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=== Answer 8 ===
 
=== Answer 8 ===
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Xi Wang
  
 
<math>X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n}</math>
 
<math>X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n}</math>
  
 
<span class="texhtml"> = ''X''(''z'') = (9''z''<sup> + 3</sup> + 4''z''<sup> + 2</sup> + ''z''. The range of the value of z is from negative infinity to positive infinity
 
<span class="texhtml"> = ''X''(''z'') = (9''z''<sup> + 3</sup> + 4''z''<sup> + 2</sup> + ''z''. The range of the value of z is from negative infinity to positive infinity

Revision as of 16:24, 12 September 2013

Practice Problem on Z-transform computation

Compute the compute the z-transform (including the ROC) of the following DT signal:

$ x[n]= n^2 \left( u[n+3]- u[n-1] \right) $

(Write enough intermediate steps to fully justify your answer.)


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Answer 1

x[n] = n2(u[n + 2] − u[n − 1]).

$ X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} $

$ = \sum_{n=-3}^{0} n^2 z^{-n} $

= 9z3 + 4z2 + z

= z3(9 + 4z − 1 + z − 2)

= X(z) = (9 + 4z − 1 + z − 2) / (z − 3), for all z in complex plane.

Answer 2

Muhammad Syafeeq Safaruddin

x[n] = n2(u[n + 3] − u[n − 1])

x[n] = n2(δ(n + 3) + δ(n + 2) + δ(n + 1) + δ(n))

$ X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} $

$ X(z) = \sum_{n=-\infty}^{+\infty} n^2(\delta(n+3)+\delta(n+2)+\delta(n+1)+\delta(n)) z^{-n} $

X(z) = 9z3 + 4z2 + z + 1 for all z in complex plane


Answer 3

Write it here.

Answer 4

Write it here.

Answer 5

Tony Mlinarich

$ X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} $

$ X(z) = n^2(\delta (n+3)+\delta (n+2)+\delta (n+1)+\delta (n)+\delta (n-1)) z^{-n} $

X(z) = 9z3 + 4z2 + z + 1/z<\span>


Back to ECE438 Fall 2013 Prof. Boutin

Answer 7

Yixiang Liu

$ X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} $

$ X(z) = \sum_{n=-\infty}^{+\infty} n^{2}[{u[n+3]-u[n-1]}]z^{-n} $

This expression equals to zero except n = -3, -2, -1

so $ X(z) = x[-3]z^{3} + x[-2]z^{2} + x[-1]z^{1} $

      = 9z^{3} + 4z^{2} + z

Answer 8

Xi Wang

$ X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} $

= X(z) = (9z + 3 + 4z + 2 + z. The range of the value of z is from negative infinity to positive infinity

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