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<math><math>Insert formula here</math></math>[[Category:ECE301]]
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<span class="texhtml"> &lt; ''m''''a''''t''''h'''''<b> &gt; ''I''</b>'''n''''s''''e''''r''''t''''f''''o''''r''''m''''u''''l''''a''''h''''e''''r''''</span>&lt;/math&gt;
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[[Category:ECE438]]
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[[Category:ECE438Fall2013Boutin]]
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[[Category:problem solving]]
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[[Category:z-transform]]
+
  
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= [[:Category:Problem_solving|Practice Problem]] on Z-transform computation =
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= [[:Category:Problem solving|Practice Problem]] on Z-transform computation =
−
Compute the compute the z-transform (including the ROC) of the following DT signal:
+
  
−
<math>x[n]=3^n u[n+3]  \ </math>
+
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===
 
−
alec green
 
  
−
[[Image:Green26_ece438_hmwrk3_power_series.png| 480x320px]]
+
=== Answer 1  ===
 +
 
 +
alec green
 +
 
 +
[[Image:Green26 ece438 hmwrk3 power series.png|480x320px]]  
  
−
<math>X(z) = \sum_{n=-\infty}^{+\infty} x[n]z^{-n}</math>
+
<math>X(z) = \sum_{n=-\infty}^{+\infty} x[n]z^{-n}</math>  
  
−
<math>= \sum_{n=-3}^{+\infty} 3^{n}z^{-n}</math>
+
<math>= \sum_{n=-3}^{+\infty} 3^{n}z^{-n}</math>  
  
−
<math>= \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n}</math>
+
<math>= \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n}</math>  
  
−
Let k = n+3:
+
Let k = n+3:  
  
−
<math>= \sum_{k=0}^{+\infty} (\frac{3}{z})^{k-3}</math>
+
<math>= \sum_{k=0}^{+\infty} (\frac{3}{z})^{k-3}</math>  
  
−
Using the geometric series property:
+
Using the geometric series property:  
  
 
<math>
 
<math>
Line 39: Line 41:
 
     \text{diverges} & \quad \text{else}
 
     \text{diverges} & \quad \text{else}
 
   \end{array} \right.
 
   \end{array} \right.
−
</math>
+
</math>  
  
−
===Answer 2===
+
=== Answer 2 ===
  
−
Muhammad Syafeeq Safaruddin
+
Muhammad Syafeeq Safaruddin  
  
−
<math>x[n] = 3^n u[n+3]</math>
+
<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} 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>  
  
−
<math>X(z) = \sum_{n=-3}^{+\infty} 3^n 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}^{+\infty} (\frac{3}{z})^{n}</math>  
  
−
Let k = n+3, n = k-3
+
Let k = n+3, n = k-3  
  
−
<math>X(z) = \sum_{k=0}^{+\infty} (\frac{3}{z})^{k-3}</math>
+
<math>X(z) = \sum_{k=0}^{+\infty} (\frac{3}{z})^{k-3}</math>  
  
−
<math>X(z) = (\frac{z}{3})^{3} \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math>
+
<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>
+
<math>X(z) = (\frac{z^3}{27}) \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math>  
  
−
By geometric series formula,
+
By geometric series formula,  
  
−
<math>X(z) = (\frac{z^3}{27}) (\frac{1}{1-(\frac{3}{z})}) </math>   ,for |z| < 3
+
<math>X(z) = (\frac{z^3}{27}) (\frac{1}{1-(\frac{3}{z})}) </math> ,for |z| &lt; 3  
  
−
X(z) = diverges, else
+
X(z) = diverges, else  
  
−
So,
+
So,  
  
−
<math>X(z) = (\frac{z}{z-3}) </math> with ROC, |z| < 3
+
<math>X(z) = (\frac{z}{z-3}) </math> with ROC, |z| &lt; 3  
  
 +
<br>
  
 +
<br>
  
 +
=== Answer 3  ===
  
−
===Answer 3===
+
<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>
+
<math>X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n}</math>  
  
−
<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} + \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>X(z) = \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} + (\frac{3}{z})^{-3} + (\frac{3}{z})^{-2} + (\frac{3}{z})^{-1}</math>
+
<math>X(z) = \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} + (\frac{z^{3}}{27}) + (\frac{z^{2}}{9}) + (\frac{z}{3})</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|&lt;1,i.e z&lt;-3 or z&gt;3,
  
−
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>
  
−
<math>(z) = (\frac{1}{1-(\frac{3}{z})}) + (\frac{z^{3}}{27}) + (\frac{z^{2}}{9}) + (\frac{z}{3})</math>
+
if -3&lt;z&lt;3,
  
−
if -3<z<3,
+
X(z) diverges
  
−
X(z) diverges
+
<br>
  
 +
<br>
  
 +
<br>
  
 +
<br>
  
 +
<br>
  
 +
<br>
  
 +
<br>
  
 +
<br>
  
 +
<br>
  
 +
<br>
  
 +
<br>
  
 +
<br>
  
 +
<br>
  
 +
<br>
  
 +
<br>
  
 +
=== 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>
  
−
 
−
 
−
 
−
 
−
 
−
 
−
 
−
 
−
===Answer 4===
 
−
 
−
 
−
<math> x[n] = 3^{n}u[n+3]  </math>
 
−
 
−
<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.
 
  or diverges else.
  
