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Energy adaptive MAC for wireless sensor networks with RF energy transfer: algorithm, analysis, and implementation

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Abstract

Radio frequency energy transfer (RET) has been proposed as a promising solution to power sensor nodes in wireless sensor networks (WSNs). However, RET has a significant drawback to be directly applied to WSNs, i.e., unfairness in the achieved throughput among sensor nodes due to the difference of their energy harvesting rates that strongly depend on the distance between the energy emitting node and the energy harvesting nodes. The unfairness problem should be properly taken into account to mitigate the drawback caused from the features of RET. To resolve this issue, in this paper, we propose a medium access control (MAC) protocol for WSNs based on RET with two distinguishing features: energy adaptive (EA) duty cycle management that adaptively manages the duty cycle of sensor nodes according to their energy harvesting rates and EA contention algorithm that adaptively manages contentions among sensor nodes considering fairness. Through analysis and simulation, we show that our MAC protocol works well under the RET environment. Finally, to show the feasibility of WSNs with RET, we test our MAC protocol with a prototype system in a real environment.

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Notes

  1. A backoff slot is a unit slot used for the backoff procedure.

  2. maxCSMABackoffs is the maximum number of performing CCA before a slave node declares the failure of its channel access.

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Acknowledgments

The authors would like to thank Min-Hwan Song in KETI, Korea for his assistance in developing the prototype slave node the RET-WSN and testing our EA-MAC in the laboratory and real field environments. This work was partly supported by Institute for Information and Communications Technology Promotion (IITP) Grant funded by the Korea Government (MSIP) (No. B0717-16-0024, Development on the core technologies of transmission, modulation and coding with low-power and low-complexity for massive connectivity in the IoT environment) and by Mid-career Researcher Program through NRF Grant funded by the MSIP, Korea 2013R1A2A2A01069053). Parts of this work were presented in ICUFN 2011 [18] and ICTC 2011 [19].

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Correspondence to Jang-Won Lee.

Appendix

Appendix

1.1 Derivations for \(\bar{T}_{c,i},\, \bar{T}_{p,i},\) and \(\bar{T}_{tx,i}\)

To calculate the throughput of slave node i in Proposition 1, we need to find \(\bar{T}_{c,i},\, \bar{T}_{p,i},\) and \(\bar{T}_{tx,i}.\) First, we analyze the backoff model of EA-MAC using DTMC to find \(\bar{T}_{c,i}\) and \(\bar{T}_{p,i}.\) In our DTMC, the unit time (a time slot) corresponds to the backoff slot whose duration is set to be W. We will consider a fixed number n of contending slave nodes.

Let \(c_i(u)\) be a stochastic process that represents the backoff time counter for slave node i at time slot u. The k-th time slot denoted by \(u = k\) means the time period from kW to \((k + 1)W.\) Let \(s_i(u)\) be a stochastic process that represents the backoff stage of slave node i at time slot u with \(s_i(u) \in \{0,\ldots , m\},\) where m is \(maxCSMABackoffs - 1.\) For simplicity, we assume that the busy probability of the channel during CCA in slave node \(i,\, q_i,\) is constant, regardless of its backoff stages as in [5, 27].

A two-dimensional DTMC will be used to model the behavior of the backoff process. The states of the DTMC are defined as combinations of two integers \(\{s_i(u),\,c_i(u)\}.\) The state transition diagram for the DTMC is depicted in Fig. 16, where \(W_{i,j}\) is defined as

$$\begin{aligned} W_{i,j} = \lceil {\omega _i 2^{minBE+j}}\rceil \quad j \in \{0, \ldots ,m\}. \end{aligned}$$
(9)
Fig. 16
figure 16

Markov chain model for the backoff process of EA-MAC

The state transition probabilities of our DTMC in Fig. 16 are obtained as follows:

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} P\{j,\,k \mid j,\,k+1\} = 1 &{}\text {if}\,k\in (0,\,W_{i,j}-2),\, j \in (0,\, m), \\ P\{0,\,k \mid j,\,0\} = \frac{(1-q_i)}{W_{i,0}} &{}\text {if}\,k\in (0,\, W_{i,0}-1),\,j \in (0,\, m-1),\\ P\{j,\,k \mid j-1,\, 0\} = \frac{q_i}{W_{i,j}} &{}\text {if}\, k\in (0,\, W_{i,j}-1),\, j \in (1,\, m), \\ P\{0,\,k \mid m,\, 0\} = \frac{1}{W_{i,0}} &{}\text {if}\,k\in (0,\,W_{i,0}-1), \end{array} \right. \end{aligned}$$
(10)

