## 4 Testing of DF in the FFA
\begin{table}
\begin{tabular}{c c c c} \hline \hline
**The Residues** & \multirow{2}{*}{**The Range Numbers**} & \multicolumn{2}{c}{**The Residues**} \\
**Values** & & **Values** & **The Range Numbers** \\ \hline \(\alpha_{1}=0\) & 0,9,18,27,36,45,54,63,72,81, & \(\alpha_{5}=4\) & 4,13,22,31,40,49,58,67,76,85,... \\ \(\alpha_{2}=1\) & 1,10,19,28,37,46,55,64,73,82, & \(\alpha_{6}=5\) & 5,14,23,32,41,50,59,68,77,86,... \\ \(\alpha_{3}=2\) & 2,11,20,29,38,47,56,65,74,83, & \(\alpha_{7}=6\) & 6,15,24,33,42,51,60,69,78,87,... \\ \(\alpha_{4}=3\) & 3,12,21,30,39,48,57,66,75,84, & \(\alpha_{8}=7\) & 7,16,25,34,43,52,61,70,79,88,... \\  & & \(\alpha_{9}=8\) & 8,17,26,35,44,53,62,71,80,89,... \\ \hline \hline \end{tabular}
\end{table} 
Table 2: The residues values for the range \(p=585\) modulo \(p_{2}=9\). i.e., we have 4 numbers \(a_{3}=7\), \(a_{2}=13\), \(a_{1}=0\), \(a_{0}=15\). Then we can write  
\[ \begin{array}{c}|32015|_{17}^{+}=|\!|7\!\cdot\!2^{12}|_{17}^{+}+|13\!\cdot\!2^{ 8}|_{17}^{+}+|\!|0\!\cdot\!2^{4}|_{17}^{+}+|15\!\cdot\!2^{0}|_{17}^{+}|_{17}^{+} =\\ =|\!|\!|7\!\cdot\!16|_{17}^{+}+|13\!\cdot\!1|_{17}^{+}+|0\!\cdot\!16|_{17}^{+}+ |15\!\cdot\!1|_{17}^{+}|_{17}^{+}|_{17}^{+}=|10+13+0+15|_{17}^{+}=4\end{array} \]  
In the case of parallel implementation of this algorithm, the time of residue calculation is  
\[T_{FC}=t_{LUT}+t_{LUT}\times log2\left[\frac{n}{k}\right] \tag{23}\]  
where \(k\)--the block size (in the example \(k=4\)); \(n=16\)--the bit width of the input number X; \(t_{LUT}\)--time of getting from the LUT-table (is taken equal to three cycles). For the considered example, the time of residue calculation considering Equation (23) is equal to nine cycles, which corresponds to the Table 1.  
To implement the considered converter (the number of hardware devices is brought to the amount of LUT-tables) eleven LUT-tables are required. For the suggested method (Figure 2), three LUT-tables are required. For the computational channel without conversion device (Figure 5), one LUT-table is required. Following Equation (21), the error introduced to the processes signal by the twelve bit ADC is \(\Delta=0.00024414\). As an example, a non-recursive DF of the forty-fifth order is taken [13]. For evaluation the samples of impulse response of the filter taking values \(N_{1}=-0.000105023\), \(N_{2}=-0.000125856\), \(N_{17}=0.0364568\), \(N_{18}=0.0328505\) are selected. To simplify the provided values the sample sign of the impulse response is not considered and it is supposed that the ADC error decreases the values of the selected samples. While calculating the output sample of the filter and representing the samples of impulse response in the binary code considering the ADC errors, their values can be written as follows: \(N_{1}=0.000139117\), \(N_{2}=0.0001185284\), \(N_{17}=0.03621266\), \(N_{18}=0.03260636\). The values of samples of impulse response in the FFA code for the module \(p_{1}=5\) without error (in decimal notation and in FFA code) will be written as follows:  
\(N_{1}=(105023)_{10}=(3)_{\rm FFA}\),  
\(N_{2}=(125856)_{10}=(1)_{\rm FFA}\),  
\(N_{17}=(36456800)_{10}=(0)_{\rm FFA}\),  
\(N_{18}=(32850500)_{10}=(0)_{\rm FFA}\). Considering the error introduced by the ADC, the values of samples of impulse response in the FFA code for the module \(p_{1}=5\) will be written (in decimal notation and in FFA code) as follows:  
\(N_{1}=(139117)_{10}=(2)_{\rm FFA}\),  
\(N_{2}=(1185284)_{10}=(4)_{\rm FFA}\),  
\(N_{17}=(362126600)_{10}=(0)_{\rm FFA}\),  
\(N_{18}=(326063600)_{10}=(0)_{\rm FFA}\). As it follows from the conducted evaluation the values of the first and the second samples in the FFA code changed because of the ADC error influence. This error will change the filter frequency response in the process of calculation of its output sample. Considering that the error introduced by the ADC can affect the values of the significant number of impulse response samples, the distortion of frequency response will be significant.