# Aperture Selection for CAP Arrays (CAPAs)

Chongjun Ouyang⋆‡, Yuanwei Liu⋆, and Xingqi Zhang† ⋆Queen Mary University of
London, ‡University College Dublin, †University of Alberta
Email: ⋆{c.ouyang<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

The concept of aperture selection is proposed for continuous aperture array
(CAPA)-based communications. The achieved performance is analyzed in an uplink
scenario by considering both line-of-sight (LoS) and non-line-of-sight (NLoS)
scenarios. In the LoS scenario, the optimal selection strategy is demonstrated
to follow the nearest neighbor criterion, and the resulting signal-to-noise
ratio (SNR) is analyzed. In the NLoS scenario, the achieved outage probability
along with the diversity order is revealed. Numerical results are provided to
demonstrate that aperture selection effectively maintains satisfactory
performance by leveraging selection diversity while simultaneously reducing
the implementation complexity of CAPAs.

##  I Introduction

Multiple-input multiple-output (MIMO) technology has evolved into a mature
technology that enhances the spectral efficiency of wireless channels. The
spatial degrees of freedom (DoFs) of a MIMO system are inherently limited by
the number of antennas it incorporates. To expand the available DoFs,
integrating more antennas into a confined space emerges as an effective
approach. It is envisaged that the ultimate evolution of existing MIMO systems
will manifest as a spatially-continuous electromagnetic (EM) aperture that is
characterized by an uncountable infinity of antennas separated by
infinitesimal distances [1], i.e., _continuous aperture array (CAPA)_.
Research in this domain is still nascent; see [2] and the references therein
for more details.

In contrast to traditional arrays, which rely on spatially discrete (SPD)
models, CAPA-based communications are rooted in continuous EM fields [3]. This
transition from discrete signaling in SPD arrays to continuous signaling in
CAPAs introduces additional complexities in both _hardware implementation and
signal processing_. Furthermore, these complexities are directly proportional
to the aperture size of the CAPA [2].

To achieve a better trade-off between leveraging a CAPA for enhanced
communication performance and managing the associated complexity, this paper
introduces the concept of _aperture selection_ , wherein only a portion of the
continuous aperture is activated for communications. By reducing the aperture
size, aperture selection enables the entire CAPA to _benefit from increased
spatial DoFs_ while simultaneously _mitigating implementation complexities_.
The main contributions of this article are summarized as follows: i) We
establish the analytical framework of aperture selection for CAPAs; ii)
Building upon the established framework, we analyze the receive signal-to-
noise ratio (SNR) achieved by aperture selection in an uplink channel under
both line-of-sight (LoS) and non-line-of-sight (NLoS) conditions; iii) For LoS
channels, we demonstrate that the optimal selection strategy adheres to the
_nearest neighbor criterion_ and derive a closed-form expression for the
corresponding SNR; iv)) In the case of NLoS channels, we propose a method
based on discrete aperture segmentation to reduce selection complexity and
characterize the resulting outage probability (OP) and diversity order; v) We
demonstrate through numerical results for both scenarios that aperture
selection effectively enhances system SNR by harnessing selection diversity.

##  II System Model

We investigate an uplink channel where a single-antenna user transmits signals
to a base station (BS) equipped with a CAPA, as shown in Figure 1. The array
is placed on the $x$-$z$ plane and centered at the origin, with physical
dimensions $L_{x}$ and $L_{z}$ along the $x$\- and $z$-axes, respectively. Let
$r$ denote the propagation distance from the center of the antenna array to
the user’s location, and $\phi\in[0,\pi]$ and $\theta\in[0,\pi]$ denote the
associated azimuth and elevation angles, respectively. Therefore, the user’s
center location can be written as
${\mathbf{s}}_{\mathsf{u}}=[r\Phi,r\Psi,r\Theta]^{\mathsf{T}}$, where
$\Phi\triangleq\cos{\phi}\sin{\theta}$, $\Psi\triangleq\sin\phi\sin\theta$,
and $\Theta\triangleq\cos{\theta}$. The user is further assumed to employ a
hypothetical isotropic antenna for signal transmission, whose transmit
aperture is denoted as ${\mathcal{A}}_{\mathsf{S}}$ with
$\lvert{\mathcal{A}}_{\mathsf{S}}\rvert=\frac{\lambda^{2}}{4\pi}$, where
$\lambda$ denotes the wavelength. The BS aperture is denoted as
${\mathcal{A}}_{\mathsf{R}}=\\{(x,0,z)|x\in[-\frac{L_{x}}{2},\frac{L_{x}}{2}],z\in[-\frac{L_{z}}{2},\frac{L_{z}}{2}]\\}$.
Throughout this paper, we assume that the aperture size of
${\mathcal{A}}_{\mathsf{R}}$ significantly exceeds that of
${\mathcal{A}}_{\mathsf{S}}$, i.e.,
$\lvert{\mathcal{A}}_{\mathsf{R}}\rvert\gg\lvert{\mathcal{A}}_{\mathsf{S}}\rvert$.

Figure 1: Illustration of a CAPA.

###  II-A Signal Model

Let us define ${\mathsf{J}}({\mathbf{s}})\in{\mathbbmss{C}}$ as the continuous
distribution of source currents generated by the user to convey information,
where ${\mathbf{s}}\in{\mathcal{A}}_{\mathsf{S}}$. The observed electric field
${\mathsf{Y}}(\mathbf{r})\in{\mathbbmss{C}}$ at point
$\mathbf{r}\in{\mathcal{A}}_{\mathsf{R}}$ is the sum of the information-
carrying electric field ${\mathsf{E}}(\mathbf{r})\in{\mathbbmss{C}}$ and a
random noise field ${\mathsf{N}}(\mathbf{r})\in{\mathbbmss{C}}$, i.e.,

$\displaystyle{\mathsf{Y}}(\mathbf{r})$
$\displaystyle={\mathsf{E}}(\mathbf{r})+{\mathsf{N}}(\mathbf{r})=\int_{{\mathcal{A}}_{\mathsf{S}}}{\mathsf{h}}(\mathbf{r},{\mathbf{s}})s{\mathsf{J}}({\mathbf{s}}){\rm{d}}{\mathbf{s}}+{\mathsf{N}}(\mathbf{r}),$
(1)

where $s\in{\mathbbmss{C}}$ represents the normalized symbol,
${\mathsf{N}}(\mathbf{r})$ accounts for the thermal noise, and
${\mathsf{h}}(\mathbf{r},{\mathbf{s}})\in{\mathbbmss{C}}$ denotes the spatial
channel response between $\mathbf{s}$ and $\mathbf{r}$. The noise field is
modeled as a zero-mean complex Gaussian random process satisfying
${\mathbbmss{E}}\\{{\mathsf{N}}(\mathbf{r}_{1}){\mathsf{N}}^{\mathsf{H}}(\mathbf{r}_{2})\\}={\sigma}^{2}\delta(\mathbf{r}_{1}-\mathbf{r}_{2})$,
where $\delta(\cdot)$ represents the Dirac delta function.

Leveraging the fact that
$\lvert{\mathcal{A}}_{\mathsf{R}}\rvert\gg\lvert{\mathcal{A}}_{\mathsf{S}}\rvert$
yields

$\displaystyle{\mathsf{Y}}(\mathbf{r})\approx
s{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}}){\mathsf{J}}({\mathbf{s}}_{\mathsf{u}})\lvert{\mathcal{A}}_{\mathsf{S}}\rvert+{\mathsf{N}}(\mathbf{r}).$
(2)

Subsequently, the CAPA-based BS recover $s$ from ${\mathsf{Y}}(\mathbf{r})$
using a linear combiner
${\mathsf{v}}(\mathbf{r})={\mathsf{h}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}})$
along with maximum-likelihood decoding, which leads to

$\displaystyle\int_{{\mathcal{S}}_{\mathsf{R}}}{\mathsf{v}}(\mathbf{r}){\mathsf{Y}}(\mathbf{r}){\rm{d}}{\mathbf{r}}=s{\mathsf{a}}_{\mathsf{R}}{\mathsf{J}}({\mathbf{s}}_{\mathsf{u}})\lvert{\mathcal{A}}_{\mathsf{S}}\rvert+\int_{{\mathcal{S}}_{\mathsf{R}}}{\mathsf{h}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}}){\mathsf{N}}(\mathbf{r}){\rm{d}}{\mathbf{r}},$
(3)

where
${\mathsf{a}}_{\mathsf{R}}\triangleq\int_{{\mathcal{S}}_{\mathsf{R}}}\lvert{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}})\rvert^{2}{\rm{d}}{\mathbf{r}}>0$,
and ${\mathcal{S}}_{\mathsf{R}}\subseteq{\mathcal{A}}_{\mathsf{R}}$ represent
the effective surface region within the BS aperture that can capture the
radiated power. Since ${\mathsf{N}}(\mathbf{r})$ is a Gaussian random field,
$\int_{{\mathcal{S}}_{\mathsf{R}}}{\mathsf{h}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}}){\mathsf{N}}(\mathbf{r}){\rm{d}}{\mathbf{r}}$
is complex Gaussian distributed, whose mean and variance is calculated as
follows.

