Throughput Maximization in Two-hop DF Multiple-Relay Network with Simultaneous Wireless Information and Power Transfer

04/24/2019 ∙ by Qi Gu, et al. ∙ BEIJING JIAOTONG UNIVERSITY Texas A&M University Corpus Christi Beijing Institute of Technology 0

This paper investigates the end-to-end throughput maximization problem for a two-hop multiple-relay network, with relays powered by simultaneous wireless information and power transfer (SWIPT) technique. Nonlinearity of energy harvester at every relay node is taken into account and two models for approximating the nonlinearity are adopted: logistic model and linear cut-off model. Decode-and-forward (DF) is implemented, and time switching (TS) mode and power splitting (PS) mode are considered. Optimization problems are formulated for TS mode and PS mode under logistic model and linear cut-off model, respectively. End-to-end throughput is aimed to be maximized by optimizing the transmit power and bandwidth on every source-relay-destination link, and PS ratio and/or TS ratio on every relay node. Although the formulated optimization problems are all non-convex. Through a series of analysis and transformation, and with the aid of bi-level optimization and monotonic optimization, etc., we find the global optimal solution of every formulated optimization problem. In some case, a simple yet optimal solution of the formulated problem is also derived. Numerical results verify the effectiveness of our proposed methods.

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I Introduction

Simultaneous wireless information and power transfer (SWIPT) is an emerging technical solution for energy-constrained wireless network and Internet of Things (IoT), which enables the transmitter to transmit power and information simultaneously to receiver via the radio frequency (RF) signal [1, 2, 3]. To realize SWIPT, there are two modes: 1) Power splitting (PS) mode; 2) Time switching (TS) mode. In PS mode, there is one power splitter at the receiver, which splits the received signal into two parts. One part is for energy harvesting (EH) and the other part is for information decoding (ID) [4]. In TS mode, the receiver switches between EH and ID alternatively, in which one round of EH and ID is called as one period [5]. By adjusting the PS ratio or TS ratio between EH and ID, the rate of data transmission and the rate of energy harvesting can be balanced. This topic has been explored in lots of literatures [6, 7, 5, 4, 8, 9, 10, 11, 12, 14, 15, 13].

A special utilization of SWIPT lies in relay network, in which one or more relay nodes with no battery extracts both energy and information from the source signal through SWIPT and then forward the received signal (in amplify-and-forward (AF) mode) or decoded information (in decode-and-forward (DF) mode) to the destination node by using the harvested energy. The SWIPT-powered relay network can save relay node from additional power supply, and has attracted a lot research attentions[15, 16, 17, 14, 18, 20, 21, 22, 24, 23, 25, 19].

Two-hop or multiple-hop relay network are considered and combinations of various system configurations, e.g., PS or TS for implementing the SWIPT, DF or AF for implementing the relay, etc., are investigated in literatures. Categorized by the research goal, two classes of literatures can be found. The first class of literature focuses on analyzing the system performance, in terms of ergodic capacity [15, 16, 17, 14], effective throughput [18]

, or outage probability

[14, 15, 16, 17, 19]. Specifically, [14] focuses on the DF relay network under PS mode; [15] considers the AF relay network under PS mode and TS mode; [16] studies the AF and DF relay network under TS mode with full-duplex relay, which brings self-interference into the system; [17] investigates the AF relay network under PS mode with multiple-antenna relay and co-channel interference; [18] looks into the AF and DF relay network under TS mode; [19] pays attention to the AF network under PS mode with multiple random distributed relay nodes in space and analyzed the associated performance under various relay selection strategies. It should be noticed that all the mentioned works in the first class investigate a two-hop relay network, among which [15, 16, 17, 18] assume one relay while [14] and [19] assume multiple relays.

The second class of literature targets at maximizing some utility including the outage capacity [20] or end-to-end throughput [21, 22, 24, 23], or minimizing some cost such as transmission time for given amount of data [25], by optimizing PS ratio, TS ratio, etc. Without specific clarification, two-hop relay network is set up in default in these literatures. In [20], PS ratio and TS ratio are optimized under PS mode and TS mode in a DF relay network, respectively. In [21], multiple antennas are assumed at an AF relay, and PS ratio and antenna selection strategy are optimized jointly. In [22]

, beamforming vector and PS/TS ratio are optimized with multiple antennas implemented at source node, AF relay, and destination node. In

[23], PS ratio is optimized over multiple channels in PS mode. In [24], PS ratio and TS ratio are optimized respectively for a multi-hop DF relay network. In [25], time for energy harvesting, information decoding, and information forwarding at the relay nodes are scheduled jointly.

