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[[Category:ECE301]]
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<center><font size= 4>
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[[Category:ECE438]]
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'''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]'''
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[[Category:ECE438Fall2013Boutin]]
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</font size>
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[[Category:problem solving]]
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[[Category:z-transform]]
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[[Category:inverse z-transform]]
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 +
Topic: Computing an inverse z-transform
  
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= [[:Category:Problem_solving|Practice Question]], [[ECE438]] Fall 2013, [[User:Mboutin|Prof. Boutin]] =
+
</center>
−
On computing the inverse z-transform of a discrete-time signal.
+
 
----
 
----
 +
==Question==
 +
 
Compute the inverse z-transform of  
 
Compute the inverse z-transform of  
  
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(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 ==
 +
 
 +
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===
+
 
 +
=== Answer 1 ===
  
 
<math>X[z] = \sum_{n=0}^{+\infty} 3^{-1-n} z^{n}</math>  
 
<math>X[z] = \sum_{n=0}^{+\infty} 3^{-1-n} z^{n}</math>  
 +
 
  <math>      = \sum_{n=-\infty}^{+\infty} u[n] 3^{-1-n} z^{n}</math>
 
  <math>      = \sum_{n=-\infty}^{+\infty} u[n] 3^{-1-n} z^{n}</math>
  
 
  NOTE: Let n=-k
 
  NOTE: Let n=-k
  
−
<math>    = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k}</math>  
+
<math>    = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k}</math> ('''compare with''' <math>\sum_{n=-\infty}^{+\infty} x[n] z^{-k}</math>)  
−
('''compare with''' <math>\sum_{n=-\infty}^{+\infty} x[n] z^{-k}</math>)  
+
  
 
<math>    = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k}</math>  
 
<math>    = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k}</math>  
  
−
Therefore, <math> x[n]= 3^{-1+n} u[-n] </math>
+
Therefore, <span class="texhtml">''x''[''n''] = 3<sup> − 1 + ''n''</sup>''u''[ − ''n'']</span>  
  
−
=== Answer 2===
+
:<span style="color:blue"> Grader's comment: Correct Answer </span>
−
<math>X(z) = \frac{1}{3-z} = \frac{1}{3} \frac{1}{1-\frac{z}{3}} </math>
+
  
−
<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{z}{3})^n </math>
 
  
−
Let n = -k
+
=== Answer 2  ===
  
−
<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[-k]3^{k}z^{-k} </math>
+
<math>X(z) = \frac{1}{3-z} = \frac{1}{3} \frac{1}{1-\frac{z}{3}} </math>  
  
−
By comparison with the x-transform formula,
+
<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{z}{3})^n </math>
  
−
<math>x[n] = 3^{n-1} u[-n] </math>
+
Let n = -k
−
===Answer 3===
+
−
Write it here.
+
−
===Answer 4===
+
−
Write it here.
+
  
 +
<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[-k]3^{k}z^{-k} </math>
  
 +
By comparison with the x-transform formula,
  
 +
<span class="texhtml">''x''[''n''] = 3<sup>''n'' − 1</sup>''u''[ − ''n'']</span>
  
−
===Answer 5===
+
:<span style="color:blue"> Grader's comment: Correct Answer </span>
−
<math>X(z) = \frac{1}{3-Z}</math>
+
 
−
<math>X(z) = \((frac{1}{3})-(1-(frac{Z}{3})))</math>
+
 
 +
=== Answer 3  ===
 +
 
 +
By Yeong Ho Lee
 +
 
 +
<math> X[z] = \sum_{n=0}^{\infty}z^{n}3^{-n-1} </math>
 +
 
 +
<math>      = \sum_{n=-\infty}^{+\infty} 3^{-n-1}z^{n}u[n] </math>
 +
 
 +
Now, let n = -k
 +
 
 +
<math>      = \sum_{n=-\infty}^{+\infty} 3^{k-1}z^{-k}u[-k] </math>
 +
 
 +
Using the z-transform formula, <span class="texhtml">''x''[''n''] = 3<sup>''n'' − 1</sup>''u''[ − ''n'']</span>
 +
 