######  Lemma 1.

Under the assumption that ${\mathsf{N}}(\mathbf{r})$ is a zero-mean complex
Gaussian process with
${\mathbbmss{E}}\\{{\mathsf{N}}(\mathbf{r}_{1}){\mathsf{N}}^{\mathsf{H}}(\mathbf{r}_{2})\\}={\sigma}^{2}\delta(\mathbf{r}_{1}-\mathbf{r}_{2})$,
we obtain
$\int_{{\mathcal{S}}_{\mathsf{R}}}{\mathsf{h}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}}){\mathsf{N}}(\mathbf{r}){\rm{d}}{\mathbf{r}}\sim{\mathcal{CN}}(0,\sigma^{2}{\mathsf{a}}_{\mathsf{R}})$.

###### Proof:

Please refer to Appendix -A for more details. ∎

The SNR for decoding $s$ is given by

$\displaystyle\gamma=\frac{{\mathsf{a}}_{\mathsf{R}}^{2}\lvert{\mathsf{J}}({\mathbf{s}}_{\mathsf{u}})\rvert^{2}\lvert{\mathcal{A}}_{\mathsf{S}}\rvert^{2}}{\sigma^{2}{\mathsf{a}}_{\mathsf{R}}}=\frac{{\mathsf{a}}_{\mathsf{R}}\lvert{\mathsf{J}}({\mathbf{s}}_{\mathsf{u}})\rvert^{2}\lvert{\mathcal{A}}_{\mathsf{S}}\rvert^{2}}{{\sigma}^{2}},$
(4)

which is related with the effective aperture ${\mathcal{S}}_{\mathsf{R}}$.

###  II-B Aperture Selection

It is evident that $\gamma$ is maximized when
${\mathcal{S}}_{\mathsf{R}}={\mathcal{A}}_{\mathsf{R}}$, which, yet, leads to
_high complexity_ in terms of _hardware implementation and signal processing_
, especially when ${\mathcal{A}}_{\mathsf{R}}$ has a large size. To alleviate
this issue, we propose the concept of _aperture selection_ : selecting an
aperture within ${\mathcal{A}}_{\mathsf{R}}$ to activate and using it for
receiving. An example is illustrated in Figure 1, where
${\mathcal{S}}_{\mathsf{R}}$ is centered at $O^{\prime}=(r_{x},0,r_{z})$ with
dimensions $A_{x}\times A_{z}$. The corresponding selection criteria is
formulated as follows:

$\displaystyle{\mathcal{S}}_{\mathsf{R}}^{\star}=\operatorname*{argmax}_{{\mathcal{S}}_{\mathsf{R}}\subseteq{\mathcal{A}}_{\mathsf{R}}}\gamma=\operatorname*{argmax}_{{\mathcal{S}}_{\mathsf{R}}\subseteq{\mathcal{A}}_{\mathsf{R}}}\int_{{\mathcal{S}}_{\mathsf{R}}}\lvert{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}})\rvert^{2}{\rm{d}}{\mathbf{r}},$
(5)

which is an NP-hard problem.

###  II-C Channel Model

We use two typical propagation models to describe
${\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}})$: i) the LoS model and ii)
the NLoS model.

####  II-C1 Line-of-Sight Channel

We commence our discussion with LoS channels which can facilitate theoretical
investigations into fundamental performance limits. In this case, we model
${\mathsf{h}}(\mathbf{r},{\mathbf{s}})$ as follows:

$\displaystyle{\mathsf{h}}(\mathbf{r},{\mathbf{s}})={\mathsf{g}}(\mathbf{r},{\mathbf{s}})\sqrt{\frac{{\mathbf{e}}_{y}^{\mathsf{T}}({\mathbf{s}}-{\mathbf{r}})}{\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}}\triangleq{\mathsf{g}}_{\mathsf{LoS}}(\mathbf{r},{\mathbf{s}}),$
(6)

where
${\mathsf{g}}(\mathbf{r},{\mathbf{s}})\triangleq\frac{{\rm{j}}k_{0}\eta{\rm{e}}^{-{\rm{j}}k_{0}\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}}{4\pi\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}$,
$\eta=120\pi$ is the impedance of free space, $k_{0}=\frac{2\pi}{\lambda}$ is
the wavenumber, and ${\mathbf{e}}_{y}=[0,1,0]^{\mathsf{T}}$ is the normal
vector of the CAPA. Here, ${\mathsf{g}}(\mathbf{r},{\mathbf{s}})$ models the
influence of free-space EM propagation, and
$\frac{{\mathbf{e}}_{y}^{\mathsf{T}}({\mathbf{s}}-{\mathbf{r}})}{\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}$
models the impact of the projected aperture of the CAPA.

####  II-C2 Non-Line-of-Sight Channel

Next, we explore the NLoS channel model by assuming that the LoS path is
blocked. In this case, we characterize ${\mathsf{h}}(\mathbf{r},{\mathbf{s}})$
as follows:

$\displaystyle{\mathsf{h}}(\mathbf{r},{\mathbf{s}})=\sum\nolimits_{n=1}^{N_{\mathsf{s}}}\alpha_{n}{\mathsf{g}}_{\mathsf{LoS}}(\mathbf{r},{\mathbf{s}}_{n}){\mathsf{g}}({\mathbf{s}}_{n},{\mathbf{s}}),$
(7)

where $N_{\mathsf{s}}$ denotes the number of scatterers,
${\mathbf{s}}_{n}=[x_{n},y_{n},z_{n}]^{\mathsf{T}}$ denotes the location of
the $n$th scatterer, and $\alpha_{n}\in{\mathbbmss{C}}$ denotes the associated
refection coefficient. Following the Swerling-I model, we model
$\alpha_{n}\sim{\mathcal{CN}}(0,\sigma_{n}^{2})$ with $\sigma_{n}^{2}$
suggesting the average reflection strength.

##  III Line-of-Sight Channel

Having established the system model, we next analyze the performance of
aperture selection in a CAPA. Particularly, we commence with the LoS scenario.

###  III-A Optimal Selection Strategy

For clarity, we assume that the activated aperture
${\mathcal{S}}_{\mathsf{R}}$ is a rectangular area. Under this assumption, we
calculate the SNR in closed-form as follows.

######  Lemma 2.

Let ${\mathcal{S}}_{\mathsf{R}}$ be an $A_{x}\times A_{z}$ rectangular area
centered at $O^{\prime}$. The SNR can be expressed follows:

$\displaystyle\gamma=\frac{\overline{\gamma}}{4\pi}\sum_{x\in{\mathcal{X}}_{O^{\prime}}}\sum_{z\in{\mathcal{Z}}_{O^{\prime}}}\arctan\left(\frac{xz}{\Psi\sqrt{\Psi^{2}+x^{2}+z^{2}}}\right),$
(8)

where
${\mathcal{X}}_{O^{\prime}}=\\{\frac{A_{x}}{2r}\pm(\Phi-\frac{r_{x}}{r})\\}$,
${\mathcal{Z}}_{O^{\prime}}=\\{\frac{A_{z}}{2r}\pm(\Theta-\frac{r_{z}}{r})\\}$,
and
$\overline{\gamma}\triangleq\frac{\lvert{\mathsf{J}}({\mathbf{s}}_{\mathsf{u}})\rvert^{2}\lvert{\mathcal{A}}_{\mathsf{S}}\rvert^{2}k_{0}^{2}\eta^{2}}{4\pi{\sigma}^{2}}$.

