RIS with Coupled Phase Shift and Amplitude: Capacity Maximization and Configuration Set Selection

This paper considers that a transmitter communicates with a receiver as shown in Fig. 1. There is also a RIS between the transmitter and the receiver. The received signal at the receiver can be written as

where $y[t]\in\mathbb{C}$ is the received signal, $x[t]\in\mathbb{C}$ is the transmitted signal, and $w[t]\sim\mathcal{N}_{\mathbb{C}}(0,N_{0})$ is the additive white Gaussian noise (AWGN). The channel between the transmitter and the receiver, denoted as $h\in\mathbb{C}$, can be expressed as [26]

In (2), $N$ is the number of RIS elements, $h_{0}\in\mathbb{C}$ is the direct channel between the transmitter and the receiver, $h^{\prime}_{n}\in\mathbb{C}$ is the channel between the transmitter and the $n$th RIS element, $\theta_{n}=\beta_{n}e^{j\alpha_{n}}\in\mathbb{C}$ is the reflection coefficient of the $n$th RIS element, $h^{\prime\prime}_{n}\in\mathbb{C}$ is the channel between the $n$th RIS element and the receiver. Since h is a complex number, it will have a magnitude and a phase. In this paper, $\angle x$ denotes the phase of the complex number $x$.

The overall channel from the transmitter to the receiver through the $n$th RIS element can be expressed as

It is also useful to define the cascaded channel coefficient $v_{n}\in\mathbb{C}$ for the $n$th RIS element as

The channel capacity from the transmitter to the receiver can be calculated as

in which $B$ is the transmitted signal bandwidth, $P$ is the transmitted signal power, $N_{0}$ is the noise power spectral density, and $\frac{P|h|^{2}}{BN_{0}}$ is the signal-to-noise ratio (SNR).

In general, the reflection coefficient of the $n$th RIS element, denoted as $\theta_{n}$, is defined by two parameters, the amplitude ($\beta_{n}$) and the phase shift ($\alpha_{n}$). In other words,

In an ideal setup, the amplitude and the phase shift are independent and can take any possible value in a particular range

However, the ideal setup is not valid for practical RIS. According to [36], for any circuit implementation of RIS, $\beta_{n}$ and $\alpha_{n}$ are not independent, and the relation between them can be expressed as

where $\beta_{\min}$, $\phi$, and $\kappa$ are all non-negative constants. Such a practical model motivated us to consider the phase shift and the amplitude to be coupled rather than independent for our system model. For better illustration, Fig. 2 demonstrates an example for the relationship between $\beta_{n}$ and $\alpha_{n}$.¹¹1According to [36], we use $\beta_{\min}=0.2$, $\kappa=1.6$, and $\phi=0.43\pi$ in the example in Fig. 2.

Moreover, assuming a continuous adjustment for the phase shift is infeasible in practical setups. A better and more realistic assumption would be to consider a finite number of discrete choices of phase shift [31]. Therefore, in our setup, for each RIS element, say the $n$th element, $\theta_{n}$ is chosen from a configuration set, i.e., a finite set of $K$ choices:

in which amplitude $\hat{\beta}_{k}$ and phase shift $\hat{\alpha}_{k}$ are constants and satisfy (8) for $k\in\{1,2,...,K\}$.

Assume $\angle h^{*}$ is known. Consider the $n$th RIS element. According to (9), there will be $K$ choices for $\theta_{n}$. Thus, there will also be $K$ choices for $g_{n}$ (expression of $g_{n}$ is given in (3)), i.e., $g_{n}\in\{g_{n,1},g_{n,2},...,g_{n,K}\}$, with

for $k=1,2,...,K$.

Let us define $\langle h^{*},g_{n,i}\rangle$ as:

If we view complex numbers $h^{*}$ and $g_{n,i}$ as vectors in a complex plane, then $\langle h^{*},g_{n,i}\rangle$ is actually the real inner product of vector $h^{*}$ and vector $g_{n,i}$.

We have the following theorem.

According to  Theorem 1, among the $K$ choices for $g_{n}$ of the $n$th RIS element, all we have to do is to find the one that has the largest $\langle h^{*},g_{n,i}\rangle$ for $i\in\{1,2,...,K\}$.

