Question

A combination of logic gates has the truth table shown below:

$P$	$Q$	$Z$
0	0	0
0	1	1
1	0	1
1	1	1

Which combination has this table?

• A.
• B.
• C.
• D.

Answer 1

Answer: (D)

(a) $Z=\overline{\bar{P}.Q}=P+\bar{Q}$
(b) $Z=(\overline{\overline{P+Q}).\bar{Q}}=(\overline{\overline{P+Q}})+Q$
$=P+Q+Q=P+Q$
(c) $Z=\overline{(\overline{P.Q}).\bar{Q}}=(\overline{\overline{P.Q}})+\overline{\stackrel{ˉ}{Q}}=P.Q+Q=Q$
(d) $Z=P.Q$
From given truth table,
$Z=\bar{P}Q+P\bar{Q}+PQ=Q(P+\bar{P})$
$+P\bar{Q}=QP+P\bar{Q}=PQ$
So, the truth table and combination of gates in option
(d) performs AND operation.