An optimization based approach to enhance the throughput and energy efficiency for cognitive unmanned aerial vehicle networks

Abstract

Unmanned aerial vehicle (UAV) is a promising technology to serve as a mission-critical communication, where human operation or intervention is risky, dangerous, impossible or critical in terms of time and cost. The performance of the UAV depends highly on the operating radio environment e.g. multipath shadowing and fading, and spectrum availability. Therefore, a cognitive radio-based UAV is considered to utilize the radio spectrum opportunistically, and minimize the network congestion during a disaster on preallocated channel or degradation of the quality of service. In this paper, we have presented an analytical model for cognitive UAV downlink communication between the UAV and the ground nodes. This model aims to maximize the throughput and improve the energy efficiency by minimizing the power consumption for the UAVs. Specifically, we developed two optimization approaches to address the energy efficiency with and without the throughput of the user as constraints. Simulation results for the energy efficiency without constraints show that $11.16\%$ and $6.43\%$ performance improvement can be achieved for throughput, and energy efficiency respectively in comparison to unoptimized one. Similar behavior has observed with the energy efficiency with constraints and a significant improvement has achieved in throughput performance e.g. $64.02\%$ with the cost of energy.

Appendices

Appendices

1.1 Proof of optimal power transmission without constraint when $\varOmega =0$

Now deriving $EE_D$ in Eq. (11) with respect to $P_s$ we get,

$$\begin{aligned}&\frac{dEE_{D}}{dP_s} = 0 \nonumber \\&\quad \Rightarrow \frac{p_r (\varOmega =0) t_d (p_f - 1) \log (1+\frac{P_{s}g_{ss}^{2}}{\sigma ^{2}})}{\log (2) (t_s + t_d) (P_{ssp}+P_{cir}+P_{s})^2} \nonumber \\&\qquad - \frac{p_r (\varOmega =0) t_d (p_f - 1) g_{ss}^{2}}{\log (2) (t_s + t_d) \sigma ^{2}(1+\frac{P_{s}g_{ss}^{2}}{\sigma ^{2}})(P_{ssp}+P_{cir}+P_{s})} = 0 \nonumber \\&\quad \Rightarrow \frac{\log (1+\frac{P_{s}g_{ss}^{2}}{\sigma ^{2}})}{P_{ssp}+P_{cir}+P_{s}} = \frac{g_{ss}^{2}}{\sigma ^{2}(1+\frac{P_{s}g_{ss}^{2}}{\sigma ^{2}})} \nonumber \\&\quad \Rightarrow \log \left( \frac{P_s g_{ss}^2+\sigma ^{2}}{\sigma ^{2}} \right) \nonumber \\&\quad = \frac{P_{ssp} g_{ss}^2+P_{cir} g_{ss}^2 + P_s g_{ss}^2}{P_s g_{ss}^2 + \sigma ^{2}} \nonumber \\&\quad \Rightarrow \log \left( \frac{P_s g_{ss}^2+\sigma ^{2}}{\sigma ^{2}} \right) -1\nonumber \\&\quad = \frac{P_{ssp} g_{ss}^2+P_{cir} g_{ss}^2 + P_s g_{ss}^2}{P_s g_{ss}^2 + \sigma ^{2}}-1 \nonumber \\&\quad \Rightarrow \log \left( \frac{P_s g_{ss}^2+\sigma ^{2}}{\sigma ^{2}} \right) + \ln {e^{-1}}\nonumber \\&\quad =\frac{P_{ssp} g_{ss}^2+P_{cir}g_{ss}^2-\sigma ^{2}}{P_s g_{ss}^2+\sigma ^2} \nonumber \\&\quad \Rightarrow \log \left( \left( \frac{P_s g_{ss}^2+\sigma ^{2}}{\sigma ^{2}} \right) \exp (-1) \right) \nonumber \\&\quad =\frac{P_{ssp} g_{ss}^2 + P_{cir}g_{ss}^2-\sigma ^{2}}{P_s g_{ss}^2+\sigma ^2} \nonumber \\&\quad \Rightarrow \left( \frac{P_s g_{ss}^2+\sigma ^{2}}{\sigma ^{2}} \right) \exp (-1) \nonumber \\&\quad = \exp \left( \frac{P_{ssp} g_{ss}^2+P_{cir}g_{ss}^2-\sigma ^{2}}{P_s g_{ss}^2+\sigma ^2} \right) \nonumber \\&\quad \Rightarrow \frac{\exp (-1)(P_{ssp} g_{ss}^2+P_{cir}g_{ss}^2-\sigma ^{2})}{\sigma ^{2}} \nonumber \\&\quad = \left( \frac{P_{ssp}g_{ss}^2+P_{cir}g_{ss}^2-\sigma ^{2}}{P_s g_{ss}^2+\sigma ^2} \right) \nonumber \\&\qquad \exp \left( \frac{P_{ssp} g_{ss}^2 + P_{cir}g_{ss}^2-\sigma ^{2}}{P_s g_{ss}^2+\sigma ^2} \right) \end{aligned}$$

