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January 22nd, 2017, 07:56 AM   #1
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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.
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January 22nd, 2017, 05:00 PM   #2
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what is $1(k)$ and $1(t)$ ?
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January 22nd, 2017, 06:37 PM   #3
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what is $1(k)$ and $1(t)$ ?
Unit step.

$\displaystyle \mathscr{Z} \{ 1(t) \} = \dfrac{1}{1 - z^{-1}}$
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January 22nd, 2017, 07:32 PM   #4
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sampled at rate $f_s = \dfrac 1 T$

$f(n) = \alpha t 1(t)= \alpha n T,~n=0,1,\dots$

$F(z) = \displaystyle{\sum_{n=0}^\infty}~\alpha n T z^{-n}$

$F(z) = \alpha T \displaystyle{\sum_{n=0}^\infty}~\ n z^{-n}$

$F(z) = \alpha T \dfrac{z}{(z-1)^2}$
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January 23rd, 2017, 03:11 AM   #5
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Quote:
Originally Posted by romsek View Post
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
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January 23rd, 2017, 04:39 AM   #6
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Quote:
Originally Posted by romsek View Post
what is $1(k)$ and $1(t)$ ?
Both are unit step if I am correct
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January 23rd, 2017, 05:20 AM   #7
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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
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January 23rd, 2017, 07:14 AM   #8
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it's just some constant
I see. Thank you very much for your explanation
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