Digital Logic

Question

F is a function which have three variable A,B and C. How many Universal Gates will be needed to implement this function. 

Answer 1

By, Min value theorem we will select, F's only those combination which is giving one. And if the input for a variable is $0$ then we will take it as $\bar{V}$ else if it $1$ then we will just take it as $V$. 

Now in above problem, F has two one, hence in expression, it will have two minterms,

$ F = \bar{A} \bar{B} C + A \bar{B} C $

$ F = \bar{B} C ( \bar{A} + A ) $

$ F = \bar{B} C $

Now to implement $ \bar{B} C $ we need one NOT and one AND gates. 

Now Since its asking about Universal Gates, We have to implement it by using NAND or NOR Gates only. 

NAND Gates: With Nand gate, we need one NAND for NOT and 2 NAND for AND hence total requirement is 3 NAND Gates.

NOR Gates: With Nor gate, we need one NOR for NOT and 3 NOR for AND  hence Total requirement is 4 NOR Gates. 

Comments

thanks

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Aren't only 2 NOR gates required?

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i think 2 NOR  gate is enough.

Answer 2

B'C can be implemented using 3 NAND gate and  2 NOR gates 

(C'+B)'=B'C    1 NOR gate

and for C' we required 1 NOR gate 

hence 2  NOR gate