Ferkans — Interactive Telecom Tutor

Why Does $N^2$ Matter?

We now arrive at the single most cited result about RIS. When the reflecting elements are phase-aligned — that is, when $\boldsymbol{\Phi}$ is chosen to coherently combine all $N$ paths at the UE — the received signal power grows as $N^2$ , not $N$ . The $N^2$ scaling is dramatic: doubling the RIS size from $256$ to $512$ elements gains $4\times$ in SNR, not $2\times$ . This is the empirical fact that justifies the optimization effort: if RIS passed only $N$ -scaling, it would barely outperform a simple scatterer. The $N^2$ scaling is what makes RIS worth the trouble.

But the $N^2$ scaling is conditional — it requires coherent combining, which requires channel state information at the RIS controller. With random or uncorrelated phases, the scaling falls back to $N$ . The distinction between these two regimes is fundamental and will reappear throughout the book.

Theorem: Coherent RIS: $\mathcal{O}(N^2)$ SNR Scaling

Consider a single-antenna BS ( $N_t = 1$ , $\mathbf{v} = 1$ ) and a single-antenna UE. Assume that $\mathbf{H}_1$ reduces to a vector $\mathbf{h}_1 \in \mathbb{C}^N$ with element amplitudes satisfying $|(\mathbf{h}_1)_n| = \alpha$ for all $n$ , and similarly $|(\mathbf{h}_2)_n| = \beta$ — that is, equal-magnitude path gains (e.g., a line-of-sight planar-wave scenario at the far field). The direct path is blocked ( $\mathbf{h}_d = \mathbf{0}$ ). Then the maximum received SNR over all $\boldsymbol{\Phi}$ with $|\phi_n| = 1$ is

$\text{SNR}^\star \;=\; \frac{P_t}{\sigma^2} \big(N \alpha \beta\big)^2 \;=\; \frac{P_t\, \alpha^2 \beta^2}{\sigma^2}\, N^2.$

The optimal phase at element $n$ is $\theta_n^\star = -\arg\!\big((\mathbf{h}_2)_n^* (\mathbf{h}_1)_n\big)$ , i.e., the RIS element compensates the round-trip phase.

Two gains multiply. First, the RIS focuses a phase-coherent beam back toward the UE, contributing $N$ in field amplitude at the user — $N^2$ in power. Second, the RIS is a coherent receive aperture of $N$ elements, so it captures $N$ times more signal energy than a single element would. These two $N$ s are not redundant: one is transmit beamforming gain (RIS → UE), the other is receive aperture (BS → RIS). Together they give $N^2$ in received SNR.

Proof

Express the signal

With $\mathbf{v} = 1$ and $\mathbf{h}_d = \mathbf{0}$ , the received signal reduces to $y = \big(\mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{h}_1\big) s + w.$ The instantaneous SNR is $\text{SNR} = (P_t/\sigma^2) \big|\sum_n (\mathbf{h}_2)_n^* \phi_n (\mathbf{h}_1)_n\big|^2$ .

Apply the triangle inequality

For any $\boldsymbol{\phi}$ with $|\phi_n| = 1$ , $\big|\sum_n (\mathbf{h}_2)_n^* \phi_n (\mathbf{h}_1)_n\big| \leq \sum_n |(\mathbf{h}_2)_n^*\phi_n (\mathbf{h}_1)_n| = \sum_n |(\mathbf{h}_2)_n| \cdot |(\mathbf{h}_1)_n|$ , with equality iff all terms have the same phase.

Identify the equality condition

Writing $(\mathbf{h}_2)_n^* (\mathbf{h}_1)_n = r_n e^{j\psi_n}$ with $r_n = |(\mathbf{h}_2)_n||(\mathbf{h}_1)_n|$ , equal phases means $\psi_n + \theta_n = \text{const}$ , i.e., $\theta_n^\star = C - \psi_n$ for any constant $C$ . Choosing $C = 0$ gives $\theta_n^\star = -\arg\!\big((\mathbf{h}_2)_n^*(\mathbf{h}_1)_n\big)$ .

