# On the Performance of Continuous Aperture Array (CAPA)-Based Wireless
Communications

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

The performance of continuous aperture array (CAPA)-based wireless
communications is analyzed in an uplink scenario. An analytical framework is
proposed to characterize uplink CAPA-based transmission using electromagnetic
field theories. On this basis, new expressions are derived for the channel
capacity in a single-user scenario and the sum-rate capacity in a multiuser
scenario, along with the capacity-achieving decoding schemes. These findings
are proved to differ greatly from those established for conventional spatially
discrete (SPD) arrays. Numerical results are provided to demonstrate that CAPA
offers significant capacity gains compared to the SPD array.

###### Index Terms:

Channel capacity, continuous aperture array (CAPA), performance analysis.

##  I Introduction

Multiple antenna technology serves as a cornerstone in the advancement of
modern wireless communication systems. At its core lies the principle of
leveraging an increased number of antenna elements to enhance spatial degrees
of freedom (DoFs) and augment channel capacity [1].

Traditionally, multiple antenna prototypes adhere to a spatially discrete
(SPD) topological structure, wherein each antenna is represented as an
individual point [2]. Within a fixed array aperture, decreasing the inter-
element distance facilitates the deployment of additional antennas, thereby
enhancing spatial DoFs and giving rise to the concept of holographic arrays.
In holographic arrays, SPD antenna elements are arranged in a sub-half-
wavelength-spaced manner [3].

This paper introduces the concept of a _continuous aperture array (CAPA)_ ,
where the entire array forms a spatially-continuous electromagnetic (EM)
aperture. In essence, a CAPA can be conceptualized as an SPD array comprising
an _infinite number_ of antennas with _infinitesimal spacing_ [2, 4]. CAPA
holds the potential to offer more spatial DoFs compared to SPD arrays, thereby
further enhancing system performance.

In contrast to traditional arrays that rely on spatially discrete modeling,
the modeling of CAPA-based wireless transmission is grounded in _EM fields_.
Specifically, while channels for a traditional SPD array are typically modeled
using discrete matrices, the spatial response for a CAPA should be represented
by a _continuous operator in the Hilbert space_ [2]. This fundamental
difference renders the traditional SPD arrays-based transmission framework
inapplicable to CAPAs. Currently, research on CAPAs is still in its infancy,
with most existing works limited to single-user scenarios; see [2, 5] for more
details. Although some authors have explored multiuser cases [6, 7], these
efforts have not fully characterized the fundamental performance limits of
CAPAs. Furthermore, an analytically tractable framework for CAPA-based
communications is still missing.

Motivated by these knowledge gaps, we analyze the performance of CAPA-based
communications by investigating an uplink system. The main contributions of
this article are summarized as follows: i) We propose an uplink transmission
framework for CAPAs leveraging EM field theories. ii) We derive the channel
capacity of a single-user uplink CAPA system and design the optimal detector
required to achieve it. iii) We extend our analysis to a multiuser uplink CAPA
system, derive the sum-rate capacity, and identify the decoding rule necessary
for achieving it. iv) We specialize the derived results to several special
aperture structures, including the SPD array, and prove the capacity achieved
by an SPD array is limited by its array occupation ratio. v) We provide
numerical results to demonstrate that CAPAs offer significant capacity
enhancements over traditional SPD arrays and that as the array occupation
ratio of an SPD array approaches $1$, its capacity converges to that of a
CAPA.

##  II System Model

We study an uplink channel where $K$ users simultaneously transmit signals to
a base station (BS) equipped with a CAPA, as shown in Figure 1. The CAPA is
centered at the origin $O\triangleq[0,0,0]^{\mathsf{T}}$, and its aperture is
denoted as $\mathcal{A}_{\mathsf{R}}\subseteq{\mathbbmss{R}}^{3\times 1}$. Let
$r_{k}$ denote the distance from the center of the CAPA to the center location
of each user $k\in{\mathcal{K}}\triangleq\\{1,\ldots,K\\}$, and
$\phi_{k}\in[0,\pi]$ and $\theta_{k}\in[0,\pi]$ denote the associated azimuth
and elevation angles, respectively. Therefore, user $k$’s center location can
be expressed as
${\mathbf{s}}_{k}=[r_{k}\Phi_{k},r_{k}\Psi_{k},r_{k}\Theta_{k}]^{\mathsf{T}}$,
where $\Phi_{k}\triangleq\cos{\phi_{k}}\sin{\theta_{k}}$,
$\Psi_{k}\triangleq\sin{\phi_{k}}\sin{\theta_{k}}$, and
$\Theta_{k}\triangleq\cos{\theta_{k}}$. Assume that user $k$ employs a
hypothetical isotropic antenna for signal transmission, whose transmit
aperture is denoted as ${\mathcal{A}}_{k}$ with
$\lvert{\mathcal{A}}_{k}\rvert=\frac{\lambda^{2}}{4\pi}$ for
$k\in{\mathcal{K}}$, where $\lambda$ denotes the wavelength. For clarity, we
assume that the aperture size of ${\mathcal{A}}_{\mathsf{R}}$ significantly
exceeds that of ${\mathcal{A}}_{k}$ ($\forall k$), i.e.,
$\lvert{\mathcal{A}}_{\mathsf{R}}\rvert\gg\lvert{\mathcal{A}}_{k}\rvert$.

Figure 1: Illustration of a CAPA-based multiuser channel.

###  II-A Signal Model

Let us define ${\mathsf{J}}_{k}({\mathbf{s}})\in{\mathbbmss{C}}$ as the
continuous distribution of source currents generated by user $k$ to convey his
data information, where ${\mathbf{s}}\in{\mathcal{A}}_{k}$. The excited
electric field ${\mathsf{E}}_{k}(\mathbf{r})\in{\mathbbmss{C}}$ due to
${\mathsf{J}}_{k}({\mathbf{s}})$ at point
$\mathbf{r}\in{\mathcal{A}}_{\mathsf{R}}$ can be expressed as follows [8]:

$\displaystyle{\mathsf{E}}_{k}(\mathbf{r})=\int_{{\mathcal{A}}_{k}}{\mathsf{h}}(\mathbf{r},{\mathbf{s}})s_{k}{\mathsf{J}}_{k}({\mathbf{s}}){\rm{d}}{\mathbf{s}},$
(1)

where $s_{k}\in{\mathbbmss{C}}$ represents the normalized data symbol subject
to ${\mathbbmss{E}}\\{\lvert s_{k}\rvert^{2}\\}=1$, and
${\mathsf{h}}(\mathbf{r},{\mathbf{s}})\in{\mathbbmss{C}}$ denotes the spatial
channel response between $\mathbf{s}$ and $\mathbf{r}$. Since
$\lvert{\mathcal{A}}_{\mathsf{R}}\rvert\gg\lvert{\mathcal{A}}_{k}\rvert$, we
simplify ${\mathsf{E}}_{k}(\mathbf{r})$ as follows:

$\displaystyle{\mathsf{E}}_{k}(\mathbf{r})\approx
s_{k}{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{k}){\mathsf{J}}_{k}({\mathbf{s}}_{k})\lvert{\mathcal{A}}_{k}\rvert.$
(2)

The total 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 fields
$\\{{\mathsf{E}}_{k}(\mathbf{r})\\}_{k=1}^{K}$, along with a random noise
field ${\mathsf{N}}(\mathbf{r})\in{\mathbbmss{C}}$, i.e.,

$\displaystyle{\mathsf{Y}}(\mathbf{r})$
$\displaystyle=\sum\nolimits_{k=1}^{K}{\mathsf{E}}_{k}(\mathbf{r})+{\mathsf{N}}(\mathbf{r})$
(3a)
$\displaystyle\approx\sum\nolimits_{k=1}^{K}s_{k}{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{k}){\mathsf{J}}_{k}({\mathbf{s}}_{k})\lvert{\mathcal{A}}_{k}\rvert+{\mathsf{N}}(\mathbf{r}),$
(3b)

where ${\mathsf{N}}(\mathbf{r})$ accounts for the thermal noise. 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.
${\mathsf{N}}(\mathbf{r})$ and $\\{s_{k}\\}_{k=1}^{K}$ are assumed to be
uncorrelated.

