2017-04-09 71 views

回答

1

下面是一个详细的解决方案,希望(i)该步骤是明确的,和(ii),我的计算是正确的:-)

该电路具有四个逻辑门:一个AND栅极与输出X = B*C(简称B AND C),一个NOT栅极它反转的A得到A'(简称NOT A)的值,一个NOR栅极与输入A'X和输出Y = (A' + X)'(简称为NOT ((NOT A) OR X)))和最后的AND门,输入为Y,AB,输出为Z = A*B*Y(简称为Z = A AND B AND Y)。对于Z的表达则是:

Z = A * B * Y = A * B * ((A' + X)') = A * B * { [A' + (B*C)]' } 

反复应用德摩根的法律来表达括号产量:

[A' + (B*C)]' = A'' * (B*C)' = A *(B' + C') 

所以

Z = A * B * [ A *(B' + C') ] = A * B * A * (B' + C') 

由于A*B*A = A*A*B = (A*A)*BA*A = A这会产生

Z = A * B * (B' + C') = A * B * B' + A * B * C' 

最后,因为B * B' = 0X * 0 = 0

Z = 0 + A * B * C' = A * B * C' = A AND B AND (NOT C) 
+0

非常感谢赞赏 – andrew

+0

不客气,保重! –