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[[Category:problem solving]] | [[Category:problem solving]] | ||
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| − | = [[ | + | <center><font size= 4> |
| − | + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | |
| + | </font size> | ||
| − | <math>x[n]=3^n u[n+3] \ </math> | + | Topic: Computing a z-transform |
| + | |||
| + | </center> | ||
| + | ---- | ||
| + | ==Question== | ||
| + | |||
| + | Compute the compute the z-transform (including the ROC) of the following DT signal: | ||
| + | |||
| + | <math>x[n]=3^n u[n+3] \ </math> | ||
(Write enough intermediate steps to fully justify your answer.) | (Write enough intermediate steps to fully justify your answer.) | ||
| + | |||
---- | ---- | ||
| − | ==Share your answers below== | + | |
| − | 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! | + | == Share your answers below == |
| − | '''No need to write your name: we can find out who wrote what by checking the history of the page.''' | + | |
| + | 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! '''No need to write your name: we can find out who wrote what by checking the history of the page.''' | ||
| + | |||
---- | ---- | ||
| − | |||
| − | |||
| − | + | === Answer 1 === | |
| − | + | [[Image:Green26 ece438 hmwrk3 power series.png|480x320px]] | |
| − | <math>= \sum_{n=- | + | <math>X(z) = \sum_{n=-\infty}^{+\infty} x[n]z^{-n}</math> |
| − | <math>= \sum_{n=-3}^{+\infty} | + | <math>= \sum_{n=-3}^{+\infty} 3^{n}z^{-n}</math> |
| − | + | <math>= \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n}</math> | |
| − | + | Let k = n+3: | |
| − | Using the geometric series property: | + | <math>= \sum_{k=0}^{+\infty} (\frac{3}{z})^{k-3}</math> |
| + | |||
| + | Using the geometric series property: | ||
<math> | <math> | ||
| Line 39: | Line 47: | ||
\text{diverges} & \quad \text{else} | \text{diverges} & \quad \text{else} | ||
\end{array} \right. | \end{array} \right. | ||
| − | </math> | + | </math> |
| − | = | + | :<span style="color:red"> TA's comment: Simple and clear derivation! </span> |
| − | + | === Answer 2 === | |
| − | + | Muhammad Syafeeq Safaruddin | |
| − | < | + | <span class="texhtml">''x''[''n''] = 3<sup>''n''</sup>''u''[''n'' + 3]</span> |
| − | <math>X(z) = \sum_{n=-\infty}^{+\infty} | + | <math>X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n}</math> |
| − | <math>X(z) = \sum_{n=- | + | <math>X(z) = \sum_{n=-\infty}^{+\infty} 3^n u[n+3] z^{-n}</math> |
| − | <math>X(z) = \sum_{n=-3}^{+\infty} | + | <math>X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n}</math> |
| − | + | <math>X(z) = \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n}</math> | |
| − | + | Let k = n+3, n = k-3 | |
| − | <math>X(z) = | + | <math>X(z) = \sum_{k=0}^{+\infty} (\frac{3}{z})^{k-3}</math> |
| − | <math>X(z) = (\frac{z | + | <math>X(z) = (\frac{z}{3})^{3} \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math> |
| − | <math>X(z) = (\frac{z^3}{27}) \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math> | + | <math>X(z) = (\frac{z^3}{27}) \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math> |
| − | + | <math>X(z) = (\frac{z^3}{27}) \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math> | |
| − | + | By geometric series formula, | |
| − | X(z) = | + | <math>X(z) = (\frac{z^3}{27}) (\frac{1}{1-(\frac{3}{z})}) </math> ,for |z| < 3 |
| − | + | X(z) = diverges, else | |
| − | + | So, | |
| − | = | + | <math>X(z) = (\frac{z}{z-3}) </math> with ROC, |z| < 3 |
| − | < | + | <br> |
| − | < | + | <br> |
| − | < | + | :<span style="color:red"> TA's comment: You could just use geometric series to calculate the summation in step 4, and dont'need to change variable. ROC means region of CONVERGENCE. So ROC should be|z|>3.</span> |
| − | + | === Answer 3 === | |
| − | <math>= \sum_{n= | + | <math>X(z) = \sum_{n=-\infty}^{+\infty} 3^n u(n+3) z^{-n}</math> |
| − | <math>= \sum_{n= | + | <math>X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n}</math> |
| − | === | + | <math>X(z) = \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} + \sum_{n=-3}^{-1} (\frac{3}{z})^{n}</math> |
