Ferkans — Interactive Telecom Tutor

Two Hops, Two Path Losses

A direct link suffers one instance of free-space path loss: the energy spreads over a sphere of radius $d_0$ . A RIS link suffers two: from the BS to the RIS over distance $d_1$ , then from the RIS to the UE over distance $d_2$ . Crucially, the two losses multiply, not add. This is the notorious product path loss and it is the single biggest reason RIS deployment is hard. Without understanding where the $d_1^2 d_2^2$ factor comes from and why it is so much worse than $d_0^2$ , one cannot make sensible deployment decisions.

Definition:
Free-Space Product Path Loss for a Passive RIS

Consider a BS–RIS–UE link in free space at carrier wavelength $\lambda$ , with BS-to-RIS distance $d_1$ and RIS-to-UE distance $d_2$ , both much larger than the RIS aperture (far-field regime). The RIS has $N$ isotropic elements with area $A_e$ each. Let $G_t, G_r$ denote the transmit and receive antenna gains. Then, under coherent phase alignment, the received power through the RIS path is

$P_r = P_t \, G_t\, G_r\, \frac{N^2 A_e^2}{(4\pi)^2\, d_1^2\, d_2^2}.$

This is the product path loss law. The hallmark is the denominator $d_1^2 d_2^2$ , not $d_1 d_2$ and certainly not $(d_1 + d_2)^2$ .

Compared with a direct free-space link of total distance $d_0 = d_1 + d_2$ , which gives $P_r^{\text{direct}} = P_t G_t G_r \lambda^{2} / (4\pi d_0)^2$ , the RIS-link-over-direct ratio contains the crucial factor $(d_1 + d_2)^2 / (d_1 d_2)^2$ . This is minimized when $d_1 = d_2$ — perfectly in between — and is actually smaller (worse!) than the direct path for nearly all geometries.

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Theorem: Product Path Loss Is Minimized at the Endpoints

Fix the total distance $d_0 = d_1 + d_2$ and consider placing the RIS along the BS–UE line. The product $d_1 d_2$ is maximized at the midpoint $d_1 = d_2 = d_0/2$ and minimized as $d_1 \to 0$ or $d_1 \to d_0$ . Since received power through the RIS is proportional to $1/(d_1^2 d_2^2)$ , the optimal RIS placement is as close as possible to either the BS or the UE.

The product $d_1^2 d_2^2$ with a fixed total $d_1 + d_2$ is maximized when $d_1 = d_2$ . But what we want to minimize is the full path-loss ratio relative to a direct link, which pushes the RIS toward one endpoint or the other.

Proof

Constrained maximization of $d_1 d_2$

With $d_2 = d_0 - d_1$ , the product is $f(d_1) = d_1(d_0 - d_1) = d_0 d_1 - d_1^2$ . Differentiating and setting to zero: $f'(d_1) = d_0 - 2d_1 = 0 \implies d_1 = d_0/2$ . $f''(d_1) = -2 < 0$ , confirming a maximum.

Behaviour at endpoints

$f(0) = f(d_0) = 0$ . The function is a downward parabola, zero at both ends and maximum in the middle.

Consequence for received power

Since received power $\propto 1/f(d_1)^2$ , it blows up at the endpoints and is smallest at the midpoint. Formally, the ratio of RIS to direct power is $\rho(d_1) = \text{const} \cdot \frac{d_0^2}{d_1^2(d_0 - d_1)^2},$ which is minimized at $d_1 = d_0/2$ (worst ratio) and diverges at $d_1 \in \{0, d_0\}$ (best ratio). In deployment: put the RIS on a wall adjacent to the BS or on the UE's building facade. $\blacksquare$

Key Takeaway

A RIS in the middle of nowhere is a RIS in the worst place. To counteract the product path loss, deploy the RIS as close as possible to either the BS or the UE cluster. In practice, walls adjacent to a BS, building facades in a dense UE area, or lamp-post-mounted panels near street-level users all dominate the midway-between placement.

Received Power vs. RIS Placement

Slide the RIS between the BS ( $d_1/d_0 = 0$ ) and the UE ( $d_1/d_0 = 1$ ). The plot shows the ratio of RIS-path power to direct-path power in dB. Notice the U-shape: the worst location is exactly in the middle. Increase $N$ to see when the RIS path finally overtakes the direct path on its own.

Parameters

RIS elements

N

256

Coherent RIS with $N$ elements.

BS–UE distance

d_0

(m)100

Total link distance.

Carrier

f_c

(GHz)3.5

Frequency sets the wavelength $\ntn{wl}$.

Element area /

\lambda^2

0.25

Effective area per RIS element in units of $\lambda^2$.

