FDMA divides the bandwidth of the channel into separate non-overlapping frequency sub-channels and allocates each sub-channel to a separate user. The capacity region of FDMA with two users satiesfies: $$R_1 < \alpha W \log_{2}\left(1+\frac{P_1}{\alpha N_0}\right)$$ $$R_2 < (1-\alpha)W\log_{2}\left(1+\frac{P_2}{(1-\alpha)N_0}\right)$$ $$0 < \alpha< 1$$ where $\alpha$ indicates the degrees of freedom allocated to user 1 and hence $(1-\alpha)$ refers to the rest of degrees of freedom to user 2.
TDMA allows several users to share the same frequency channel by dividing the signal into different time slots. The capacity region of TDMA with two users satiesfies: $$R_1 < \alpha W \log_{2}\left(1+\frac{P_1}{N_0}\right)$$ $$R_2 < (1-\alpha)W\log_{2}\left(1+\frac{P_2}{N_0}\right)$$ $$0 < \alpha< 1$$ where $\alpha$ indicates the degrees of time allocated to user 1 and hence $(1-\alpha)$ refers to the rest of degrees of time to user 2.
With orthogonal CDMA, signals of several users are encoded with orthogonal scheme so that they can send information simultaneously and share whole band of frequencies over a single communication channel. With $K$ users in the system, the capacity of user $k$ is: $$R_k = \frac{1}{K} W \log_{2}\left(1+\frac{K P_k}{N_0}\right)$$ When applies to two user scenario, the capacity region is shown as below: $$R_1 = \frac{1}{2} W \log_{2}\left(1+\frac{2 P_1}{N_0}\right)$$ $$R_2 = \frac{1}{2} W \log_{2}\left(1+\frac{2 P_2}{N_0}\right)$$
With conventional CDMA, several users send information simultaneously and share whole band of frequencies over a single communication channel. They treat the interference of all other users as noises and include them into the calculation of SNR. The capacity region with two users satisfies: $$R_1 = W \log_{2}\left(1+\frac{P_1}{P_2+N_0}\right)$$ $$R_2 = W \log_{2}\left(1+\frac{P_2}{P_1+N_0}\right)$$