Ferkans — Interactive Telecom Tutor

Not All Antennas Deserve the Same Number of Bits

The mixed-ADC architecture of Section 19.3 answers "how many antennas get high resolution?" with a single number $\alpha$ and a binary split. Section 19.4 generalizes the question: allow each antenna $n$ its own resolution $b_n \in \{0, 1, 2, \ldots, b_{\max}\}$ , and optimize the vector $\mathbf{b} = (b_1, \ldots, b_{N_r})$ subject to a total ADC-power budget $\sum_n c_0\,2^{b_n}\,f_s \leq P_{\text{bud}}.$ The optimal allocation assigns more bits to the antennas with larger channel gains (strong signals deserve precise digitization) and fewer bits to weak antennas — an ADC-domain analogue of water-filling.

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Definition:
The Optimal Bit Allocation Problem

Let $g_n = \mathbb{E}[|\mathbf{H}_{n}|^2]$ denote the mean per-antenna channel gain, $P_{\text{bud}}$ the ADC power budget, and $P_{\text{ADC}}(b) = c_0\,2^b f_s$ the per-ADC power cost at resolution $b$ . The bit allocation problem is

$\begin{aligned} \max_{\mathbf{b} \in \{0,\ldots,b_{\max}\}^{N_r}} \quad & \sum_{n=1}^{N_r} \log_2\!\left(1 + \kappa_{b_n} g_n\,\text{SNR}\right) \\ \text{s.t.} \quad & \sum_{n=1}^{N_r} 2^{b_n} \leq B_{\max} \triangleq \frac{P_{\text{bud}}}{c_0 f_s}, \end{aligned}$

where $\kappa_{b}$ is the Bussgang effective-gain factor of Definition DBussgang Distortion Factor. Assigning $b_n = 0$ means the antenna is turned off (contributes nothing and consumes one "budget unit" for the pilot stream).

The objective is the sum of per-antenna rates under MRC with perfect CSI, which decouples across antennas once the Bussgang linearization is applied. This decoupling is what makes the problem tractable — without it we would have to jointly optimize the resolution and the combiner, a mixed-integer nonlinear program.

Theorem: Continuous Bit Allocation is Water-Filling on Log Gains

Relax $b_n \in \mathbb{R}_{\geq 0}$ and approximate $\kappa_{b_n} \approx 1 - c_1\cdot 4^{-b_n}$ with $c_1 \approx 2 - 2/\pi$ so that $\log_2(1 + \kappa_{b_n} g_n\,\text{SNR}) \approx \log_2(1 + g_n\,\text{SNR}) - \epsilon(b_n)$ where $\epsilon(b_n) \approx c_2 \cdot 4^{-b_n} g_n\,\text{SNR}/ (1 + g_n\,\text{SNR})$ . The optimal continuous allocation satisfies

$b_n^\star = \frac{1}{2}\log_2\!\left( \frac{g_n\,\text{SNR}}{\mu\,\ln 2}\right)_+ = \tfrac{1}{2}\log_2(g_n\,\text{SNR}) - \tfrac{1}{2}\log_2(\mu\,\ln 2),$

where $\mu \geq 0$ is the Lagrange multiplier for the power budget and is chosen so that $\sum_n 2^{b_n^\star} = B_{\max}$ . The integer solution is obtained by rounding $b_n^\star$ and redistributing any power slack among the antennas with the largest gradient.

As with classical water-filling, we allocate more bits to antennas with larger log-gains; antennas below the water level are shut off ( $b_n = 0$ ) because the marginal rate return is smaller than the marginal power cost. The $\tfrac{1}{2}\log_2$ slope comes from the $4^{-b}$ decay of the Bussgang distortion factor — each additional bit cuts the residual distortion by a factor of 4 and buys roughly $\tfrac{1}{2}$ bit of rate per channel use.

Proof

Lagrangian

Form the Lagrangian $\mathcal{L} = \sum_n \log_2(1 + \kappa_{b_n} g_n\,\text{SNR}) - \mu\sum_n (2^{b_n} - B_{\max}/N_r)$ .

Stationarity

Differentiate w.r.t. $b_n$ (real relaxation): $\tfrac{\partial \kappa_{b_n}/\partial b_n}{1 + \kappa_{b_n} g_n\,\text{SNR}} \cdot g_n\,\text{SNR}/\ln 2 = \mu\, 2^{b_n} \ln 2.$ At moderate $b_n$ where $\kappa_{b_n}\approx 1$ , the numerator becomes $2 c_1 \ln 2 \cdot 4^{-b_n}$ and the equation simplifies to $c_1 g_n\,\text{SNR}/(1 + g_n\,\text{SNR}) \cdot 4^{-b_n} = \mu\, 2^{b_n}$ .

Solve

$2^{3 b_n} = c_1 g_n\,\text{SNR}/(\mu(1 + g_n\,\text{SNR}))$ . Taking logs and simplifying gives the stated $b_n^\star \propto \tfrac{1}{2}\log_2(g_n\,\text{SNR})$ form after absorbing constants into $\mu$ .

