# Optimized Gateway Placement for Interference Cancellation in Transmit-Only LPWA Networks
## 3 Single GW Placement for Two SNs
### IC and the Decoding Crescent  
Considering interfering packets from SNs \(s_{1}\) and \(s_{2}\), assuming \(s_{1}\) is the stronger packet, the optimized GW location should satisfy both the conditions for capture of \(s_{1}\) and ICD of \(s_{2}\). Therefore, combining Equations (6) and (11), we can get  
\[\frac{1}{\beta}\leq\frac{d_{2}}{d_{1}}\leq\beta^{\prime}\]  
\[\Longleftrightarrow\tau^{\frac{1}{\alpha}}\leq\frac{d_{2}}{d_{1}}\leq(z* \tau)^{-\frac{1}{\alpha}}, \tag{14}\]  
if and only if  
\[\tau^{\frac{1}{\alpha}}\leq(z*\tau)^{-\frac{1}{\alpha}}, \tag{15}\]  
i.e.,  
\[z\leq\frac{1}{\tau^{2}}. \tag{16}\]  
The inequality in Equation (16) has some interesting practical implications. \(z\) and \(\tau\) are controlled by separate physical mechanisms and separate parts of a packet. The minimum required SIR \(\tau\) depends mainly on the choice of modulation and coding of the data. For example, a bandwidth efficient high-order quadratic amplitude modulation (QAM) with no error correction code can necessitate a high value for \(\tau\), whereas a power efficient high-order frequency shift keying modulation coupled with an error correction code can enable a low value for \(\tau\)[31]. On the other hand, \(z\) is mainly controlled by the quality of the cancellation, which in turn, depends on the quality of synchronization and channel estimation, which is usually performed based on the packet's preamble and training sequence or pilot symbols [32; 33]. For example, a long preamble and a long training sequence (and a low-Doppler channel, i.e., long coherence time) can enable a low value of \(z\), whereas short versions of these will cause poor cancellation and a high value of \(z\). We next consider three cases involving \(z\) and \(\tau\geq 1\), especially: (1) \(z\leq\frac{1}{\tau^{2}}\); (2) \(\frac{1}{\tau^{2}}<z\leq\frac{1}{\tau}\); and (3) \(z>\frac{1}{\tau}\). Three corresponding lemmas will prove that only the first case yields a non-empty set of GW locations, such that capture and ICD are possible. Define the ordered pair \((s_{1},s_{2})\) to indicate that the packet from \(s_{1}\) is to be captured, in the presence of interference from the packet from only \(s_{2}\). Lemma 1: _For any ordered pair of SNs \((s_{1},s_{2})\), if \(z\leq\frac{1}{\tau^{2}}\)and\(\tau\geq 1\), the ICD circle will be inside the capture circle and there exists a region which satisfies both capture and IC requirements for GWs._  
Proof: According to definition of capture circle and ICD circle separately in Equations (8) and (12), we can get the center and radius of them as follows. Capture circle: center \(C_{cc}=\frac{s_{1}-\beta^{2}s_{2}}{1-\beta^{2}}\) and radius \(R_{cc}=\frac{\beta}{1-\beta^{2}}\|s_{1}-s_{2}\|\) and ICD circle: center \(C_{ic}=\frac{s_{2}-\beta^{2}s_{1}}{1-\beta^{2}}\) and radius \(R_{ic}=\frac{\beta^{\prime}}{\beta^{2}-1}\|s_{2}-s_{1}\|\) with \(\beta=\tau^{-1/n}\) and \(\beta^{\prime}=(z\tau)^{-\frac{1}{\alpha}}\). The distance between the two circle centers is  
\[D_{c_{1},c_{2}}=\|\frac{s_{1}-\tau^{-2/n}s_{2}}{1-\tau^{-2/n}}-\frac{s_{2}-(z \tau)^{-2/n}s_{1}}{1-(z\tau)^{-2/n}}\|. \tag{17}\]  
Set \(\alpha=\tau^{2/n},\gamma=z^{2/n}\), then Equation (17) can be expressed  
\[D_{c_{1},c_{2}}=\|\frac{\alpha s_{1}-s_{2}}{\alpha-1}-\frac{s_{2}\alpha\gamma -s_{1}}{\alpha\gamma-1}\|.\]Then, after some algebra operations, we have  
\[D_{c_{1},c_{2}}=\|\frac{\alpha^{2}\gamma-1}{(\alpha-1)(\alpha\gamma-1)}\|*\|s_{1 }-s_{2}\|. \tag{18}\]  
On the other hand, the difference between the radii of the two circles is  
\[D_{r_{1},r_{2}}=\|(\frac{\tau^{-1/n}}{1-\tau^{-2/n}}-\frac{z^{-1/n}\tau^{-1/n}}{ z^{-2/n}\tau^{-2/n}-1})\|*\|s_{1}-s_{2}\|. \tag{19}\]  
After substituting \(z\) and \(\tau\), and more algebra, we have  
\[D_{r_{1},r_{2}}=\|\frac{\alpha^{1/2}(\alpha\gamma-1)+\alpha^{1/2}\gamma^{1/2}( \alpha-1)}{(\alpha-1)(\alpha\gamma-1)}\|*\|s_{1}-s_{2}\|. \tag{20}\]  
If we can prove that \(D_{r_{1},r_{2}}\geq D_{c_{1},c_{2}}\) and \(R_{cc}\geq R_{ic}\), we can prove that the ICD circle will be inside of the capture circle.