Digital Electronics - Code Converters and Multiplexers - Discussion
Discussion Forum : Code Converters and Multiplexers - General Questions (Q.No. 29)
29.
For the following conditions on a 7485 magnitude comparator, what will be the state of each of the three outputs?
| A0 = 0 | B0 = 1 | IA < B = 0 |
| A1 = 1 | B1 = 0 | IA = B = 1 |
| A2 = 1 | B2 = 0 | IA > B = 0 |
| A3 = 0 | B3 = 0 |
Discussion:
4 comments Page 1 of 1.
Kelsy21 said:
1 decade ago
Can anybody please explain me how the answer A came?
Avy said:
9 years ago
A --> 0110 = 6.
B --> 0001 = 1.
A>B.
B --> 0001 = 1.
A>B.
Karan said:
9 years ago
@Avy.
Why 0001?
Why 0001?
Sukriti said:
9 years ago
Bit position:
i.e 15 14 13 12 ...................................... 1 0
The LSB (least significant bit) is 0.
MSB (most significant bit ) is 15.
So,
A3 A2 A1 A0 => 0110
B3 B2 B1 B0=> 0001
The decimal Values for A is 6 & B is 1
Now compare for each state.
State 1: A=B
=> No A!=B(A is not equal to B)
=> So, A=B=0
State 2: A<B
=> No A!<B(A is not less than B)
=> So, A<B=0 ()
State 3: A>B
=> A>B(A is greater than B)
=> So, A>B=1 (6>1)
Therefore option (A) is correct.
i.e 15 14 13 12 ...................................... 1 0
The LSB (least significant bit) is 0.
MSB (most significant bit ) is 15.
So,
A3 A2 A1 A0 => 0110
B3 B2 B1 B0=> 0001
The decimal Values for A is 6 & B is 1
Now compare for each state.
State 1: A=B
=> No A!=B(A is not equal to B)
=> So, A=B=0
State 2: A<B
=> No A!<B(A is not less than B)
=> So, A<B=0 ()
State 3: A>B
=> A>B(A is greater than B)
=> So, A>B=1 (6>1)
Therefore option (A) is correct.
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