Transmission delay minimization in wireless powered communication systems

Abstract

We study transmission delay minimization of a wireless powered communication (WPC) system in a point-to-point scenario with one hybrid access point (HAP) and one WPC node. In this type of communications, the HAP sends energy to the node at the downlink (DL) for a given time duration and the WPC node harvests enough radio frequency power. Then, at the uplink (UL) channel, the WPC node transmits its collected data in a given time duration to the HAP. Minimizing such round trip delay is our concern here. So, we have defined four optimization problems to minimize this delay by applying the optimal DL and UL time durations and also the optimal power at the HAP. These optimization problems are investigated here with thorough comparison of the obtained results. After that, we extend our study to the multiuser case with one HAP and K nodes and two different optimization problems are studied again in these cases.

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Notes

  1. 1.

    The HAP may have some information for the nodes and can transmit them in the DL in different time slot. However, without lose of generality, information transmission from the HAP to the nodes is not our concern here.

  2. 2.

    A complete survey can be found in [2, 43].

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Correspondence to Mohammad Lari.

Appendices

Appendix 1: convergence of (9)

We introduce a continuous and differentiable function \(z_1(x)\) over \((0,\infty )\) as

$$\begin{aligned} z_1(x) = \underbrace{\dfrac{2R_0m^m}{B\ln 2\varGamma (m)}}_{c_1}\dfrac{x^{m-1}{\text {e}}^{-mx}}{\ln (1+a_0x^2)} \end{aligned}$$
(37)

with \(a_0>0,\,m>0\) and redefine (9) as

$$\begin{aligned} \overline{\textsf {TD}}=\int _0^{\infty }z_1(x)dx=\underbrace{\int _0^1 z_1(x)dx}_{\overline{\textsf {TD}}_1}+\underbrace{\int _1^{\infty }z_1(x)dx}_{\overline{\textsf {TD}}_2} \end{aligned}$$
(38)

and we prove that for \(m>2\), \(\overline{\textsf {TD}}_1\) and \(\overline{\textsf {TD}}_2\) are convergent and so does \(\overline{\textsf {TD}}\).

For convergence testing of \(\overline{\textsf {TD}}_1\), we choose the comparison function \(p_1(x)\) as

$$\begin{aligned} p_1(x)=\dfrac{x^{m-1}}{a_0x^2}=\dfrac{1}{a_0x^{3-m}} \end{aligned}$$
(39)

such that

$$\begin{aligned} \lim _{x\rightarrow 0}\dfrac{z_1(x)}{p_1(x)}=c_1. \end{aligned}$$
(40)

The limit comparison test [48] says that \(\overline{\textsf {TD}}_1\) converges if and only if \(\int _0^1p_1(x)dx\) converges. We now use the direct comparison test (some times called p-test) for the convergence of \(\int _0^1p_1(x)dx\), which says that \(\int _0^1p_1(x)dx\) is a type II improper integral and converges if and only if \(3-m<1\). Therefore, choosing \(m>2\) leads to the convergence of \(\overline{\textsf {TD}}_1\) as well.

Then, for testing \(\overline{\textsf {TD}}_2\), from the Taylor expansion of \({\text {e}}^{mx}\) at large x, we know that

$$\begin{aligned} {\text {e}}^{mx}>\frac{(mx)^{m+1}}{(m+1)!}. \end{aligned}$$
(41)

So, for large x we can write

$$\begin{aligned} z_1(x)<\underbrace{\dfrac{c_1(m+1)!}{m^{m+1}}}_{c^\prime _1}\dfrac{x^{m-1}}{x^{m+1}\ln (1+a_0x^2)}<\underbrace{c^\prime _1\dfrac{1}{x^2}}_{q_1(x)}. \end{aligned}$$
(42)

\(\int _{1}^{\infty }q_1(x)\) is a type I improper integral which is always convergent. Therefore, \(\overline{\textsf {TD}}_2\) is convergent too.

Finally, we can conclude that choosing \(m>2\) leads to the convergence of \(\overline{\textsf {TD}}_1\) and \(\overline{\textsf {TD}}_2\) and consequently \(\overline{\textsf {TD}}\).