 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
 +
 
 +
=== Answer 4  ===
 +
 
 +
Gena Xie
 +
 
 +
<math>X(z) = \frac{1}{3-Z} </math>  
 +
 
 +
since |z|&lt;3,
 +
 
 +
|z|/3 &lt; 1
 +
 
 +
<math>X(z) = \frac{1}{3} \frac{1} { 1-\frac{Z}{3}} </math>
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{n=0}^{+\infty} (\frac{z}{3})^n = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n](\frac{z}{3})^n </math>
 +
 
 +
substitute n by -n,
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[-n](\frac{z}{3})^{-n}  = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[-n](\frac{1}{3})^{-n} Z^{-n}</math>
 +
 
 +
based on the definition
 +
 
 +
<math>x[n] = \frac{1}{3} u[-n](\frac{1}{3})^{-n} = (\frac{1}{3})^{-n+1} u[-n] </math>
 +
 
 +
<br>
 +
 
 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
 +
 
 +
=== Answer 5  ===
 +
 
 +
By Yixiang Liu
 +
 
 +
<math>X(z) = \frac{1}{3-Z}</math>
 +
 
 +
<br> <math>X(z) = \frac{\frac{1}{3}} { 1-\frac{Z}{3}} </math>
 +
 
 +
<math>X(z) = \frac{1}{3} * \frac{1} { 1-\frac{Z}{3}} </math>
 +
 
 +
by geometric series
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{n=0}^{+\infty} (\frac{z}{3})^n </math>
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n](\frac{z}{3})^n </math>
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{k=-\infty}^{+\infty} u[-k](\frac{z}{3})^{-k} </math>
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{k=-\infty}^{+\infty} u[-k](\frac{1}{3})^{-k} Z^{-k}</math>
 +
 
 +
By comparison with the x-transform formula
 +
 
 +
<math>x[n] = \frac{1}{3} u[-n](\frac{1}{3})^{-n} </math>
 +
 
 +
<math>x[n] = (\frac{1}{3})^{-n+1} u[-n] </math>
 +
 
 +
<span class="texhtml">''x''[''n''] = 3<sup>''n'' − 1</sup>''u''[ − ''n'']</span>
 +
 
 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
  
 
----
 
----
 +
 +
Answer 6 - Ryan Atwell
 +
 +
<math>X(z) =\frac{1}{3-z}, \quad \text{ROC} \quad |z|<3 </math>
 +
 +
<math>X(z) = \frac{1}{3} \frac{1}{1-\frac{z}{3}} </math>
 +
 +
<math>X(z) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{z}{3})^{n} </math>
 +
 +
<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{\infty} u[n](\frac{z}{3})^{n}</math>
 +
 +
n=-k
 +
 +
<br> <math>X(z) = \frac{1}{3} \sum_{k=-\infty}^{\infty} u[-k](\frac{z}{3})^{-k}</math>
 +
 +
<math>X(z) = \frac{1}{3} \sum_{k=-\infty}^{\infty} u[-k]{3}^{k}{z}^{-k}</math>
 +
 +
<br> <math>X(z) = \sum_{k=-\infty}^{\infty} u[-k]{3}^{k-1}{z}^{-k}</math>
 +
 +
by formula
 +
 +
<span class="texhtml">''X''[''n''] = 3<sup>''n'' − 1</sup>''u''[ − ''n'']</span>
 +
 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
 +
 
----
 
----
−
[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013 Prof. Boutin]]
+
 
 +
Answer 7
 +
 
 +
<math>X(z) = \frac{1}{3-z}</math>
 +
 
 +
<math>X(z) = \frac{1}{3} \frac{1}{1-\frac{z}{3}}</math>
 +
 
 +
<math>X(z) = \frac{1}{3} \frac{1-\left( \frac{z}{3} \right) ^n}{1- \frac{z}{3}} </math> <math>, \quad \text{As n goes to} \quad \infty \quad \text{since} \quad |z|<3</math>
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{n=0}^{\infty} \left( \frac{z}{3} \right)^n</math>
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{0} \left( \frac{1}{3} \right)^{-n} z^{-n}</math>
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{\infty} \left(\frac{1}{3} \right)^{-n} u[-n] z^{-n}</math> <math>, \quad \text{By comparing with DTFT equation, we get} \quad</math>
 +
 
