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

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- Jun 19th 2010, 07:06 AMmoonnightingaleShifting from Laplace to time domian and then 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 - Jun 19th 2010, 07:13 AMAckbeet
- Jun 19th 2010, 07:20 AMmoonnightingale
Sorry it is Capital T

I have corrected it

exp(sT/2) / s^2 - Jun 21st 2010, 10:37 PMchisigma
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

$\displaystyle \chi$ $\displaystyle \sigma$