數(shù)字電路與邏輯設(shè)計(jì)課后習(xí)題答案蔡良偉(第三版)

1、數(shù)字電路答案數(shù)字電路答案第一章習(xí)題(1) 2210= 2*8!+ 6*8°= 26$26s = 2 6 = 10110.010110 10110】=00010110= 1616* 1 6(2) 10810 = 1*82+ 5*8X+ 4車8°= 154s154s= 1 5 4 = 1101100.00110110011011002= 01101100= 6C166 C(3) 13.12510= 1訶+58°+廣8=15.ls15.18= 1 5. 1 = 1101.001,001101 0011101.001= 1101.0010= D.26D 2(4) 131.

2、62510= 28，+ 08】+ 38°+ 58= 203.5$203.5s = 2 0 3.5 = 10000011.1012010 000 011 10110000011.1012 = 10000011.1010= 83.A83 A1-2(1) 101101.= 101101= 55855101101.= 00101101= 2%2 D55s= 5唧+ 5*8°= 4510(2) 11100101.= 011100101= 345s34511100101,= 11100101= E56 E 5345$ = 3*82 + 4*8l+ 5*8°= 22910數(shù)字電

3、路答案(3) 101.00112 = 101.001100= 5.14s514101.0011, = 0101.0011= 5兀535.1嚴(yán) 51 卻8石 8)(4) 100111.101,= 100111.101= 47.48"474100111.101,= 00100111.1010= 27.A27 A47.58 =4*81+7*8° + 5*8'1 = 39.625101-3(1) 16$ =廣8】+68°= 1410 = 1 6 = 11®001110 1 1 ip= 1 1N(EE(2) 1728 = 1*82 + 7*8!+ 2*8&

4、#176;= 12210172s= 17 2 = 1111010.001111010 1111010 . =01111010 =7A67 A(3) 61.53s= 6*8】+ 1*8°+ 5*8*3*8 2= 49.672106153s= 6 1 5 3 = 110001.101011.110 001 101011110001.101011. = 00110001.10101100= 31.AC16"31 A C(4) 126.74s = 1*82+ 2*8l+ 6*8°+ 7 車 8+ 4*8"2 = 86.937510126.74s = 1 2 6.

5、7 4 = 1010110.1111001010110 1111001010110.1111,= 01010110.1111= 56.F1656 F1-4(1) 2入=2 A= 101010.00101010 101010.= 101010= 52s5252g= 5*8+ 2*8°= 42l0(2) B2F】6= B 2 F = 10110010111L101100101111“101100101111,= 101100101111= 5457g54575457$ = 5*83+ 4*82+ 5*8!+ 7*8°= 286310(3) 。3.耳6= D 3 E = 1101

6、0011.Ul211010011 1110 11010011.111,= 011010011.111= 323.7g3237323.7S = 3*8+ 2*8J+ 3*8°+ 7*8= 2U.8751O(4) 1C3F9“= 1 C 3 F 9 = 111000011.11111001200011100 0011 11111001111000011.11111001. = 111000011.111110010= 703.762s703762703.7628 = 7*8:+ O*8!+ 3*8°+ 7*8*l+ 6*82+ 28= 451.972610-5(1) A(B +

7、 C)=AB+ACABC左式右式0 0 0000 0 1000 1 0000 1 1001 0 0001 0 1111 1 0111 1 111左式=右式.得證。(2) A+EC = (A+B)(A+C)ABC左式右式0 0 0000 0 1000 1 0000 1 1111 0 0111 0 1111 1 0111 1 111左式=右式.得證。(3) A+B= ABAB左式右式0011010010001100左式=右式.得證。(4) AB= A+ BAB左式右式0011011110111100左式=右式，得證。(5) A+ BC + ABC = 1ABC左式右式0 0 0110 0 1110

8、 1 0110 1 1111 0 0111 0 1111 1 0111 1 111左式=右式，得證。(6) AB +AB= AB+ABAB左式右式0000011110111100左式=右式，得證。(7) AB= A®AB左式右式0000011110111100左式=右式.得證。7數(shù)字電路答案(8) AB + BC+CA= AB+BC+CAABC左式右式0 0 0000 0 1110 1 0110 1 1111 0 0111 0 1111 1 0111 1 100左式=右式.得證.1-6(1) A+ AB+ B= 1證：A+ AB+ B= A+ B+ B= A+ 1= 1(2) A+

