Digital Logic

Question

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?

 

2

3

4

5

Comments

Getting 3 :)

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Answer given is 4

Pls share your approach

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make kmap of 4 variable with 5,6,7,8,9 and from 10-15 as dont care we will get 3 term each has 2 element . try to minimize that and realize using gate.

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let four bit no. given as $A_3A_2A_1A_0$

minimized expression = $A_3+A_2A_0+A_2A_1=A_3+A_2(A_1+A_0)$

only 3 gates are required viz, 1 AND and 2 OR