After observing ${\mathsf{Y}}(\mathbf{r})$, the BS should employ a properly
designed detector along with maximum-likelihood (ML) decoding to recover the
data information contained in $\\{s_{k}\\}_{k=1}^{K}$, which shall be detailed
in the following pages.

###  II-B Channel Model

To facilitate theoretical investigations into fundamental performance limits
and asymptotic behaviors, we focus our discussion on line-of-sight (LoS)
channels. Under this consideration, ${\mathsf{h}}(\mathbf{r},{\mathbf{s}})$ is
modeled as follows [9]:

$\displaystyle{\mathsf{h}}(\mathbf{r},{\mathbf{s}})={\mathsf{h}}_{\mathsf{em}}(\mathbf{r},{\mathbf{s}}){\mathsf{h}}_{\mathsf{pa}}(\mathbf{r},{\mathbf{s}}).$
(4)

The terms appearing in (4) are defined as follows:

  * •

${\mathsf{h}}_{\mathsf{pa}}(\mathbf{r},{\mathbf{s}})\triangleq\sqrt{\frac{\lvert{\mathbf{e}}_{{\mathbf{r}}}^{\mathsf{T}}({\mathbf{s}}-{\mathbf{r}})\rvert}{\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}}$
models the impact of the projected aperture of the BS array, which is
reflected by the projection of the normal vector onto the wave propagation
direction at each local point $\mathbf{r}$.

  * •

${\mathbf{e}}_{{\mathbf{r}}}\in{\mathbbmss{R}}^{3\times 1}$ is the normal
vector of the CAPA at $\mathbf{r}\in{\mathcal{A}}_{\mathsf{R}}$.

  * •

The function

$\displaystyle{\mathsf{h}}_{\mathsf{em}}(\mathbf{r},{\mathbf{s}})\triangleq\frac{{\rm{j}}k_{0}\eta{\rm{e}}^{-{\rm{j}}k_{0}\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}}{4\pi\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}\left(1+\frac{{\rm{j}}/k_{0}}{\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}-\frac{1/k_{0}^{2}}{\lVert{\mathbf{r}}-{\mathbf{s}}\rVert^{2}}\right)$
(5)

models the influence of free-space EM propagation, where $\eta=120\pi$ is the
impedance of free space and $k_{0}=\frac{2\pi}{\lambda}$ is the wavenumber.

The function ${\mathsf{h}}_{\mathsf{em}}(\mathbf{r},{\mathbf{s}})$ comprises
three terms: the first term corresponds to the radiating near-field and far-
field regions, while the remaining two terms correspond to the reactive near-
field region. The squared magnitude of the parenthesis in (5) is
$1-\frac{1/k_{0}^{2}}{\lVert{\mathbf{r}}-{\mathbf{s}}\rVert^{2}}+\frac{1/k_{0}^{4}}{\lVert{\mathbf{r}}-{\mathbf{s}}\rVert^{4}}$,
which equals $0.97$ at distance
$\lVert{\mathbf{r}}-{\mathbf{s}}\rVert=\lambda$. Hence, when considering
practical systems with $r_{k}\gg\lambda$, the last two terms in (5) can be
neglected; see [9] for a more detailed theoretical justification. We thus
neglect the reactive terms and approximate (5) as
${\mathsf{h}}_{\mathsf{em}}(\mathbf{r},{\mathbf{s}})\approx\frac{{\rm{j}}k_{0}\eta{\rm{e}}^{-{\rm{j}}k_{0}\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}}{4\pi\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}$.

##  III Single-User Case

Having established the system model, we next analyze the performance of CAPA-
based uplink communications. For clarity, we commence with the single-user
case.

By setting $K=1$ and dropping the user index $k$, the observed electric field
degenerates into

$\displaystyle{\mathsf{Y}}(\mathbf{r})=s{\mathsf{h}}(\mathbf{r},{\mathbf{s}}){\mathsf{J}}_{k}({\mathbf{s}})\lvert{\mathcal{A}}\rvert+{\mathsf{N}}(\mathbf{r}).$
(6)

We then design a detector ${\mathsf{v}}(\mathbf{r})$ to recover the data
information contained in $s$. Applying this detector to
${\mathsf{Y}}(\mathbf{r})$ yields

$\begin{split}\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}^{\mathsf{H}}(\mathbf{r}){\mathsf{Y}}(\mathbf{r}){\rm{d}}{\mathbf{r}}&=s{\mathsf{J}}_{k}({\mathbf{s}})\lvert{\mathcal{A}}\rvert\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}^{\mathsf{H}}(\mathbf{r}){\mathsf{h}}(\mathbf{r},{\mathbf{s}}){\rm{d}}{\mathbf{r}}\\\
&+\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}^{\mathsf{H}}(\mathbf{r}){\mathsf{N}}(\mathbf{r}){\rm{d}}{\mathbf{r}}.\end{split}$
(7)

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. ∎

Taken together, the single-to-noise ratio (SNR) for decoding $s$ is given by

$\begin{split}\gamma=\frac{\lvert{\mathsf{J}}({\mathbf{s}})\rvert^{2}\lvert{\mathcal{A}}\rvert^{2}\lvert\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}^{\mathsf{H}}(\mathbf{r}){\mathsf{h}}(\mathbf{r},{\mathbf{s}}){\rm{d}}{\mathbf{r}}\rvert^{2}}{\sigma^{2}\int_{{\mathcal{A}}_{\mathsf{R}}}\lvert{\mathsf{v}}(\mathbf{r})\rvert^{2}{\rm{d}}{\mathbf{r}}}.\end{split}$
(8)

The subsequent task is to find a detector that can maximize the SNR shown in
(11). By observing the mathematical structure of $\gamma$, we note that
$\gamma$ is independent of the norm of ${\mathsf{v}}(\mathbf{r})$, i.e.,
$\int_{{\mathcal{A}}_{\mathsf{R}}}\lvert{\mathsf{v}}(\mathbf{r})\rvert^{2}{\rm{d}}{\mathbf{r}}$.
On this basis, the problem of designing ${\mathsf{v}}(\mathbf{r})$ is
equivalently transformed as follows:

$\operatorname*{argmax}\nolimits_{{\mathsf{v}}(\mathbf{r})}{\left\lvert\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}^{\mathsf{H}}(\mathbf{r}){\mathsf{h}}(\mathbf{r},{\mathbf{s}}){\rm{d}}{\mathbf{r}}\right\rvert^{2}},$
(9)

whose optimal solution, i.e., the optimal detector, is given by

${\mathsf{v}}(\mathbf{r})={\mathsf{h}}(\mathbf{r},{\mathbf{s}}).$ (10)

######  Remark 1.

The result in (10) suggests that the capacity-achieving detector for single-
user CAPA communications aligns with the spatial channel response
${\mathsf{h}}(\mathbf{r},{\mathbf{s}})$, which can be regarded as continuous
version of the maximal-ratio combiner.

