# High-Reliability Underwater Acoustic Communication Using an M-ary Cyclic Spread Spectrum
## 2 Proposed Method
### Estimation of the Approximate Doppler Shift Based on Two Peaks  
The signal is compressed or extended as a result of Doppler distortion in the time domain. As a result of the Doppler's influence, UWA signals are frequently modeled in the following way.  
\[\mathbf{r}(t)=\mathbf{s}((1+\Delta)t)\ \ [\ \mathbf{s}_{1}(1+\Delta_{1})t),\ \mathbf{s}_{2}(1+ \Delta_{2})t)\,\ldots,\mathbf{s}_{i}(1+\Delta_{i})t)] \tag{8}\]  
where \(r(\mathbf{t})\) is the received signal and can be expressed as the sum of each symbol. \(\Delta\) is the Doppler scale, which is defined as the ratio of the relative velocity between the source and receiver to the propagation wave velocity. The received signal with a Doppler scale (\(\Delta\)) is equivalent to a scaling of the sampling period in a discrete-time processing system with interpolation.  
\[f_{s+\Delta}\ \ (1+\Delta)f_{s}\ \equiv\ \ f_{\sphericalangle\Delta}\ (1+\Delta)f_{c} \tag{9}\]  
The sampling frequency and the carrier frequency are both affected by the same ratio of Doppler spread and \(f_{s}\) indicates the sampling frequency. The underwater vehicle's top speed is approximately 45 knots (12.86 m/s). The sound speed is 1500 m/s, with a range of \(\Delta=-0.0085\) to 0.0085. As a result, if \(f_{c}\) = 2.5 kHz, the Doppler frequency range is \(-21.4\) Hz to 21.4 Hz. We want to find Doppler by the approximate sample difference of the chip in the sample area since the Doppler spread is highly unpredictable.  
\[\mathrm{Chip}_{\mathrm{ns}}=\frac{f_{s}}{sps\times L} \tag{10}\]  
Here, \(\mathrm{Chip}_{\mathrm{ns}}\) indicates the number of samples that one chip has. The symbol rate (symbol per second, sps) of the transmission signal is 10 and one symbol includes 127 chips. As a result, the chip duration is 0.787 ms and each chip has 78 samples when the sampling frequency \(f_{s}\) is set to 100 k. The number of samples that can be changed by Doppler ranges from \(-84\) to 84 when \(\Delta=-0.0085\) to 0.0085 and one symbol has 9906 samples. Because there is a sample change greater than the number of samples on a single chip, a stable system can be achieved by setting the minimum distance m to 3 or more.  
Signals such as Equation (11) are generated to determine the \(\mathbf{C}_{q}\) sequence with the highest peak through MSS, and Equation (12) can be written as a reference signal to find the difference between the peaks of CSK as follows:  
\[\mathbf{MS}_{f_{\mathrm{c}}}^{q}(t)=\mathrm{Re}\Bigg{\{}\left[\sum_{l=1}^{L}\mathbf{ C}_{q}(l)g\Big{(}t-IT_{\mathbf{C}_{q}}\Big{)}\right]e^{2\pi f_{c}t}\Bigg{\}}\qquad \textbf{for}\ q\ \in\ [1,\ 2,\ \ldots,\ Q]. \tag{11}\]  
\[\mathbf{CSK}_{f_{\mathrm{c}}}(t)=\mathrm{Re}\Bigg{\{}\left[\sum_{l=1}^{L}\mathbf{C}_{ 0}(l)g\Big{(}t-IT_{\mathbf{C}_{0}}\Big{)}\right]e^{j2\pi f_{c}t}\Bigg{\}}. \tag{12}\]  
As shown in Figure 3, the sequence with the largest amplitude can be found by correlating \(\mathbf{MS}_{f_{\mathrm{c}}}^{q}\), and the correlation result of \(q\) (\(q\in[1,\ 2...,\ Q]\)) is shown. It was also possible to check how much cyclic shift there was by correlating \(\mathbf{CSK}_{f_{\mathrm{c}}}\) in the same symbol. Each chip contains approximately 78 samples, as previously stated. The desired value of \(n\) can be calculated because the minimum distance \(m\) is determined when transmitting. The peak-to-peak distance in Figure 3 differs by 1186 samples. Each chip has 78 samples and the cyclic shift is as much as the product of the minimum distance \(m\) and bit information \(n\). As a result, 1186 divided by 78 \(\times m\) equals \(n\).