## 4 A Transmission Strategy for Minimizing Energy Consumption
### Energy Consumption Analysis  
Let us consider a certain link of a network like the one described in Section 3, consisting of a transmitter node, which will periodically send packets, and a receiver node, which will measure the RSS or the TOA of all received packets. The receiver will average all the measurements (\(n\)) obtained during the localization period \(T_{0}\), so that the standard deviation of the measurements can be reduced by a factor of \(\sqrt{n}\). Therefore, a desired accuracy in the measurement of the RSS or the TOA imposes a minimum number of received packets \(n_{obj}\). On the other hand, when \(k\) packets are transmitted, the average number of packets that are finally received is:  
\[n=k\cdot PRP=\frac{T_{0}}{T}\cdot PRP \tag{4}\]  
where \(T\) is the time interval between consecutive transmitted packets. Consequently, a desired accuracy in the RSS or the TOA imposes a relationship between \(T\) and \(PRP\), given \(T_{0}\). As shown in Section 3.1, the \(PRP\) depends on the \(snr\) at the receiver. As a result, a desired accuracy imposes a relationship between \(snr\) and \(T\). Among all \(snr\)-\(T\) pairs that ensure the required number of received packets, we want to find the one with the lowest energy consumption. The consumed energy at transmission during \(T_{0}\) is given by:  
\[E_{TX}=k\cdot E_{p}=k\cdot p_{tx}\cdot T_{p}=\frac{n\cdot p_{tx}\cdot T_{p}}{PRP} \tag{5}\]  
where \(k\) is the number of transmitted packets during the period \(T_{0}\), \(E_{p}\) is the consumed energy when transmitting a single packet, \(p_{tx}\) is the transmission power (in natural units), \(T_{p}\) is the time it takes to send a packet and \(n\) is the average number of received packets. Since the localization procedure is periodic with period \(T_{0}\), this consumed energy is an appropriate parameter to analyze the energy efficiency. If we assume that the receiver is awake and listening to possible receptions only at given periods (previously negotiated with the transmitter), the consumed energy at reception during \(T_{0}\) is given by:  
\[E_{RX}=C_{rx}\cdot T_{r} \tag{6}\]  
where \(T_{r}\) is the reception period and \(C_{rx}\) is the consumption when the node is in receiving mode. \(E_{RX}\) is constant, as \(C_{rx}\) is a constant value that depends on the hardware characteristics. Therefore, in order to minimize the total energy consumption, we will only need to minimize the energy consumption during transmission. Using Equations (1) and (3), \(E_{TX}\) can be expressed as a function of the signal to noise ratio at the receiver:  
\[E_{TX}=\frac{n\ T_{p}\ v_{b}\ d^{\eta}\ n_{rx}}{2\ WA}\cdot\frac{x^{2}}{(1-Q(x ))^{8l}}\cdot\mathcal{X} \tag{7}\]  
where \(n_{rx}\) and \(a\) are \(N_{RX}\) and \(A\) expressed in natural units, \(\mathcal{X}\) is a log-normal random variable with parameters \(\mu_{\mathcal{X}}=0\) and \(\sigma_{\mathcal{X}}=\sigma\cdot\ln 10/10\), and \(x\) is given by:  
\[x=\sqrt{2 snr\frac{W}{v_{b}}} \tag{8}\]  
where \(snr\) is the mean value of the \(SNR\) expressed in natural units. It can be obviously noticed that, for a given distance, the greater the required accuracy (described by \(n\)), the more energy will be consumed. On the other hand, as the distance between transmitter and receiver increases, the consumed energy increases as well, for a given desired accuracy.