# Electronic Fourier-Galois Spectrum Analyzer for the Field \(Gf(31)\)
## 3 Results
### Algorithm of the Fourier-Galois Spectrum Analyzer  
To ensure that the Fourier-Galois spectrum analyzer works using typical binary elements, it is convenient to use the following considerations: The prime number 31 is a special case of Mersenne prime numbers, represented as \(2^{n}-1\). Such numbers have the following property:  
\[2\times a_{4}a_{3}a_{2}a_{1}a_{0}=_{(31)}a_{3}a_{2}a_{1}a_{0}a_{4} \tag{32}\]  
where \(a_{i}\) denote binary characters. Hence, to list all numbers corresponding to non-zero elements of the field \(GF(31)\), we can use Table 1.  
The expression for the spectral component (in the sense of the Galois field used) can be represented in the following form:  
\[Z_{k}=\sum_{i=1}^{i=30}U_{i}g_{1}^{ik}g_{2}^{ik}g_{3}^{ik} \tag{34}\]  
Recall that we consider the scheme using threshold sensors, i.e., \(U_{i}\) takes a value of either zero or one. Let us take into account that the product of functions \(g_{1}^{ik}g_{2}^{ik}\) has a period of 6, and the function \(g_{3}^{ik}\) has a period equal to 5. Figure 1 illustrates the following peculiarity that arises due to such values of periods. Dark rectangles, directed for clarity in the negative area, mark the same values of the product of functions \(g_{1}^{ik}g_{2}^{ik}\). They are really the same, since the function under consideration has period 6. Due to the specificity of the difference in periods, each such position corresponds to a different function value \(g_{3}^{ik}\). In particular, for any \(k\) and any \(q\), the following takes place:  
\[g_{1}^{6 qk}g_{2}^{6 qk}=1 \tag{35}\]  
Let us distinguish the next five sums in Formula (34) corresponding to Formula (35). Then:  
\[Z_{k}^{(6)}=U_{6}g_{3}^{6k}+U_{12}g_{3}^{12k}+U_{18}g_{3}^{18k}+U_{24}g_{3}^{24 k}+U_{30}g_{3}^{30k} \tag{36}\]  
Due to the fact that \(g_{3}^{5}=1(31)\), in this formula, it makes sense to consider only the case of \(k=1,2,3,4,5\) and the case \(k=5\) is degenerate, i.e. :  
\[Z_{5}^{(6)}=U_{6}+U_{12}+U_{18}+U_{24}+U_{30} \tag{37}\]  
Figure 1: Partitioning in the computation of spectral components in the considered Galois field. \begin{table}
\begin{tabular}{c c c c c c c} \hline \(m=1\) & **1** & **2** & **3** & \(m=2\) & **1** & **2** & **3** \\ \hline \(a_{4}\) & 0 & 1 & 0 & \(a_{4}\) & 1 & 0 & 0 \\ \hline \(a_{3}\) & 0 & 1 & 0 & \(a_{3}\) & 1 & 0 & 0 \\ \hline \(a_{2}\) & 1 & 0 & 0 & \(a_{2}\) & 0 & 1 & 0 \\ \hline \(a_{1}\) & 0 & 0 & 0 & \(a_{1}\) & 0 & 0 & 0 \\ \hline \(a_{0}\) & 1 & 1 & 1 & \(a_{0}\) & 1 & 1 & 1 \\ \hline \end{tabular}
\end{table} 
Table 3: Values of binary characters in binary recording of elements \(g_{2}^{nk}\). For the nondegenerate case, since the calculation of degree indices is carried out modulo 5, Formula (36) takes the following form:  
\[Z_{k}^{(6)}=U_{6}g_{3}^{k}+U_{12}g_{3}^{2k}+U_{18}g_{3}^{3k}+U_{24}g_{3}^{4k}+U_{ 30} \tag{38}\]  
The upper index in this formula means that the sums starting from the sixth are considered. This formula, in particular, shows that all the degrees \(g_{3}^{4n}\) appearing in it are different, which is also illustrated in Figure 1.  
Returning to Table 2, we can use the fact that in the binary representation, all elements \(g_{3}^{3k}\) are a sequence of binary characters \(a_{4}a_{3}a_{2}a_{1}a_{0}\), only one of which is different from zero. Let us also take into account that, for the considered construction, all numbers \(U_{i}\) are binary. This means that from expression (38) we can immediately pass to the positional notation of the number \(Z_{k}^{(6)}\). Specifically, for the case \(k=1\) as it follows from Table 2, we can write down the following:  
\[Z_{1}^{(6)}=U_{6}\times 2+U_{12}\times 2^{2}+U_{18}\times 2^{3}+U_{24}\times 2^{4 }+U_{30} \tag{39}\]  
Or in the usual binary representation:  
\[Z_{1}^{(6)}=U_{24}U_{18}U_{12}U_{6}U_{30} \tag{40}\]  
For other values of \(k\), Formula (40) will be preserved; the only difference is that the symbols \(U_{i}\) in this entry will be interchanged.