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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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= [[:Category:Problem_solving|Practice Question]], [[ECE438]] Fall 2013, [[User:Mboutin|Prof. Boutin]] =
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Topic: Computing an inverse z-transform
−
On computing the inverse z-transform of a discrete-time signal.
+
 
 +
</center>
 
----
 
----
 +
==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===
 
−
<math>X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math>
 
  
−
<math>= -\frac{1}{3-z}-\frac{1}{2-z}</math>
+
=== Answer 1 ===
  
−
<math>= -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}})</math>
+
<math>X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math> <span style="color:green"> (Instructor's comment: You can skip this step.)</span>
 +
 
 +
<math>= -\frac{1}{3-z}-\frac{1}{2-z}</math>
 +
 
 +
<math>= -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}})</math>  
  
 
<math>= -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n</math>  
 
<math>= -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n</math>  
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<math>= \sum_{n=0}^{+\infty}[(-\frac{1}{3}) (\frac{1}{3})^n + (-\frac{1}{2})(\frac{1}{2})^n]z^n</math>  
 
<math>= \sum_{n=0}^{+\infty}[(-\frac{1}{3}) (\frac{1}{3})^n + (-\frac{1}{2})(\frac{1}{2})^n]z^n</math>  
  
−
Let k=-n
+
Let k=-n  
  
 
<math>= \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + (-\frac{1}{2})2^k]z^{-k}</math>  
 
<math>= \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + (-\frac{1}{2})2^k]z^{-k}</math>  
  
−
by comparison with z-transform formula
+
by comparison with z-transform formula  
  
−
<math>x[n]=u[-n](-3^{n-1}-2^{n-1})</math>
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<span class="texhtml">''x''[''n''] = ''u''[ − ''n'']( − 3<sup>''n'' − 1</sup> − 2<sup>''n'' − 1</sup>)</span>
−
=== Answer 2===
+
 
−
Write it here.
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:<span style="color:blue"> Grader's comment: Partial Fractions splitting is wrong </span>
−
===Answer 3===
+
 
−
Write it here.
+
=== Answer 2  ===
−
===Answer 4===
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Using a partial fraction expansion, we can change the original equation to
−
Write it here.
+
 
−
----
+
<math>X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math> Where A = 1, B = -1, so we get <span style="color:green"> (Instructor's comment: You can skip this explanation and write the expansion directly)</span>
−
----
+
 
−
[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013 Prof. Boutin]]
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<math>= -\frac{1}{3-z}-\frac{1}{2-z}</math>
 +
 
 +
By factoring out 1/3 for the first term, and 1/2 for the second term, we can have both terms in form of <span style="color:green"> (Instructor's comment: No need to explain this.)</span>
 +
 
 +
<math> \frac{1}{1-r} </math>, which is equal to <math> \sum_{n=0}^{+\infty} (\frac{1}{r})^n </math> <span style="color:green"> (Instructor's comment: This is not true in general: only when |r|<1)</span>
 +
 
 +
<math>= -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}})</math>
 +
 
 +
<math>= -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n</math>
 +
 
 +
<math>= \sum_{n=0}^{+\infty}[(-\frac{1}{3}) (\frac{1}{3})^n + (-\frac{1}{2})(\frac{1}{2})^n]z^n</math>
 +
 
 +
Then let k=-n
 +
 
 +
<math>= \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + (-\frac{1}{2})2^k]z^{-k}</math>
 +
 
 +
Comparing it with z-transform formula, we can get
 +
 
 +
<span class="texhtml">''x''[''n''] = ''u''[ − ''n'']( − 3<sup>''n'' − 1</sup> − 2<sup>''n'' − 1</sup>)</span>
 +
 
 +
:<span style="color:blue"> Grader's comment: Partial Fractions splitting is wrong </span>
 +
 
 +
=== Answer 3  ===
 +
First, using partial fraction we get..
 +
 
 +
<math> X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math> <span style="color:green"> (Instructor's comment: You can skip this step.)</span>
 +
 
 +
A(2-z) + B(3-z) = 1 
 +
 
 +
let z=2, then B=1
 +
 
 +
let z=3, then A=-1 <span style="color:green"> (Instructor's comment: You do not need to explain how you got the A and the B. )</span>
 +
 
 +
<math> = -\frac{1}{3-z}+\frac{1}{2-z}</math>
 +
 
 +
<math>= -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})+\frac{1}{2}(\frac{1}{1-\frac{z}{2}})</math>
 +
 
 +
<math>= -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n +\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n</math>
 +
 
 +
<math>= -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{1}{3})^n(z)^n +\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{1}{2})^n(z)^n</math>  
 +
 