Substituting (10) back into (11) gives

$\begin{split}\gamma=\frac{\lvert{\mathsf{J}}({\mathbf{s}})\rvert^{2}\lvert{\mathcal{A}}\rvert^{2}}{\sigma^{2}}\int_{{\mathcal{A}}_{\mathsf{R}}}\lvert{\mathsf{h}}(\mathbf{r},{\mathbf{s}})\rvert^{2}{\rm{d}}{\mathbf{r}}=\overline{\gamma}\int_{{\mathcal{A}}_{\mathsf{R}}}\lvert{\mathsf{g}}(\mathbf{r},{\mathbf{s}})\rvert^{2}{\rm{d}}{\mathbf{r}},\end{split}$
(11)

where
$\overline{\gamma}\triangleq\frac{\lvert{\mathsf{J}}({\mathbf{s}})\rvert^{2}\lvert{\mathcal{A}}\rvert^{2}k_{0}^{2}\eta^{2}}{4\pi\sigma^{2}}$
and
${\mathsf{g}}(\mathbf{r},{\mathbf{s}})\triangleq\frac{{\rm{e}}^{-{\rm{j}}k_{0}\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}}{\sqrt{4\pi}\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}{\mathsf{h}}_{\mathsf{pa}}(\mathbf{r},{\mathbf{s}})$.
Note that
$\frac{{\rm{e}}^{-{\rm{j}}k_{0}\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}}{\sqrt{4\pi}\lVert{\mathbf{r}}-{\mathbf{s}}\rVert}$
characterizes the spherical-wave propagation in arbitrary homogeneous mediums,
and thus we term
$\int_{{\mathcal{A}}_{\mathsf{R}}}\lvert{\mathsf{g}}(\mathbf{r},{\mathbf{s}})\rvert^{2}{\rm{d}}{\mathbf{r}}\triangleq{\mathsf{a}}_{\mathsf{R}}$
as the array gain. The remaining term $\overline{\gamma}$ is termed as the
transmit SNR.

Building upon (11), the channel capacity under the single-user scenario can be
written as follows:

$\displaystyle{\mathsf{C}}=\log_{2}(1+\gamma)=\log_{2}(1+{\overline{\gamma}}{\mathsf{a}}_{\mathsf{R}}).$
(12)

We note that the channel capacity is mainly influenced by the receive aperture
$\mathsf{A}_{\mathsf{R}}$ as well as the user’s center location $\mathbf{s}$.

##  IV Multiuser Case

This section extends the single-user scenario to the more general multiuser
scenario, with the _sum-rate capacity_ as the performance metric of interest.
For brevity, our analyses focus on the two-user case. The extension to
scenarios with an arbitrary number of users will be left to future work.

###  IV-A Optimal Multiuser Detection

When $K=2$, the observed electric field in (3) can be rewritten as follows:

$\begin{split}{\mathsf{Y}}(\mathbf{r})&=s_{2}{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{2}){\mathsf{J}}_{2}({\mathbf{s}}_{2})\lvert{\mathcal{A}}_{2}\rvert\\\
&+s_{1}{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{1}){\mathsf{J}}_{1}({\mathbf{s}}_{1})\lvert{\mathcal{A}}_{1}\rvert+{\mathsf{N}}(\mathbf{r}).\end{split}$
(13)

The sum-rate capacity of an uplink multiuser channel can be achieved by using
_successive interference cancellation (SIC) decoding_ [1]. More specifically,
the message sent by one user is first decoded by treating the message from the
other user as interference and then removed, and the other message is decoded
without inter-user interference (IUI).

Given the uplink signal model (13), there exist two SIC orders: $1\rightarrow
2$ and $2\rightarrow 1$. In the sequel, we discuss the sum-rate capacity
achieved by the decoding order $2\rightarrow 1$. In this case, $s_{2}$ is
decoded from ${\mathsf{Y}}(\mathbf{r})$ by treating
$s_{1}{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{1}){\mathsf{J}}_{1}({\mathbf{s}}_{1})\lvert{\mathcal{A}}_{1}\rvert$
as interference. After $s_{2}$ is decoded and subtracted from
${\mathsf{Y}}(\mathbf{r})$, $s_{1}$ is decoded without IUI from the following
signal:

$\displaystyle{\mathsf{Z}}(\mathbf{r})\triangleq
s_{1}{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{1}){\mathsf{J}}_{1}({\mathbf{s}}_{1})\lvert{\mathcal{A}}_{1}\rvert+{\mathsf{N}}(\mathbf{r}),$
(14)

which is the same as that given in (6). Following the derivation steps in
obtaining (10), we know that the BS can use
${\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{1})$ as the detector to recover $s_{1}$
from (14). As a result, the SNR in decoding the message sent by user $1$ is
given by

$\displaystyle\gamma_{1}={\overline{\gamma}}_{1}{\mathsf{a}}_{1},$ (15)

where
$\overline{\gamma}_{1}=\frac{\lvert{\mathsf{J}}_{1}({\mathbf{s}}_{1})\rvert^{2}\lvert{\mathcal{A}}_{1}\rvert^{2}k_{0}^{2}\eta^{2}}{4\pi\sigma^{2}}$
is the transmit SNR of user $1$ and
${\mathsf{a}}_{1}=\int_{{\mathcal{A}}_{\mathsf{R}}}\lvert{\mathsf{g}}(\mathbf{r},{\mathbf{s}}_{1})\rvert^{2}{\rm{d}}{\mathbf{r}}$
is the associated channel gain.

We then investigate the decoding of $s_{2}$ from ${\mathsf{Y}}(\mathbf{r})$,
where the interference-plus-noise term is given by ${\mathsf{Z}}(\mathbf{r})$
shown in (14). Since ${\mathsf{N}}(\mathbf{r})$ and $s_{1}$ are uncorrelated,
the autocorrelation function of the random field ${\mathsf{Z}}(\mathbf{r})$ is
given by

$\begin{split}{\mathbbmss{E}}\\{{\mathsf{Z}}(\mathbf{r}_{1}){\mathsf{Z}}^{\mathsf{H}}(\mathbf{r}_{2})\\}&={\mathsf{g}}(\mathbf{r}_{1},{\mathbf{s}}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r}_{2},{\mathbf{s}}_{1})\overline{\gamma}_{1}\sigma^{2}\\\
&+{\sigma}^{2}\delta(\mathbf{r}_{1}-\mathbf{r}_{2})\triangleq{\mathscr{R}}_{{\mathsf{Z}}{\mathsf{Z}}}(\mathbf{r}_{1},\mathbf{r}_{2}).\end{split}$
(16)

In order to obtain the optimal detector that maximizes the achievable rate of
user $2$, we need to first design an invertible linear transformation
${\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r})$ to whiten
${\mathsf{Z}}(\mathbf{r})$, i.e., finding a
${\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r})$ that makes the
autocorrelation function of
$\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r}){\mathsf{Z}}(\mathbf{r}){\rm{d}}\mathbf{r}\triangleq{\mathsf{Z}}_{\mathsf{w}}(\mathbf{r}^{\prime})$
proportional to the Dirac delta function. To this end, we introduce the
following lemmas.

######  Lemma 2.

Let
${\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r})=\delta(\mathbf{r}^{\prime}-\mathbf{r})+\lambda{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{1})$
and
$\overline{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r})=\delta(\mathbf{r}^{\prime}-\mathbf{r})-\overline{\lambda}{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{1})$
with $\overline{\lambda}=\frac{\lambda}{1+\lambda{\mathsf{a}}_{1}}$. Then, it
has

$\int_{{\mathcal{A}}_{\mathsf{R}}}\overline{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime\prime},\mathbf{r}^{\prime})\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r}){\mathsf{u}}(\mathbf{r}){\rm{d}}\mathbf{r}{\rm{d}}\mathbf{r}^{\prime}={\mathsf{u}}(\mathbf{r}^{\prime\prime})$
(17)

for an arbitrary function ${\mathsf{u}}(\mathbf{r})$ defined in
$\mathbf{r}\in{\mathcal{A}}_{\mathsf{R}}$.

###### Proof:

Please refer to Appendix -B for more details. ∎

######  Lemma 3.

Let
${\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r})=\delta(\mathbf{r}^{\prime}-\mathbf{r})+\lambda{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{1})$
with
$\lambda=-\frac{1}{{\mathsf{a}}_{1}}\pm\frac{1}{{\mathsf{a}}_{1}\sqrt{1+\overline{\gamma}_{1}{\mathsf{a}}_{1}}}\triangleq\lambda^{\star}$.
Then, the autocorrelation function of
${\mathsf{Z}}_{\mathsf{w}}(\mathbf{r}^{\prime})$ satisfies
${\mathbbmss{E}}\\{{\mathsf{Z}}_{\mathsf{w}}(\mathbf{r}_{1}){\mathsf{Z}}_{\mathsf{w}}^{\mathsf{H}}(\mathbf{r}_{2})\\}={\sigma}^{2}\delta(\mathbf{r}_{1}-\mathbf{r}_{2})$.

