(50 intermediate revisions by 10 users not shown) | |||
Line 7: | Line 7: | ||
− | = [[ | + | <center><font size= 4> |
− | + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | |
+ | </font size> | ||
+ | |||
+ | Topic: Computing an inverse z-transform | ||
+ | |||
+ | </center> | ||
---- | ---- | ||
+ | ==Question== | ||
Compute the inverse z-transform of | Compute the inverse z-transform of | ||
Line 20: | Line 26: | ||
---- | ---- | ||
===Answer 1=== | ===Answer 1=== | ||
− | <math> X(z) =\frac{1}{\frac{3z}{z} - z} </math> | + | <math> X(z) =\frac{1}{\frac{3z}{z} - z} = \frac{-1}{z} \frac{1}{1-\frac{3}{z}} </math> |
− | <math> =\frac{-1}{z} \frac{1}{1-\ | + | <math> =\frac{-1}{z} \sum_{n=0}^{+\infty} (\frac{3}{z})^n = \sum_{n=0}^{+\infty} (-z^{-1}) (3z^{-1})^n </math> |
+ | NOTE: <math> (3z^{-1})^n = (3^n) (z^{-n}) </math> | ||
+ | <math> = \sum_{n=0}^{+\infty} -3^n z^{-n-1} = \sum_{n=-\infty}^{+\infty} -3^n u[n] z^{-n-1} </math> | ||
+ | NOTE: Let k=n+1 | ||
+ | <math> = \sum_{n=-\infty}^{+\infty} -3^{k-1} u[k-1] z^{-k}</math> ('''compare with''' <math>\sum_{n=-\infty}^{+\infty} x[n] z^{-k}</math>) | ||
+ | |||
+ | Therefore, <math> x[n]= -3^{n-1} u[n-1] </math> | ||
+ | |||
+ | :<span style="color:blue"> Grader's comment: Correct Answer </span> | ||
=== Answer 2=== | === Answer 2=== | ||
− | + | <math> X(z) = \frac{1}{3-z} = -\frac{1}{z} \frac{1}{1-\frac{3}{z}} </math> | |
+ | |||
+ | <math> X(z) = \frac{1}{z} \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} = \sum_{n=-\infty}^{+\infty} -u[n]3^{n}z^{-n-1} </math> | ||
+ | |||
+ | Let k = n+1, so n = k-1 | ||
+ | |||
+ | <math> X(z) = \sum_{n=-\infty}^{+\infty} -u[k-1]3^{k-1}z^{-k} </math> | ||
+ | |||
+ | By comparison with the z-transform equation | ||
+ | |||
+ | <math> x[n] = -u[n-1] 3^{n-1} </math> | ||
+ | |||
+ | :<span style="color:blue"> Grader's comment: Correct Answer </span> | ||
+ | |||
===Answer 3=== | ===Answer 3=== | ||
− | + | ||
+ | <math> X(z) = \frac{1}{3-z} = -\frac{1}{z} \frac{1}{1-\frac{3}{z}} </math> | ||
+ | |||
+ | <math> X(z) = \frac{1}{z} \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} = \sum_{n=0}^{+\infty} (-3)^{n}({z})^{-n-1} = \sum_{n=-\infty}^{+\infty} -u[n]3^{n}z^{-(n+1)} </math> | ||
+ | |||
+ | Let k = n+1 then n = k-1 | ||
+ | |||
+ | <math> X(z) = \sum_{n=-\infty}^{+\infty} -u[k-1]3^{k-1}z^{-k} </math> | ||
+ | |||
+ | By comparison, | ||
+ | |||
+ | <math> x[n] = -u[n-1] 3^{n-1} </math> | ||
+ | |||
+ | :<span style="color:blue"> Grader's comment: Correct Answer </span> | ||
+ | |||
===Answer 4=== | ===Answer 4=== | ||
− | + | <math>X(Z) = \frac{1}{3-Z}</math> | |
+ | |||
+ | <math>X(Z) = \frac{-1}{Z} \frac{1}{1-\frac{3}{Z}}</math> | ||
+ | |||
+ | Since,<math>|3|<Z</math> | ||
+ | |||
+ | <math>\frac{1}{1-\frac{3}{Z}} = \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n}</math> | ||
+ | |||
+ | Thus, | ||
+ | |||
+ | <math>X(Z) = \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n}</math> | ||
+ | |||
+ | <math>X(Z) = -Z^{-1} \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n}</math> | ||
+ | |||
+ | <math>X(Z) = -\sum_{n=-\infty}^{+\infty} u[n] 3^{n}Z^{-n-1}</math> | ||
+ | |||
+ | Let k=n+1, then -k=-n-1,n=k-1 | ||
+ | |||
+ | <math>X(Z) = -\sum_{n=-\infty}^{+\infty} u[k-1] 3^{k-1}Z^{-k}</math> | ||
+ | |||