 +
<math>x[n] = \frac{1}{3} * \frac{1}{3}^{-n} u[-n]</math>
 +
 
 +
<math>x[n] = \left( \frac{1}{3} \right) ^{-n+1}  u[-n]</math>
 +
 
 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
 +
 
 +
----
 +
 
 +
Answer 8
 +
 
 +
<math>X(z) = \frac{1}{3-z}</math>
 +
 
 +
<math>=\frac{1}{3}\frac{1}{1-\frac{z}{3}}</math>
 +
 
 +
<math>\frac{1}{3}\sum_{n=-\infty}^{\infty} u[n](\frac{z}{3})^n</math>
 +
 
 +
let n=-k
 +
 
 +
<math>X(z)=\frac{1}{3} \sum_{k=-\infty}^{\infty} u[-k]3^k z^{-k}</math>
 +
 
 +
by comparison to inverse z-transform formula,
 +
 
 +
<span class="texhtml">''x''[''n''] = 3<sup> − 1 + ''k''</sup>''u''[ − ''k'']</span>
 +
 
 +
:<span style="color:blue"> Grader's comment: Right hand side is in 'k' while the left hand side in 'n' </span>
 +
 
 +
----
 +
[[2013 Fall ECE 438 Boutin|Back to ECE438 Fall 2013 Prof. Boutin]] &lt;/math&gt;
 +
 
 +
 
 +
----
 +
 
 +
Answer 8
 +
 
 +
<math>X(z) = \frac{1}{3-z}</math>
 +
 
 +
we have |z| &lt; 3, so
 +
 
 +
<math>X(z) = \frac{1}{3} \frac{1}{1-\frac{z}{3}} </math>
 +
 
 +
sum will look like this:
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{z}{3})^{n} </math>
 +
 
 +
with unit step:
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{\infty} u[n](\frac{z}{3})^{n}</math>
 +
 
 +
substituting n with -k we get:
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{k=-\infty}^{\infty} u[-k](\frac{z}{3})^{-k}</math>
 +
 
 +
<br> finally we get:
 +
 
 +
<math>X(z) = \sum_{k=-\infty}^{\infty} u[-k]{3}^{k-1}{z}^{-k}</math>
 +
 
 +
using the formula we get:
 +
 
 +
<math>x[n] = (\frac{1}{3})^{-n+1} u[-n] </math>
 +
 
 +
<br>
 +
 
 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
 +
 
 +
=== Answer 9  ===
 +
 
 +
<math>  X(z) = \frac{(\frac{1}{3})}{1-(\frac{z}{3})}                    </math>
 +
 
 +
<br> <math>  X(z) = (\frac{1}{3})*\sum_{n=0}^{\infty} \frac{1}{1-(\frac{z}{3})}  </math>
 +
 
 +
<br> <math>  X(z) = (\frac{1}{3})*\sum_{n=-\infty}^{\infty} u[n] (\frac{z}{3})^{n}  </math>
 +
 
 +
<br> let -k = n,
 +
 
 +
<br> <math>  X(z) = (\frac{1}{3})*\sum_{n=-\infty}^{\infty} u[-k] (\frac{z}{3})^{-k}  </math>
 +
 
 +
<br> <math>  X(Z) = (\frac{1}{3})*\sum_{n=-\infty}^{\infty} u[n] (\frac{1}{3})^{-k}* Z^{-k}  </math>
 +
 