The Far-Field Assumption

The product-path-loss formula above is a far-field result: the RIS is modelled as a point scatterer at distances $d_1, d_2$ . For very large RIS and/or short link distances, the UE can fall inside the RIS's Fraunhofer region (near-field), where the far-field $1/d^2$ scaling breaks down and the path loss becomes closer to $1/d$ per hop. Chapter 3 develops the near-field extension. For design at hundreds of meters and moderate $N$ , the far-field formula is an adequate starting point.

Historical Note: The $1/d^2$ Confusion

2019–2021

The early RIS literature (2019–2020) produced a small academic scandal. Several widely-cited papers wrote the RIS-link path loss as $\propto 1/(d_1 d_2)^2$ , while others wrote $1/(d_1 + d_2)^2$ , and yet others $1/d_1^\alpha d_2^\alpha$ with fitted exponents. Who was right? The resolution, clarified by Ellingson (2021) and independently by Tang et al. (2021), is that both limiting behaviours are correct, at different RIS sizes relative to the Fresnel zone. A small RIS (much smaller than the Fresnel zone) behaves as a point scatterer with $1/d_1^2 d_2^2$ (the "RCS-like" regime, full product path loss). A large RIS (spanning the Fresnel zone or more) behaves like an anomalous mirror with $1/(d_1 + d_2)^2$ — the same as a flat mirror — because the multiple reflection points from a large surface constructively combine to compensate one of the distance factors. This book works in the "small-RIS" regime (point-scatterer, far-field) except where explicitly noted; the large-RIS anomalous-mirror regime is explored in Chapter 3 and Chapter 11 (array-fed RIS).

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Example: Product Path Loss at 3.5 GHz

At $f_c = 3.5\text{ GHz}$ ( $\lambda = 8.57\text{ cm}$ ), a BS transmits $P_t = 30\text{ dBm}$ with $G_t G_r = 20\text{ dBi}$ combined antenna gain. A $256$ -element RIS with $\lambda^{2}/4$ per element sits at $d_1 = 30\text{ m}$ from the BS and $d_2 = 30\text{ m}$ from the UE. What is the received power? Compare to a direct link at the same total distance $d_0 = 60\text{ m}$ .

Solution

RIS link path loss

$P_r^{\text{RIS}} = P_t G_t G_r \frac{N^2 A_e^2}{(4\pi)^2 d_1^2 d_2^2}$ . Plugging numbers: $N^2 A_e^2 = 256^2 \cdot (\lambda^{2}/4)^2 = 65\,536 \cdot (0.0857^2/4)^2 = 65\,536 \cdot 3.37 \cdot 10^{-6} = 0.221\text{ m}^4$ . $(4\pi)^2 d_1^2 d_2^2 = 158 \cdot 900 \cdot 900 = 1.28 \cdot 10^8 \text{ m}^4$ . Hence $P_r^{\text{RIS}} / (P_t G_t G_r) = 1.73 \cdot 10^{-9}$ , or $-87.6\text{ dB}$ .

Direct link path loss

$P_r^{\text{direct}} = P_t G_t G_r \lambda^{2} / (4\pi d_0)^2 = P_t G_t G_r \cdot (0.0857)^2 / (4\pi \cdot 60)^2$ . $= P_t G_t G_r \cdot 7.35 \cdot 10^{-3} / 5.68 \cdot 10^5 = P_t G_t G_r \cdot 1.29 \cdot 10^{-8}$ , or $-78.9\text{ dB}$ .

Compare

The direct link is about $9\text{ dB}$ better than the symmetric RIS path in free space. If the direct link is unblocked, the RIS helps only when it is close to one endpoint or when $N$ is large enough that $N^2$ dominates. At midpoint placement, the RIS is most useful as a redundancy path, not a primary path — unless blockage kills the direct link entirely, which is often the case in urban mmWave and the raison d'être of most RIS deployments.

Common Mistake: Linear-in- $d$ vs. Quadratic-in- $d$ Confusion

Mistake:

"Since the two hops are independent, the total loss is $d_1 + d_2 = d_0$ , so the RIS link isn't worse than the direct link."

Correction:

Path loss in free space is $\propto d^2$ (inverse square law). The two-hop system chains two $1/d^2$ factors, giving $1/d_1^2 d_2^2$ — not $1/(d_1 + d_2)^2$ . To see the gap, fix $d_0 = 100\text{ m}$ and compare: direct $1/d_0^2 = 10^{-4}$ ; symmetric RIS $1/d_1^2 d_2^2 = 1/2500^2 = 1.6 \cdot 10^{-7}$ . The RIS link is three orders of magnitude worse in free-space path loss terms, which is why the $N^2$ coherent gain is necessary to be competitive.

The Product Path Loss