Digital Electronics - Boolean Algebra and Logic Simplification - Discussion
Discussion Forum : Boolean Algebra and Logic Simplification - General Questions (Q.No. 11)
11.
An AND gate with schematic "bubbles" on its inputs performs the same function as a(n)________ gate.
Discussion:
12 comments Page 1 of 2.
Nitesh said:
10 years ago
AND inverted NOR.
OR inverted NAND.
NAND inverted OR.
NOR inverted AND.
OR inverted NAND.
NAND inverted OR.
NOR inverted AND.
(2)
Jayamala Pakhare said:
4 years ago
For bubbled input to AND gate;
A B (A' . B')
0 0 1
0 1 0
1 0 0
1 1 0
And for NOR gate;
A B (A+B)'
0 0 1.
0 1 0.
1 0 0.
1 1 0.
So bubbled AND will have the same output as that of NOR gate.
A B (A' . B')
0 0 1
0 1 0
1 0 0
1 1 0
And for NOR gate;
A B (A+B)'
0 0 1.
0 1 0.
1 0 0.
1 1 0.
So bubbled AND will have the same output as that of NOR gate.
(2)
Sam said:
3 years ago
Agree, it's NOR.
(2)
Kathiresan said:
8 years ago
I want an explanation for this question. Can anyone provide it?
(1)
Athiban eswar said:
8 years ago
No, the correct answer is NAND.
Otherwise, I want an explanation for the given answer.
Otherwise, I want an explanation for the given answer.
(1)
Ram suraj said:
8 years ago
NO. The correct answer is NOR GATE.
We will see a example.
1. In case of bubbled inputs of AND gate,
A B A'.B'
0 0 1
0 1 0
1 0 0
1 1 0
2. In case of normal NOR gate,
A B (A+B)'
0 0 1
0 1 0
1 0 0
1 1 0
Hence, the answer is NOR GATE.
We will see a example.
1. In case of bubbled inputs of AND gate,
A B A'.B'
0 0 1
0 1 0
1 0 0
1 1 0
2. In case of normal NOR gate,
A B (A+B)'
0 0 1
0 1 0
1 0 0
1 1 0
Hence, the answer is NOR GATE.
(1)
Jigar said:
1 decade ago
Bubble performs the inversion function so consider
A=1,B=0 ->A*B=0
NOW LOGI OUTPUT WITH BUBBL:
A=0,B=1-> A*B=0.............................(1)
CONSIDER SAME VALUES FOR NOR GATE:
FIRST ORED OUT PUT:
A+B=0+1=1
NOW NORED OUTPUT:
NOR OF A+B=0.................................(2)
FROM (1) AND (2) AN GATE WITH BUBBLE INPUT =NOR GATE.
A=1,B=0 ->A*B=0
NOW LOGI OUTPUT WITH BUBBL:
A=0,B=1-> A*B=0.............................(1)
CONSIDER SAME VALUES FOR NOR GATE:
FIRST ORED OUT PUT:
A+B=0+1=1
NOW NORED OUTPUT:
NOR OF A+B=0.................................(2)
FROM (1) AND (2) AN GATE WITH BUBBLE INPUT =NOR GATE.
Neethu said:
1 decade ago
Bubbled input makes A = A' and B = B'.
Normally output of AND gate is A.B.
In bubbled input case (A'.B')' = (A+B)'.
i.e NOR GATE.
Normally output of AND gate is A.B.
In bubbled input case (A'.B')' = (A+B)'.
i.e NOR GATE.
Trupti said:
7 years ago
Answer is NAND gate.
Bright Zambia said:
7 years ago
NOR gate is the answer.
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