Utility-Based Optimal Precoding for SWIPT in MIMO Broadcasting Systems

Abstract

In this paper, we focus on optimal precoding for simultaneous wireless information and power transfer in multiple-input multiple-output broadcasting systems. A utility function, which is a nonnegative weighted sum of the achievable data rate and the received energy, is designed to effectively allocate the available power. Then, the problem of optimal precoding is formulated as a utility maximization problem, and a closed-form solution is provided. In order to clearly illustrate the rate-utility tradeoff, a two-tier optimization structure is adopted, and the corresponding two-tier optimization algorithm, for which the outer-tier optimization is accomplished by the golden section search method, is also provided. Our results indicate that the maximum information transmission and the energy transfer must be well balanced in order to maximize the total payoff, and the method employed by this paper can obtain the unique precoding matrix for the optimal operating point.

Appendices

Appendix 1: Proof of Theorem 1

Proof

The Lagrangian of the utility maximization problem (4) can be written as

$$\begin{aligned} \begin{aligned} L({\mathbf{Q}},\mu )&=\log |{\mathbf{I}}+{\mathbf{HQH}}^H|+ \zeta {\text {Tr}}({\mathbf{GQG}}^H)\\&\quad- \mu [{\text {Tr}}({\mathbf{Q}}) - {P_t}], \end{aligned} \end{aligned}$$

(12)

where $\mu \ge 0$ is the dual variable.

Then, the Lagrange dual function of the problem (4) is defined as

$$\begin{aligned} g(\mu ) = \mathop {\max }\limits _{\mathbf{Q} \succeq 0} L(\mathbf{Q},\mu ). \end{aligned}$$

(13)

The dual problem of the problem (4) is defined as

$$\begin{aligned} \mathop {\min }\limits _{\mu \ge 0} g(\mu ). \end{aligned}$$

(14)

Since the original optimization problem (4) can be solved equivalently by solving the dual problem (14), we can first maximized the Lagrangian to obtain the dual function with fixed $\mu \ge 0$, and then find the optimal dual solution $\mu ^*$ to minimize the dual function. The transmit covariance ${\mathbf{Q}}^*$ that maximize the Lagrangian to obtain $g(\mu ^*)$ is thus the optimal solution of the problem (4).

Conversely, we can also first find $\mu ^*$ that minimize the Lagrangian, and then find $\mathbf{Q}^*$ by solving the following problem

$$\begin{aligned} \mathop {\max }\limits _{{\mathbf{Q}} \succeq 0}\, \log |{\mathbf{I}} + {\mathbf{HQ}}{{\mathbf{H}}^H}| - \mathrm{{Tr[}}(\mu ^* {\mathbf{I}} - \zeta {{\mathbf{G}}^H}{\mathbf{G}}){\mathbf{Q}}], \end{aligned}$$

(15)

where the constant term has been discarded.

Since the objective function in (15) must be bounded, it is shown as follows that the matrix ${\mathbf{A}}\triangleq \mu ^* {\mathbf{I}} - \zeta {{\mathbf{G}}^H}{\mathbf{G}}$ must be positive definite. Otherwise, suppose that ${\mathbf{A}}$ is not positive definite, then we can choose some $\mathbf{Q}=\alpha {\mathbf{w}} {\mathbf{w}}^H, \alpha >0$, such that $\mathrm{{tr}}({\mathbf{A}}{{\mathbf{Q}}})<0$. Then, with $\alpha \rightarrow +\infty$, the optimal value in (15) will be unbounded, and this contradicts to the optimality of $\mu ^*$.

