高二物理竞赛课件逻辑电路的变换

逻辑电路的变换逻辑电路的变换为了提高电路的速度，提高器件的利用率，从而减少IC的数量、也减少外部的连接线和提高电路的可靠性，需要对从逻辑表达式直接画出的逻辑电路图进行变换，尽可能使其用同一类型的输出端带非的门来实现。一、“与—或”电路变换为“与非—与非”电路例F=A+BC+DEH+G=A+BC+DEH+G(原函数二次求反)=ABCDEHG(运用反演规则)=ABCDEHG思考题：设计一个两位二进制数乘法器。分析：输入变量X=x2x1

Y=y2y1

输出变量Z=z4z3z2z11x2x1z4y2y11111111111111z3z1z2z4=

x2x1y2y1z3=

x2x1y2+x2y2y1z2=

x2x1y1+x2y2y1

+x2x1y2+x1y2y1

z1=x1y1输出函数:思考题：血液判别器。分析：输入变量：输血者血型Ai、Bi、ABi、Oi

受血者血型Ao、Bo、ABo、Oo输出变量：配血成功F输血者血型受血者血型OiAiBiABiOoAoBoABo

F=Oi+Ai(Ao+ABo)+Bi(Bo+ABo)+ABi

ABo思考题：开关控制电路

某一集体宿舍，共住有7人，公用一套照明设施，每人床头都有一个控制开关，要求每按动开关一下，就改变一次灯的状态(即亮→灭、灭→亮)。试设计出此灯控开关电路。分析：输入变量K1、K2、K3、K4、K5、K6、和K7

输出变量F

设开关初态F=0当K1K2K3K4K5K6K6均为0可以从四变量输入分析入手，找出生成输出函数的特征和规律。思考题：

码制转换电路输入为2421码ABCD输出为余3码Y3Y2Y1Y0分析列出真值表：十进制数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

0110

1000

1010

1100

111dddddd1

0001

0011

0101

0111

100ABCDY3Y2Y1Y000000001001000110100010101100111100010011010101111001101111011110

0110

1000

1010

1100

111dddddd1

0001

0011

0101

0111

10011dd1d1dd1dABCDY3Y2Y1Y000000001001000110100010101100111100010011010101111001101111011110

0

1

10

1

000

1

010

1

100

1

11dddddd1

0

001

0

011

0

101

0

111

1

00ABCDY3Y2Y1Y0000000010010001101000101011001111000100110101011110011011110111100110100010101100111dddddd10001001101010111100Y3=AY2=AB+AC+AD+BCDY1=A⊕C⊕DY0=D1dd1dd11d1dABCDY31d1dd1d11ddABCDY211dd1d1dd1dABCDY1111dddd1d1dABCDY0对应的二个不同的电路如下：F=A+BC+DEH+G=ABCDEHGA1&≥1&BCDEHGFA&≥1BCDEHGF1&二、“或—与”电路变换为“或非—或非”电路例F=(A+B)(C+D)(E+H+G)=(A+B)(C+D)(E+H+G)(原函数二次求反)=(A+B)+(C+D)+(E+H+G)(运用反演规则)=(A+B)•(C+D)•(E+H+G)(运用反演规则)A≥1BCDEHF&≥1G≥1A≥1BCDEHFG&≥1≥1三、“与—或”电路变换为“与或非”电路例F=AC+AB