: without reducing implement the following Boolean expression using basic gates
Y = A(B+C)'+AC'
Answers
Explanation:
Firstly, we need to know how to realize basic logic gates ( AND, OR, NOT) with NOR gate.
Coming to the Boolean Expression, ((A+B) C)'D. You need to segregate various basic gates that are required to make the whole expression. Follow the steps. Realize the following with NOR gate.
A+B (OR gate with NOR, inputs are A and B)
(A+B) C (AND gate with NOR, inputs are A+B [From step 1] and C)
D’ (NOT gate with NOR, inputs are D and D)
((A+B) C)'D) (AND gate with NOR, inputs are (A+B) C [From step 2] and D’ [From step 3]
Therefore, the circuit in below image is the required design to make ((A+B) C)'D)
After designing the expression with basic gates, we can remove the redundant NOR gates that are solving the same purpose. For example, after the Gate No. 8 we will obtain the expression as D, which we are giving input to Gate No. 7. So we can, directly give input D to Gate No. 9.