阎石数字电路课后答案第一章习题答案

第一章

1.1二进制到十六进制、十进制

(1)(10010111)2=(97)16=(151)10 (2)(1101101)2=(6D)16=(109)10

(3)(0.01011111)2=(0.5F)16=(0.37109375)10 (4)(11.001)2=(3.2)16=(3.125)10 1.2十进制到二进制、十六进制

(1)(17)10=(10001)2=(11)16 (2)(127)10=(1111111)2=(7F)16

011)(0.63D70A)(3)(0.39) 0 1101 0111 0000 101010(0.0110216011)(19.B3)(4)(25.7)1 010(11001.1012161.8用公式化简逻辑函数

(1)Y=A+B (3)Y=1

(2)YABCABC (4)YABCDABDACD解：YAD(BCBC)AD(BCC)AD

解：YBCABCCABC（1A＋A＝1）(5)Y=0 (7)Y=A+CD

(6)YAC(CDAB)BC(BADCE)解：YBC(BADCE)BC(BAD)CEABCD(CE)ABCDE（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)YBCADAD (10)YACADAEFBDEBDE

1.9 (a) YABCBC (b) YABCABC

(c) Y1ABACD,Y2ABACDACDACD

(d) Y1ABACBC,Y2ABC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 (2)YACD

(3)Y(AB)(AC)ACBC解：Y(AB)(AC)ACBC[(AB)(AC)AC]BC (4)YABC

(ABACBCAC)(BC)BC(5)YADACBCDC解：Y(AD)(AC)(BCD)CAC(AD)(BCD)ACD(BCD)ABCD (6)Y0

1.11 将函数化简为最小项之和的形式

(1)YABCACBC解：YABCACBCABCA(BB)C(AA)BCABCABCABCABCABCABCABCABCABC（2）YABCDABCDABCDABCDABCDABCD

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数字电路 习题答案 （第一章）

（3）YABCD解：YA(BCDBCDBCDBCDBCDBCDBCDBCD)B(ACDACDACDACDACDACDACDACD)(ABABABAB)CDABCDABCDABCDABCDABCDABCDABCDABCDABCDABCDABCDABCDABCD(13)

(4)YABCDABCDABCDABCDABCDABCDABCDABCD (5)YLMNLMNLMNLMNLMNLMN

1.12 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) (2)Y(ABC)(ABC)(ABC) (3)YM0M3M4M6M7 (4)YM0M4M6M9M12M13 (5)YM0M3M5

1.13 用卡诺图化简法将下列函数化为最简与或形式：

(1)YAD (3)Y1

(2)YABACBCCD (4)YABACBC

YCDAB YABAC

(5)YBCD (6)YABACBC

(7)YC (9)YBDADBCACD (8)Y(A,B,C,D)m(0,1,2,3,4,6,8,9,10,11,14) (10)Y(A,B,C)(m1,m4,m7)

YBCDAD YABCABCABC

1.14化简下列逻辑函数

(1)YABCD (2)YCDACD (3)YABDAC (4)YBCBD (5)YABDECEBDEADACDE

1.20将下列函数化为最简与或式

(1)YACDBCDAD (2)YBADAC (3)YABC (4)YABD

(5)Y1 (6)YCDBDAC

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