| + | <math>X(z) = \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} + (\frac{3}{z})^{-3} + (\frac{3}{z})^{-2} + (\frac{3}{z})^{-1}</math> | ||
| − | <math> | + | <math>X(z) = \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} + (\frac{z^{3}}{27}) + (\frac{z^{2}}{9}) + (\frac{z}{3})</math> |
| − | + | if |3/z|<1,i.e z<-3 or z>3, | |
| + | <math>(z) = (\frac{1}{1-(\frac{3}{z})}) + (\frac{z^{3}}{27}) + (\frac{z^{2}}{9}) + (\frac{z}{3})</math> | ||
| + | if -3<z<3, | ||
| + | X(z) diverges | ||
| + | <br> | ||
| + | :<span style="color:red"> TA's comment: You can get a closed form expression by applying the geometric series property to the whole equation rather than split them into to parts.</span> | ||
| + | === Answer 4 === | ||
| + | <span class="texhtml">''x''[''n''] = 3<sup>''n''</sup>''u''[''n'' + 3]</span> | ||
| + | <math> X[Z] = \sum_{n=-\infty}^{+\infty} 3^{n}u[n+3] Z^{-n} </math> | ||
| + | |||
| + | <math> X[Z] = \sum_{n=-3}^{+\infty} 3^{n}Z^{-n} </math> | ||
| + | |||
| + | <math> X[Z] = \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n} </math> | ||
| + | |||
| + | <math> X[Z] = \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n} + \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} </math> | ||
| + | |||
| + | <math> for \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n}, no effect, because this converges everywhere on plane. </math> | ||
| + | |||
| + | <math> for \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} = \frac{1}{1-\frac{3}{z}}, if |\frac{3}{z}|<1, |z|>3 </math> | ||
| + | |||
| + | or diverges else. | ||
| + | |||
| + | <br> | ||
| + | |||
| + | :<span style="color:red"> TA's comment: The ROC of the Z transform is the intersection of each part's ROC, which is |z|>3. You should state that explicitly. </span> | ||
---- | ---- | ||
| + | <br> | ||
| + | === Answer 5 === | ||
| − | + | Yixiang Liu | |
| − | + | <span class="texhtml">''x''[''n''] = 3<sup>''n''</sup>''u''[''n'' + 3]</span> | |
| − | <math> | + | <math>X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n}</math> |
| − | + | <math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{n} u[n+3] z^{-n}</math> | |
| − | + | ||
| − | <math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{n} u[n+3] z^{-n}</math> | + | |
| − | Let k = n + 3 | + | Let k = n + 3 |
| − | Now <math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{k-3} u[k] z^{3-k}</math> | + | Now <math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{k-3} u[k] z^{3-k}</math> |
| − | <math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{k=3} z^{3-k}</math> | + | <math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{k=3} z^{3-k}</math> |
<math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{k} 3^{-3} z^{-k} z^{3}</math> | <math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{k} 3^{-3} z^{-k} z^{3}</math> | ||
| Line 130: | Line 163: | ||
<math>X(z) = \sum_{n=-\infty}^{+\infty} (\frac{z}{3})^{3} (\frac{3}{z})^{k}</math> | <math>X(z) = \sum_{n=-\infty}^{+\infty} (\frac{z}{3})^{3} (\frac{3}{z})^{k}</math> | ||
| − | <math>X(z) = (\frac{z}{3})^{3}\sum_{n=-\infty}^{+\infty} (\frac{3}{z})^{k}</math> | + | <math>X(z) = (\frac{z}{3})^{3}\sum_{n=-\infty}^{+\infty} (\frac{3}{z})^{k}</math> |
| + | |||
| + | using geometric series formula | ||
| + | |||
| + | <math> | ||
| + | X(z) = \left\{ | ||
| + | \begin{array}{l l} | ||
| + | (\frac{z}{3})^3 \frac{1}{1-\frac{3}{z}} &, if \quad |\frac{3}{z}| < 1\\ | ||
| + | \text{diverges} &, \quad \text{else} | ||
| + | \end{array} \right. | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | X(z) = \left\{ | ||
| + | \begin{array}{l l} | ||
| + | (\frac{z}{3})^3 \frac{1}{1-\frac{3}{z}} &, if \quad |z| > 3\\ | ||
| + | \text{diverges} &, \quad \text{else} | ||
| + | \end{array} \right. | ||
| + | </math> | ||
| + | |||
| + | |||
| + | :<span style="color:red"> TA's comment: Once you drop the unit step function u[k] in step 4, the summation should be from 0 to infinity.</span> | ||
| + | |||
| + | [[2013 Fall ECE 438 Boutin|Back to ECE438 Fall 2013 Prof. Boutin]] | ||