(6.31)

Using the Lambert method, we rewrite in Eq. (6.31)

$$\begin{aligned}&W\left( \frac{\exp (-1)(P_{ssp}g_{ss}^2 + P_{cir} g_{ss}^2 - \sigma ^{2})}{\sigma ^{2}} \right) \nonumber \\&\quad = \frac{P_{ssp} g_{ss}^2 + P_{cir}g_{ss}^2 - \sigma ^{2}}{P_s g_{ss}^2 + \sigma ^2} \nonumber \\&\Rightarrow P_s g_{ss}^2 + \sigma ^{2} = \frac{P_{ssp} g_{ss}^2+P_{cir}g_{ss}^2-\sigma ^{2}}{W\left( \frac{\exp (-1)(P_{ssp} g_{ss}^2 + P_{cir} g_{ss}^2-\sigma ^{2})}{\sigma ^{2}} \right) } \nonumber \\&\therefore P_s^1 =\frac{P_{ssp} g_{ss}^2 + P_{cir} g_{ss}^2 - \sigma ^{2}}{W\left( 0,\frac{\exp (-1)(P_{ssp} g_{ss}^2 + P_{cir}g_{ss}^2 - \sigma ^{2})}{\sigma ^{2}} \right) g_{ss}^2} - \frac{\sigma ^{2}}{g_{ss}^2} \end{aligned}$$

(6.32)

1.2 Proof of optimal power transmission without constraint when $\varOmega =1$

Here deriving $EE_I$ in Eq. (12) with respect to $P_s$ we get,

$$\begin{aligned}&\frac{dEE_{I}}{dP_s} = 0 \nonumber \\&\quad \Rightarrow \log \left( 1+\frac{P_s g_{ss}^2}{\sigma ^{2}(1+\gamma _{p} g_{s}p^2)} \right) \nonumber \\&\quad =\frac{P_{ssp} g_{ss}^2+P_{cir} g_{ss}^2+P_s g_{ss}^2}{ P_s g_{ss}^2+(1+\gamma _{p} g_{ps}^2)\sigma ^2 } \nonumber \\&\quad \Rightarrow \log \left( 1+\frac{P_s g_{ss}^2}{\sigma ^{2}(1+\gamma _{p} g_{ps}^2)}\right) + \ln {e^{-1}} \nonumber \\&\quad = \frac{P_{ssp} g_{ss}^2+P_{cir} g_{ss}^2 - (1+\gamma _{p} g_{ps}^2)\sigma ^2}{P_s g_{ss}^2+(1+\gamma _{p} g_{ps}^2)\sigma ^2 } \nonumber \\&\quad \Rightarrow \left( \frac{P_{ssp} g_{ss}^2 + P_{cir} g_{ss}^2-(1+\gamma _{p} g_{ps}^2)\sigma ^2}{\sigma ^{2}(1+\gamma _{p} g_{ps}^2)}\right) \,\exp (-1) \nonumber \\&\quad =\left( \frac{P_{ssp} g_{ss}^2 + P_{cir} g_{ss}^2 - (1+\gamma _{p} g_{ps}^2)\sigma ^2}{ P_s g_{ss}^2 + (1+\gamma _{p} g_{ps}^2)\sigma ^2 }\right) \nonumber \\&\qquad \exp \left( \frac{P_{ssp} g_{ss}^2 + P_{cir} g_{ss}^2-(1+\gamma _{p} g_{ps}^2) \sigma ^2}{ P_s g_{ss}^2+(1+\gamma _{p} g_{ps}^2)\sigma ^2 }\right) \end{aligned}$$

(6.33)