 +
<br> so by comparison <math> , x[n] = (\frac{1}{3})^{-n+1} u[-n]  </math>
 +
 
 +
<br>
 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
 +
 
 +
=== Answer 10  ===
 +
 
 +
<math>X(z) =\frac{1}{3-z} </math>.
 +
 
 +
<math>X(z) =(\frac{1}{3}) \frac{1}{1-\frac{z}{3}} </math>
 +
 
 +
Since z/3 &lt; |1|, base on geometric series:
 +
 
 +
<math> \frac{1}{1-\frac{z}{3}} = \sum_{n=0}^{\infty} (\frac{z}{3})^{n} </math>
 +
 
 +
<math>X(z) =(\frac{1}{3}) \sum_{n=0}^{\infty} (\frac{z}{3})^{n} = \sum_{n=0}^{\infty} z^{n} (\frac{1}{3})^{n+1}</math>
 +
 
 +
<math>X(z) =\sum_{n=-\infty}^{\infty} u[n] z^{n} (\frac{1}{3})^{n+1}</math>
 +
 
 +
Let n = -m, <math>X(z) =\sum_{-m=-\infty}^{\infty} u[-m] z^{-m} (\frac{1}{3})^{-m+1}</math>
 +
 
 +
<math>X(z) =\sum_{m=-\infty}^{\infty} {z}^{-m} u[-m]{3}^{m-1}</math>
 +
 
 +
base on observation: <span class="texhtml">''x''[''n''] = ''u''[ − ''n'']3<sup>''n'' − 1</sup></span>
 +
 
 +
<br>
 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
 +
 
 +
=== Answer 11<br>  ===
 +
 
 +
 
 +
<math>X(z) =\frac{1}{3-z} </math> <math>X(z) =(\frac{1}{3}) \frac{1}{1-\frac{z}{3}} </math>
 +
 
 +
for <math>\quad \text{ROC} \quad |z|<3 </math>
 +
 
 +
we can use geometric series
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{n=0}^{+\infty} (\frac{z}{3})^n </math><br> <math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n](\frac{z}{3})^n </math>
 +
 
 +
<math>X(z) = \frac{1}{3} \sum_{k=-\infty}^{+\infty} u[-k](\frac{1}{3})^{-k} Z^{-k}</math>
 +
 
 +
so, by comparing to z-transform formula,we have
 +
 
 +
<br> <math> x[n] = (\frac{1}{3})^{-n+1} u[-n]  </math> <br>
 +
 
 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
 +
 
 +
<br>
 +
 
 +
 
 +
[[Category:ECE301]] [[Category:ECE438]] [[Category:ECE438Fall2013Boutin]] [[Category:Problem_solving]] [[Category:Z-transform]] [[Category:Inverse_z-transform]]

Latest revision as of 11:54, 26 November 2013

Practice Question on "Digital Signal Processing"

Topic: Computing an inverse z-transform

Question

Compute the inverse z-transform of

$ X(z) =\frac{1}{3-z}, \quad \text{ROC} \quad |z|<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!

Answer 1

$ X[z] = \sum_{n=0}^{+\infty} 3^{-1-n} z^{n} $ 

$       = \sum_{n=-\infty}^{+\infty} u[n] 3^{-1-n} z^{n} $

NOTE: Let n=-k

$ = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k} $ (compare with $ \sum_{n=-\infty}^{+\infty} x[n] z^{-k} $)

$ = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k} $

Therefore, x[n] = 3 − 1 + nu[ − n]

Grader's comment: Correct Answer


Answer 2

$ X(z) = \frac{1}{3-z} = \frac{1}{3} \frac{1}{1-\frac{z}{3}} $

$ X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{z}{3})^n $

Let n = -k

$ X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[-k]3^{k}z^{-k} $

By comparison with the x-transform formula,

x[n] = 3n − 1u[ − n]

Grader's comment: Correct Answer


Answer 3

By Yeong Ho Lee

$ X[z] = \sum_{n=0}^{\infty}z^{n}3^{-n-1} $

$ = \sum_{n=-\infty}^{+\infty} 3^{-n-1}z^{n}u[n] $ 

Now, let n = -k

$ = \sum_{n=-\infty}^{+\infty} 3^{k-1}z^{-k}u[-k] $

Using the z-transform formula, x[n] = 3n − 1u[ − n]

