(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.

 

 
 
 
 
 

Answers (1)
S safeer

(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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