Using the Lambert method, we rewrite in Eq. (6.33) as follows

$$\begin{aligned}&W\left( \left( \frac{P_{ssp}g_{ss}^2+P_{cir}g_{ss}^2-(1+\gamma _{p} g_{ps}^2)\sigma ^2}{\sigma ^{2}(1+\gamma _{p} g_{ps}^2)}\right) \,e^{-1}\right) \nonumber \\&\quad = \left( \frac{P_{ssp} g_{ss}^2+P_{cir} g_{ss}^2-(1+\gamma _{p} g_{ps}^2)\sigma ^2}{ P_s g_{ss}^2+(1+\gamma _{p} g_{ps}^2)\sigma ^2 }\right) \nonumber \\&\quad \Rightarrow \, \frac{1}{W\left( \left( \frac{P_{ssp} g_{ss}^2 + P_{cir} g_{ss}^2-(1+\gamma _{p} g_{ps}^2)\sigma ^2}{\sigma ^{2}(1+\gamma _{p} g_{ps}^2)}\right) \,e^{-1}\right) } \nonumber \\&\quad = \frac{P_{s} g_{ss}^2+(1+\gamma _{p} g_{ps}^2)\sigma ^2}{P_{ssp} g_{ss}^2+ P_{cir} g_{ss}^2-(1+\gamma _{p} g_{ps}^2)\sigma ^2 } \nonumber \\&\quad \Rightarrow \, \frac{P_{ssp} g_{ss}^2 + P_{cir} g_{ss}^2 - (1+\gamma _{p} g_{ps}^2)\sigma ^2}{ W\left( \left( \frac{P_{ssp} g_{ss}^2 + P_{cir} g_{ss}^2-(1+\gamma _{p} g_{ps}^2)\sigma ^2}{\sigma ^{2} (1+\gamma _{p} g_{ps}^2)}\right) \,e^{-1}\right) }\nonumber \\&\quad = P_{s} g_{ss}^2+(1+\gamma _{p} g_{ps}^2)\sigma ^2 \nonumber \\&\quad \Rightarrow P_{s}^{2}= \frac{\left( P_{ssp} g_{ss}^2 + P_{cir} g_{ss}^{2}-\left( 1+ \gamma _{p} g_{ps}^{2} \right) \sigma ^{2} \right) }{W\left( 0,\left( \frac{P_{ssp} g_{ss}^2+P_{cir} g_{ss}^{2}-\left( 1+\gamma _{p} g_{ps}^{2} \right) \sigma ^{2}}{\sigma ^{2}\left( 1+ \gamma _{p} g_{ss}^{2} \right) } \right) \exp (-1) \right) g_{ss}^{2}} \nonumber \\&\quad - \frac{\left( 1+\gamma _{p} g_{ps}^{2} \right) \sigma ^{2}}{g_{ss}^{2}} \end{aligned}$$

(6.34)

1.3 Proof of optimal power transmission with constraint when $\varOmega =0$

The Lagrangian approach [40] providing the solution to the optimization problem stated above in Eqs. (20a) and (20b) for the inactive $PT_x$ ($\varOmega = 0$) is as follows:

$$\begin{aligned} l(P_s,\delta _a) = P_s + \delta _a (S_{D.min}-S_D ) \end{aligned}$$

(6.35)

here $\delta _a$ is positive variable mutual to the constraint given in Eq. (20b). Now deriving $l(P_s,\delta _a)$ in Eq. (6.35) with respect to $\delta _a$, we get:

$$\begin{aligned} \frac{\mathrm {d} l(P_s,\delta _a)}{\mathrm {d} \delta _a} = S_{D.min} +\frac{t_d p_r(\varOmega = 0) (p_f-1) \log \left( 1+\frac{g_{ss}^2 P_s}{\sigma ^2} \right) }{\log (2)(t_s+t_d)} \end{aligned}$$

(6.36)

Now taking Eq. (6.36) and equaling it to zero, corresponding $P_s^{3}$ is:

$$\begin{aligned}&\frac{dl(P_s,\delta _a)}{d\delta _a} = 0 \nonumber \\&\quad \Rightarrow S_{D.min}+\frac{t_d p_r(\varOmega = 0) (p_f-1) \log \left( 1+\frac{g_{ss}^2P_s}{\sigma ^2}\right) }{\log (2)(t_s+t_d)}=0 \nonumber \\&\quad \Rightarrow \log (2) (t_s+t_d) S_{D.min} + t_d p_r(\varOmega = 0) (p_f-1) \log \left( 1+\frac{g_{ss}^2P_s}{\sigma ^2} \right) = 0 \nonumber \\&\quad \Rightarrow \, \log \left( 1+\frac{g_{ss}^2P_s}{\sigma ^2} \right) \nonumber \\&\quad = - \frac{\log (2) (t_s+t_d) S_{D.min}}{t_d p_r(\varOmega = 0) (p_f-1)} \nonumber \\&\quad \Rightarrow \, 1+\frac{g_{ss}^2P_s}{\sigma ^2} \nonumber \\&\quad =\exp {\left( - \frac{\log (2)(t_s+t_d)S_{D.min}}{t_d p_r(\varOmega = 0) (p_f-1)} \right) } \nonumber \\&\quad \Rightarrow \, P_{s}^{3} = \frac{\sigma ^2}{g_{ss}^2}\nonumber \\&\quad \left( \exp \left( - \frac{\log (2)(t_s+t_d)S_{D.min}}{t_d p_r(\varOmega = 0) (p_f-1)} \right) - 1 \right) \end{aligned}$$

