## 3 BER Considering Leaky Feeder Transmitter
### BER for M-QAM Modulation Technique
For the LF, the transmitted signal for \(n^{th}\) slot will be given by  
\[S_{n}\left(t\right)=\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\,A_{x}\,\cos \left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x}\,\sin \left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{\frac{\left|2\,\pi\,Dn \,\beta}{\lambda}}. \tag{29}\]  
The total signal which is a sum of all slots signals is given by  
\[S_{T}\left(t\right)=\sum_{n=1}^{N}\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\, A_{x}\,\cos\left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x} \,\sin\left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{ \frac{\left|2\,\pi\,Dn\,\beta}{\lambda}}. \tag{30}\]  
Assume \(A_{x}=B_{x}=C_{x}\), and use the following identity  
\[\cos\left(z\right)-\sin\left(z\right)=\sqrt{\left(2\right)}\,\sin\left(\frac{ \pi}{4}-z\right). \tag{31}\]  
The total signal will be written as  
\[S_{T}\left(t\right)=\sum_{n=1}^{N}\bigg{\{}2\,\sqrt{\left(\frac{E}{T}\right)} \,C_{x}\,\sin\left(\frac{\pi}{4}-2\,\pi\,f_{c}\,t\right)10^{\frac{-Dnk}{20}} \,e^{\frac{\left|2\,\pi\,Dn\,\beta\right|}{\lambda}}\bigg{\}}. \tag{32}\]  
By using [PERSON]'s identity, \(S_{T}\left(t\right)\) can be written as  
\[S_{T}\left(t\right)=\sum_{n=1}^{N}\bigg{\{}2\,\sqrt{\left(\frac{N\,E}{T} \right)}\,C_{x}\,\cos\left(2\,\pi\,f_{c}\,t+\frac{\pi}{4}+\arctan\left(\frac{ 2\,\pi\,n\,D\,\beta}{\lambda}\right)\right)\bigg{\}}. \tag{33}\]  
Then, the total amplitude of all signals transmitted from all slots \(A_{T}\) will be as  
\[A_{T}=\sum_{n=1}^{N}\bigg{(}2\,\sqrt{\left(\frac{E_{b}\,N}{T}\right)}\,C_{x} \,\left(10^{\frac{-Dnk}{20}}\right)\bigg{)}\,. \tag{34}\]  
To calculate the BER of the LF which carries M-QAM signal, we have to find the conditional probability first, which is given by [13]  
\[P\left(E\setminus\{\gamma_{l}\}_{l=1}{}^{L}\right)=4\left(1-\frac {1}{\sqrt{\left(M\right)}}\right)\,Q\left(\sqrt{\left(2\,g_{QAM}\,\gamma_{t} \right)}\right)\\ -4\left(1-\frac{1}{\sqrt{\left(M\right)}}\right)^{2}\,Q^{2}\left( \sqrt{\left(2\,g_{QAM}\,\gamma_{t}\right)}\right), \tag{35}\]where \(g_{QAM}=\frac{3}{2M-2}\), [PERSON] and [PERSON] in [18], generalized the alternative representation of the Gaussian \(Q\) function to the two-dimensional case and showed in particular  
\[Q\left(x\right)=\frac{1}{\pi}\,\int_{0}^{\left(\frac{\pi}{2}\right)}\exp\left( \frac{-x^{2}}{2\,sin^{2}\left(\phi\right)}\right)\mathrm{d}\phi \tag{36}\]  
and  
\[Q^{2}\left(x\right)=\frac{1}{\pi}\,\int_{0}^{\left(\frac{\pi}{2}\right)}\exp \left(\frac{-x^{2}}{2\,sin^{2}\left(\phi\right)}\right)\mathrm{d}\phi \tag{37}\]  
so the conditional probability \(P\left(E\setminus\{\gamma_{l}\}_{l=1}^{L}\right)\) under the AWGN channel can be written as  
\[S_{n}\left(t\right)=\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\,A_{x}\,\cos \left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x}\,\sin \left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{\frac{\left|2\,\pi\,Dn \,\beta}{\lambda}}. \tag{29}\]0  
The BER for a single antenna transmitted a M-QAM signal in Rayleigh fading channel is given by [13]  
\[S_{n}\left(t\right)=\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\,A_{x}\,\cos \left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x}\,\sin \left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{\frac{\left|2\,\pi\,Dn \,\beta}{\lambda}}. \tag{29}\]1  
By using (34), and (17), \(\gamma_{t}\) is given by  
\[S_{n}\left(t\right)=\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\,A_{x}\,\cos \left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x}\,\sin \left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{\frac{\left|2\,\pi\,Dn \,\beta}{\lambda}}. \tag{29}\]2  
and \(\overline{\gamma_{t}}\) is given by  
\[S_{n}\left(t\right)=\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\,A_{x}\,\cos \left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x}\,\sin \left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{\frac{\left|2\,\pi\,Dn \,\beta}{\lambda}}. \tag{29}\]3  
