# (a) Write the truth tables of (i) AND gate and (ii) NOT gate.(b) Show how an OR gate may be obtained with the combination of NAND gates.

(a) AND gate

A logic gate which performs the Boolean function Y = A.B

 A B AB 0 0 0 0 1 0 1 0 0 1 1 1

(ii) NOT gate

A logic which performs the Boolean function $Y = \overline{A}$

 A $Y = \overline{A}$ 0 1 1 0

(b) NAND gate

A logic gate which performs the Boolean function $Y = \overline{A.B}$

NAND gate is a universal gate

 A B $Y = \overline{A.B}$ 0 0 1 0 1 1 1 0 1 1 1 0

By using NAND gate we can make any gate.

For OR logic

OR logic is C = A + B

NAND logic C = $\overline{AB}$

By applying Demorgan's thereom

$\overline{AB}= \overline{A}+\overline{B}$

$\overline{\overline{A}.\overline{B}}= \overline{\overline{A}}+\overline{\overline{B}}=A+B$              OR gate

By using 3 NAND gate we can make an OR gate.

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