Then we can define $\widehat{\mathbf{Q}} = {\mathbf{A}^{1/2}}\mathbf{Q}{\mathbf{A}^{1/2}}$, and the problem (13) is then reformulated in terms of $\widehat{\mathbf{Q}}$ as

$$\begin{aligned} \mathop {\max }\limits _{\widehat{{\mathbf{Q}}} \succeq 0} \log |{\mathbf{I}} + {\mathbf{H}}{\mathbf{A}^{ - 1/2}}{\widehat{\mathbf{Q}}}{\mathbf{A}^{ - 1/2}}{{\mathbf{H}}^H}| - \mathrm{{Tr(}}{\widehat{\mathbf{Q}}}). \end{aligned}$$

(16)

According to [13], the above problem (16) is equivalent to the standard point-to-point MIMO channel capacity optimization problem subject to a single sum-power constraint, and its optimal solution can be expressed as

$$\begin{aligned} {\widehat{\mathbf{Q}}}^* = \mathbf{V}\varSigma {\mathbf{V}^H} \end{aligned}$$

(17)

where $\mathbf{V}$ is obtained from the reduced SVD of ${\mathbf{H}}{\mathbf{A}}^{ - 1/2}=\mathbf{U}\varLambda ^{1/2} \mathbf{V}^H$, with $\mathbf{U}\in {\mathbb {C}}^{N_s \times M}$, $\mathbf{V}\in {\mathbb {C}}^{N_r \times M}$, $\varLambda ={\text {diag}}(\lambda _1, \lambda _2,\cdots , \lambda _M)$, $M=\min (N_s, N_r)$, and $\lambda _1\ge \lambda _2\ge \cdots \ge \lambda _M\ge 0$; $\varSigma ={\text {diag}}(\sigma _1, \sigma _2, \cdots , \sigma _M)$ is obtained by the standard water-filling algorithm, that is

$$\begin{aligned} {\sigma _{i}} = {\left( {1 - {1}/{{{\lambda _{i}}}}} \right) ^ + },\quad i = 1,2, \cdots ,{M}. \end{aligned}$$

(18)

Thus, we can immediately get the optimal solution for the problem (4), which has the following forms

$$\begin{aligned} {\mathbf{Q}}^* = {\mathbf{A}^{ - 1/2}}\mathbf{V}\varSigma {\mathbf{V}^H}{\mathbf{A}^{ - 1/2}} \end{aligned}$$

(19)

Hence, the proof of Theorem 1 is completed. $\square$

Appendix 2: Proof of Theorem 2

Proof

Denote $\beta =2^t-1$, the Lagrangian of the inner-tier optimization problem (10) is given by

$$\begin{aligned} \begin{aligned} L(\mathbf{Q},\lambda ,\mu )&= \zeta {\text {Tr}}(\widehat{\mathbf{G}}\mathbf{Q}) + \lambda [{\text {Tr}}(\widehat{\mathbf{H}}\mathbf{Q}) - \beta ]\\&\quad - \mu [{\text {Tr}}(\mathbf{Q}) - {P_t}], \end{aligned} \end{aligned}$$

(20)

where $\lambda \ge 0$ and $\mu \ge 0$ are the dual variables, and the constant term has been discarded.

Then the Lagrange dual function of the problem (10) is defined as

$$\begin{aligned} g(\lambda ,\mu ) = \mathop {\max }\limits _{\mathbf{Q} \succeq 0} L(\mathbf{Q},\lambda ,\mu ). \end{aligned}$$

(21)

The dual problem of the problem (10) is written as

$$\begin{aligned} \mathop {\min }\limits _{\lambda \ge 0,\mu \ge 0} g(\lambda ,\mu ). \end{aligned}$$

(22)

Since the problem (10) can can be solved equivalently by solving the problem (22), we can first maximized the Lagrangian to obtain the dual function with fixed $\lambda \ge 0$ and $\mu \ge 0$, and then find the optimal dual solutions $\lambda ^*$ and $\mu ^*$ to minimize the dual function. The transmit covariance $\mathbf{Q}^*$ that maximize the Lagrangian to obtain $g(\lambda ^*, \mu ^*)$ is thus the optimal solution of the problem (10).