###### Proof:

Please refer to Appendix -B for more details. ∎

Substituting (8) into (5) gives

$\displaystyle{\mathcal{S}}_{\mathsf{R}}^{\star}=\operatorname*{argmax}_{{\mathcal{S}}_{\mathsf{R}}\subseteq{\mathcal{A}}_{\mathsf{R}}}\sum_{x\in{\mathcal{X}}_{O^{\prime}}}\sum_{z\in{\mathcal{Z}}_{O^{\prime}}}\arctan\left(\frac{xz/\Psi}{\sqrt{\Psi^{2}+x^{2}+z^{2}}}\right).$
(9)

When ${\mathcal{S}}_{\mathsf{R}}$ is fixed to be an $A_{x}\times A_{z}$
rectangular, it is solely determined by its center location $O^{\prime}$,
whose feasible set is given by
${\mathcal{O}}=\\{(x,0,z)|x\in[\frac{A_{x}-L_{x}}{2},\frac{L_{x}-A_{x}}{2}],z\in[\frac{A_{z}-L_{z}}{2},\frac{L_{z}-A_{z}}{2}]\\}$.
Therefore, the problem formulated in (9) can be rewritten as follows:

$\displaystyle\max_{(r_{x},0,r_{z})\in{\mathcal{O}}}\sum_{x\in{\mathcal{X}}_{O^{\prime}}}\sum_{z\in{\mathcal{Z}}_{O^{\prime}}}\arctan\left(\frac{xz/\Psi}{\sqrt{\Psi^{2}+x^{2}+z^{2}}}\right).$
(10)

Generally, obtaining the optimal $(r_{x},0,r_{z})$ requires a brute-force two-
dimensional search within the feasible set ${\mathcal{O}}$, which is time-
consuming. Yet, using the special mathematical structure of the SNR, Lemma 3
can be found, based on which a closed-form solution to problem (10) is
available.

######  Lemma 3.

Given $(r,\Phi,\Theta)$, the objective function of (10) is monotone decreasing
with either $\lvert r\Phi-r_{x}\rvert$ or $\lvert r\Theta-r_{z}\rvert$.

###### Proof:

Please refer to Appendix -C for more details. ∎

######  Theorem 1.

Let ${\mathcal{S}}_{\mathsf{R}}$ be an $A_{x}\times A_{z}$ rectangular area
centered at $O^{\prime}$. The optimal $O^{\prime}$ that maximizes the SNR
$\gamma$ is given by

$\displaystyle{{r}}_{x}^{\star}$
$\displaystyle=\operatorname*{argmin}\nolimits_{x\in[\frac{A_{x}-L_{x}}{2},\frac{L_{x}-A_{x}}{2}]}\lvert
r\Phi-x\rvert,$ (11) $\displaystyle{{r}}_{z}^{\star}$
$\displaystyle=\operatorname*{argmin}\nolimits_{z\in[\frac{A_{z}-L_{z}}{2},\frac{L_{z}-A_{z}}{2}]}\lvert
r\Theta-z\rvert,$ (12)

where $\operatorname*{argmin}_{x\in[a,b]}|c-x|$ can be calculated in closed-
form:

$\displaystyle\operatorname*{argmin}\nolimits_{x\in[a,b]}|c-x|=\left\\{\begin{array}[]{rcl}a&&{c<a}\\\
c&&{c\in[a,b]}\\\ b&&{c>b}\end{array}\right..$ (16)

###### Proof:

By using the result in Lemma 3, the final result follows immediately. ∎

The function $\operatorname*{argmin}_{x\in[a,b]}|c-x|$ returns the point in
the interval $[a,b]$, which is closest to $c$. This means that
$({{r}}_{x}^{\star},0,{{r}}_{z}^{\star})$ corresponds to the point in the
feasible set that is closest to the projection of $\mathbf{s}_{\mathsf{u}}$ on
the receive aperture, i.e., $(r\Phi,0,r\Theta)$.

######  Remark 1.

The results in Theorem 1 suggest that the optimal aperture selection criterion
in our considered LoS scenario is essentially the nearest neighbour criterion.
This makes intuitive sense, as the nearest neighbor point minimizes both free-
space path loss and projected aperture loss.

######  Remark 2.

According to Theorem 1, if the projection of ${\mathbf{s}}_{\mathsf{u}}$ on
the receive aperture, i.e., $(r\Phi,0,r\Theta)$, is located in the feasible
set $\mathcal{O}$, we have
$r\Phi\in[\frac{A_{x}-L_{x}}{2},\frac{L_{x}-A_{x}}{2}]$ and
$r\Theta\in[\frac{A_{z}-L_{z}}{2},\frac{L_{z}-A_{z}}{2}]$, which yields
${{r}}_{x}^{\star}=r\Phi$ and ${{r}}_{z}^{\star}=r\Theta$. This gives the
performance upper bound of aperture selection, and the resulting SNR is given
by

$\displaystyle\gamma=\frac{\overline{\gamma}}{\pi}\arctan\left(\frac{{A_{x}A_{z}}}{2r\Psi\sqrt{4r^{2}\Psi^{2}+A_{x}^{2}+A_{z}^{2}}}\right).$
(17)

We next consider some special cases to unveil more system insights on aperture
selection.

###  III-B Further Discussions

####  III-B1 Impact of the Selected Aperture Size

Let us first investigate the impact of the selected aperture size on the
received SNR. For clarity, we consider the case where
${\mathcal{S}}_{\mathsf{R}}$ is a square area subject to $A_{x}=A_{z}$. We
further assume that the projection of ${\mathbf{s}}_{\mathsf{u}}$ on the
receive aperture, i.e., $(r\Phi,0,r\Theta)$, is located in the feasible set
$\mathcal{O}$. In this case, the received SNR is given by

$\displaystyle\gamma=\frac{\overline{\gamma}}{\pi}\arctan\left({{\lvert{\mathcal{S}}_{\mathsf{R}}\rvert}}{(2r\Psi)^{-1}(4r^{2}\Psi^{2}+2\lvert{\mathcal{S}}_{\mathsf{R}}\rvert)^{-\frac{1}{2}}}\right).$
(18)

The SNR achieved by activating the entire receive aperture is obtained by
inserting ${\mathcal{S}}_{\mathsf{R}}={\mathcal{A}}_{\mathsf{R}}$ into (18),
which yields

$\displaystyle\gamma_{\mathsf{f}}=\frac{\overline{\gamma}}{\pi}\arctan\left({{\lvert{\mathcal{A}}_{\mathsf{R}}\rvert}}{(2r\Psi)^{-1}({4r^{2}\Psi^{2}+2\lvert{\mathcal{A}}_{\mathsf{R}}\rvert})^{-\frac{1}{2}}}\right).$
(19)

An intriguing question to ask in this case is “for a given $\beta\in(0,1)$,
how large the active aperture size under the nearest neighbour criterion are
required to achieve an SNR of $\beta\gamma_{\mathsf{f}}$?” To this end, we
calculate the derivative of
$\arctan\left(\frac{x}{2r\Psi\sqrt{4r^{2}\Psi^{2}+2x}}\right)$ with respect to
$x$ and note that
$\frac{{\rm{d}}^{2}}{{\rm{d}}x}\arctan\left(\frac{x}{2r\Psi\sqrt{4r^{2}\Psi^{2}+2x}}\right)<0$
($\forall r,\Psi$). This suggests that the achieved SNR involves a gradually
decreasing incremental rate as the activated aperture size increases, which
leads to the following conclusion.

######  Remark 3.

A receive SNR of $\beta\gamma_{\mathsf{f}}$ is achieved by less than $\beta$
of the receive aperture ${\mathcal{A}}_{\mathsf{R}}$ being active.

