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when n=0,1,2, x[n] is n; otherwise x[n]=0. So:
 
when n=0,1,2, x[n] is n; otherwise x[n]=0. So:
  
<math>x(z)=0z^{-0}+1z^{-1}+2z^{-2}=\frac{1}{z}+\frac{2}{z^2}</math>
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<math>x(z)=0z^{-0}+1z^{-1}+2z^{-2}=\frac{1}{z}+\frac{2}{z^2}</math> with ROC=all finite complex number.
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test for infinity:
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<math>X(\frac{1}{z})=z+z^2</math>
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when z=0,<math>X(\frac{1}{z})</math>converges
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X(z) converges at <math>z=\infty</math>
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so ROC of X(z) is all complex number.
 
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Revision as of 09:24, 10 September 2011

Z-transform computation

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

$ x[n]= n u[n]-n u[n-3] $

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


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Answer 1

Begin with the definition of a Z-Transform.

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

Simplify a little. (pull out the n and realize $ u[n]-u[n-3] $ is only non-zero for 0, 1, and 2.)

$ X(z) = \sum_{n=0}^{2}n z^{-n} $

Then we have a simple case of evaluating for 3 points.

$ \begin{align} X(z) &= 0 z^{-0} + 1 z^{-1} + 2 z^{-2} \\ &= \frac{z+2}{z^2} \end{align} $

Answer 2

$ Z(x[n])= \sum_{n=-\infty}^{\infty}x[n]z^{-n}= \sum_{n=-\infty}^{\infty}n(u[n]- u[n-3])z^{-n} $

when n=0,1,2, x[n] is n; otherwise x[n]=0. So:

$ x(z)=0z^{-0}+1z^{-1}+2z^{-2}=\frac{1}{z}+\frac{2}{z^2} $ with ROC=all finite complex number.

test for infinity:

$ X(\frac{1}{z})=z+z^2 $

when z=0,$ X(\frac{1}{z}) $converges

X(z) converges at $ z=\infty $

so ROC of X(z) is all complex number.


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