From 7a0e1b0e76e6bf9a42afe5dac26cdfaf13a3e2fd Mon Sep 17 00:00:00 2001 From: Robin Date: Mon, 14 Jun 2010 07:44:14 +0100 Subject: [PATCH] . --- pt100/pt100.tex | 7 ++++--- 1 file changed, 4 insertions(+), 3 deletions(-) diff --git a/pt100/pt100.tex b/pt100/pt100.tex index b875bf8..926a252 100644 --- a/pt100/pt100.tex +++ b/pt100/pt100.tex @@ -620,7 +620,7 @@ We can use the equation \ref{eqn:correctedccps2}, reproduced below to verify thi \begin{equation} |{\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)!}} } - \label{eqn:correctedccps2} + %\label{eqn:correctedccps2} \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 ! -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} |{\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)!}} } - \label{eqn:correctedccps2} + %\label{eqn:correctedccps2} \end{equation} $|{\mathcal{P}_{2}SU}|$ is the number of valid combinations of faults to check