The receive SNR in (18) can be rewritten as follows:

$\displaystyle\gamma=\frac{\overline{\gamma}}{\pi}\arctan\left(\tau_{\mathcal{S}}{(1+2\tau_{\mathcal{S}})^{-\frac{1}{2}}}\right).$
(20)

where
$\tau_{\mathcal{S}}\triangleq\frac{\lvert{\mathcal{S}}_{\mathsf{R}}\rvert}{4r^{2}\Psi^{2}}$.
Considering the case that $\tau_{\mathcal{S}}\ll 1$, which arises when the
user is located in the far-field region of the CAPA, we use the Taylor
expansion of $\arctan(\cdot)$ to approximate (20) as follows:

$\displaystyle\gamma\approx\frac{\overline{\gamma}}{\pi}\tau_{\mathcal{S}}\frac{1}{\sqrt{1+2\tau_{\mathcal{S}}}}\approx\frac{\overline{\gamma}}{\pi}\tau_{\mathcal{S}}=\frac{\overline{\gamma}}{\pi}\frac{\lvert{\mathcal{S}}_{\mathsf{R}}\rvert}{4r^{2}\Psi^{2}},$
(21)

where the second approximation is due to the fact that
$\frac{1}{\sqrt{1+2\tau_{\mathcal{S}}}}\approx 1$ when $\tau_{\mathcal{S}}\ll
1$. The results in (21) suggest that the receive SNR approximately scales
linearly with the activated aperture size
$\lvert{\mathcal{S}}_{\mathsf{R}}\rvert$ when the user is located in the far-
field region, which leads to the following conclusion.

######  Remark 4.

The phenomenon in Remark 3 is less highlighted when the user is located in the
far-field region of the CAPA.

####  III-B2 Impact of the Aperture Shape

We next study the impact of the aperture shape on the receive SNR. For
simplicity, we focus on the case that the projection of
${\mathbf{s}}_{\mathsf{u}}$ on the receive aperture is located in the feasible
region of ${\mathcal{S}}_{\mathsf{R}}$’s center location. In this case, the
receive SNR achieved by activating a rectangular area is given by (18).
Following the same principle, the SNR achieved by activating a circle area
with a radius of $\mathsf{R}$ can be calculated as follows:

$\begin{split}\gamma^{\prime}=\frac{\overline{\gamma}}{4\pi}\int_{0}^{2\pi}\int_{0}^{\mathsf{R}}\frac{(r\Psi)\rho{\rm{d}}\rho{\rm{d}}\theta}{(\rho^{2}+r^{2}\Psi^{2})^{\frac{3}{2}}}=\frac{\overline{\gamma}}{2}\Big{(}1-\Big{(}1+\frac{4\tau_{\mathcal{S}}}{\pi}\Big{)}^{-\frac{1}{2}}\Big{)},\end{split}$
(22)

where
$\tau_{\mathcal{S}}=\frac{\lvert{\mathcal{S}}_{\mathsf{R}}\rvert}{4r^{2}\Psi^{2}}=\frac{\pi{\mathsf{R}}^{2}}{4r^{2}\Psi^{2}}$.
Let $\tau_{\mathcal{S}}\ll 1$ and obtain
$\frac{4\tau_{\mathcal{S}}}{\pi}\approx 0$, which, together with the Taylor
expansion of $(1+x)^{-\frac{1}{2}}=1-\frac{1}{2}x+o(x)$, yields

$\gamma^{\prime}\approx\frac{\overline{\gamma}}{2}\Big{(}1-\Big{(}1-\frac{4\tau_{\mathcal{S}}}{2\pi}\Big{)}\Big{)}=\frac{\overline{\gamma}}{\pi}\tau_{\mathcal{S}}.$
(23)

We next consider the case that $\tau_{\mathcal{S}}\gg 1$, which happens when
the user is located in the near-field region of the CAPA. On this condition,
we have

$\begin{split}\gamma^{\prime}=\frac{\overline{\gamma}}{2}\Big{(}1-\Big{(}1+\frac{4\tau_{\mathcal{S}}}{\pi}\Big{)}^{-\frac{1}{2}}\Big{)}\approx\frac{\overline{\gamma}}{2}\Big{(}1-\frac{\sqrt{\pi}}{2}\frac{1}{\sqrt{\tau_{\mathcal{S}}}}\Big{)}.\end{split}$
(24)

By further using the expansion in [4, Eq. (4.24.4)], we have

$\gamma\approx\frac{\overline{\gamma}}{\pi}\Big{(}\frac{\pi}{2}-\frac{(1+2\tau_{\mathcal{S}})^{\frac{1}{2}}}{\tau_{\mathcal{S}}}\Big{)}\approx\frac{\overline{\gamma}}{2}\Big{(}1-\frac{2\sqrt{2}}{\pi}\frac{1}{\sqrt{\tau_{\mathcal{S}}}}\Big{)}$
(25)

for $\tau_{\mathcal{S}}\gg 1$. By comparing (23) with (21) as well as (24)
with (25), we arrive at the following conclusions.

######  Remark 5.

The results in (23) and (21) suggest that in our considered scenarios, the
shape of the activated aperture has no clear impact on the receive SNR when
the user is located in the far-field region.

######  Remark 6.

The results in (24) and (25) as well as the fact that
$\frac{2\sqrt{2}}{\pi}>\frac{\sqrt{\pi}}{2}$ suggest that in the near-field
region, activating a circle area achieves a receive SNR than activating a
square area. However, since
${\frac{\sqrt{\pi}}{2}}/{\frac{2\sqrt{2}}{\pi}}\approx 0.98$, the
aforementioned SNR gap in the near-field region is quite limited.

####  III-B3 Discrete Aperture Segmentation

Theorem 1 suggests that the optimal selected aperture is a function of the
user’s location. In practice, users may move freely, which necessitates
frequent switching of the resulting optimal receive aperture
${\mathcal{S}}_{\mathsf{R}}^{\star}$. However, such frequent changes can lead
to a high overhead in control signaling. One effective method to address this
challenge is by discretizing the entire receive aperture into $K$ aperture
segments denoted as
$\\{{\mathcal{S}}_{\mathsf{R}}^{(k)}\\}_{k=1}^{K}\triangleq{\mathbbmss{S}}$
such that
$\bigcup_{k=1}^{K}{\mathcal{S}}_{\mathsf{R}}^{(k)}={\mathcal{A}}_{\mathsf{R}}$,
and selecting the aperture from this candidate set. Under this circumstance,
the optimal aperture index is determined as follows:

$k^{\star}=\operatorname*{argmax}\limits_{k=1,\ldots,K}\int_{{\mathcal{S}}_{\mathsf{R}}^{(k)}}\lvert{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}})\rvert^{2}{\rm{d}}{\mathbf{r}}.$
(26)

By definition, we obtain

$\int_{{\mathcal{S}}_{\mathsf{R}}^{(k^{\star})}}\lvert{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}})\rvert^{2}{\rm{d}}{\mathbf{r}}\leq\int_{{\mathcal{S}}_{\mathsf{R}}^{\star}}\lvert{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}})\rvert^{2}{\rm{d}}{\mathbf{r}},$
(27)

which indicates the sub-optimality of the aperture segmentation. A simple
aperture segmentation strategy involves partitioning the receive aperture into
four disjoint parts, each corresponding to a quadrant. Specifically, we have

${\mathbbmss{S}}=\\{\\{(\pm x,0,\pm
z)|x\in[0,{L_{x}}/{2}],z\in[0,{L_{z}}/{2}]\\}\\}\triangleq{\mathbbmss{S}}_{\mathsf{q}},$

which yields $O^{\prime}\in\\{(\pm{L_{x}}/{4},0,\pm{L_{z}}/{4})\\}$. By using
the results in Theorem 1, the following conclusion is drawn.

######  Remark 7.

When ${\mathbbmss{S}}={\mathbbmss{S}}_{\mathsf{q}}$, the optimal candidate
aperture that can maximize the receive SNR is the quadrant where the
projection of $\mathbf{s}_{\mathsf{u}}$ on ${\mathcal{A}}_{\mathsf{R}}$ is
located in.

