數(shù)字電路與數(shù)字電子技術(shù)課后答案

第四章邏輯函數(shù)及其符號簡化列出下述問題的真值表，并寫出邏輯表達式：有A、B、C三個輸入信號，如果三個輸入信號中出現(xiàn)奇數(shù)個1時，輸出信號F=1，其余情況下，輸出F=0.有A、B、C三個輸入信號，當三個輸入信號不一致時，輸出信號F=1，其余情況下，輸出為0.列出輸入三變量表決器的真值表.解:(1)ABCFABCF00000011010101101001101011001111 F=C+B+A+ABC(2)ABCFABCF00000011010101111001101111011111F=(A+B+C)(++)(3)ABCFABCF00000010010001111000101111011111 F=BC+AC+AB+ABC2.對下列函數(shù)指出變量取哪些組值時,F的值為“1”：(1)F=AB+(2)F=AB+C(3)F=(A+B+C)(A+B+)(A++C)(A++)解:AB=00或AB=11時F=1ABC110或111,或001,或011時F=1ABC=100或101或110或111時F=13.用真值表證明下列等式.(1)A+BC=(A+B)(A+C)(2)BC+AC+AB=BC+AC+AB(3)=ABC+(4)AB+BC+AC=(A+B)(B+C)(A+C)(5)ABC+++=1證:(1)ABCA+BC(A+B)(A+C)0000000100010000111110011101111101111111(2)ABCABC+ABC+ABCBCABC+ACABC+ABABCABCABC+ABC+ABCBCABC+ACABC+ABABC0000000100010000111110000101111101111100(3)ABCAB+BC+ACABC+ABCABCAB+BC+ACABC+ABC0001100100010000110010000101001100011111ABCAB+BC+AC(A+B)(B+C)(A+C)00000ABCAB+BC+AC(A+B)(B+C)(A+C)0000000100010000111110000101111101111111(5)ABCABC+A+B+C000100110101011110011011110111114.直接寫出下列函數(shù)的對偶式F′及反演式的函數(shù)表達式.(1)F=[B(C+D)][B+B(+D)](2)F=A+(+)(A+C)(3)F=AB++(4)F=解:F`=[+B+CD]+[(B++)B+D]]=[A++]+[(+C+D)+C]]F`=(A+)=(+)F`=+=+5.若已知x+y=x+z，問y=z嗎?為什么?解:y不一定等于z,因為若x=1時,若y=0,z=1,或y=1,z=0,則x+y=x+z=1,邏輯或的特點,有一個為1則為1。6.若已知xy=xz，問y=z嗎?為什么?解:y不一定等于z，因為若x=0時，不論取何值則xy=xz=0,邏輯與的特點,有一個為0則輸出為0。7.若已知 x+y=x+z Xy=xz問y=z嗎?為什么?解:y等于z。因為若x=0時，0+y=0+z，∴y=z，所以xy=xz=0，若x=1時,x+y=x+z=1，而xy=xz式中y=z要同時滿足二個式子y必須等于z。8.用公式法證明下列個等式(1)++BC+=+BC證:左=+BC+=+BC+=(1+)+BC=+BC=右邊(2)C+BD+ACD+B+CD+B+BCD=C+B+BD證:左 =(C+CD+ACD)+(ABCD+BCD+BD)+(BD+B+B) =C(+D+AD)+BD(AC+C+)+B(D++) =C+B+BD(3)++=1證:左 =(+D)+()+(C+) =[(+)(+)+D](+)+C+ =[++++D][+]+C+ =[++D][+]+C+ =+++D+C+ =++C+ =1x+wy+uvz=(x+u+w)(x+u+y)(x+v+w)(x+v+y)(x+z+w)(x+z+y)證:對等式右邊求對偶,設(shè)右邊=F,則F` =xuw+xuy+xvw+xvy+xzw+xzy =xu(w+y)+xv(w+y)+xz(w+y) =(w+y)(xu+xv+xz)F`` =F=wy+[(x+u)(x+v)(x+z)] =wy+[(x+xu+xv+uv)(x+z)] =wy+[(x+uv)(x+z)] =wy+[x+xuv+xz+uvz] =wy+[x+uvz] =wy+x+uvzA⊕B⊕C=A⊙B⊙C證:左 =(A⊕B)⊕C =+(A⊕B) =(A⊙B)C+() =A⊙B⊙C(6)=⊙⊙證:左 = =[(A⊕B)+](A⊙B)+C] =(A⊙B)+[(A⊕B)C] =+AB+BC+AC右 =(⊙)⊙ =[(⊙)+] =[(+AB)+] =+AB+ =+AB+(A⊕B)C =+AB+BC+AC9.