Also, according to (11), we have:

Using (13), the right-hand side of (12) can now be updated as:

in which $\angle g_{n,i}=\angle(v_{n}\hat{\beta}_{i}e^{j\hat{\alpha}_{i}})=\angle v_{n}+\hat{\alpha}_{i}$ (from (11)).

As seen in (14), $|h^{*}|$ and $|v_{n}|$ are the same for all the members of $\{\langle h^{*},g_{n,1}\rangle,\langle h^{*},g_{n,2}\rangle,...,\langle h^{*},g_{n,K}\rangle\}$. Thus, finding the maximum $\langle h^{*},g_{n,i}\rangle$ would be equivalent to finding the largest $\hat{\beta}_{i}\cos(\angle h^{*}-\angle g_{n,i})$. So if $\angle h^{*}$ is known, we can determine the optimal reflection coefficient for each RIS element (say the $n$th RIS element) by finding the $g_{n,i}$ with the largest $\hat{\beta}_{i}\cos(\angle h^{*}-\angle g_{n,i})$. From here on, we refer to the largest $\hat{\beta}_{i}\cos(\angle h^{*}-\angle g_{n,i})$ as $\beta_{n}^{*}\cos(\angle h^{*}-\angle g_{n}^{*})$ and the corresponding $g_{n,i}$ as $g_{n}^{*}$.

In Section III-A, we find the optimal reflection coefficient for each RIS element with given known $\angle h^{*}$. However, in reality, $\angle h^{*}$ is unknown at the beginning. One method is to go through all possibilities of $\angle h^{*}$ to find the best one. Since there are infinite possibilities for $\angle h^{*}$, this method will not be feasible. But as we will soon show, there is a way to go through a finite number of possibilities for $\angle h^{*}$ and still be able to find the global optimal solution. This is possible because $g_{n}^{*}~{}(n=1,2,...,N)$ keep unchanged over a range of $\angle h^{*}$.

To get a better picture, Fig. 3 shows an example for the $n$th RIS element. In this example, there are $K=4$ choices for $\theta_{n}$, where the four choices are selected randomly from the curve in Fig. 2. The resulting $g_{n,1},g_{n,2},g_{n,3},g_{n,4}$ in this example are 1.7279, 3.0369, 4.0841, 5.6549 rad, respectively. Recall that we should find the largest one among $\hat{\beta}_{i}\cos(\angle h^{*}-\angle g_{n,i})$, $i\in\{1,2,3,4\}$. Fig. 3 shows how the four curves $\hat{\beta}_{i}\cos(\angle h^{*}-\angle g_{n,i})$ ($i=1,2,3,4$) change with $\angle h^{*}\in[0,2\pi)$. For presentation brevity, we call curve $\hat{\beta}_{i}\cos(\angle h^{*}-\angle g_{n,i})$ as “curve $\angle g_{n,i}$” in the legend of Fig. 3 and in the following discussion.

According to Section III-A, in Fig. 3 we are looking for the maximal of the four curves $\angle g_{n,i}$ ($i=1,2,3,4$) at each $\angle h^{*}$ value. The maximal of the four curves is the red curve shown in Fig. 3. In Fig. 3, $o_{1},o_{2},o_{3},o_{4},p_{1},p_{2},p_{3},p_{4}$ are $\angle h^{*}$ values of the intersections of the curves. From Fig. 3, when $\angle h^{*}$ changes from 0 to $p_{1}$, curve $\angle g_{n,4}$ is always above the other three curves, and thus, $g^{*}_{n}$ remains unchanged (keeps as $g_{n,4}$). Similarly, when $\angle h^{*}$ changes within $[p_{1},p_{2}]$, $[p_{2},p_{3}]$, $[p_{3},p_{4}]$, or $[p_{4},2\pi)$, $g^{*}_{n}$ remains as $g_{n,1}$, $g_{n,2}$, $g_{n,3}$, or $g_{n,4}$, respectively.