####  III-B4 Linear Array

We consider the case when $\frac{L_{z}}{r}\ll 1$, where the variation of the
channel response along the $z$-axis is negligible. This makes the entire
_planar array_ degenerate into a _linear array_ along the $x$-axis. The
receive SNR achieved by selecting the interval
$[r_{x}-\frac{A_{x}}{2},r_{x}+\frac{A_{x}}{2}]$ is given by

$\displaystyle\zeta$
$\displaystyle=\overline{\gamma}\frac{A_{z}}{4\pi}\int_{r_{x}-\frac{A_{x}}{2}}^{r_{x}+\frac{A_{x}}{2}}\frac{r{\Psi}{\rm{d}}x}{((x-r\Phi)^{2}+r^{2}{\Psi}^{2}+r^{2}\Theta^{2})^{\frac{3}{2}}}$
(28)
$\displaystyle=\left.\frac{\overline{\gamma}\frac{A_{z}}{4\pi}\frac{x-r_{x,\Phi}}{r^{2}{\Psi}^{2}+r^{2}\Theta^{2}}r{\Psi}}{\sqrt{(x-r_{x,\Phi})^{2}+r^{2}{\Psi}^{2}+r^{2}\Theta^{2}}}\right\rvert_{-A_{x}/2}^{A_{x}/2}$
(29)

with $r_{x,\Phi}=r\Phi-r_{x}$. Following the steps in obtaining Theorem 1, the
following corollary can be found.

######  Corollary 1.

Given the above linear array, the optimal $r_{x}$ that maximizes the SNR
$\gamma$ is given by

$\displaystyle{{r}}_{x}^{\star}$
$\displaystyle=\operatorname*{argmin}\nolimits_{x\in[\frac{A_{x}-L_{x}}{2},\frac{L_{x}-A_{x}}{2}]}\lvert
r\Phi-x\rvert,$ (30)

where $\operatorname*{argmin}_{x\in[a,b]}|c-x|$ is calculated by (16).

The function $\operatorname*{argmin}_{x\in[a,b]}|c-x|$ returns the point in
the interval $[a,b]$, which is closest to $c$. Thus, ${{r}}_{x}^{\star}$
corresponds to the point in the feasible set
$[\frac{A_{x}-L_{x}}{2},\frac{L_{x}-A_{x}}{2}]$, which is closest to the
projection of $\mathbf{s}_{\mathsf{u}}$ on the linear array, i.e., $r\Phi$.

######  Remark 8.

The results in Corollary 1 suggest that the optimal interval selection
criterion in our considered scenario is the nearest neighbour criterion, which
is akin to Theorem 1.

##  IV Non-Line-of-Sight Channel

###  IV-A Optimal Aperture Selection

Having analyzed the performance of aperture selection for the LoS channel, we
now shift our attention to NLoS channels. For the sake of brevity, we
concentrate our attention on the case where the activated aperture
${\mathcal{S}}_{\mathsf{R}}$ is an $A_{x}\times A_{z}$ rectangular area
centered at $(r_{x},0,r_{z})$. Given ${\mathcal{S}}_{\mathsf{R}}$, the receive
SNR under the NLoS channel model shown in (7) is calculated as follows:

$\begin{split}&\gamma_{\mathsf{NLoS}}={\overline{\gamma}}\sum\nolimits_{n=1}^{N_{\mathsf{s}}}\lvert\alpha_{n}\rvert^{2}\lvert{\mathsf{g}}({\mathbf{s}}_{n},{\mathbf{s}}_{\mathsf{u}})\rvert^{2}{\mathsf{a}}_{\mathsf{LoS}}^{(n)}+{\overline{\gamma}}\sum\nolimits_{n=1}^{N_{\mathsf{s}}}\\\
&\times\sum\nolimits_{n^{\prime}\neq
n}\alpha_{n}\alpha_{n^{\prime}}^{\mathsf{H}}{\mathsf{g}}({\mathbf{s}}_{n},{\mathbf{s}}_{\mathsf{u}}){\mathsf{g}}^{\mathsf{H}}({\mathbf{s}}_{n^{\prime}},{\mathbf{s}}_{\mathsf{u}})\sqrt{y_{n}y_{n^{\prime}}}{{\rho}}_{n,n^{\prime}},\end{split}$
(31)

where
${\mathsf{a}}_{\mathsf{LoS}}^{(n)}\triangleq\frac{1}{4\pi}\int_{{\mathcal{S}}_{\mathsf{R}}}\frac{y_{n}{\rm{d}}x{\rm{d}}z}{((x-x_{n})^{2}+y_{n}^{2}+(z-z_{n})^{2})^{{3}/{2}}}$
and
${{\rho}}_{n,n^{\prime}}\triangleq\frac{1}{4\pi}\int_{{\mathcal{S}}_{\mathsf{R}}}\frac{{\rm{e}}^{{\rm{j}}k_{0}(({(x-x_{n^{\prime}})^{2}+y_{n^{\prime}}^{2}+(z-z_{n^{\prime}})^{2}})^{\frac{1}{2}}-({(x-x_{n})^{2}+y_{n}^{2}+(z-z_{n})^{2}})^{\frac{1}{2}})}{\rm{d}}x{\rm{d}}z}{((x-x_{n})^{2}+y_{n}^{2}+(z-z_{n})^{2})^{{3}/{4}}((x-x_{n^{\prime}})^{2}+y_{n^{\prime}}^{2}+(z-z_{n^{\prime}})^{2})^{{3}/{4}}}$.
Note that ${\mathsf{a}}_{\mathsf{LoS}}^{(n)}$ can be calculated in closed-form
as (8), whereas ${{\rho}}_{n,n^{\prime}}$ can be approximated using the
Gauss–Legendre quadrature rule [4] twice, which yields

$\displaystyle{{\rho}}_{n,n^{\prime}}\approx\sum_{i=1}^{T}\sum_{j=1}^{T}\frac{{\rm{e}}^{{\rm{j}}k_{0}((\frac{A_{x}\xi_{j}}{2}+r_{x,n^{\prime}})^{2}+y_{n^{\prime}}^{2}+(\frac{A_{z}\xi_{i}}{2}+r_{z,n^{\prime}})^{2})^{\frac{1}{2}}}}{((\frac{A_{x}\xi_{j}}{2}+r_{x,n^{\prime}})^{2}+y_{n^{\prime}}^{2}+(\frac{A_{z}\xi_{i}}{2}+r_{z,n^{\prime}})^{2})^{\frac{3}{4}}}$
$\displaystyle~{}~{}\times\frac{\frac{w_{i}w_{j}}{4\pi}\frac{A_{x}A_{z}}{4}{\rm{e}}^{-{\rm{j}}k_{0}((\frac{A_{x}\xi_{j}}{2}+r_{x,n})^{2}+y_{n}^{2}+(\frac{A_{z}\xi_{i}}{2}+r_{z,n})^{2})^{\frac{1}{2}}}}{((\frac{A_{x}\xi_{j}}{2}+r_{x,n})^{2}+y_{n}^{2}+(\frac{A_{z}\xi_{i}}{2}+r_{z,n})^{2})^{\frac{3}{4}}},$
(32)

where $r_{x,n}\triangleq r_{x}-x_{n}$, $r_{z,n}\triangleq r_{z}-z_{n}$, where
$\\{w_{i}\\}$ and $\\{\xi_{i}\\}$ denote the weight and abscissa factors of
Gauss–Legendre integration, and $T$ is a complexity-vs-accuracy tradeoff
parameter. Under the NLoS scenario, the aperture selection criteria is
formulated as follows:

$\displaystyle{\mathcal{S}}_{\mathsf{R}}^{\star}=\operatorname*{argmax}\nolimits_{{\mathcal{S}}_{\mathsf{R}}\subseteq{\mathcal{A}}_{\mathsf{R}}}\gamma_{\mathsf{NLoS}}.$
(33)

Upon comparing (8) with (31), we observe that the receive SNR in the NLoS
scenario is more intractable than that in the LoS scenario due to the
existence of multiple scatterers.

######  Remark 9.

The intractability of (31) makes the nearest neighbour criterion not optimal
for the NLoS channels.

