Boolean Algebra

Question: Implement XNOR gate using NOR only

Q- Implement 2 variable XNOR gate using NOR only in minimum number of gates.

F= x XNOR y = (x XOR y)’

F’= (x XOR y) = x’y+xy’ = x’y+xy’+xx’+yy’ = (x+y) (x’+y’)

F’= x’(x+y) + y’(x+y)

Take compliment

F= (x’(x+y) + y’(x+y))’ = (x’(x+y) )’ . (y’(x+y))’ =(x+(x+y)’). (y + (x+y)’)  

Take compliment again

F’= ( (x+(x+y)’). (y + (x+y)’) )’ = (x+(x+y)’)’ + (y + (x+y)’)’

Take compliment again

F= [ (x+(x+y)’)’ + (y + (x+y)’)’ ]’

Now we can implement the circuit of XNOR gate using NOR gates

Hence we need only 4 NOR GATES to implement XNOR gate and we require 5 NOR gates to implement XOR gate with 5th NOR gate used as inverter and placed in front of 4th NOR gate.

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