9、BA+ CD= A證：A+BA+CD= A+ACD= A+?CD= ?(1 + CD)= A(3) AB+ A2+ BC= AB+C證：AB+ A?+ BC= AB+ (A+ B)C = AB+ 3C = AB+ C(4) AB+ A+C+B(D+E)C= AB+A?證：AB+ A+C+B(D+E)C= AB+A?+BC(D+E)= AB+A?(5) A? B AB= A+ B證：A? B AB= AB+ AB+ AB= A+ AB= A+ B(6) AB+ BC+ CA= ABC+ ABC證：AB+ BC+ CA= (A+ B)(B+ C)(C+ BC+ ABC(7) ABD+ BCD+ A

10、D+ ABC+ ABCD= AB+ AD+ BC證：原式=ABD -k ABCD + BCD + AD-b ABC + ABCD (再加一次最后一項(xiàng))=BD(A+ A?)+ BCD+ AD+ BC(A+ AD)=BD(A+ C)+ BCD+ AD+ BCAD=BD(A+ C)+ B(C+ CD) + 兀=B(CD+ C+ D)+ AD=B(AD+ D)+ CB+ AD=(AB+ BD+ AD) + CB=AB+AD+BC(8) A©B+B©C + CD= AB-fBC+ CD + DA證：原式=AB +AB+BC+BC + CD+CDAB-fBC + CD-f AB-fBC

11、+CD +AD + DA=AB + BC +CD + DA+AB+BC + CD+AD=AB + BC + CD + DA+ AB + BC + CD + DA=AB + BC 4-CD-fDA15(1)耳=ABC+ ABC耳=(A+B + C)(A+B + C)(2)F2= A(B+ C)+ C(B+ D)F =(A+BC)(C-FBD)(3)f3 =(a+bxc+d)(4) F4 = (AB+ CD)(B+ AD)巧=(A+ Bp + D)+ B(A+ D)(5) F5= AB+ A?B+ Df5=(a+b)(a+c + bd)(6) F6= A+ BC+ B+CDf, = a(bTc)b

12、cTd(7) F7= AC+ BDC+ A+ BDf7=(a+c)(bVB)-fcabd(8) Fs= (A+ D)(B+ C)+ (A+ C+ B)AB+ CDf3 = ad+bcacb+(a+bJc + d)1-8(1) F = AB+ CD(2) F2 = (A+ BXC+ D)F> = AB+ CD(3) F§= NB+b)+B(A+©F5 = (A+ BDXB+ AC)(4) F4 = (A+ BCDXABC + D)F4 =理 B+ C+ D) + (A+ B+ C)D(5) F5= A+ B+C+ DF； = TSCD(6) F6= BC+CDB(PD+

13、C)F6 = (B+ CXC+ D)+ B+ (A+ D)C(7) F7= BC+ ADAC+C+ ABF* = (B+ C)(A+ D) + A+ CCA+ B(8) Fs= ABC+ A+ CD(BD+ C)+ (BC+ A+ D)B+ A+ BCFs = (A+ B+ C)&C+ D)+ (B+ D)C(B+ C)疋 + BA(B+C)1-9(1)ABC+ABC+ABC+ ABCABCF0 0 010 0 100 1000 1 1110 001 0 111 1 001 1 1110100110Ap 00 01 11 10(2) F,= A+ BC + CDA B C DFA B

14、C DF0 0 0 0010 0 010 0 0 1010 0 110 0 10010 1010 0 11110 1110 10 00110 010 10 10110 110 1101111010 11111111100001000111111111100011110011110(3) F;= AB+ 政C+ AD)A B C DFA B C DF0 0 0 0010 0 000 0 0 1010 0 110 0 10110 1010 0 11110 1110 10 01110 000 10 11110 100 1101111000 111111110AB? 00011110000011011