Grader's comment: Correct Answer

Answer 4

Gena Xie

$ X(z) = \frac{1}{3-Z} $

since |z|<3,

|z|/3 < 1

$ X(z) = \frac{1}{3} \frac{1} { 1-\frac{Z}{3}} $

$ X(z) = \frac{1}{3} \sum_{n=0}^{+\infty} (\frac{z}{3})^n = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n](\frac{z}{3})^n $

substitute n by -n,

$ X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[-n](\frac{z}{3})^{-n} = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[-n](\frac{1}{3})^{-n} Z^{-n} $

based on the definition

$ x[n] = \frac{1}{3} u[-n](\frac{1}{3})^{-n} = (\frac{1}{3})^{-n+1} u[-n] $


Grader's comment: Correct Answer

Answer 5

By Yixiang Liu

$ X(z) = \frac{1}{3-Z} $


$ X(z) = \frac{\frac{1}{3}} { 1-\frac{Z}{3}} $

$ X(z) = \frac{1}{3} * \frac{1} { 1-\frac{Z}{3}} $

by geometric series

$ X(z) = \frac{1}{3} \sum_{n=0}^{+\infty} (\frac{z}{3})^n $

$ X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n](\frac{z}{3})^n $

$ X(z) = \frac{1}{3} \sum_{k=-\infty}^{+\infty} u[-k](\frac{z}{3})^{-k} $

$ X(z) = \frac{1}{3} \sum_{k=-\infty}^{+\infty} u[-k](\frac{1}{3})^{-k} Z^{-k} $

By comparison with the x-transform formula

$ x[n] = \frac{1}{3} u[-n](\frac{1}{3})^{-n} $

$ x[n] = (\frac{1}{3})^{-n+1} u[-n] $

x[n] = 3n − 1u[ − n]

Grader's comment: Correct Answer

Answer 6 - Ryan Atwell

$ X(z) =\frac{1}{3-z}, \quad \text{ROC} \quad |z|<3 $

$ X(z) = \frac{1}{3} \frac{1}{1-\frac{z}{3}} $

$ X(z) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{z}{3})^{n} $

$ X(z) = \frac{1}{3} \sum_{n=-\infty}^{\infty} u[n](\frac{z}{3})^{n} $

n=-k


$ X(z) = \frac{1}{3} \sum_{k=-\infty}^{\infty} u[-k](\frac{z}{3})^{-k} $

$ X(z) = \frac{1}{3} \sum_{k=-\infty}^{\infty} u[-k]{3}^{k}{z}^{-k} $


$ X(z) = \sum_{k=-\infty}^{\infty} u[-k]{3}^{k-1}{z}^{-k} $

by formula

X[n] = 3n − 1u[ − n]

Grader's comment: Correct Answer

Answer 7

$ X(z) = \frac{1}{3-z} $

$ X(z) = \frac{1}{3} \frac{1}{1-\frac{z}{3}} $

$ X(z) = \frac{1}{3} \frac{1-\left( \frac{z}{3} \right) ^n}{1- \frac{z}{3}} $ $ , \quad \text{As n goes to} \quad \infty \quad \text{since} \quad |z|<3 $

$ X(z) = \frac{1}{3} \sum_{n=0}^{\infty} \left( \frac{z}{3} \right)^n $

$ X(z) = \frac{1}{3} \sum_{n=-\infty}^{0} \left( \frac{1}{3} \right)^{-n} z^{-n} $

$ X(z) = \frac{1}{3} \sum_{n=-\infty}^{\infty} \left(\frac{1}{3} \right)^{-n} u[-n] z^{-n} $ $ , \quad \text{By comparing with DTFT equation, we get} \quad $

