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Can someone tell me how to find the Z Tranform of the sequence:

$x(n)=n, \quad n=0,1,2,3,4,5 \Rightarrow X(z)=\sum\limits_{n=0}^{N-1}nz^{-n} $

I have searched everywhere I could, but in every single example there is a coefficient raised in $n$ which made Z Transform too easy. I have found the result for something similar to this one in a lot of places but not the proof!

Adam
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  • Is the $z$ in $x(n) = nz^{-n}$ the same $z$ used in the $z$ transform? If so, that variable is overused. Also, is that for $n \ge 0$ only, or is that sequence two sided? – JimmyK4542 Jun 19 '14 at 21:22
  • @jimmyK4542 sorry I didn't notice it! I will edit and correct it! – Adam Jun 19 '14 at 21:25
  • Ok, now this is a finite sum, specifically $X(z) = 1z^{-1}+2z^{-2}+3z^{-3}+4z^{-4}+5z^{-5}$. Do you want this in a different form. – JimmyK4542 Jun 19 '14 at 21:34
  • This is the obvious! I want it in a close form. Thanks for your answer though. – Adam Jun 19 '14 at 21:36
  • That is in a closed form. – JimmyK4542 Jun 19 '14 at 21:40
  • Sorry the problem obviously is the constant 5. I read something somilar in a book so I thought that it would be easier to plave a constant and not N-1 as in there. I will edit again! – Adam Jun 19 '14 at 21:42

1 Answers1

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Let's try to manipulate that sum as follows:

[1] $X(z) = z^{-1}+2z^{-2}+3z^{-3}+\cdots+(N-1)z^{-(N-1)}$

[2] $z^{-1}X(z) = z^{-2}+2z^{-3}+3z^{-4}+\cdots+(N-1)z^{-N}$ (Multiply [1] by $z^{-1}$)

[3] $(1-z^{-1})X(z) = z^{-1}+z^{-2}+z^{-3}+\cdots+z^{-(N-1)} - (N-1)z^{-N}$ (Subtract [2] from [1])

[4] $z^{-1}(1-z^{-1})X(z) = z^{-2}+z^{-3}+z^{-4}+\cdots+z^{-N} - (N-1)z^{-(N+1)}$ (Multiply [3] by $z^{-1}$)

[5] $(1-z^{-1})^2X(z) = z^{-1}-Nz^{-N}+(N-1)z^{-(N+1)}$ (Subtract [4] from [3])

[6] $X(z) = \dfrac{z^{-1}-Nz^{-N}+(N-1)z^{-(N+1)}}{(1-z^{-1})^2}$ (Divide both sides of [5] by $(1-z^{-1})^2$)

You should double check the algebra.

JimmyK4542
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