###### Proof:

Please refer to Appendix -C for more details. ∎

######  Remark 2.

The results in Lemma 2 suggest that
${\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r})=\delta(\mathbf{r}^{\prime}-\mathbf{r})+\lambda{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{1})$
is an invertible linear transformation. Applying this transformation to
${\mathsf{Y}}(\mathbf{r})$ is information lossless and has no influence on the
channel capacity [1].

######  Remark 3.

The results in Lemma 3 suggest that setting $\lambda=\lambda^{\star}$ makes
${\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r})=\delta(\mathbf{r}^{\prime}-\mathbf{r})+\lambda{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{1})$
capable of whitening the interference-plus-noise term
$s_{1}{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{1}){\mathsf{J}}_{1}({\mathbf{s}}_{1})\lvert{\mathcal{A}}_{1}\rvert+{\mathsf{N}}(\mathbf{r})$.

Leveraging the insights unveiled in Remarks 2 and 3, we exploit
${\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r})=\delta(\mathbf{r}^{\prime}-\mathbf{r})+\lambda^{\star}{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{1})$
to transform ${\mathsf{Y}}(\mathbf{r})$ as follows:

$\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r}){\mathsf{Y}}(\mathbf{r}){\rm{d}}\mathbf{r}=s_{2}{\mathsf{J}}_{2}({\mathbf{s}}_{2})\lvert{\mathcal{A}}_{2}\rvert\overline{\mathsf{h}}(\mathbf{r}^{\prime},{\mathbf{s}}_{2})+{\mathsf{Z}}_{\mathsf{w}}(\mathbf{r}^{\prime}),$
(18)

where
$\overline{\mathsf{h}}(\mathbf{r}^{\prime})\triangleq\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r}){\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{2}){\rm{d}}\mathbf{r}$.
Based on Lemma 3, we have
${\mathbbmss{E}}\\{{\mathsf{Z}}_{\mathsf{w}}(\mathbf{r}_{1}){\mathsf{Z}}_{\mathsf{w}}^{\mathsf{H}}(\mathbf{r}_{2})\\}={\sigma}^{2}\delta(\mathbf{r}_{1}-\mathbf{r}_{2})$.
As a result, the model presented in (18) is similar to that (6). Motivated by
this, we next design the maximal-ratio combining (MRC)-based detector
$\overline{\mathsf{v}}(\mathbf{r}^{\prime})$ by treating
$\overline{\mathsf{h}}(\mathbf{r}^{\prime})$ as the equivalent channel
response, which yields

$\overline{\mathsf{v}}(\mathbf{r}^{\prime})=\overline{\mathsf{h}}(\mathbf{r}^{\prime})=\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r}){\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{2}){\rm{d}}\mathbf{r}.$
(19)

The resultant SNR is given by

$\begin{split}\gamma_{2}=\frac{\lvert{\mathsf{J}}_{2}({\mathbf{s}}_{2})\rvert^{2}\lvert{\mathcal{A}}_{2}\rvert^{2}}{\sigma^{2}}\int_{{\mathcal{A}}_{\mathsf{R}}}\lvert\overline{\mathsf{h}}(\mathbf{r}^{\prime})\rvert^{2}{\rm{d}}{\mathbf{r}}^{\prime}.\end{split}$
(20)

A closed-form expression for $\gamma_{2}$ is given as follows.

######  Theorem 1.

By first whitening ${\mathsf{Z}}(\mathbf{r})$ with
${\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r})$ and then
employing the MRC detector $\overline{\mathsf{v}}(\mathbf{r}^{\prime})$, the
resultant SNR in decoding $s_{2}$ can be written as follows:

$\begin{split}\gamma_{2}=\overline{\gamma}_{2}({\mathsf{a}}_{2}+\lvert\rho\rvert^{2}({\lambda^{\star}}^{2}{\mathsf{a}}_{1}+2\lambda^{\star})),\end{split}$
(21)

where
$\overline{\gamma}_{2}=\frac{\lvert{\mathsf{J}}_{2}({\mathbf{s}}_{2})\rvert^{2}\lvert{\mathcal{A}}_{2}\rvert^{2}k_{0}^{2}\eta^{2}}{4\pi\sigma^{2}}$
is the transmit SNR of user $2$,
${\mathsf{a}}_{2}=\int_{{\mathcal{A}}_{\mathsf{R}}}\lvert{\mathsf{g}}(\mathbf{r},{\mathbf{s}}_{2})\rvert^{2}{\rm{d}}{\mathbf{r}}$
is the associated channel gain, and
$\rho=\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{g}}^{\mathsf{H}}({\mathbf{r}},{\mathbf{s}}_{1}){\mathsf{g}}({\mathbf{r}},{\mathbf{s}}_{2}){\rm{d}}{\mathbf{r}}$.

###### Proof:

Please refer to Appendix -D for more details. ∎

After obtaining the capacity-achieving detectors under the SIC order
$2\rightarrow 1$, we summarize the entire decoding procedure in Table I on the
bottom of next page for ease of reference. The presented decoding procedure
directly extends to the other SIC order $1\rightarrow 2$ by exchanging the
user indices.

1:Whiten the interference-plus-noise term ${\mathsf{Z}}(\mathbf{r})$:
$\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r}){\mathsf{Y}}(\mathbf{r}){\rm{d}}\mathbf{r}=s_{2}{\mathsf{J}}_{2}({\mathbf{s}}_{2})\lvert{\mathcal{A}}_{2}\rvert\overline{\mathsf{h}}(\mathbf{r}^{\prime},{\mathbf{s}}_{2})+{\mathsf{Z}}_{\mathsf{w}}(\mathbf{r}^{\prime})$
2:Use the MRC detector and ML decoder to recover $s_{2}$:
$\int_{{\mathcal{A}}_{\mathsf{R}}}{\overline{\mathsf{v}}}^{\mathsf{H}}(\mathbf{r}^{\prime})\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r}){\mathsf{Y}}(\mathbf{r}){\rm{d}}\mathbf{r}{\rm{d}}\mathbf{r}^{\prime}\rightarrow
s_{2}$ 3:Employ SIC to subtract
$s_{2}{\mathsf{J}}_{2}({\mathbf{s}}_{2})\lvert{\mathcal{A}}_{2}\rvert\overline{\mathsf{h}}(\mathbf{r}^{\prime},{\mathbf{s}}_{2})$
from ${\mathsf{Y}}(\mathbf{r})$ 4:Use the MRC detector and ML decoder to
recover $s_{1}$:
$\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{h}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{1})(s_{1}{\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{1}){\mathsf{J}}_{1}({\mathbf{s}}_{1})\lvert{\mathcal{A}}_{1}\rvert+{\mathsf{N}}(\mathbf{r})){\rm{d}}\mathbf{r}\rightarrow
s_{1}$
---
TABLE I: SIC decoding for CAPA communications.

###  IV-B Sum-Rate Capacity Analysis

After obtaining the decoding SNRs for $s_{1}$ and $s_{2}$, the achievable
rates of user $1$ and user $2$ can be calculated as

$\displaystyle{\mathsf{R}}_{1}$
$\displaystyle=\log_{2}(1+\gamma_{1})=\log_{2}(1+{\overline{\gamma}}_{1}{\mathsf{a}}_{1}),$
(22) $\displaystyle{\mathsf{R}}_{2}$ $\displaystyle=\log_{2}(1+\gamma_{2})$
$\displaystyle=\log_{2}(1+\overline{\gamma}_{2}({\mathsf{a}}_{2}+\lvert\rho\rvert^{2}({\lambda^{\star}}^{2}{\mathsf{a}}_{1}+2\lambda^{\star}))),$
(23)

respectively. The sum-rate capacity is given as follows.

