Chapitre 6

In this method, we perform the inverse Laplace transform of Y\((p)\), which yields y(t). Then, we sample \(y(t)\) to obtain y(n) (numeric values), and finally, we take the Z-transform of \(y(n)\).

\(G_{anal}\left(p\right)=\frac{1}{1+\tau p\ }\) \(\implies\) \(Y\left(p\right)=G_{anal}\left(p\right)\Gamma\left(p\right)=\frac{1}{p}.\frac{1}{1+\tau p}\)

\(\quad\) \(\quad\) \(\quad\) \(\quad\) \(\xrightarrow{\text{ Laplace Transform}} \) \(Y(t)=(1- e^{-\frac{1}{\tau}})y(t)\)

\(\quad\) \(\quad\) \(\quad\) \(\quad\) \(\quad\) \(\xrightarrow{\text{ sampling }}\) \(y(n)=(1- e^{-\frac{nT_e}{\tau}})u(n)\)

\(\quad\) \(\quad\) \(\xrightarrow{\text{ Z Transform }}\) \(Y(Z)=\frac{1}{1-Z^-1}=\frac{1}{1- e^-\frac{T_e}{\tau } Z^-1}=H(Z).U(Z)=H(Z). \frac{1}{1-Z^-1}\)

Therefore, we obtain the expression for \(H(z)\) :

\(H\left(Z\right)=\left(1-Z^{-1}\right)Y\left(Z\right)=\frac{1-Z^{-1}}{1-e^{-\frac{T_e}{\tau}}Z^{-1}}=\frac{\left(1-e^{-\frac{T_e}{\tau}}\right)Z^{-1}}{1-e^{-\frac{T_e}{\tau}}Z^{-1}}\)