## Appendix A
\[\textit{SNR}=\frac{\int\limits_{-\infty}^{+\infty}(e^{-a\lambda^{2}}\ \cos\lambda\ )^{2}\ d\lambda}{\sqrt{2\int\limits_{-\infty}^{+\infty}(e^{-a\lambda^{2}}\cos \lambda\ )^{2}\ d\lambda)\ \sigma_{N}^{2}+(\int\limits_{-\infty}^{+\infty}w_{R}(\lambda)\ ^{2}\ d \lambda)\sigma_{N}^{2}}}\] (A-25)  
By assuming that the variance of the noise is much smaller than the power of signal:  
\[\int\limits_{-\infty}^{+\infty}(e^{-a\lambda^{2}}\ \cos\lambda\ )^{2}\ d\lambda>>\int\limits_{-\infty}^{+\infty}w_{R}(\lambda)\ ^{2}\ d\lambda\] (A-26)  
Equation A-25 can be approximated to:  
\[\textit{SNR}\approx\frac{\int\limits_{-\infty}^{+\infty}(e^{-a\lambda^{2}}\ \cos\lambda\ )^{2}\ d\lambda}{\sqrt{(2\int\limits_{-\infty}^{+\infty}(e^{-a\lambda^{2}} \cos\lambda\ )^{2}\ d\lambda)\ \sigma_{N}^{2}}}=\frac{\int\limits_{-\infty}^{+\infty}(e^{-a\lambda^{2}}\cos \lambda\ )^{2}\ d\lambda}{\sqrt{2\ \sigma_{N}}}\] (A-27)  
Again, _SNR_ of \(i_{S}(t)\) (or \(i_{R}(t)\)) before the cross-correlation is:  
\[\textit{SNR}=\frac{i_{NF,MAX}}{\sigma_{N}}\] (A-28)  
so cross-correlation has improved the _SNR_ by less than a factor of:  
\[\textit{SNR}_{\textit{improve}}=\frac{\sqrt{\int\limits_{-\infty}^{+\infty}(e ^{-a\lambda^{2}}\ \cos\lambda\ )^{2}\ d\lambda}}{\sqrt{2\ \sigma_{N}}}\times\frac{\sigma_{N}}{i_{NF, MAX}}=\frac{1}{\sqrt{2}}\frac{\sqrt{\int\limits_{-\infty}^{+\infty}e^{-2a\lambda^{2}}\cos ^{2}\lambda\ d\lambda}}{i_{NF,MAX}}\] (A-29)  
It can be shown that energy of the (noise-free) interference signal \(\int\limits_{-\infty}^{+\infty}e^{-2a\lambda^{2}}\cos^{2}\lambda\ d\lambda\) in Equation A-29 is given by \(\frac{1}{2}\frac{1}{\sqrt{2}}\sqrt{\frac{\pi}{a}}=\frac{1}{2}\frac{1}{\sqrt{ 2}}\sqrt{\pi^{3}N_{f}^{2}}\) as follows:  
\[i_{NF}\ast i_{NF}(0)=\int\limits_{-\infty}^{\infty}e^{-2\alpha x^{2}}\cos^{2 }xdx=\frac{1}{2}\int\limits_{-\infty}^{\infty}e^{-2\alpha x^{2}}\big{(}1+\cos 2 x\big{)}dx\] (A-30)  
Useful relations are given by:  
\[\int\limits_{-\infty}^{\infty}e^{-2\alpha x^{2}}dx=\frac{1}{\sqrt{2}}\sqrt{ \frac{\pi}{a}}\] (A-31)  
\[\int\limits_{-\infty}^{\infty}e^{-2\alpha x^{2}}\cos 2 xdx=\frac{1}{\sqrt{2}}\sqrt{ \frac{\pi}{a}}e^{\frac{1}{2a}}\] (A-32)Then Equation A-30 becomes:  
\[\int\limits_{-\infty}^{+\infty}e^{-2\alpha x^{2}}\cos^{2}x\ dx\ =\frac{1}{2}\frac{1}{ \sqrt{2}}\sqrt{\frac{\pi}{a}}+\frac{1}{2}\frac{1}{\sqrt{2}}\sqrt{\frac{\pi}{a}}e ^{\frac{-1}{2a}}\] (A-33)  
But, the second term in Equation A-29 is negligible because \(e^{\frac{1}{2a}}\) has a negligible value and Equation A-33 becomes:  
\[\textit{SNR}=\frac{\int\limits_{-\infty}^{+\infty}(e^{-a\lambda^{2}}\ \cos\lambda\ )^{2}\ d\lambda}{\sqrt{2\int\limits_{-\infty}^{+\infty}(e^{-a\lambda^{2}}\cos \lambda\ )^{2}\ d\lambda)\ \sigma_{N}^{2}+(\int\limits_{-\infty}^{+\infty}w_{R}(\lambda)\ ^{2}\ d \lambda)\sigma_{N}^{2}}}\]0 (A-34)  
Then, finally \(i_{NF,\ MAX}\) is \"1\" and combining Equation A-29 and Equation A-34 we get:  
\[\textit{SNR}=\frac{\int\limits_{-\infty}^{+\infty}(e^{-a\lambda^{2}}\ \cos\lambda\ )^{2}\ d\lambda}{\sqrt{2\int\limits_{-\infty}^{+\infty}(e^{-a\lambda^{2}}\cos \lambda\ )^{2}\ d\lambda)\ \sigma_{N}^{2}+(\int\limits_{-\infty}^{+\infty}w_{R}(\lambda)\ ^{2}\ d \lambda)\sigma_{N}^{2}}}\]1 (A-35)  
which is 14 dB when \(N_{f}=26\)