Appendix 2: derivation of (12) and (13)

Using Lagrangian method, the cost function is written as

$$\begin{aligned} J=-2T_0+\lambda \left( -R_0+BT_0\log _2 \left( 1+\beta a_0{\mathsf {h}}^2\right) \right) +\mu \left( 1-{{\mathbb {E}}}\left\{ \beta \right\} \right) \end{aligned}$$
(43)

where \(\lambda \ge 0\), \(\mu \ge 0\) represent the Lagrange multipliers corresponding to the constraints (11a) and (11b) respectively. Taking the partials with respect to \(T_0\) and \(\beta \), we obtain

$$\begin{aligned} \dfrac{\partial J}{\partial T_0}=-2+\lambda B\log _2\left( 1+\beta a_0{\mathsf {h}}^2\right) \end{aligned}$$
(44)

and

$$\begin{aligned} \dfrac{\partial J}{\partial \beta }=\dfrac{\lambda B}{\ln 2}\dfrac{T_0a_0{\mathsf {h}}^2}{1+\beta a_0{\mathsf {h}}^2}-\mu . \end{aligned}$$
(45)

Now, we have to find \(T_0^*\), \(\beta ^*\), \(\lambda ^*\) and \(\mu ^*\) such that

$$\begin{aligned}&-2+\lambda ^* B\log _2\left( 1+\beta ^* a_0{\mathsf {h}}^2\right) =0 \end{aligned}$$
(46a)
$$\begin{aligned}&\quad \dfrac{\lambda ^* B}{\ln 2}\dfrac{T_0^*a_0{\mathsf {h}}^2}{1+\beta ^* a_0{\mathsf {h}}^2}-\mu ^*=0 \end{aligned}$$
(46b)
$$\begin{aligned}&\quad \lambda ^*\left( -R_0+B T_0^*\log _2\left( 1+\beta ^*a_0{\mathsf {h}}^2\right) \right) =0 \end{aligned}$$
(46c)
$$\begin{aligned}&\quad \mu ^*\left( 1-{{\mathbb {E}}}\left\{ \beta ^*\right\} \right) =0. \end{aligned}$$
(46d)

From (46a), we can conclude that \(\lambda ^*\ne 0\) and then from (46b), we find that \(\mu ^*\ne 0\). Therefore, we can rewrite (46) as

$$\begin{aligned}&\lambda ^* B\log _2\left( 1+\beta ^* a_0{\mathsf {h}}^2\right) =2 \end{aligned}$$
(47a)
$$\begin{aligned}&\quad \dfrac{\lambda ^* B}{\ln 2}\dfrac{T_0^*a_0{\mathsf {h}}^2}{1+\beta ^* a_0{\mathsf {h}}^2}=\mu ^* \end{aligned}$$
(47b)
$$\begin{aligned}&\quad B T_0^*\log _2\left( 1+\beta ^*a_0{\mathsf {h}}^2\right) =R_0 \end{aligned}$$
(47c)
$$\begin{aligned}&\quad {{\mathbb {E}}}\left\{ \beta ^*\right\} =1. \end{aligned}$$
(47d)

From (47a), we write \(\lambda ^*\) in term of \(\beta ^*\) and from (47b) we write \(T_0^*\) in term of \(\beta ^*\) and insert them into (47c) to obtain

$$\begin{aligned} \left( 1+\beta ^*a_0{\mathsf {h}}^2\right) \ln ^2\left( 1+\beta ^*a_0{\mathsf {h}}^2\right) =\dfrac{2\ln 2a_0{\mathsf {h}}^2R_0}{B\mu ^*}. \end{aligned}$$
(48)

Finally, (48) can be solved according to

$$\begin{aligned} \beta ^*=\dfrac{{\text {e}}^{2{{\mathscr {W}}}_0\left( \sqrt{\dfrac{2\ln 2a_0{\mathsf {h}}^2R_0}{B\mu ^*}}\Bigg /2\right) }-1}{a_0{\mathsf {h}}^2} \end{aligned}$$
(49)

where \({{\mathscr {W}}}_0(.)\) represents Lambert-W function [42]. Then, by inserting (49) into (47c), \(T_0^*\) is attained and \(\mu ^*\) is calculated from (47d).