We refer to the interval of $\angle h^{*}$ (within $[0,2\pi)$) over which one curve stays above all other curves as the active interval of the curve. Thus, in Fig. 3, the active intervals of curves $\angle g_{n,1}$, $\angle g_{n,2}$, and $\angle g_{n,3}$ are $[p_{1},p_{2}]$, $[p_{2},p_{3}]$, and $[p_{3},p_{4}]$, respectively, and the active interval of curve $\angle g_{n,4}$ is the union of two sub-intervals at the two sides of $[0,2\pi)$: $[p_{4},2\pi)\cup[0,p_{1}]$. For presentation simplicity, we represent $[p_{4},2\pi)\cup[0,p_{1}]$ as $[p_{4},p_{1}]$. As a summary, we have the following definition for an interval of $\angle h^{*}$ written as $[x_{1},x_{2}]$ where $x_{1},x_{2}\in[0,2\pi)$:

• •

  if $x_{1}<x_{2}$, then the interval is a continuous interval from $x_{1}$ to $x_{2}$ (e.g., the active intervals of curves $\angle g_{n,1}$, $\angle g_{n,2}$, and $\angle g_{n,3}$);

• •

  if $x_{1}>x_{2}$, then the interval is the union of two continuous sub-intervals located at the two sides of $[0,2\pi)$, which is $[x_{1},2\pi)\cup[0,x_{2}]$ (e.g., the active interval of $\angle g_{n,4}$).

In either case, $x_{1}$ and $x_{2}$ are called the left and right boundary of interval $[x_{1},x_{2}]$.

We call the right boundary ($\angle h^{*}$ value) of the active interval of a curve $\angle g_{n,i}$ as the active intersection of the curve. For example, $p_{2}$ is the active intersection for curve $\angle g_{n,1}$, and $p_{1}$ is the active intersection for curve $\angle g_{n,4}$.

Since $g_{n}^{*}$ (and $\theta_{n}^{*}$) remains unchanged if $\angle h^{*}$ is within an active interval, we do not need to go through the infinite possibilities of $\angle h^{*}$. As long as we determine the active intervals, we can get $g_{n}^{*}$ (and $\theta_{n}^{*}$) in each active interval. For example, for Fig. 3, when $\angle h^{*}$ is within active interval $[p_{1},p_{2}]$, $[p_{2},p_{3}]$, $[p_{3},p_{4}]$, and $[p_{4},p_{1}]$, $g_{n}^{*}$ is $g_{n,1}$, $g_{n,2}$, $g_{n,3}$, and $g_{n,4}$, respectively.

We will show how to determine the active intervals of one RIS element in Section III-C, and extend the result to get the active intervals of all RIS elements in Section III-D.

For each RIS element, say the $n$th element, its $g_{n}$ has $K$ choices: $g_{n,1},g_{n,2},...,g_{n,K}$, corresponding to $K$ curves called curve $\hat{\beta}_{i}\cos(\angle h^{*}-\angle g_{n,i})$ ($i=1,2,...,K$) over $\angle h^{*}\in[0,2\pi)$, similar to the curves in Fig. 3. Recall that $\angle g_{n,i}=\angle v_{n}+\hat{\alpha}_{i}$.

To find the intersections ($\angle h^{*}$ values) of curves $\hat{\beta}_{i}\cos(\angle h^{*}-\angle g_{n,i})$ and $\hat{\beta}_{l}\cos(\angle h^{*}-\angle g_{n,l})$, we need to solve the following equation:

From (15), the two curves intersect at two points in the range of $[0,2\pi)$: one at $\angle h^{*}=\arctan(\frac{\hat{\beta}_{l}\cos(\angle g_{n,l})-\hat{\beta}_{i}\cos(\angle g_{n,i})}{\hat{\beta}_{i}\sin(\angle g_{n,i})-\hat{\beta}_{l}\sin(\angle g_{n,l})})\mod 2\pi$ and the other at $\angle h^{*}=(\pi+\arctan(\frac{\hat{\beta}_{l}\cos(\angle g_{n,l})-\hat{\beta}_{i}\cos(\angle g_{n,i})}{\hat{\beta}_{i}\sin(\angle g_{n,i})-\hat{\beta}_{l}\sin(\angle g_{n,l})}))\mod 2\pi$. Note that the two intersections are $\pi$ radians apart.