For all the previously surveyed works in SWIPT-powered relay network, linear model is assumed for the energy harvester at the relay node, which indicates that the output power of the energy harvesting circuit grows linearly with the power of input RF signal. However, measurements show that the practical energy harvesting circuit is subject to a non-linear model. Hence the mismatch of energy harvesting model in surveyed literatures will lead to the degradation of system performance. In [26], a nonlinear EH model based on logistic function is built, which fits the measurement data well. Some literatures related to SWIPT [12, 27] have also taken use of this non-linear model. For the ease of discussion, we will call this kind of model as logistic model. In [28], a linear cut-off model is used to approximate the nonlinear feature of energy harvester, which goes with the power of input RF signal constantly, then linearly, and at last constantly in [28]. The linear cut-off model is also shown to be a good approximation. It should be noticed that when logistic model or linear cut-off model is adopted for a SWIPT-powered relay network, the methods in existing literatures cannot offer a solution.

In this paper, we investigate the two-hop DF relay network with a consideration of nonlinear energy harvester under TS mode and PS mode for the first time. For the nonlinearity of energy harvester, both logistic model and linear cut-off model will be taken into account. The scenario with multiple relay nodes is considered, which is more general and beneficial since more copies of source signal can be utilized. Thus there are multiple links from source to destination through a relay node. End-to-end throughput is targeted to be maximized by optimizing transmit power and bandwidth on every link and PS ratio or TS ratio on every relay node. Optimization problems are formulated for TS mode and PS mode, respectively.

  • For TS mode under two nonlinear models of energy harvester, the associated optimization problem is non-convex. To find the global optimal solution, the original optimization problem is decomposed into two levels. In the lower level, with some further transformations and by exploring the special properties of investigated problem, closed-form optimal solution is derived. In the upper level, the associated problem is transformed to be a standard monotonic optimization problem, whose global optimal solution is achievable.

  • For PS mode under logistic model, with some transformations, the original optimization is also transformed to be a standard monotonic optimization problem. Hence the global optimal solution is also achievable.

  • For PS mode under linear cut-off model, the method for PS mode under logistic model also applies. However, to further save the computation complexity, we transform the original optimization problem to be an equivalent form and then derive the semi-closed-form solution for the transformed problem, which is also global optimal.

The rest of this paper is organized as follows. In Section II, the system model is presented and the research problems are formulated. Section III and Section IV present the optimal solution of the formulated problem in TS mode and PS mode, respectively. Section V shows the numerical results, followed by concluding remarks in Section VI.

Ii System Model and Problem Formulation

Consider a two-hop DF multiple-relay network as shown in Fig. 1, in which the source node would like to transmit information to the destination node via relay node , who has no power supply, for . The source, destination, and relays all have single antenna. Denote the channel gain from to as , the channel gain from to as , and the path from node to node through node as link , for . A direct link from the source node to the destination node does not exist due to physical obstacles [29, 15]. All the links also constitute the set . In the system, all the channel gains keep stable in one fading block, and are randomly and independently distributed over fading blocks with continuous distribution function.

Fig. 1: Illustration of a two-hop multiple-relay network.

The information is transmitted with the help of relay nodes in the following way. Denote the bandwidth allocated to link as , suppose the transmit power of source node as for link . By assuming the total system bandwidth as , and total transmit power of source node as , there are

(1)
(2)
(3)

and

(4)

For every relay node, it should be noticed that they all have no power supply. Thus for has to harvest energy from the signal transmitted by node . SWIPT technique is utilized, and two modes are considered: TS mode and PS mode.

In TS mode,

  • Step 1: As shown in Fig. 2, time is divided into multiple frames with equal length . The is smaller than the coherence time, hence channel gains and for within in keeps invariant. Within one frame, first harvests energy from ’s RF signal in the time duration between , where . In this step, the harvested energy can be written as , where indicates the power of harvested energy of every relay nodes’s energy harvester when the power of received energy is 111Without loss of generality, the feature of of the every relay node’s energy harvester is assumed to be identical..

  • Step 2: In the rest of time of one frame, i.e., within time duration . The received signal is left for information decoding.

