文档介绍：数字电路****题答案(第一章)
第一章
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)()2=()16=()10 (4)()2=()16=()10
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
(3) ()10 = ( 0011 1101 0111 0000 1010)2 = ()16 (4) ()10 = ( 0011)2 = ()16
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 =BC+ A + B+C =C + A + B+C =(1 A+A=1) 解:Y = AD(BC + B + C ) = AD(B + C + C) = AD
(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
(a) Y = ABC + BC (b) Y = ABC + ABC
(c)
Y1 = AB + AC D,Y2 = AB + ACD + ACD + ACD
(d)
Y1 = AB + AC + BC,Y2 = ABC + ABC + ABC + ABC
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
(4)
解:Y = (A + B)(A + C)AC + BC = [(A + B)(A + C) + AC]⋅ BC Y = A + B + C
= (AB + AC + BC + AC)(B + C) = B + C
(5)Y = AD + AC + BCD + C
(6)Y = 0
解:Y = (A + D)(A + C)(B + C + D)C = AC(A + D)(B + C + D)
= ACD(B + C + D) = ABCD
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****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 = LM N + LMN + LMN + LMN + L MN + LMN
将下列各函数式化为最大项之积的形式
(1)Y = (A + B + C )(A