## 3 Model-Free Control
### Online Estimation of \(F\)  
Various techniques have been introduced in the literature for constructing an estimation of \(F\)[37]. In this work, the method developed in [25] is used. Where the unknown function \(F\) is approximated by a step function (a piecewise constant function) \(\Phi\), which can be reliably performed over a sufficiently small time interval. Therefore, Equation (11) becomes  
\[\dot{y}(t)=\Phi+\alpha u(t) \tag{13}\]  
By applying a Laplace transform (13), since \(\Phi\) is a step function, it yields [25]  
\[sY=\frac{\Phi}{s}+\alpha U+y_{0} \tag{14}\]  
where \(y_{0}\) is the initial condition and \(s\) is the Laplace domain variable.  
Both sides of (14) are multiplied by \(d/ds\), thereby eliminating the initial condition and obtaining  
\[Y+s\frac{dY}{ds}=-\frac{\Phi}{s^{2}}+\alpha\frac{dU}{ds} \tag{15}\]  
By multiplying both sides of (15) by a sufficiently negative power of \(s\) to acquire a minimum of one integrator (\(\frac{1}{s}\)) for each term in (15) in order to mitigate both noise and numerical computation inaccuracies in the output estimation, in this case \(s^{-2}\), the following equation is thus obtained:  
\[\frac{Y}{s^{2}}+\frac{dY}{sds}=-\frac{\Phi}{s^{4}}+\frac{\alpha}{s^{2}}\frac{ dU}{ds} \tag{16}\]  
From (16), the estimate of \(\Phi\) can be deduced, in the time domain, using inverse Laplace transforms, where [38]\[s^{-1}F(s)\rightarrow\int\limits_{0}^{t}f(t)dt \tag{17}\]  
\[s^{-n}F(s)\rightarrow\int^{(n)}f(t)dt \tag{18}\]  
\[\frac{d}{ds}\rightarrow-t \tag{19}\]  
\[\frac{d^{n}}{ds^{n}}\rightarrow(-1)^{n}t^{n} \tag{20}\]  
\[\frac{1}{s^{n}}\rightarrow\frac{t^{n-1}}{(n-1)!} \tag{21}\]  
From (18), we can obtain  
\[\frac{Y}{s^{2}}\rightarrow\int^{(2)}y(t)dt \tag{22}\]  
and from (17) and (19), we obtain  
\[\frac{dY}{ds}\rightarrow\int-ty(t)dt \tag{23}\]  
and from (21), we obtain  
\[\frac{\Phi}{s^{4}}\rightarrow\frac{L^{3}}{6}\Phi \tag{24}\]  
and from (18) and (19), we obtain  
\[\frac{\alpha}{s^{2}}\frac{dU}{ds}\rightarrow\alpha\int^{(2)}-tu(t)dt \tag{25}\]  
Using (22)-(25), we can transform (16) to a time domain as follows:  
\[\int^{(2)}y(t)dt-\int ty(t)dt=-\frac{L^{3}}{6}\Phi-\alpha\int^{(2)}tu(t)dt \tag{26}\]  
Using [PERSON]'s theorem in (27) to reduce multiple integrals into a simple one, we obtain  
\[\int\limits_{0}^{T}\int\limits_{0}^{t_{1}}\ldots\int\limits_{0}^{t_{k-1}}v(t_ {k})dt_{k}\ldots dt_{1}=\int\limits_{0}^{T}\frac{(T-t)^{k-1}}{(\kappa-1)!}v(t)dt \tag{27}\]  
Applying (27) to the terms in (26) that contain multiple integrals, we obtain  
\[\int\limits_{0}^{T}\int\limits_{0}^{t_{1}}y(\sigma)d\sigma dt_{1}=\int\limits _{0}^{T}\frac{(T-t)}{1!}y(t)dt \tag{28}\]  
\[\int\limits_{0}^{T}\int\limits_{0}^{t_{1}}\sigma u(\sigma)d\sigma dt_{1}=\int \limits_{0}^{T}\frac{(T-t)}{1!}tu(t)dt \tag{29}\]  
From (28) and (29), and by integrating during a quite small interval \(L\), (26) becomes  
\[\int\limits_{t-L}^{t}(L-\sigma)y(\sigma)d\sigma-\int\limits_{t-L}^{t}\sigma y (\sigma)d\sigma=-\frac{L^{3}}{6}F_{est}-\alpha\int\limits_{t-L}^{t}\sigma(L- \sigma)u(\sigma)d\sigma \tag{30}\]Finally, we obtain the expression of \(F_{est}\):  
\[F_{est}=-\frac{6}{L^{3}}\int\limits_{t-L}^{t}\;((L-2\sigma)y(\sigma)+\alpha \sigma(L-\sigma)u(\sigma))d\sigma \tag{31}\]  
In practice, the integral in (31) is substituted with finite impulse response (FIR) filters as presented in Figure 3.  
The coefficients of the FIRs determined according to (31), and they are as follows:  
\[\text{FIR1}=\begin{bmatrix}1&1&1&\dots&1\end{bmatrix} \tag{32}\]  
\[\text{FIR2}=\begin{bmatrix}1&2&3&\dots&N\end{bmatrix} \tag{33}\]  
\[\text{FIR3}=\begin{bmatrix}1&4&9&\dots&N^{2}\end{bmatrix} \tag{34}\]  
where \(N\) denotes the number of samples used in the FIR, such as \(L=N\times T_{s}\), and \(T_{s}\) is the sampling time.