Fig. 2: Time frame structure of TS mode for relay node .

In PS mode, time is also divided into multiple frames with equal length , within which and for keeps invariant. But different from TS mode, as shown in Fig. 3, a fraction where , of the received signal’s power is left for energy harvesting, and a fraction of received signal’s power is left for information decoding.

Fig. 3: Time frame structure of PS mode for relay node .

Denote the transmit power of as for . In TS mode, the transmit power . In PS mode, the transmit power . Thus the end-to-end throughput in TS mode can be written as

(5)

and the end-to-end throughput in PS mode can be written as

(6)

where is the power spectrum density of noise 222When taking into security issue, a different throughput can be expressed and achieved as shown in [30, 31]. Due to the limit of space, we will only look into the ideal case without consideration of security in this work, which is also a general case in most of related literatures. . On the other hand, should be also subject to a limit on the maximal transmit power, denoted as , due to the physical limit of the relay node . Hence there is

(7)

in TS mode, and

(8)

in PS mode.

For the feature of energy harvester, as shown in Fig. 4, experimental measurements in [26] shows that the power of harvested energy first grows with the power of received energy when the power of received energy is larger than a threshold, and then the grows slowly and slowly until it reaches up to an upper bound. To approximate this feature, two models are adopted.

  • Logistic Model: In this model,

    (9)

    where represents the maximal power the energy harvester can harvest, and are parameters for nonlinearity. This model is broadly used when taking into account the nonlinearity of the energy harvester [12, 27].

  • Linear Cut-off Model: In this model,

    (10)

Note that both the function in (9) and the function in (10) are monotonic increasing functions with , which is in coordination with such an intuition: More power can be harvested when more power is received. Fig. 4 also plots versus under logistic model and linear cut-off model under selected parameter setup. It can be seen that both of these two models can achieve a good approximation of measurement data.

Fig. 4: Harvested power vs. input power.

Collecting the formulated constraints, the associated optimization problem under TS mode and PS mode can be given as follows.

In TS mode, the associated optimization problem is

Problem 1
s.t. (11a)
(11b)
(11c)
(11d)
(11e)
(11f)

In PS mode, the associated optimization problem is

Problem 2
s.t. (12a)
(12b)
(12c)
(12d)
(12e)
(12f)

In the following, we will show how to solve Problem 1 and Problem 2 under two energy harvester models, i.e., logistic model and linear cut-off model, respectively.

Iii Optimal Solution in TS Mode

In this section, Problem 1 will be solved. Note that Problem 1 is a non-convex optimization problem given that the function is a non-concave function with the vector of , , and . Thus the global optimal solution of Problem 1 is hard to achieve. In the following, we will do some transformation and simplification on Problem 1, and find the global optimal solution of Problem 1. Attention that the presented solution in this section works for both the case under logistic model and the case under cut-off model.

To solve Problem 1 optimally, we decompose it into two levels 333This method is referred to as bi-level optimziation.. In the lower level, is fixed, and the following optimization problem need to be solved

Problem 3
s.t. (13a)
(13b)
(13c)
(13d)

For the upper level, look into the constraint (11d), which is equivalent with

(14)

Define . Note that . Thus the constraint (11d) and constraint (11a) can be combined to be

(15)

In the upper level, we need to optimize so as to solve the following optimization problem

Problem 4
s.t. (16a)

It can be checked that Problem 1 is equivalent with the upper level optimization problem, i.e., Problem 4.

Iii-a Optimal Solution for the Lower Level Optimization Problem

In this subsection, we will solve the lower level optimization problem, i.e., Problem 3. To simplify the solving of Problem 3, we impose one additional constraint

(17)

then the objective function of Problem 3 reduces to

In addition, by relaxing the equality constraint (13d) in Problem 3 to be an inequality, Problem 3 turns to be the following optimization problem

Problem 5
s.t. (18a)
(18b)
(18c)
(18d)
(18e)

It should be noticed that maximal achievable utility of Problem 5 equals the maximal achievable utility of Problem 3. The reason is as follows: Even the optimal solution of Problem 3 does not obey the constraint (17), i.e., , the throughput on link is still , which can be achieved by setting . In other words, in the feasible region such that constraint (17) holds, the maximal achievable utility of Problem 3 is also achievable. To be consistent with the constraint (17), the equality constraint (13d) in Problem 3 is relaxed to be the inequality constraint (18c), which has no influence on equality between the maximal achievable utility of Problem 3 and the maximal achievable utility of Problem 5. Therefore solving Problem 3 is equivalent with solving Problem 5.

