令X表示异或(XOR)运算。令“ 1”和“ 0”表示二进制常量。考虑以下针对两个变量P和Q的F的布尔表达式:
F(P, Q) = ( ( 1 X P) X (P X Q) ) X ( (P X Q) X (Q X 0) )
F的等效表达式为
(A) P + Q
(B) (P + Q)’
(C) PXQ
(D) (PXQ)’答案: (D)
解释:
我们需要简化上面的表达式。由于给定的运算是XOR,因此我们将看到XOR的属性。
令A和B为布尔变量。
在A XOR B中,如果两个位/输入都不同,则结果为1,否则为0。
现在,
( ( 1 X P) X (P X Q) ) X ( (P X Q) X (Q X 0) )
( P' X P X Q ) X ( P X Q X Q ) ( as 1 X P = P' and Q X 0 = Q )
(1 X Q) X ( P X 0) ( as P' X P = 1 , and Q X Q = 0 )
Q' X P ( as 1 X Q = Q' and P X 0 = P )
PQ + P'Q' ( XOR Expansion, A X B = AB' + A'B )
This is the final simplified expression.
Now we need to check for the options.
If we simplify option D expression.
( P X Q )' = ( PQ' + P'Q )' ( XOR Expansion, A X B = AB' + A'B )
((PQ')'.(P'Q)') ( De Morgan's law )
( P'+ Q).(P + Q') ( De Morgan's law )
P'P + PQ + P'Q' + QQ'
PQ + P'Q' ( as PP' = 0 and QQ' = 0 )
Hence both the equations are same. Therefore Option D.
这个问题的测验