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[[Category:inverse z-transform]]
 
[[Category:inverse z-transform]]
  
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= [[:Category:Problem_solving|Practice Question]], [[ECE438]] Fall 2013, [[User:Mboutin|Prof. Boutin]] =
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<center><font size= 4>
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On computing the inverse z-transform of a discrete-time signal.
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'''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]'''
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</font size>
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Topic: Computing an inverse z-transform
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</center>
 
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==Question==
 
Compute the inverse z-transform of  
 
Compute the inverse z-transform of  
  
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===Answer 1===
 
===Answer 1===
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Gena Xie
 
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<math>X(z) = e^{-2z}. </math>
 
<math>X(z) = e^{-2z}. </math>
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By Taylor Series,
 
By Taylor Series,
  
 
<math>X(z) = e^{-2z} = \sum_{n=0}^{+\infty}\frac{{(-2 z)} ^ {n}}{n!}</math>
 
<math>X(z) = e^{-2z} = \sum_{n=0}^{+\infty}\frac{{(-2 z)} ^ {n}}{n!}</math>
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substitute n by -n
 
substitute n by -n
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<math>X(z) = \sum_{n=-\infty}^{0}\frac { { (-2) } ^ {-n} } { (-n)! } {z^{-n}} = \sum_{n=-\infty}^{+\infty} \frac { { (-2) } ^ {-n} } { (-n)! } u[-n]{z^{-n}}</math>
 
<math>X(z) = \sum_{n=-\infty}^{0}\frac { { (-2) } ^ {-n} } { (-n)! } {z^{-n}} = \sum_{n=-\infty}^{+\infty} \frac { { (-2) } ^ {-n} } { (-n)! } u[-n]{z^{-n}}</math>
  
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based on the definition,
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based on the definition, <span style="color:green"> (Instructor's comment: Definition of what? Be clear.)</span>
  
 
<math>X(z) = \frac{ {(-2)} ^ {-n} } { (-n)! } u[-n] </math>
 
<math>X(z) = \frac{ {(-2)} ^ {-n} } { (-n)! } u[-n] </math>
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 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
  
 
=== Answer 2===
 
=== Answer 2===
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Write it here.
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 +
alec green
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 +
an exponential can be expanded into the series:
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<math>e^{x} = \sum_{n=0}^{+\infty}\frac{x^{n}}{n!}</math>
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<math>X(z) = e^{-2z} = \sum_{n=0}^{+\infty}(\frac{(-2z)^{n}}{n!} = \frac{(-2)^{n}}{n!}z^{n})</math>
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<math> = \sum_{n=-\infty}^{+\infty}u[n]\frac{(-2)^{n}}{n!}z^{n}</math>
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letting k = -n:
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<math> = \sum_{k=-\infty}^{+\infty}u[-k]\frac{(-2)^{-k}}{(-k)!}z^{-k}</math>
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and by comparison with:
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<math>X(z) = \sum_{n=-\infty}^{+\infty}x[n]z^{-n}</math>
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<math>x[n] = u[-n]\frac{(-2)^{-n}}{(-n)!}</math>
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 +
due to the step function in the x[n], the factiorial in x[n] is never evaluated on a negative argument (which would be undefined).
 +
 
 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
 +
 
 
===Answer 3===
 
===Answer 3===
−
Write it here.
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 +
<math>X(z) = e^{-2z}. </math>
 +
 
 +
By Taylor Series,
 +
 
 +
<math>X(z) = e^{-2z} = \sum_{n=0}^{+\infty}\frac{{(-2 z)} ^ {n}}{n!} = 1 - 2 z + \frac{4 z^2}{2!} - \frac{8 z^3}{3!} ...</math>
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 +
We also know that the Z transform of an impulse <math>\delta (n - n0)</math> is:
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<math> X(z) = \sum_{n=-\infty}^{+\infty}\delta (n - n0) z^{-n} = z^{-n0}</math>
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 +
Therefore the inverse Z Transform of the signal will be given by:
 +
 
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<math>n[n] = \delta[n] - 2 \delta[n+1] + \frac{4 \delta[n+2]}{2!} - \frac{8 \delta[n+3]}{3!} ... = \sum_{k=0}^{+\infty}\frac{(-2)^k}{k!} \delta[n+k]</math>
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 +
Since we know that a discrete signal can be written as a weighted impulse train, we can alternatively rewrite x[n] as:
 +
 
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<math>x[n] = \frac{(-2)^n}{(-n)!} u[-n]</math>
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 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
 +
 
 
===Answer 4===
 
===Answer 4===
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Write it here.
+
Xiang Zhang
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 +
From the formula of exponential function of Taylor series we can find that
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<math>  e^x = \sum_{ n = 0 }^{+ \infty} \frac{x^n}{n!} </math>
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 +
Hence we can find in our expression that
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<math> x = -2z </math>
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 +
Let's expand the original signal to the expression below
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 +
<math>  e^{-2z} = \sum_{ n = 0 }^{+ \infty} \frac{ (- 2 z) ^n}{n!} </math>
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 +
Replace <math>  n = 0  </math>  to <math>  n = - \infty </math> by introducing u[n].
 +
 
 +
We can get that,
 +
 
 +
<math>  e^{-2z} = \sum_{ n = - \infty }^{+ \infty} \frac{ (- 2 z) ^n}{n!} u[n] </math>
 +
 
