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} $

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

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