Ferkans — Interactive Telecom Tutor

How Much Does Discretization Cost?

The bottom-line question: at $B$ -bit resolution, what fraction of the continuous-phase performance do we retain? Section 8.4 answers this precisely for different scenarios — single-user vs multi-user, coherent vs random-phase baselines, and different utility functions (rate, fairness).

Theorem: Coherent SNR Loss at $B$ -bit Resolution

For the single-user coherent-combining problem,

$\eta_B = \frac{\text{SNR}^{\text{disc}}_B}{\text{SNR}^{\text{cont}}} = \text{sinc}^2\!\left(\frac{\pi}{2^B}\right).$

Equivalent bit-loss table:

$B$	$\eta_B$	dB loss
1	0.405	3.92
2	0.811	0.91
3	0.949	0.22
4	0.987	0.056
5	0.997	0.014
$\infty$	1.000	0.000

Diminishing returns after $B = 3$ : going from 3 to 4 bits saves only $0.16$ dB. This is why the industry default is $B = 3$ .

Under coherent alignment, each element's quantization error is uniform on $[-\Delta\theta/2, \Delta\theta/2]$ . The coherent sum has complex mean $\text{sinc}(\pi/2^B)$ times the continuous magnitude; squaring gives the SNR loss.

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Theorem: Multi-User Rate Loss with $B$ -bit RIS

For $K$ -user RIS with coherent alignment at $B$ bits:

$\Delta R_{\text{sum}} \approx K \log_2(\eta_B^{-1}), \qquad \Delta R_{\text{mm}} \approx \log_2(\eta_B^{-1}).$

At $B = 3$ and $K = 4$ : $\Delta R_{\text{sum}} \approx 4 \cdot 0.074 \approx 0.3$ bits/s/Hz. Essentially negligible. At $B = 1$ and $K = 4$ : $\Delta R_{\text{sum}} \approx 4 \cdot 1.3 \approx 5.2$ bits/s/Hz. Noticeable.

The quantization penalty scales linearly with $K$ for sum rate: more users share the penalty per user, cumulated across the aggregate throughput.

For multi-user scenarios, the per-user SNR loss is $\eta_B$ as before. Sum rate at high SNR behaves like $K \log_2(\text{SNR})$ , so the rate loss is $K \log_2(\eta_B^{-1})$ . Max-min is a single per-user rate, loss $\log_2(\eta_B^{-1})$ . Both scale with $B$ like the single-user case.

Discrete-Phase SNR Loss vs. $B$

Plot the coherent SNR loss (dB) and effective-element-count ratio $N_{\text{eff}}/N = \eta_B$ as $B$ varies from 1 to 6. Compare the analytical $\text{sinc}^2(\pi/2^B)$ prediction with Monte-Carlo empirical results from random problem instances.

Parameters

Max bits to sweep6

RIS elements

N

64

Monte-Carlo trials50

Example: Effective RIS Size from Quantization

A 1-bit RIS with $N = 1024$ elements and a 3-bit RIS with $N = 400$ elements — which has the higher effective coherent SNR? Use the $\eta_B$ factors.

Solution

1-bit, $N = 1024$

Coherent SNR $\propto \eta_1 \cdot N^2 = 0.405 \cdot 1024^2 = 4.25 \cdot 10^5$ .

3-bit, $N = 400$

Coherent SNR $\propto \eta_3 \cdot N^2 = 0.949 \cdot 400^2 = 1.52 \cdot 10^5$ .

Compare

The 1-bit RIS is 2.8× higher SNR (4.4 dB) despite the higher quantization loss, because of its larger $N$ . The $N^2$ scaling dominates.

Lesson

For a given hardware budget, 1-bit with many elements often beats 3-bit with few. The crossover is governed by $B \log_2 L / \log_2(1/\eta_B)$ vs. the pin cost per bit. This is the Section 8.1 bit-budget analysis made concrete. $\blacksquare$

Deployment: What Bit Count Do We Actually See?

Empirically, production RIS panels are:

1-bit (PIN diode): rare commercially; common in research prototypes (e.g., RFocus MIT 2020).
2-bit: early sub-6 GHz commercial panels. Cheap PIN-diode hardware.
3-bit: current mmWave commercial panels (2024). Balance of hardware complexity and quantization loss.
Continuous (varactor + 6-bit DAC): research demonstrators and high-end prototypes. Offers no-quantization-loss operation at the cost of calibration overhead.

The roadmap suggests 3-bit panels will dominate commercial deployment at sub-6 GHz and mmWave, with continuous phase reserved for high-accuracy applications (radar sensing, ISAC).

Quick Check

A fixed control-bit budget of 1024 pins is available for a RIS panel. Which configuration yields the highest coherent SNR?

$B = 1, N = 1024$

$B = 2, N = 512$

$B = 3, N = 341$

$B = 4, N = 256$

Correction:

B = 1, N = 1024

Coherent SNR scales as $\eta_B N^2$ . For $B=1, N=1024$ : $0.405 \cdot 1024^2 \approx 4.25 \cdot 10^5$ . For $B=2, N=512$ : $0.811 \cdot 512^2 \approx 2.13 \cdot 10^5$ . For $B=3, N=341$ : $0.949 \cdot 341^2 \approx 1.10 \cdot 10^5$ . Even after the 1-bit $3.92$ dB penalty, quadrupling $N$ wins.

⚠️Engineering Note

Design Choice: Which Bit Depth?

A checklist for bit-depth selection:

Objective: Sum rate or max-min? Max-min is more quantization-sensitive (per-user rate loss matters).
User count: More users → more aggregate penalty. Reserve higher $B$ for high- $K$ deployments.
Hardware constraint: Control pins per element scale with $B$ . Budget carefully.
Frequency band: Sub-6 GHz has less strict phase resolution needs (wider coherence time); mmWave / sub-THz benefits more from higher $B$ .
Application: Pure communication → $B = 2$ - $3$ . ISAC / radar → $B = 4$ + for phase accuracy. Sensing → continuous.

Typical sweet spot: $B = 3$ . Below that, more elements might be worth the tradeoff (see Section 8.1); above that, marginal returns are tiny.

Practical Constraints

•
Sub-6 GHz commercial panels: $B = 2$ .
•
mmWave commercial panels (2024): $B = 3$ .
•
Research: $B = 1$ or continuous, depending on agenda.
•
ISAC: $B = 4$ + typically required.

Quantifying the Discrete-Phase Loss