电子技术基础数字部分(第四版)答案.doc
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1 2 3 4 * 5 6 7 8解:0101 10101.2.1试按表121所列的数字集成电路的分类依据，指出下列器件属于何种集成度器件：(1)微处理器；⑵IC计算器：(3) IC加法器；⑷逻辑门; (5) 4兆位存储器IC解：⑴ 微处理器属于超大规模；⑵IC计算器属于大规模；(3)IC加法器属于中规模；⑷ 逻轲门属于小规模：(5)4兆位存储器IC属于萇大规 模1.3.1将下列十进制数转换为二进制数、八进制数、十六进制数和8421 BCD码(要求转换误差不大于2'4)：(1) 43 (2) 127 (3)254.25 (4) 2.718解：(1) 43D=101011 B=53O=2BH: 43 的 BCD 编码为 0100 0011 BCD2) 127D=I111111B=I77O=7FH； 127 的 BCD 编码为 0001 0010 0111BCD3) 254.25D= III III 10.01 B=376.2O=FE.4H: 00100101 0100.00100101 BCD4) 2.718D=10.1011 0IllB=2.56O=2.B7H： 0010.01 II 0001 lOOOBCDo1.3.3将下列每一二进制数转换为十六进制码：(I) I0100IB (2) 11.01101B解:(I) IOIOO1B=29H (2) 11.01101B=3.68H1.3.4将下列十进制转换为十六进制数：(1) 500D (2) 59D (3) 0.34D (4) I002.45D解：(l)500D=IF4H (2) 59D=3BH (3) 0.34D=0.570AH (4) I002.45D=3EA.7333H13.5将下列十六进制数转换为二进制数：(1)23F.45H (2)A040.51H解：(l)23E45H=10 00H 1111.01000101B(2) A040.51H=IO1O 0000 0100 0000.0101 0001B13.6将下列十六进制数转换为十进制数：(1) I03.2H (2)A45D.0BCH解：(1) 103・2H=259・125D (2) A45D.0BCH=41024.046D2.43解：(l)LSTTL驱动同类门；OL(max)二 加11 /Umax) = 加1N°l_ 8mA0.4mA=20’OH(max)N=20=OAmAI W(max) = 0・°2 加 ANoh_ OAmA0.02mA=20(2) LSTTL驱动基木TTL门；OL(max) =I /L(max)= L6 皿Nol_ 8mA1.6mA=5=OAmA1 /H(max) = 0・°4 加 A_ OAmA0.04mA=10’OH(max)N=52.4.5 解:L = ABBCDE = AB + BC + D + E263B=()时,传输门开通,L=A：B=l时，传输门关闭，A相当于经过3个反相器到达输出L, L=AA B L0 0 00 I II 0 II I 0所以，L - AB + AB = A ㊉ B2.7A 解:C , BC = BCD . ED = DEBC・ DE A BC・ DE = A(BC + DE)AF・ GF . E AF GF 二 E(AF + GF)二 EF(A + G)L = A(BC + DE\ EF(A + G) = A(BC + DE) + EF(A + G)2.7.2 解：AB • B • A = AB • (A + B) = AB + A fi = A ㊉ BL = A ㊉ B =a0b2.9.11 解：当没有车辆行驶时，道路的状态设为0,有车辆行驶时，道路的状态为I：通道允许行驶时的状态设为1，不允许行驶时的状态设为0。
设A表示通道A有无车辆的状态，Bl. B2表示通道Bl、B2有无车辆的情况，LA表示通道A的允许行驶状态，LB表示通道B的允许行驭 状态山此列出真值表ABlB2LALB000100010101001011011XX10LA = A + A • B] • = A + B[ •LB = LA = A + B[ • B? = A(B)+ 艮)3-1.2用逻辑代数证明下列不等式(a) A + A5 = A + B山交换律 A + BC = (A + B)G4 + C),得A + AB = (A + A)(A + B) = A + B(b) ABC + ABC + ABC = AB + ACABC + ABC + ABC = A(BC + BC + BC) = A(C + SC)= A(C + B) = AB + AC(c) A + ABC+ACD + (f+D)E = A-^-CD + EA + ABC+ACD + (^ + D)E = A + ACD + (C+D)E=A + CD + ~CD E = A + CD + E3.1.3用代数法化简下列等式(a) AB{BC + A)AB(BC + A) = ABC + AB = AB(b) (A + B)(AB)(A + B)(AB) = AB(c) Xbc(b+c)入 BC(B + /) = (A + P + C)(B + C)=AB + BC + AC + B C + C = AB + C(d) A + ABC + A BC + CB + CBA + ABC + ABC+ CB + CB = A + C(e)AB + AB + AB + AB= A + A =0⑴(A + B)+ (A + B)+ (AB)-(AB)(A + B)+ (A + B)+(AB)- (AB) = (A + 2?) • (A + ^) •(AB\(AB)=(AB + + B)(AB + AB) = B(AB + AB) = AB(&)(A + B + C )(A + B + C)(A + B + C)(A + B + C) = A + B(h)ABC + ABC + ABC + A+ BCABC + ABC + ABC + A + BC=A + ABC + BC=A + BC +BC = A + C⑴ AB + (4 + B)AB + (A7fi) = AB-(A + B) = (A + B)(A + B) = A ㊉ 3(j) B + ABC + 4C ABB + ABC + AC+AB = B+ ABC + AC=B + AC + AC (k)ABCD + ABD + BCD + A BCD + BC AB(D+ ABCh- BCD+ABCZ> BC = ABC+ ABCh- B(CD + C)=ABC+ABCh- B© + 万)=ABC+ABEh- BC + BD = B(AC+AZ)+C+D) = B(A+C+A + D) = AB+ BC + BD ⑴ AC + ABC + BC + ABCAC+ABC+BC+A£C=(AC+ABQ(B+C)(A+B+C) =(ABC+ABC)(A +B+C) = BC(A +B + C) = ABC+ BC= BC (in) AB + ABC + A(B + AB)AB + ABC + A(B + AB) = A( B + BC) + AB + AB=A(B+C) + A = A+B+C^ A =03.1.4将下列各式转换成与-或形式⑴当A㊉B = 0, C㊉D = 1时，真值为|。
于是AB=01, CD二00 或 CD= 11 时，真值为 I；AB=10, CD=00 CD= 11 时，真值为 1则有四个最小项不为o,即ABCD、A BCD . ABCD、AB CD(2)当力㊉B = 1, C㊉D =0时，真值为iAB=00, CD二 10 或 CD=01 时，真值为 1；AB=1I, CD=IO 或 CD=01 时，真值为 1则有四个最小项不为0,即ABCD、ABCD、ABCD、ABCD 人㊉ B㊉ C ㊉ D =工加(1,2,4,7,X,11,13,14)(b) A + B+ C + D+ C + D+ A + Z)A + B+C^D+C+D+A + D = (A + B)(C + D) + (C + D)(A + D) = (C + D)(A + B + D) = AC + AD + BC + BD + CD + D =AC+BC+D(C)~AC ~BD^BC ~ABAC- BD・ BC・ AB = AC- BD+ BC・ AB= (A+ C)(B +^) + (B+C)(A + B)= AB + BC+AD-^-CD+AB+AC+B + BC=B+AD+CD+AC3.1.7利用与非门实现下列函数(a) L=AB+ACL= ~AB-JC(b) L = O(A + C)L=D(A + C)= D\C(c) L = (4 + B)(C + D)L = (A + B)(C + D)=3.2.2用卡诺图法化简下列各式(a) AC + ABC + BC+ ABCAC + ABC + BC+ ABC = AC + BC + lc+ ABC=AC + C+ AflC = C +ABC= C _(b) AB CD + ABCD + AB + AD + AB C ABCCh-ABCD+AB+AD+ABC=AB^ABCD-}-Ab=A(B + BCD)+AD = AB+ACD+AD = AB+A(D+DC) =AB^-AD+AC(c)(AB + BD)C + BD (AC)+ D(A+B)(AB +BD)C +BD(AC)+D(A + B)=ABC + BCD + BD(A + C) + DAB=ABC + BCD + ABD + BCD + ABD=ABC+ BCD+AB + BCD(⑴ AB CD + D(BCD) + (A + C)BD + A (B+C) ABCD + D(BCD) + (A + C)BD + A (B+C)=ABCD + BCD + ABD + BCD + ABC=j + 7?Z] + m9 + mn + ml4 + m6 + m]4 +m4 + m5=工加(1,4,5,6,9,11,12,14)BD +ACD + ABDL(A, B, C, D)=工加(3,4,5,6,7,&9,10,12,13,14,15)AB + BC + BD + ACD + ACD + ABCD ⑴ L(A,B,C,D)=工加(0,1,2,5,6,7,8,913,14)CBADBC +CD + ABC + ACD + BCDL(A, B, C,D)=工加(0,1,4,6,9,13) + 工〃 (2,3,5,7,11,15)CBADA + D(h) LG4,5C,D) =》/(0,13,14,15) +》d(l,2,3,9,10,ll)AB^AD^AC3.3.4试分析图题3.3・4所示逻辑电路的功能。
C = （A ㊉ B）C AB = AB （。