15、111110000100111(4) F4 = (A+ B+ CXA+ B+ CXA+ B+ C)0001111011100011(5) F5= (BD+CXC+ AD)oooooooo oooo11OO11OO1O1O1O1OABCDooooooooF1111OOOO11OO11OO1O1O1O1OABCDooooooooFF7H A+ BC+ BA+C+ Doooooooo 1111OOOOooooooooABCD11OO11OoFooooO1O1O1oABCDF(6) F6H (AB+CDXBC+D娶k+BD)數(shù)字電路答案11ABCDFABCDF0 0 0 0110 0 010 0 0

16、1110 0 110 0 10010 1010 0 11010 1110 10 00110 010 10 10110 110 1100111010 11111111100（8） Fs = ABD-fBC-fC（A-fD）-fD2cZbD）ABCDFABCDF0 0 0 0110 0 01Q 001110 0 110 0 10110 1010 0 1111 01 L10 10 01110 010 10 11110 110 1101111000 111011110標(biāo)準(zhǔn)與或式:110000101111111100011110011110oo1111110111001111011110011110F

17、= ABCD + ABCD + ABCD + ABCD+ ABCD + ABCD + ABCD數(shù)字電路答案F =(A+B + C + D)(A+B + U +D)(A-fB + C + D Ka+B-fC +D)(A+B + C + D)(A+B + C + D)(A+B + C + D)(A+B + C + D)1-11 F=(0,2,4,5,6,7)(2) F2 =另(0,2,3,4,5,12,13,15) F3=(0,1,245) E =(24567,14,15)(5) F5 =藝(0,345,6)(6) F6 = X(0 J,3,4,5,8 JO J 1,12,13,1445)(7)

18、F7=Y(025.6)(8) F8=2：(1A5,7JU4)片=口(0,7) Fj =口(2,3,5,6,7,13,15)(3) 巧=口(6仏11,14,15)(4) F4 = n(04,23,4,6,7,10,12,14)(5) F5 = 11(345,7)(6) F6 = n(05,8,9J0J1J3)(7疔7=口(2,3,5,7) F$ =H(024,791215)(a) F(1,3610.11,12)F =口(024"&9,13,14.15)(b) F =Q(03,46&9,13,15)數(shù)字電路再案F = 口(125 億 10,11,12,14)1-14(1)

19、 A+ AB+ BCD= A+ B+ BCD= A+ B(2) AB+ BC+ A+ C= A+ B+ C+ B= 1 A+ B+ AB(C+ D)= AB+ AB(C+ D)= A+ BAB+ AC + BC + A? B=AB+ AC+ BC+ AB+ AB(4)_=AB+ AB+ (AC+ AB) + BC=A+ A+ BC= 1BD+ ABCD+ A+ B+ C=BD+ ABCD+ ABC=BD+ AC(D+ B)=BD+ ACA§+ (A+ C)(D+ A+ B)+ (C+ D)E(6) = AB+ AD+ CD+ ABC+ CDE=廳 + AD+ CD+ BC+ EC+

20、ABCD+ CD+ (A+ BXCD+ CD)(7) = C+ AB(CD+ CD)+ AB(CD+ CD)=C+ CD+ CD= C+ D(8) 丘 + BC+ CA+ AB= AB+ BC+ CA+ A+ B= A+ C + B+ C = 1(1) AB+ BC+ AC>100,01, u ,1000A0c wAAB+ BC+ A+ BC(2) ABC+ BC+ AD+ CD13數(shù)字電路答案ABC+ BC+ AD+ CD= BC+ AC+ D忑 C+ ABCD+ ABCD+ A? = AB+ A?+ BCD+ BCD(4) B(C+ 疋)+ C(B+ D)00 01 11 1000

21、0 0 0 0B(C+ AD) + AC(B+ D)= AB+ M+ BD(5) B(C? D) ACD+ BCD+ BCDB(C? D) ACD+ BCD+ BCD= CD+ BCD(6) ABC+ BCD+ AD+ A(B+ CD)ABC+ BCD+ AD+ A(B+ CD)= ABD+ ABD+ 疋(7) ABC+CD+ AC+ BD+ ACDABC+ CD+ AC+ BD+ ACD= ?C+ BC+ BC+ D(8) ACD+ ABCD+ AB(C+ D)+ D17ACD+ ABCD+ AB<C+ D)+ DM A+ B+ c+ Dli-615)f1(abqd)m 閡 + ABC