+ | Therefore, <math>x(n) = -u[n-1] 3^{n-1}</math> | ||
+ | |||
+ | :<span style="color:blue"> Grader's comment: Correct Answer </span> | ||
+ | |||
+ | === Answer 5 === | ||
+ | By Yixiang Liu | ||
+ | |||
+ | <math>X(z) = \frac{1}{3-Z}</math> | ||
+ | |||
+ | |||
+ | <math>X(z) = \frac{1}{-z}* \frac{1} {\frac{3}{-Z}+1} </math> :<span style="color:blue"> Grader's comment: Use x instead of convolution symbol </span> | ||
+ | |||
+ | <math>X(z) = \frac{1}{-z}* \frac{1} {1-\frac{3}{Z}} </math> | ||
+ | |||
+ | <math>X(z) = \frac{1}{-Z} \sum_{n=0}^{+\infty} (\frac{3}{3})^n </math> | ||
+ | |||
+ | <math>X(z) = \sum_{n=0}^{+\infty} 3^{n} Z^{-n-1} </math> | ||
+ | |||
+ | <math>X(z) = \sum_{-\infty}^{+\infty} u[n] 3^{n} Z^{-n-1} </math> | ||
+ | |||
+ | Let -k = -n-1, so n = k-1 | ||
+ | |||
+ | <math>X(z) = \sum_{-\infty}^{+\infty} u[k-1] 3^{k-1} Z^{-k} </math> | ||
+ | |||
+ | By comparison with the x-transform formula | ||
+ | |||
+ | <math>x[n] = 3^{n-1} u[n-1] </math> | ||
+ | |||
+ | :<span style="color:blue"> Grader's comment: Forgot the negative sign </span> | ||
+ | |||
---- | ---- | ||
+ | Answer 6 | ||
+ | |||
+ | <math>X(z) = \frac{1}{3-z}</math> | ||
+ | |||
+ | we have |z| > 3, | ||
+ | |||
+ | <math>X(z) = \frac{1}{3} \frac{1}{1-\frac{z}{3}} </math> | ||
+ | |||
+ | |||
+ | <math>x[n] = (\frac{1}{3})^{-n+1} u[-n] </math> | ||
+ | |||
+ | :<span style="color:blue"> Grader's comment: Wrong approach </span> | ||
+ | |||
+ | |||
---- | ---- | ||
+ | |||
+ | Answer 7 | ||
+ | |||
+ | |||
+ | |||
+ | <math>X(z) = \frac{1}{3-z}</math> | ||
+ | |||
+ | <math>X(z) = \frac{1}{\frac{3}{z} z -z}</math> | ||
+ | |||
+ | <math>X(z) = \frac{1}{z} \frac{1}{\frac{3}{z} - 1}</math> | ||
+ | |||
+ | <math>X(z) = \frac{-1}{z} \frac{1}{1-\frac{3}{z}}</math> | ||
+ | |||
+ | <math>X(z) = \frac{-1}{z} \frac{1-\left( \frac{3}{z} \right) ^{n}}{1-\frac{3}{z}}, \quad \text{As n goes to} \quad \infty \text{ since ROC} \quad |z|>3</math> | ||
+ | |||
+ | <math>X(z) = \frac{-1}{z} \sum_{n=0}^{\infty} \left( \frac{3}{z} \right)^{n}</math> | ||
+ | |||
+ | <math>X(z) = \frac{-1}{z} \sum_{n=0}^{\infty} 3^n z^{-n}</math> | ||
+ | |||
+ | <math>X(z) = -\sum_{n=0}^{\infty} 3^n z^{-\left(n+1 \right)}, \quad \text{Replace n+1 with k}</math> | ||
+ | |||
+ | <math>X(z) = -\sum_{k=1}^{\infty} 3^{k-1} z^{-k}</math> | ||
+ | |||
+ | <math>X(z) = -\sum_{-\infty}^{\infty} 3^{k-1} u[k-1] z^{-k}, \quad \text{By comparing with DTFT equation we get}</math> | ||
+ | |||
+ | <math>x[n] = -3^{\left( n-1 \right) } u[n-1]</math> | ||
+ | |||
+ | :<span style="color:blue"> Grader's comment: Correct Answer </span> | ||
+ | |||
+ | === Answer 5 === | ||
+ | |||
+ | <math>X(z) = \frac{1}{\frac{3z}{z}-z}</math> | ||
+ | |||
+ | <math> = -(\frac{1}{z})(\frac{1}{1-\frac{3}{z}})</math> | ||
+ | |||
+ | Using the geometric series formula, | ||
+ | |||
+ | <math> = -(\frac{1}{z})\sum_{n=0}^{\infty}(\frac{3}{z})^n</math> | ||
+ | |||
+ | <math> = -(\frac{1}{z})\sum_{n=0}^{\infty}3^nz^{-n}</math> | ||
+ | |||
+ | <math> = \sum_{n=-\infty}^{\infty}-u[n]3^nz^{-n-1}</math> | ||
+ | |||
+ | Let k = n+1, | ||
+ | |||
+ | <math> = \sum_{k=-\infty}^{\infty}-u[k-1]3^{k-1}z^{-k}</math> | ||
+ | |||
+ | By comparison with the Z-transform formula, | ||
+ | |||
+ | x[n] = -u[n-1]3<sup>n-1</sup> | ||