It should be also noticed that the solution of Problem 5 may not serve as the optimal solution of Problem 3 directly, since the optimal solution of Problem 5 may have . In the real application, to get the optimal solution of Problem 3, we only need to find the optimal solution of Problem 5 in the first step, and then keeps unchanged for , and enlarge for calculated by solving Problem 5 such that constraint (13d) holds.

Next we turn to solve Problem 5. It can be checked that Problem 5 is a convex optimization problem since the constraints of Problem 5 are all linear and the objective function is concave with . Although existing method can help to find the global optimal solution, in the next we will explore some special property of Problem 5’s optimal solution so as to simplify the solving of Problem 5.

It can be checked that Problem 5 satisfies the Slater’s condition. Hence the KKT condition of Problem 5 can serve as the sufficient and necessary condition of its optimal solution [32], which can be given as follows

(19a)
(19b)
(19c)
(19d)
(19e)
(19f)
(19g)
(19h)
(19i)
(19j)

where , , , , and are the Lagrange multipliers associated with the constraints (18a), (18b), (18c), (18d), (18e), respectively.

Before we start the investigation on the KKT condition listed in (19), two facts about the optimal solutions of Problem 5 are claimed.

  • Define and . Then there is . This fact indicates that the case with (or the case ) will not happen for the optimal solution of Problem 5. This is because the case with (or the case ) indicates a wasteful use of power resource (or spectrum resource ). Higher utility can be achieved by transferring the wasted resources to the other links with positive bandwidth allocation or power allocation.

  • The constraint (18d) is active, which means that , for the optimal solution of Problem 5. This is due to the fact that the objective function of Problem 5 is an increasing function with for . So it is better to increase for as much as possible.

Then we turn to investigate the KKT condition  in (19), which can help to prove the following lemma.

Lemma 1

Define , the term equals a constant for .

Proof:

For , there is , thus it can be inferred that from (19e). Define , then (19b) can be rewritten as

(20)

The function is actually a strictly increasing function with for . Hence from (20) it can be concluded that for equals a common value, which is denoted as for the ease of presentation in the following.

This completes the proof.

According to the claim in Lemma 1, there is . Combining with the two claimed facts for the optimal solution of Problem 5, it can be derived that

(21)

which further indicates that

(22)

Therefore the objective function of Problem 5 can be rewritten as

(23)

Since maximizing is equivalent with maximizing , then solving Problem 5 is equivalent with solving the following optimization problem

Problem 6
s.t. (24a)
(24b)
(24c)

For Problem 6, it is straightforward to see that the optimal policy is to allocate more power resource to the link with higher channel gain, i.e., to set the with higher as large as possible. Specifically, the optimal allocation of for can be found as follows.

1:  Order for in descending order, such that .
2:  Define . Set for , for , and for , , …, . Note that when , does not exist and the optimal solution is , .
Algorithm 1 Searching procedure for the optimal solution of Problem 6.

In the end of this subsection, the optimal solution of the lower level optimization problem, i.e., Problem 3, can be summarized as follows.

1:   By following Algorithm 1, find the optimal for of Problem 6.
2:   Set where is calculated in Step 1 of Algorithm 2 for .
3:   Increase calculated in Step 1 to be for such that .
4:   Output and for .
Algorithm 2 Searching procedure for the optimal solution of Problem 3.

Iii-B Optimal Solution for the Upper Level Optimization Problem

In this subsection, we will solve the upper level optimization problem, i.e., Problem 4. In the first step, there is such a lemma.

Lemma 2

The function , which is defined in Problem 3, is monotonically increasing with .

Proof:

Suppose there is . Define the optimal solution of Problem 3 associated with and are and , and and , respectively, for . Then there is

where holds is due the fact that the coefficient for , and holds since the set of and for is the optimal solution of Problem 3 when .

This completes the proof.

