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 January 22nd, 2017, 07:56 AM #1 Member     Joined: Nov 2014 From: SG Posts: 55 Thanks: 1 Sampled signal - Z transform I managed to solve Q1(a) but how do I solve for Q1(b)? Last edited by KaiL; January 22nd, 2017 at 08:13 AM.
 January 22nd, 2017, 05:00 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 what is $1(k)$ and $1(t)$ ?
January 22nd, 2017, 06:37 PM   #3
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Quote:
 Originally Posted by romsek what is $1(k)$ and $1(t)$ ?
Unit step.

$\displaystyle \mathscr{Z} \{ 1(t) \} = \dfrac{1}{1 - z^{-1}}$

 January 22nd, 2017, 07:32 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 sampled at rate $f_s = \dfrac 1 T$ $f(n) = \alpha t 1(t)= \alpha n T,~n=0,1,\dots$ $F(z) = \displaystyle{\sum_{n=0}^\infty}~\alpha n T z^{-n}$ $F(z) = \alpha T \displaystyle{\sum_{n=0}^\infty}~\ n z^{-n}$ $F(z) = \alpha T \dfrac{z}{(z-1)^2}$
January 23rd, 2017, 03:11 AM   #5
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Quote:
 Originally Posted by romsek sampled at rate $f_s = \dfrac 1 T$ $f(n) = \alpha t 1(t)= \alpha n T,~n=0,1,\dots$ $F(z) = \displaystyle{\sum_{n=0}^\infty}~\alpha n T z^{-n}$ $F(z) = \alpha T \displaystyle{\sum_{n=0}^\infty}~\ n z^{-n}$ $F(z) = \alpha T \dfrac{z}{(z-1)^2}$
Hi Thank you very much for your help but could I check with you what does the alpha α means in the equation ? I can't really find any information online with regard to the α alpha

January 23rd, 2017, 04:39 AM   #6
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Quote:
 Originally Posted by romsek what is $1(k)$ and $1(t)$ ?
Both are unit step if I am correct

January 23rd, 2017, 05:20 AM   #7
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Quote:
 Originally Posted by KaiL Hi Thank you very much for your help but could I check with you what does the alpha α means in the equation ? I can't really find any information online with regard to the α alpha
it's just some constant

January 23rd, 2017, 07:14 AM   #8
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Quote:
 Originally Posted by romsek it's just some constant
I see. Thank you very much for your explanation

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