(6.37)

Finally putting the value of $P_s^{3}$ in Eq. (6.35), we get

$$\begin{aligned} l(P_s^{3},\delta _a) = P_s^{3} \end{aligned}$$

(6.38)

1.4 Proof of optimal power transmission with constraint when $\varOmega =1$

The Lagrangian approach [40] providing the solution to the optimization problem stated above in Eqs. (22a), and (22b) for the active $PT_x$ ($\varOmega = 1$) is as follows:

$$\begin{aligned} l(P_s,\delta _b) = P_s + \delta _{b} (S_{I.min} - S_I) \end{aligned}$$

(6.39)

here $\delta _b$ is positive variable mutual to the constraint given in Eq. (22b). Now deriving $l(P_s,\delta _b)$ in Eq. (6.39) with respect to $\delta _b$, we get:

$$\begin{aligned}&\frac{\mathrm {d} l(P_s,\delta _b)}{\mathrm {d} \delta _b}\nonumber \\&\quad = S_{I.min} - \frac{t_d p_r(\varOmega = 1) (1-p_d) \log \left( 1+\frac{g_{ss}^2 P_s}{(1+\gamma _p g_{ps}^2)\sigma ^2 } \right) }{\log (2)(t_s+t_d)} \end{aligned}$$

(6.40)

Now taking Eq. (6.40) and equaling it to zero, corresponding $P_s^{4}$ is:

$$\begin{aligned}&\frac{dl(P_s,\delta _b)}{d\delta _b} = 0 \nonumber \\&\quad \Rightarrow S_{I.min}+\frac{t_d p_r(\varOmega = 1) (p_d-1) \log \left( 1+\frac{g_{ss}^2 P_s}{(1+\gamma _p g_{ps}^2)\sigma ^2} \right) }{\log (2)(t_s+t_d)} = 0 \nonumber \\&\quad \Rightarrow \log (2)(t_s+t_d) S_{I.min} \nonumber \\&\qquad +t_d p_r(\varOmega = 1) (p_d-1) \log \left( 1+\frac{g_{ss}^2 P_s}{(1+\gamma _p g_{ps}^2)\sigma ^2} \right) =0 \nonumber \\&\quad \Rightarrow \log \left( 1+\frac{g_{ss}^2 P_s}{(1+\gamma _p g_{ps}^2)\sigma ^2} \right) =- \frac{\log (2)(t_s+t_d)S_{I.min}}{t_d p_r(\varOmega = 1)(p_d-1)} \nonumber \\&\quad \Rightarrow \, 1+\frac{g_{ss}^2 P_s}{(1+\gamma _p g_{ps}^2) \sigma ^2 } =\exp {\left( - \frac{\log (2)(t_s+t_d)S_{I.min}}{t_d p_r(\varOmega = 1)(p_d-1)} \right) } \nonumber \\&\quad \Rightarrow \, P_s^{4} = \frac{(1+\gamma _p g_{ps}^2)\sigma ^2}{g_{ss}^2} \left( \exp {\left( -\frac{\log (2)(t_s+t_d) S_{I.min}}{t_d p_{r}\left( \varOmega =1 \right) (p_d-1)} \right) } -1 \right) \end{aligned}$$

(6.41)

Finally putting the value of $P_s^{4}$ in Eq. (6.39), we get

$$\begin{aligned} l(P_s^{4},\delta _b) = P_s^{4} \end{aligned}.$$

(6.42)

1.5 Proof of optimal power transmission for interference minimization when $\varOmega =1$

To solve the optimization problem in Eq. (28), we assume the transmit power $P_s$ on the interval $[0,P_{max}]$. As shown in Fig. 15, we can see that $S_{I}(P_s)$ increases at first and then stables as $P_s$ increases. So $S_{I}(P_s)$ achieves the maximum value between $P_s$ and $P_{max}$. For this reason, we find the optimization solution due to the consideration of $P_s \le P_{max}$ (Eq. (28c)).

Fig. 15

Interference throughput and minimum required interference throughput at various transmit power considering $n = 2$ and $p_r (\varOmega =1) = 0.5$

The Lagrangian approach [40] providing the solution to the optimization problem stated above in Eqs. (28a), (28b) and (28c) for the active $PT_x$ ($\varOmega = 1$) is as follows:

$$\begin{aligned} l(P_s,\delta _b,\delta _c) = P_s g_{sp}^{2} + \delta _{b} (S_{I.min} - S_I)+\delta _c(P_s-P_{max}) \end{aligned}$$

(6.43)

here $\delta _b$ and $\delta _c$ are positive variable mutual to the constraints given in Eqs. (28b) and (28c).

