Ferkans — Interactive Telecom Tutor

The Two-Hop Signal Path

An RIS does not transmit; it only reflects. So the signal reaching the user arrives via two hops: a first hop from the base station (BS) to the RIS, and a second hop from the RIS to the user equipment (UE). This two-hop geometry is not a detail — it is the reason RIS channels behave differently from conventional single-hop links, and it is the reason every optimization problem in this book takes the shape it does.

Definition:
Canonical RIS-Aided System Model

Consider a downlink narrowband system with:

a base station equipped with $N_t$ antennas,
an RIS with $N$ passive reflecting elements,
a single-antenna user equipment (the extension to $N_r$ receive antennas is straightforward and treated in Chapter 3).

Define the three physical channels:

$\mathbf{H}_1 \in \mathbb{C}^{N \times N_t}$ : BS-to-RIS channel (the incident wavefront at the RIS).
$\mathbf{h}_2 \in \mathbb{C}^{N}$ : RIS-to-UE channel (one column, since the UE has a single antenna).
$\mathbf{h}_d \in \mathbb{C}^{N_t}$ : direct BS-to-UE channel (often blocked in RIS deployments).

Let $\mathbf{v} \in \mathbb{C}^{N_t}$ , $\|\mathbf{v}\| = 1$ , be the BS beamforming vector, let $\boldsymbol{\Phi} = \text{diag}(e^{j\theta_1}, \ldots, e^{j\theta_N})$ be the RIS phase-shift matrix, and let $s \in \mathbb{C}$ be the data symbol with $\mathbb{E}[|s|^2] = P_t$ . The received signal at the UE is

$y = \underbrace{\big(\mathbf{h}_d^H + \mathbf{h}_2^H \boldsymbol{\Phi}\, \mathbf{H}_1\big)}_{\mathbf{h}_{\text{eff}}^H} \mathbf{v}\, s + w, \qquad w \sim \mathcal{CN}(0, \sigma^2).$

The bracketed $1 \times N_t$ row vector

$\boxed{\;\mathbf{h}_{\text{eff}}^H \;=\; \mathbf{h}_d^H \;+\; \mathbf{h}_2^H \boldsymbol{\Phi}\, \mathbf{H}_1\;}$

is the effective end-to-end channel seen by the BS beamformer.

Notice the crucial structural fact: the RIS phase shifts enter $\mathbf{h}_{\text{eff}}$ linearly via the diagonal matrix $\boldsymbol{\Phi}$ , but $\boldsymbol{\Phi}$ is constrained to the non-convex unit-modulus torus. The system is linear in $\mathbf{v}$ , linear in $\boldsymbol{\Phi}$ , but the joint feasibility set of $(\mathbf{v}, \boldsymbol{\Phi})$ is not convex. This is the root of nearly every difficulty and nearly every algorithm in this book.

,

Theorem: Diagonal Factorization of the Cascaded Channel

For any $\mathbf{h}_2 \in \mathbb{C}^N$ and any diagonal $\boldsymbol{\Phi} = \text{diag}(\phi_1, \ldots, \phi_N)$ ,

$\mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1 \;=\; \boldsymbol{\phi}^T \big( \mathbf{h}_2^* \odot \mathbf{H}_1 \big) \;=\; \boldsymbol{\phi}^T \text{diag}(\mathbf{h}_2^*)\, \mathbf{H}_1,$

where $\boldsymbol{\phi} = [\phi_1, \ldots, \phi_N]^T$ and $\odot$ denotes row-wise element-wise multiplication (i.e., $\mathbf{h}_2^*$ is broadcast against the rows of $\mathbf{H}_1$ ).

This identity is a simple but pivotal algebraic rearrangement: it turns the expression $\mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1$ , which looks trilinear in $(\mathbf{h}_2, \boldsymbol{\phi}, \mathbf{H}_1)$ , into something linear in $\boldsymbol{\phi}$ . Every RIS optimization algorithm exploits this.

Proof

Write out the triple product

$\mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1 = \sum_{n=1}^N (\mathbf{h}_2)_n^* \phi_n\, (\mathbf{H}_1)_{n,:}$ , the sum of the rows of $\mathbf{H}_1$ weighted by $(\mathbf{h}_2)_n^* \phi_n$ .

