How to do power allocation in NOMA? (with fairness to far/weak user)

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Power allocation is very important in non orthogonal multiple access (NOMA). In the previous post "Does power allocation affect NOMA?", we saw how the bit error rate (BER) of NOMA has a strong relationship with the values of power allocation coefficients.

We used fixed power allocation till now. That is, we fixed the values of $\alpha_1$ and $\alpha_2$ irrespective of the channel conditions. But there are better ways to optimize $\alpha_1$ and $\alpha_2$ dynamically based on the values of channel state information (CSI). 

There are a few different dynamic power allocation schemes each trying to accomplish a specific goal. The goal could be maximizing the sum rate, maximizing the energy efficiency, etc., The power allocation scheme we are going to see in this post is a simple one whose goal is to provide user fairness. Let's call this fair power allocation scheme (fair PA, in short). 

Our fair PA gives priority to the weak/far user. That is, the power allocation coefficients are calculated such that the far user's target rate is met. Only after meeting the target rate of far user, all the remaining available power is allocated to the near user. Let's go ahead and derive the power allocation coefficients to meet this specification. 

In this post, we will derive the power allocation coefficients for a dynamic power allocation scheme called fair PA. We will also simulate and study the outage and sum rate performance of fair PA using MATLAB. You can download the MATLAB codes used in this post below:

NOMA Power allocation (Outage vs target rate) MATLAB code

NOMA Power allocation (Sum rate vs transmit power) MATLAB code

NOMA Capacity Equations

As we saw in this post, the capacity equations for NOMA far user and near user can be written as follows:$$R_f = log_2(1+\dfrac{|h_f|^2P\alpha_f}{|h_f|^2P\alpha_n+\sigma^2})$$ $$R_n = log_2(1+\dfrac{|h_n|^2P\alpha_n}{\sigma^2})$$ $R_n$ is obtained after removing the interference from far user transmission by successive interference cancellation (SIC).

• $\alpha_n$ - power allocation coefficient for near user
• $\alpha_f$ - power allocation coefficient for far user
• $h_n$ - Rayleigh fading coefficient for near user
• $h_f$ - Rayleigh fading coefficient for far user
• $P$ - Total transmit power
• $\sigma^2$ - Noise power
• $\alpha_n + \alpha_f = 1$
• $\alpha_f \gt \alpha_n$

For more details, please see this post. How to simulate BER, capacity and outage probability of NOMA (Rayleigh fading) using MATLAB?

Derivation of power allocation coefficients $\alpha_n$ and $\alpha_f$

Let $R^*$ denote the target rate of far user. Our goal is to choose $\alpha_n$ and $\alpha_f$ such that $R_f \geq R^*$.

Let's set $R_f = R^*$. $$log_2(1+\dfrac{|h_f|^2P\alpha_f}{|h_f|^2P\alpha_n+\sigma^2}) = R^*$$ First, let's remove the $log_2$ by taking $2^x$ on both sides. $$1+\dfrac{|h_f|^2P\alpha_f}{|h_f|^2P\alpha_n+\sigma^2} = 2^{R^*}$$ $$\dfrac{|h_f|^2P\alpha_f}{|h_f|^2P\alpha_n+\sigma^2} = 2^{R^*} - 1$$ Let's denote $\xi = 2^{R^*} - 1$. 

$\xi$ is the target SINR for the far user who has target rate $R^*$. $$\dfrac{|h_f|^2P\alpha_f}{|h_f|^2P\alpha_n+\sigma^2} = \xi$$ $$|h_f|^2P\alpha_f = \xi|h_f|^2P\alpha_n + \xi\sigma^2$$ Since $\alpha_n + \alpha_f = 1$, $\alpha_n = 1 - \alpha_f$. $$|h_f|^2P\alpha_f = \xi|h_f|^2P(1 - \alpha_f) + \xi\sigma^2$$ $$|h_f|^2P\alpha_f = \xi|h_f|^2P - \xi|h_f|^2P\alpha_f + \xi\sigma^2$$ Collecting all the $\alpha_f$ terms to the LHS, $$|h_f|^2P\alpha_f + \xi|h_f|^2P\alpha_f = \xi|h_f|^2P + \xi\sigma^2$$ $$\alpha_f|h_f|^2P(1 + \xi) = \xi(|h_f|^2P + \sigma^2)$$ $$\alpha_f = \dfrac{\xi(|h_f|^2P + \sigma^2)}{|h_f|^2P(1 + \xi)}$$ We don't want $\alpha_f$ to exceed 1. So let's set a limit as, $$\alpha_f = min(1, \dfrac{\xi(|h_f|^2P + \sigma^2)}{|h_f|^2P(1 + \xi)}$$ Once $\alpha_f$ is computed using the above equation, we can easily calculate $\alpha_n$ as, $$\alpha_n = 1 - \alpha_f$$ That's it! We have derived the power allocation coefficients for our dynamic power allocation scheme, fair PA. Next, let's see how our fair PA performs.

MATLAB Simulation Results

Let's see some interesting plots, comparing the performances of fixed PA and fair PA. First, let's compare the outage performance of fixed and fair PA schemes.