| + | |||
| + | <br> | ||
| + | |||
| + | === Answer 6 === | ||
| + | |||
| + | Xi Wang | ||
| + | |||
| + | <math>X[z] = \sum_{n = -\infty}^{+\infty} 3^n u[n+3] z ^{-n} </math> <span class="texhtml">''k'' = ''n'' + 3</span> | ||
| + | |||
| + | <math>X[z] = \sum_{n = -\infty}^{+\infty} 3^{k-3} u[k] z ^{3-k} </math> | ||
| + | <math>X[z] = \sum_{n = -\infty}^{+\infty} (\frac{3}{z})^{k-3} </math> | ||
| + | |||
| + | if z > 3 | ||
| + | |||
| + | <math>X[z] = (\frac{1}{1-(\frac{3}{z})})</math> | ||
| + | |||
| + | if z < 3 | ||
| + | |||
| + | <span class="texhtml">''D''''i''''v''''e''''r''''g''''e''''s'''''</span> | ||
| + | |||
| + | |||
| + | :<span style="color:red"> TA's comment: Once you drop the unit step function u[k], the summation should be from 0 to infinity. The last step is wrong. Where is the (3/z)^3? .</span> | ||
| + | |||
| + | === Answer 7 === | ||
| + | |||
| + | <math>X[z] = \sum_{n = -\infty}^{+\infty} 3^n u[n+3] z ^{-n}</math> | ||
| + | |||
| + | Let k = n + 3, thus n = k - 3 | ||
| + | |||
| + | With that we obtain, | ||
| + | |||
| + | <math>X[z] = \sum_{n = -\infty}^{+\infty} 3^{k-3} u[k] z ^{-k+3}</math> | ||
| + | |||
| + | <math>X(z) = (\frac{z}{3})^{3}\sum_{n=-\infty}^{+\infty} (\frac{3}{z})^{k}</math> | ||
| + | |||
| + | <math> | ||
| + | X(z) = \left\{ | ||
| + | \begin{array}{l l} | ||
| + | (\frac{z}{3})^3 \frac{1}{1-\frac{3}{z}} &, if \quad |z| > 3\\ | ||
| + | \text{diverges} &, \quad \text{else} | ||
| + | \end{array} \right. | ||
| + | </math> | ||
| + | |||
| + | Thus, ROC is |z| > 3 because it is restricted by geometric series. | ||
| + | |||
| + | :<span style="color:red"> TA's comment: Once you drop the unit step function u[k], the summation should be from 0 to infinity.</span> | ||
| + | |||
| + | <br> | ||
| + | |||
| + | === Answer 8 === | ||
| + | |||
| + | Cary Wood | ||
| + | |||
| + | <span class="texhtml">''x''[''n''] = 3<sup>''n''</sup>''u''[''n'' + 3]</span> | ||
| + | |||
| + | <math>X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n}</math> | ||
| + | |||
| + | <math>X(z) = \sum_{n=-\infty}^{+\infty} 3^n u[n+3] z^{-n}</math> | ||
| + | |||
| + | <span class="texhtml">''X''(''z'') = 3<sup>''n''</sup>''z''<sup> − ''n''</sup>''f''''o''''r''''a''''l''''l''''n'' > − 3</span> | ||
| + | |||
| + | and | ||
| + | |||
| + | <span class="texhtml">''X''(''z'') = 0,''e''''l''''s''''e'''''</span> | ||
| + | |||
| + | Thus, we re-write X(z) as... | ||
| + | |||
| + | <math>= \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n}</math> | ||
| + | |||
| + | By the geometric series formula, | ||
| + | |||
| + | <math>X(z) = (\frac{1}{1-(\frac{3}{z})}) </math> , for |3/z| < 1 | ||
| + | |||
| + | X(z) = diverges, elsewhere | ||
| + | |||
| + | ROC, |z| > 3 | ||
| + | |||
| + | <br> | ||
| + | |||
| + | :<span style="color:red"> TA's comment: In step 3: X(z) is not a function of n, so it's not proper say it that way. You probably apply the geometric series in a wrong way. If the summation doesn't start from 0, you should multiply the sum by the first term (3/z)^3? .</span> | ||
| + | |||
| + | === Answer 9 === | ||
| + | |||
| + | Shiyu Wang | ||
| + | |||
| + | <math>X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n}= \sum_{n=-3}^{+\infty} (3/z)^{n}</math> | ||
| − | + | when |3/z| < 1, |z| > 3 | |
| − | + | <math>X(z) = (\frac{z^3}{27}) (\frac{1}{1-(\frac{3}{z})}) = (\frac{z^3}{27}) (\frac{z}{z-3}) </math> , for |z|>3; else, diverges. | |
| − | [[ | + | :<span style="color:red"> TA's comment: Simple and clear derivation! </span> |
| + | [[Category:ECE301]] [[Category:ECE438]] [[Category:ECE438Fall2013Boutin]] [[Category:Problem_solving]] [[Category:Z-transform]] | ||
Latest revision as of 14:19, 1 May 2016
Practice Question on "Digital Signal Processing"
Topic: Computing a z-transform
Contents
Question
Compute the compute the z-transform (including the ROC) of the following DT signal:
$ x[n]=3^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! No need to write your name: we can find out who wrote what by checking the history of the page.