 +
now let n = -k
 +
 
 +
<math>= -\frac{1}{3}\sum_{n=0}^{+\infty} 3^{k} z^{-k} +\frac{1}{2}\sum_{n=0}^{+\infty} 2^{k}z^{-k}</math> <span style="color:green"> (Instructor's comment: Your sum is over n, but the expression depends on k.)</span>
 +
 
 +
by comparison with z-transfrom formula
 +
 
 +
<span class="texhtml">''x''[''n''] =  − 3<sup>''n'' − 1</sup>''u''[ − ''n''] + 2<sup>''n'' − 1</sup>''u''[ − ''n'']</span>
 +
 
 +
<span class="texhtml">''x''[''n''] = ( − 3<sup>''n'' − 1</sup> + 2<sup>''n'' − 1</sup>)''u''[ − ''n'']</span>
 +
 
 +
<br>
 +
 
 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
 +
 
 +
=== Answer 4  ===
 +
 
 +
<math>X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math> <span style="color:green"> (Instructor's comment: You can skip this step.)</span>
 +
 
 +
<math>= -\frac{1}{3-z} - \frac{1}{2-z}</math>
 +
 
 +
<math>= -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}})</math>
 +
 
 +
<math>= -\frac{1}{3}\sum_{k=0}^{+\infty} (\frac{z}{3})^k -\frac{1}{2}\sum_{k=0}^{+\infty} (\frac{z}{2})^k</math>
 +
 
 +
<math>= \sum_{k=0}^{+\infty}[(-\frac{1}{3})(\frac{1}{3})^k + (-\frac{1}{2})(\frac{1}{2})^k]u[k] * z^k</math>
 +
 
 +
Substitute k with -n
 +
 
 +
<math>= \sum_{n=-\infty}^{+\infty}[(-\frac{1}{3})3^{-n} + (-\frac{1}{2})2^{-n}]u[-n] * z^{-n}</math>
 +
 
 +
Look up Z transform equation on RHEA table and see that X(z) becomes...  
 +
 
 +
<span class="texhtml">''x''[''n''] = ( − 3<sup>''n'' − 1</sup> − 2<sup>''n'' − 1</sup>)''u''[ − ''n'']</span>
 +
 
 +
 
 +
<br>
 +
 
 +
:<span style="color:blue"> Grader's comment: Partial Fractions splitting is wrong </span>
 +
 
 +
=== Answer 5  ===
 +
 
 +
by partical fraction, we get,
 +
 
 +
<math>X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math> <span style="color:green"> (Instructor's comment: You can skip this step.)</span>
 +
 
 +
<math>= -\frac{1}{3-z}+\frac{1}{2-z}</math>
 +
 
 +
For <math>\quad \text{ROC} \quad |z|<2 </math>
 +
 
 +
<math>X(z)= -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})+\frac{1}{2}(\frac{1}{1-\frac{z}{2}})</math>
 +
 
 +
<br> <math>= -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n +\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n</math>
 +
 
 +
assume n=-k.  
 +
 
 +
<math>X(z)= -\frac{1}{3}\sum_{k=-\infty}^{0} 3^{k} z^{-k} +\frac{1}{2}\sum_{k=-\infty}^{0} 2^{k}z^{-k}</math>
 +
 
 +
<br> <math>= \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + \frac{1}{2}2^k]z^{-k}</math>
 +
 
 +
So, x[n] = (−3<sup>n-1</sup>+2<sup>n-1</sup>)u[-n] <br>
 +
 
 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
 +
 
 +
 
 +
=== Answer 6  ===
 +
 
 +
<math>X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math> <span style="color:green"> (Instructor's comment: You can skip this step.)</span>
 +
 
 +
<math> = \frac{1}{2-z}-\frac{1}{3-z}</math>
 +
 
 +
<math> = \frac{1}{2}\frac{1}{1-\frac{z}{2}}-\frac{1}{3}\frac{1}{1-\frac{z}{3}}</math>
 +
 
 +
By the geometric series formula,<br>
 +
 
 +
<math>X(z) = \frac{1}{2}\sum_{n=0}^{+\infty}(\frac{z}{2})^n - \frac{1}{3}\sum_{n=0}^{+\infty}(\frac{z}{3})^n</math>
 +
 
 +
<math> = \sum_{n=0}^{+\infty}(\frac{1}{2}(\frac{1}{2})^n - \frac{1}{3}(\frac{1}{3})^n)z^n</math>
 +
 
 +
<math> = \sum_{n=-\infty}^{+\infty}u[n]((\frac{1}{2})^{n+1} - (\frac{1}{3})^{n+1})z^n</math>
 +
 
 +
Substituting k = -n for n gives,<br>
 +
 
 +
<math>X(z) = \sum_{k=-\infty}^{+\infty}u[-k](\frac{1}{2}^{-k+1} - \frac{1}{3}^{k+1})z^{-k}</math>
 +
 