######  Theorem 2.

The sum-rate capacity achieved under the SIC order $2\rightarrow 1$ is given
by

$\begin{split}{\mathsf{C}}=\log_{2}(1+{\overline{\gamma}}_{1}{\mathsf{a}}_{1}+{\overline{\gamma}}_{2}{\mathsf{a}}_{2}+{\overline{\gamma}}_{1}{\overline{\gamma}}_{2}{\mathsf{a}}_{1}{\mathsf{a}}_{2}(1-\lvert\rho_{\mathsf{u}}\rvert^{2})),\end{split}$
(24)

where $\rho_{\mathsf{u}}=\frac{\rho}{\sqrt{{\mathsf{a}}_{1}{\mathsf{a}}_{2}}}$
represents the channel correlation factor between user $1$ and user $2$.

###### Proof:

Please refer to Appendix -E for more details. ∎

Following the same derivation steps in obtaining (24), we can also obtain the
sum-rate capacity achieved under the SIC order $1\rightarrow 2$, which is
given as follows.

######  Corollary 1.

Under the SIC order $1\rightarrow 2$, the achievable rates of user 1 and user
2 as well as the sum-rate capacity can be expressed as follows:

$\displaystyle{\mathsf{R}}_{1}$
$\displaystyle=\log_{2}(1+\overline{\gamma}_{1}({\mathsf{a}}_{1}+\lvert\rho\rvert^{2}({\lambda^{\bullet}}^{2}{\mathsf{a}}_{2}+2\lambda^{\bullet}))),$
(25) $\displaystyle{\mathsf{R}}_{2}$
$\displaystyle=\log_{2}(1+{\overline{\gamma}}_{2}{\mathsf{a}}_{2}),$ (26)
$\displaystyle{\mathsf{C}}$
$\displaystyle=\log_{2}(1+{\overline{\gamma}}_{1}{\mathsf{a}}_{1}+{\overline{\gamma}}_{2}{\mathsf{a}}_{2}+{\overline{\gamma}}_{1}{\overline{\gamma}}_{2}{\mathsf{a}}_{1}{\mathsf{a}}_{2}(1-\lvert\rho_{\mathsf{u}}\rvert^{2})),$
(27)

where
$\lambda^{\bullet}=-\frac{1}{{\mathsf{a}}_{2}}\pm\frac{1}{{\mathsf{a}}_{2}\sqrt{1+\overline{\gamma}_{2}{\mathsf{a}}_{2}}}$.

###### Proof:

Similar to the proof of Theorem 2. ∎

By comparing (24) with (27), we find the following result.

######  Theorem 3.

The sum-rate capacity of the considered uplink CAPA-based channel is always
the same, i.e.,

$\displaystyle{\mathsf{C}}=\log_{2}(1+{\overline{\gamma}}_{1}{\mathsf{a}}_{1}+{\overline{\gamma}}_{2}{\mathsf{a}}_{2}+{\overline{\gamma}}_{1}{\overline{\gamma}}_{2}{\mathsf{a}}_{1}{\mathsf{a}}_{2}(1-\lvert\rho_{\mathsf{u}}\rvert^{2})),$
(28)

no matter which decoding order is used.

######  Remark 4.

By taking a further look at the expression of the sum-rate capacity (28), we
note that it is determined by the channel gain of each user and the channel
correlation factor, all of which are influenced by the receive aperture
${\mathcal{A}}_{\mathsf{R}}$ and the user locations
$\\{{\mathbf{s}}_{k}\\}_{k=1}^{2}$. This expression applies to an arbitrary
aperture regardless of its location, shape, and size.

###  IV-C Special Cases

The sum-rate capacity characterized in (24) or Corollary 1 applies to an
arbitrary aperture regardless of its location, shape, and size, among others.
In the sequel, we specialize ${\mathcal{A}}_{\mathsf{R}}$ to several special
cases to unveil more system insights.

####  IV-C1 Planar CAPAs

Let us first consider the case where the CAPA is a contiguous-aperture planar
array placed on the $x$-$z$ plane with edges parallel to the axes and physical
dimensions $L_{x}$ and $L_{z}$ along the $x$\- and $z$-axes, as shown in [9,
Figure 2(b)]. In this case, we have
${\mathcal{A}}_{\mathsf{R}}=\\{[x,0,z]|x\in[-\frac{L_{x}}{2},\frac{L_{x}}{2}],z\in[-\frac{L_{z}}{2},\frac{L_{z}}{2}]\\}$
and ${\mathbf{e}}_{{\mathbf{r}}}=[0,1,0]^{\mathsf{T}}$, which yields

$\displaystyle{\mathsf{g}}(\mathbf{r},{\mathbf{s}}_{k})=\frac{\sqrt{r_{k}\Psi_{k}}{\rm{e}}^{-{\rm{j}}k_{0}((x-r_{k}\Phi_{k})^{2}+{\Psi}_{k}^{2}+(z-\Theta_{k})^{2})^{\frac{1}{2}}}}{\sqrt{4\pi}((x-r_{k}\Phi_{k})^{2}+{\Psi}_{k}^{2}+(z-\Theta_{k})^{2})^{\frac{3}{2}}}.$
(29)

And the channel gain is calculated as follows.

######  Corollary 2.

When
${\mathcal{A}}_{\mathsf{R}}=\\{[x,0,z]|x\in[-\frac{L_{x}}{2},\frac{L_{x}}{2}],z\in[-\frac{L_{z}}{2},\frac{L_{z}}{2}]\\}$
and ${\mathbf{e}}_{{\mathbf{r}}}=[0,1,0]^{\mathsf{T}}$, the channel gain is
give by

$\displaystyle{\mathsf{a}}_{k}=\frac{1}{4\pi}\sum_{x\in{\mathcal{X}}_{k}}\sum_{z\in{\mathcal{Z}}_{k}}\arctan\bigg{(}\frac{xz/\Psi_{k}}{\sqrt{\Psi_{k}^{2}+x^{2}+z^{2}}}\bigg{)}\triangleq{\mathsf{a}}_{k}^{\mathsf{c}},$
(30)

where ${\mathcal{X}}_{k}\triangleq\\{\frac{L_{x}}{2r_{k}}\pm\Phi_{k}\\}$ and
${\mathcal{Z}}_{k}\triangleq\\{\frac{L_{z}}{2r_{k}}\pm\Theta_{k}\\}$.

###### Proof:

Please refer to [2] for more details. ∎

However, deriving a closed-form expression for the correlation factor
$\rho_{\mathsf{u}}$ is a challenging task. As is widely known, the correlation
factor decreases with the aperture size and
$\lim_{\lvert{\mathcal{A}}_{\mathsf{R}}\rvert\rightarrow\infty}\lvert\rho_{\mathsf{u}}\rvert\approx
0$ [10]. Upon neglecting the channel correlation, we approximate the sum-rate
capacity as follows:

$\begin{split}{\mathsf{C}}&\approx\log_{2}(1+{\overline{\gamma}}_{1}{\mathsf{a}}_{1}+{\overline{\gamma}}_{2}{\mathsf{a}}_{2}+{\overline{\gamma}}_{1}{\overline{\gamma}}_{2}{\mathsf{a}}_{1}{\mathsf{a}}_{2})\\\
&=\log_{2}(1+{\overline{\gamma}}_{1}{\mathsf{a}}_{1})+\log_{2}(1+{\overline{\gamma}}_{2}{\mathsf{a}}_{2}),\end{split}$
(31)

which serves as an upper bound of the sum-rate capacity. By considering a
limiting case where the aperture is infinitely large, i.e.,
$L_{x},L_{z}\rightarrow\infty$, we have
$\lim_{L_{x},L_{z}\rightarrow\infty}{\mathsf{a}}_{k}^{\mathsf{c}}=\frac{1}{4\pi}\frac{4\pi}{2}=\frac{1}{2}$,
and the asymptotic sum-rate capacity is given as follows:

$\begin{split}{\mathsf{C}}\approx\log_{2}(1+{\overline{\gamma}}_{1}/2)+\log_{2}(1+{\overline{\gamma}}_{2}/2).\end{split}$
(32)