$ x[n] = \frac{1}{3} * \frac{1}{3}^{-n} u[-n] $

$ x[n] = \left( \frac{1}{3} \right) ^{-n+1} u[-n] $

Grader's comment: Correct Answer

Answer 8

$ X(z) = \frac{1}{3-z} $

$ =\frac{1}{3}\frac{1}{1-\frac{z}{3}} $

$ \frac{1}{3}\sum_{n=-\infty}^{\infty} u[n](\frac{z}{3})^n $

let n=-k

$ X(z)=\frac{1}{3} \sum_{k=-\infty}^{\infty} u[-k]3^k z^{-k} $

by comparison to inverse z-transform formula,

x[n] = 3 − 1 + ku[ − k]

Grader's comment: Right hand side is in 'k' while the left hand side in 'n'

Back to ECE438 Fall 2013 Prof. Boutin </math>


Answer 8

$ X(z) = \frac{1}{3-z} $

we have |z| < 3, so

$ X(z) = \frac{1}{3} \frac{1}{1-\frac{z}{3}} $

sum will look like this:

$ X(z) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{z}{3})^{n} $

with unit step:

$ X(z) = \frac{1}{3} \sum_{n=-\infty}^{\infty} u[n](\frac{z}{3})^{n} $

substituting n with -k we get:

$ X(z) = \frac{1}{3} \sum_{k=-\infty}^{\infty} u[-k](\frac{z}{3})^{-k} $


finally we get:

$ X(z) = \sum_{k=-\infty}^{\infty} u[-k]{3}^{k-1}{z}^{-k} $

using the formula we get:

$ x[n] = (\frac{1}{3})^{-n+1} u[-n] $


Grader's comment: Correct Answer

Answer 9

$ X(z) = \frac{(\frac{1}{3})}{1-(\frac{z}{3})} $


$ X(z) = (\frac{1}{3})*\sum_{n=0}^{\infty} \frac{1}{1-(\frac{z}{3})} $


$ X(z) = (\frac{1}{3})*\sum_{n=-\infty}^{\infty} u[n] (\frac{z}{3})^{n} $


let -k = n,


$ X(z) = (\frac{1}{3})*\sum_{n=-\infty}^{\infty} u[-k] (\frac{z}{3})^{-k} $


$ X(Z) = (\frac{1}{3})*\sum_{n=-\infty}^{\infty} u[n] (\frac{1}{3})^{-k}* Z^{-k} $


so by comparison $ , x[n] = (\frac{1}{3})^{-n+1} u[-n] $


Grader's comment: Correct Answer

Answer 10

$ X(z) =\frac{1}{3-z} $.

$ X(z) =(\frac{1}{3}) \frac{1}{1-\frac{z}{3}} $

Since z/3 < |1|, base on geometric series:

$ \frac{1}{1-\frac{z}{3}} = \sum_{n=0}^{\infty} (\frac{z}{3})^{n} $

$ X(z) =(\frac{1}{3}) \sum_{n=0}^{\infty} (\frac{z}{3})^{n} = \sum_{n=0}^{\infty} z^{n} (\frac{1}{3})^{n+1} $

$ X(z) =\sum_{n=-\infty}^{\infty} u[n] z^{n} (\frac{1}{3})^{n+1} $

Let n = -m, $ X(z) =\sum_{-m=-\infty}^{\infty} u[-m] z^{-m} (\frac{1}{3})^{-m+1} $

$ X(z) =\sum_{m=-\infty}^{\infty} {z}^{-m} u[-m]{3}^{m-1} $

base on observation: x[n] = u[ − n]3n − 1


Grader's comment: Correct Answer

Answer 11

$ X(z) =\frac{1}{3-z} $ $ X(z) =(\frac{1}{3}) \frac{1}{1-\frac{z}{3}} $

for $ \quad \text{ROC} \quad |z|<3 $

we can use geometric series

$ X(z) = \frac{1}{3} \sum_{n=0}^{+\infty} (\frac{z}{3})^n $
$ X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n](\frac{z}{3})^n $

$ X(z) = \frac{1}{3} \sum_{k=-\infty}^{+\infty} u[-k](\frac{1}{3})^{-k} Z^{-k} $

so, by comparing to z-transform formula,we have


$ x[n] = (\frac{1}{3})^{-n+1} u[-n] $

Grader's comment: Correct Answer


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Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood