## 4 Simulation
In addition, please note that according to Reference [49], \"\(MaxEnt_{ANEW}\)\" is not applicable for \(SNR=0\) dB. According to Figures 2-4, the new proposed algorithm (\"\(MaxEnt_{BNEW}\)\" (46)) has a better equalization performance from the residual ISI point of view compared to \"\(MaxEnt_{ANEW}\)\" and \"\(Godard\)\". Next, I tested the proposed equalization method (\"\(MaxEnt_{BNEW}\)\" (46)) with Channel 2 and Channel 3 where \(\sum_{k}|h_{k}|^{2}\
eq 1\). Figures 5 and 6 show the equalization performance comparison between the new proposed equalization method (\"\(MaxEnt_{BNEW}\)\" (46)) and [PERSON]'s [53] algorithm for the 16 QAM constellation input sent via Channel 2 and Channel 3, respectively, with \(SNR=7\) dB. According to Figures 5 and 6, the new proposed algorithm (\"\(MaxEnt_{BNEW}\)\" (46)) has a better equalization performance from the residual ISI point of view compared to \"\([PERSON]\)\". As a matter of fact, for the case of \(\sum_{k}|h_{k}|^{2}>1\) (Channel 2), the improvement in the residual ISI compared to the results obtained by \"\([PERSON]\)\" is approximately 5 dB while the improvement in the residual ISI compared to the results obtained by \"\(Godard\)\" for \(\sum_{k}|h_{k}|^{2}<1\) (Channel 3) is only approximately 2 dB. Thus, we may say that the proposed algorithm (\"\(MaxEnt_{BNEW}\)\" (46)) has a promising equalization performance from the residual ISI point of view for channels with \(\sum_{k}|h_{k}|^{2}\geq 1\). Figure 3: Performance comparison between equalization algorithms for a 16 QAM source input going through Channel 1. The averaged results were obtained in 50 Monte Carlo trials for an SNR = 7 dB. \(\mu_{G}=2.5\times 10^{-5}\), \(\mu_{ANEW}=0.00008\), \(\beta_{ANEW}=1\times 10^{-5}\), \(\mu_{BNEW}=0.00008\), \(\beta_{BNEW}=1\times 10^{-4}\), \(\varepsilon=0.85\). Figure 2: Performance comparison between equalization algorithms for a 16 QAM source input going through Channel 1. The averaged results were obtained in 50 Monte Carlo trials for a signal-to-noise ratio (SNR) = 10 dB. \(\mu_{G}=7\times 10^{-5}\), \(\mu_{ANEW}=0.00009\), \(\beta_{ANEW}=1\times 10^{-5}\), \(\mu_{BNEW}=0.00008\), \(\beta_{BNEW}=1\times 10^{-4}\), \(\varepsilon=0.85\). Figure 4: Performance comparison between equalization algorithms for a 16 QAM source input going through Channel 1. The averaged results were obtained in 50 Monte Carlo trials for an SNR = 0 dB. \(\mu_{G}=4\times 10^{-6}\), \(\mu_{BNEW}=0.00002\), \(\beta_{BNEW}=6\times 10^{-5}\), \(\varepsilon=0.05\). Figure 5: Performance comparison between equalization algorithms for a 16 QAM source input going through Channel 2. The averaged results were obtained in 50 Monte Carlo trials for an SNR = 7 dB. \(\mu_{G}=2.5\times 10^{-5}\), \(\mu_{BNEW}=0.00008\), \(\beta_{BNEW}=1\times 10^{-4}\), \(\varepsilon=0.85\). Up to now, I have assumed that the SNR as well as the channel power are known. Thus, we could calculate the required Lagrange multipliers via (48). When the SNR and the channel power are unknown, the \(\widehat{m}_{2}\), \(\widehat{m}_{4}\), and \(\widehat{m}_{6}\) cannot be calculated anymore via (48). However, on the basis of (33), we may calculate the required \(\widehat{m}_{2}\), \(\widehat{m}_{4}\), and \(\widehat{m}_{6}\) needed for the Lagrange multipliers in (48) from  
\[\widehat{m}_{2}=\bar{m}_{2}+\sigma_{\widehat{w}_{r}}^{2},\]  
\[\widehat{m}_{4}=3\sigma_{\widehat{w}_{r}}^{4}+6\widehat{m}_{2}\sigma_{\widehat {w}_{r}}^{2}+\bar{m}_{4}, \tag{53}\]  
\[\widehat{m}_{6}=15\sigma_{\widehat{w}_{r}}^{6}+45\sigma_{\widehat{w}_{r}}^{4} \,\bar{m}_{2}+15\sigma_{\widehat{w}_{r}}^{2}\bar{m}_{4}+\bar{m}_{6},\]  
where \(\sigma_{\widehat{w}_{r}}^{2}\) is the variance of the real part of \(\widehat{w}[n]\) and may be simulated as  
\[\sigma_{\widehat{w}_{r}}^{2}=\sigma_{p_{1}}^{2}. \tag{54}\]  
Please note that for the ideal case, when the equalizer has converged, the convolutional noise power tends to zero. Thus, this makes (54) reasonable.