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<math>X(z) = Z\left(x[n]\right) =Z\left(\delta[n-1]+2\delta[n-2]\right) = Z\left(\delta[n-1]\right)+Z\left(2\delta[n-2]\right) = z^{-1}+2z^{-2}, ROC = C/[0] | <math>X(z) = Z\left(x[n]\right) =Z\left(\delta[n-1]+2\delta[n-2]\right) = Z\left(\delta[n-1]\right)+Z\left(2\delta[n-2]\right) = z^{-1}+2z^{-2}, ROC = C/[0] | ||
</math> | </math> | ||
| + | |||
| + | ===Answer 4=== | ||
| + | <math>X[n] = nu[n] - nu[n-3]</math> | ||
| + | |||
| + | <math> X(z) = \sum_{n=0}^{2}n z^{-n}</math> | ||
| + | <math> = 0 + z^-1 + 2*Z^-2 | ||
| + | ROC z not equal to 1 | ||
---- | ---- | ||
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | ||
Revision as of 10:42, 19 September 2011
Contents
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.)
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
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} $
TA's comments: What about the ROC?
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 except 0.
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 except 0.
Answer 3
First the axiom need to be prove:
$ Z(\delta [n- n_0]) = \sum_{n=-\infty}^{\infty}\delta[n-n_0]z^{-n} = \sum_{n=-\infty}^{\infty}\delta[n-n_0]z^{-n_0} = z^{-n_0}, ROC = C/[0] $
Observe the original function
$ x\left[ n \right]= n u[n]-n u[n-3] = n(u[n] - u[n-3]) = n(\delta[n] + \delta[n-1] + \delta[n-2]) = 0\delta[n] + 1\delta[n-1] + 2\delta[n-2] $
so by two axioms proved above, with the linearity property,
$ X(z) = Z\left(x[n]\right) =Z\left(\delta[n-1]+2\delta[n-2]\right) = Z\left(\delta[n-1]\right)+Z\left(2\delta[n-2]\right) = z^{-1}+2z^{-2}, ROC = C/[0] $
Answer 4
$ X[n] = nu[n] - nu[n-3] $
$ X(z) = \sum_{n=0}^{2}n z^{-n} $ $ = 0 + z^-1 + 2*Z^-2 ROC z not equal to 1 ---- [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] $