 +
Substitute in n = -k (k = -n)
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<math>  e^{-2z} = \sum_{ k = + \infty }^{- \infty} \frac{ (- 2 z) ^{-k}}{(-k)!} u[-k] </math>
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 +
Change the integration and reorder the expression
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<math>  e^{-2z} = \sum_{ k = - \infty }^{+ \infty} \frac{  u[-k] (-2) ^{-k}}{(-k)!}  z^{-k} </math>
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 +
By comparison with original expression
 +
 
 +
<math>X(z) = \sum_{n=-\infty}^{+\infty}x[n]z^{-n}</math>
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 +
Substitute k back to n (k = n)
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<math>  e^{-2z} = \sum_{ n = - \infty }^{+ \infty} \frac{  u[-n] (-2) ^{-n}}{(-n)!}  z^{-n} </math>
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 +
Then we can recover back x[n]
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<math> x[n] = \frac{  (-2) ^{-n}}{(-n)!} u[-n]</math>
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 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
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===Answer 5 ===
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 +
By taylor series
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<math>  e^{-2z} = \sum_{k = 0}^{+ \infty} \frac{(-2z)^k}{k!} </math>
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<math>  = \sum_{k = 0}^{+ \infty} \frac{ (-2z) ^k}{k!} </math>
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Substitute -n for k,
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<math>  = \sum_{ n = - \infty }^{+ \infty} \frac{ (-2z)^{-n}}{(-n)!} u[-n] </math>
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Pull z^-n out of expression,
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<math>  = \sum_{ k = - \infty }^{+ \infty} \frac{ (-2) ^{-n}}{(-n)!} z^{-n} u[-n] </math>
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Compare with the Z transform equation from RHEA table to get...
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<math> x[n] = \frac{  (-2) ^{-n}}{(-n)!} u[-n]</math>
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 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
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 +
 
 +
 
 +
 
 +
 
 +
 
 +
 
 +
 
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----
 
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[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013 Prof. Boutin]]
 
[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013 Prof. Boutin]]

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) = e^{-2z}. $

(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) = e^{-2z}. $


By Taylor Series,

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


substitute n by -n

$ X(z) = \sum_{n=-\infty}^{0}\frac { { (-2) } ^ {-n} } { (-n)! } {z^{-n}} = \sum_{n=-\infty}^{+\infty} \frac { { (-2) } ^ {-n} } { (-n)! } u[-n]{z^{-n}} $


based on the definition, (Instructor's comment: Definition of what? Be clear.)

$ X(z) = \frac{ {(-2)} ^ {-n} } { (-n)! } u[-n] $

Grader's comment: Correct Answer

Answer 2

alec green

an exponential can be expanded into the series:

$ e^{x} = \sum_{n=0}^{+\infty}\frac{x^{n}}{n!} $

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

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

letting k = -n:

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

and by comparison with:

$ X(z) = \sum_{n=-\infty}^{+\infty}x[n]z^{-n} $

$ x[n] = u[-n]\frac{(-2)^{-n}}{(-n)!} $

due to the step function in the x[n], the factiorial in x[n] is never evaluated on a negative argument (which would be undefined).

Grader's comment: Correct Answer

Answer 3

$ X(z) = e^{-2z}. $

By Taylor Series,

$ X(z) = e^{-2z} = \sum_{n=0}^{+\infty}\frac{{(-2 z)} ^ {n}}{n!} = 1 - 2 z + \frac{4 z^2}{2!} - \frac{8 z^3}{3!} ... $

We also know that the Z transform of an impulse $ \delta (n - n0) $ is:

$ X(z) = \sum_{n=-\infty}^{+\infty}\delta (n - n0) z^{-n} = z^{-n0} $

Therefore the inverse Z Transform of the signal will be given by:

$ n[n] = \delta[n] - 2 \delta[n+1] + \frac{4 \delta[n+2]}{2!} - \frac{8 \delta[n+3]}{3!} ... = \sum_{k=0}^{+\infty}\frac{(-2)^k}{k!} \delta[n+k] $

Since we know that a discrete signal can be written as a weighted impulse train, we can alternatively rewrite x[n] as:

$ x[n] = \frac{(-2)^n}{(-n)!} u[-n] $

Grader's comment: Correct Answer

Answer 4

Xiang Zhang

From the formula of exponential function of Taylor series we can find that

$ e^x = \sum_{ n = 0 }^{+ \infty} \frac{x^n}{n!} $

Hence we can find in our expression that

$ x = -2z $

Let's expand the original signal to the expression below

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

Replace $ n = 0 $ to $ n = - \infty $ by introducing u[n].

We can get that,

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

Substitute in n = -k (k = -n)

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

Change the integration and reorder the expression

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

By comparison with original expression

$ X(z) = \sum_{n=-\infty}^{+\infty}x[n]z^{-n} $

Substitute k back to n (k = n)

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

Then we can recover back x[n]

$ x[n] = \frac{ (-2) ^{-n}}{(-n)!} u[-n] $

Grader's comment: Correct Answer

Answer 5

By taylor series

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

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

Substitute -n for k,

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

Pull z^-n out of expression,

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

Compare with the Z transform equation from RHEA table to get...

$ x[n] = \frac{ (-2) ^{-n}}{(-n)!} u[-n] $

Grader's comment: Correct Answer





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