# Development of Post Processing for Wave Propagation Problem: Response Filtering Method
## 3 Response Filtering Method for Improved Transient Stress
### Analysis of the Accuracy Improvement with the Response Filtering Method  
As shown in the previous section, the simple averaging procedure with the RFM results in significant improvements in the prediction compared to the analytical solution by generating piecewise constant solutions. For the sake of conducting an accuracy analysis of the proposed method, this subsection investigates the solutions using the Fourier transformation. Without the loss of generality, a simple signal, \(y\), with a 50 Hz frequency component was generated for 0.1 s with 100 samplings in Figure 6. In other words, \(y(t)=\sin(2\pi 50t)\), and the sampling time, \(T_{s}=0.001\) s. Figure 6a (left) and (right) shows the original signal and its Fourier transformation, respectively, with the peak at 50 Hz. Figure 6b-d shows the filtered signals with different filtering lengths and their transformations. These figures illustrate that the Fourier transformations of the filtered signals contain some additional frequency components. For example, with \(FL=2\), the filtered signal bears additional frequency components at 450 Hz. With \(FL=4\), frequencies around 200, 300, and 450 Hz are added, while with \(FL=5\), frequencies around 150, 250, 350, and 450 Hz are added. Furthermore, the signals in the time domain become rectangular in shape owing to these added frequency components. These analyses indicate that the filtering procedures used for the time signals actually generate additional frequencies. In addition, it is found that there is a rule for the emerging frequencies. For example, Figure 6a shows that the original signal has a single frequency component of 50 Hz, whereas the filtered signal (\(FL=2\)) has an additional frequency component at 450 Hz in addition to the frequency component (50 Hz) of the original signal, as shown in Figure 6b. Prior to determining the relationship between the filtering length and the added frequency components, it should be noted that the discrete Fourier transform (DFT) defines the following relationship between the analysis time and frequency:  
\[T_{s}=\frac{T_{0}}{N},\;f_{0}=\frac{f_{s}}{N},\;f_{0}=\frac{1}{T_{0}},\;f_{s} =\frac{1}{T_{s}}, \tag{22}\]  
where the number of time step, total analysis time and the sampling time (\(\Delta t\)) are denoted by \(N\), \(T_{0}\), and \(T_{s}\), respectively, and the frequency step (\(\Delta f\)) and the sampling frequency are denoted by \(f_{0}\) and \(f_{s}\), respectively. Using the above notations, this study formulates the following theory based on the DFT. Figure 5: Comparison of the mechanical stresses at \(x=0.25\) in (1 st element) for the benchmark problem in Figure 2.  
**Theorem 1**. : _Consider the signal, \(\overline{y}\), filtered from original signal, \(y\), which has only a single frequency component, \(f\), with a filtering length \(FL\). With sampling frequency, \(f_{s}\), the Fourier transformation of the filtered signal has the following frequency component:_  
\[\widetilde{f}=\frac{f_{s}}{FL}\cdot m\pm f\text{ and }0\leq\widetilde{f}\leq \frac{f_{s}}{2}, \tag{23}\]  
_where \(m\) and \(\widetilde{f}\) denote an arbitrary integer and the frequency components generated by the frequency filtering method (i.e., the RFM), respectively. The frequency components generated by the filtering scheme in DFT can be mathematically found through the following proof._  
Figure 6: Modification of the original signal depending on the filtering length (\(FL\)): (**a**) without RFM, with RFM for (**b**) \(FL=2\), (**c**) \(FL=4\), and (**d**) \(FL=5\). **Proof. ** The \(N\)-point DFT of discrete signal \(y_{k}\) is obtained as follows:  
\[Y_{n}=\frac{1}{N}{\sum\limits_{k=0}^{N-1}{y_{k}e^{-j\frac{2\pi n}{N}\cdot k}}}. \tag{24}\]  
Assuming \(y_{k}=\exp(j\frac{2\pi f^{*}}{N}k)\) (a signal with a single frequency component, \(f\), where \(f^{*}=\frac{f}{f_{0}}\)), the above DFT can be summarized as follows:  
\[Y_{n}=\frac{1}{N}{\sum\limits_{k=0}^{N-1}{e^{-j\frac{2\pi f}{N}(n-f^{*})\cdot k }}}=\delta_{f}(n),\ \delta_{f}(n)=\left\{