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Question

Minimum number of 2-input NAND gates required to implement the Boolean function given below is _______.

f=(¯¯¯¯A+¯¯¯¯B)(C+D)
  1. 4

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Solution

The correct option is A 4
f=(¯¯¯¯A+¯¯¯¯B)(C+D)

=¯¯¯¯¯¯¯¯AB(C+D)=¯¯¯¯¯¯¯¯AB.C+¯¯¯¯¯¯¯¯ABD

=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ABC+¯¯¯¯¯¯¯¯ABD

=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ABC.¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ABD


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