####  IV-C2 SPD Planar Arrays

We then consider a case where the above planar CAPA is partitioned into
$M=M_{z}M_{x}$ SPD elements, where $M_{x}=2\tilde{M}_{x}+1$ and
$M_{z}=2\tilde{M}_{z}+1$ denote the number of antenna elements along the $x$\-
and $z$-axes, as depicted in [9, Figure 2(a)]. The physical dimensions of each
element along the $x$\- and $z$-axes are indicated by $\sqrt{A}$, and the
inter-element distance is denoted as $d$, where $d\geq\sqrt{A}$. In this case,
we have $L_{x}\approx M_{x}d$, $L_{z}\approx M_{z}d$, and
${\mathcal{A}}_{\mathsf{R}}=\\{(m_{x}d+\ell,0,m_{z}d+\ell)|\ell\in[-\frac{\sqrt{A}}{2},\frac{\sqrt{A}}{2}],m_{x}\in{\mathcal{M}}_{x},m_{z}\in{\mathcal{M}}_{z}\\}$,
where ${\mathcal{M}}_{x}\triangleq\\{0,\pm 1,\ldots,\pm\tilde{M}_{x}\\}$ and
${\mathcal{M}}_{z}\triangleq\\{0,\pm 1,\ldots,\pm\tilde{M}_{z}\\}$. Under this
condition that $\sqrt{A}\ll r_{k}$ and $d\ll r_{k}$, we can approximate the
corresponding channel gain as follows [9]:

${\mathsf{a}}_{k}\approx\mu_{\mathsf{oc}}{\mathsf{a}}_{k}^{\mathsf{c}}\triangleq{\mathsf{a}}_{k}^{\mathsf{s}},$
(33)

where $\mu_{\mathsf{oc}}\triangleq\frac{A}{d^{2}}\leq 1$ represents the array
occupation ratio. On this basis, we obtain
$\lim_{L_{x},L_{z}\rightarrow\infty}{\mathsf{a}}_{k}^{\mathsf{s}}=\frac{\mu_{\mathsf{oc}}}{2}$,
and the asymptotic sum-rate capacity is given as follows:

$\begin{split}{\mathsf{C}}&\approx\log_{2}(1+\mu_{\mathsf{oc}}{\overline{\gamma}}_{1}/2)+\log_{2}(1+\mu_{\mathsf{oc}}{\overline{\gamma}}_{2}/2)\\\
&\leq\log_{2}(1+{\overline{\gamma}}_{1}/2)+\log_{2}(1+{\overline{\gamma}}_{2}/2).\end{split}$
(34)

Comparing (32) with (34) yields the following observations.

######  Remark 5.

The capacity achieved an SPD array converges to that achieved by a CAPA when
$\mu_{\mathsf{oc}}=1$. This makes intuitive sense as a CAPA is a special case
of an SPD array when the array occupation ratio equals one.

##  V Numerical Results

In this section, we conduct computer simulations to analyze the performance of
CAPA communications. All simulations utilize planar arrays with all edges
parallel to the axes. Unless explicitly stated otherwise, we set the
parameters as follows: ${\mathbf{e}}_{{\mathbf{r}}}=[0,1,0]^{\mathsf{T}}$,
$\overline{\gamma}_{1}=30$ dB, $\overline{\gamma}_{2}=40$ dB,
$\theta_{1}=\theta_{2}=\frac{\pi}{6}$, $\phi_{1}=\phi_{2}=\frac{\pi}{3}$,
$r_{1}=10$ m, $r_{2}=20$ m, $\lambda=0.0107$ m, $d=\frac{\lambda}{2}$,
$A=\frac{\lambda^{2}}{4\pi}$, $M_{x}=M_{z}$, and $L_{x}=L_{z}$.

(a) CAPAs.

(b) SPD arrays.

Figure 2: Transmission rates achieved by CAPAs and SPD arrays.

Figure 2(a) and Figure 2(b) plot the transmission rates of user $1$
(${\mathsf{R}}_{1}$) and user $2$ ($\mathsf{R}_{2}$) under SIC decoding order
$1\rightarrow 2$ and $2\rightarrow 1$, as well as the sum-rate capacity
($\mathsf{C}$) and its upper bound (sum-rate without considering the IUI)
achieved by the CAPA and the SPD array, respectively. It can be observed that
with the increase in the aperture size
$\lvert{\mathcal{A}}_{\mathsf{R}}\rvert$, all presented rates experience an
increase. This phenomenon occurs because a larger aperture size leads to a
higher channel gain and a lower channel correlation factor, thereby enhancing
the dedicated signal while diminishing the IUI. This also causes the sum-rate
capacity to approach its upper bound, as derived in (32) and (34), which
aligns with the findings illustrated in Figure 2.

(a) Sum-rate capacity vs. $\mu_{\mathsf{oc}}$.

(b) Sum-rate capacity vs. $\lvert{\mathcal{A}}_{\mathsf{R}}\rvert$.

Figure 3: Sum-rate capacity of CAPAs and SPD arrays.

Upon comparing Figure 2(a) with Figure 2(b), we note that the CAPA achieves a
higher sum-rate capacity than an SPD array. Our attention then turns to
examining the changing trend of the sum-rate capacity achieved by SPD arrays
versus the array occupation ratio $\mu_{\mathsf{oc}}$, which is illustrated in
Figure 3(a) by fixing $d=\frac{\lambda}{2}$ and changing $A$. We observe that
as the array occupation ratio increases, the sum-rate capacity achieved by an
SPD array gradually converges to that achieved by a CAPA. This observation
further validates the conclusion drawn in Remark 5. Additionally, the results
depicted in Figure 3(a) suggest that increasing the aperture size can enhance
the channel capacity. To delve deeper into the capacity limit as
$\lvert{\mathcal{A}}_{\mathsf{R}}\rvert$ approaches infinity, we plot the sum-
rate capacity against the aperture size in Figure 3(b). As expected, the
channel capacity converges to a constant value, i.e., the asymptotic value
(calculated by (32) or (34)), in the larger limit of
$\lvert{\mathcal{A}}_{\mathsf{R}}\rvert$, which adheres to the law of energy
conservation [2].

(a) Capacity region vs. $\mu_{\mathsf{oc}}$. $d=\frac{\lambda}{2}$

(b) Capacity region vs. $\lvert{\mathcal{A}}_{\mathsf{R}}\rvert$.

Figure 4: Capacity regions of CAPAs and SPD arrays.

Figure 4 illustrates the capacity regions achieved by the CAPA and SPD array,
where the red and green points represent the achieved rates by _SIC decoding_
order $1\rightarrow 2$ and $2\rightarrow 1$, respectively. The rate tuple on
the line segment connecting these two points is achieved by the _time-sharing
strategy_. As depicted in Figure 4(a), the CAPA yields a broader capacity
region than the SPD array, and their gap decreases with increasing
$\mu_{\mathsf{oc}}$, which is consistent with the results shown in Figure
3(a). In Figure 4(b), the capacity regions achieved by the CAPA for various
values of $\lvert{\mathcal{A}}_{\mathsf{R}}\rvert$ are illustrated. It is
observed that the capacity region gradually extends as
$\lvert{\mathcal{A}}_{\mathsf{R}}\rvert$ increases, which transitions from a
pentagon to a rectangular shape. This phenomenon occurs because the IUI
gradually decreases as the aperture size increases, which aligns with the
results shown in Figure 2(a). These findings underscore the superiority of
CAPAs over traditional SPD arrays in terms of channel capacity.

##  VI Conclusion

We have proposed an analytically tractable framework for CAPA-based wireless
communications. Utilizing this framework, we derived the channel capacity for
an uplink CAPA-based system, along with the associated decoding principle. We
have demonstrated through both theoretical analyses and numerical simulations
that CAPA exhibits a higher sum-rate capacity and a broader capacity region
compared to the traditional SPD array. These results highlight CAPA’s
potential as a promising transmission paradigm for future wireless networks.

