Digital Electronics - Combinational Logic Analysis - Discussion
Discussion Forum : Combinational Logic Analysis - General Questions (Q.No. 18)
18.
Implementing the expression 
using NAND logic, we get:
Discussion:
6 comments Page 1 of 1.
POOJA said:
1 decade ago
For (a'+b') , second gate is there but what about complement of this?
Can anyone clarify?
Can anyone clarify?
Avinash said:
1 decade ago
Answer is A.
((AB)')'+C'+D'+E' = ((AB)'CDE)' = ((A'+B')CDE)'.
((AB)')'+C'+D'+E' = ((AB)'CDE)' = ((A'+B')CDE)'.
Ganesh said:
1 decade ago
(C'D'E')(A'B') = (A'B')=(A'+B')......By De-morgans law.
Therefore ((A'+B')CDE)'.
Therefore ((A'+B')CDE)'.
Kishore said:
9 years ago
Option A is the correct answer.
Muthulakshmi said:
9 years ago
No, option A is the correct answer.
KIRAN V said:
5 years ago
((A' + B') . C. D.E)'.
By De-Morgan's law.
(A' + B')' + C' + D' + E'.
(A" . B") + C' + D' + E'.
(A . B) + C' + D' + E'.
Apply this Expression to Option A, it will give you the Answer.
By De-Morgan's law.
(A' + B')' + C' + D' + E'.
(A" . B") + C' + D' + E'.
(A . B) + C' + D' + E'.
Apply this Expression to Option A, it will give you the Answer.
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