tell me what you think!

When ATPG errors

suppose an ATPG errors with slight probability p p->0

now suppose it is used to calculate untestability of a fault.

Let T be 1 if fault is testable, let T be 0 if fault is untestable.

Now, suppose we use an errorneous ATPG be T OR A, where A = and(x1,x2......x_n)

where n->infinity.

the average for an untestable fault if n->infinity = = 0 in RTG

T = T OR A for exactly for every case except 1 case, for an untestable fault.

Now, T OR A can be solved by deterministic ATPG, T OR A = 1.

= = 0 , is untestable by RTG ATPG. = 0 , therefore has 0 solutions.


= 1/2^n

no. of solutions = * 2^n = ( 1/2^n )*2^n = lim episoln1,2->0 n->infinity (1/2^n -episoln1+episoln2)*2^n as n->infinity select episoln1=1/2^n = (0 +episoln2)*2^n Select episoln2=0, such that episoln2*2^n =0 = 0 no. of solutions = 0;

T OR A has 1 solution by deterministic ATPG.

Therefore solutions= 0 = 1

Suppose T is the output T + 0 = T + solutions = T + 1, if T is 0 = 1, if T- 0 = = T - solutions = T - 1 if T is 1 = 0

untestable is testable, testable is untestable! Such effects may be heuristically observable.

ATPG will remain unsolved

Suppose a cripple found a solution and says it is solved, since a cripple found it , it is not solved.

Reply to
Loading thread data ...


Reply to

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.