I want to have Z transform of This function
It is in laplac at present
exp(sT/2) / s^2
can it be directly converted to Z transform or first i have to make in time domain by taking Laplac inverse and then taking its Z transform
I want to have Z transform of This function
It is in laplac at present
exp(sT/2) / s^2
can it be directly converted to Z transform or first i have to make in time domain by taking Laplac inverse and then taking its Z transform
Your approach is correct and we have...
$\displaystyle \displaystyle f(t)= \mathcal{L}^{-1} \{\frac{e^{s \frac{T}{2}}}{s^{2}}\} = t + \frac{T}{2}$ (1)
... so that is...
$\displaystyle \displaystyle \mathcal {Z} \{f(t)\} = T \sum_{n=0}^{\infty} (n+\frac{1}{2})\ z^{-n} = T\ \{ \frac{z^{-1}}{(1-z^{-1})^{2}} + \frac{1}{2\ (1-z^{-1})}\} = \frac{T}{2}\ \frac{1+z^{-1}}{(1-z^{-1})^{2}} $ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$