GATE | GATE-CS-2004 | Question 90

A circuit outputs a digit in the form of 4 bits. 0 is represented by 0000, 1 by 0001, …, 9 by 1001. A combinational circuit is to be designed which takes these 4 bits as input and outputs 1 if the digit ≥ 5, and 0 otherwise. If only AND, OR and NOT gates may be used, what is the minimum number of gates required?
(A) 2
(B) 3
(C) 4
(D) 5


Answer: (B)

Explanation: We should get output 1 for values>=5

Making truth table for problem

A B C D Op
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 0
0 1 0 1 1
0 1 1 0 1
0 1 1 1 1
1 0 0 0 1
1 0 0 1 1
1 0 1 0 X
1 0 1 1 X
1 1 0 0 X
1 1 0 1 X
1 1 1 0 X
1 1 1 1 X

Putting this in kmap and solving



 

49

Here crucial point is that we need to make pair of 8 elements using don’t cares also…so final expression is

A+BD+BC

  • A+B(C+D)

Hence we’ll use two OR gate and one AND gate so total 3 gates.

Ans (B) part.

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