Answer 1
$ X(z) = \sum_{n=-\infty}^{+\infty} x[n]z^{-n} $
$ = \sum_{n=-3}^{+\infty} 3^{n}z^{-n} $
$ = \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n} $
Let k = n+3:
$ = \sum_{k=0}^{+\infty} (\frac{3}{z})^{k-3} $
Using the geometric series property:
$ X(z) = \left\{ \begin{array}{l l} (\frac{z}{3})^3 \frac{1}{1-\frac{3}{z}} & \quad |z| > 3\\ \text{diverges} & \quad \text{else} \end{array} \right. $
- TA's comment: Simple and clear derivation!
Answer 2
Muhammad Syafeeq Safaruddin
x[n] = 3nu[n + 3]
$ X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} $
$ X(z) = \sum_{n=-\infty}^{+\infty} 3^n u[n+3] z^{-n} $
$ X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n} $
$ X(z) = \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n} $
Let k = n+3, n = k-3
$ X(z) = \sum_{k=0}^{+\infty} (\frac{3}{z})^{k-3} $
$ X(z) = (\frac{z}{3})^{3} \sum_{k=0}^{+\infty} (\frac{3}{z})^{k} $
$ X(z) = (\frac{z^3}{27}) \sum_{k=0}^{+\infty} (\frac{3}{z})^{k} $
$ X(z) = (\frac{z^3}{27}) \sum_{k=0}^{+\infty} (\frac{3}{z})^{k} $
By geometric series formula,
$ X(z) = (\frac{z^3}{27}) (\frac{1}{1-(\frac{3}{z})}) $ ,for |z| < 3
X(z) = diverges, else
So,
$ X(z) = (\frac{z}{z-3}) $ with ROC, |z| < 3
- TA's comment: You could just use geometric series to calculate the summation in step 4, and dont'need to change variable. ROC means region of CONVERGENCE. So ROC should be|z|>3.
Answer 3
$ X(z) = \sum_{n=-\infty}^{+\infty} 3^n u(n+3) z^{-n} $
$ X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n} $
$ X(z) = \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} + \sum_{n=-3}^{-1} (\frac{3}{z})^{n} $
$ X(z) = \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} + (\frac{3}{z})^{-3} + (\frac{3}{z})^{-2} + (\frac{3}{z})^{-1} $
$ X(z) = \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} + (\frac{z^{3}}{27}) + (\frac{z^{2}}{9}) + (\frac{z}{3}) $
if |3/z|<1,i.e z<-3 or z>3,
$ (z) = (\frac{1}{1-(\frac{3}{z})}) + (\frac{z^{3}}{27}) + (\frac{z^{2}}{9}) + (\frac{z}{3}) $
if -3<z<3,
X(z) diverges
- TA's comment: You can get a closed form expression by applying the geometric series property to the whole equation rather than split them into to parts.
Answer 4
x[n] = 3nu[n + 3]
$ X[Z] = \sum_{n=-\infty}^{+\infty} 3^{n}u[n+3] Z^{-n} $
$ X[Z] = \sum_{n=-3}^{+\infty} 3^{n}Z^{-n} $
$ X[Z] = \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n} $
$ X[Z] = \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n} + \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} $
$ for \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n}, no effect, because this converges everywhere on plane. $
$ for \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} = \frac{1}{1-\frac{3}{z}}, if |\frac{3}{z}|<1, |z|>3 $
or diverges else.
