## 4 Design of Disturbance Estimator  
In this section, the disturbance estimator (DE) for \(d(t)\) in system (2) is designed such that \(\hat{d}(t)\to d(t)\). In order to undertake this, Equation (2) is re-written as follows. \[\left\{ \begin{array}{l}\dot{x}_{1}=x_{2}\\ \vdots\\ \dot{x}_{n-1}=x_{n}\\ \dot{x}_{n}=x^{(n)}=f_{0}(\mathbf{x},t)+g_{01}(\mathbf{x},t)u(t)+g_{02}(t)d(t) \\ y=x_{1}\end{array} \right. \tag{37}\]  
In a compact form, this equation is re-written as follows. \[\left\{ \begin{array}{l}\dot{\mathbf{x}}=\mathbf{F}(\mathbf{x},t)+\mathbf{G_ {1}}(\mathbf{x},t)u(t)+\mathbf{G_{2}}(\mathbf{x},t)d(t)\\ y=x_{1}\\ \mathbf{F}(\mathbf{x},t)=\left[x_{2},\,x_{3},\ldots,\,x_{n},f_{0}(\mathbf{x},t)\right]^{T}\in R^{n}\\ \mathbf{G_{1}}(\mathbf{x},t)=\left[0,\ldots,\,0,\,g_{01}(\mathbf{x},t)\right] ^{T}\in R^{n}\\ \mathbf{G_{2}}(\mathbf{x},t)=\left[0,\ldots,\,0,\,g_{02}(t)\right)\right]^{T} \in R^{n}.\end{array} \right. \tag{38}\]  
Then, the estimate \(\hat{d}(t)\) is expressed as follows [33]. \[\hat{d}(t)=z(\mathbf{x},t)+\mathbf{1}(\mathbf{x})\mathbf{x} \tag{39}\]  
In the above, a vector of constant parameters \(\mathbf{1}(\mathbf{x})\) needs to be chosen so that the following error function converges to zero. \[E(t)=d(t)-\hat{d}(t) \tag{40}\]  
In Equation (41), \(z(\mathbf{x},t)\) is the internal state of the nonlinear observer. The function \(z(\mathbf{x},t)\) is then estimated as follows:  
\[ \begin{array}{l}\dot{z}(\mathbf{x},t)=-\mathbf{1}(\mathbf{x})(\mathbf{G_{1 }}(\mathbf{x},t)u(t)+\mathbf{G_{2}}(\mathbf{x},t)p(\mathbf{x})+\mathbf{F}( \mathbf{x},t))-\mathbf{1}(\mathbf{x})\mathbf{G_{2}}(\mathbf{x},t)z(\mathbf{ x},t)\\ p(\mathbf{x})=\mathbf{1}(\mathbf{x})\,\mathbf{x}.\end{array}  \tag{41}\]  
In order to design DE, consider control system (2), which is re-written as (40). If the estimate \(\hat{d}(t)\) given in (41) with a constant parameter vector \(\mathbf{1}(\mathbf{x})=\left[l_{1},\ldots,l_{n}\right]\) is used, then \(\forall l_{i}\,(i=1\ldots n)\in\Re\) and \(l_{n}>0\), \(E(t)\to 0\) when \(t\rightarrow\infty\). Thus, this is a stable process. This can be proved as follows. Through considering the time derivative of (42) with Equations (41) and (40), the following equation is obtained. \[ \begin{array}{l}\dot{E}(t)=\dot{d}(t)-\dot{\hat{d}}(t)=\dot{d}(t)-\dot{z}( \mathbf{x},t)-\frac{\partial p}{\partial\mathbf{x}}\dot{\mathbf{x}}\\ =\dot{d}(t)-\dot{z}(\mathbf{x},t)-\mathbf{1}(\mathbf{x})\dot{\mathbf{x}}\\ =\dot{d}(t)-\dot{z}(\mathbf{x},t)-\mathbf{1}(\mathbf{x})(\mathbf{F}(\mathbf{x },t)+\mathbf{G_{1}}(\mathbf{x},t)u(t)+\mathbf{G_{2}}(\mathbf{x},t)d(t))\end{array}  \tag{42}\]  
By substituting \(\dot{z}(\mathbf{x},t)\) from (43) into (44) and taking (42), the following equation is achieved. \[ \begin{array}{l}\dot{E}(t)=\dot{d}(t)+\mathbf{1}(\mathbf{x})\mathbf{G_{2}} (\mathbf{x},t)(z(\mathbf{x},t)+p(\mathbf{x})-d(t))\\ \dot{E}(t)=\dot{d}(t)-\mathbf{1}(\mathbf{x})\mathbf{G_{2}}(\mathbf{x},t)E(t) \end{array}  \tag{43}\]  
Figure 3 presents a quarter suspension system consisting of four main equipment groups as follows. The suspension system consists of the linear spring \(k_{s}=2.8\times 10^{4}\) N/m and the MRD with the damping coefficient of \(c_{s}=3000\) Ns/m corresponding to the zero current. The hydraulic unit generates excitations with different frequencies and amplitudes. The mechanical structure is constituted of the upper bed, lower bed, and four parallel vertical cylindrical pillars which are used to fix the suspension system, wheel, sprung mass, and sensors. The control system essentially consists of a computer, an analog to digital/digital to analog (AD/DA) converter (dSPASE DS1104, Paderborn, Germany), a current amplifier, and sensors. The sprung mass is set by \(m_{s}=246.5\pm 35\) kg, including the mass of the upper bed. The relative displacement between the sprung and unsprung mass is measured by LVDT (linear variable differential transformer), while the acceleration of the chassis mass is measured by an accelerometer.