###  IV-B Discrete Aperture Segmentation

Finding the optimal solution to problem (33) asks for a brute-force two-
dimensional search within the receive aperture ${\mathcal{A}}_{\mathsf{R}}$,
which is computationally intensive. As a compromise, we propose to portion the
entire aperture into several segments
$\\{{\mathcal{S}}_{\mathsf{R}}^{(k)}\\}_{k=1}^{K}={\mathbbmss{S}}$, and
selecting the aperture from this candidate set, i.e.,
$k^{\star}=\operatorname*{argmax}_{k=1,\ldots,K}\gamma_{k}$, where
$\gamma_{k}\triangleq\int_{{\mathcal{S}}_{\mathsf{R}}^{(k)}}\lvert\sum\nolimits_{n=1}^{N_{\mathsf{s}}}\alpha_{n}{\mathsf{g}}_{\mathsf{LoS}}(\mathbf{r},{\mathbf{s}}_{n}){\mathsf{g}}({\mathbf{s}}_{n},{\mathbf{s}})\rvert^{2}{\rm{d}}{\mathbf{r}}$.
Since $\alpha_{n}\sim{\mathcal{CN}}(0,\sigma_{n}^{2})$ for
$n=1,\ldots,N_{\mathsf{s}}$, we have $\gamma_{k}\overset{d}{=}\lvert
h_{k}\rvert$, where $\overset{d}{=}$ denotes the equality in distribution and
$h_{k}$ is complex Gaussian distributed. By defining
${\mathbf{h}}\triangleq[h_{1},\ldots,h_{K}]^{\mathsf{T}}$, we have
${\mathbf{h}}\sim{\mathcal{CN}}({\mathbf{0}},{\mathbf{R}})$ where
${\mathbf{R}}={\mathbbmss{E}}\\{{\mathbf{h}}{\mathbf{h}}^{\mathsf{H}}\\}$
represents the correlation matrix. To describe $\mathbf{R}$, we introduce the
following lemma.

######  Lemma 4.

Let ${\mathbf{r}}_{1},{\mathbf{r}}_{2}\in{\mathcal{A}}_{\mathsf{R}}$. The
correlation function
${\mathscr{R}}({\mathbf{r}}_{1},{\mathbf{r}}_{2})\triangleq{\mathbbmss{E}}\\{\int_{{\mathcal{A}}_{\mathsf{S}}}{\mathsf{h}}(\mathbf{r}_{1},{\mathbf{s}}){\mathsf{h}}^{\mathsf{H}}(\mathbf{r}_{2},{\mathbf{s}}){\rm{d}}{\mathbf{s}}\\}$
for the NLoS spatial response (7) is given by

$\begin{split}&{\mathscr{R}}({\mathbf{r}}_{1},{\mathbf{r}}_{2})={\overline{\gamma}}\sum\nolimits_{n=1}^{N_{\mathsf{s}}}\sigma_{n}^{2}\lvert{\mathsf{g}}({\mathbf{s}}_{n},{\mathbf{s}}_{\mathsf{u}})\rvert^{2}\frac{{\rm{e}}^{-{\rm{j}}k_{0}\lVert{\mathbf{r}}_{1}-{\mathbf{s}}_{n}\rVert}}{\sqrt{4\pi}\lVert{\mathbf{r}}_{1}-{\mathbf{s}}_{n}\rVert}\\\
&\times\sqrt{\frac{{\mathbf{e}}_{y}^{\mathsf{T}}({\mathbf{s}}_{n}-{\mathbf{r}}_{1})}{\lVert{\mathbf{r}}_{1}-{\mathbf{s}}_{n}\rVert}}\frac{{\rm{e}}^{{\rm{j}}k_{0}\lVert{\mathbf{r}}_{2}-{\mathbf{s}}_{n}\rVert}}{\sqrt{4\pi}\lVert{\mathbf{r}}_{2}-{\mathbf{s}}_{n}\rVert}\sqrt{\frac{{\mathbf{e}}_{y}^{\mathsf{T}}({\mathbf{s}}_{n}-{\mathbf{r}}_{2})}{\lVert{\mathbf{r}}_{2}-{\mathbf{s}}_{n}\rVert}}.\end{split}$
(34)

###### Proof:

This lemma is proved by substituting (7) into the expression of
${\mathscr{R}}({\mathbf{r}}_{1},{\mathbf{r}}_{2})$ and using the fact that
${\mathbbmss{E}}\\{\lvert\alpha_{n}\rvert^{2}\\}=\sigma_{n}^{2}$ for
$n=1,\ldots,N_{\mathsf{s}}$ and
${\mathbbmss{E}}\\{\alpha_{n}^{\mathsf{H}}\alpha_{n^{\prime}}\\}=0$ for $n\neq
n^{\prime}$. ∎

After obtaining ${\mathscr{R}}({\mathbf{r}}_{1},{\mathbf{r}}_{2})$, the
correlation matrix $\mathbf{R}$ can be directly constructed. Particularly, the
$(k,k^{\prime})$th element of $\mathbf{R}$ can be calculated as follows:

$\begin{split}[\mathbf{R}]_{k,k^{\prime}}=\int_{{\mathcal{S}}_{\mathsf{R}}^{(k)}}\int_{{\mathcal{S}}_{\mathsf{R}}^{(k^{\prime})}}{\mathscr{R}}({\mathbf{r}}_{1},{\mathbf{r}}_{2}){\rm{d}}{\mathbf{r}}_{1}{\rm{d}}{\mathbf{r}}_{2}.\end{split}$
(35)

We comment that the double integral in (35) can be approximated using the
Gauss–Legendre quadrature rule, which is akin to (31). In summary, the
following lemma is found.

######  Lemma 5.

Under the considered model, it has
$\gamma_{k^{\star}}\overset{d}{=}\max_{k=1,\ldots,K}\lvert h_{k}\rvert^{2}$
with
$[h_{1},\ldots,h_{K}]^{\mathsf{T}}={\mathbf{h}}\sim{\mathcal{CN}}({\mathbf{0}},{\mathbf{R}})$.

Since ${\mathbf{h}}\sim{\mathcal{CN}}({\mathbf{0}},{\mathbf{R}})$, we have
${\mathbf{h}}\overset{d}{=}{\mathbf{R}}^{\frac{1}{2}}\tilde{\mathbf{h}}$ with
$\tilde{\mathbf{h}}\sim{\mathcal{CN}}({\mathbf{0}},{\mathbf{I}}_{K})$. We next
characterize the OP of $\gamma_{k^{\star}}$, which is defined as
$\Pr(\gamma_{k^{\star}}<\gamma_{\mathsf{th}})$ with $\gamma_{\mathsf{th}}$
representing the target SNR. Let
${\mathbf{U}}^{\mathsf{H}}{\bm{\Lambda}}{\mathbf{U}}$ denote the
eigendecomposition of ${\mathbf{R}}$, where
${\mathbf{U}}\in{\mathbbmss{C}}^{{\mathsf{r}}\times{K}}$ is a semi-unitary
matrix with ${\mathbf{U}}{\mathbf{U}}^{\mathsf{H}}={\mathbf{I}}_{\mathsf{r}}$,
and
${\bm{\Lambda}}=\mathsf{diag}([\lambda_{1},\ldots,\lambda_{\mathsf{r}}]^{\mathsf{T}})\in{\mathbbmss{C}}^{{\mathsf{r}}\times{\mathsf{r}}}$
with $\\{\lambda_{i}\\}_{i=1}^{\mathsf{r}}$ being the positive eigenvalues of
$\mathbf{R}$, and $\mathsf{r}$ is the rank of $\mathbf{R}$. The statistical
distribution of $\mathbf{h}$ is thus given by [5, pp. 527–528]

$f_{\mathbf{h}}(\mathbf{x})={\rm{e}}^{-{\mathbf{x}}^{\mathsf{H}}{\mathbf{R}}^{+}{\mathbf{x}}}/({{\pi^{\mathsf{r}}{{\det}^{*}}({\mathbf{R}})}}),$
(36)

where
${\mathbf{R}}^{+}={\mathbf{U}}^{\mathsf{H}}{\bm{\Lambda}}^{-1}{\mathbf{U}}\in{\mathbbmss{C}}^{K\times
K}$ is the generalized inverse of $\mathbf{R}$, and
$\det^{*}({\mathbf{R}})=\prod_{i=1}^{\mathsf{r}}\lambda_{i}$ is the pseudo-
determinant of $\mathbf{R}$. Using the moment generating function-based
strategy, an exact expression for
$\Pr(\gamma_{k^{\star}}<\gamma_{\mathsf{th}})$ is available, which is yet
quite articulated and does not provide important insights. As a compromise, we
focus on the high-SNR asymptotic behavior of
$\Pr(\gamma_{k^{\star}}<\gamma_{\mathsf{th}})$ by letting
$\overline{\gamma}\rightarrow\infty$.