Appendix 3: convergence of (14) and (15)

The proof is similar to “Appendix 1” and we explain the process for both (15) and (14) respectively. First we use \(z_2(x)\)

$$\begin{aligned} z_2(x)=\underbrace{\dfrac{R_0\ln 2m^m}{B\varGamma (m)}}_{c_{21}}\dfrac{x^{m-1}{\text {e}}^{-mx}}{{\mathscr {W}}_0\left( \underbrace{\sqrt{\dfrac{\ln 2a_0R_0}{2B\mu ^*}}}_{c_{22}}\,x\right) }=c_{21}\dfrac{x^{m-1}{\text {e}}^{-mx}}{{\mathscr {W}}_0\left( c_{22}x\right) } \end{aligned}$$
(50)

over \((0,\infty )\) with \(c_{21}>0,\,c_{22}>0\) and redefine (15) as

$$\begin{aligned} \overline{\textsf {TD}}=\int _0^{\infty }z_2(x)dx=\underbrace{\int _0^1 z_2(x)dx}_{\overline{\textsf {TD}}_1}+\underbrace{\int _1^{\infty }z_2(x)dx}_{\overline{\textsf {TD}}_2}. \end{aligned}$$
(51)

Regarding the Taylor expansion of \({{\mathscr {W}}}_0(x)\) near 0 as [42]

$$\begin{aligned} {{\mathscr {W}}}_0(x)=\sum _{n=1}^{\infty }\dfrac{(-n)^{n-1}}{n!}x^n, \end{aligned}$$
(52)

and in a quite similar way to the “Appendix 1”, we can find that \(\overline{\textsf {TD}}\) in (15) is convergent when \(m>1\).

Now, we go on to the proof of convergence for (14) and introduce \(z^\prime _2(x)\) as

$$\begin{aligned} z^{\prime }_2(x)&= \underbrace{\dfrac{m^m}{\varGamma (m)}}_{c_{23}}\left( \dfrac{{\text {e}}^{2{{\mathscr {W}}}_0(c_{22}x)}x^{m-1}{\text {e}}^{-mx}}{a_0x^2}-\dfrac{x^{m-1}{\text {e}}^{-mx}}{a_0x^2}\right) \nonumber \\&= \underbrace{c_{23}\dfrac{{\text {e}}^{2{{\mathscr {W}}}_0(c_{22}x)}x^{m-1}{\text {e}}^{-mx}}{a_0x^2}}_{z^\prime _{21}(x)}-\underbrace{c_{23}\dfrac{x^{m-1}{\text {e}}^{-mx}}{a_0x^2}}_{z^\prime _{22}(x)}, \end{aligned}$$
(53)

then redefine (14) as

$$\begin{aligned} {{\mathbb {E}}}\{\beta ^*\}=\underbrace{\int _0^{1}z^\prime _{21}(x)dx}_{I_1}+\underbrace{\int _1^{\infty }z^\prime _{21}(x)dx}_{I_2}-\underbrace{\int _0^{1}z^\prime _{22}(x)dx}_{I_3}-\underbrace{\int _1^{\infty }z^\prime _{22}(x)dx}_{I_4}. \end{aligned}$$
(54)

\(I_1\) is a type II improper itegral and we can choose comparison function \(p^{\prime }_{21}(x)=\dfrac{x^{m-1}}{a_0x^2}=\dfrac{1}{a_0x^{3-m}}\) such that

$$\begin{aligned} \lim _{x\rightarrow 0}\dfrac{z^{\prime }_{21}(x)}{p^{\prime }_{21}(x)}=c_{23}, \end{aligned}$$
(55)

so, using the limit comparison test [48], we find that \(I_1\) is convergent when \(m>2\). Again, \(I_2\) is a type I improper integral and we know that for large x, \({\text {e}}^{{\mathscr {W}}_0(x)}<x\) [42]. Therefore,

$$\begin{aligned} z^{\prime }_{21}(x)<c_{23}\dfrac{2c_{22}xx^{m-1}{\text {e}}^{-mx}}{a_0x^2}, \end{aligned}$$
(56)

So, \(I_2\) is always convergent and does not depend on m. In a similar way, \(I_3\) is a type II improper integral and it is convergent for \(m>2\) and \(I_4\) is a type I improper integral, but always convergent.

Finally, we can conclude that choosing \(m>2\), leads to the convergence of both (14) and (15).

