Wireless Communication

We saw how non-orthogonal multiple access (NOMA) can somehow magically break the capacity limitation faced by the other traditional orthogonal multiple access (OMA) schemes. OMA schemes like TDMA, FDMA, CDMA, OFDMA separates the users in time, frequency, code, subcarrier domains respectively. No two users are allowed to share the same resource simultaneously. If this condition is violated, interference would occur and both users would lose their data. For a moment, let's consider an OFDMA network with 64 subcarriers. If we assign one orthogonal subcarrier per user, the maximum number of users we can simultaneously serve is 64. If the 65th user requests a connection, his call must be dropped or he must wait for one of the other users to finish transmitting. This is because we don't have any more orthogonal subcarrier to assign to the new user. This orthogonality bottleneck imposes a capacity limitation on the number of

Next post: Single Carrier NOMA - How many users can it support? In a previous post , we saw how to plot the bit error rate (BER) of a two user NOMA. There, we used BPSK modulation for both the users. Can we multiplex more than two users in a single carrier in NOMA? Of course. In this post, we are going to multiplex three users, each following QPSK modulation, in a single frequency carrier. Download MATLAB code for 3 user NOMA here System Model Let us consider a wireless network consisting of three NOMA users, numbered U1, U2 and U3. Let $d_1$, $d_2$ and $d_3$ denote their respective distances from the base station (BS) such that, $d_1 \gt d_2 \gt d_3$. Based on their distances, U1 is the weakest/farthest user and U3 is the strongest/nearest user to the BS.  Let $h_1$, $h_2$, and $h_3$ denote their corresponding Rayleigh fading coefficients such that, $|h_1|^2 \lt |h_2|^2 \lt |h_3|^2$ . (The channels are ordered this way because