 +
<math> = \sum_{k=-\infty}^{+\infty}u[-k](2^{k-1} - 3^{k-1})z^{-k}</math>
 +
 
 +
By comparison with the Z-transform formula,<br>
 +
 
 +
x[n] = u[-n](2<sup>n-1</sup>-3<sup>n-1</sup>)<br>
 +
 
 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
 +
 
 +
=== Answer 7  ===
 +
<math>X(z) =\frac{1}{(3-z)(2-z)}</math>
 +
 
 +
<math>= \frac{A}{3-z}+\frac{B}{2-z}</math> <span style="color:green"> (Instructor's comment: You can skip this step.)</span>
 +
<math>= -\frac{1}{3-z}-\frac{1}{2-z}</math>
 +
 
 +
<math>= -\frac{1}{3}*(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}*(\frac{1}{1-\frac{z}{2}})</math>  <span style="color:green"> (Instructor's comment: Be careful! You do not mean convolution here, do you? Then you should use <math>\times</math> instead of <math>*</math>.)</span>
 +
 
 +
<math>= -\frac{1}{3}\sum_{n=0}^{+\infty}(\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty}(\frac{z}{2})^n</math>
 +
<math>= -\frac{1}{3}\sum_{n=0}^{+\infty}(\frac{1}{3})^n(z)^n -\frac{1}{2}\sum_{n=0}^{+\infty}(\frac{1}{2})^n(z)^n</math>
 +
 
 +
substituting k for -n:
 +
 
 +
<math>= -\frac{1}{3}\sum_{n=0}^{+\infty}3^{k} z^{-k}-\frac{1}{2}\sum_{n=0}^{+\infty}2^{k}z^{-k}</math>
 +
 
 +
Using the Z transform tables to find the common transformation: <span style="color:green"> (Instructor's comment: I don't understand what you mean. You should rephrase this.)</span>
 +
 
 +
x[n] = (−3^(n−1))u[−n] - (2^(n−1))u[− n]
 +
 
 +
:<span style="color:blue"> Grader's comment: Partial Fractions splitting is wrong </span>
 +
 
 +
 
 +
[[2013 Fall ECE 438 Boutin|Back to ECE438 Fall 2013 Prof. Boutin]]
 +
 
 +
[[Category:ECE301]] [[Category:ECE438]] [[Category:ECE438Fall2013Boutin]] [[Category:Problem_solving]] [[Category:Z-transform]] [[Category:Inverse_z-transform]]

Latest revision as of 11:55, 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)(2-z)}, \quad \text{ROC} \quad |z|<2 $.

(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) = \frac{A}{3-z}+\frac{B}{2-z} $ (Instructor's comment: You can skip this step.)

$ = -\frac{1}{3-z}-\frac{1}{2-z} $

$ = -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $

$ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n $

$ = \sum_{n=0}^{+\infty}[(-\frac{1}{3}) (\frac{1}{3})^n + (-\frac{1}{2})(\frac{1}{2})^n]z^n $

Let k=-n

$ = \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + (-\frac{1}{2})2^k]z^{-k} $

by comparison with z-transform formula

x[n] = u[ − n]( − 3n − 1 − 2n − 1)

Grader's comment: Partial Fractions splitting is wrong

Answer 2

Using a partial fraction expansion, we can change the original equation to

$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $ Where A = 1, B = -1, so we get (Instructor's comment: You can skip this explanation and write the expansion directly)

$ = -\frac{1}{3-z}-\frac{1}{2-z} $

By factoring out 1/3 for the first term, and 1/2 for the second term, we can have both terms in form of (Instructor's comment: No need to explain this.)

$ \frac{1}{1-r} $, which is equal to $ \sum_{n=0}^{+\infty} (\frac{1}{r})^n $ (Instructor's comment: This is not true in general: only when |r|<1)

$ = -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $

$ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n $

$ = \sum_{n=0}^{+\infty}[(-\frac{1}{3}) (\frac{1}{3})^n + (-\frac{1}{2})(\frac{1}{2})^n]z^n $

Then let k=-n

$ = \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + (-\frac{1}{2})2^k]z^{-k} $

Comparing it with z-transform formula, we can get

x[n] = u[ − n]( − 3n − 1 − 2n − 1)

Grader's comment: Partial Fractions splitting is wrong

Answer 3

First, using partial fraction we get..

$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $ (Instructor's comment: You can skip this step.)