The Karush–Kuhn–Tucker (KKT) conditions are necessary to obtain optimal solution, and the following KKT conditions are

$$ \left\{ {\begin{array}{*{20}c} {} & {\delta _{b} \ge 0,\delta _{c} \ge 0} \\ {} & {\delta _{b} (S_{{I.min}} - S_{I} ) = 0} \\ {} & {\delta _{c} (P_{s} - P_{{max}} ) = 0} \\ {} & {\frac{{{\text{d}}l(P_{s} ,\delta _{b} ,\delta _{c} )}}{{{\text{d}}P_{s} }} = 0,\frac{{{\text{d}}l(P_{s} ,\delta _{b} ,\delta _{c} )}}{{{\text{d}}\delta _{b} }} = 0,\frac{{{\text{d}}l(P_{s} ,\delta _{b} ,\delta _{c} )}}{{{\text{d}}\delta _{c} }} = 0} \\ \end{array} } \right. $$

(6.44)

Now deriving $l(P_s,\delta _b,\delta _c)$ in Eq. (6.43) with respect to $\delta _b$, we get:

$$\begin{aligned}&\frac{\mathrm {d} l(P_s,\delta _b,\delta _c,\delta _d)}{\mathrm {d} \delta _b}\nonumber \\&\quad = S_{I.min} - \frac{t_d p_r(\varOmega = 1) (1-p_d)\log \left( 1+\frac{g_{ss}^2P_s}{(1+\gamma _p g_{ps}^2)\sigma ^2 } \right) }{\log (2)(t_s+t_d)} \end{aligned}$$

(6.45)

Now taking Eq. (6.45) and equaling it to zero, corresponding $P_s$ is:

$$\begin{aligned}&\frac{dl(P_s,\delta _b,\delta _c)}{d\delta _b} = 0 \nonumber \\&\quad \Rightarrow S_{I.min}+\frac{t_d p_r(\varOmega = 1) (p_d-1) \log \left( 1+\frac{g_{ss}^2P_s}{(1+\gamma _p g_{ps}^2)\sigma ^2} \right) }{\log (2)(t_s+t_d)} = 0 \nonumber \\&\quad \Rightarrow \log \left( 1+\frac{g_{ss}^2 P_s}{(1+\gamma _p g_{ps}^2)\sigma ^2} \right) \nonumber \\&\quad =- \frac{\log (2)(t_s+t_d)S_{I.min}}{t_d p_r(\varOmega = 1)(p_d-1)} \nonumber \\&\quad \Rightarrow \, 1+\frac{g_{ss}^2 P_s}{(1+\gamma _p g_{ps}^2) \sigma ^2 }\nonumber \\&\quad =\exp {\left( - \frac{\log (2)(t_s+t_d)S_{I.min}}{t_d p_r(\varOmega = 1)(p_d-1)} \right) } \nonumber \\&\quad \Rightarrow \, P_s = \frac{(1+\gamma _p g_{ps}^2)\sigma ^2}{g_{ss}^2} \nonumber \\&\qquad \left( \exp {\left( -\frac{\log (2)(t_s+t_d) S_{I.min}}{t_d p_{r}\left( \varOmega =1 \right) (p_d-1)} \right) } -1 \right) \end{aligned}$$

(6.46)

Next, deriving $l(P_s,\delta _b,\delta _c)$ in Eq. (6.43) with respect to $\delta _c$, we get:

$$\begin{aligned}&\frac{dl(P_s,\delta _b,\delta _c)}{d\delta _c}= P_s - P_{max} = 0\nonumber \\&\therefore P_s = P_{max} \end{aligned}$$

(6.47)

Finally combining Eqs. (6.46) and (6.47), we get

$$\begin{aligned} P_s^{5}= & {} P_{max} \nonumber \\= & {} \frac{(1+\gamma _p g_{ps}^2)\sigma ^2}{g_{ss}^2} \left( \exp {\left( -\frac{\log (2)(t_s+t_d) S_{I.min}}{t_d p_{r}\left( \varOmega =1 \right) (p_d-1)} \right) } -1 \right) \end{aligned}$$

(6.48)

The optimal solution $P_s^{5}$ is satisfy to the Lagrangian function in Eq. (6.43) when the KKT conditions are $S_{I.min} - S_I = 0$ and $P_s - P_{max} = 0$.