数字电子技术基础 阎石 第四版课后习题答案详解 .docx

第一章 1.1 二进制到十六进制、十进制 (1) 2 = (97 )16= (151)10 (3) (0.2= (0.5F)i6= (0io 1. 2 十进制到二进制、十六进制  (2) (1101101)2 =(6D)1 6=(109)1 0 (4) (11.001)2 =(3. 2)16=(3. 125) 10 (1) (17)10=(10001)2 = (11 )16 (2) (127)10=(1111111)2 = (7F )16 (3)(0.039=)(10.010100111101011100001010)=2 1 . 8 用 公 式 化 简 逻 辑 函 数 (1)Y=A+B 2 (0.63D70AX ⑶ Y=1 (4)(25.7)01 =(11001.1011 001 i)2 =(19.53)! ( )Y=ABC+A+B+C 角率Y：=BC+A+B+C=C+A+B+C=\iA+A=\) (5)Y=0 (4)7 = ABCD + ABD + ACD 解：Y = AD(FC + B + U) = AD(B + 己 + C) = AD (7)Y=A+CD (6)7 =+AC(CD AB) + BC(B+AD + CE) (6)7 = + 角率：Y =BC(B-AD + CE) = BC(B+AD)-CE = ABCD(C +E)= ABODE (8)Y = A + (B + C)(A + B +C)(A+B + C) 解 ：Y = A +(B-C)(A + B+C)(A + B+C) = A + (ABC+BC)(A + B+C) =A + BC(A + B+C) = A + ABC +BC = A+BC (9)7 =BC +AD + AD 9 (a ) Y = A B C + BC (b ) (10)7 =AC + AD + AEF + BDE + BDE Y = A B C + AB C Yl= A B + AC D ,Y 2 = A B + a TH5 + ACD + ACD Y1=AB+AC+ BC,Y=2ABC + ABC + ABC + ABC 1. 10 求下列函数的反函数并化简为最简与或式 (1)Y = AC+BC  ⑵ F =：i + C + 万 (3)7+={A+E)(A + C)AC BC (3)7 + + + + = {AB + AC+BC+AC){B+C) =B+C (^ Y = A + B + C (5)Y = AD +AC +BCD +C _ --------------- — (6) 7 = 0 =ACD(B +C+D) = ABCD 11 将函数化简为最小项之和的形式 Q )Y = A B C + AC + BC 解： F = +WC + 互 + + 互)C + (W + Z)互C = ABC +ABC +ABC +ABC +ABC = ABC +ABC + ABC + ABC Y = ABCD + ABC D + ABCD + ABC D + ABCD + A B C D 数字电路习题答案 数字电路习题答案 （第一章) Y = A + B + CD 解 ：Y = A(BC D + BCD + BCD + BCD + B C D + BCD + BCD + BCD) + B (1 C D + ACD + AC D + AC D + ACD + AC D + ACD + ACD) + (AB + AB + AB + AB)CD = ABC D + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD (13) Y = A B C D + A B C D ~ A B C D + A B C D + A B C D + A B C D + A B C D + A B C D Y = LM N + LMN + LM N + LMN + L M N + LM N 1 .1 2 将下列各函数式化为最大项之积的形式 (1) Y = (A +B +C )(A +B +C)(A + B + C ) Y = M 0 M 3 M 4 M 6 M 7 (2) Y = (A + B + C)(A + B + C)(A + B + C) (4) Y = M 0 M 4 M 6 M 9 M 12 M 13 .1 3 用卡诺图化简法将下列函数化为最简与或形式 : (1 )Y = A + D (2) Y = AB + A C + BC + CD  (3 )Y = 1 Y = AB +AC +BC Y = C + D