Electronics and Communication Engineering - Digital Electronics - Discussion
Discussion Forum : Digital Electronics - Section 2 (Q.No. 29)
29.
The minimum number of NAND gates required to implement the Boolean function A +AB + ABC is equal to
Answer: Option
Explanation:
A + AB +A B C = A + AB ( 1 + C) = A + AB = A(1 + B) = A.
Discussion:
4 comments Page 1 of 1.
Ankit kumar said:
6 years ago
let Y= a+ab'+ab'c.
= a(1+b'+b'c),
= a.
We get the value "a" but my question is that we can draw the and gate by using one input line, 1*a we can draw the Nand gate can anyone explain why it is 0?
= a(1+b'+b'c),
= a.
We get the value "a" but my question is that we can draw the and gate by using one input line, 1*a we can draw the Nand gate can anyone explain why it is 0?
Lalit Kawadase said:
7 years ago
After converting the above boolean function in minimized expression we get f=A.
Now we can implement A by using 2 nand gate.{i.e Nand A=A' again Nand A'= A}.
As above we have asked for a minimum number of Nand gates And just have to implement
f=A.
Can be done by using single wire. So 0 Nand gate required
Now we can implement A by using 2 nand gate.{i.e Nand A=A' again Nand A'= A}.
As above we have asked for a minimum number of Nand gates And just have to implement
f=A.
Can be done by using single wire. So 0 Nand gate required
(1)
CHANDRA said:
7 years ago
Please anyone explain it.
Sandy said:
8 years ago
Please, anyone explain it.
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