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Robin 2010-06-25 20:30:27 +01:00
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pt100/pt100.tex
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@ -675,7 +675,7 @@ $$ NoOfTestCasesToCheck = 6 + 15 - ( 1 + 1 + 1 ) = 18 $$
As the test case are all different and are of the correct cardinalities (6 single faults and (15-3) double)
we can be confident that we have looked at all `double combinations', of the possible faults
in the pt100 circuit. The next task is to investigate
these test cases in more detail to prove the failure mode hypothese set out in table \ref{tab:ptfmea2}.
these test cases in more detail to prove the failure mode hypothesis set out in table \ref{tab:ptfmea2}.
\subsection{Proof of Double Faults Hypothesis }