# New Results for the Error Rate Performance of LoRa Systems over Fading Channels
## 3 Main Results
### Symbol Error Probability for Hoyt Channels  
Under Hoyt (Nakagami-\(q\)) fading, the RV \(h^{2}\) follows a non-central chi-square distribution with PDF given as [28]  
\[f_{h^{2}}(y) =\sqrt{\frac{1}{2}+\frac{1}{4\eta}+\frac{\eta}{4}}\exp\biggl{[}- \biggl{(}\frac{1}{2}+\frac{1}{4\eta}+\frac{\eta}{4}\biggr{)}y\biggr{]}\] \[\times I_{0}\biggl{[}\biggl{(}\frac{1}{4\eta}-\frac{\eta}{4} \biggr{)}y\biggr{]}, \tag{29}\]  
where \(\eta=q^{2}\), with \(0<q\leq 1\) being a parameter related to the fade intensity. For \(q=1\), (29) reduces to the exponential distribution (Rayleigh fading).  
An exact expression for \(P_{\rm s}\) can be deduced using the following proposition.  
**Proposition 6**.: _Under Hoyt fading, \(P_{\rm s}\) can be expressed in terms of a single integral as_  
\[P_{\rm s}^{\rm Hoyt} =\frac{\eta+1}{2\sqrt{A}}\int_{0}^{\infty}e^{-x/2}[1-(1-e^{-x/2})^ {L}]\] \[\times e^{\frac{N\eta x([1+\eta]^{2}+4N\eta)}{2A}}\,I_{0}\biggl{[} \frac{N(\eta^{2}-1)\gamma x}{2A}\biggr{]}\mathrm{d}x \tag{30}\]  
_where_  
\[A=1+2N\gamma+\eta^{2}(1+2N\gamma)+2\eta(1+2N\gamma+2N^{2}\gamma^{2}) \tag{31}\]  
Proof.: The proof can be concluded by following a similar line of arguments as in the proof of Proposition 1. Specifically, by substituting (29) into (9) and changing the order of integration, \(P_{\rm s}\) can be expressed as  
\[P_{\rm s}=\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{4\eta}+\frac{\eta }{4}}\int_{0}^{\infty}e^{-x/2}[1-(1-e^{-x/2})^{L}]\] \[\times\biggl{\{}\int_{0}^{\infty}e^{-\bigl{(}N\gamma+\frac{1}{2} +\frac{1}{4\eta}+\frac{\eta}{4}\bigr{)}y}I_{0}\biggl{[}\biggl{(}\frac{1}{4\eta }-\frac{\eta}{4}\biggr{)}y\biggr{]}\] \[\times\,I_{0}\biggl{(}\sqrt{2 Nxy}\biggr{)}\mathrm{d}y\biggr{\}} \mathrm{d}x. \tag{32}\]  
The inner integral, i.e., with respect to \(y\), can be evaluated in closed form by employing [29] (Equation (3.15.17/15)), yielding (30), thus completing the proof.  
An accurate approximation for \(P_{\rm s}\) can be obtained using the following proposition. The following result holds.  
**Proposition 7**.: _Under Hoyt fading, an accurate approximation for \(P_{\rm s}\) can be deduced as_  
\[P_{\rm s}^{\rm Hoyt}\approx 1-\frac{1}{\gamma}\sqrt{\frac{1}{4 \eta}+\frac{\eta}{4}+\frac{1}{2}}\] \[\times\mathcal{I}\biggl{[}\sqrt{2N},\sqrt{\bar{x}_{N^{\prime}}}, \frac{1}{\gamma}\biggl{(}\frac{1}{4\eta}-\frac{\eta}{4}\biggr{)},\frac{1}{ \gamma}\biggl{(}\frac{1}{4\eta}+\frac{\eta}{4}+\frac{1}{2}\biggr{)}\biggr{]} \tag{33}\]_where_  
\[\mathcal{I}(a,b,c,p) =\frac{a^{2}}{2p+a^{2}}\exp\left[-\frac{b^{2}p}{2p+a^{2}}\right]\] \[\times\sum_{k=0}^{\infty}\sum_{n=0}^{2k}\frac{\Gamma(2k+1)c^{2k}}{ p^{2k+1}(kt)^{2}4^{k}}\] \[\times\xi_{n,k}\bigg{(}\frac{2p}{2p+a^{2}}\bigg{)}^{n}\mathrm{L}_ {n}\bigg{[}-\frac{b^{2}a^{2}}{4p+2a^{2}}\bigg{]} \tag{34}\]  
_and_  
\[\zeta_{n,k}=\begin{cases}1&\text{if }n<2k\\ 1+\frac{2p}{a^{2}}&\text{if }n=2k\end{cases} \tag{35}\]  
Proof.: Using (18) and (29), \(P_{s}\) can be approximated as  
\[P_{s}\approx 1-\sqrt{\frac{1}{2}+\frac{1}{4\eta}+\frac{\eta}{4}}\int_{0}^{ \infty}e^{-\left(\frac{1}{2}+\frac{1}{4\eta}+\frac{\eta}{4}\right)y}\] \[\times I_{0}\bigg{[}\bigg{(}\frac{1}{4\eta}-\frac{\eta}{4} \bigg{)}y\bigg{]}Q_{1}\Big{(}\sqrt{2N\gamma y},\sqrt{\xi_{N}}\Big{)}\mathrm{d}y \tag{36}\]  
which can be written as (33) with  
\[\mathcal{I}(a,b,c,p)=\int_{0}^{\infty}\exp(-px)I_{0}(cx)Q_{1}(a\sqrt{x},b) \mathrm{d}x \tag{37}\]  
To the best of our knowledge, however, this integral is not available in related works such as [31; 35; 36]. Nevertheless, as shown in the Appendix A, \(\mathcal{I}(a,b,c,p)\) can be evaluated as (34), thus completing the proof.