Appendix 4: derivation of (23), (24) and (25)

The cost function can be written as

$$\begin{aligned} J=-(T_1+T_2)+\lambda \left( -R_0+BT_2\log _2 \left( 1+\beta a_0{\mathsf {h}}^2\dfrac{T_1}{T_2}\right) \right) +\mu \left( 1-{{\mathbb {E}}}\left\{ \beta \right\} \right) \end{aligned}$$
(57)

where \(\lambda \ge 0\), \(\mu \ge 0\) represent the Lagrange multipliers corresponding to the constraints (22a) and (22b) respectively. Taking the partials with respect to \(T_1\), \(T_2\) and \(\beta \), we will have

$$\begin{aligned} \dfrac{\partial J}{\partial T_1}&= -1+\lambda \dfrac{B}{\ln 2}\dfrac{\beta a_0{\mathsf {h}}^2}{1+\beta a_0{\mathsf {h}}^2\dfrac{T_1}{T_2}} \end{aligned}$$
(58)
$$\begin{aligned} \dfrac{\partial J}{\partial T_2}&= -1+\lambda \left( B\log _2\left( 1+\beta a_0{\mathsf {h}}^2\dfrac{T_1}{T_2}\right) -\dfrac{B}{\ln 2}\dfrac{\beta a_0{\mathsf {h}}^2\dfrac{T_1}{T_2}}{1+\beta a_0{\mathsf {h}}^2\dfrac{T_1}{T_2}}\right) \end{aligned}$$
(59)

and

$$\begin{aligned} \dfrac{\partial J}{\partial \beta }=\dfrac{\lambda B}{\ln 2}T_2\dfrac{a_0{\mathsf {h}}^2\dfrac{T_1}{T_2}}{1+\beta a_0{\mathsf {h}}^2\dfrac{T_1}{T_2}}-\mu . \end{aligned}$$
(60)

Now, we have to find \(T_1^*\), \(T_2^*\), \(\beta ^*\), \(\lambda ^*\) and \(\mu ^*\) such that

$$\begin{aligned}&-1+\lambda ^*\dfrac{B}{\ln 2}\dfrac{\beta ^*a_0{\mathsf {h}}^2}{1+\beta ^*a_0{\mathsf {h}}^2\dfrac{T_1^*}{T_2^*}}=0 \end{aligned}$$
(61a)
$$\begin{aligned}&\quad -1+\lambda ^*\left( B\log _2\left( 1+\beta ^* a_0{\mathsf {h}}^2\dfrac{T_1^*}{T_2^*}\right) -\dfrac{B}{\ln 2}\dfrac{\beta ^*a_0{\mathsf {h}}^2\dfrac{T_1^*}{T_2^*}}{1+\beta ^*a_0{\mathsf {h}}^2\dfrac{T_1^*}{T_2^*}}\right) =0 \end{aligned}$$
(61b)
$$\begin{aligned}&\quad \dfrac{\lambda ^*B}{\ln 2}T_2^*\dfrac{a_0{\mathsf {h}}^2\dfrac{T_1^*}{T_2^*}}{1+\beta ^* a_0{\mathsf {h}}^2\dfrac{T_1^*}{T_2^*}}-\mu ^*=0 \end{aligned}$$
(61c)
$$\begin{aligned}&\quad \lambda ^*\left( -R_0+BT_2^*\log _2\left( 1+\beta ^*a_0{\mathsf {h}}^2\dfrac{T_1^*}{T_2^*}\right) \right) =0 \end{aligned}$$
(61d)
$$\begin{aligned}&\quad \mu ^*\left( 1-{{\mathbb {E}}}\left\{ \beta ^*\right\} \right) =0. \end{aligned}$$
(61e)

