Questions and Answers

1. Using D-algorithm show that the circuit given below is un-testable:

Answer:

In the first step, the signal values are initialized to X and the s-a-0 net is assigned to D.  So the D-frontier is {2} and J-frontier is {1}.

Now, as there is an unique D-drive,  D is prorogated to gate-2, making signal F=D. By backward implications on gate-1, A=1,d=1 and B=1.

Also, by backward implication on gate-2, e=0. This is an inconsistency because, e=0 would make B=0, by backward implication.

It may be noted that this fault is not testable, because there is no alternative to backtrack, as there were no choices at any step.