Digital Electronics - Combinational Logic Analysis - Discussion
Discussion Forum : Combinational Logic Analysis - General Questions (Q.No. 8)
8.
Implementing the expression 
with NOR logic, we get:
Discussion:
4 comments Page 1 of 1.
KIRAN V said:
5 years ago
According to De-Morgan's Theorem.
(A'.B')' = (A+B).
So, ((A'.B') + (C+D+E))' ==> (A+B) + (C+D+E)'.
(A'.B')' = (A+B).
So, ((A'.B') + (C+D+E))' ==> (A+B) + (C+D+E)'.
Foysal said:
4 years ago
A nor B= (A+B)'=A'B' (According to de morgan theorems).
(A'B') NOR C NOR D NOR E= ((A'B')+(C+D+E))'
(A'B') NOR C NOR D NOR E= ((A'B')+(C+D+E))'
FIRY said:
1 decade ago
A nor B = A'B'.
A'B' nor C nor D nor E= (A'B' + (C+D+E))'.
So Answer is A.
A'B' nor C nor D nor E= (A'B' + (C+D+E))'.
So Answer is A.
T.K said:
8 years ago
A'.B'=(A+B)'.
So ((A+B)'+(C+D+E))'.
So, Ans is A.
So ((A+B)'+(C+D+E))'.
So, Ans is A.
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