I am unable to calculate Z transform of t^2 ( t Square)
Can any body show me step by step method to calculate its Z transform
The [one side] z-transform is usually applicable to a discrete sequence $\displaystyle a_{n}$ and is defined as...
$\displaystyle \displaystyle \mathcal{Z} \{a_{n}\} = \sum_{n=0}^{\infty} a_{n} z^{-n}$ (1)
According to Z-Transform -- from Wolfram MathWorld, if You have a continous function $\displaystyle f(t)$ , its z-transform is the z-tranform of the sequence $\displaystyle a_{n} = f(nT)$ , where T is the so called 'sampling time'. In your case is $\displaystyle f(t)= t^{2}$ so that is...
$\displaystyle \displaystyle \mathcal{Z} \{f(nT)\} = T^{2} \sum_{n=0}^{\infty} n^{2} z^{-n}= T^{2} \frac{z (z+1)}{(z-1)^{3}}$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
You differentiate:
$\displaystyle
\displaystyle
f(z)=\frac{z}{z-1}=1+z^{-1}+z^{-2}+...+z^{-n}+ ... \ \ ;\ z \in (-1,1)
$
twice and manipulate the results to get the series $\displaystyle 1+z^{-1}+2 z^{-2}+...+n z^{-n}+ ... \ \ $ on the right hand side.
CB
In detail You start from the expansion...
$\displaystyle \displaystyle \frac{1}{1-z} = \sum_{n=0}^{\infty} z^{n}$ , $\displaystyle |z|<1$ (1)
... and then You derivate (1) and obtain ...
$\displaystyle \displaystyle \frac{d}{dz} \frac{1}{1-z} = - \frac{1}{(1-z)^{2}} = \frac{1}{z} \sum_{n=0}^{\infty} n \cdot z^{n}$ (2)
... so that is...
$\displaystyle \displaystyle \sum_{n=0}^{\infty} n\cdot z^{n} = -\frac{z}{(1-z)^{2}}$ (3)
Now You derivate (3) and obtain...
$\displaystyle \displaystyle \frac{d}{dz} \{- \frac{z}{(1-z)^{2}}\} = - \frac{1+z}{(1-z)^{3}} = \frac{1}{z} \sum_{n=0}^{\infty} n^{2} z^{n}$ (4)
... so that is ...
$\displaystyle \displaystyle \sum_{n=0}^{\infty} n^{2} z^{n} = \frac{z\cdot (z+1)}{(z-1)^{3}}$ (5)
... and, replacing $\displaystyle z$ with $\displaystyle z^{-1}$, ...
$\displaystyle \displaystyle \sum_{n=0}^{\infty} n^{2} z^{-n} = \frac{z^{-1}\cdot (1+ z^{-1})}{(z^{-1}-1)^{3}}$ (6)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$