###  -A Proof of Lemma 1

The mean of
$\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}^{\mathsf{H}}(\mathbf{r}){\mathsf{N}}(\mathbf{r}){\rm{d}}{\mathbf{r}}$
is calculated as follows:

$\begin{split}{\mathbbmss{E}}\left\\{\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}^{\mathsf{H}}(\mathbf{r}){\mathsf{N}}(\mathbf{r}){\rm{d}}{\mathbf{r}}\right\\}=\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}^{\mathsf{H}}(\mathbf{r}){\mathbbmss{E}}\\{{\mathsf{N}}(\mathbf{r})\\}{\rm{d}}{\mathbf{r}},\end{split}$
(35)

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

$\begin{split}&{\mathbbmss{E}}\left\\{\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}^{\mathsf{H}}(\mathbf{r}){\mathsf{N}}(\mathbf{r}){\rm{d}}{\mathbf{r}}\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}(\mathbf{r}^{\prime}){\mathsf{N}}^{\mathsf{H}}(\mathbf{r}^{\prime}){\rm{d}}{\mathbf{r}}^{\prime}\right\\}\\\
&=\int_{{\mathcal{A}}_{\mathsf{R}}}\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}^{\mathsf{H}}(\mathbf{r}){\mathsf{v}}(\mathbf{r}^{\prime}){\mathbbmss{E}}\\{{\mathsf{N}}(\mathbf{r}){\mathsf{N}}^{\mathsf{H}}(\mathbf{r}^{\prime})\\}{\rm{d}}{\mathbf{r}}{\rm{d}}{\mathbf{r}}^{\prime}\\\
&=\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}(\mathbf{r}^{\prime})\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}^{\mathsf{H}}(\mathbf{r})\sigma^{2}\delta(\mathbf{r}-\mathbf{r}^{\prime}){\rm{d}}{\mathbf{r}}{\rm{d}}{\mathbf{r}}^{\prime}.\end{split}$
(36)

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

$\eqref{Noise_Field_Covariance}=\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{v}}(\mathbf{r}^{\prime})(\sigma^{2}{\mathsf{v}}^{\mathsf{H}}(\mathbf{r}^{\prime})){\rm{d}}{\mathbf{r}}^{\prime}=\sigma^{2}\int_{{\mathcal{A}}_{\mathsf{R}}}\lvert{\mathsf{v}}(\mathbf{r})\rvert^{2}{\rm{d}}{\mathbf{r}}.$
(37)

The lemma is thus proved.

###  -B Proof of Lemma 2

By virtue of changing the order of integration in
$\int_{{\mathcal{A}}_{\mathsf{R}}}\overline{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime\prime},\mathbf{r}^{\prime})\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r}){\mathsf{u}}(\mathbf{r}){\rm{d}}\mathbf{r}{\rm{d}}\mathbf{r}^{\prime}$,
we obtain

$\begin{split}&\int_{{\mathcal{A}}_{\mathsf{R}}}\overline{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime\prime},\mathbf{r}^{\prime})\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r}){\mathsf{u}}(\mathbf{r}){\rm{d}}\mathbf{r}{\rm{d}}\mathbf{r}^{\prime}\\\
&=\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{u}}(\mathbf{r})\int_{{\mathcal{A}}_{\mathsf{R}}}\overline{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime\prime},\mathbf{r}^{\prime}){\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r}){\rm{d}}\mathbf{r}^{\prime}{\rm{d}}\mathbf{r}.\end{split}$
(38)

By inserting
${\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r})=\delta(\mathbf{r}^{\prime}-\mathbf{r})+\lambda{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{1})$
and
$\overline{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r})=\delta(\mathbf{r}^{\prime}-\mathbf{r})-\overline{\lambda}{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{1})$
into (38) as well as calculating the resultant integral using the fact that
$\int_{{\mathcal{A}}_{\mathsf{R}}}\delta({\mathbf{x}}-{\mathbf{x}}_{0})f(\mathbf{x}){\rm{d}}{\mathbf{x}}=f({\mathbf{x}}_{0})$
and
${\mathsf{a}}_{1}=\int_{{\mathcal{A}}_{\mathsf{R}}}\lvert{\mathsf{g}}(\mathbf{r},{\mathbf{s}}_{1})\rvert^{2}{\rm{d}}{\mathbf{r}}$,
we obtain

$\begin{split}&\int_{{\mathcal{A}}_{\mathsf{R}}}\overline{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime\prime},\mathbf{r}^{\prime}){\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r}){\rm{d}}\mathbf{r}^{\prime}=\delta({\mathbf{r}}^{\prime\prime}-{\mathbf{r}})\\\
&-(\overline{\lambda}-\lambda+\lambda\overline{\lambda}{\mathsf{a}}_{1}){\mathsf{g}}(\mathbf{r}^{\prime\prime},\mathbf{s}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r}^{\prime},\mathbf{s}_{1}).\end{split}$
(39)

Inserting $\overline{\lambda}=\frac{\lambda}{1+\lambda{\mathsf{a}}_{1}}$ into
$\overline{\lambda}-\lambda+\lambda\overline{\lambda}{\mathsf{a}}_{1}$ gives

$\overline{\lambda}-\lambda+\lambda\overline{\lambda}{\mathsf{a}}_{1}=\overline{\lambda}(1+\lambda{\mathsf{a}}_{1})-\lambda=0,$
(40)

which, together with the fact in (39), yields

$\begin{split}\int_{{\mathcal{A}}_{\mathsf{R}}}\overline{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime\prime},\mathbf{r}^{\prime}){\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r}){\rm{d}}\mathbf{r}^{\prime}=\delta({\mathbf{r}}^{\prime\prime}-{\mathbf{r}}).\end{split}$
(41)

Substituting (41) into (38) gives

$\begin{split}&\int_{{\mathcal{A}}_{\mathsf{R}}}\overline{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime\prime},\mathbf{r}^{\prime})\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r}){\mathsf{u}}(\mathbf{r}){\rm{d}}\mathbf{r}{\rm{d}}\mathbf{r}^{\prime}\\\
&=\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{u}}(\mathbf{r})\delta({\mathbf{r}}^{\prime\prime}-{\mathbf{r}}){\rm{d}}\mathbf{r}={\mathsf{u}}(\mathbf{r}^{\prime\prime}).\end{split}$
(42)

The lemma is thus proved.

###  -C Proof of Lemma 3

The autocorrelation function of ${\mathsf{Z}}_{\mathsf{w}}(\cdot)$ satisfies

$\displaystyle{\mathbbmss{E}}\\{{\mathsf{Z}}_{\mathsf{w}}(\mathbf{r}_{1}){\mathsf{Z}}_{\mathsf{w}}^{\mathsf{H}}(\mathbf{r}_{2})\\}$
$\displaystyle=\int_{{\mathcal{A}}_{\mathsf{R}}}\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}_{1},\mathbf{r}){\mathscr{K}}_{{\mathsf{Z}}}^{\mathsf{H}}(\mathbf{r}_{2},\mathbf{r}^{\prime}){\mathbbmss{E}}\\{{\mathsf{Z}}(\mathbf{r}){\mathsf{Z}}^{\mathsf{H}}(\mathbf{r}^{\prime})\\}{\rm{d}}\mathbf{r}{\rm{d}}\mathbf{r}^{\prime}$
$\displaystyle=\sigma^{2}\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}^{\mathsf{H}}(\mathbf{r}_{2},\mathbf{r}^{\prime})\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}_{1},\mathbf{r}){\mathscr{R}}_{{\mathsf{Z}}{\mathsf{Z}}}(\mathbf{r},\mathbf{r}^{\prime}){\rm{d}}\mathbf{r}{\rm{d}}\mathbf{r}^{\prime}.$
(43)

By substituting
${\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}^{\prime},\mathbf{r})=\delta(\mathbf{r}^{\prime}-\mathbf{r})+\lambda{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{1})$
and (16) into (43) and then calculating the resultant double integral with
respect to $\mathbf{r}$, we obtain