where \(P\{j_1,\,k_1 \mid j_0,\,k_0\}\) indicates \(P\{s_i(u+1)=j_1,\, c_i(u+1)=k_1 \mid s_i(u)=j_0,\, c_i(u)=k_0\},\) and \(q_i\) is the probability that at least one of the \(n-1\) other slave nodes except slave node i is in the transmit state in a randomly chosen time. The first equation in (10) represents the decrement of the backoff time counter, which happens with probability 1. The second equation states that slave node i in backoff stages from 0 to \(m-1\) starts a new transmission with backoff stage 0 after successful transmission. The initial backoff value is uniformly chosen in the range (\(0,\,W_{i,0}-1\)). The third equation indicates that when a channel access failure occurs at backoff stage \(j-1,\) the backoff stage increases, and a new initial backoff value is uniformly chosen in the range (\(0,\,W_{i,j}-1\)). Finally, the fourth equation accounts for the fact that slave node i in the backoff stage m starts a new transmission with backoff stage 0 regardless of the result during CCA. An initial backoff value is uniformly chosen in the range (\(0,\,W_{i,0}-1\)).

Let \(b_{j,k}(i) = \lim _{u \rightarrow \infty } P\{s_i(u)=j,\, c_i(u)=k\},\, j \in (0,\,m),\,k \in (0,\, W_{i,j}-1)\) be the stationary distribution of the state \(\{j,\,k\}\) in slave node i. By the balance condition, the probability that slave node i is in the state \(\{j,\,0\},\,b_{j,0}(i),\) can be expressed as

$$\begin{aligned} b_{j,0}(i) = b_{j-1, 0}(i) q_i, \quad 1\le j\le m. \end{aligned}$$
(11)

From (11), \(b_{j,0}(i)\) can be rewritten with respect to \(b_{0,0}(i)\) as

$$\begin{aligned} b_{j,0}(i) = q_i^j b_{0,0}(i), \quad 0\le j\le m. \end{aligned}$$
(12)

Again, with the balance condition, we can represent \(b_{j, k}(i)\) for each \(k \in (1,\, W_{i,j}-1)\) as

$$\begin{aligned}&b_{j, k}(i)\nonumber \\&\quad = {\left\{ \begin{array}{ll} {\frac{W_{i,j} - k}{W_{i,j}}\times } {(1-q_i) \sum _{l=0}^{m-1}b_{l,0}(i) + b_{m,0}(i)}, &{} \text {if}\,j = 0,\\ {\frac{W_{i,j} - k}{W_{i,j}}\times } {q_i b_{j-1, 0}(i)}, &{} \text {if}\,1\le j\le m. \end{array}\right. } \end{aligned}$$
(13)

From (11) and (12), if \(j = 0\) in (13),

$$\begin{aligned}&\left( 1-q_i\right) \sum _{l=0}^{m-1}b_{l,0}(i) + b_{m,0}(i) \nonumber \\&\quad = \left( 1-q_i\right) \frac{1-q_i^{m}}{1-q_i} b_{0,0}(i) + q_i^{m} b_{0,0}(i) \nonumber \\&\quad = \left( 1-q_i^{m}\right) b_{0,0}(i) + q_i^{m} b_{0,0}(i) = b_{0,0}(i), \end{aligned}$$
(14)

and if \(1\le j\le m,\)

$$\begin{aligned} q_i b_{j-1, 0}(i) = b_{j,0}(i) = q_i^j b_{0,0}(i). \end{aligned}$$
(15)

By (12)–(15), the stationary distribution of slave node \(i,\,b_{j, k}(i),\) can be expressed as a function of \(b_{0,0}(i)\) and \(q_i\) as

$$\begin{aligned} b_{j, k}(i)= & {} \frac{W_{i,j} - k}{W_{i,j}} q_i^j b_{0,0}(i), \nonumber \\&j \in (0,\,m),\quad k \in \left( 0,\, W_{i,j}-1\right) . \end{aligned}$$
(16)