- TA's comment: The ROC of the Z transform is the intersection of each part's ROC, which is |z|>3. You should state that explicitly.
Answer 5
Yixiang Liu
x[n] = 3nu[n + 3]
$ X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} $
$ X(z) = \sum_{n=-\infty}^{+\infty} 3^{n} u[n+3] z^{-n} $
Let k = n + 3
Now $ X(z) = \sum_{n=-\infty}^{+\infty} 3^{k-3} u[k] z^{3-k} $
$ X(z) = \sum_{n=-\infty}^{+\infty} 3^{k=3} z^{3-k} $
$ X(z) = \sum_{n=-\infty}^{+\infty} 3^{k} 3^{-3} z^{-k} z^{3} $
$ X(z) = \sum_{n=-\infty}^{+\infty} (\frac{z}{3})^{3} (\frac{3}{z})^{k} $
$ X(z) = (\frac{z}{3})^{3}\sum_{n=-\infty}^{+\infty} (\frac{3}{z})^{k} $
using geometric series formula
$ X(z) = \left\{ \begin{array}{l l} (\frac{z}{3})^3 \frac{1}{1-\frac{3}{z}} &, if \quad |\frac{3}{z}| < 1\\ \text{diverges} &, \quad \text{else} \end{array} \right. $
$ X(z) = \left\{ \begin{array}{l l} (\frac{z}{3})^3 \frac{1}{1-\frac{3}{z}} &, if \quad |z| > 3\\ \text{diverges} &, \quad \text{else} \end{array} \right. $
- TA's comment: Once you drop the unit step function u[k] in step 4, the summation should be from 0 to infinity.
Back to ECE438 Fall 2013 Prof. Boutin
Answer 6
Xi Wang
$ X[z] = \sum_{n = -\infty}^{+\infty} 3^n u[n+3] z ^{-n} $ k = n + 3
$ X[z] = \sum_{n = -\infty}^{+\infty} 3^{k-3} u[k] z ^{3-k} $ $ X[z] = \sum_{n = -\infty}^{+\infty} (\frac{3}{z})^{k-3} $
if z > 3
$ X[z] = (\frac{1}{1-(\frac{3}{z})}) $
if z < 3
D'i'v'e'r'g'e's
- TA's comment: Once you drop the unit step function u[k], the summation should be from 0 to infinity. The last step is wrong. Where is the (3/z)^3? .
Answer 7
$ X[z] = \sum_{n = -\infty}^{+\infty} 3^n u[n+3] z ^{-n} $
Let k = n + 3, thus n = k - 3
With that we obtain,
$ X[z] = \sum_{n = -\infty}^{+\infty} 3^{k-3} u[k] z ^{-k+3} $
$ X(z) = (\frac{z}{3})^{3}\sum_{n=-\infty}^{+\infty} (\frac{3}{z})^{k} $
$ X(z) = \left\{ \begin{array}{l l} (\frac{z}{3})^3 \frac{1}{1-\frac{3}{z}} &, if \quad |z| > 3\\ \text{diverges} &, \quad \text{else} \end{array} \right. $
Thus, ROC is |z| > 3 because it is restricted by geometric series.
- TA's comment: Once you drop the unit step function u[k], the summation should be from 0 to infinity.
Answer 8
Cary Wood
x[n] = 3nu[n + 3]
$ X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} $
$ X(z) = \sum_{n=-\infty}^{+\infty} 3^n u[n+3] z^{-n} $
X(z) = 3nz − nf'o'r'a'l'l'n > − 3
and
X(z) = 0,e'l's'e
Thus, we re-write X(z) as...
$ = \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n} $
By the geometric series formula,
$ X(z) = (\frac{1}{1-(\frac{3}{z})}) $ , for |3/z| < 1
X(z) = diverges, elsewhere
ROC, |z| > 3
- TA's comment: In step 3: X(z) is not a function of n, so it's not proper say it that way. You probably apply the geometric series in a wrong way. If the summation doesn't start from 0, you should multiply the sum by the first term (3/z)^3? .
Answer 9
Shiyu Wang
$ X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n}= \sum_{n=-3}^{+\infty} (3/z)^{n} $
when |3/z| < 1, |z| > 3
$ X(z) = (\frac{z^3}{27}) (\frac{1}{1-(\frac{3}{z})}) = (\frac{z^3}{27}) (\frac{z}{z-3}) $ , for |z|>3; else, diverges.
- TA's comment: Simple and clear derivation!