22、+ ACB (2) f(abcdncdo豆 2368913.15) Aoo 01 11 10f(abcd)h 斗6+ ABD+(3) r(abcd)hp 豆02468101315)6數(shù)字電路答案110 WA o9E(AB.C,D)= BD+ AD+ ABD(4) F4(AB,C,D)= a m(0丄2,3,5,7,9,10,13)巧(A BC,D) = AB+ CD+ BDC+(5) E(ABC,D)= 0M(h3,7,&9,10J4)F<(ABCD)= AD+ BC+ ACD(6) F6(AB.C,D)= 0M(347,&11J2,15)F6(AB.C,D)= CD+

23、CD+ ABC(7) F,ABC,D)= 0M(2,5,6、10、12,13,14)F?(AB,C,D)= CD+ BC+ CD(8) Fs(AB.C,D)= 0M(12367,13,14,15)Fs(ABCD)= CD+ AB+ ABC(1) F1(AB,C)= a m(02.3,7)數(shù)字電路答案A弋 0001 11 10111311<cF(ABC)=(B+ C)( A+ C)(2) F2(AB,CD)= a 1瑣0丄7,&10,12,13)F(AB,C,D)= (A+ B+ C)(A+ B+ CXB+ C+ DXA+ C+ D)(A+ B+ D)(3) F3(ABX,D)=

24、a 11X123.7,&9,12,14)Fs(ABCD)= (A+ C+ D)(A+ B+ DXB+ C+ D)(A+ C+ DXA+ B+ C)(4) F4(AB,C,D)= a m(0,25,7&1O.13J5)19FdABasH (OI+ DXA+ c+ BX0+ c+ BXA+WI+21 (6) F6(ABCD)H 0M(046glpllE5)Agoo 01 11 10F6(ABCD)H (A+ c+ D)(ffll+ c+ DXA+ 口 + D)(>l+ B+2I(>I+ B+ 6X>I+ 6+s'(7) f7(ab,qd)mom(234-1

25、P11S13JP15)10 1FHABPDI(B+ DXB+ D)(5) F5(ABCD)H om(12567101314)AB 00 01 11 101131111110001111000 01 11 10數(shù)字電路答案25F7(AB,C,D)= (A+ B+ C+ DXB+ CXA+ C)(A+ B+ D)(S 耳(AB.C,D)= 0M(O,3$6,&1O,12J5)oo y11011®1®11A11101101 11 10Fs(AB,C,D)= (B+ C+ D)( A+ C + D)(A+ B+ D)(A+ B+ C+ D)=(A+ B+ C+ D)(A+

26、B+ C+ D)(A+ B+ C+ D)1-18F = ABCD+ ABCD+ ABCD約束條件AB+AC=O(1)0101 11300a0000XXXX0Xr111數(shù)字電路答案(2) F,= ABD+ABD+BCD約束條件AB+AC=O01 少0 0 qI-F: = BD+ BD(3)F3 = BCD+ BCD+ ABCD約束條件BC+CD= 0F；= BD+ AD+ BD(4) F4 = ABCD+ BCD+ ABCD約束條件BD+ BD= 0F4 = BC + AD約束條件ABCD+ ABCD= 0(5) F- = AC+ BD+ ABC+CD數(shù)字電路答案D+BC+AC(6) F6= A

27、B+ BC+CDd(2,4,9,11,14)約束條件5 d（0丄26）=0F6 = B+ CD(1) F(AB,C,D)=堪円(0丄3,5,1OJ5)+F(AB.C,D)= AB+ &+ 7c(2) Fj(ABC,D)=遺p如5,7,& 11J4)+d(3,9,15)(IIOO.WO)PoAooo<XIoVOio01 II 10 呂+ (201690頤 MQoHvrd a) av +OD+OV + JaIK)C00/?|(二 6oo,eto)pQ H(crooav)vF6(ABCD)H CD+ BC(7) fjabgdt篋a3>6710)+ dpl24815)F7(

