## 4 Application of the \(EDNN\) in Denoising
### Realization of \(EDNN\)  
The single-input, single-output \(EDNN\) is shown in Figure 3. There are eight nodes in the hidden layer, and the activation function of the hidden layer is a sigmoid function. The output of the sample error calculator (\(SEC\)) is:  
\[a^{o}(x)=f\Big{[}(\sum_{i=1}^{8}f\Big{(}W_{i,1}^{1,I}*x+b_{i}^{1}\Big{)})\cdot W _{1,j}^{2,1}+b_{1}^{2}\Big{]} \tag{25}\]where \(W_{i,1}^{1,I}\) is the weight of the input to the \(i\)th node of the first hidden layer; \(W_{1,i}^{2,1}\) is the weight of the \(i\)th node in the first hidden layer to output; \(b_{\text{i}}^{1}\) is the bias of the \(i\)th node of the first hidden layer; \(b_{1}^{2}\) is the bias of the output.  
The \(k\)th derivative of the output with respect to the input is shown in (26).  
\[\frac{d^{k}a^{o}(x)}{dx^{k}}=\sum_{i=1}^{n}f^{(k)}\Big{(}W_{i,1}^{1,I}*x+b_{i} ^{1}\Big{)}\cdot W_{1,i}^{2,1}*W_{i,1}^{1,I} \tag{26}\]  
where \(f^{(k)}\Big{(}W_{i,1}^{1,I}*x+b_{i}^{1}\Big{)}\) represents the \(k\)th-order derivative of the activation function corresponding to the \(i\)th node of the hidden layer.  
The highest derivative order of the voltage with respect to the time of the second-order damped free oscillation signal is two. Thus, the highest derivative order of the _EDNN_ could be set to four, which satisfies the noise reduction demand. Therefore, the maximum value of \(k\) in expression (26) is four. Additionally, only the first- through fourth-order derivatives of the activation function are required [41]. The first- through fourth-order derivatives of the activation function are shown in expressions (27)-(30):  
\[f^{\prime}(x)=f(x)-[f(x)]^{2} \tag{27}\]  
\[f^{\prime\prime}(x)=f(x)-3[f(x)]^{2}+2\cdot[f(x)]^{3} \tag{28}\]  
\[f^{(3)}(x)=f(x)-7\cdot[f(x)]^{2}+12\cdot[f(x)]^{3}-6\cdot[f(x)]^{4} \tag{29}\]  
\[f^{(4)}(x)=\Big{\{}f(x)-15\cdot[f(x)]^{2}+50\cdot[f(x)]^{3}-60\cdot[f(x)]^{4}+2 4\cdot[f(x)]^{5}\Big{\}} \tag{30}\]  
where \(f(x)=\frac{1}{1+e^{-x}}\) is the sigmoid function; \(f^{\prime}(x),f^{\prime\prime}(x),f^{(3)}(x)\),\(f^{(4)}(x)\) are the first, second, third, and fourth derivatives of \(f(x)\) with respect to \(x\).  
The \(k\)th derivative of the output with respect to the input is \(\frac{d^{k}a^{o}(x)}{dx^{k}}\). Additionally, the solution of \(\frac{d^{k}a^{o}(x)}{dx^{k}}=0\) is:  
\[Sol_{k-1}=\left\{x\bigg{|}\frac{d^{k}a^{o}(x)}{dx^{k}}=0\right\} \tag{31}\]  
Figure 3: Schematic of the single-input, single-output, single-hidden-layer _EDNN_.  
The number of extreme points \(R_{k-1}\) corresponding to the derivative \(\frac{d^{k-1}a^{o}(x)}{dx^{k-1}}\) is the rank of the set \(Sol_{k-1}\), i.e.:  
\[R_{k-1}=Rank\Bigg{(}x\Bigg{|}\frac{d^{k}a^{o}(x)}{dx^{k}}=0\Bigg{)} \tag{32}\]  
when the number of extreme points of the real physical process corresponding to the derivative \(\frac{d^{k-1}a^{o}(x)}{dx^{k-1}}\) is \(TCE_{k-1}\).  
Therefore, the extreme points distribution feature error corresponding to the \((k-1)\)th derivative \(\frac{d^{k-1}a^{o}(x)}{dx^{k-1}}\) is:  
\[Er_{k-1}=R_{k-1}-TCE_{k-1} \tag{33}\]  
where \(TCE_{k-1}\) is determined using known physical process information.  
The total loss function of the _EDNN_ is:  
\[Loss=S_{S}\cdot LossSp+S_{E}\cdot LossEV=S_{S}\cdot\sum_{s=1}^{N_{t}}[t_{s}-a_ {s}^{o}(x)]^{2}+S_{E}\cdot\sum_{k=0}^{k=4}(Er_{k-1})^{2} \tag{34}\]  
where \(LossEV\) is the extreme points distribution feature error loss function; \(LossSp\) is the sample error loss function; \(S_{S}\) is the weight of the sample error loss function \(LossSp\); \(S_{E}\) is the weight of the extreme points distribution feature error loss function \(LossEV\); \(t_{s}\) is target of the \(sth\) training sample; \(a_{s}^{o}(x)\) is the output corresponding to the \(sth\) training sample; \(N_{t}\) is the number of training samples.