MIMO - NOMA

NOMA is a versatile technology because it can be combined with lots of other techniques like cooperative communication, SWIPT, MIMO etc. We already discussed the intersection of NOMA with cooperative and SWIPT networks. In this post, let's see how NOMA works in a simple MIMO network

Download the MATLAB code here

System model

Consider a 2 x 1 downlink MIMO system as shown in Fig. 1. Let $d_1$ and $d_2$ denote the distances of U1 and U2 respectively from the MIMO transmitter. Here, we assume $d_1 \gt d_2$. That is, U1 is the weak user and U2 is the strong user. 

We know that MIMO can be used for either spatial multiplexing (increase achievable rate) or diversity gain (decrease BER). Here, we are using MIMO for achieving diversity gain. Hence, both the transmit antennas 1 and 2 transmit the same information. 

Let $x_1$ and $x_2$ denote the information intended for U1 and U2. Following the notation conventions of MIMO, let $h_{rt}$ denote the Rayleigh fading channel between the $t^{th}$ transmit antenna and $r^{th}$ receiver.

Fig. 1: MIMO-NOMA system model

Signal model

Transmit signal

Signal transmitted by both the transmit antennas are given by, $$x = \sqrt{P}\left(\sqrt{\alpha_1}x_1+\sqrt{\alpha_2}x_2 \right)$$ 

(1)

where $\alpha_1$ and $\alpha_2$ are the NOMA power allocation coefficients. Since U1 is the weak user, we have $\alpha_1 \gt \alpha_2$.

Received signals

$x$ is transmitted simultaneously by both the transmit antennas. Therefore, the received signal at U1 is,

$$y_1 = xh_{11}+xh_{12}+n_1=x\left(h_{11}+h_{12}\right)+n_1$$

(2)

Similarly, the signal received by U2 is given by,

$$y_2=xh_{21}+xh_{22}+n_2=x\left(h_{21}+h_{22}\right)+n_2$$

(3)

Here, $n_1$ and $n_2$ are AWGN noise samples with mean zero and variance $\sigma^2$

Decoding at user 1

Now, U1 has to decode $x_1$ from $y_1$. Since U1 is the weak user, his signal, $x_1$ is allocated more power. That is, $\alpha_1 \gt \alpha_2$. Therefore, he can directly decode $x_1$ from $y_1$, treating the $x_2$ term as interference.

Substituting for $x$ in (2), we get, $$y_1 = \sqrt{P}\left(\sqrt{\alpha_1}x_1+\sqrt{\alpha_2}x_2\right)\left(h_{11}+h_{12}\right)+n_1$$

(4)

Expanding, $$y_1 = \underbrace{\sqrt{P}\sqrt{\alpha_1}x_1\left(h_{11}+h_{12}\right)}_\textrm{desired}+\underbrace{\sqrt{P}\sqrt{\alpha_2}x_2\left(h_{11}+h_{12}\right)}_\textrm{interference}+n_1$$

(5)

Now, we can write the SINR equation for U1 in decoding $x_1$ as follows,

$$\gamma_1 = \dfrac{P\alpha_1|h_{11}+h_{12}|^2}{P\alpha_2|h_{11}+h_{12}|^2+\sigma^2}$$

(6)

Therefore, the achievable rate at U1 is given by,

$$R_1 = log_2\left(1+\gamma_1 \right)$$

(7)

Decoding at user 2

U2 must decode $x_2$ from $y_2$. Since U2 is the strong user, his signal, $x_2$ is allocated less power. Therefore, in $y_2$, the power of the $x_1$ term will be dominating. So, U2 will first perform direct decoding on $y_2$ to obtain $x_1$. Then successive interference cancellation (SIC) is carried out to remove $x_1$. Then, $x_2$ is decoded.

Substituting for $x$ in (3) and expanding, we get,

$$y_2 = \underbrace{\sqrt{P}\sqrt{\alpha_1}x_1\left(h_{21}+h_{22}\right)}_\textrm{undesired & dominating}+\underbrace{\sqrt{P}\sqrt{\alpha_2}x_2\left(h_{21}+h_{22}\right)}_\textrm{desired}+n_2$$

(8)

The SINR at U1 for direct decoding of $x_1$ is,

$$\gamma_{12} = \dfrac{P\alpha_1|h_{21}+h_{22}|^2}{P\alpha_2|h_{21}+h_{22}|^2+\sigma^2}$$

(9)

After SIC, the first term of (8) will be removed and the remaining signal will be,

$$y'_2 = \underbrace{\sqrt{P}\sqrt{\alpha_2}x_2\left(h_{21}+h_{22}\right)}_\textrm{desired}+n_2$$

(10)

Now, the SNR for U2 to decode its own signal is given by,

$$\gamma_2 = \dfrac{P\alpha_2|h_{21}+h_{22}|^2}{\sigma^2}$$

(11)

Finally, the achievable rates at U2 for decoding $x_1$ and $x_2$ are given by,

$$R_{12} = log_2\left(1+\gamma_{12} \right)$$

(12)

$$R_2 = log_2\left(1+\gamma_2 \right)$$

(13)

MIMO-OMA for comparison

To see how well our MIMO-NOMA network performs, we are going to use a MIMO-OMA network as our baseline. In MIMO-OMA, let's divide our transmission into two equal time slots. In the first time slot, both the antennas transmit to U1 and in the second time slot, both the antennas transmit to U2.

