A Pole at DC

Since both of the conditions

$$\begin{eqnarray*} e^\tau ( 1 - \tau) &=& \cos(\nu) \\ e^\tau &=& \frac{\sin(\nu)}{\nu} \end{eqnarray*}$$

are clearly satisfied for $\tau=\nu=0$ , we see that there is in fact a pole in the reflectance at dc ($s=0$ ), provided it is not canceled by a zero at dc in the numerator $N(s)$ .