## 5 Simulation Results--BER Performance
### Rayleigh Channel  
Let us consider the energy costs for algorithm (9) for FTN signal processing in Rayleigh fading channel. At the same time, we will take into account that it is advisable to use an equalizer when receiving FTN signals in the presence of fading. The MMSE (Minimum Mean Square Error) equalizer was used in this work. Figure 7 shows the dependences of the BER on the \(E_{b}/N_{0}\) ratio for bit rates up to \(R=2.5/T\). Let us analyze the dependence in Figure 7. It can be concluded that in the presence of Rayleigh fading, the application of algorithm (9) makes it possible to ensure the reception of FTN signals at least up to bit rates of \(R=2.5/T\). Energy costs are equal to 12 dB for a bit rate of \(R=2/T\) at \(BER=10^{-2}\). When bit rates increase above the values of \(R=2.22/T\), a well-known effect of stabilization of BER performance with increasing \(E_{b}/N_{0}\) is observed in case of fading. This can be explained by the fact that at such values of bit rates, the level of intersymbol interference, which is caused by the formation of FTN and interference signals and passing through the transmission channel, is quite high. This effect is especially noticeable in the area of high signal-to-noise ratios. Simulation Results--Comparison of Energy Efficiency and Computational Complexity of Decision Feedback Detection Algorithm with Estimation of Subsequent Symbols in Observation Interval by Maximum Likelihood Ratio  
Let us estimate the increase in the computational complexity of decision feedback algorithm (9) caused by an increase in data transmission rate. We try to answer the following questions:  
1. To what extent must the computational complexity of the digital demodulator of spectrally efficient signals be increased under conditions of a fixed BER and increased symbol rate? 2. What is the value of the energy loss caused by an increased symbol rate under conditions of a fixed BER and computational complexity? Figure 7: BER performance. Rayleigh channel. Processing algorithm (9) in case of high rates. By the energy loss, we mean the difference between the signal-to-noise ratios \(E_{b}/N_{0}\) in the case of the ISI-free BER performance and in the case of using proposed algorithm (9) at \(BER=10^{-4}\). Then, it is possible to construct dependences between the energy losses of algorithm (9) in relation to the potential BER depending on the number of addition and multiplication operations for different bit rates. Figure 8 shows the values of the energy loss for the bit rates from \(R=1.11/T\) to \(R=2/T\). Figure 8 shows the energy loss values obtained as a result of simulation modeling for bit rates from \(R=1.11/T\) to \(R=2/T\) for FTN signals, the parameters of which are given in Section 5. In this figure, the X-axis shows the number of addition and multiplication operations (computational complexity). Moreover, each point along this axis corresponds to the value of parameter \(q\) in formula (9). This parameter determines the number of subsequent signals falling within the observation interval, which are taken into account when implementing the scheme in Figure 5. For example, for \(R=1.11/T\) and \(q=1\) (one previous symbol is used in the feedback), the energy loss is equal to 0.6 dB at the minimum values of the computational complexity. When \(q=4\) (four previous values of symbols are used in the feedback ), the energy loss practically becomes zero, that is, the BER coincides with the potential one. By increasing the transmission rate to \(R=1.25/T\), approaching the potential BER is achieved by increasing the computational complexity, which reaches a value of \(0.4\times 10^{5}\) addition and multiplication operations. With a further increase in the bit rate up to \(R=1.42/T\), the minimum energy loss will be equal to 1.2 dB with the number of operations \(3\times 10^{5}\). When the bit rate \(R=2/T\) is reached, the number of operations increases up to \(6.5\times 10^{5}\) with a minimum energy loss of 3.5 dB.