## 3 Multichannel Signal Acquisition
### Maximum Power Selection  
Let \(\Omega_{k}=E[|r_{k}[i]|^{2}]\), \(k=1,2,3\), be the power of \(r_{k}[i]\). In addition, let \(r_{m}[i]\) denote the one that has the maximum power, i.e., \(\Omega_{m}=E[\big{|}r_{m}[i]\big{|}^{2}]=\max\{\Omega_{1},\ \Omega_{2},\ \Omega_{3}\}\). The signal acquisition decision statistic \(\Lambda_{m}\) in response to this selected signal \(r_{m}[i]\) is given by  
\[\Lambda_{m}=|z_{m}|^{2},\ \ z_{m}=\sum_{i=0}^{N-1}s^{*}[i]r_{m}[i], \tag{15}\]  
where \(z_{m}\) is the output of the matched filter for the strongest receiver channel, sampled at \(i=N-1\). A flowchart of this method is presented in Figure 2b. Following the  
Figure 2: Flowcharts for the presented methods: (**a**) multichannel combining, (**b**) maximum power selection.  
same approach as the previous subsection, the following performance probabilities can be obtained for this detector  
\[P_{A,m}=\Pr(\Lambda_{m}>\gamma|H_{1})=\exp(-\gamma/\rho_{m}), \tag{16}\]  
\[P_{FA,m}=\Pr(\Lambda_{m}>\gamma|H_{0})=\exp(-\gamma/\xi_{m}), \tag{17}\]  
where \(\rho_{m}=E[|z_{m}|^{2}|H_{1}]=(d_{m}+1)\sigma_{m}^{2}\overline{\mathrm{E}}\), \(d_{m}=(\eta_{m}^{2}/\sigma_{m}^{2})\overline{\mathrm{E}}\) and \(\xi_{m}=E[|z_{m}|^{2}|H_{0}]=\sigma_{m}^{2}\overline{\mathrm{E}}\).  
To theoretically compare the performance of the maximum power selection signal acquisition method with the multichannel combining method, first we consider the case where the noise powers are equal, i.e., \(\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma_{3}^{2}=\sigma^{2}\) and consequently \(\xi_{1}=\xi_{2}=\xi_{3}=\xi\) (this assumption is dropped in the next section). Then, we obtain this CF for \(\Lambda\) under \(H_{0}\), \(\Theta_{\Lambda}(\omega)=(1-j\xi\omega)^{-3}\), which, upon inverse Fourier transform and integration of the resulting PDF, a chi-squared PDF [20] with six degrees of freedom, provides the following probability  
\[P_{FA}=\exp(-\gamma/\xi)(1+(\gamma/\xi)+0.5(\gamma/\xi)^{2}). \tag{18}\]  
For any given value of \(P_{FA}\), Equation (18) can be solved and the resulting \(\gamma\) needs to be substituted in (13) to compute \(P_{A}\) for the multichannel combining method. For the maximum power selection method and by combining (16) and (17), we obtain, \(P_{A,m}=P_{FA,m}^{1/(d_{m}+1)}\). Using these results, the signal acquisition probabilities of these two methods are graphed and compared in Figure 3, where, without loss of generality, it is assumed that receiver channel 1 has the maximum power.  
Figure 3 shows \(P_{A}\) and \(P_{A,m}\) plotted as functions of \(d_{1}\), signal-to-noise ratio (SNR) of the receiver channel 1, with \((d_{2},d_{3})=(0.01,0.02)\) in the top panel and \((d_{2},d_{3})=(1,2)\) in the bottom panel, \(P_{FA}=0.01\), and unit energy signal. We note that when channels 2 and 3 are not strong (top panel), the maximum power selection method that picks channel 1 outperforms the multichannel combining method, which is a reasonable result. When channels 2 and 3 become stronger (bottom panel), the selection method converges to the combining method, as \(d_{1}\) increases.  
Figure 3: Probability of signal acquisition for the two methods.