Ferkans — Interactive Telecom Tutor

From 'Add Up the Signals' to a Power-Control Problem

Section §16.1 imposed a single design rule — $b_k h_k = \eta$ for all $k$ — and read off the MSE $\sigma^2/|\eta|^2$ . Making the MSE small means making $|\eta|$ large. But $|\eta|$ is bounded by the weakest channel: to satisfy $|b_k|^2 \leq P_k / \mathbb{E}[|s_k|^2]$ for every user, one sets $|\eta| \leq |h_k| \sqrt{P_k / \mathbb{E}[|s_k|^2]}$ across all $k$ .

The point is that AirComp's MSE is a water-filling problem in reverse — the receive amplitude is limited by the worst channel, not the average. Strong channels sit below their power budgets. The section formalizes this, derives the zero-forcing and MMSE receivers, and characterizes when — if ever — dropping weak users helps.

,

Theorem: Zero-Forcing AirComp MSE

Fix a common receive amplitude target $\eta > 0$ . Users transmit with $b_k = \eta / h_k$ , subject to $|b_k|^2 \sigma_s^2 \leq P_k$ where $\sigma_s^2 = \mathbb{E}[|s_k|^2]$ . Write $\gamma_k = |h_k|^2 P_k / \sigma_s^2$ (per-user effective channel gain).

Feasibility. The magnitude-alignment choice $b_k = \eta / h_k$ is feasible for every $k$ iff $\eta^2 \leq \min_k \gamma_k$ .
MSE. Under feasibility, the zero-forcing post-processor $\hat{y} = r / \eta$ achieves $\mathsf{MSE}(\eta) \;=\; \frac{\sigma^2}{\eta^2}.$
Optimal receive amplitude. The MSE is minimized by $\eta^{\star} = \sqrt{\min_k \gamma_k}$ , giving $\mathsf{MSE}^{\star} \;=\; \frac{\sigma^2}{\min_k \gamma_k}.$

Proof

Express $|b_k|^2$

With $b_k h_k = \eta$ and $h_k \neq 0$ : $|b_k|^2 = \eta^2 / |h_k|^2$ . The transmit-power constraint becomes $\eta^2 \sigma_s^2 / |h_k|^2 \leq P_k$ , i.e., $\eta^2 \leq |h_k|^2 P_k / \sigma_s^2 = \gamma_k$ .

Feasibility across all users

The constraint must hold for every user. The binding user is the one with smallest $\gamma_k$ ; the joint feasibility region is $\eta^2 \leq \min_k \gamma_k$ .

MSE under alignment

Substituting $b_k h_k = \eta$ into the signal model of §16.1: $r = \eta \sum_k s_k + \mathbf{w}$ . The estimator $\hat{y} = r/\eta$ gives $\hat{y} = \sum_k s_k + \mathbf{w}/\eta$ , whence $\mathsf{MSE} = \sigma^2/\eta^2$ .

Optimizing $\eta$

$\mathsf{MSE}(\eta)$ decreases in $\eta$ , so the optimum saturates the feasibility constraint: $\eta^{\star} = \sqrt{\min_k \gamma_k}$ .

Operational interpretation

The minimum $\gamma_k$ dominates — not the average and not the maximum. A single weak user (deep fade, low transmit budget, large source variance) sets the MSE for everyone. This is the fundamental worst-user bottleneck of AirComp.

,

Example: Computing the ZF MSE for Four Heterogeneous Users

Four users share an AirComp MAC. Channel gains $|h_k|^2 = \{1.0, 0.6, 0.3, 0.1\}$ . Per-user budget $P_k = 1$ . Source variance $\sigma_s^2 = 1$ . Noise variance $\sigma^2 = 0.01$ . Compute $\eta^{\star}$ and $\mathsf{MSE}^{\star}$ under zero-forcing.

