# Feedback Stabilization of First Order Neutral Delay Systems Using the Lambert W Function
## 4 Examples  
A few examples and numerical simulation results have been presented here to validate the analytical results of the previous section. **Example 1**. : _Consider the NDDE_  
\[\dot{x}(t)-0.3\dot{x}(t-1.7)+2.1x(t-0.3)=0, \tag{32}\]  
_where, \(a=-0.3,b=2.1,\tau_{1}=1.7,\tau_{2}=0.3\). Because \(-1<a<0\), from Theorem 1 we get:_  
\[2-W(-ae^{-1})+ae^{W(-ae^{-1})-1}+b\tau_{1}=3.29266.\]  
_Equation (32) satisfies the condition of Theorem 1. Hence, every solution of this NDDE is oscillatory. The results are depicted in Figure 0(a)._  
**Example 2**. : _Consider the NDDE [24]_  
\[\dot{x}(t)-\frac{1}{2e}\dot{x}(t-1)+\frac{1}{2e}x(t-2)=0, \tag{33}\]  
_where, \(a=-\frac{1}{2e},b=\frac{1}{2e},\tau_{1}=1,\tau_{2}=1\). As \(-1<a<0\), from Theorem 1 we have:_  
\[2-W(-ae^{-1})+ae^{W(-ae^{-1})-1}+b\tau_{1}=-0.31606.\]  
_Equation (33) does not satisfy the condition of Theorem 1, but satisfies the condition of Theorem 4. Therefore, there are two negative real roots. The results are depicted in Figure 0(b)._  
**Example 3**. : _Consider the NDDE [28]_  
\[\dot{x}(t)-2e\dot{x}(t-1)+\sqrt{e}x(t-0.5)=0, \tag{34}\]  
_where, \(a=-2e,b=\sqrt{e},\tau_{1}=1,\tau_{2}=0.5\). As \(a<-1\), from Theorem 3 we obtain:_  
\[2-W(-ae^{-1})+ae^{W(-ae^{-1})-1}+\tau_{1}b(-a)^{-\frac{\tau_{2}}{\tau_{1}}}=-5.4366.\]  
_Equation (34) does not satisfy the condition of Theorem 3, but satisfies the condition of Theorem 5. Thus, there are two positive real roots. The results are shown in Figure 0(c)._  
The solid red, blue and black lines in Figure 1 represent \(f(\lambda)\), \(g(\lambda)\), and the characteristic equation \(p(\lambda)\), respectively. The intersections of the two functions (red and blue lines), denoted by the blue dots, are the real roots of the corresponding equation. The intersection points between the characteristic equation and the real axis exactly equal the blue dots. Figure 0(a) shows that there is actually no intersection between \(p(\lambda)\) and the real axis, which implies that every solution of Equation (3) oscillates. Figure 0(b),\(c\) establishes the existence of real roots. **Example 4**. : _Consider a NDDS with state feedback:_  
\[\dot{x}(t)+a\dot{x}(t-\tau_{1})+bx(t-\tau_{2})=u(t), \tag{35}\]  
_and \(u(t)=k_{d}\dot{x}(t-\tau_{1})+k_{p}x(t-\tau_{2})\), where \(a=-3\), \(b=1.31014\), \(\tau_{1}=0.7\), \(\tau_{1}=1.2\) and the continuous initial history is given by \(x(t)=-1\)._  
_The characteristic equation of Equation (35) in the absence of control becomes a simple quasi-polynomial:_  
\[\lambda-3\lambda e^{-0.7\lambda}+1.31014e^{-1.2\lambda}=0. \tag{36}\]  
_Applying Equation (27) to Equation (36), we obtain \(-0.271786\). This indicates that there are two positive real roots based on Theorem 5. Applying the QPmR (Quasi-Polynomial Mapping based Root-finder [29]) algorithm to Equation (36), the rightmost spectrum of the characteristic equation (36) is depicted in Figure 2a. Here, the four rightmost roots are \(1.2902\), \(0.6683\), \(1.5766+i7.9386\), and \(1.5766-i7.9386\). Figure 3a shows the time response in the absence of feedback control._  
Figure 2: Rightmost spectrum of the characteristic equation. The spectrum values are denoted as black dots. The blue and red lines are the isolines of the real and imaginary parts of the characteristic equation to zero, respectively. (**a**) Without state feedback. A few eigenvalues are located in the right-half plane of the imaginary axis. (**b**) With a state feedback controller. All eigenvalues are located in the open left-half plane of the imaginary axis. We used a state feedback controller where the controller gains were analytically determined from Theorem 6, as one of the goals of this study was to stabilize the NDDS (35). First, we selected the derivative gain \(k_{d}=-2.5\) that satisfies the condition \(a<k_{d}<a+1\), such that \(-1<a_{f}(=a-k_{d})<0\). Next, we can obtain \(\lambda_{m}=\frac{1}{\tau_{1}}(1-W(-ea_{f}^{-1}))=-0.5355\) from Equation (13), and finally the proportional gain was determined using Equation (31).