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Ewan D. Barr edited considerations.tex
over 8 years ago
Commit id: f1035075d6c734548e1ad93a5a50efd28fccc375
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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 (both in MHz). By describing $\delta\nu$ in terms of its reciprocal time resolution $t_{\rm
res}=1/\delta\nu$, res}=10^6/\delta\nu$, we can determine the DM at which the dispersive smearing in a frequency channel is equal to that channel's time resolution (i.e. $t_{\rm res} = \delta t$). This leads to the following requirements:
\begin{itemize}