Ferkans — Interactive Telecom Tutor

The Blockage Problem at mmWave

At mmWave (28 GHz and above), a human body creates a 20 dB shadow; a brick wall, 30+ dB. The coverage problem is not about path loss but blockage probability: the chance that no LOS path exists between BS and UE. RIS solves this by providing a programmable non-line-of-sight (NLOS) reflector that re-routes the signal around the blockage. This is the single most compelling commercial motivation for RIS at mmWave.

Definition:
Coverage Probability with RIS

Let $\text{SNR}(\mathbf{u})$ be the received SNR at UE position $\mathbf{u}$ with RIS enabled. The coverage probability for target threshold $\tau$ is $p_{\text{cov}}(\tau) = \mathbb{P}[\text{SNR}(\mathbf{u}) \geq \tau]$ where the probability is over UE location (uniform in $\Omega$ ) and small-scale fading. With RIS enabled, both the direct (if unblocked) and RIS-reflected path contribute.

Theorem: Coverage Improvement from RIS Deployment

Let $p_{\text{blk}}$ be the probability that the direct BS-UE path is blocked. Let $p_{\text{RIS-cov}}$ be the probability that a deployed RIS at position $\mathbf{p}_{\text{RIS}}$ provides SNR $\geq \tau$ via the reflected path (not blocked and RIS gain sufficient). Then the total coverage probability is $p_{\text{cov}}^{\text{total}} = p_{\text{LOS-cov}}(1 - p_{\text{blk}}) + p_{\text{RIS-cov}} p_{\text{blk}}$ assuming the RIS path and the direct path are not both simultaneously blocked.

Proof

Disjoint cases

Case 1 (probability $1 - p_{\text{blk}}$ ): direct LOS available — UE covered iff SNR via LOS exceeds $\tau$ .

Blocked case

Case 2 (probability $p_{\text{blk}}$ ): direct blocked — UE covered iff RIS path SNR exceeds $\tau$ .

Total probability

By law of total probability: $p_{\text{cov}}^{\text{total}} = (1 - p_{\text{blk}}) p_{\text{LOS-cov}} + p_{\text{blk}} p_{\text{RIS-cov}}$ . $\blacksquare$

Practical implication

If the direct path is rarely blocked ( $p_{\text{blk}}$ small), RIS adds little. If blockage is common ( $p_{\text{blk}} = 0.5$ ), RIS can double the coverage from $0.5 \to 1.0$ .

Example: 28 GHz Urban Blockage Mitigation

An urban 28 GHz mmWave cell has $p_{\text{blk}} = 0.6$ (common in dense urban). Without RIS, coverage probability is 40% (60% of UEs blocked). A RIS deployed on a corner building covers 80% of the blocked region with sufficient SNR. Compute the new coverage.

Solution

Baseline

$p_{\text{cov}}^{\text{no-RIS}} = 0.4$ (40% coverage).

With RIS

Of the 60% blocked UEs, 80% are now covered by RIS. Total: $0.4 + 0.6 \cdot 0.8 = 0.4 + 0.48 = 0.88$ .

Delta

RIS boosts coverage from 40% to 88% — a $2.2 \times$ improvement. Users notice the difference immediately. This is the commercial pitch for RIS at mmWave. $\blacksquare$

Coverage Heatmap with RIS

Visualize the coverage heatmap in a city block. BS at one corner, RIS on a nearby building facade. Colored areas show SNR above threshold (green) vs. below (red). Move RIS and BS positions; count covered fraction.

Parameters

City block size (m)150

RIS elements256

Building blockage density0.5

SNR threshold (dB)5

Definition:
RIS Deployment Density

The RIS deployment density $\lambda_{\text{RIS}}$ (panels/km²) characterizes how many RIS panels are installed per unit area. Typical urban 6G targets are $5$ - $50$ panels/km² — compared to BS density $\sim 5$ - $10$ /km². The density is the output of the deployment optimization; the input is the per-panel cost and desired coverage.

Theorem: Coverage vs. RIS Density (PPP Model)

Under a Poisson point process (PPP) model for UEs and RIS placement with density $\lambda_{\text{RIS}}$ , the probability that at least one RIS panel is within "useful range" $R$ of a given UE is $p_{\text{cov}}^{\text{RIS}}(\lambda) = 1 - \exp(-\lambda_{\text{RIS}} \pi R^2).$ The "useful range" $R$ depends on $N$ , the RIS gain efficiency, and blockage statistics: typically $R \in [20, 100]$ m for urban mmWave deployments.

Proof

PPP void probability

In a PPP with intensity $\lambda_{\text{RIS}}$ , the number of panels in an area $A$ is Poisson-distributed with mean $\lambda_{\text{RIS}} A$ . Probability of zero panels in area $A = \pi R^2$ is $e^{-\lambda \pi R^2}$ .

At least one

Probability at least one is $1 - e^{-\lambda \pi R^2}$ . $\blacksquare$

Target 95% coverage

For $p_{\text{cov}} = 0.95$ : $\lambda \pi R^2 = 3$ . If $R = 50$ m: $\lambda = 3/(\pi \cdot 2500) = 3.8 \times 10^{-4}$ /m² $\approx 380$ panels/km² — dense. If $R = 100$ m: $\lambda \approx 95$ panels/km².

⚠️Engineering Note

Practical Deployment Densities

Current operator deployment plans (2024 trials) suggest $\lambda_{\text{RIS}} \sim 10$ - $50$ /km² for urban 28 GHz cells. This implies useful range $R \sim 80$ - $180$ m per panel — achievable with $N = 256$ - $1024$ element panels. For indoor enterprise deployments (shopping malls, stadiums, airports), the density is $\sim 50$ - $200$ panels/km² (much higher), reflecting the premium coverage requirement.

Multi-RIS Diversity

With $M$ RIS panels deployed, even if one is blocked or misaligned, others can fill the coverage hole. This is spatial diversity: the probability that ALL panels are blocked scales as $p_{\text{blk}}^M$ . For $p_{\text{blk}} = 0.5$ and $M = 4$ panels per cell, the all-blocked probability is $0.0625$ — less than 7%. This argues for distributed, redundant RIS deployment, not one large central panel.

Coverage Analysis and Blockage Mitigation