Ferkans — Interactive Telecom Tutor

The Simplest Scheme That Works

If we cannot excite the RIS elements with pilots directly, can we at least excite them one at a time? The ON/OFF protocol enables each element in turn while disabling the rest, extracts that element's cascaded channel by subtraction, and loops over the $N$ elements. It is conceptually the simplest RIS estimation scheme; it is also the least efficient. Understanding it is the baseline for all more sophisticated schemes.

Definition:
The ON/OFF Subset-Switching Protocol

Partition the pilot-training phase into $\tau_p = N + 1$ slots:

Slot 0: all RIS elements OFF ( $\phi_n = 0$ for all $n$ , where "OFF" means the element absorbs rather than reflects). The BS receives only the direct path: $\mathbf{y}_0 = \mathbf{h}_d x + \mathbf{w}_0$ .
Slot $n$ ( $n = 1, \ldots, N$ ): element $n$ ON, others OFF ( $\phi_n = 1$ , $\phi_m = 0$ for $m \neq n$ ). Received signal: $\mathbf{y}_n = (\mathbf{h}_d + \mathbf{g}_n) x + \mathbf{w}_n$ , where $\mathbf{g}_n$ is the $n$ -th row of the cascaded channel $\mathbf{G}^H$ .

Subtracting slot 0 from slot $n$ recovers the individual RIS channel: $\hat{\mathbf{g}}_n = (\mathbf{y}_n - \mathbf{y}_0)/x$ . After $N + 1$ slots, the entire cascaded channel $\mathbf{G}^H$ is estimated, plus $\mathbf{h}_d$ .

"Turning an RIS element OFF" physically means setting its reflection coefficient to approximately zero. In reality, most passive RIS hardware cannot reach exactly $\phi_n = 0$ — the element always reflects something. A common workaround is to reverse-bias the diode into a high-loss state, achieving $|\phi_n| \approx 0.1$ or lower. We revisit this limitation in the pitfall below.

Theorem: Per-Element MSE of the ON/OFF Estimator

With unit-power pilots ( $|x|^2 = P_t$ ) and i.i.d. Gaussian noise $\mathbf{w}_t \sim \mathcal{CN}(\mathbf{0}, \sigma^2 \mathbf{I}_{N_t})$ , the ON/OFF estimator $\hat{\mathbf{g}}_n = (\mathbf{y}_n - \mathbf{y}_0)/x$ is unbiased with per-element MSE

$\mathbb{E}\big[\|\hat{\mathbf{g}}_n - \mathbf{g}_n\|^2\big] = \frac{2 N_t\, \sigma^2}{P_t}.$

The total estimation energy scales as $N$ : each element uses one pilot slot, so the total time spent is $N \cdot T_s$ and the per-element energy is fixed.

The subtraction $\hat{\mathbf{g}}_n = (\mathbf{y}_n - \mathbf{y}_0)/x$ is clean for the signal but doubles the noise: two noisy slots combine, so the variance of $\hat{\mathbf{g}}_n$ is twice the single-slot noise variance, divided by the pilot power. This is the $3\text{ dB}$ penalty that more sophisticated schemes (DFT codebook, Section 4.3) avoid.

Proof

Noise in the subtraction

$\hat{\mathbf{g}}_n = \mathbf{g}_n + (\mathbf{w}_n - \mathbf{w}_0)/x$ . The noise term $(\mathbf{w}_n - \mathbf{w}_0)/x$ is $\mathcal{CN}(\mathbf{0}, 2\sigma^2/|x|^2 \cdot \mathbf{I})$ .

Per-element MSE

$\mathbb{E}\|\hat{\mathbf{g}}_n - \mathbf{g}_n\|^2 = \text{tr}(2\sigma^2/|x|^2 \cdot \mathbf{I}_{N_t}) = 2N_t\sigma^2/P_t$ . $\blacksquare$

ON/OFF Channel Estimation

Complexity:

\tau_p = N + 1

pilot slots;

O(N N_t)

arithmetic

Input: BS antennas

N_t

, RIS elements

N

, pilot power

P_t

Output: direct channel

\hat{\mathbf{h}}_d

and cascaded channel

\hat{\mathbf{G}} \in \mathbb{C}^{N \times N_t}

.

1. Slot 0: all elements OFF. Transmit pilot

x

. Record

\mathbf{y}_0

.

2.

\hat{\mathbf{h}}_d \leftarrow \mathbf{y}_0 / x

.

3. for

n = 1, \ldots, N

do

4.

