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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 $t_{\rm res}=10^6/\delta\nu$, res}=(\delta\nu \times 10^6)$,  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}