證明(1)如果a+b=c，則a+c=b，反之亦成立(2)如果+ab=0，則=a+b證:a+c=a()+(a+b)=a(ab+)+b=ab+b=b(2)+ab=0說明a=或b= == =(+)(a+) =a++ =a+ =a+b10.寫出下列各式F和它們的對偶式，反演式的最小項表達式(1)F=ABCD+ACD+B(2)F=A+B+BC(3)F=+解:F=∑m=∑m(0,1,2,3,5,6,7,8,9,10,13,14) F`=∑m(15,14,13,12,10,9,8,7,6,5,2,1)F=∑m(2,3,4,5,7)=∑m(0,1,6)F`=∑m(7,6,1)F=∑m(1,5,6,7,8,913,14,15)=∑m(0,1,3,4,10,11,12)F`=∑m(15,13,12,11,5,4,3)11.將下列函數(shù)表示成最大項之積(1)F=(A⊙B)(A+B)+(A⊙B)AB(2)F=(A⊕B)+(B⊕C)解:F=(A⊙B)A+B+AB)=(+AB)(A+B)=AB+AB=AB=∑m(3)=ΠM(0,1,2)F=(A⊕B)+(C+B)=B+A+C+B=B+A+C=∑m(1,2,3,4,5)=ΠM(0,6,7)12.用公式法化簡下列各式(1)F=A+AB+ABC+BC+B解:F=A(1+B+BC)+B(C+1)=A+B(2)F=AC+D+A解:F=A+A+DF=(A+B)(A+B+C)(+C)(B+C+D)解:F`=AB+ABC+C+BCD =AB+C+BCD =AB+CF``=F=(A+B)(+C)F=解:F=AB++BC+=AB+C+F=解:F=C+ACF=(x+y+z+)(v+x)(+y+z+)解:F`=xyz+vx+yz=vx+yz+xyz=vx+yzF``=F=(v+x)(+y+z+)13.指出下列函數(shù)在什么輸入組合時使F=0(1)F=∑m(0,1,2,3,7)(2)F=∑m(7,8,9,10,11)解:F在輸入組合為4,5,6時使F=0F在輸入組合為0,1,2,3,8,10,11,13,14,15時使F=014.指出下列函數(shù)在什么組合時使F=1(1)F=ΠM(4,5,6,7,8,9,12)(2)F=ΠM(0,2,4,6)解:F在輸入組合為0,1,2,3,8,10,11,13,14,15時使F=1;F在輸入組合為1,3,5,7時使F=115.變化如下函數(shù)成另一種標準形式(1)F=∑m(1,3,7)(2)F=∑m(0,2,6,11,13,14)(3)F=ΠM(0,3,6,7)(4)F=ΠM(0,1,2,3,4,6,12)解:F=ΠM(0,2,4,5,6)F=ΠM(1,3,4,5,7,8,9,10,12,15)F=∑m(1,2,4,5)F=∑m(5,7,8,9,10,11,13,14,15)16.用圖解法化簡下列各函數(shù)(1)化簡題12中(1),(3),(5)(2)F=∑m(0,1,3,5,6,8,10,15)(3)F=∑m(4,5,6,8,10,13,14,15)(4)F=ΠM(5,7,13,15)(5)F=ΠM(1,3,9,10,11,14,15)(6)F=∑m(0,2,4,9,11,14,15,16,17,19,23,25,29,31)(7)F=∑m(0,2,4,5,7,9,13,14,15,16,18,20,21,23,25,29,30,31)解:化簡題12中(1),(3),(5)ABABCD0001111000011110(b)0100010001110111ABC000111ABC0001111001Fˊ=A+B(a)01110111=+A