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<math><math>Insert formula here</math></math>[[Category:ECE301]]
 
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[[Category:ECE438]]
 
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[[Category:ECE438Fall2013Boutin]]
 
 
[[Category:problem solving]]
 
[[Category:problem solving]]
−
[[Category:z-transform]]
 
  
−
= [[:Category:Problem_solving|Practice Problem]] on Z-transform computation =
+
<center><font size= 4>
−
Compute the compute the z-transform (including the ROC) of the following DT signal:
+
'''[[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===
 
−
alec green
 
  
−
[[Image:Green26_ece438_hmwrk3_power_series.png| 480x320px]]
+
=== Answer 1  ===
 +
 
 +
[[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 47:
 
     \text{diverges} & \quad \text{else}
 
     \text{diverges} & \quad \text{else}
 
   \end{array} \right.
 
   \end{array} \right.
−
</math>
+
</math>  
  
−
===Answer 2===
+
:<span style="color:red"> TA's comment: Simple and clear derivation! </span>
  
−
Muhammad Syafeeq Safaruddin
+
=== Answer 2  ===
  
−
<math>x[n] = 3^n u[n+3]</math>
+
Muhammad Syafeeq Safaruddin
  
−
<math>X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n}</math>
+
<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=-\infty}^{+\infty} x[n] z^{-n}</math>  
  
−
<math>X(z) = \sum_{n=-3}^{+\infty} 3^n 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} (\frac{3}{z})^{n}</math>
+
<math>X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n}</math>  
  
−
Let k = n+3, n = k-3
+
<math>X(z) = \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n}</math>
  
−
<math>X(z) = \sum_{k=0}^{+\infty} (\frac{3}{z})^{k-3}</math>
+
Let k = n+3, n = k-3  
  
−
<math>X(z) = (\frac{z}{3})^{3} \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math>
+
<math>X(z) = \sum_{k=0}^{+\infty} (\frac{3}{z})^{k-3}</math>  
  
−
<math>X(z) = (\frac{z^3}{27}) \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>  
  
−
By geometric series formula,
+
<math>X(z) = (\frac{z^3}{27}) \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math>
  
−
<math>X(z) = (\frac{z^3}{27}) (\frac{1}{1-(\frac{3}{z})}) </math>  ,for |z| < 3
+
By geometric series formula,  
  
−
X(z) = diverges, else
+
<math>X(z) = (\frac{z^3}{27}) (\frac{1}{1-(\frac{3}{z})}) </math> ,for |z| &lt; 3
  
−
So,
+
X(z) = diverges, else
  
−
<math>X(z) = (\frac{z}{z-3}) </math> with ROC, |z| < 3
+
So,  
  
−
===Answer 3===
+
<math>X(z) = (\frac{z}{z-3}) </math> with ROC, |z| &lt; 3  
  
−
<math>X(z) = \sum_{n=-\infty}^{+\infty} 3^n u(n+3) z^{-n}</math>
+
<br>  
  
−
<math>X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n}</math>
+
<br>  
  
−
<math>x(z) = (\frac{z^{3}}{27}) + (\frac{z^{2}}{9}) + (\frac{z}{3})</math>
+
:<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>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=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{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,
  
 +
<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,
  
 +
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>
  
−
===Answer 4===
+
<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[n] = 3^{n}u[n+3]  </math>
+
Let k = n + 3  
  
−
<math> X[Z] = \sum_{n=-\infty}^{+\infty} 3^{n}u[n+3] Z^{-n}   </math>
+
Now <math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{k-3} u[k] z^{3-k}</math>  
  
−
<math> X[Z] = \sum_{n=-3}^{+\infty} 3^{n}Z^{-n}   </math>
+
<math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{k=3} z^{3-k}</math>  
  
−
<math> X[Z] = \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n}   </math>
+
<math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{k} 3^{-3} z^{-k} z^{3}</math>  
  
−
<math> X[Z] = \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n} + \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} </math>
+
<math>X(z) = \sum_{n=-\infty}^{+\infty} (\frac{z}{3})^{3} (\frac{3}{z})^{k}</math>  
  
−
<math> for \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n}, no effect, because this converges everywhere on plane. </math>
+
<math>X(z) = (\frac{z}{3})^{3}\sum_{n=-\infty}^{+\infty} (\frac{3}{z})^{k}</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>
+
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 &gt; 3
  
 +
  <math>X[z] = (\frac{1}{1-(\frac{3}{z})})</math>
  
 +
if z &lt; 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  ===
  
−
===Answer 5===
+
<math>X[z] = \sum_{n = -\infty}^{+\infty} 3^n u[n+3] z ^{-n}</math>
  
−
Yixiang Liu
+
Let k = n + 3, thus n = k - 3
  
−
<math>x[n] = 3^{n} u[n+3] </math>
+
With that we obtain,
  
−
<math>X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n}</math>
+
<math>X[z] = \sum_{n = -\infty}^{+\infty} 3^{k-3} u[k] z ^{-k+3}</math>  
−
         
+
−
<math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{n} u[n+3] z^{-n}</math>
+
  
−
Let k = n + 3
+
<math>X(z) = (\frac{z}{3})^{3}\sum_{n=-\infty}^{+\infty} (\frac{3}{z})^{k}</math>
  
−
Now <math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{k-3} u[k] z^{3-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>  
  
−
<math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{k=3} z^{3-k}</math> 
+
Thus, ROC is |z| &gt; 3 because it is restricted by geometric series.
  
−
<math>X(z) = \sum_{n=-\infty}^{+\infty} 3^{k} 3^{-3} z^{-k} z^{3}</math>  
+
:<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>
  
−
<math>X(z) = \sum_{n=-\infty}^{+\infty} (\frac{z}{3})^{3} (\frac{3}{z})^{k}</math>  
+
<br>  
  
−
<math>X(z) = (\frac{z}{3})^{3}\sum_{n=-\infty}^{+\infty} (\frac{3}{z})^{k}</math>
+
=== Answer 8  ===
  
−
using geometric series formula
+
Cary Wood
  
−
<math>
+
<span class="texhtml">''x''[''n''] = 3<sup>''n''</sup>''u''[''n'' + 3]</span>  
−
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>
+
<math>X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n}</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>  
+
  
−
[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013 Prof. Boutin]]
+
<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'' &gt;  − 3</span>
  
−
===Answer 6===
+
and
  
−
Xi Wang
+
<span class="texhtml">''X''(''z'') = 0,''e''''l''''s''''e'''''</span>
  
−
<math>X[z] = \sum_{n = -\infty}^{+\infty} 3^n u[n+3] z ^{-n} </math>
+
Thus, we re-write X(z) as...
−
<math>k = n + 3 </math>
+
 
−
  <math>X[z] = \sum_{n = -\infty}^{+\infty} 3^{k-3} u[k] z ^{3-k} </math>
+
<math>= \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n}</math>  
−
  <math>X[z] = \sum_{n = -\infty}^{+\infty} (\frac{3}{z})^{k-3} </math>
+
 
−
if z > 3
+
By the geometric series formula,
−
  <math>x[z] = (\frac{1}{1-(\frac{3}{z})})</math>
+
 
 +
<math>X(z) = (\frac{1}{1-(\frac{3}{z})}) </math> , for |3/z| &lt; 1
 +
 
 +
X(z) = diverges, elsewhere
 +
 
 +
ROC, |z| &gt; 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&nbsp; |3/z| &lt; 1,&nbsp;|z| &gt; 3  
 +
 
 +
<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.
 +
 
 +
:<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

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.)

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

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. $

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!
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