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The logic shown in the given figure works as:
- a)decoder
- b)binary to XS-3 converter
- c)priority encoder
- d)binary to gray converter
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The logic shown in the given figure works as:a)decoderb)binary to XS-3...
Priority encoder:-
If more than one input is high then encoder produce an output which may not be correct to overcome this we use priority encoder.
We considered one more output, V in order to know, whether the code available at outputs is valid or not.
If at least one input of the encoder is ‘1’, then the code available at outputs is a valid one. In this case, the output, V will be equal to 1.
If all the inputs of encoder are ‘0’, then the code available at outputs is not a valid one. In this case, the output, V will be equal to 0.
Analysis:-
Logic shown in given figure will work as
Priority encoder (4 × 2)
Inputs = D3, D2, D1, D0
Output:
y = D3 + D2’ D1
X = D3 + D2
V = D3 + D2 + D1 + D0
Most Upvoted Answer
The logic shown in the given figure works as:a)decoderb)binary to XS-3...
Priority encoder:-
If more than one input is high then encoder produce an output which may not be correct to overcome this we use priority encoder.
We considered one more output, V in order to know, whether the code available at outputs is valid or not.
If at least one input of the encoder is ‘1’, then the code available at outputs is a valid one. In this case, the output, V will be equal to 1.
If all the inputs of encoder are ‘0’, then the code available at outputs is not a valid one. In this case, the output, V will be equal to 0.
Analysis:-
Logic shown in given figure will work as
Priority encoder (4 × 2)
Inputs = D3, D2, D1, D0
Output:
y = D3 + D2’ D1
X = D3 + D2
V = D3 + D2 + D1 + D0
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Question Description
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