For the $K$ curves of the $n$th RIS element, the total number of intersections will be ${2}{{K}\choose{2}}$. Among all the intersections, only the active intersections matter to us. For example, as seen in Fig. 3, intersections $o_{1}$ and $o_{2}$ are not active and therefore not of any importance to our optimization goal.

Let us consider the intersections on a single curve. It has two intersections with any of the other $K-1$ curves. Thus, the total number of intersections on a single curve will be $2(K-1)$. For example, in Fig. 3, since there are 4 curves in total, each curve has $2(4-1)=6$ intersections on it.

Considering the $2(K-1)$ intersections on each curve, there will be an interval between every two consecutive intersections, making the total number of intervals on each curve being $2(K-1)$.²²2The $2(K-1)$ intersections partition range $[0,2\pi)$ of $\angle h^{*}$ into $2(K-1)+1$ intervals. As aforementioned, the union of the interval from $0$ to the left-most intersection and the interval from the right-most intersection to $2\pi$ is viewed as a single interval. Note that by an “interval”, we mean an interval of $\angle h^{*}$.

In the following theorem, we prove that the considered curve can have at most one active interval among all $2(K-1)$ intervals.

According to  Theorem 2, each curve will have at most one active interval. Recall that there are $K$ curves associated with each RIS element. Therefore, there will be at most $K$ active intervals for each RIS element. This means that the number of active intersections for each RIS element can at most be $K$. Next, we will develop a method for finding the active intersections of each RIS element, say the $n$th RIS element.

For the $n$th RIS element, it has $K$ curves. Suppose $I_{n,i,l}$ is the interval at which curve $\hat{\beta}_{i}\cos(\angle h^{*}-\angle g_{n,i})$ is above curve $\hat{\beta}_{l}\cos(\angle h^{*}-\angle g_{n,l})$, i.e., $\hat{\beta}_{i}\cos(\angle h^{*}-\angle g_{n,i})>\hat{\beta}_{l}\cos(\angle h^{*}-\angle g_{n,l})$. $I_{n,i,l}$ can be determined with the help of the intersection points of the two curves as discussed at the beginning of Section III-C. For the example in Fig. 3, we have $I_{n,2,3}=[o_{1},p_{3}]$ and $I_{n,3,2}=[p_{3},o_{1}]$.

For curve $\hat{\beta}_{i}\cos(\angle h^{*}-\angle g_{n,i})$, its active interval denoted as $I_{n,i}$ can be determined as

Note that $\bigcap$ refers to the common range of intervals. The following function helps us find the common range of two intervals.

Common range calculator (CRC): The CRC function takes two intervals $[l_{1},r_{1}]$ and $[l_{2},r_{2}]$ as input intervals, and gets interval $[l_{3},r_{3}]$ as the common range (CR) of the two input intervals. Here the intervals follow the interval definition in Section III-B, and ‘$l$’ and ‘$r$’ mean the left and right boundary of an interval, respectively. Table I summarizes the CRC calculation results for different cases (in which “No CR” means that the two input intervals do not have a range in common), and Fig. 4 provides a demonstration for each case in Table I.

By using the CRC, we can get $I_{n,1},I_{n,2},...,I_{n,K}$, i.e., the $K$ active intervals of the $K$ curves associated with the $n$th RIS element. Based on the active intervals, we can get the active intersections for the active intervals of the $n$th RIS element.

After we get the active intersections for the $n$th RIS element, we can quickly get active intersections for any other RIS element, say the $m$th RIS element, as follows.

