question_answer

Following diagram performs the logic function of

A)  OR gate                              

B)  AND gate

C)  XOR gate                           

D)  NAND gate

Correct Answer: B

Solution :

                 For our convenience, the output of first NAND gate is chosen as X as shown, Output of first NAND gate, \[X=\overline{A.B}\] Using De-Morgan's theorem,                 \[\overline{A.B}=\overline{\overline{A}.\overline{B}}\] So,            \[X=\overline{A}+\overline{B}\] Now, output of 2nd NAND gate, \[Y=\overline{X}=\overline{\overline{A}+\overline{B}}\] again       \[\overline{\overline{A}+\overline{B}}=\overline{\overline{A}}.\overline{\overline{B}}=A.B\] \[(\because \overline{\overline{A}}=A)\] Hence,          \[Y=A.B\] This is the logic function of AND gate.