Budget

Feed $b_n^\star$ back into $\sum_n 2^{b_n^\star} = B_{\max}$ to find $\mu$ . Integer rounding and residual reallocation give the final discrete solution. $\blacksquare$

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Water-Filling Bit Allocation

Complexity:

\mathcal{O}(N_r \log(1/\epsilon) + N_r b_{\max})

Input: per-antenna gains g_1..g_Nr, per-antenna SNR rho,

total bit budget B_max = P_bud / (c_0 f_s),

maximum per-antenna resolution b_max

Output: integer bit allocation b_1..b_Nr

// Phase 1 — continuous relaxation

mu_lo, mu_hi <- 1e-12, 1e12

for iter = 1 .. 60: // bisection on mu

mu <- sqrt(mu_lo * mu_hi)

for n = 1 .. Nr:

b_cont[n] <- 0.5 * log2( (g[n] * rho) / (mu * ln 2) )

b_cont[n] <- clip(b_cont[n], 0, b_max)

used <- sum_n 2^b_cont[n]

if used > B_max: mu_lo <- mu

else: mu_hi <- mu

// Phase 2 — integer rounding

for n = 1 .. Nr:

b[n] <- floor(b_cont[n])

// Phase 3 — reallocate leftover budget

leftover <- B_max - sum_n 2^b[n]

while leftover > 0:

// compute marginal rate gain of giving each antenna one more bit

for n = 1 .. Nr:

gain[n] <- log2(1 + kappa[b[n]+1] * g[n] * rho)

- log2(1 + kappa[b[n]] * g[n] * rho)

cost[n] <- 2^(b[n]+1) - 2^b[n] // = 2^b[n]

n_star <- argmax_n (gain[n] / cost[n]) among n with cost[n] <= leftover

if n_star is empty: break

b[n_star] <- b[n_star] + 1

leftover <- leftover - cost[n_star]

return b_1..b_Nr

Phase 1 is a one-dimensional bisection and converges in $\sim 60$ iterations to double-precision tolerance. Phases 2 and 3 together form a greedy integer rounding that is provably within $\log_2 b_{\max}$ of the continuous optimum (Roth et al. 2018, Proposition 4).

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Water-Filling Bit Allocation Across Antennas

Generate a random per-antenna gain profile (sorted in decreasing order), run the water-filling allocator, and plot the resulting integer bit assignment $b_n$ vs antenna index. Try different power budgets to see antennas getting switched off or upgraded as the water level moves.

Parameters

Number of antennas

N_r

64

Per-antenna SNR (dB)0

Gain spread across antennas (dB)15

Average budget per antenna (

2^{\bar b}

units)4

Maximum resolution

b_{\max}

6

Example: Water-Filling on 8 Antennas

An 8-antenna receiver sees per-antenna gains $g_1,\ldots,g_8 = (10, 8, 6, 4, 3, 2, 1, 0.5)$ at per-antenna SNR $\text{SNR} = 1$ (0 dB). The total bit budget is $B_{\max} = 32$ (equivalent to an average of 2 bits per antenna or $2^{\bar b} = 4$ per-antenna). Find the continuous and integer allocations, rounded to the nearest bit, with $b_{\max} = 5$ .

Solution

Compute the relaxed allocation

$b_n^\star = \tfrac{1}{2}\log_2(g_n) - \tfrac{1}{2}\log_2(\mu\ln 2)$ . We solve numerically: set $\mu$ so that $\sum_n 2^{b_n^\star} = 32$ . Starting guess $\mu = 1.0$ gives $b_n^\star = (1.66, 1.50, 1.29, 1.00, 0.79, 0.50, 0.00, -0.50)$ ; clip at 0 and rebudget. Iterating the bisection gives $\mu \approx 0.31$ and $b^\star \approx (2.50, 2.34, 2.13, 1.84, 1.63, 1.34, 0.84, 0.34)$ (antennas 7 and 8 get small values).

Integer rounding

Floor: $b = (2, 2, 2, 1, 1, 1, 0, 0)$ , cost $\sum 2^b = 4+4+4+2+2+2+1+1 = 20$ . Leftover $12$ .

Greedy reallocation

Add one bit to antenna 1 (highest $g$ ), cost $4$ , leftover $8$ . Add one bit to antenna 2, cost $4$ , leftover $4$ . Add one bit to antenna 3, cost $4$ , leftover $0$ . Final: $\mathbf{b} = (3, 3, 3, 1, 1, 1, 0, 0)$ , total $\sum 2^b = 8+8+8+2+2+2+1+1 = 32$ .

