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Robin 2010-06-14 07:44:14 +01:00
parent b94e189119
commit 7a0e1b0e76
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pt100/pt100.tex
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@ -620,7 +620,7 @@ We can use the equation \ref{eqn:correctedccps2}, reproduced below to verify thi
\begin{equation} \begin{equation}
|{\mathcal{P}_{cc}SU}| = {\sum^{k}_{1..cc} \frac{|{SU}|!}{k!(|{SU}| - k)!}} |{\mathcal{P}_{cc}SU}| = {\sum^{k}_{1..cc} \frac{|{SU}|!}{k!(|{SU}| - k)!}}
- \sum^{p}_{2..cc}{{\sum^{j}_{j \in J} \frac{|FM({C_j})|!}{p!(|FM({C_j})| - p)!}} } - \sum^{p}_{2..cc}{{\sum^{j}_{j \in J} \frac{|FM({C_j})|!}{p!(|FM({C_j})| - p)!}} }
\label{eqn:correctedccps2} %\label{eqn:correctedccps2}
\end{equation} \end{equation}
@ -629,12 +629,13 @@ $|FM(C_j)|$ is always 2 here, as all the components are resistors and have two
% %
% Factorial of zero is one ! You can only arrange an empty set one way ! % Factorial of zero is one ! You can only arrange an empty set one way !
Populating this equation with $|SU| = 6$ and $|FM(C_j)|$ is always 2 here as all the components are resistors and have two failure modes. Populating this equation with $|SU| = 6$ and $|FM(C_j)|$ = 2.
%is always 2 for this circuit, as all the components are resistors and have two failure modes.
\begin{equation} \begin{equation}
|{\mathcal{P}_{2}SU}| = {\sum^{k}_{1..2} \frac{6!}{k!(6 - k)!}} |{\mathcal{P}_{2}SU}| = {\sum^{k}_{1..2} \frac{6!}{k!(6 - k)!}}
- \sum^{p}_{2..2}{{\sum^{j}_{1..3} \frac{2!}{p!(2 - p)!}} } - \sum^{p}_{2..2}{{\sum^{j}_{1..3} \frac{2!}{p!(2 - p)!}} }
\label{eqn:correctedccps2} %\label{eqn:correctedccps2}
\end{equation} \end{equation}
$|{\mathcal{P}_{2}SU}|$ is the number of valid combinations of faults to check $|{\mathcal{P}_{2}SU}|$ is the number of valid combinations of faults to check