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1、1.1 二进制到十六进制、十进制数字电路习题答案（第一章）第一章(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)101.2 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(3) (0.39)10= (0.01100011 1101 0111 0000 1010)2= (0.63D70A)16(4) (25.
2、7)10= (11001.1011 0011)2=1.8 用公式化简逻辑函数(1)Y=A+B (3)Y=1 Y = ABC + A +(4)Y = ABCD + ABD + ACD(2)BC(19.B3)16Y = BC + A + B+C =C + A +解：BC1解：Y=)=（1 AA）AD(B C + B + C ) = AD(B +C + CAD(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+
3、 (B + C )(A + B + C )(A + BC) =+ (+)解：Y = A+ (BC )(A + B + C )( A+ B + CA= A + BC (A + B + C) = A + ABC + BC = A + BC(9)Y= BC+ AD + ADABC + BC)( A+ BC(10)Y = AC + AD + AEF + BDE + BDE1.9 (a) Y =ABC + BC (b) Y = ABC + ABC(c)Y1=,AB + AC D Y2=AB + AC D+ACDACD(d)Y1= AB + AC + BC,Y2=ABC + ABC + ABC + ABC
4、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)ACBCY= A + B + C解：Y = ( A + B)(A + C)AC + BC = (A + B)( A + C) + ACBC= (+AB + AC + BC + AC)(B + C) = BC+(5)Y = AD + AC + BCDC (4)+)解：Y = ( A + D)(A + C)(B + C + D)C = AC( A + D)(B + CD= ACD(B + C + D) = ABCD1.11将函数化简为最小项之和的形式
5、(1)Y = A BC + AC + B C (6)Y= 0解：Y = A BC + AC + BC = A BC + A( B + B )C + (A + A )BC= A BC + ABC + AB C + AB C + ABC = A BC + ABC + AB C + ABC（2）Y =ABCD + ABCD + ABCD + AB CD + ABCD + ABC D13+Y = A + BCD数字电路习题答案（第一章）（ ）解：Y = A(BC D + BCD + BCD + BCD + BC D + BC D +B( AC D + ACD + ACD + ACD + AC D +
6、AC D +BCDBCD)+=ABC D +ABCD +ABCD + ABCD +ABC D +ACDACD)(ABC D + ABCD +AB + AB + AB + AB)CDABCD +ABC D +ABCD + ABCD + ABCD + ABCD(13)(4)Y = ABCD+ ABCD+ ABCD+ABC D+ABCD + ABCD +ABCD + ABCD(5)Y = LM N + LMN + LMN +LMN + L M N + L MN1.12 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
7、)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 M3M51.13 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 + C D (4)Y = AB + AC + BCY = C + D + ABY = AB + AC(5)Y= B + C + D (6)Y = AB+ AC+ BC(7)Y = C (9)Y = B D + AD + BC + ACD(8)Y ( A, B,
8、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 + ADY = ABC +)1.14 化简下列逻辑函数(1)Y = A + B + C + D (2)Y = C D + A CD(3)Y = AB + D + AC (4)Y = BC + B D(5)Y = AB + DE + CE + BDE + A D + ACDE1.20 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 +
9、 B D(5)Y =1 (6)Y = CD + B D + AC2ABCABC2.1 解:数字电路习题答案（第二章）第二章(a)当vi0V时， vB= 10+5.120 5.1 = 2VT截止 vo10V当vi5VI时，B50.75.110.720=0.3mAI10BS302=0.17 mAIBT饱和vo0.2V (0 0.3V都行）悬空时， vB负值， T截止，vo10V(bv)当 vi0V时，为负值5vVBT截止I当vi5V时，50.78.7o= 0.42mA5=B54 7。18IBS5020.05mAIBT饱和vo0.2V (0 0.3V都行）I悬空时，50.78.7= 0.08mAI=
10、0.05B4.718 T0.2(0 0.3mAI，vVVBSB饱和，都行）2.3 解：s 闭合时，输入低电平，此时VIL= R2 5I IL 0.4VR2o0.4=0.4V= 200R 的最大允许值为200s 断开时，输入为高电平，此时5IIL2mAV4V21V= (+ R) 5IVV4V+ Rcc= 10KIHccR12IH10KRR125IIH0.1mAR 最大允许值为2.4 解：12I(max)8GM输出为低电平时，扇出系数 NOLIIL=0.420I(max)0.42.5 解：GM输出为高电平时，扇出系数 NOH所以，N20 IIH=0.02=20GMN输出为低电平时，扇出系数IOL(max)IILIOH(max)16= 101.60.4GM所以，N5N