F=(A+B)(+C)⑤F=(AC+C)(+AC+) =AC+C+ACABABC0001111001(C)00001011F=AC+C圖P4.A16(1)(2)F=∑m(0,1,3,5,6,8,10,15)ABABCD0001111000011110圖P4.A16(2)11111111F=+D+D +A+ABCD+BC(3)F=∑m(4,5,6,8,9,10,13,14,15)ABABCD0001111000011110圖P4.A16(3)111111111F=B+A+ABD +BC+AC(4)F=ΠM(5,7,13,15)ABABCD0001111000011110圖P4.A16(4)0000=BD F=+(5)F=ΠM(1,3,9,10,11,14,15)ABABCD0001111000011110圖P4.A16(5)0000000=AC+D F=(+)(B+)(6)F=∑m(0,2,4,9,11,14,15,16,17,19,23,25,29,31)ABABCD00000101101011011110110000011110圖P4.A16(6)11111111111111F=++BCD+BE+ABE+ACDE+A+AE(7)F=∑m(0,2,4,5,7,9,13,14,15,16,18,20,21,23,25,29,30,31)ABABCD00000101101011011110110000011110圖P4.A16(7)111111111111111111F=ACE+BE+BCD+C+17.將下列各函數(shù)化簡成與非一與非表達式,并用與非門實現(xiàn)(1)F=∑m(0,1,3,4,6,7,10,11,13,14,15)(2)F=∑m(0,2,3,4,5,6,7,12,14,15)(3)F=∑m(0,1,4,5,12,13)(4)F=ΠM(4,5,6,7,9,10,11,12)解:圈“1”格化簡(1)F=∑m(0,1,3,4,6,7,10,11,13,14,15)ABABCD0001111000011110(a)11111111111(b)圖P4.A17(1)F=AC+BC+D++ABD=(2)F=∑m(0,2,3,4,5,6,7,12,14,15)ABABCD0001111000011110(a)1111111111(b)圖P4.A17(2)F=C+BC++B+B=(3)F=∑m(0,1,4,5,12,13)ABABCD0001111000011110(a)111111(b)F=+B=圖P4.A17(3)ABCD0001111000ABCD0001111000011110(a)1001101010101010(b)圖P4.A17(4)F=+ABD+ABC+=18.將下列各函數(shù)化簡成或非一或非表達式并用或非門實現(xiàn)(1)F=∑m(0,1,2,4,5)(2)F=∑m(0,2,8,10,14,15)(3)F=A+C+CD(4)F=AB+C+C解:圈“0”格化簡(1)F=∑m(0,1,2,4,5)ABCABC0001111001(a)11011001(b)圖P4.A18(1)=AB+BCF=(+)(+)=ABCD00011110ABCD0001111000011110(a)1001000000101011(b)圖P4.A18(2)=D+B+D+BF=(A+)(+C)B+)A+)=(3)F=A+C+CDABABCD0001111000011110(a)0001000111011101(b)圖P4.A18(3)=+ABF=(A+C)+)=(4)F=AB+C+CABCABC0001111001(a)00101111(b)圖P4.A18(4)=+F=(A+C)(B+C)=19.將下列各函數(shù)化簡為與或非表達式,并用與或非門實現(xiàn).(1)F=A+C+C+A+B+DF=∑m(1,2,6,7,8,9,10,13,14,15)F=∑m(0,1,3,7,8,9,13,15,17,19,23,24,25,28,30)解:圈“0”格化簡(1)F=A+C+C+A+B+DABCDABCD0001111000011110(a)0111111111011111(b)圖P4.A19(1)=+ABCDF=(2)F=∑m(1,2,6,7,8,9,10,13,14,15)ABCDABCD0001111000011110(a)0001101101101111(b)圖P4.A19(2)=CD+B+B+F=F=∑m(0,1,3,7,8,9,13,15,17,19,23,24,25,28,30)ABABCD00000101101011011110110000011110(a)10111100101110011110001100000111(b)圖P4.A19(3)=C+A+BD+D+C+ABCEF=20.