The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+22 votes
4.7k views

What is the minimum number of gates required to implement the Boolean function $\text{(AB+C)}$ if we have to use only $2\text{-input NOR}$ gates?

  1. $2$
  2. $3$
  3. $4$
  4. $5$
asked in Digital Logic by Veteran (59.7k points)
edited by | 4.7k views

5 Answers

+50 votes
Best answer

Given boolean function is 

 f$ = AB + C$

   $= (A+C) . (B +C)$ 

   $=((A+C)' +(B+C)')'$

Therefore, $3$ NOR gate required .

answered by Loyal (6.1k points)
edited by
0
Silly question but
Where is this rule written AB+C = (A+c)(B+C)

I just can figure out by looking at the question also never found this rule before anywhere.
0

It is a rule you can check it by using truth table or look into morris mano postulate 4, distributive (b)

0
beautiful explanation.
+6 votes
Answer should be 3 nor gates...A.B + C can be written as (A+C).(B+C) which is a OR-AND realization and can be made with 3 NOR gates as OR- AND is similar to a NOR-NOR realization...
answered by (81 points)
edited by
+2 votes

Option : B 

Solution: 

answered by (57 points)
+2 votes

Almost same solution,

answered by Loyal (8k points)
0 votes
answer - D

AB + C = ((AB + C)')' = ((AB)'C')' = ((A' + B')C')' = (A' + B')' + C

two NOR gates for complementing A and B

one for computing (A' + B')'

one to compute ((A' + B')' + C)'

one to negate the last result
answered by Loyal (8.8k points)
Answer:

Related questions



Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

44,284 questions
49,776 answers
164,293 comments
65,856 users