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Unformatted text preview: the probability of obtaining a logic 1 at the output of this AND gate. Another issue regarding random test generation is the number of random vectors needed. Given a circuit with n primary inputs, there are clearly 2 n possible input vectors. One can express the probability of detecting fault f by any random vector to be: d f = T f 2 n where T f is the set of vectors that can detect fault f . Consequently, the probability that a random vector will not detect f ( i.e. , f escapes a random vector) is: e f = 1 − d f Therefore, given N random vectors, the probability that none of the N vectors detects fault f is: e N f = ² 1 − d f ³ N In other words, the probability that at least one out of N vectors will detect fault f is: 1 − ² 1 − d f ³ N...
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 Spring '08
 elbarki
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