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1、数字电子技术基础习题集项目一习题1 .将下列二进制数转换为十进制数(1) 10101(2) 0.10101(3) 1010.1012 .写出下列八进制数的按权展开式(1) (247) 8(2) (0.651 ) 8(3) (465.43) 83 .将下列十六进制数转换为十进制数(1) (6BD 16(2) (0.7A) 16(3) (8E.D) 164 .将下列十进制数转换为二进制数，小数部分精确到小数点后第四位(1) (47) 10(2) (0.786) 10(3) (53.634) 105 .将下列二进制数转换为八进制数(1) (10111101) 2(2) (0.11011 )2(3)
2、(1101011.1101)26.将下列二进制数转换为十六进制数(1) (1101111011) 2(2) (0.10111 ) 2(3) (110111.01111)27 .指出下列逻辑函数式中A、B、C取哪些值时，F=1。(1) F (A.B.C) =AB+AC(2) F(A.B.C.尸A+BC(A+B)(3)F (A.B.C) =A B+ABCA BC8 .用公式法化简下列函数，使之为最简与或式。(1) F=AB+A C+B C+AB CD(2) F= (A+B AB(3) F=AC+ABC+BC+ABC F=A B (C+D +BC+A B+AC+BC+B C D(5) F=(A+BC
3、)(A+DE)9 .直接画出逻辑函数F=AB+B (AC)的实现电路10 .有三个输入信号A、B、C,若三个同时为0或只有两个信号同时为1时,输出F为1,否则F为0列出其真值表。11 .用真值表证明下列等式(1) A+B=A B- 一(2) AB+AB= ( A + B) (A+B12.直接根据对偶规则和反演规则，写出下列逻辑函数的对偶函数和反函数(1) F=A+BC+A(B+CD) F= A B +BC+AC(3) F=(A+B)(B+C)(A+C)(4) F=AB(C+BC)+A(B+C)(1)已知逻辑函数(2)已知逻辑函数(3)已知逻辑函数(4)已知逻辑函数13 .判断下列命题是否正确A
4、+B=A+C 则 B=CA+B=AB 贝U A=BAB=AC 贝U B=CA+B=A+C AB=AC 则 B=C14 .用卡诺图化简下列函数，并写出最简与或表达式(1) F (A.B.C.D) =A B C+AB D+ABCB D+A BCD(2) F (A.B.C) =AC+B C+ABC(3)F (A.B.C.D) =Em (0, 2, 3, 7)(4) F (A.B.C.D) =Em (1, 2(5) F (A.B.C.D) =Em (0, 1,(6) F (A.B.C.D) =Em (0, 2, F (A.B.C.D) =Em (1, 3,4, 6, 10, 12, 13, 14)4,
5、 5, 6, 7, 9, 10, 13, 14, 15)4, 7, 8, 10, 12, 13)4, 7, 13, 14) +Z2d (2, 5, 12(8) F(A.B.C.D) =Em(0, 1, 12, 13,14)+ 汇 d (6, 7,15)(9) F(A.B.C.D) =Em(0, 1, 4, 7, 9,10,13) + 2 d (2,5,8,12, 15)(10) F (A.B.C.D) =Em (0, 2, 7, 13, 15)且 ABC+ABD + A B D=0第一章习题答案1. (1)(21) 10(2)(0.9375) 10(3)(10.625) 10_, . . ，一
6、 一 _2102. (1)(247) 8=2X 8 +4X 8 +7X 8(2) (0.651 ) 8=6X 8 1+5X 8 2+1X 8 F(A.B.C尸A+BC(A+B)=A BC(A+B) =A(B+C)(A+B) =(AB+AC)(A+B) =ABC(465.43) 8=4X 82 +6X 81+5X 80 +4X 8 1 +3X 8(3)3. (1)(1725) 104. (1)(101111) 25. (1)(275) 86. (1)(77B) 16(2)(0.4765625) 10(2)(0.1100) 2(2)(0.66) 8(2) (0.B8) 16(3) (142.812
7、5) 10(3)(110101.1010) 2(3)(153.64) 8(3)(37.78) 167.解此题时应把F表达式展开成最小项标准与或式，每个最小项所对应的输入便 是问题的答案。(1) F (A.B.C) =AB+AC=AB(C+C) +AC (B+B)=ABC+AB C +A BC+A B C=m7 + m6 + m3 + m1当ABCXJ输入组合111, 110, 011, 001中任一种时，F=1当 ABO 011 时，F=1(3) F (A.B.C) = AB+ABCABC=AB (C+C) +ABC+A BC=ABC+ABC+ABC当ABC为输入组合111, 011, 010
8、中任一种时，F=18. (1) F=AB+C(2) F=AB(3) F=CF=1(5) F= A B+A C +AD+AE9.电路图如下图所示A1B=110.ABCF0001001001000111100010111101111011. (1)令 Fi=A+B12.(1) F =A(B+C)A+B(C+D)F=A(B+C)A+B(C+D)(2) F =(A+B)(B+C)(A+C)F=(A+B)(B+C)(A+C)(3) F =AB+ ( bc+ACF=AB+(BC+AC(4) F =A+B+C(B+C)(A+BC)F=A+B+C(B+C)(A+BC)ABF1F2001101001000110013.(D X(3) x(4) V(2)令 F仁AB+AB F 2=(A + B)(A+BABF1F2000001111011110014.(1)F = AB + AC+ BD +BC(2)F = A+ BC(3)F = BC + AC (4)F = CD + ABC + BD + ABCD (5)F = AC +CD + BC + ACD(6)F = CD + BD + ABC + ABCD(7)F = BC + AB + AD(8)F = AB+ ABC(9)F = C + BD+ BD(10)F = A + BD卡诺图如下(2)CDCD(8)CD(9)CD(10)