Conversely, we can also first find $(\lambda ^*, \mu ^*)$ that minimize the Lagrangian, and then find $\mathbf{Q}^*$ by solving the following problem

$$\begin{aligned} \mathop {\max }\limits _{\mathbf{Q} \succeq 0} \,\zeta {\text {Tr}}(\widehat{\mathbf{G}}\mathbf{Q}) - {\text {Tr}}[({\mu ^*}{\mathbf{I}} - {\lambda ^*}\widehat{\mathbf{H}})\mathbf{Q}], \end{aligned}$$

(23)

where the constant term has been discarded.

Since the objective function in (14) must be bounded, it is shown as follows that the matrix $\mathbf{B}={\mu ^*}{\mathbf{I}} - {\lambda ^*}\widehat{\mathbf{H}}$ must be positive definite. Otherwise, suppose that $\mathbf{B}$ is not positive definite, then we can choose some $\mathbf{Q}=\alpha {\mathbf{w}} {\mathbf{w}}^H, \alpha >0$, such that $\mathrm{{tr}}({\mathbf{B}}{{\mathbf{Q}}})<0$. Then, with $\alpha \rightarrow +\infty$, the optimal value in (23) will be unbounded, this contradicts to the optimality of $(\lambda ^*,\mu ^*)$.

Let $\widehat{\mathbf{Q}}={\mathbf{B}}^{1/2}{{\mathbf{Q}}}{\mathbf{B}}^{1/2}$, then, the problem (23) can be rewritten as

$$\begin{aligned} \mathop {\min }\limits _{\widehat{\mathbf{Q}}\succeq 0}\,\zeta ({\mathbf{B}}^{-1/2}{\mathbf{g}})^H{{\widehat{\mathbf{Q}}}}({\mathbf{B}}^{-1/2}{\mathbf{g}}) - {\mathrm{{Tr}}}({\widehat{\mathbf{Q}}}). \end{aligned}$$

(24)

Now we can argue that the optimal solution of (24) is always rank-one. Otherwise, suppose that the optimal solution $\widehat{\mathbf{Q}}^*$ is not rank-one, the rank of it is assumed to be $r (2 \le r \le M)$, and we have

$$\begin{aligned} \widehat{\mathbf{Q}}^*=\sum \limits _{j = 1}^r \lambda _j \mathbf{q}_j \mathbf{q}_j^H \end{aligned}$$

(25)

denotes the eigen-decomposition of $\widehat{\mathbf{Q}}$, where $\lambda _1 \ge \lambda _2 \ge \cdots \ge \lambda _r > 0$ are the eigenvalues and $\mathbf{q}_1,\mathbf{q}_2,\cdots ,\mathbf{q}_r \in {\mathbb {C}}^M$ are the associating eigenvectors. Then we can choose another

$$\begin{aligned} \widehat{\mathbf{Q}}^\star =\left( \sum \limits _{j = 1}^r \lambda _j\right) \mathbf{q}_m \mathbf{q}_m^H \end{aligned}$$

(26)

where

$$\begin{aligned} m=\arg \mathop {\max }\limits _{j\in (1,\cdots ,r)} \left\| {\mathbf{B}^{-1/2}\mathbf{g}^H\mathbf{q}_j} \right\| . \end{aligned}$$

(27)

Thus, compared to $\widehat{\mathbf{Q}}^*$, $\widehat{\mathbf{Q}}^\star$ can make the objective function in (24) larger, which presents a contradiction to the optimality of $\widehat{\mathbf{Q}}^*$. Then, we can conclude that $\widehat{\mathbf{Q}}^*$ is always rank-one.

Since ${{\mathbf{Q}}}^*={\mathbf{B}}^{-1/2}\widehat{\mathbf{Q}}^*{\mathbf{B}}^{-1/2}$, we can legitimately argue that the optimal solution $\mathbf{Q}^*$ for the problem (10) is also rank-one, which completes the proof of Theorem 2. $\square$