Ferkans — Interactive Telecom Tutor

The Product Path Loss and Placement Intuition

The RIS two-hop link attenuates as $d_1^2 d_2^2$ (product path loss; Chapter 1). For a fixed total distance $d_1 + d_2 = D$ , the product $d_1^2 d_2^2$ is minimized when $d_1 = d_2$ (midpoint) and maximized at the endpoints. This is the opposite of one-way path loss, which is invariant in midpoint vs. endpoint under a sum constraint. The point is: placement is not intuitive from point-to-point communication reasoning; the RIS does NOT want to be at the midpoint in general — we'll see why below.

Theorem: RIS Belongs Near an Endpoint

Consider a RIS aiding a link with $d_0$ direct BS-UE distance, placed at a point with $d_1 + d_2 \geq d_0$ (the triangle inequality with equality iff the RIS is on the direct path). Under the product path-loss model, the effective SNR is $\text{SNR}_{\text{eff}} = \text{SNR}_{\text{LOS}}(d_0) + \frac{\text{SNR} \cdot N^2 |\Gamma|^2}{(4\pi/\lambda)^4 \, d_1^2 d_2^2}.$ For fixed $d_1 + d_2$ , $d_1^2 d_2^2$ is minimized at $d_1 = d_2$ (midpoint) BUT the direct LOS term is unaffected; thus the incremental RIS gain is largest at the midpoint. In the blocked-direct-LOS case, placement near either BS or UE is equivalent up to minor geometric factors.

Proof

Lagrangian

Let $d_1 + d_2 = D$ be fixed. $f(d_1) = d_1^2 (D - d_1)^2$ . $f'(d_1) = 2 d_1 (D - d_1)(D - 2d_1)$ , zero at $d_1 = D/2$ . $f(D/2) = (D/2)^4$ .

Second-order

$f''(D/2) < 0$ $\Rightarrow$ maximum of $f$ , minimum of $1/f$ . So incremental SNR gain is maximized at midpoint for fixed total path.

LOS-dominated vs. blocked

If direct LOS dominates: midpoint RIS gives the biggest boost in dB but the boost is tiny compared to LOS — RIS is wasted. If direct is blocked: the RIS is the only path. Midpoint is still optimal, but geometrically the RIS must be visible from both BS and UE. Typically this means near-BS (fewer UEs served, higher per-UE gain) or near-UE (one UE served with maximum gain). $\blacksquare$

In Practice, Place RIS Near the User

The midpoint analysis assumes a single UE and fixed geometry. In real deployments: (i) multiple UEs share the same RIS panel — placing near BS serves many UEs with lower per-UE gain, (ii) building walls constrain which surfaces are reflective, and (iii) UEs are mobile, so a panel near the UE cluster (e.g., shopping mall entrance) sees the most traffic. Empirically, near-UE placement dominates in published field trials.

Definition:
RIS Placement Region

The RIS placement region $\Omega_{\text{RIS}}$ is the set of candidate positions where a panel can be installed. Usually a discrete set: building facades, lamp posts, billboards. Each candidate has an orientation $\hat{\mathbf{n}}$ and maximum aperture $A$ . The optimization is: $\mathbf{p}^\star = \arg\max_{\mathbf{p} \in \Omega_{\text{RIS}}} \sum_{\text{UE } k} w_k \cdot G(\mathbf{p}, \mathbf{p}_k, \mathbf{p}_{\text{BS}})$ where $w_k$ is the priority weight of UE $k$ and $G$ is the geometry-dependent RIS gain formula.

Candidate Placement Search

Complexity: O(|Ω| · K)

Input: Ω_RIS = {p_1, ..., p_|Ω|}, UEs = {u_1, ..., u_K},

w_k priorities

Output: Best placement p★ and achieved utility

1. best_util = -∞

2. for each candidate p in Ω_RIS:

3. util_p = 0

4. for each UE k = 1, ..., K:

5. g_k = compute_RIS_gain(p, u_k, BS)

6. util_p += w_k · log(1 + g_k · SNR_BS)

7. end for

8. if util_p > best_util:

9. best_util = util_p

10. p★ = p

11. end if

12. end for

13. return (p★, best_util)

Example: Shopping Mall RIS Placement

A BS is at one corner of a 100m × 100m shopping mall. UEs are clustered in three zones: entrance (far corner, 141 m away, LOS blocked by walls), atrium (center, 70 m, LOS), and food court (opposite side, 100 m, LOS blocked). Three candidate RIS positions: wall near BS (5m), atrium center, and wall near food court (95m). Which is best?

Solution

Blocked-link targets

Entrance and food court need RIS (LOS blocked). Atrium has LOS — RIS gives marginal gain.

Entrance (141 m, blocked)

Placing RIS near BS (5 m from BS, 136 m to UE): $d_1^2 d_2^2 = 25 \cdot 18496 = 4.6 \times 10^5$ . Placing RIS near entrance (136 m from BS, 5 m to UE): $d_1^2 d_2^2 = 18496 \cdot 25 = 4.6 \times 10^5$ (same by symmetry). Placing RIS in atrium (70 m, 71 m): $d_1^2 d_2^2 = 4900 \cdot 5041 = 2.5 \times 10^7$ .

Verdict

Near-BS or near-UE placement delivers $\sim 50 \times$ better RIS gain than the atrium midpoint for the entrance user. The asymmetry in the product $d_1^2 d_2^2$ is the physics behind near-endpoint placement.

Multi-UE strategy

For the food-court user, a second RIS near the food court is needed. Total: two panels (one near BS for broad coverage, one near food court for the blocked user), total cost $\sim 2 \times \$200 = \$400$ — much less than an indoor small cell ( $\sim \$2000$ ). $\blacksquare$

RIS Position vs. SNR Gain

Sweep the RIS position along a line between BS and UE. Plot the achieved SNR gain. The product $d_1^2 d_2^2$ is minimized at midpoint but real-world placement depends on blockage and aperture orientation.

Parameters

BS-UE distance (m)100

RIS elements

N

256

Direct BS-UE path blocked

Carrier (GHz)28

⚠️Engineering Note

Facade Selection Heuristics

In urban deployment, a few simple heuristics cover 80% of decisions:

Visibility: The RIS must have LOS to both BS and a useful UE cluster. Check with a 3D building model.
Orientation: Prefer facades with normal pointing toward bisector of BS and UE. Near-broadside reduces $\eta_{\text{ang}}$ loss (Chapter 16).
Access: Lamp posts and bus stops are commercially accessible. Private building walls require leases.
Power: RIS needs only $\sim 5$ -10 W for control electronics — easy from existing street lighting lines.

Common Mistake: Don't Naively Deploy RIS at the Geometric Midpoint

Mistake:

Placing a RIS at the geometric midpoint between BS and typical UE based on "shortest total path."

Correction:

For one UE, this is correct (product minimized). For multiple UEs, the optimal placement is near the priority-weighted centroid — usually near the BS or near a UE cluster, NOT midway. Additionally, the midpoint often has no buildable surface; real deployments are constrained by where installation is possible. Use the candidate-search algorithm above.

Multi-RIS Placement Is NP-Hard

Placing multiple RIS panels to maximize aggregate utility is the facility location problem — NP-hard in general. Two practical approaches: (i) greedy, add one RIS at a time maximizing marginal utility; (ii) submodular, if utility is submodular (it is approximately), greedy gives a $(1 - 1/e) \approx 63\%$ optimality guarantee. For dense urban deployments with 100+ candidates and 10+ panels, greedy submodular is the practical choice.

Optimal RIS Placement