Interpretation

The strongest three antennas get 3 bits each, three middling ones get 1 bit each, and the weakest two are turned off. Roughly 40% of the bits concentrate on the top 3 antennas, where the marginal rate gain is highest. $\blacksquare$

Bit Allocation is a Refinement of Mixed-ADC

Notice that the mixed-ADC architecture of Section 19.3 is a special case of bit allocation with the constraint $b_n \in \{1, b_H\}$ . Allowing $b_n$ to take intermediate integer values gains a small amount of rate at the same power budget but complicates the RF front-end (multiple ADC reference designs on the same chip). In practice deployments usually pick two resolutions; the bit-allocation formulation is a useful upper bound on what mixed-ADC can achieve.

Common Mistake: Don't Use Long-Term Gains for Fast-Fading Channels

Mistake:

A common mistake is to compute the bit allocation once, based on the long-term average gains $g_n = \mathbb{E}[|\mathbf{H}_{n}|^2]$ , and then freeze it for all coherence blocks.

Correction:

The water-filling analysis depends on the instantaneous effective channel gain when MRC is applied, which fluctuates rapidly in fast-fading environments. Freezing the allocation wastes budget on antennas that are currently in a deep fade and starves antennas that have just come out of one. Practical systems either (i) compute the allocation once per coherence interval $T_c$ using the short-term estimates, or (ii) use the long-term allocation but permute which antennas hold which resolution via the same RF-switch hardware of Section 19.3. The switch-based approach is preferred when $T_c$ is comparable to the switching latency.

⚠️Engineering Note

Can We Actually Build Per-Antenna Different ADCs?

Strictly per-antenna resolution is hard to implement: it requires either physically different ADC chips on different ports or a reconfigurable ADC whose output can be re-quantized on the fly. Reconfigurable ADCs exist (cost a power premium) but are usually limited to 2–3 resolution settings. A pragmatic compromise is: assign each antenna to one of three classes — 1 bit, 4 bits, 8 bits — and cast the optimization as a discrete mixed-ADC with three levels. The Roth-Nossek algorithm specializes cleanly to this three-level case and is the basis of most published 1-bit/mixed-ADC testbeds.

Practical Constraints

•
Typical deployments use $\{1,4\}$ -bit or $\{1,4,8\}$ -bit sets, not arbitrary $b_n$
•
Per-antenna calibration cost grows with the number of distinct resolutions
•
RF switch matrix permutes antenna-to-ADC mapping dynamically

📋 Ref: Prototyping literature, Studer lab 2018-2020

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Historical Note: From Signal Processing to Massive MIMO

1990s-2020s

Bit-allocation methods have a long history in audio/video source coding, where each transform coefficient gets its own precision (classical rate-distortion allocation). Widrow and Kollár's 1996 textbook on quantization theory applied the framework to signal processing, and the Massey–Goodman type inequalities made the allocation rule look like $b_n \propto \tfrac{1}{2}\log_2\sigma_n^2$ . The translation to MIMO receivers came roughly 20 years later: Choi et al. (2016) and Roth and Nossek (2018) formulated the bit allocation across antennas, connecting it to water-filling. The interesting twist specific to MIMO is that each antenna carries a sum over the users (rather than a single coefficient), so the allocation interacts with the multi-user combiner design — a subject still under active research.

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Quick Check

In the continuous bit-allocation water-filling formula $b_n^\star \propto \tfrac{1}{2}\log_2(g_n\,\text{SNR})$ , what does the factor $1/2$ represent?

Division between I and Q rails

Bandwidth factor from Nyquist sampling

Slope relating bits to $\log_2$ of effective SNR (6 dB per bit / 2)

Half-rate coding overhead

Correction:

Slope relating bits to

\log_2

of effective SNR (6 dB per bit / 2)

Each bit cuts the quantization distortion by a factor $4$ , which corresponds to a factor $2$ in effective SNR amplitude, and hence $\\tfrac{1}{2}$ in $\\log_2$ SNR. This is the classic 6 dB/bit SQNR heuristic in the bit-allocation domain.

Key Takeaway

Optimal bit allocation water-fills over per-antenna log gains: strong antennas get more bits, weak ones get turned off. Under the relaxed continuous formulation the optimal resolution is $b_n^\star \propto \tfrac{1}{2}\log_2(g_n\,\text{SNR})$ , which integer-rounds to 2–5 bits for the top antennas and 0–1 bits for the rest. Mixed-ADC with a small $\alpha$ is the best-known discrete implementation of this principle.

Optimal Bit Allocation

Not All Antennas Deserve the Same Number of Bits

Definition: The Optimal Bit Allocation Problem

Theorem: Continuous Bit Allocation is Water-Filling on Log Gains

Lagrangian

Stationarity

Solve

Budget

Water-Filling Bit Allocation

Water-Filling Bit Allocation Across Antennas

Parameters

Example: Water-Filling on 8 Antennas

Compute the relaxed allocation

Integer rounding

Greedy reallocation

Interpretation

Bit Allocation is a Refinement of Mixed-ADC

Common Mistake: Don't Use Long-Term Gains for Fast-Fading Channels

Can We Actually Build Per-Antenna Different ADCs?

Historical Note: From Signal Processing to Massive MIMO

Quick Check

Key Takeaway

Definition:
The Optimal Bit Allocation Problem