be reflected along the $\theta = \pi /M$ line, to give the quantization for the wedge $[0, 2\pi /M]$, which when repeated for the remaining $M-1$ wedges will cover the entire $(\Gamma , \Theta)$ plane.

Channel Quantization for the $M$-PSK case {#sec:chan_quan}
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In this subsection we propose a technique to obtain the quantization of the $(\Gamma, \Theta)$ plane, when both users use $M$-PSK signal sets. From Lemma 1, when both users use $M$-PSK signal sets at the input, the minimum distance in $\mathcal{S}_{\text{eff}}$ for any value of $(\gamma,\theta)$, $d_{min}(\gamma,\theta) \leq d_{min}(\mathcal{S}) =\sqrt{2\left[1-\cos (\frac{2 \pi}{M})\right]}$. Now the following lemma gives the number of singular fade states in the wedge $[0,\pi /M]$.

![image](qpsksum.eps){width="3.45in"}

\[periodic\] When both users use $M$-PSK signal sets at the input, the number of singular fade states in the wedge $[0, \pi /M]$, is given by $\frac{M^2}{8}-\frac{M}{4}+1$. Further, these singular fade states lie along the two lines $\theta=0$ and $\theta=\pi /M$.

From [@Vijay], the total number of singular fade states other than zero is $\frac{M^3}{4}-\frac{M^2}{2}+M$. Out of these, $M$ lie on the circle $\gamma=1$. It is also known from [@Vijay], that if $\gamma e^{j \theta}$ is a singular fade state, then $\frac{1}{\gamma}e^{-j \theta}$ is also a singular fade 