######  Theorem 2.

Let $\overline{\gamma}\rightarrow\infty$, and the OP satisfies

$\Pr(\gamma_{k^{\star}}<\gamma_{\mathsf{th}})={\mathcal{O}}(\overline{\gamma}^{-{\mathsf{r}}}\gamma_{\mathsf{th}}^{\mathsf{r}}/{{{\det}^{*}}({\mathbf{R}})}).$
(37)

###### Proof:

Please refer to [6, 7] for more details. ∎

######  Remark 10.

Theorem 2 suggests that under the NLoS condition, the diversity order achieved
by discrete aperture selection is given by $\mathsf{r}$, i.e., the rank of the
correlation matrix $\mathbf{R}$.

According to the properties of the rank of a matrix, we have
${\mathsf{r}}\leq\min\\{K,N_{\mathsf{s}}\\}$. This means that the diversity
order is mainly limited by the number of scatterers in a poorly scattered
environment where $N_{\mathsf{s}}$ is a finite value. By contrast, for a
richly scattered environment where $N_{\mathsf{s}}=\infty$, the diversity
order will be mainly limited by $K$, i.e., the size of the candidate set
$\mathbbmss{S}$.

##  V Numerical Results

In this section, we employ numerical results to validate the effectiveness of
aperture selection. Unless explicitly stated otherwise, we set the following
parameters: $\overline{\gamma}=40$ dB, $\theta=\frac{\pi}{3}$,
$\phi=\frac{\pi}{6}$, $r=10$ m, $\lambda=0.0107$ m, $A_{x}=A_{z}$, and
$L_{x}=L_{z}$.

(a) $A_{x}=L_{x}/8$.

(b) $L_{x}=2$ m.

Figure 2: Aperture selection for LoS channels.

Figure 2(a) illustrates the receive SNR for LoS channels plotted against the
aperture size $\lvert{\mathcal{A}}_{\mathsf{R}}\rvert$ when the selected
aperture is an $A_{x}\times A_{x}$ square. For comparison, we present the
results achieved by the optimal selection strategy (the nearest neighbor
criterion), discrete aperture segmentation, and no selection
(${\mathcal{S}}_{\mathsf{R}}$ fixed to be centered at the origin). In the
discrete aperture segmentation, we consider a finite candidate containing five
square areas centered at
${\mathbbmss{o}}\triangleq\\{[0,0,0]^{\mathsf{T}},[\pm\frac{L_{x}}{2},0,\pm\frac{L_{z}}{2}]^{\mathsf{T}}\\}$.
It is observed that both proposed aperture selection methods effectively
enhance the receive SNR compared to the scheme without aperture selection.
This improvement stems from the utilization of selection diversity. To further
explore the effectiveness of aperture selection, Figure 2(b) investigates the
impact of the selected aperture size $\lvert{\mathcal{S}}_{\mathsf{R}}\rvert$,
where the selection is based on the nearest neighbor criterion and the entire
aperture size is fixed to
$\lvert{\mathcal{A}}_{\mathsf{R}}\rvert=4~{}{\text{m}}^{2}$. Figure 2(b)
depicts $\frac{\gamma}{\gamma_{\mathsf{f}}}$ (the ratio of the SNR) as a
function of
$\frac{\lvert{\mathcal{S}}_{\mathsf{R}}\rvert}{\lvert{\mathcal{A}}_{\mathsf{R}}\rvert}$
(the ratio of the aperture size) for selected values of $r$, where
${\gamma_{\mathsf{f}}}$ represents the SNR achieved by setting
$\frac{\lvert{\mathcal{S}}_{\mathsf{R}}\rvert}{\lvert{\mathcal{A}}_{\mathsf{R}}\rvert}=1$.
As shown, for small values of $r$, i.e., when the user is located in near-
field region, 60% of ${\gamma_{\mathsf{f}}}$ is achieved by less than 40% of
the receive aperture being active. However, as $r$ increases,
$\frac{\gamma}{\gamma_{\mathsf{f}}}$ approximately increases linearly with
$\frac{\lvert{\mathcal{S}}_{\mathsf{R}}\rvert}{\lvert{\mathcal{A}}_{\mathsf{R}}\rvert}$.
These results align with our discussion in Section III-B1, which suggests that
aperture selection is more efficient in the near-field region for LoS
channels.

We then turn our focus to examining the impact of the aperture shape on
aperture selection in LoS channels. Figure 3(a) presents the SNRs achieved by
selecting square and circular areas in terms of
$\tau_{\mathcal{S}}=\frac{\lvert{\mathcal{S}}_{\mathsf{R}}\rvert}{4r^{2}\Psi^{2}}$,
where the explicit SNRs are calculated using (20) and (22), respectively.
Additionally, asymptotic results for $\tau_{\mathcal{S}}\rightarrow\infty$ and
$\tau_{\mathcal{S}}\rightarrow 0$ are presented, which are computed by (23)
((21)) and (24) ((25)), respectively. These asymptotic results closely follow
the explicit results as $\tau_{\mathcal{S}}$ approaches $\infty$ or $0$, which
verifies our previous derivations. Furthermore, from Figure 3(a), we observe
that selecting a circular area yields a slightly higher SNR than selecting a
square one. To illustrate the performance gap further, Figure 3(b) plots the
SNR ratio between activating a circle and activating a square in terms of
$\tau_{\mathcal{S}}$. It is observed that the performance gap between these
two schemes is relatively small, which is consistent with our discussion in
Remark 6.

(a) Achieved SNRs for various shapes.

(b) SNR ratio.

Figure 3: Impact of the aperture shape for LoS channels.

Finally, we utilize Figure 4 to discuss the performance of aperture selection
under NLoS channels. Specifically, we assume there are $N_{\mathsf{s}}=4$
radonly distributed scatterers. Additionally, we set
$\sigma_{n}^{2}=\frac{4\pi}{k_{0}^{2}\eta^{2}}$ for
$n=1,\ldots,N_{\mathsf{s}}$. Figure 4(a) illustrates the receive SNR versus
$\lvert{\mathcal{A}}_{\mathsf{R}}\rvert$ when the selected aperture is an
$A_{x}\times A_{x}$ square. For comparison, we present the results achieved by
the optimal selection strategy (brute-force search), discrete segmentation,
and no selection (${\mathcal{S}}_{\mathsf{R}}$ centered at the origin).
Concerning the discrete aperture segmentation, we consider $K=5$ segments
centered at ${\mathbbmss{o}}$. As observed, the proposed aperture selection
methods effectively enhance the receive SNR. Figure 4(b) further demonstrates
the OP in terms of $\overline{\gamma}$ for various values of $K$ ($K\leq
N_{\mathsf{s}}$), each corresponding to a random subset of ${\mathbbmss{o}}$.
In the high-SNR regime, the curves for explicit OP are parallel to the
asymptotic ones
($\overline{\gamma}^{-{\mathsf{r}}}\gamma_{\mathsf{th}}^{\mathsf{r}}/{{{\det}^{*}}({\mathbf{R}})}$).
This indicates that the achievable diversity order obtained in Theorem 2 is
accurate. Since we examine the case where $K\leq N_{\mathsf{s}}$, the
diversity order is constrained by $K$, which is consistent with the results
shown in Figure 4(b).

(a) $A_{x}=L_{x}/2$.

(b) OP. $\gamma_{\mathsf{th}}=-30$ dB.

Figure 4: Aperture selection for NLoS channels.

##  VI Conclusion

We have analyzed the receive SNR achieved by aperture selection in an uplink
CAPA-based system. For LoS channels, we discussed the impact of both the size
and shape of the selected aperture. For NLoS channels, we obtained the
relationship between diversity order, scattering extent, and selection
granularity. Our findings underscore the effectiveness of aperture selection
in enhancing communication performance for both scenarios, which serves as a
promising low-complexity transmission paradigm for future CAPA systems.