Plug in the equal-magnitude assumption

With $|(\mathbf{h}_1)_n| = \alpha$ and $|(\mathbf{h}_2)_n| = \beta$ , the optimal amplitude is $\sum_n \alpha \beta = N\alpha\beta$ , so

$\text{SNR}^\star = \frac{P_t}{\sigma^2} (N\alpha\beta)^2 = \frac{P_t\,\alpha^2 \beta^2}{\sigma^2} N^2.$

This is the celebrated quadratic-in- $N$ scaling. $\blacksquare$

,

Dissecting the Two Factors of $N$

The $N^2$ scaling is the product of two independent $N$ s:

Aperture gain (BS → RIS): The RIS collects energy over $N$ elements. In field amplitude, this is a factor of $\sqrt{N}$ per element × $N$ elements = $N\sqrt{\cdot}$ when properly phased — or just $N$ in the coherent sum.
Beamforming gain (RIS → UE): The $N$ elements cohere in re-transmission to focus the reflected beam at the UE, contributing another factor of $N$ in received amplitude.

Power scales as amplitude squared, so the two $N$ s multiply. This is the same structure as the $(N\alpha\beta)^2$ bound from the proof, which can also be read as the square of a coherent sum of $N$ equal contributions.

Theorem: Random / Incoherent RIS: $\mathcal{O}(N)$ SNR Scaling

Under the same setup as $\mathcal{O}(N^2)$ $O (N^{2})$ SNR Scaling" data-ref-type="theorem">TCoherent RIS: $\mathcal{O}(N^2)$ SNR Scaling, if the RIS phase shifts are chosen uniformly at random independent of the channel ( $\theta_n \sim \text{Unif}[0, 2\pi)$ ), then the expected received SNR is

$\mathbb{E}_{\boldsymbol{\theta}}\big[\text{SNR}\big] = \frac{P_t\, \alpha^2 \beta^2}{\sigma^2}\, N.$

The $N$ -scaling comes from the law of large numbers on incoherent sums — each path contributes an independent random phase, and the sum behaves like a random walk with variance proportional to $N$ .

Proof

Expected squared sum

Let $Z = \sum_{n=1}^N X_n$ where $X_n = (\mathbf{h}_2)_n^* e^{j\theta_n}(\mathbf{h}_1)_n$ . Given the channels, $\theta_n$ uniform makes $X_n$ a zero-mean random variable with $\mathbb{E}|X_n|^2 = \alpha^2 \beta^2$ and independent across $n$ .

Apply independence

$\mathbb{E}|Z|^2 = \sum_n \mathbb{E}|X_n|^2 + \sum_{n \neq m} \mathbb{E}[X_n X_m^*] = N \alpha^2 \beta^2 + 0,$ since the cross-terms vanish: $\mathbb{E}[e^{j\theta_n} e^{-j\theta_m}] = 0$ for $n \neq m$ with independent uniform phases.

Conclude

$\mathbb{E}[\text{SNR}] = (P_t/\sigma^2) \cdot N \alpha^2 \beta^2 = \mathcal{O}(N).$ Compared to the $N^2$ coherent case, the random RIS loses a factor of $N$ in SNR — a significant gap that motivates the entire channel estimation and optimization machinery of the following chapters. $\blacksquare$

SNR vs. $N$ : Coherent ( $N^2$ ) versus Random ( $N$ )

Compare the SNR scaling law for three cases: coherent (optimally phased RIS), random-phase RIS, and a no-RIS baseline of matched transmission on the direct path. The coherent curve has slope 2 in log–log; the random curve has slope 1. The crossover with the direct path shows the RIS size at which the indirect path overtakes the direct.

Parameters

Max

N

swept256

Upper end of the RIS-size sweep.

Direct vs. indirect path loss (dB)0

Positive = direct path stronger per unit; 0 = equal; negative = direct weaker (blockage).

Reference SNR at

N = 1

(dB)0

Anchor for the vertical axis.

Coherent Combining at the RIS: Why $N^2$ ?

The RIS phases rotate to align each reflected contribution with the others. As alignment progresses, the length of the summed phasor grows linearly in

N

; received power grows as the square of the length, giving the

N^2

law.

Example: A 256-Element RIS in a Blocked Urban Link

A mmWave BS at 28 GHz ( $\lambda = c/f = 1.07\text{ cm}$ ) communicates with a UE whose direct path is blocked by a concrete wall. An RIS of $N = 256$ elements ( $16 \times 16$ , half-wavelength spacing) is mounted at a nearby building. Under a two-ray LoS model with $\alpha = \beta = 10^{-4}$ (path-loss amplitudes for each hop) and transmit SNR $P_t/\sigma^2 = 60\text{ dB}$ , compute the received SNR in the coherent-RIS case and the random-RIS case. How much did coherent combining buy us?