For MRC with \(L\) independent identical channels (i.i.d), the BER for M-QAM can be obtained by averaging (38) over (20) as in (2)  
\[S_{n}\left(t\right)=\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\,A_{x}\,\cos \left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x}\,\sin \left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{\frac{\left|2\,\pi\,Dn \,\beta}{\lambda}}. \tag{29}\]4 \[S_{n}\left(t\right)=\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\,A_{x}\,\cos \left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x}\,\sin \left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{\frac{\left|2\,\pi\,Dn \,\beta}{\lambda}}. \tag{29}\]5The integral in (42) can be divided into two integrals such as  
\[S_{n}\left(t\right)=\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\,A_{x}\,\cos \left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x}\,\sin \left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{\frac{\left|2\,\pi\,Dn \,\beta}{\lambda}}. \tag{29}\]6  
where \(I_{1}\) and \(I_{2}\) is given by  
\[S_{n}\left(t\right)=\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\,A_{x}\,\cos \left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x}\,\sin \left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{\frac{\left|2\,\pi\,Dn \,\beta}{\lambda}}. \tag{29}\]7  
\[S_{n}\left(t\right)=\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\,A_{x}\,\cos \left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x}\,\sin \left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{\frac{\left|2\,\pi\,Dn \,\beta}{\lambda}}. \tag{29}\]8  
According to [19; 20], \(I_{1}\) and \(I_{2}\) can be evaluated as  
\[S_{n}\left(t\right)=\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\,A_{x}\,\cos \left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x}\,\sin \left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{\frac{\left|2\,\pi\,Dn \,\beta}{\lambda}}. \tag{29}\]9  
where  
\[S_{T}\left(t\right)=\sum_{n=1}^{N}\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\, A_{x}\,\cos\left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x} \,\sin\left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{ \frac{\left|2\,\pi\,Dn\,\beta}{\lambda}}. \tag{30}\]0  
and  
\[S_{T}\left(t\right)=\sum_{n=1}^{N}\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\, A_{x}\,\cos\left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x} \,\sin\left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{ \frac{\left|2\,\pi\,Dn\,\beta}{\lambda}}. \tag{30}\]1  
and  
\[S_{T}\left(t\right)=\sum_{n=1}^{N}\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\, A_{x}\,\cos\left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x} \,\sin\left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{ \frac{\left|2\,\pi\,Dn\,\beta}{\lambda}}. \tag{30}\]2  
\(I_{2}\) given by  
\[S_{T}\left(t\right)=\sum_{n=1}^{N}\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\, A_{x}\,\cos\left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x} \,\sin\left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{ \frac{\left|2\,\pi\,Dn\,\beta}{\lambda}}. \tag{30}\]3  
where  
\[S_{T}\left(t\right)=\sum_{n=1}^{N}\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\, A_{x}\,\cos\left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x} \,\sin\left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{ \frac{\left|2\,\pi\,Dn\,\beta}{\lambda}}. \tag{30}\]4  
\[S_{T}\left(t\right)=\sum_{n=1}^{N}\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\, A_{x}\,\cos\left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x} \,\sin\left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{ \frac{\left|2\,\pi\,Dn\,\beta}{\lambda}}. \tag{30}\]5  
\[S_{T}\left(t\right)=\sum_{n=1}^{N}\left\{\sqrt{\left(\frac{2\,E}{T}\right)}\, A_{x}\,\cos\left(2\,\pi\,f_{c}\,t\right)-\sqrt{\left(\frac{2\,E}{T}\right)}\,B_{x} \,\sin\left(2\,\pi\,f_{c}\,t\right)\right\}\cdot 10^{\frac{-Dnk}{20}}\,e^{ \frac{\left|2\,\pi\,Dn\,\beta}{\lambda}}. \tag{30}\]6  
where \({}_{2}F_{1}\left(0.5+l,1+l;1.5+l;-b\right)\) is the hypergeometric function.