Factor out the phase vector

Collect $\phi_n$ to the front: $= \sum_n \phi_n \big[ (\mathbf{h}_2)_n^* (\mathbf{H}_1)_{n,:} \big] = \boldsymbol{\phi}^T \mathbf{M}$ , where $\mathbf{M} \in \mathbb{C}^{N \times N_t}$ has rows $\mathbf{M}_{n,:} = (\mathbf{h}_2)_n^* (\mathbf{H}_1)_{n,:}$ . This is exactly $\mathbf{M} = \text{diag}(\mathbf{h}_2^*)\,\mathbf{H}_1$ , which also equals $\mathbf{h}_2^* \odot \mathbf{H}_1$ under row-wise broadcasting. $\blacksquare$

Key Takeaway

The RIS channel is linear in $\boldsymbol{\phi}$ — even though the optimization is non-convex. The difficulty of RIS beamforming does not come from the objective (which is a simple quadratic form in $\boldsymbol{\phi}$ ) but from the feasibility set $\{|\phi_n| = 1\}$ . This separation of "easy objective" and "hard constraint" is why SDR, manifold optimization, and alternating methods all apply.

Example: Received SNR for a Single-User MISO-RIS Link

A BS with $N_t$ antennas transmits $s$ with beamformer $\mathbf{v}$ , $\|\mathbf{v}\| = 1$ , transmit power $P_t$ . The direct path $\mathbf{h}_d$ is blocked ( $\mathbf{h}_d = \mathbf{0}$ ). The RIS has $N$ elements with phase-shift matrix $\boldsymbol{\Phi}$ . Noise is $\mathcal{CN}(0, \sigma^2)$ . Derive an expression for the received SNR.

Solution

Write the received signal

With $\mathbf{h}_d = \mathbf{0}$ , $y = \mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1 \mathbf{v}\, s + w$ . The signal component has power $P_t \big|\mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1 \mathbf{v}\big|^2$ and the noise has power $\sigma^2$ .

Compute the SNR

$\text{SNR}(\mathbf{v}, \boldsymbol{\Phi}) = \frac{P_t}{\sigma^2} \big|\mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1 \mathbf{v}\big|^2.$ $

Linearize in $\boldsymbol{\phi}$ via the product identity

Using the factorization from the previous theorem, $\mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1 \mathbf{v} = \boldsymbol{\phi}^T (\mathbf{h}_2^* \odot \mathbf{H}_1 \mathbf{v}) = \boldsymbol{\phi}^T \mathbf{a}$ , where $\mathbf{a} = \text{diag}(\mathbf{h}_2^*)\mathbf{H}_1 \mathbf{v} \in \mathbb{C}^N$ is a fixed vector once $\mathbf{v}$ is chosen. Hence

$\text{SNR}(\mathbf{v}, \boldsymbol{\Phi}) = \frac{P_t}{\sigma^2} \big|\boldsymbol{\phi}^T \mathbf{a}\big|^2.$

The passive-beamforming problem is now: maximize the squared inner product $|\boldsymbol{\phi}^T \mathbf{a}|^2$ subject to $|\phi_n| = 1$ . This is the RIS analog of matched filtering — and it has the beautiful closed-form solution $\phi_n^\star = e^{-j \arg(a_n)}$ , analyzed in Section 1.3.

Common Mistake: The $\boldsymbol{\phi}^T$ vs $\boldsymbol{\phi}^H$ Trap

Mistake:

A student writes $\mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1 = \boldsymbol{\phi}^H \mathbf{M}$ and proceeds to optimize in $\boldsymbol{\phi}^H$ .

Correction:

The factorization uses $\boldsymbol{\phi}^T$ , not $\boldsymbol{\phi}^H$ . This is because $\boldsymbol{\Phi}$ is a diagonal matrix of the non-conjugated phase shifts, so the sum $\sum_n (\mathbf{h}_2)_n^* \phi_n \cdots$ pulls out $\phi_n$ (not $\phi_n^*$ ). Getting the transpose wrong flips the sign of the phase-compensation term and sends the optimization in the opposite direction. Always double-check: is the symbol you are extracting conjugated in the original expression, or not?

Effective Channel Gain vs. RIS Phase Alignment

Visualize how the magnitude $|\boldsymbol{\phi}^T \mathbf{a}|^2$ varies as the RIS phases progressively align with the coefficient vector $\mathbf{a}$ . At the "aligned" end ( $\phi_n^\star = e^{-j\arg(a_n)}$ ) the magnitude equals $\|\mathbf{a}\|_1^2$ , the peak achievable by any unit-modulus $\boldsymbol{\phi}$ .

Parameters

Number of RIS elements

N

32

More elements widen the gap between random and aligned.

Fraction of optimal phase applied0.5

$0$ = random phases, $1$ = matched-filter phases.

The Cascaded Channel and the System Model