 +
<br>
  
 +
<br>
  
 +
<br>
  
−
 
+
<br>
−
 
+
−
 
+
−
 
+
  
 
----
 
----
  
 +
<br>
  
 +
=== Answer 5  ===
  
−
===Answer 5===
+
Yixiang Liu
  
−
Yixiang Liu
+
<span class="texhtml">''x''[''n''] = 3<sup>''n''</sup>''u''[''n'' + 3]</span>
  
−
<math>x[n] = 3^{n} u[n+3] </math>
+
<math>X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n}</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 178: Line 181:
 
<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
+
using geometric series formula  
  
 
<math>
 
<math>
Line 188: Line 191:
 
     \text{diverges} &, \quad \text{else}
 
     \text{diverges} &, \quad \text{else}
 
   \end{array} \right.
 
   \end{array} \right.
−
</math>  
+
</math>  
  
 
<math>
 
<math>
Line 196: Line 199:
 
     \text{diverges} &, \quad \text{else}
 
     \text{diverges} &, \quad \text{else}
 
   \end{array} \right.
 
   \end{array} \right.
−
</math>  
+
</math>  
  
−
[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013 Prof. Boutin]]
+
[[2013 Fall ECE 438 Boutin|Back to ECE438 Fall 2013 Prof. Boutin]]  
  
 +
<br>
  
−
===Answer 6===
+
=== Answer 6 ===
  
−
Xi Wang
+
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^n u[n+3] z ^{-n}  </math> 
 
−
<math>k = n + 3 </math>
 
 
   <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} u[k] z ^{3-k} </math>
−
  <math>X[z] = \sum_{n = -\infty}^{+\infty} (\frac{3}{z})^{k-3} </math>
+
<math>X[z] = \sum_{n = -\infty}^{+\infty} (\frac{3}{z})^{k-3} </math>
−
if z > 3
+
 
 +
if z &gt; 3  
 +
 
 
   <math>X[z] = (\frac{1}{1-(\frac{3}{z})})</math>
 
   <math>X[z] = (\frac{1}{1-(\frac{3}{z})})</math>
−
if z < 3
 
−
  <math>Diverges</math>
 
  
−
===Answer 7===
+
if z &lt; 3
 +
 
 +
  <span class="texhtml">''D''''i''''v''''e''''r''''g''''e''''s'''''</span>
 +
 
 +
=== Answer 7 ===
  
−
<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, thus n = k - 3
+
Let k = n + 3, thus n = k - 3  
  
−
With that we obtain,
+
With that we obtain,  
  
−
<math>X[z] = \sum_{n = -\infty}^{+\infty} 3^{k-3} u[k] z ^{-k+3}</math>
+
<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) = (\frac{z}{3})^{3}\sum_{n=-\infty}^{+\infty} (\frac{3}{z})^{k}</math>  
  
 
<math>
 
<math>
Line 232: Line 240:
 
     \text{diverges} &, \quad \text{else}
 
     \text{diverges} &, \quad \text{else}
 
   \end{array} \right.
 
   \end{array} \right.
−
</math>
+
</math>  
  
−
Thus, ROC is |z| > 3 because it is restricted by geometric series.
+
Thus, ROC is |z| &gt; 3 because it is restricted by geometric series.  
  
 +
<br>
  
−
=== Answer 8 ===
+
=== Answer 8 ===
  
−
Cary Wood
+
Cary Wood  
  
−
<math>x[n] = 3^n u[n+3]</math>
+
<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} 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>  
  
−
<math>X(z) = 3^{n} z^{-n} for all n > -3 </math>
+
<span class="texhtml">''X''(''z'') = 3<sup>''n''</sup>''z''<sup> − ''n''</sup>''f''''o''''r''''a''''l''''l''''n'' &gt;  − 3</span>  
  
 
and  
 
and  
  
−
<math>X(z) = 0, else </math>  
+
<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| &lt; 1
 +
 
 +
X(z) = diverges, elsewhere
 +
 
 +
ROC, |z| &gt; 3
 +
 
 +
<br>  
  
−
Thus, we re-write X(z) as...
+
=== Answer 9  ===
  
−
<math>= \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n}</math>
+
Shiyu Wang
  
−
By the geometric series formula,
+
<math>X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n}= \sum_{n=-3}^{+\infty} (3/z)^{n}</math>
  
−
<math>X(z) = (\frac{1}{1-(\frac{3}{z})}) </math>  , for |3/z| < 1
+
when&nbsp; |3/z| &lt; 1,&nbsp;|z| &gt; 3
  
−
X(z) = diverges, elsewhere
+
<math>X(z) = (\frac{z^3}{27}) (\frac{1}{1-(\frac{3}{z})}) = (\frac{z^3}{27}) (\frac{z}{z-3}) </math>&nbsp;, for |z|&gt;3; else, diverges.
  
−
ROC, |z| > 3
+
[[Category:ECE301]] [[Category:ECE438]] [[Category:ECE438Fall2013Boutin]] [[Category:Problem_solving]] [[Category:Z-transform]]

Revision as of 22:38, 12 September 2013

< m'a't'h > In's'e'r't'f'o'r'm'u'l'a'h'e'r'</math>

Practice Problem on Z-transform computation

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.)

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! No need to write your name: we can find out who wrote what by checking the history of the page.

Answer 1

alec green

Green26 ece438 hmwrk3 power series.png

$ 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. $

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



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
















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.






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. $

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

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.


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


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.

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

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