Digital Electronics - Combinational Logic Analysis - Discussion
Discussion Forum : Combinational Logic Analysis - General Questions (Q.No. 5)
5.
Implementing the expression AB + CDE using NAND logic, we get:
Discussion:
11 comments Page 1 of 2.
Kranti said:
5 years ago
The bubbled OR gate is equivalent to NAND gate, so(ab* (.)cde*)*= ab + cde.
Raji oruganti said:
5 years ago
A is correct answer as per rule A"=A.
(AB)"+(CDE)" =(AB)+(CDE).
(AB)"+(CDE)" =(AB)+(CDE).
Paul said:
6 years ago
Yes, A is correct.
(A+B)' + (CDE)' = from the output of NAN Gates.
(A+B)+(CDE) = to the input of OR gate.
(A+B)' + (CDE)' = from the output of NAN Gates.
(A+B)+(CDE) = to the input of OR gate.
Asharaf ansari said:
7 years ago
No, we have to use only NAND gates.
Rohit said:
7 years ago
I am not understanding this. Please explain.
Vineet said:
8 years ago
First of all, "read" the question. They asks using NAND gate. So option c having OR gate is canceled out, option D having AND gate also cancel out, now there are A and B. We are to find first AB we see option A has AB. So, accurately answer is A.
Gaurav Chavda said:
9 years ago
A is the correct answer. Continuous 2 bubble cancels each other. So circuit in option A is simple to implement AB + CDE.
Md Nazim said:
10 years ago
Bubble at the end of and gate and bubble at the starting of or gate will cancel each other so A is correct answer. Because 1st and gate will give ab and 2nd and gate will give cde and or gate taking input 1st input=ab, 2nd input=cde will give the result ab+cde.
Jai said:
10 years ago
But there is answer A? Is right or wrong?
Priya said:
1 decade ago
Any one can explain me?
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