Signal transmitted by both antennas in time slot 1 to U1 is $Px_1$. Signal received at U1 is, $$y_{1,oma} = \sqrt{P}x_1\left(h_{11}+h_{12}\right)+n_1$$

Similarly, signal transmitted by both the antennas in time slot 2 to U2 is $Px_2$ and the received signal at U2 is, $$y_{2,oma} = \sqrt{P}x_2\left(h_{21}+h_{22}\right)+n_1$$

The SNRs at U1 and U2 are, $\gamma_{1,oma} = \dfrac{P|h_{11}+h_{12}|^2}{\sigma^2}$ and $\gamma_{2,oma} = \dfrac{P|h_{21}+h_{22}|^2}{\sigma^2}$. 

Therefore, the achievable rates of MIMO-OMA for U1 and U2 are,

$$R_{1,oma} = \dfrac{1}{2}log_2\left(1+\gamma_{1,oma}\right)$$

(14)

$$R_{2,oma} = \dfrac{1}{2}log_2\left(1+\gamma_{2,oma}\right)$$

(15)

The factor $\dfrac{1}{2}$ in (14) and (15) is due to the fact that only half of the time slot is used for communicating to each user. (whereas in MIMO-NOMA, the entire time slot is used for simultaneous transmission to both users).

MATLAB simulation

Download the MATLAB code here

In this section, let's study the sum rate and outage performance of our MIMO-NOMA network. The following simulation parameters were used: $d_1 = 500m$, $d_2 = 200m$. Path loss exponent, $\eta = 4$. $\alpha_1 = 0.75$, $\alpha_2 = 0.25$. Bandwidth = 1 MHz.

Sum rates

The achievable sum rate of MIMO-NOMA is, $R_1 + R_2$, while that of MIMO-OMA is, $R_{1,oma}+R_{2,oma}$.

It is very clear that MIMO-NOMA provides greater sum rate than MIMO-OMA, due to the fact that both users are served simultaneously with the same frequency resource. 

If we plot the individual achievable rates, we get the following interesting graph:

We can notice that the weak user suffers from a saturation in its achievable rate after a transmit power of 10 dBm. This is a characteristic theme that we observe in all NOMA networks. The interference experienced by weak user translates to a saturation in its achievable rate. This saturation of achievable rate won't be a problem if the required data rate of the weak user is less than the saturation limit. This problem is not present in OMA because, the weak user does not suffer from interference due to simultaneous transmissions. 

Outage probabilities

Next, let's plot the outage graphs of the users for both MIMO-NOMA and MIMO-OMA schemes. Since we are using fixed power allocation, it is very important to choose the target rates and power allocation coefficients properly. Let's choose the target rate for the weak user (U1) to be $R^*_1$=1 bps/Hz and that of the strong user (U2) to be $R^*_2$=3 bps/Hz. 

MIMO-NOMA

The weak user (U1) is in outage if his achievable rate, $R_1$ is less than his target rate, $R^*_1$. Mathematically, we can denote this as, $$P^1_{noma}=Pr\left(R_1 \lt R^*_1 \right)$$

The strong user must decode both U1's message and his own message correctly. That is, the target rates of both U1 and U2 must be met by the strong user. If the target rate of U1 is not met OR if the target rate of U1 is met but that of U2 is not met, then U2 will be in outage. Mathematically,

$$P^2_{noma} = Pr\left(R_{12}<R^*_1 \right)+Pr\left(R_{12}>R^*_1, R_2 \lt  R^*_2 \right)$$

MIMO-OMA

Outage equations for MIMO-OMA are straightforward.

$$P^1_{oma}=Pr\left(R_{1,oma} \lt R^*_1 \right)$$

$$P^2_{oma}=Pr\left(R_{2,oma} \lt R^*_2 \right)$$

The outage graph looks like this:

It is clear that MIMO-NOMA achieves less outage than MIMO-OMA for both the users, as expected. 

In the following posts, we will see more about MIMO-NOMA, such as, how beam forming can be used to further enhance the performance. Thank you, See you soon! 

Reference:

This excellent tutorial on transmit diversity by Dr. Raviraj Adve: [available online] https://www.comm.utoronto.ca/~rsadve/Notes/DiversityTransmit.pdf

For further reading about MIMO NOMA:

"The application of MIMO to non-orthogonal multiple access," Z. Ding et al., IEEE transactions on wireless communications, vol. 15, no. 1, 2016

"On the ergodic capacity of MIMO NOMA systems," Q. Sun et al.,IEEE wireless communication letters, vol. 4, no. 4, 2015

Bonus Code

What happens when we apply our fair power allocation technique to MIMO-NOMA? Download MIMO NOMA with Fair Power allocation here