28、ABC9D)H A+ BD(8) fs(abqd)h 篋a0481112915)+d(2K13)5FdABCDNCD+ CD+ AD(6) F6(ABCD)H jl<236,10一匚 4)+ d(0p431213) hD數(shù)字電路答案AB 0001111000代0AX0110HX111Xh010V0V0Fs(ABCD)= CD+CD第二章習(xí)題2-1a) Zi= AB+ BC = B(A+ C)真值花ABCF0 0 000 0 100 1 000 1 111 0 001 0 101 1 011 1 11b) Z2= A+ BC + D = ABCD27A B C DF0 0 0 010 0

29、0 100 0 1000 0 1100 10 000 10 100 11000 1110A B C DF10 0 0010 0 1010 10010 110110 00110 101110011110c）Z3= C+ D+ B+ A? E C+ D+ B+ AG E=（C+ D）B+ AJ E 真值表：A B C D EF0 0 0 0 000 0 0 0 110 0 0 1 010 0 0 1 100 0 10 010 0 10 100 0 11010 0 1110ABcDEF010000010011010100010111011000011011011100011111ABcD EFABC

30、DEF100001110001100010110010100101110101100110110110101001111001101010111010101101111101101110111110a）表達(dá)式：Z= B+CACA? B C? A ABC 1I（BC） AABCF0 0 010 0 110 1 010 1 111 0 011 0 101 1 001 1 10X= B 排C Ab）表達(dá)式： _Y= AB+ BC+ PC真值表:ABCX Y0 0 00 00 0 11 10 1 01 10 1 10 11 0 01 01 0 10 01 1 00 01 1 11 12-3農(nóng)達(dá)式：Z=

31、 ADBBCDADC= BCD真值表：數(shù)字電路？?案ABCDZABcDz00001100010001110011001001010000111101110100111001010111101101101111010111111111波形圖:3)29數(shù)字電路答案4)CBD5)336)F6 =(B+CD)AB+ E= EB + (A+ B)CDE2-51) F= ABCDD 1燈F2)F = ABC AD3)F = ATDBCD4)F = ACBDBD數(shù)字電路并案5)F = ABBCCA6)F = ABBCCA2 61)F = A+ B+ C+ D35數(shù)字電路答案2) F = A+ D+ B+ C

32、55D 1 >3) F = A+ C+ B+ D+ C+ A+ C+ A+ D4) F = A+ B+ A+ B+ C+ D+ C+ DAABB>11>1nil5) F = A+ B+ B+ C+ C+ A6)F = A+ B+ C+ A+ D0101101001011010000111102-7（1）卡諾圖及表達(dá)式:ABX 00011110F = (A+ B+C+ D)(A+ B+ C+ D)(A+ B+C+ 5)( A+ B+C+ D)(A+ B+C+ D)(A+ B+ C+ D)(A+ B+C+ D)(A+ B+C+ D)(2)電路圖：2 8（1）真值表:S2SiSoA

33、BF000001000010000100000110001000001011001100001110010000010010010101010110011000011010011100011111（2）卡諾圖及農(nóng)達(dá)式:s2SiSoABF1000011000111001011001101010011010111011001011111100011100101101011101111110001110101111001111110001工00000000000000000C0000 001 011 010 110 111 101 1001110F= 'SOAB+ S.SQB+ 心瓦 + B

34、S】So+ABS1S0+ ASQ+(3)電路圖:2-9(1)真值表:ABCDFABCDF00001100010001110011001011010100110101100100011000010111101001100111000111111110(2)卡諾圖及表達(dá)式:F = BC+ BD+ ABD= BCBDABD(3) 電路圖：BDCF2-10(1)真值表:ABCZ0 0 010 0 110 1 010 1 101 0 011 0 101 1 001 1 10(2)卡諾圖及衣達(dá)式：A弋 00 01 11 100C©000得乙=ABBCAC(3) 電路圖：2-11 真值衣:AiAoBiBoFAiAoBiBoF00001100000001010010001001010100110101100100011000010111101001100111000111011111 卡諾圖及表達(dá)式:A眇。10000100001000010111000111F = AABi