deletions | additions       diff --git a/considerations.tex b/considerations.tex  index 102f1b9..2c9079a 100644  --- a/considerations.tex  +++ b/considerations.tex 
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\begin{equation}  \delta t = 4.15 \times 10^3 \left( \frac{1}{\nu^2} - \frac{1}{(\nu+\delta\nu)^2} \right) {\rm DM},  \end{equation}  where $\delta\nu$ is the channel width and $\nu$ is the frequency of the bottom of the channel. By describing $\delta\nu$ in terms of its reciprocal time resolution $t_{\rm res}=1/\delta\nu$, we can determine the DM at which$t_{\rm res} = \delta t$. Using this, we can determine the DM at which  the dispersive smearing in a channel is equal to that channels time resolution. resolution (i.e. $t_{\rm res} = \delta t$).  This leads to the following requirements: \begin{itemize}