With Lemma 2, the objective function of Problem 4 is actually the difference between two monotonically increasing function with , i.e., the difference between and . Thus solving Problem 4 is equivalent with solving the following optimization problem

Problem 7
s.t. (25a)
(25b)

For Problem 7, since both in its objective function and in its objective function are increasing functions with and respectively, the maximum of Problem 7 can be achieved by increasing both and as large as possible. Looking into the constraint (25a), both and are increasing functions with and respectively, thus the maximum of Problem 7 will be achieved when both and reach their maximal allowable value in the feasible region of Problem 7, in which case there is

(26)

which indicates that

(27)

Replace with the expression in (27), the objective function of Problem 7 turns to be . Since maximizing is equivalent with maximizing , solving Problem 7 is equivalent with solving Problem 4.

Then we focus on solving Problem 7. Although being non-convex, Problem 7 actually falls into the standard form of Monotonic Optimization Problem, whose standard form can be given as follows.

Problem 8
s.t. (28a)
(28b)

where the variable is a multiple dimensional vector, and represent the lower bound and upper bound of respectively, and both and are monotonically increasing functions with . For a standard monotonic optimization problem, there is a polyblock algorithm to achieve the -optimal solution of a standard monotonic optimization problem, where indicates the gap between the achieved utility and the global optimal utility is bounded by . The is a predefined parameter before running the polyblock algorithm. By following the polyblock algorithm, the detailed procedure for solving Problem 7 is given as follows.

1:  Initialize a point set by a two-dimensional point , where the and indicate the maximal achievable value of the variable and , respectively.
2:  while  do
3:     for  do
4:         Find the such that by utilizing the bisection search method, where and are the 1st and 2nd element of the vector respectively. Set .
5:     Find , where and are the 1st and 2nd element of the vector respectively.
6:     For , if there is , then delete the point from the set .
7:     if  then
8:         Search
9:         Find such that by utilizing the bisection search method. Set .
10:         Generate two new points and , where is the operation of Hadamard product.
11:         Add and into , delete from the set .
12:  Output the last before is subtracted to be an empty set.
Algorithm 3 -optimal solution for Problem 7.

By following Algorithm 3, the optimal solution of Problem 7 can be achieved, which also paves the way for working out the optimal solution of Problem 4. To this end, Problem 4 can be solved optimally, which also indicates the optimal solving of Problem 1.

Iv Optimal Solution in PS Mode

In this section, Problem 2 will be solved. It can be checked that Problem 2 is a non-convex optimization problem either, considering the non-convexity of under both logistic model and linear cut-off model in the objective function of Problem 2. In the following, we will show how to find the global optimal solution of Problem 2 under logistic model and linear cut-off model, respectively.

Iv-a The Case under Logistic Model

In this subsection, logistic model is adopted for the energy harvester, i.e., is set to be the function in (9). To solve Problem 2, look into the objective function of Problem 2, the term is monotonically decreasing function with , and the term is monotonically increasing function with considering the increasing monotonicity of the function . Hence the maximal value of the term and the term is achieved when these two terms are equal, equivalently, there is

(29)

which further indicates

(30)

Taking into account the fact that for , and combine the constraint (30), there is an implicit constraint

(31)

which imposes an upper bound on , denoted as , for . The can be found by following bi-section search method such that

(32)

It can be easily derived that since is bounded for . Combining with the constraint (12d), which indicates that , and the constraint (12a), which indicates that , define for , should satisfy

(33)

Then by following the similar discussion for Problem 5 in Section III, solving Problem 2 is equivalent with solving the following optimization problem

Problem 9
s.t. (34a)
(34b)
(34c)
(34d)

By following the similar transformation from Problem 5 to Problem 6, Problem 9 is equivalent with the following optimization problem

Problem 10
s.t. (35a)
(35b)

Recalling defined in (9) is a monotonically increasing function, thus both the objective function of Problem 10 and the left-hand side function of (35b) are increasing functions with the vector . Hence when , the optimal solution is just set as large as possible, i.e., set for . In general case, i.e., when , Problem 10 also falls into the standard form of monotonic optimization problem. Then by following the similar procedure in Algorithm 3444Algorithm 3 works for a two-dimensional vector. The general solving algorithm can be found in [33] and is omitted due to the limit of space., the -optimal solution of Problem 10 can be achieved.

In summary, the optimal solution of the original optimization problem in PS mode, i.e., Problem 2, can be achieved by following the steps in Algorithm 4, i.e.,

1:  if   then
2:     Set for .
3:  else
4:      By following the similar procedure in Algorithm 3, find the optimal for .
5:   Set