## 5 Simulation Results and Discussion
The first system consists of turbo equalization that follows the classic scheme, according to the theory mentioned in Section 3. The recursive systematic convolutional (RSC) encoder was used in this study using the octal generating polynomials G0 = 7 and G1 = 5, at a rate R = 1/2 and constraint K = 3. Figure 6: The contribution of the Log-MAP equalizer to the BER performance vs. SNR, the considered h1 = [0.18 0.85 0.32]: **(a)** First scenario—the iterative receiver makes it0 and it1, while the LDPC decoder makes ten iterations at each. Second scenario—the iterative receiver makes only a single iteration, while the LDPC decoder makes 20 iterations and **(b)** First scenario—the iterative receiver makes it0 and it1, while the LDPC decoder makes twenty iterations at each. Second scenario—the iterative receiver makes only a single iteration, while the LDPC decoder makes 40 iterations. Figure 8: _Cont_. Figure 7: BER vs. SNR, LDPC decoder–10 iterations: (**a**) h1 = [0.18 0.85 0.32]; (**b**) h2 = [0.302 0.725 0.456]; and (**c**) h3 = [0.407 0.815 0.407]. The second system consists of the equalization performed separately before the decoding process using MMSE equalizer and LDPC decoder at a rate of \(\
icefrac{{1}}{{2}}\), similar to [13]. For example, for a channel with h1, at the SNR of 5 dB, in the case of our proposed system, as was specified earlier, the BER is situated between \(10^{-4}\) and \(10^{-5}\) if ten iterations are made in the LDPC decoder and for 20 iterations within the LDPC decoder at the SNR of 5 dB for iterations 1-5 no more errors were found. In the case of turbo equalization at the same SNR of 5 dB, for a channel with h1, the BER is \(10^{-3}\), as can be observed in Figure 9a. But, for a channel with h2 or h3, the BER is situated between \(10^{-2}\) and \(10^{-3}\), as can be noticed in Figure 9b,c. In Figure 10a for h1 in the case of the separate equalization performed before the decoding process, using MMSE and LDPC decoder, at the SNR of 5 dB, the BER is located between \(10^{-5}\) and \(10^{-6}\) (if the decoder makes 40 iterations). Constant performance was shown over 40 iterations with a BER between \(10^{-5}\) and \(10^{-6}\), as demonstrated in Figure 10b. Repeating the previous scenario, Figures 11a and 12a show that the BER deteriorates considerably for h2 and h3 at the SNR of 5 dB, being within \(10^{0}\) and \(10^{-1}\). The graphics in Figures 11b and 12b demonstrate a relatively constant level of BER over 40 iterations. Table 3 summarizes the BER vs. SNR performances of turbo equalization and separate MMSE equalization with LDPC decoding. \begin{table}
\begin{tabular}{c c c} \hline \hline
**Graphic** & **h** & **BER (Range)** \\ \hline Figure 7a & h1 = [0.18 0.85 0.32] & \(10^{-4}\)-\(10^{-5}\) \\ Figure 7b & h2 = [0.302 0.725 0.456] & \(10^{-1}\) \\ Figure 7c & h3 = [0.407 0.815 0.407] & \(10^{-1}\)-\(10^{-2}\) \\ \hline \hline \end{tabular}
\end{table} 
Table 2: BER vs. SNR in an ISI channel, at the SNR of 5 dB, using the proposed system for different impulse responses h.  
Figure 8: BER vs. SNR, LDPC decoder–20 iterations: (**a**) h1 = [0.18 0.85 0.32]; (**b**) h2 = [0.302 0.725 0.456]; (**c**) h3 = [0.407 0.815 0.407]. Figure 10: BER vs. SNR, using separately MMSE equalization and LDPC decoder, h1 = [0.18 0.85 0.32]; (**a**) LDPC decoder makes 40 it. and (**b**) LDPC decoder makes 80 it. Figure 9: BER vs. SNR, using turbo equalization: (**a**) h1 = [0.18 0.85 0.32]; (**b**) h2 = [0.302 0.725 0.456]; and (**c**) h3 = [0.407 0.815 0.407]. Figure 11: BER vs. SNR, using separately MMSE equalization and LDPC decoder, h2 = [0.302 0.725 0.456]; (a) LDPC decoder makes 40 it. and (b) LDPC decoder makes 80 it.