Now, we can determine \(b_{0,0}(i)\) by applying the normalization condition and (9) as follows:

$$\begin{aligned} 1&= \sum _{j=0}^{m} \sum _{k=0}^{W_{i,j}-1} b_{j,k}(i) \\&= \sum _{j=0}^{m} q_i^j b_{0,0}(i) \sum _{k=0}^{W_{i,j}-1} \frac{W_{i,j} - k}{W_{i,j}}\\&= \sum _{j=0}^{m} q_i^j b_{0,0}(i) \frac{W_{i,j}+1}{2} \\&= \frac{1}{2} \left( \sum _{j=0}^{m} q_i^j b_{0,0}(i) W_{i,j} + \sum _{j=0}^{m} q_i^j b_{0,0}(i) \right) \\&= \frac{1}{2} \left( \sum _{j=0}^{m} q_i^j b_{0,0}(i) \omega _i 2^{j+minBE} + \sum _{j=0}^{m} q_i^j b_{0,0}(i) \right) \\&= \frac{b_{0,0}(i)}{2} \left( \omega _i 2^{minBE} \sum _{j=0}^{m} \left( 2q_i\right) ^j + \sum _{j=0}^{m} q_i^j \right) \\&= \frac{b_{0,0}(i)}{2} \left( \omega _i 2^{minBE} \frac{1-(2q_i)^{m+1}}{1-2q_i} + \frac{1-q_i^{m+1}}{1-q_i} \right) . \end{aligned}$$

Therefore, \(b_{0,0}\) can be written as

$$\begin{aligned}&b_{0, 0}(i) \nonumber \\&\quad = \frac{2(1-2q_i) (1-q_i)}{\omega _i 2^{minBE} (1-(2q_i)^{m+1})(1 -q_i) + (1-q_i^{m+1})(1-2q_i) }. \end{aligned}$$
(17)

Now, let \(\tau _i\) be the conditional probability that slave node i performs CCA in a randomly chosen time slot under the condition that it is in the contention state. A CCA occurs when the backoff time counter is equal to zero, i.e., \(k = 0,\) regardless of its backoff stage. Hence, from (12) and (17), \(\tau _i\) can be given as

$$\begin{aligned} \tau _i&= \sum _{j=0}^{m} b_{j,0}(i) = \sum _{j=0}^{m} q_i^j b_{0,0}(i) = \frac{1-q_i^{m+1}}{1-q_i} b_{0,0}(i) \nonumber \\&= \frac{2(1-2q_i) (1-q_i^{m+1})}{\omega _i 2^{minBE} (1 -(2q_i)^{m+1})(1-q_i) + (1-q_i^{m+1})(1-2q_i) }. \end{aligned}$$
(18)

Note that \(\tau _i\) depends on the probability of the busy channel during CCA by slave node i and the probability that at least one of the \(n-1\) other slave nodes except slave node i is in the transmit state in a randomly chosen time, \(q_i.\) Let \(q_{tx,j}\) be the probability that slave node i transmits a packet in the steady state. Then,

$$\begin{aligned} q_i = 1 - \prod _{j\in N,\,j\ne i}\left( 1-q_{tx,j}\right) , \end{aligned}$$
(19)

and by (4),

$$\begin{aligned} q_{tx,j}&= \lim _{h \rightarrow \infty } \frac{ \sum _{t=1}^{h} T_{tx,j}(t)}{\sum _{t=1}^{h} T_{c,j}(t)+ \sum _{t=1}^{h} T_{tx,j}(t)+ \sum _{t=1}^{h} T_{s,j}(t)} \nonumber \\&= \frac{\bar{T}_{tx,j}}{(1+\alpha _j) \bar{T}_{c,j}+ (1+\beta _j)\bar{T}_{tx,j}}. \end{aligned}$$
(20)

Now, we return to determine \(\bar{T}_{c,i},\, \bar{T}_{p,i},\) and \(\bar{T}_{tx,i}.\) The average duration of the transmit state, \(\bar{T}_{tx,i},\) and the average duration used by the successfully transmitted data, \(\bar{T}_{p,i},\) of the slave node i in a round can be expressed as

$$\begin{aligned}&\bar{T}_{tx,i} = T_{data} \left( 1-q_i^{m+1}\right) \quad \text {and} \end{aligned}$$
(21)
$$\begin{aligned}&\bar{T}_{p,i} = q_{s,i}T_{data} \left( 1-q_i^{m+1}\right) , \end{aligned}$$
(22)

where \(T_{data}\) is the packet transmission time that is assumed to be fixed, \((1-q_i^{m+1})\) is the probability that slave node i goes into the transmit state from the contention state and \(q_{s,i}\) is the probability that a transmission is completed without collision.