###  -A Proof of Lemma 1

The mean of
$\int_{{\mathcal{S}}_{\mathsf{R}}}{\mathsf{h}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}}){\mathsf{N}}(\mathbf{r}){\rm{d}}{\mathbf{r}}$
is calculated as

$\begin{split}{\mathbbmss{E}}\left\\{\int_{{\mathcal{S}}_{\mathsf{R}}}{\mathsf{h}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}}){\mathsf{N}}(\mathbf{r}){\rm{d}}{\mathbf{r}}\right\\}=\int_{{\mathcal{S}}_{\mathsf{R}}}{\mathsf{h}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}}){\mathbbmss{E}}\\{{\mathsf{N}}(\mathbf{r})\\}{\rm{d}}{\mathbf{r}},\end{split}$
(38)

which, together with the fact that
${\mathbbmss{E}}\\{{\mathsf{N}}(\mathbf{r})\\}=0$, yields
${\mathbbmss{E}}\\{\int_{{\mathcal{S}}_{\mathsf{R}}}{\mathsf{h}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}}){\mathsf{N}}(\mathbf{r}){\rm{d}}{\mathbf{r}}\\}=0$.
The variance is given by

$\begin{split}&{\mathbbmss{E}}\left\\{\int_{{\mathcal{S}}_{\mathsf{R}}}{\mathsf{h}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}}){\mathsf{N}}(\mathbf{r}){\rm{d}}{\mathbf{r}}\int_{{\mathcal{S}}_{\mathsf{R}}}{\mathsf{h}}^{\mathsf{H}}(\mathbf{r}^{\prime},{\mathbf{s}}_{\mathsf{u}}){\mathsf{N}}(\mathbf{r}^{\prime}){\rm{d}}{\mathbf{r}}^{\prime}\right\\}\\\
&=\int_{{\mathcal{S}}_{\mathsf{R}}}\int_{{\mathcal{S}}_{\mathsf{R}}}{\mathsf{h}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}}){\mathsf{h}}(\mathbf{r}^{\prime},{\mathbf{s}}_{\mathsf{u}}){\mathbbmss{E}}\\{{\mathsf{N}}(\mathbf{r}){\mathsf{N}}(\mathbf{r}^{\prime})\\}{\rm{d}}{\mathbf{r}}{\rm{d}}{\mathbf{r}}^{\prime}\\\
&=\int_{{\mathcal{S}}_{\mathsf{R}}}{\mathsf{h}}(\mathbf{r}^{\prime},{\mathbf{s}}_{\mathsf{u}})\int_{{\mathcal{S}}_{\mathsf{R}}}{\mathsf{h}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{\mathsf{u}})\sigma^{2}\delta(\mathbf{r}-\mathbf{r}^{\prime}){\rm{d}}{\mathbf{r}}{\rm{d}}{\mathbf{r}}^{\prime}.\end{split}$
(39)

Using the fact that
$\int_{{\mathcal{S}}_{\mathsf{R}}}\delta({\mathbf{x}}-{\mathbf{x}}_{0})f(\mathbf{x}){\rm{d}}{\mathbf{x}}=f({\mathbf{x}}_{0})$,
where $f(\cdot)$ is an arbitrary function defined on
${\mathcal{S}}_{\mathsf{R}}$, we obtain

$\eqref{Noise_Field_Covariance}=\int_{{\mathcal{S}}_{\mathsf{R}}}\sigma^{2}{\mathsf{h}}(\mathbf{r}^{\prime},{\mathbf{s}}_{\mathsf{u}}){\mathsf{h}}^{\mathsf{H}}(\mathbf{r}^{\prime},{\mathbf{s}}_{\mathsf{u}}){\rm{d}}{\mathbf{r}}^{\prime}=\sigma^{2}{\mathsf{a}}_{\mathsf{R}}.$
(40)

Lemma 1 is thus proved.

###  -B Proof of Lemma 2

As per (6), the channel gain achieved by activating the aperture
${\mathcal{S}}_{\mathsf{R}}$ can be calculated as follows:

$\begin{split}{\mathsf{a}}_{\mathsf{R}}=\int_{-\frac{A_{z}}{2}}^{\frac{A_{z}}{2}}\int_{-\frac{A_{x}}{2}}^{\frac{A_{x}}{2}}\frac{{k_{0}^{2}\eta^{2}}/{(4\pi)^{2}}r{\Psi}{\rm{d}}x{\rm{d}}z}{((x-r_{x,\Phi})^{2}+r^{2}{\Psi}^{2}+(z-r_{z,\Theta})^{2})^{\frac{3}{2}}},\end{split}$
(41)

where $r_{x,\Phi}\triangleq r\Phi-r_{x}$ and $r_{z,\Theta}\triangleq r\Theta-
r_{z}$. Through employing the integral identities [8, Eq. (2.264.5)] and [8,
Eq. (2.284)], the final results follow immediately.

###  -C Proof of Lemma 3

We use (41) to prove this theorem, as the variables $(x,z)$ therein are
decoupled. Given $z$, we have

$\displaystyle\left.{\mathsf{a}}_{\mathsf{R}}\right|_{z}$
$\displaystyle=\int_{r_{x}-\frac{A_{x}}{2}}^{r_{x}+\frac{A_{x}}{2}}\frac{{k_{0}^{2}\eta^{2}}/{(4\pi)^{2}}r{\Psi}{\rm{d}}x{\rm{d}}z}{((x-r\Phi)^{2}+r^{2}{\Psi}^{2}+(z-r\Theta)^{2})^{\frac{3}{2}}}$
(42)
$\displaystyle=\frac{k_{0}^{2}\eta^{2}}{(4\pi)^{2}}\frac{r{\Psi}(f_{z}(-r_{x,\Phi})+f_{z}(r_{x,\Phi}))}{r^{2}{\Psi}^{2}+(z-r\Theta)^{2}},$
(43)

where $r_{x,\Phi}\triangleq r\Phi-r_{x}$, $r_{z,\Theta}\triangleq r\Theta-
r_{z}$,
$f_{z}(x)=\frac{{A_{x}}/{2}+x}{(({A_{x}}/{2}+x)^{2}+r^{2}{\Psi}^{2}+(z-r\Theta)^{2})^{{1}/{2}}}$.
Note that $f_{z}(-r_{x,\Phi})+f_{z}(r_{x,\Phi})$ is an even function of
$r_{x,\Phi}$, and thus

$\begin{split}f_{z}(-r_{x,\Phi})+f_{z}(r_{x,\Phi})=f_{z}(-\lvert
r_{x,\Phi}\rvert)+f_{z}(\lvert r_{x,\Phi}\rvert).\end{split}$ (44)

Furthermore, given an arbitrary real number $b$, we have

$\begin{split}&\frac{{\rm{d}}}{{\rm{d}}x}\left(\frac{\frac{A_{x}}{2}+x}{((\frac{A_{x}}{2}+x)^{2}+b^{2})^{\frac{1}{2}}}+\frac{\frac{A_{x}}{2}-x}{((\frac{A_{x}}{2}-x)^{2}+b^{2})^{\frac{1}{2}}}\right)\\\
&=\frac{-b^{2}}{((\frac{A_{x}}{2}-x)^{2}+b^{2})^{\frac{3}{2}}}+\frac{b^{2}}{((\frac{A_{x}}{2}+x)^{2}+b^{2})^{\frac{3}{2}}}\leq
0\end{split}$ (45)

for $x>0$. Since $r^{2}{\Psi}^{2}+(z-r\Theta)^{2}\geq 0$, we conclude from
(45) that $f_{z}(-\lvert r_{x,\Phi}\rvert)+f_{z}(\lvert r_{x,\Phi}\rvert)$ is
monotone decreasing with $\lvert r_{x,\Phi}\rvert$. Taken together, for a
given $z$, $\left.{\mathsf{a}}_{\mathsf{R}}\right|_{z}$ decreases with $\lvert
r_{x,\Phi}\rvert$. Following the similar steps, we can also prove from (41)
that for a given $x$, the integrated function decreases with $\lvert
r_{z,\Theta}\rvert$. Since the variables $(x,z)$ in (41) are decoupled, we
conclude that the definite integral in (41) is maximized when $\lvert
r_{z,\Theta}\rvert$ and $\lvert r_{x,\Phi}\rvert$ are both minimized. The
final results follow immediately, and Lemma 3 is thus proved.

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