A(2-z) + B(3-z) = 1

let z=2, then B=1

let z=3, then A=-1 (Instructor's comment: You do not need to explain how you got the A and the B. )

$ = -\frac{1}{3-z}+\frac{1}{2-z} $

$ = -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})+\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $

$ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n +\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n $

$ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{1}{3})^n(z)^n +\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{1}{2})^n(z)^n $

now let n = -k

$ = -\frac{1}{3}\sum_{n=0}^{+\infty} 3^{k} z^{-k} +\frac{1}{2}\sum_{n=0}^{+\infty} 2^{k}z^{-k} $ (Instructor's comment: Your sum is over n, but the expression depends on k.)

by comparison with z-transfrom formula

x[n] = − 3n − 1u[ − n] + 2n − 1u[ − n]

x[n] = ( − 3n − 1 + 2n − 1)u[ − n]


Grader's comment: Correct Answer

Answer 4

$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $ (Instructor's comment: You can skip this step.)

$ = -\frac{1}{3-z} - \frac{1}{2-z} $

$ = -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $

$ = -\frac{1}{3}\sum_{k=0}^{+\infty} (\frac{z}{3})^k -\frac{1}{2}\sum_{k=0}^{+\infty} (\frac{z}{2})^k $

$ = \sum_{k=0}^{+\infty}[(-\frac{1}{3})(\frac{1}{3})^k + (-\frac{1}{2})(\frac{1}{2})^k]u[k] * z^k $

Substitute k with -n

$ = \sum_{n=-\infty}^{+\infty}[(-\frac{1}{3})3^{-n} + (-\frac{1}{2})2^{-n}]u[-n] * z^{-n} $

Look up Z transform equation on RHEA table and see that X(z) becomes...

x[n] = ( − 3n − 1 − 2n − 1)u[ − n]



Grader's comment: Partial Fractions splitting is wrong

Answer 5

by partical fraction, we get,

$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $ (Instructor's comment: You can skip this step.)

$ = -\frac{1}{3-z}+\frac{1}{2-z} $

For $ \quad \text{ROC} \quad |z|<2 $

$ X(z)= -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})+\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $


$ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n +\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n $

assume n=-k.

$ X(z)= -\frac{1}{3}\sum_{k=-\infty}^{0} 3^{k} z^{-k} +\frac{1}{2}\sum_{k=-\infty}^{0} 2^{k}z^{-k} $


$ = \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + \frac{1}{2}2^k]z^{-k} $

So, x[n] = (−3n-1+2n-1)u[-n]

Grader's comment: Correct Answer


Answer 6

$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $ (Instructor's comment: You can skip this step.)

$ = \frac{1}{2-z}-\frac{1}{3-z} $

$ = \frac{1}{2}\frac{1}{1-\frac{z}{2}}-\frac{1}{3}\frac{1}{1-\frac{z}{3}} $

By the geometric series formula,

$ X(z) = \frac{1}{2}\sum_{n=0}^{+\infty}(\frac{z}{2})^n - \frac{1}{3}\sum_{n=0}^{+\infty}(\frac{z}{3})^n $

$ = \sum_{n=0}^{+\infty}(\frac{1}{2}(\frac{1}{2})^n - \frac{1}{3}(\frac{1}{3})^n)z^n $

$ = \sum_{n=-\infty}^{+\infty}u[n]((\frac{1}{2})^{n+1} - (\frac{1}{3})^{n+1})z^n $

Substituting k = -n for n gives,

$ X(z) = \sum_{k=-\infty}^{+\infty}u[-k](\frac{1}{2}^{-k+1} - \frac{1}{3}^{k+1})z^{-k} $

$ = \sum_{k=-\infty}^{+\infty}u[-k](2^{k-1} - 3^{k-1})z^{-k} $

By comparison with the Z-transform formula,

x[n] = u[-n](2n-1-3n-1)

Grader's comment: Correct Answer

Answer 7

$ X(z) =\frac{1}{(3-z)(2-z)} $

$ = \frac{A}{3-z}+\frac{B}{2-z} $ (Instructor's comment: You can skip this step.) $ = -\frac{1}{3-z}-\frac{1}{2-z} $

$ = -\frac{1}{3}*(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}*(\frac{1}{1-\frac{z}{2}}) $ (Instructor's comment: Be careful! You do not mean convolution here, do you? Then you should use $ \times $ instead of $ * $.)

$ = -\frac{1}{3}\sum_{n=0}^{+\infty}(\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty}(\frac{z}{2})^n $ $ = -\frac{1}{3}\sum_{n=0}^{+\infty}(\frac{1}{3})^n(z)^n -\frac{1}{2}\sum_{n=0}^{+\infty}(\frac{1}{2})^n(z)^n $

substituting k for -n:

$ = -\frac{1}{3}\sum_{n=0}^{+\infty}3^{k} z^{-k}-\frac{1}{2}\sum_{n=0}^{+\infty}2^{k}z^{-k} $

Using the Z transform tables to find the common transformation: (Instructor's comment: I don't understand what you mean. You should rephrase this.)

x[n] = (−3^(n−1))u[−n] - (2^(n−1))u[− n]

Grader's comment: Partial Fractions splitting is wrong


Back to ECE438 Fall 2013 Prof. Boutin

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