A line L in a circuit is said to have a stuck-at-0 fault if the line p...
Answer should be 3N-1.
This is because the total possible combinations (i.e a line may either be at fault (in 2 ways i.e stuck at fault 0 or 1) or it may not be , so there are only 3 possibilities for a line) is 3N. In only one combination the circuit will have all lines to be correct (i.e not at fault.) Hence 3N-1. (as it has been said that circuit is said to have multiple stuck up fault if one or more line is at fault)
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A line L in a circuit is said to have a stuck-at-0 fault if the line p...
Answer:
The given question is related to the concept of stuck-at faults in digital circuits. A stuck-at fault occurs when a signal line in a circuit is permanently stuck at either logic 0 or logic 1.
To find the number of distinct multiple stuck-at faults in a circuit with N lines, we need to consider all possible combinations of stuck-at faults.
Let's consider the number of possible stuck-at-0 faults first. For each line in the circuit, there are two possibilities: either it has a stuck-at-0 fault or it doesn't. Therefore, the total number of possible stuck-at-0 faults is 2^N.
Similarly, the number of possible stuck-at-1 faults is also 2^N.
To find the number of distinct multiple stuck-at faults, we need to subtract the cases where all lines have either stuck-at-0 or stuck-at-1 faults from the total number of stuck-at faults.
- For all lines stuck-at-0: In this case, there is only one possible combination because all lines have the same stuck-at fault. So, the number of cases is 1.
- For all lines stuck-at-1: Similar to the previous case, there is only one possible combination, so the number of cases is 1.
Therefore, the number of distinct multiple stuck-at faults is given by:
Total number of stuck-at faults - Number of cases where all lines have either stuck-at-0 or stuck-at-1 faults.
Total number of stuck-at faults = Number of stuck-at-0 faults + Number of stuck-at-1 faults
= 2^N + 2^N
= 2^(N+1)
Number of cases where all lines have either stuck-at-0 or stuck-at-1 faults = 2
Number of distinct multiple stuck-at faults = Total number of stuck-at faults - Number of cases where all lines have either stuck-at-0 or stuck-at-1 faults
= 2^(N+1) - 2
Simplifying the expression, we get:
Number of distinct multiple stuck-at faults = 2^N - 2
However, the correct answer given is 3N - 1, which does not match the derived expression. Therefore, the correct answer should be 2^N - 2, not 3N - 1.
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