From (61a), we can conclude that \(\lambda ^*\ne 0\) and then from (61c), we find that \(\mu ^*\ne 0\). Therefore, we can rewrite (61) as

$$\begin{aligned}&\lambda ^*\dfrac{B}{\ln 2}\dfrac{\beta ^*a_0{\mathsf {h}}^2}{1+\beta ^*a_0{\mathsf {h}}^2\tau ^*}=1 \end{aligned}$$
(62a)
$$\begin{aligned}&\quad \lambda ^*B\log _2\left( 1+\beta ^* a_0{\mathsf {h}}^2\tau ^*\right) =1+\lambda ^*\dfrac{B}{\ln 2}\dfrac{\beta ^*a_0{\mathsf {h}}^2\tau ^*}{1+\beta ^*a_0{\mathsf {h}}^2\tau ^*} \end{aligned}$$
(62b)
$$\begin{aligned}&\quad \lambda ^*\dfrac{B}{\ln 2}T_2^*\dfrac{a_0{\mathsf {h}}^2\tau ^*}{1+\beta ^*a_0{\mathsf {h}}^2\tau ^*}=\mu ^* \end{aligned}$$
(62c)
$$\begin{aligned}&\quad BT_2^*\log _2\left( 1+\beta ^*a_0{\mathsf {h}}^2\tau ^*\right) =R_0 \end{aligned}$$
(62d)
$$\begin{aligned}&\quad {{\mathbb {E}}}\left\{ \beta ^*\right\} =1 \end{aligned}$$
(62e)

where \(\tau ^*=T_1^*/T_2^*\). From (62a), we write \(\lambda ^*\) in term of \(\beta ^*\) and \(\tau ^*\) to reduce the equations as

$$\begin{aligned}&\dfrac{1+\beta ^*a_0{\mathsf {h}}^2\tau ^*}{\beta ^*a_0{\mathsf {h}}^2}\ln \left( 1+\beta ^*a_0{\mathsf {h}}^2\tau ^*\right) =1+\tau ^* \end{aligned}$$
(63a)
$$\begin{aligned}&\quad T_2^*\tau ^*=\beta ^*\mu ^* \end{aligned}$$
(63b)
$$\begin{aligned}&\quad BT_2^*\log _2\left( 1+\beta ^*a_0{\mathsf {h}}^2\tau ^*\right) =R_0 \end{aligned}$$
(63c)
$$\begin{aligned}&\quad {{\mathbb {E}}}\left\{ \beta ^*\right\} =1. \end{aligned}$$
(63d)

Moreover from (63c) we solve \(T_2^*\) in term of \(\beta ^*\) and \(\tau ^*\). Therefore we will have

$$\begin{aligned}&\dfrac{1+\beta ^*a_0{\mathsf {h}}^2\tau ^*}{\beta ^*a_0{\mathsf {h}}^2}\ln \left( 1+\beta ^*a_0{\mathsf {h}}^2\tau ^*\right) =1+\tau ^* \end{aligned}$$
(64a)
$$\begin{aligned}&\quad \dfrac{R_0\ln 2}{B}\tau ^*=\beta ^*\mu ^*\ln \left( 1+\beta ^*a_0{\mathsf {h}}^2\tau ^*\right) \end{aligned}$$
(64b)
$$\begin{aligned}&\quad {{\mathbb {E}}}\left\{ \beta ^*\right\} =1. \end{aligned}$$
(64c)

Next, from (64a) and Lambert-W function definition [42], we can write

$$\begin{aligned} 1+\beta ^*a_0{\mathsf {h}}^2\tau ^*={\text {e}}^{1+{{\mathscr {W}}}_0\left( \dfrac{\beta ^*a_0{\mathsf {h}}^2-1}{\text {e}}\right) } \end{aligned}$$
(65)

and insert it into (64b) to obtain (23). In addition, \(\mu ^*\) is calculated from (64c). Then, \(\tau ^*\) is derived from (65) as

$$\begin{aligned} \tau ^*=\dfrac{{\text {e}}^{1+{{\mathscr {W}}}_0\left( \dfrac{\beta ^*a_0{\mathsf {h}}^2-1}{\text {e}}\right) }-1}{\beta ^*a_0{\mathsf {h}}^2} \end{aligned}$$
(66)

and \(T_2^*\) is obtained from (63b) as

$$\begin{aligned} T_2^*=\dfrac{\beta ^*\mu ^*}{\tau ^*}. \end{aligned}$$
(67)

Now, we can use (66) and (67) to extract (24) and (25) respectively.

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Lari, M. Transmission delay minimization in wireless powered communication systems. Wireless Netw 25, 1415–1430 (2019). https://doi.org/10.1007/s11276-018-1778-0

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Keywords

  • Delay minimization
  • Nakagami-m fading channel
  • Optimization
  • Power allocation
  • Time allocation
  • Wireless powered communications