Ferkans — Interactive Telecom Tutor

Green Communications with Massive MIMO

Beyond spectral efficiency, massive MIMO offers a transformative advantage in energy efficiency — the number of bits delivered per joule of energy consumed. With $M$ antennas, the transmit power per user can be reduced by a factor of $M$ (or $\sqrt{M}$ with imperfect CSI) while maintaining the same rate. However, each additional antenna brings circuit power costs: RF chains, ADCs/DACs, oscillators, and baseband processing. There exists an optimal number of antennas $M^{\star}$ that maximises energy efficiency by balancing the beamforming gain against the circuit power overhead. Understanding this trade-off is essential for designing green 5G/6G networks.

Definition:
Energy Efficiency Metric

The energy efficiency (EE) of a massive MIMO system is defined as the ratio of the total achievable sum rate to the total power consumption:

$\text{EE} = \frac{\sum_{k=1}^{K} R_k}{P_{\text{total}}} \quad [\text{bits/joule}]$

The total power consumption includes:

$P_{\text{total}} = \frac{1}{\eta_{\text{PA}}}\sum_{k=1}^{K}p_k^{\text{DL}} + M \cdot P_{\text{circuit}} + P_{\text{fixed}}$

where:

$\eta_{\text{PA}} \in (0,1]$ is the power amplifier efficiency
$p_k^{\text{DL}}$ is the radiated power for user $k$
$P_{\text{circuit}}$ is the circuit power per antenna element (RF chain, ADC, mixing, filtering)
$P_{\text{fixed}}$ is fixed power consumption (cooling, backhaul, baseband processing independent of $M$ )

For the uplink analysis, $P_{\text{total}}$ includes the users' transmit power plus the BS circuit power.

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Theorem: Optimal Number of Antennas Maximising Energy Efficiency

Consider a massive MIMO system with MR combining, equal power users, and i.i.d. Rayleigh fading. The per-user rate is:

$R(M) = K\log_2\!\left(1 + \frac{pM\beta}{(K-1)p\beta + \sigma^2}\right)$

and the total power is $P(M) = Kp/\eta_{\text{PA}} + M P_{\text{circuit}} + P_{\text{fixed}}$ .

The energy efficiency $\text{EE}(M) = R(M)/P(M)$ is a quasi-concave function of $M$ with a unique maximiser $M^{\star}$ satisfying:

$\frac{\partial R}{\partial M}\Big|_{M^{\star}} \cdot P(M^{\star}) = R(M^{\star}) \cdot P_{\text{circuit}}$

This yields the implicit equation:

$\frac{Kp\beta/\!\ln 2}{(K-1)p\beta + \sigma^2 + M^{\star}p\beta} = \frac{R(M^{\star})}{P(M^{\star})} \cdot P_{\text{circuit}}$

The solution trades off the marginal rate gain from one more antenna against the marginal power cost $P_{\text{circuit}}$ .

Adding antennas has two competing effects:

Rate gain: Each antenna contributes approximately $K/(M\ln 2)$ bits/s/Hz (diminishing as $\log M$ ).
Power cost: Each antenna costs $P_{\text{circuit}}$ watts.

When $M$ is small, the rate gain per antenna is large and dominates the circuit cost — EE increases. When $M$ is large, the rate gain is marginal (logarithmic saturation) while the circuit cost grows linearly — EE decreases. The optimal $M^{\star}$ is where the marginal rate-per-watt equals the average rate-per-watt.

Proof

EE as a ratio

$\text{EE}(M) = \frac{R(M)}{P(M)} = \frac{K\log_2\!\left(1 + \frac{pM\beta}{(K-1)p\beta+\sigma^2}\right)} {Kp/\eta_{\text{PA}} + MP_{\text{circuit}} + P_{\text{fixed}}}$ $This is a ratio of a concave (logarithmic) numerator and a linear (affine) denominator, hence quasi-concave in$ M$.

First-order optimality condition

The maximum of $R(M)/P(M)$ satisfies (Dinkelbach's theorem):

$\frac{dR}{dM}\Big|_{M^{\star}} = \text{EE}(M^{\star}) \cdot \frac{dP}{dM}$

Since $dP/dM = P_{\text{circuit}}$ :

$\frac{dR}{dM}\Big|_{M^{\star}} = \text{EE}(M^{\star}) \cdot P_{\text{circuit}}$

$\frac{Kp\beta}{((K-1)p\beta + \sigma^2 + M^{\star}p\beta)\ln 2} = \frac{R(M^{\star})}{P(M^{\star})} \cdot P_{\text{circuit}}$

Properties of the optimum

The optimal $M^{\star}$ increases when:

$P_{\text{circuit}}$ decreases (cheaper antennas favour more of them)
$K$ increases (more users benefit from additional antennas)
SNR decreases (lower SNR means each antenna helps more)

For typical 5G parameters ( $P_{\text{circuit}} = 100$ mW, $P_{\text{fixed}} = 10$ W, $K = 10$ , SNR = 0 dB), $M^{\star}$ is typically in the range 50--200. $\blacksquare$

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Energy Efficiency vs. Number of Antennas

Explore the energy efficiency trade-off. The EE curve first rises (beamforming gains dominate) then falls (circuit power dominates), with an optimal $M^{\star}$ . Vary the circuit power, number of users, and SNR to see how the optimum shifts.

