Digital Electronics - Boolean Algebra and Logic Simplification - Discussion
Discussion Forum : Boolean Algebra and Logic Simplification - General Questions (Q.No. 29)
29.
The Boolean expression 
is logically equivalent to what single gate?
is logically equivalent to what single gate?Discussion:
8 comments Page 1 of 1.
Richal said:
1 decade ago
Answer should be NOR C rite?
Sushil said:
1 decade ago
No,
bubbled OR = NAND
bubbled OR = NAND
(1)
Chandeep said:
1 decade ago
According to De-Morgan's Law, A'+B'= (A. B) '.
So, applying this to the question,
A'+B'+C'= (A. B) '+C'.
(A. B) '+C'= (A. B. C) ' [Applying DeMorgans theorem again!].
Now, (A. B. C) '=NAND gate. So in short, bubbled OR gate= NAND gate and Bubbled AND gate=NOR gate.
So, applying this to the question,
A'+B'+C'= (A. B) '+C'.
(A. B) '+C'= (A. B. C) ' [Applying DeMorgans theorem again!].
Now, (A. B. C) '=NAND gate. So in short, bubbled OR gate= NAND gate and Bubbled AND gate=NOR gate.
Karan kumar said:
1 decade ago
X = A'+B'+C'.
X' = (A'+B'+C')'.
= (A')'. (B')'. (C')'.
=A. B. C.
Then:
X= (ABC)'.
Hence NAND gate is required.
X' = (A'+B'+C')'.
= (A')'. (B')'. (C')'.
=A. B. C.
Then:
X= (ABC)'.
Hence NAND gate is required.
Vijayalaxmi choudhury said:
1 decade ago
If NOR gate will not be in option then we choose NAND gate. Because it is an universal gate.
Another reason is using a single NAND gate input we can't represent this function. It need more then one gate.
So the answer will be NOR gate.
Another reason is using a single NAND gate input we can't represent this function. It need more then one gate.
So the answer will be NOR gate.
(2)
Aditya said:
1 decade ago
Let A, B, C are the inputs for a NAND gate.
Then output is (ABC)'.
APPLYING DE MORGAN'S LAW to output we get,
A'+B'+C'
Then output is (ABC)'.
APPLYING DE MORGAN'S LAW to output we get,
A'+B'+C'
(2)
Cheralia said:
8 years ago
Using a single NAND gate we can represent;
X = A'+B'+C'= (A'+B'+C')'.
So, NAND is the correct answer.
X = A'+B'+C'= (A'+B'+C')'.
So, NAND is the correct answer.
Ravi kumar said:
3 years ago
Yes, the right answer is NAND gate.
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