Since $\angle g_{n,i}=\angle v_{n}+\hat{\alpha}_{i}$, we have:

So $\hat{\beta}_{i}\cos(\angle h^{*}-\angle v_{n}-\hat{\alpha}_{i})$ is the $i$th curve of the $n$th RIS element. If we shift this curve to the left by $\angle v_{n}-\angle v_{m}$ (i.e., we replace $\angle h^{*}$ with $\angle h^{*}+\angle v_{n}-\angle v_{m}$), we will get curve $\hat{\beta}_{i}\cos(\angle h^{*}+\angle v_{n}-\angle v_{m}-\angle v_{n}-\hat{\alpha}_{i})$ = $\hat{\beta}_{i}\cos(\angle h^{*}-\angle v_{m}-\hat{\alpha}_{i})$ = $\hat{\beta}_{i}\cos(\angle h^{*}-\angle g_{m,i})$, which is the $i$th curve of the $m$th RIS element. Therefore, if we shift all the curves of the $n$th RIS element by the constant value of $\angle v_{n}-\angle v_{m}$, we will end up with the curves of the $m$th RIS element. As a result, the active intersections of the $m$th RIS element would be the active intersections of the $n$th RIS element shifted by $\angle v_{n}-\angle v_{m}$. Thus, simply by shifting, we can calculate the active intersections of the rest of the elements.

In Section III-D, we have determined the active intersections for all RIS elements. Since each RIS element has at most $K$ active intervals and at most $K$ active intersections, there will be at most $NK$ active intersections in total for all $N$ RIS elements. The active intersections partition the range $[0,2\pi)$ of $\angle h^{*}$ into at most $NK$ regions. As discussed in Section III-B, when $\angle h^{*}$ is within one region, $\theta_{1}^{*},\theta_{2}^{*},...,\theta_{N}^{*}$ remain unchanged. Thus, we only need to go through at most $NK$ regions of $\angle h^{*}$, find $\theta_{1}^{*},\theta_{2}^{*},...,\theta_{N}^{*}$ and the corresponding channel capacity $C$ when $\angle h^{*}$ is within each region, and pick up the largest channel capacity and select the $\theta_{1}^{*},\theta_{2}^{*},...,\theta_{N}^{*}$ in the corresponding region as the overall optimal reflection coefficients for the $N$ RIS elements. Since we only need to go through at most $NK$ regions, our method has a complexity linear with $N$ and $K$.

Before we present our method, we introduce two functions $F_{n}(\angle h^{*})$ and $S(\angle h^{*})$ as follows.

In Section III, for the $n$th RIS element, we have shown that we should get the maximal of curves $\hat{\beta}_{i}\cos(\angle h^{*}-\angle g_{n,i})$, $i\in\{1,2,...,K\}$ for any $\angle h^{*}\in[0,2\pi)$ . Accordingly, for the $n$th RIS element, we can define $F_{n}(\angle h^{*})$ as the maximal of the curves for a given $\angle h^{*}$ value, as

Fig. 5 shows an example of function $F_{n}(\angle h^{*})$ with $K=4$. For an RIS with $N$ elements, we have $N$ different functions: $F_{1}(\angle h^{*}),F_{2}(\angle h^{*}),...,F_{N}(\angle h^{*})$.

Also define function $S(\angle h^{*})$ as

Accordingly, we have

Equation (21) means that if we shift curve $F_{n}(\angle h^{*})$, for any $n$, to the left by $\angle v_{n}$, then we can get curve $S(\angle h^{*})$. From (21), we also have

Next, we introduce our method for configuration set selection.

Consider $C_{\Psi}^{*}$ (the maximal capacity over a channel realization of the system with the configuration set $\Psi$). For presentation simplicity, we omit subscript ‘$\Psi$’ and write $C_{\Psi}^{*}$ as $C^{*}$ in the sequel. For $C^{*}$, its expectation over channel realizations is expressed as

in which $R$ (a very large number) is the number of channel realizations, and $h_{r}^{*}$ means the optimal overall channel from the transmitter to the receiver in the $r$th realization.

The term $|h_{r}^{*}|$ in (23) is expressed as:

We assume $|v_{1}|,|v_{2}|,...,|v_{N}|$ are approximately the same and are equal to constant $c$.³³3This is a common assumption when RIS is at the far field of the transmitter and the receiver [26]. Also assuming a weak direct path ($|h_{0}|\approx 0$), $|h_{r}^{*}|$ can be further expressed from (24) as

Here step (i) is from (22), and step (ii) is to use the integral to replace Riemann sum. As we can see in (25), $|h_{r}^{*}|$ is proportional to $|\int_{0}^{2\pi}S(x)\,dx|$.