Parameters

K

(number of users)10

SNR [dB]0

P_{\mathrm{circuit}}

per antenna [mW]100

Example: Computing the Optimal Number of Antennas

A massive MIMO system has $K = 10$ users with $\beta = 1$ and $p = 1$ W. The noise variance is $\sigma^2 = 1$ (SNR = 0 dB). The system parameters are: $\eta_{\text{PA}} = 0.4$ , $P_{\text{circuit}} = 0.1$ W per antenna, $P_{\text{fixed}} = 10$ W.

(a) Compute the EE at $M = 50$ , $M = 100$ , and $M = 200$ . (b) Determine which of these three is closest to $M^{\star}$ .

Solution

Rate and power at each M

The per-user SINR with MR is $\text{SINR} = M/(K-1+\sigma^2/p\beta) = M/10$ .

$M$	SINR	Sum Rate (bits/s/Hz)	$P_{\text{total}}$ (W)
50	5	$10\log_2(6) = 25.85$	$25 + 5 + 10 = 40$
100	10	$10\log_2(11) = 34.59$	$25 + 10 + 10 = 45$
200	20	$10\log_2(21) = 43.93$	$25 + 20 + 10 = 55$

Energy efficiency comparison

$M$	EE (bits/joule)
50	$25.85/40 = 0.646$
100	$34.59/45 = 0.769$
200	$43.93/55 = 0.799$

EE is still increasing at $M = 200$ but with diminishing returns. The optimality condition requires:

$\frac{10}{(10 + M^{\star})\ln 2} = \text{EE}(M^{\star}) \times 0.1$

At $M = 200$ : LHS $= 10/(210 \times 0.693) = 0.0687$ , RHS $= 0.799 \times 0.1 = 0.0799$ .

Since LHS $<$ RHS, $M^{\star} < 200$ . Checking $M = 150$ : SINR $= 15$ , $R = 10\log_2(16) = 40$ , $P = 50$ , EE $= 0.800$ . LHS $= 10/(160 \times 0.693) = 0.0902$ , RHS $= 0.080$ . Since LHS $>$ RHS, $M^{\star} > 150$ .

The optimum is around $M^{\star} \approx 170$ antennas. $\blacksquare$

Quick Check

If the circuit power per antenna $P_{\text{circuit}}$ is halved (e.g., due to more efficient hardware), what happens to the optimal number of antennas $M^{\star}$ that maximises energy efficiency?

$M^{\star}$ decreases

$M^{\star}$ increases

$M^{\star}$ stays the same

EE becomes independent of $M$

Correction:

M^{\star}

increases

When $P_{\text{circuit}}$ decreases, the marginal power cost of each antenna drops while the marginal rate gain remains unchanged. The balance point shifts to a larger $M^{\star}$ , allowing more antennas before circuit power dominates.

Why This Matters: Energy Savings in 5G Massive MIMO Deployments

Field measurements from 5G NR massive MIMO deployments have confirmed the energy efficiency benefits predicted by theory:

Ericsson reported that massive MIMO with 64T64R at sub-6 GHz achieves 10 $\times$ better energy efficiency (bits/joule) than 4G macro cells, primarily through concentrated beamforming that reduces the per-user transmit power.
Sleep modes: With channel hardening, the BS can predict traffic demand from channel statistics and put unused RF chains to sleep, saving 30--50% additional power.
Power scaling: Massive MIMO can reduce the total radiated power by $1/M$ while maintaining the same per-user rate, enabling a 64-antenna BS to radiate 18 dB less total power than a single-antenna system at the same QoS.

The energy efficiency gains are particularly important for operators in emerging markets, where energy costs dominate the total cost of ownership of cellular networks.

See full treatment in Chapter 19

Key Takeaway

Energy efficiency in massive MIMO is quasi-concave in $M$ : there exists a unique optimal $M^{\star}$ that balances the logarithmic rate gain from beamforming against the linear circuit power cost. Deploying more antennas than $M^{\star}$ wastes energy; deploying fewer leaves spectral efficiency on the table. As hardware becomes cheaper and more power-efficient ( $P_{\text{circuit}} \downarrow$ ), the optimum shifts to larger arrays — a virtuous cycle driving antenna count upward with each technology generation.

Energy Efficiency (EE)

The ratio of the achievable sum rate to the total power consumption: $\text{EE} = \sum_k R_k / P_{\text{total}}$ , measured in bits per joule. In massive MIMO, the EE is quasi-concave in $M$ , with an optimal $M^{\star}$ balancing beamforming gain against circuit power overhead.

Circuit Power

The power consumed by each antenna element's RF chain, including the ADC/DAC, mixer, filter, and local oscillator. Typical values range from 50 mW (low-power IoT) to 1 W (high-performance macro BS) per antenna. Circuit power creates a linear cost in $M$ that limits the EE-optimal number of antennas.

Energy Efficiency of Massive MIMO