# Help for Z transform of t^2

• Jun 18th 2010, 09:27 PM
moonnightingale
Help for Z transform of t^2
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
• Jun 18th 2010, 11:04 PM
CaptainBlack
Quote:

Originally Posted by moonnightingale
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 sided) z-transform is defined as:

$
\displaystyle
\mathcal{Z}(t^2)=\sum_{t=0}^{\infty} \frac{t^2}{z^n}
$

Now you proceed by finding the sum of the series on the right (or just looking $n^2$ up in a table of z-transforms).

CB
• Jun 19th 2010, 12:30 AM
chisigma
The [one side] z-transform is usually applicable to a discrete sequence $a_{n}$ and is defined as...

$\displaystyle \mathcal{Z} \{a_{n}\} = \sum_{n=0}^{\infty} a_{n} z^{-n}$ (1)

According to Z-Transform -- from Wolfram MathWorld, if You have a continous function $f(t)$ , its z-transform is the z-tranform of the sequence $a_{n} = f(nT)$ , where T is the so called 'sampling time'. In your case is $f(t)= t^{2}$ so that is...

$\displaystyle \mathcal{Z} \{f(nT)\} = T^{2} \sum_{n=0}^{\infty} n^{2} z^{-n}= T^{2} \frac{z (z+1)}{(z-1)^{3}}$ (2)

Kind regards

$\chi$ $\sigma$
• Jun 19th 2010, 01:45 AM
moonnightingale
yes i know answer but how did u perform last step
Kindly explain how u wrote (z-1)^3 in denominator
I am well conversant with Z transform but this step involves some math series
Kindly explain ur last step
• Jun 19th 2010, 02:33 AM
CaptainBlack
You differentiate:

$
\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 $1+z^{-1}+2 z^{-2}+...+n z^{-n}+ ... \ \$ on the right hand side.

CB
• Jun 19th 2010, 03:16 PM
chisigma
In detail You start from the expansion...

$\displaystyle \frac{1}{1-z} = \sum_{n=0}^{\infty} z^{n}$ , $|z|<1$ (1)

... and then You derivate (1) and obtain ...

$\displaystyle \frac{d}{dz} \frac{1}{1-z} = - \frac{1}{(1-z)^{2}} = \frac{1}{z} \sum_{n=0}^{\infty} n \cdot z^{n}$ (2)

... so that is...

$\displaystyle \sum_{n=0}^{\infty} n\cdot z^{n} = -\frac{z}{(1-z)^{2}}$ (3)

Now You derivate (3) and obtain...

$\displaystyle \frac{d}{dz} \{- \frac{z}{(1-z)^{2}}\} = - \frac{1+z}{(1-z)^{3}} = \frac{1}{z} \sum_{n=0}^{\infty} n^{2} z^{n}$ (4)

... so that is ...

$\displaystyle \sum_{n=0}^{\infty} n^{2} z^{n} = \frac{z\cdot (z+1)}{(z-1)^{3}}$ (5)

... and, replacing $z$ with $z^{-1}$, ...

$\displaystyle \sum_{n=0}^{\infty} n^{2} z^{-n} = \frac{z^{-1}\cdot (1+ z^{-1})}{(z^{-1}-1)^{3}}$ (6)

Kind regards

$\chi$ $\sigma$
• Jun 20th 2010, 03:57 AM
moonnightingale
Thanks a lot for detailed reply