用卡諾圖將下列含有無關(guān)項的邏輯函數(shù)化簡為最簡“與或”式和最簡“或與”式。(1)F=∑m(0,1,5,7,8,11,14)+∑m(3,9,15)(2)F=∑m(1,2,5,6,10,11,12,15)+∑m(3,7,8,14)(3)F=AB+A+C+AC,變量A,B,C,D不可能出現(xiàn)相同的取值.(4)F=+ABC+C,約束條件A⊕B=0解:化簡為最簡“與或”式 圈“1”格,化簡為最簡“或與”式圈“0”格(1)F=∑m(0,1,5,7,8,11,14)+∑m(3,9,15)ABABCD0001111000011110(a)1111xx1x11ABCD0001111000011110(b)000xx000圖P4.A20(1)F=+D+CD+ABC=C+C+AB+BF=(A++D)B++D)++C)(+C+D)(2)F=∑m(1,2,5,6,10,11,12,15)+∑m(3,7,8,14)ABABCD0001111000011110(a)1x11xx1111x1ABCD0001111000011110(b)00x00xx圖P4.A20(2)F=C+D+A=+ADF=(A+C+D)+C+)ABCD0001111000011110(a)x1111x11ABCD000111100001ABCD0001111000011110(a)x1111x11ABCD0001111000011110(b)x00000x000圖P4.A20(3)F=A+ =+CD+BC F=(A+C)(+)(+)(4)F=+ABC+C,約束條件A⊕B=0ABABCD0001111000011110(a)1xx1xxx1x1x11ABCD0001111000011110(b)x0xx0x0xxxx圖P4.A20(4)F=AC++ =B+CD F=(+C)A++)21.在輸入只有原變量條件下，用最少與非門實現(xiàn)下列函數(shù)。(1)F=A+B+C(2)F=∑m(1,3,4,5,6,7,9,12,13)(3)F=∑m(1,2,4,5,10,12)(4)F=∑m(1,5,6,7,9,11,12,13,14)解:可利用禁止原理(1)F=A+B+CABCABC000111100101111111(a)(b)圖P4.A21(1)F=A+B+C=F=∑m(1,3,4,5,6,7,9,12,13)ABABCD0001111000011110(a)0110111111001100(b)圖P4.A21(2)F=B+D=(3)F=∑m(1,2,4,5,10,12) ABCDABCD0001111000011110(a)0110110000001001(b)圖P4.A21(3)F=B）+C）+D（=B+C+D=(4)F=∑m(1,5,6,7,9,11,12,13,14)ABCDABCD00011110000111100010111101010110(a)(b)圖P4.A21(4)F=AB+BC+AD+D=AB+D+BC+AD=22.輸入只有原變量條件下,用或非門實現(xiàn)下列函數(shù)(1)F=∑m(0,6,7)(2)F=∑m(0,1,2,3,4,6,7,8,9,11,15)(3)F=∑m(0,4,5,7,11,12,13,15)解:輸入只有原變量條件下用或非門實現(xiàn)邏輯函數(shù)時,應(yīng)先求出F的對偶式F`,將F`化為與非一與非表達式,再求一次對偶F``=F,即可得出F的或非一或非表達式.(1) F=∑m(0,6,7) =∑m(1,2,3,4,5) F`=∑m(6,5,4,3,2)ABABC0001111001圖P4.A22(1)(a)01110101F′=A+B=F″=F=圖P4.A22(1)(b)(2)F=∑m(0,1,2,3,4,6,7,8,9,11,15)=∑m(5,10,12,13,14)F′=∑m(10,5,3,2,1)ABABCD0001111000011110圖P4.A22(2)(a)0000110