Solution

Plug the numbers into the coherent formula

$\text{SNR}^\star = 10^6 \cdot (256)^2 \cdot (10^{-4})^2 \cdot (10^{-4})^2 = 10^6 \cdot 65\,536 \cdot 10^{-16} = 6.55 \cdot 10^{-6}$ , i.e., about $-52\text{ dB}$ . This is painfully low — the blocked link on its own is essentially useless.

Compare with the random-phase case

$\mathbb{E}[\text{SNR}_{\text{rand}}] = 10^6 \cdot 256 \cdot 10^{-16} = 2.56 \cdot 10^{-8}$ , about $-76\text{ dB}$ . The coherent RIS gains $24\text{ dB} = 10 \log_{10}(256)$ over random — exactly the factor of $N$ predicted by theory.

Interpret the practical consequence

Even with $256$ elements, the double-hop path loss is brutal. To close the link at a useful rate, one needs either (i) more elements (recall $N^2$ : doubling $N$ adds $6\text{ dB}$ ), (ii) narrower directivity per element, or (iii) an active RIS that can amplify. This example is what Chapter 4 on channel estimation and Chapter 9 on active RIS are reacting to.

🎓CommIT Contribution(2023)

Array-Fed RIS: Motivation from the $N^2$ Law

G. Caire, I. Atzeni — IEEE Trans. Signal Process. (preprint 2023)

The $N^2$ SNR scaling, impressive as it is, cannot overcome the double path loss on its own at mmWave frequencies where the unit path loss is enormous. Caire and collaborators (2023) proposed the array-fed RIS architecture precisely to combine two kinds of gain: aperture gain from a large passive RIS with spatial multiplexing from a small active array that illuminates it. The BS-to-RIS link becomes a controlled near-field spot beam, avoiding the compounding uncertainty of the far-field BS-RIS channel. We develop the architecture in detail in Chapter 11; the motivation starts here, with the recognition that even $N^2$ isn't enough at sub-THz without a careful system-level co-design.

array-fed-rishigh-frequencycaire-2023

⚠️Engineering Note

What Does $N^2$ Mean for Deployment?

The $N^2$ scaling is seductive but has to be tempered by three practical realities:

Physical size. At sub-6 GHz, half-wavelength spacing implies $N = 1024$ elements occupies roughly $1\text{m}^2$ of surface. At mmWave (28 GHz), the same element count fits in about $50\text{cm} \times 50\text{cm}$ .
Channel estimation overhead. Estimating the cascaded channel naïvely requires $\mathcal{O}(N)$ or $\mathcal{O}(N^2)$ pilot symbols (Chapter 4). The $N^2$ SNR gain is only realized if the RIS is actually configured optimally — and that requires CSI.
Coherence time. In mobile scenarios, the channel changes faster than the RIS controller can update. The "effective $N$ " is limited by the number of elements that remain correctly phased during a coherence interval.

Practical Constraints

•
Each RIS element requires a dedicated biasing circuit — wiring complexity scales as $\mathcal{O}(N)$ .
•
Control link bandwidth: for 3-bit phase shifters, $3N$ bits per update — at $N=1024$ and 100 Hz update rate, ~300 kbps.
•
Element failures degrade coherent gain faster than random-phase gain (quadratic vs linear).

📋 Ref: ETSI GR RIS 001 (2023)

Common Mistake: $N^2$ Is Not Free Amplification

Mistake:

"If RIS gives $N^2$ gain, let's just use $N = 10\,000$ elements and have a $40\text{ dB}$ SNR boost for free."

Correction:

The $N^2$ law is relative to a single RIS element. The single RIS element already has dreadful path loss because it is a sub-wavelength scatterer. The absolute SNR of an $N$ -element RIS is compared against a direct link, not against a single element, and the direct link — if it exists — has its own $1/d_0^2$ scaling. The RIS becomes attractive when the direct path is weak or blocked, or when the per-element path loss is so small that even the squared gain is competitive. Chapter 4 discusses when RIS "wins" the system-level comparison.

Quick Check

Doubling the number of RIS elements from $N = 256$ to $N = 512$ changes the coherent received SNR by:

+3 dB

+6 dB

+12 dB

Depends on the direct path

Correction:

+6 dB

Coherent scaling is $N^2$ , so doubling $N$ multiplies SNR by 4, which is $10 \log_{10}(4) \approx 6\text{ dB}$ .

SNR Scaling Laws: From O(N)\mathcal{O}(N)O(N) to O(N2)\mathcal{O}(N^2)O(N2)

Why Does N2N^2N2 Matter?

Theorem: Coherent RIS: O(N2)\mathcal{O}(N^2)O(N2) SNR Scaling