Solution

Compute $\gamma_k$

$\gamma_k = |h_k|^2 P_k / \sigma_s^2 = \{1.0, 0.6, 0.3, 0.1\}$ .

Identify the weakest user

$\min_k \gamma_k = 0.1$ (user 4). This user sets the feasibility boundary.

Optimal receive amplitude

$\eta^{\star} = \sqrt{0.1} \approx 0.316$ .

MSE

$\mathsf{MSE}^{\star} = 0.01 / 0.1 = 0.1$ . A full order of magnitude above the noise floor, entirely because of user 4.

What if we drop user 4?

With users $\{1, 2, 3\}$ : $\min \gamma_k = 0.3$ , $\mathsf{MSE} = 0.01/0.3 \approx 0.033$ . Three-fold reduction — but the aggregate is now $\sum_{k=1}^{3} s_k$ , not the full sum. Whether dropping helps depends on what the aggregate is for (§16.4 returns to this for federated learning).

Definition:
MMSE AirComp Receiver

Fix transmit scalings $b_k$ (not necessarily aligned). Let $\alpha = \sum_k h_k b_k$ denote the sum of channel-scaled gains. The receiver's MMSE estimator for $y = \sum_k s_k$ takes the form $\hat{y}_{\text{MMSE}} = c \cdot r$ , with $c$ chosen to minimize $\mathbb{E}[|\hat{y}_{\text{MMSE}} - y|^2]$ .

Assuming i.i.d.\ sources ( $s_k$ zero- mean with variance $\sigma_s^2$ ), the optimal linear estimator is $c^{\star} \;=\; \frac{\alpha^{*} \sigma_s^2}{|\alpha|^2 \sigma_s^2 + \sigma^2}.$ When the transmit scalings satisfy magnitude alignment ( $b_k h_k = \eta$ , $\alpha = n \eta$ ) and $\sigma_s^2$ is large, $c^{\star} \to 1/\alpha$ — recovering zero-forcing. For small $\sigma_s^2$ (low source energy), MMSE shrinks the estimate toward zero to trade bias for variance reduction.

When ZF Is Optimal and When It Is Not

Zero-forcing inverts the channel perfectly: $\hat{y} = \sum_k s_k + \mathbf{w}/\eta$ — no bias, but the noise is amplified by $1/\eta$ . MMSE accepts a small bias in exchange for lower noise, giving uniformly lower total MSE.

The bias-free property of ZF matters when the aggregated value feeds into a downstream estimator that assumes unbiased inputs (e.g., an FL gradient update). For applications that only care about total squared error, MMSE dominates. §16.4 returns to this choice through the FL lens of Chapter 17.

AirComp MSE vs. Transmit Power and User Count

Explore how the zero-forcing AirComp MSE scales with the common per-user transmit budget $P$ and the number of users $n$ . Channels are drawn from a Rayleigh fading distribution; the minimum channel gain governs the MSE (Theorem 16.2.1). As $n$ grows, the minimum fades deteriorate in distribution — so MSE increases with $n$ , despite more signals superimposing. This exposes the worst-user bottleneck.

Parameters

P_{\max}

(dB) — maximum transmit power30

n

— number of users10

\sigma^2

(dB) — noise variance-20

Theorem: Threshold User Selection

Let $\mathcal{A} \subseteq [n]$ be any subset of users. Restricting AirComp to $\mathcal{A}$ yields aggregate $\sum_{k \in \mathcal{A}} s_k$ with $\mathsf{MSE}(\mathcal{A}) \;=\; \frac{\sigma^2}{\min_{k \in \mathcal{A}} \gamma_k}.$ The threshold scheduling rule $\mathcal{A}(\tau) = \{k : \gamma_k \geq \tau\}$ is Pareto-optimal: no other subset achieves both a smaller MSE and a larger $|\mathcal{A}|$ .

