.1将下列函数化简,并用“与非”、“或非”门画出逻辑电路图,解1解1解:解:解：&≥11解：3.2将下列函数化简,并用“与或非门”画出逻辑电路图,解：解：ABCZ000000110100011110001010110111113.3解：由时间图得真值表如下：3.4解：3.51解：一位二进制数全减器：2全加器3.6解：当A=B时等效F1=F2=F3=0XYY3Y2Y1Y00000000100011001001110013.7解1&&11XYY4Y3Y2Y1Y000000000100001100100011110113.7解2&1C1C0XY000000100100011110011010110111113.8解：真值表如下：&1&&3.9依题意得真值表如下：B8B4B2B1F7F6F5F4F3F2F1F0000000000000000100000101001000010000001100010101010000100000010100100101011000110000011100110101100001000000100101000101依真值表得:3.10依题意得真值表如下：y1y0x1x0Z1Z00000110001010010010011010100100101110110010111011000101001101010111011011100101101101110101111113.11依题意得真值表如下：B1B0B1B0F00001000100010000111010000101101101011101000010011101011011011001110101110011111=1=1=1=16.1：用两个4位二进制并行加法器实现两位十进制8421BCD码到余3码的转换高位低位A3A2A1A0B3B2B1B0A>BAba=baBAba=ba