\quad

Turn element

n

ON, others OFF. Transmit pilot

x

. Record

\mathbf{y}_n

.

5.

\quad \hat{\mathbf{g}}_n \leftarrow (\mathbf{y}_n - \mathbf{y}_0)/x

(the

n

-th row of

\hat{\mathbf{G}}^H

)

6. end for

7. return

\hat{\mathbf{h}}_d, \hat{\mathbf{G}}

The protocol is trivially parallelizable across BS antennas: each pilot slot produces $N_t$ independent channel components simultaneously.

Example: ON/OFF Overhead for a 256-Element RIS

A mobile-access RIS has $N = 256$ elements and operates in a coherence block of $T = 500$ symbols. The BS has $N_t = 8$ antennas, $P_t/\sigma^2 = 20\text{ dB}$ pilot SNR. What fraction of the coherence block is spent on pilots? What is the per-element estimation MSE?

Solution

Pilot length

$\tau_p = N + 1 = 257$ pilot slots. Overhead $\tau_p/T = 257/500 = 51.4\%$ . Over half the coherence block is spent estimating the channel — a punishingly high figure for moderate $N$ .

Per-element MSE

$\text{MSE} = 2N_t/\text{SNR} = 2 \cdot 8/100 = 0.16$ . Relative to a unit-variance channel, the estimation is accurate to $\pm 40\%$ per complex entry.

Effective rate

Effective throughput is scaled by $(1 - \tau_p/T) = 48.6\%$ . Even if optimization gives a $4\times$ rate improvement, the net gain is $1.94\times$ — modest. This motivates every more-sophisticated scheme in this chapter.

Common Mistake: ON/OFF Wastes Half the RIS Aperture

Mistake:

"Turning $N-1$ elements OFF while estimating element $n$ is just bookkeeping — all the other elements reflect zero, so they don't cost us anything."

Correction:

During ON/OFF pilots, only one element contributes to the received signal — the signal power is $|\mathbf{g}_n|^2$ , no $N^2$ coherent gain. Compared with schemes where all elements reflect simultaneously (DFT codebook), ON/OFF has signal power $|\mathbf{g}_n|^2$ per slot versus $\sim N |\mathbf{g}_n|^2$ — an $N$ -fold loss in effective estimation SNR. This is the $3\text{ dB}$ penalty (for large $N$ , more like $10 \log_{10} N$ dB) that DFT codebooks recover.

⚠️Engineering Note

Practical Limits of the OFF State

Physical RIS hardware cannot reach exactly $\phi_n = 0$ . A PIN-diode element in reverse bias typically achieves $|\phi_n|_{\text{off}} \approx 0.1$ – $0.3$ ; varactors saturate with $|\phi_n|_{\text{off}} \approx 0.3$ – $0.5$ depending on the resonant design. The ON/OFF protocol degrades gracefully: the subtraction $\mathbf{y}_n - \mathbf{y}_0$ still isolates the target element's contribution, but a residual "bias" from the non-zero OFF state of the other elements contaminates the estimate.

The residual bias is approximately $\|\mathbf{G}^H\|_F \cdot |\phi|_{\text{off}}$ , which does not vanish with higher pilot power. For $|\phi|_{\text{off}} = 0.3$ , the bias can dominate at moderate SNR — one of several reasons DFT codebook estimation (Section 4.3) is preferred in practice.

Practical Constraints

•
Best-case OFF state in commercial panels: $|\phi|_{\text{off}} \leq 0.15$ .
•
Switching time between ON and OFF: $\sim 5\,\mu\text{s}$ for PIN-diode, limits pilot rate.
•
OFF-state insertion loss contributes $\sim 0.5$ – $1$ dB to the overall link budget.

ON/OFF Protocol: Sweeping the Active Element

Animation showing each RIS element being activated in turn while others remain OFF. The BS isolates the contribution of each element by subtraction. After

N+1

slots, the full cascaded channel

\mathbf{G}

is known.

The ON/OFF Switching Protocol

The Simplest Scheme That Works

Definition: The ON/OFF Subset-Switching Protocol

Theorem: Per-Element MSE of the ON/OFF Estimator

Noise in the subtraction

Per-element MSE

ON/OFF Channel Estimation

Example: ON/OFF Overhead for a 256-Element RIS

Pilot length

Per-element MSE

Effective rate

Common Mistake: ON/OFF Wastes Half the RIS Aperture

Practical Limits of the OFF State

ON/OFF Protocol: Sweeping the Active Element

Definition:
The ON/OFF Subset-Switching Protocol