Proof

MSE of a subset

Applying Theorem 16.2.1 to users $\mathcal{A}$ gives $\mathsf{MSE}(\mathcal{A}) = \sigma^2/\min_{k\in\mathcal{A}}\gamma_k$ .

Threshold subset optimality

Suppose $\mathcal{A}$ is not a threshold set. Then there exist $j \in \mathcal{A}, k \notin \mathcal{A}$ with $\gamma_k > \gamma_j$ . Replacing $j$ by $k$ weakly increases the minimum $\gamma$ — weakly reducing MSE — while preserving $|\mathcal{A}|$ . Iterating, we transform $\mathcal{A}$ into a threshold set without degrading either objective.

Pareto frontier

The frontier is a staircase: as $\tau$ increases, users drop out one at a time in reverse-sorted order of $\gamma_k$ , and $\mathsf{MSE} = \sigma^2/\tau$ decreases correspondingly.

Trade-off interpretation

Each user dropped decreases the aggregated sample size by one and decreases the MSE. For an FL application, the downstream effect depends on whether the drop introduces bias — if the dropped user's data is non-i.i.d., bias is introduced and the MSE reduction may not translate into a better global model.

⚠️Engineering Note

Deploying AirComp Scheduling

Practical guidance for AirComp user selection:

Choose $\tau$ from the target MSE. Given a tolerance $\mathsf{MSE}^{\text{tol}}$ , set $\tau = \sigma^2/\mathsf{MSE}^{\text{tol}}$ . Users with $\gamma_k < \tau$ are excluded.
Fairness concerns. Threshold scheduling systematically excludes weak-channel users. In FL, this can introduce statistical bias (weak users often correlate with specific data distributions — rural clients, edge devices). Consider rotating the excluded set across rounds.
CSIT quality. Channel estimation error $\delta_k = \hat{h}_k - h_k$ translates into alignment error $\delta_k \cdot b_k$ . A quick rule of thumb: keep $|\delta_k|^2 \leq |h_k|^2 / 10$ to stay within $0.5$ dB of the ideal-CSIT MSE.
Truncate aggressively or run more rounds? An aggressive threshold yields lower MSE per round but excludes users; running more rounds with a lax threshold includes everyone but costs channel uses. The optimal trade-off is FL-application- specific (§17.3).

Practical Constraints

•
CSIT error must be $\leq |h|/\sqrt{10}$ for $0.5$ dB MSE loss
•
Threshold $\tau$ determines included users
•
Rotation across rounds mitigates bias
•
Trade round count against per-round threshold

📋 Ref: Yang-Jiang-Shi 2020; Cao-Yang-Jiang 2020

Key Takeaway

The AirComp MSE is dominated by the weakest user's effective channel gain $\gamma_k = |h_k|^2 P_k / \sigma_s^2$ . The zero-forcing receiver — the workhorse of AirComp in practice — achieves $\mathsf{MSE}^{\star} = \sigma^2/\min_k \gamma_k$ . Threshold user selection traces a Pareto-optimal MSE-vs.-sample-size curve. In FL contexts (Chapter 17), this trade-off is a core design knob: decreasing MSE may cost statistical representativeness.

Quick Check

In zero-forcing AirComp with $n = 4$ users, the effective channel gains are $\gamma_k = \{0.8, 0.5, 0.2, 0.4\}$ and $\sigma^2 = 0.02$ . What is $\mathsf{MSE}^{\star}$ ?

$0.02 / 0.8 = 0.025$

$0.02 / 0.2 = 0.1$

$0.02 / \bar{\gamma} = 0.02 / 0.475 \approx 0.042$

$0.02 \cdot n / \sum_k \gamma_k$

Correction:

0.02 / 0.2 = 0.1

The minimum $\gamma_k$ is $0.2$ (user 3). By Theorem 16.2.1, $\mathsf{MSE}^{\star} = \sigma^2/\min_k \gamma_k = 0.02/0.2 = 0.1$ .

Power Control and MSE Analysis