Impulse Invariance
In this method, we proceed as follows :
\(G_{anal}\left(p\right)= \frac{1}{\tau p}\xrightarrow{\text{ Laplace Transform}}=g(t)= \frac{1}{\tau}e^-\frac{\tau}{\tau } \gamma (t)\)
\(\quad\) \(\quad\) \(\quad\) \(\quad\) \(\quad\) \(\quad\) \(\xrightarrow{\text{ Sampling }}=T_eg(nT_e)=\frac{T_e}{\tau}\left(e^{-\frac{T_e}{\tau}}\right)^nu(n)\)
\(\quad\) \(\quad\) \(\quad\) \(\quad\) \(\xrightarrow{\text{ Z Transform }}=H(Z)=\)\(\frac{T_e}{\tau}\frac{1}{1-e^{-\frac{T_e}{\tau}}Z^{-1}}\)
Noticed : \(G_{anal}\left(p\right)\) has a pole (value of p makes \(G_{anal}\left(p\right) infinity)\) : \(p=-\frac{1}{\tau}\)
\(H(Z)\) has a pole : \({Z=e}^{-\frac{1}{\tau}}=e^{T_ep}\) , where p is the pole of \(G_{anal}\left(p\right)\) .