When we plot the outage probabilities as a function of far user's target rate ($R^*$), we get this plot. Here, we have fixed the total transmit power as 30 dBm. Also, to get the outage probability, we have set the same target rate for both near and far users. (You can download the MATLAB code here)

Fig. 1: Outage vs Target rate for fixed and fair PA

We can immediately see that fixed PA is performing very poorly and its outage probability saturates to 1 all the time when $R^* \gt 1.5$ bps/Hz. In other words, the receiver is ALWAYS in outage if we use fixed PA with $R^* \gt 1.5$ bps/Hz. This is because, fixed PA neither exploits the instantaneous CSI, nor takes the target rate requirements into account. Thus, fixed PA, although simple to implement, is not so good after all.

But our fair PA has lower outage probability because $\alpha_n$ and $\alpha_f$ and dynamically adjusted based on target rate requirement and CSI.

In our fair PA, we can see that the outage of far user increases gradually as his target rate requirement increases. This is expected because, as the target rate becomes higher and higher, the chances of far user achieving that target rate becomes lower and lower. This would lead to increase in its outage probability.

The outage of near user shows quite a sharp transition around $R^*$ values of 4 to 7 bps/Hz. And beyond that, the near user is always in outage. This is still better than fixed PA, but why does this happen? Can we do something to reduce the near user's outage further?

 

We did come up with a better power allocation than fixed PA. But can we improve our fair PA further?

Problems with our fair PA scheme

The far user has a weak channel with the BS. When we did the limiting operation, $$\alpha_f = min(1, \dfrac{\xi(|h_f|^2P + \sigma^2)}{|h_f|^2P(1 + \xi)}$$ what we actually did was, whenever $\alpha_f$ was calculated to be greater than 1, we limited $\alpha_f$ to be equal to 1. 

For example, if we get $\dfrac{\xi(|h_f|^2P + \sigma^2)}{|h_f|^2P(1 + \xi)} = 50$, we set $\alpha_f = min(1,50) = 1$. 

But what does $\dfrac{\xi(|h_f|^2P + \sigma^2)}{|h_f|^2P(1 + \xi)} = 50$ mean? This means that only if we set $\alpha_f = 50$, we will be able to meet the far user's targe rate $R^*$. 

In that case, limiting $\alpha_f$ to 1 does not help. Because, any value of $\alpha_f \lt 50$ will lead to the outage of far user. 

In other words, when $\dfrac{\xi(|h_f|^2P + \sigma^2)}{|h_f|^2P(1 + \xi)} \gt 1$, even if we allocate the entire power to the far user (i.e., $\alpha_f = 1$), he will still be in outage. 

 Another consequence is, by setting $\alpha_f=1$, we are automatically setting $\alpha_n = 0$, which is bad because now the near user is also in outage since we are not allocating any power to him at all.

Let's tweak our fair PA further

To address this problem, let's add a simple tweak our fair PA. 

Whenever $\dfrac{\xi(|h_f|^2P + \sigma^2)}{|h_f|^2P(1 + \xi)}$ exceeds 1, instead of limiting $\alpha_f$ to 1, let's set $\alpha_f = 0$. Which automatically sets $\alpha_n = 1$. 

Will setting $\alpha_f=0$ affect the far user's outage? NO. Because even by setting $\alpha_f = 1$ (allocating him the entire power), we cannot bring him out of outage. So why bother wasting all the power on him? 

After adding this little tweak, our fair PA shows an interesting outage trend. (Download the MATLAB code here)
 

Fig. 2: Outage vs Target rate for fixed and improved fair PA

Outage of far user follows the trend that we saw in Fig. 1. This indicates that our tweak of setting $\alpha_f = 0$, when required, did not affect the outage of far user at all. Now let's look at the graph of near user's outage. The outage probability increases, peaks and then starts to decrease. Why do we observe this trend?

When $R^*$ lies in the range of  0 to 6.5 bps/Hz, it looks like we are favoring the far user by allocating more and more power to him, at the cost of sacrificing the performance of near user. But beyond 6.5 bps/Hz, any value of $\alpha_f$ may not fully satisfy $R^*$. When this happens, we favor the near user, instead of wasting all our power on the far user.

This leads to a decrease in the outage of near user, for $R^* \gt 6.5$ bps/Hz, without affecting the outage of far user, which is a good thing. 

Next, let's compare the sum rate of our improved fair PA with fixed PA that we have been using so far. By sum rate, we mean $R_n + R_f$. When we plot the sum rate as a function of transmit power, we will get a graph like this. (Download the MATLAB code here)

Fig. 3: Sum rate vs transmit power of fixed and fair PA

We can clearly see that our fair PA outperforms fixed PA in terms of achievable capacity. 

Conclusion

We know that the wireless channel is extremely dynamic in nature. Fixed PA does not worry about the instantaneous channel conditions of users. Whatever be channel condition, $\alpha_n$ and $\alpha_f$ are fixed. But our fair PA is a dynamic scheme. That is, whenever the channel changes, the values of $\alpha_n$ and $\alpha_f$ are updated to meet the specification. That's why fair PA is able to achieve higher sum rate and lower outage than fixed PA.

References:

1. "The impact of power allocation on cooperative non-orthogonal multiple access with SWIPT", Z. Yang, Z. Ding, P. Fan and N. Al-Dhahir, IEEE transactions on wireless communications, vol.6, no.7, July 2017
2. F. Kara, H. Kaya, BER Performances of Downlink and Uplink NOMA in the Presence of SIC Errors over Fading Channels. IET Communications. Vol. 12, no. 15, pp. 1834-1844.