edited by
8,247 views
22 votes
22 votes

Let $\oplus$ and $\odot$ denote the Exclusive OR and Exclusive NOR operations, respectively. Which one of the following is NOT CORRECT?

  1. $\overline{P \oplus Q} = P \odot Q$
  2. $\overline{P} \oplus Q = P \odot Q$
  3. $\overline{P} \oplus \overline{Q} = P \oplus Q$
  4. $P \oplus \overline{P} \oplus Q = ( P \odot \overline{P} \odot \overline{Q})$ 
edited by

7 Answers

Best answer
23 votes
23 votes

Consider Option(D). LHS can be simplified as,

$$(P \oplus \overline{P}) \oplus Q = 1 \oplus Q = \overline{Q}$$

Similarly, for RHS 

$$(P \odot \overline{P}) \odot \overline{Q} = 0 \odot \overline{Q} = Q$$

$LHS \neq RHS$ therefore, Option (D) is the correct answer.


Other options can be simplified as follows.

  1. $\overline{P \oplus Q} = \overline{P\overline{Q} + \overline{P}Q} = \left(\overline{P\overline{Q}}\right) \left(\overline{\overline{P}Q}\right) = \left(\overline{P}+Q \right) \left( P+\overline{Q}\right) = PQ + \overline{P}\ \overline{Q} = P \odot Q$
  2. $\overline{P} \oplus Q = \left( \overline {P}\right)\overline{Q} + \overline{\left( \overline {P}\right)}Q = PQ + \overline{P}\ \overline{Q} =  P \odot Q$
  3. $\overline{P} \oplus\overline{Q} = \left( \overline {P}\right) \overline{\left( \overline {Q}\right)} + \overline{\left( \overline {P}\right)}\left( \overline {Q}\right) = \overline{P}Q + P\overline{Q} = P \oplus Q $
selected by
15 votes
15 votes
P XOR Q = P'Q + PQ'
(a) ~(P XOR Q) = ~(P'Q + PQ') = (P + Q') (P' + Q) = P'Q' + PQ = P XNOR Q
(b) ~P XOR Q = PQ + P'Q' = P XNOR Q
(c)  ~P XOR ~Q = P'Q + PQ' = P XOR Q

(d) (P XOR ~P) XOR Q = (PP + P'P') XOR Q = 1 XOR Q = Q'
     (P XNOR ~P) XNOR ~Q = (PP' + P'P) XNOR ~Q = 0 XNOR ~Q = Q
     
     (P XOR ~P) XOR Q != (P XNOR ~P) XNOR ~Q

Answer: D
edited by
3 votes
3 votes

P ⊕ Q = P'Q + PQ'

P XNOR Q = PQ + P'Q'

  1. P' ⊕ Q = (P')'Q + P'Q' = PQ + P'Q' = P XNOR Q [ OPTION B IS CORRECT ]

   

      2. P' ⊕ Q' = (P')'Q' + P'(Q')' = PQ' + P'Q = P XOR Q [ OPTION C IS CORRECT ]

   

      3. ( P ⊕ Q )' = ( P'Q + PQ' )' = (P'Q)'.(PQ')' = (P + Q' ) ( P' + Q ) = P.P' + P.Q + P'Q' + Q'Q = PQ + P'Q' = P XNOR Q  [OPTION A IS CORRECT ]

   

      4. ( P ⊕ P' ) ⊕ Q = 1 ⊕ Q =  Q' [ 1 ⊕ A = A' and A ⊕ A' = 1 ]

        ( P XNOR P') XNOR Q' = 0 XNOR Q' = Q  [ A XNOR A' = 0 and 0 XNOR A' = A]

         Q' ≠ Q

SO OPTION D IS NOT CORRECT

ANSWER: D

Answer:

Related questions

41 votes
41 votes
9 answers
1
32 votes
32 votes
6 answers
3