To determine (21), we need to find the probability that a transmission of the slave node i is completed without collision, \(q_{s,i}.\) A collision occurs only if at least one of the other slave nodes is in the CCA state during slave node i is performing CCA, since ideal channel condition and no hidden terminal problem are assumed. Hence, \(q_{s,i}\) can be obtained as

$$\begin{aligned} q_{s,i} = \prod _{j\in N,\, j\ne i}\left( 1-\tau _j q_{c,j}\right) , \end{aligned}$$
(23)

where \(q_{c,j}\) is the probability that slave node j is in the contention state. Thus, \(\tau _j q_{c,j}\) is the probability that slave node j performs CCA in a randomly chosen time. Here, we can express the probability \(q_{c,j}\) that slave node j is in the contention state in a randomly chosen time as

$$\begin{aligned} q_{c,j}= & {} \lim _{h \rightarrow \infty } \frac{ \sum _{t=0}^{h-1} T_{c,j}(t)}{\sum _{t=0}^{h-1} T_{c,j}(t)+ \sum _{t=0}^{h-1} T_{tx,j}(t)+ \sum _{t=0}^{h-1} T_{s,j}(t)} \nonumber \\= & {} \frac{\bar{T}_{c,j}}{(1+\alpha _j) \bar{T}_{c,j}+ (1+\beta _j)\bar{T}_{tx,j}}. \end{aligned}$$
(24)

Now, the average duration for the contention state of the slave node i in a \(round,\,\bar{T}_{c,i},\) can be determined by calculating of the average time of backoff experienced by the node as

$$\begin{aligned} \bar{T}_{c,i}&= W\left( \sum _{v=0}^{m-1}\left( q_i^v \left( 1 -q_i\right) \sum _{j=0}^v \frac{W_{i,j}-1}{2} \right) + q_i^{m} \sum _{j=0}^{m} \frac{W_{i,j}-1}{2}\right) \nonumber \\&= \frac{W}{2}\left( \sum _{v=0}^{m-1}\left( q_i^v \left( 1-q_i\right) \sum _{j=0}^v \left( \omega _i 2^{minBE+j}-1\right) \right) \right. \nonumber \\&\quad \left. +\, q_i^{m} \sum _{j=0}^{m} \left( \omega _i 2^{minBE+j}-1\right) \right) \nonumber \\&= \frac{W}{2}\left( \sum _{v=0}^{m-1}\left( q_i^v \left( 1-q_i\right) \left( \omega _i 2^{minBE} \sum _{j=0}^v{2^j} -v-1\right) \right) \right. \nonumber \\&\quad \left. +\, q_i^{m} \left( \omega _i 2^{minBE} \sum _{j=0}^m{2^j} -m-1\right) \right) \nonumber \\&= \frac{W}{2}\left( \sum _{v=0}^{m-1}\left( q_i^v\left( 1-q_i\right) \left( \omega _i 2^{minBE} \left( 2^{v+1} -1\right) -v-1\right) \right) \right. \nonumber \\&\quad \left. +\, q_i^{m} \left( \omega _i 2^{minBE} \left( 2^{m+1}-1\right) -m-1\right) \right) , \end{aligned}$$
(25)

where W is the duration of the backoff slot, \(q_i^v (1-q_i) \sum _{j=0}^v \frac{W_{i,j}-1}{2}\) is the average time of backoff experienced by the slave node i in the case that it successfully gets the channel in the backoff stage v,  and \(q_i^{m} \sum _{j=0}^{m} \frac{W_{i,j}-1}{2}\) is the average time of backoff experienced by the slave node i in the backoff stage m regardless of the result of CCA.

Finally, we found all required parameters, \(\bar{T}_{p,i},\,\bar{T}_{c,i},\) and \(\bar{T}_{tx,i},\) to analyze throughput. The values of the parameters can be obtained iteratively. The iterative algorithm works as follows. First, we choose initial values of \(q_i\) and \(q_{s,i}\) appropriately and calculate \(\bar{T}_{tx,i},\, \bar{T}_{p,i}\) and \(\bar{T}_{c,i}\) using (22), (21) and (25). Next, we can find \(S_i,\, \tau _i,\,q_{tx,i}\) and \(q_{c,i}\) by (6), (18), (20) and (24). Once all variables are obtained, we update \(q_i\) and \(q_{s,i}\) in (19) and (23) using them. After that, we recalculate other variables in the same order as above. In this way, the procedures are repeated until the algorithm converges.

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Kim, J., Lee, JW. Energy adaptive MAC for wireless sensor networks with RF energy transfer: algorithm, analysis, and implementation. Telecommun Syst 64, 293–307 (2017). https://doi.org/10.1007/s11235-016-0176-0

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