## 4 Results
### Application of the Digital Convolution Theorem to Partial Convolutions  
Let us show, with a specific example, that the use of algebraic extensions, which allow us to reduce the calculations of partial convolutions to Fourier-Galois bases containing the same number of cycles, allows us to extend the theorem on digital convolution to the case when the convolution is represented as a sum of partial convolutions. Let us consider the convolution  
\[U_{1}(j)=\sum_{1}^{24}K(j-i)U_{0}(i) \tag{34}\]  
We will assume that the maximum value of the functions \(U_{1}\), \(K\), and \(U_{0}\) does not exceed 34; i.e., it is permissible to assume that these functions take values in the ring of residue classes modulo 35. We will also assume that the function \(U_{0}(i)\) is periodic, and its period is 24 cycles, and that the domain of the definition of the function \(K\) fits into this interval. In the ring under consideration, each number can be represented in the form  
\[u\equiv 21\cdot u_{1}+15\cdot u_{2}(mod\ 35) \tag{35}\]  
where \(u_{1}=0,1,2,4\), \(u_{2}=0,1,\ldots,6\); and the numbers 21 and 15 are idempotent elements, which, in particular, satisfy relation (4), which can be verified by direct verification. We apply the general Formulas (11)-(34), representing all the functions appearing in (34) in the form (35). Then  
\[U_{1}(j)\equiv 21\cdot\sum_{1}^{24}K_{1}(j-i)U_{01}(i)+15\cdot\sum_{1}^{24}K_{2 }(j-i)U_{02}(i)(mod\ 35) \tag{36}\]  
In this formula, the functions \(K_{1,2}\) and \(U_{01,2}\) are defined as  
\[U_{01}\equiv U_{0}(mod\ 5);\ U_{02}\equiv U_{0}(mod\ 7);\ K_{1}\equiv K(mod\ 5); \ K_{2}\equiv K(mod\ 7) \tag{37}\]  
The possibility of such a representation follows directly from general Formula (6). It is essential that the convolution \(\sum_{1}^{24}K_{1}(j-i)U_{01}(i)\) is calculated in the field \(GF\big{(}5^{2}\big{)}\), and the convolution \(\sum_{1}^{24}K_{2}(j-i)U_{02}(i)\) is calculated in the field \(GF\big{(}7^{2}\big{)}\), i.e., the digital convolution theorem in the form [40] is applicable to both of them. Specifically, the Fourier-Galois transform of the first of the convolutions under consideration, \(F_{1}(K_{1}*U_{01})\), is  
\[F_{1}(K_{1}*U_{01})=q_{1}\sum_{j=1}^{24}\xi^{\overline{k}j}\sum_{i=1}^{24}K_{ 1}(j-i)U_{01}(i)=q_{1}\sum_{i=1}^{24}U_{01}(i)\sum_{j=1}^{24}\xi^{\overline{k} j}K_{1}(j-i) \tag{38}\]  
where the bar over the symbol \(\overline{k}\) means that the calculations use the number (degree) corresponding to the conjugate sequence, i.e., the sequence determined by condition (15). In the right part of Formula (38) the order of summation is changed. The next identity is valid:  
\[\sum_{i=0}^{23}\xi^{\overline{k}j}K_{1}(j-i)=\xi^{\overline{k}j}\sum_{j=0}^{ 23}\xi^{\overline{k}(j-i)}K_{1}(j-i) \tag{39}\]  
The sum on the right side of Formula (39) is exactly the result of the Fourier-Galois transform of function \(K_{1}\), since all the functions under consideration change cyclically. Substituting (39) into (38), we obtain  
\[F_{1}(K_{1}*U_{01})=q_{1}^{-1}F_{1}(K_{1})F_{1}(U_{01}) \tag{40}\]  
It is easy to show that a similar formula is also valid for the second of the convolutions under consideration. \[F_{2}(K_{2}*U_{02})=q_{2}^{-1}F_{2}(K_{2})F_{2}(U_{02}) \tag{41}\]  
The difference is that the Fourier-Galois transforms for these convolutions are performed using different bases corresponding to algebraic extensions of different Galois fields. It can be seen that in the case under consideration it is permissible to speak of partial transfer functions \(F_{1}(K_{1})\) and \(F_{2}(K_{2})\), which exhaustively describe the result of the convolution operation. In particular, to return to the function \(U_{1}(j)\), it is necessary to perform inverse Fourier-Galois transforms in the corresponding fields. It is easy to see that the obtained result can be generalized to an arbitrary number of terms in the formula for partial convolutions (11).