3第2章逻辑门与逻辑代数基础习题与参考答案【题2-1】试画出图题2-1(a)所示电路在输入图题2-1(b)波形时的输出端B、C的波形。图题2-1解:A.....B、C【题2-2】试画出图题2-2(a)所示电路在输入图题2-2(b)波形时的输出端X、Y的波形。图题2-2解:..AB..XY...【题2-3】试画出图题2-3(a)所示电路在输入图题2-3(b)波形时的输出端X、Y的波形。图题2-3解:.AB.YX....4【题2-4】试画出图题2-4(a)所示电路在输入图题2-4(b)波形时的输出端X、Y的波形。图题2-4解:.AB.YX.....【题2-5】试设计一逻辑电路,其信号A可以控制信号B,使输出Y根据需要为Y=B或Y=B。解:可采用异或门实现,BABAY,逻辑电路如下:=1ABY....【题2-6】某温度与压力检测装置在压力信号A或温度信号B中有一个出现高电平时,输出低电平的报警信号,试用门电路实现该检测装置。解:压力信号、温度信号与报警信号之间的关系为:BAY,有如下逻辑图。1≥AB.Y...【题2-7】某印刷裁纸机,只有操作工人的左右手同时按下开关A与B时,才能进行裁纸操作,试用逻辑门实现该控制。解:开关A、B与裁纸操作之间的关系为BAY,逻辑图如下:&AB.Y...【题2-8】某生产设备上有水压信号A与重量信号B,当两信号同时为低电平时,检测电路输出高电平信号报警,试用逻辑门实现该报警装置。解:水压信号A、重量信号B与报警信号之间的关系为BAY,逻辑图如下:1≥AB.Y...【题2-9】如果如下乘积项的值为1,试写出该乘积项中每个逻辑变量的取值。5(1)AB;(2)ABC;(3)ABC;(4)ABC解:(1)A=1,B=1(2)A=1、B=1、C=0(3)A=0,B=1,C=0(4)A=1,B=0或C=1【题2-10】如果如下和项的值为0,试写出该和项中每个逻辑变量的取值。(1)AB;(2)ABC;(3)ABC;(4)ABC解:(1)A=0,B=0(2)A=0,B=1或C=1(3)A=1,B=0,C=1(4)A=0,B=1或C=0【题2-11】对于如下逻辑函数式中变量的所有取值,写出对应Y的值。(1)YABCAB;(2)()()YABAB解:(1)YABCAB)(BCAABCY00000010010001101001101111001111(2)()()YABABA当A取1时,输出Y为1,其他情况Y=0。【题2-12】试证明如下逻辑函数等式。(1)ABABCAB;(2)ABCCACABAC();(3)()()ABCBCACABCAC解:(1)左边BACBACBABA)(1右边(2)左边=ACABACCCAB)(右边(3)左边=右边)()(ACBCAACBCBCA【题2-13】对如下逻辑函数式实行摩根定理变换。(1)1YAB;(2)2YAB;(3)3YABCD();(4)4YABCCDBC()解:(1)BABAY1(2)BABAY26(3)DCBADCBADCBAY)()(3(4)BCBABCCBABCDCBACBABCDCCBABCCDCBAY)(()(4【题2-14】试用代数法化简如下逻辑函数式。(1)1()YAAB;(2)2YBCBC;(3)3()YAAAB解:(1)1()YAAB=A(2)2YBCBC=C(3)3()YAAAB=A【题2-15】试用代数法将如下逻辑函数式化简成最简与或式。(1)1YABABCABCDABCDE;(2)2YABABCA;(3)3YABABCAB()解:(1)1YABABCABCDABCDEBA(2)2YABABCA=CA(3)3YABABCAB()=CAB【题2-16】试用代数法将如下逻辑函数式化简成最简与或式。(1)1()YABCABCABCD;(2)2YABCDABCDABCD;(3)3(())YABCABCBCAC解:(1)1()YABCABCABCD=BA(2)2YABCDABCDABCD=CDAB(3)3(())YABCABCBCAC=ABC【题2-17】将如下逻辑函数式转换成最小项之和形式。(1)1()()YABCB;(2)2()YABCC;(3)3YABCDABCD();(4)4()YABBCBD解:(1)1()()YABCB=),,,(7651m7(2)2()YABCC=),(75m(3)3YABCDABCD()=),,,,,,(151413121173m(4)4()YABBCBD),(1513m【题2-18】试用卡诺图化简如下逻辑函数式。(1)1YABCABCB;(2)2YAABCAB;(3)3YACABAB;(4)4YABCACC解:(1)1YABCABCBBCA0100011110.111...111...BAY1(2)2YAABCAB;BCA0100011110111..1...AY2(3)3YACABABBCA0100011110111..1...AY3(4)4YABCACCBCA0100011110.111...111...CAY48【题2-19】试用卡诺图化简如下逻辑函数式。解:(1)(,,,)(0,1,2,8,9,10,12,13,14,15)FABCDm;ABCD0001111000011110..111111111.1....CBDBABY1(2)(,,,)(2,4,5,6,7,11,12,14,15)FABCDm;.ABCD0001111000011110.11111111.1..ACDDCADBBAY2(3)(,,,)(0,2,4,6,7,8,12,14,15)FABCDmABCD0001111000011110..11111111.1..BCDADCY3【题2-20】试用卡诺图化简如下具有任意项的逻辑函数式。解:(1)(,,,)(3,5,8,9,10,12)(0,1,2,13)FABCDmd;ABCD0001111000011110..XXX11111.1X..DCCADBBAY19(2)(,,,)(4,5,6,13,14,15)(8,9,10,12)FABCDmd;ABCD0001111000011110..111X11X1X.X...ABDBCBY2(3)(,,,)(0,2,9,11,13)(4,8,10,15)FABCDmdABCD0001111000011110..1111XXX1.X...DBADY3【题2-21】将如下逻辑函数式画成真值表。解:(1)1YABBC;ABCY100000010010001111000101011011111(2)2()YABC;ABCY0000001001000111100010111011001111(3)3()()YABBCABCY00000010010001111001101111001111【题2-22】将如下逻辑函数式画成真值表。解:(1)1FABCABCABC;ABCY00000010010101101000101111001111(2)2FABCDABCDABCDABCDABCDY0000100010001000011001000010110110101110100001001110100101101100011010111001111011【题2-23】写出图题2-23所示逻辑电路的逻辑函数式。图题2-23解:(1)BABAY(2)CBCACBAY)(【题2-24】画出如下逻辑函数式的逻辑电路图。(1)ABAB;&&1≥11ABY1....(2)ABABABC;.11&&&1≥ABCY2....(3)()ABCD;11&1≥&ABCDY3....(4)(())ABCDBC1≥1≥&&1ABCD1≥Y4....12【题2-25】写出表题2-25的与或逻辑函数式。ABCCBACBACBAY【题2-26】用与非门实现如下逻辑函数。(1)FABC=ABC.ABC.F.......&&(2)FABCD=ABCDCDAB&ABCD.F....(3)()()FABCD=BDADBCACBDADBCACBDADBCAC&&&&&ABCD.F....表题2-25ABCY0000001101000110100110111100111113【题2-27】用或非门实现题2-26中的逻辑函数。(1)FABC=CBAABC1111≥ABCF....(2)FABCD=DCBACDAB111ABC..1D.1≥1F....(3)()()FABCD=BDADBCACBDADBCACBDADBCAC)()()()())()()((DBDACBCADBDACBCA111ABC..1D.1≥F..1≥1≥1≥1≥...14