$\begin{split}&\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathscr{K}}_{{\mathsf{Z}}}(\mathbf{r}_{1},\mathbf{r}){\mathscr{R}}_{{\mathsf{Z}}{\mathsf{Z}}}(\mathbf{r},\mathbf{r}^{\prime}){\rm{d}}\mathbf{r}=\sigma^{2}(\delta({\mathbf{r}}_{1}-{\mathbf{r}}^{\prime})\\\
&+(\overline{\gamma}_{1}+\lambda+\lambda\overline{\gamma}_{1}{\mathsf{a}}_{1}){\mathsf{g}}(\mathbf{r}_{1},\mathbf{s}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r}^{\prime},\mathbf{s}_{1})).\end{split}$
(44)

We next calculate the integral in terms of ${{\mathbf{r}}}^{\prime}$, which
yields

$\begin{split}&\int_{\mathcal{A}_{\mathsf{S}}}\eqref{Denominator_Calculation_Forward_1}\times(\delta(\mathbf{r}_{2}-\mathbf{r}^{\prime})+\lambda{\mathsf{g}}^{\mathsf{H}}(\mathbf{r}_{2},{\mathbf{s}}_{1}){\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{1})){\rm{d}}{{\mathbf{r}}^{\prime}}\\\
&=\sigma^{2}(\delta({\mathbf{r}}_{1}-{\mathbf{r}}_{2})+{\mathsf{g}}(\mathbf{r}_{1},\mathbf{s}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r}_{2},\mathbf{s}_{1})\times(2\lambda+\overline{\gamma}_{1}\\\
&+2\lambda\overline{\gamma}_{1}{{\mathsf{a}}_{1}}+\lambda^{2}{{\mathsf{a}}_{1}}+\overline{\gamma}_{1}\lambda^{2}{{\mathsf{a}}_{1}^{2}})).\end{split}$
(45)

It is readily shown that the solutions to the equation
$2\lambda+\overline{\gamma}_{1}+2\lambda\overline{\gamma}_{1}{{\mathsf{a}}_{1}}+\lambda^{2}{{\mathsf{a}}_{1}}+\overline{\gamma}_{1}\lambda^{2}{{\mathsf{a}}_{1}^{2}}=0$
are given by
$\lambda=-\frac{1}{{\mathsf{a}}_{1}}\pm\frac{1}{{\mathsf{a}}_{1}\sqrt{1+\overline{\gamma}_{1}{\mathsf{a}}_{1}}}$.
Hence, setting
$\lambda=-\frac{1}{{\mathsf{a}}_{1}}\pm\frac{1}{{\mathsf{a}}_{1}\sqrt{1+\overline{\gamma}_{1}{\mathsf{a}}_{1}}}$
gives
${\mathbbmss{E}}\\{{\mathsf{Z}}_{\mathsf{w}}(\mathbf{r}_{1}){\mathsf{Z}}_{\mathsf{w}}^{\mathsf{H}}(\mathbf{r}_{2})\\}=\sigma^{2}\delta({\mathbf{r}}_{1}-{\mathbf{r}}_{2})$.
The lemma is thus proved.

###  -D Proof of Theorem 1

By definition, $\overline{\mathsf{h}}(\mathbf{r}^{\prime})$ can be calculated
as follows:

$\displaystyle\overline{\mathsf{h}}(\mathbf{r}^{\prime})$
$\displaystyle=\int_{{\mathcal{A}}_{\mathsf{R}}}(\delta(\mathbf{r}^{\prime}-\mathbf{r})+\lambda^{\star}{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{1}){\mathsf{g}}^{\mathsf{H}}(\mathbf{r},{\mathbf{s}}_{1})){\mathsf{h}}(\mathbf{r},{\mathbf{s}}_{2}){\rm{d}}\mathbf{r}$
$\displaystyle={{\rm{j}}k_{0}\eta}/{\sqrt{4\pi}}({\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{2})+\lambda^{\star}{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{1})\rho).$
(46)

Inserting (46) into (20) gives

$\begin{split}\gamma_{2}&=\overline{\gamma}_{2}\int_{{\mathcal{A}}_{\mathsf{R}}}\lvert{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{2})+\lambda^{\star}\rho{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{1})\rvert^{2}{\rm{d}}{\mathbf{r}}^{\prime}\\\
&=\overline{\gamma}_{2}\int_{{\mathcal{A}}_{\mathsf{R}}}\lvert{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{2})\rvert^{2}{\rm{d}}{\mathbf{r}}^{\prime}+\overline{\gamma}_{2}\int_{{\mathcal{A}}_{\mathsf{R}}}\lvert{\mathsf{g}}(\mathbf{r}^{\prime},{\mathbf{s}}_{2})\rvert^{2}{\rm{d}}{\mathbf{r}}^{\prime}\\\
&\times{\lambda^{\star}}^{2}\lvert\rho\rvert^{2}+2\overline{\gamma}_{2}\lambda^{\star}\Re\left\\{\rho\int_{{\mathcal{A}}_{\mathsf{R}}}{\mathsf{g}}({\mathbf{r}},{\mathbf{s}}_{1}){\mathsf{g}}^{\mathsf{H}}({\mathbf{r}},{\mathbf{s}}_{2}){\rm{d}}{\mathbf{r}}\right\\}\\\
&=\overline{\gamma}_{2}({\mathsf{a}}_{2}+\lvert\rho\rvert^{2}({\lambda^{\star}}^{2}{\mathsf{a}}_{1}+2\lambda^{\star})).\end{split}$
(47)

The theorem is thus proved.

###  -E Proof of Theorem 2

Since $\lambda=\lambda^{\star}$ is the solution to
$2\lambda+\overline{\gamma}_{1}+2\lambda\overline{\gamma}_{1}{{\mathsf{a}}_{1}}+\lambda^{2}{{\mathsf{a}}_{1}}+\overline{\gamma}_{1}\lambda^{2}{{\mathsf{a}}_{1}^{2}}=0$,
we have

$\begin{split}2\lambda^{\star}+{\lambda^{\star}}^{2}{{\mathsf{a}}_{1}}&=-\overline{\gamma}_{1}-2\lambda^{\star}\overline{\gamma}_{1}{{\mathsf{a}}_{1}}-\overline{\gamma}_{1}{\lambda^{\star}}^{2}{{\mathsf{a}}_{1}^{2}}\\\
&=-\overline{\gamma}_{1}(1+\lambda^{\star}{\mathsf{a}}_{1})^{2},\end{split}$
(48)

which, together with (23), yields

$\displaystyle{\mathsf{R}}_{2}=\log_{2}(1+\overline{\gamma}_{2}({\mathsf{a}}_{2}-\lvert\rho\rvert^{2}\overline{\gamma}_{1}(1+\lambda^{\star}{\mathsf{a}}_{1})^{2})).$
(49)

Furthermore, inserting
$\lambda^{\star}=-\frac{1}{{\mathsf{a}}_{1}}\pm\frac{1}{{\mathsf{a}}_{1}\sqrt{1+\overline{\gamma}_{1}{\mathsf{a}}_{1}}}$
into (49) gives

$\displaystyle{\mathsf{R}}_{2}=\log_{2}(1+\overline{\gamma}_{2}({\mathsf{a}}_{2}-\lvert\rho\rvert^{2}\overline{\gamma}_{1}/(1+\overline{\gamma}_{1}{\mathsf{a}}_{1}))).$
(50)

Substituting (22) and (50) into
${\mathsf{C}}={\mathsf{R}}_{1}+{\mathsf{R}}_{2}$ gives

$\begin{split}{\mathsf{C}}&=\log_{2}(1+\overline{\gamma}_{1}{\mathsf{a}}_{1}+\overline{\gamma}_{2}({\mathsf{a}}_{2}(1+\overline{\gamma}_{1}{\mathsf{a}}_{1})-\lvert\rho\rvert^{2}\overline{\gamma}_{1}))\\\
&=\log_{2}(1+{\overline{\gamma}}_{1}{\mathsf{a}}_{1}+{\overline{\gamma}}_{2}{\mathsf{a}}_{2}+{\overline{\gamma}}_{1}{\overline{\gamma}}_{2}{\mathsf{a}}_{1}{\mathsf{a}}_{2}(1-\lvert\rho_{\mathsf{u}}\rvert^{2})).\end{split}$
(51)

The theorem is proved.

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