+ | |||
+ | :<span style="color:blue"> Grader's comment: Correct Answer </span> | ||
+ | |||
+ | |||
+ | === Answer 9 === | ||
+ | |||
+ | <math>X(z)=\frac{1}{3-Z}</math> | ||
+ | |||
+ | <math> = \frac{-1}{z}*\frac{1}{1-\frac{3}{z}}</math> | ||
+ | |||
+ | <math> = \frac{-1}{z}\sum_{n=0}^{\infty}(\frac{3}{z})^n </math> | ||
+ | |||
+ | <math> = \sum_{n=0}^{\infty}(-z^{-1})(3z^{-1})^n </math> | ||
+ | |||
+ | <math> = \sum_{n=0}^{\infty}-3^n z^{-n-1}</math> | ||
+ | |||
+ | <math> = \sum_{n=-\infty}^{\infty}-3^n u[n] z^{-n-1}</math> | ||
+ | |||
+ | substituting n+1 with k: | ||
+ | |||
+ | <math> = \sum_{n=-\infty}^{\infty}-u[k-1]3^{k-1}z^{-k}</math> | ||
+ | |||
+ | Comparing to common Z-transform tables: | ||
+ | |||
+ | x[n] = (−3^(n−1))u[n−1] | ||
+ | |||
+ | :<span style="color:blue"> Grader's comment: Correct Answer </span> | ||
+ | |||
+ | -------------------------------------------------------------------------------------- | ||
+ | |||
+ | |||
[[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 12:54, 26 November 2013
Practice Question on "Digital Signal Processing"
Topic: Computing an inverse z-transform
Contents
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.)
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{1}{\frac{3z}{z} - z} = \frac{-1}{z} \frac{1}{1-\frac{3}{z}} $
$ =\frac{-1}{z} \sum_{n=0}^{+\infty} (\frac{3}{z})^n = \sum_{n=0}^{+\infty} (-z^{-1}) (3z^{-1})^n $ NOTE: $ (3z^{-1})^n = (3^n) (z^{-n}) $ $ = \sum_{n=0}^{+\infty} -3^n z^{-n-1} = \sum_{n=-\infty}^{+\infty} -3^n u[n] z^{-n-1} $ NOTE: Let k=n+1
$ = \sum_{n=-\infty}^{+\infty} -3^{k-1} u[k-1] z^{-k} $ (compare with $ \sum_{n=-\infty}^{+\infty} x[n] z^{-k} $)
Therefore, $ x[n]= -3^{n-1} u[n-1] $
- Grader's comment: Correct Answer
Answer 2
$ X(z) = \frac{1}{3-z} = -\frac{1}{z} \frac{1}{1-\frac{3}{z}} $
$ X(z) = \frac{1}{z} \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} = \sum_{n=-\infty}^{+\infty} -u[n]3^{n}z^{-n-1} $
Let k = n+1, so n = k-1
$ X(z) = \sum_{n=-\infty}^{+\infty} -u[k-1]3^{k-1}z^{-k} $
By comparison with the z-transform equation
$ x[n] = -u[n-1] 3^{n-1} $
- Grader's comment: Correct Answer
Answer 3
$ X(z) = \frac{1}{3-z} = -\frac{1}{z} \frac{1}{1-\frac{3}{z}} $
$ X(z) = \frac{1}{z} \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} = \sum_{n=0}^{+\infty} (-3)^{n}({z})^{-n-1} = \sum_{n=-\infty}^{+\infty} -u[n]3^{n}z^{-(n+1)} $
Let k = n+1 then n = k-1
$ X(z) = \sum_{n=-\infty}^{+\infty} -u[k-1]3^{k-1}z^{-k} $
By comparison,
$ x[n] = -u[n-1] 3^{n-1} $
- Grader's comment: Correct Answer
Answer 4
$ X(Z) = \frac{1}{3-Z} $
$ X(Z) = \frac{-1}{Z} \frac{1}{1-\frac{3}{Z}} $
Since,$ |3|<Z $
$ \frac{1}{1-\frac{3}{Z}} = \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n} $
Thus,
$ X(Z) = \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n} $
$ X(Z) = -Z^{-1} \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n} $
$ X(Z) = -\sum_{n=-\infty}^{+\infty} u[n] 3^{n}Z^{-n-1} $
Let k=n+1, then -k=-n-1,n=k-1
$ X(Z) = -\sum_{n=-\infty}^{+\infty} u[k-1] 3^{k-1}Z^{-k} $
Therefore, $ x(n) = -u[n-1] 3^{n-1} $
- Grader's comment: Correct Answer
Answer 5
By Yixiang Liu
$ X(z) = \frac{1}{3-Z} $
$ X(z) = \frac{1}{-z}* \frac{1} {\frac{3}{-Z}+1} $ : Grader's comment: Use x instead of convolution symbol
$ X(z) = \frac{1}{-z}* \frac{1} {1-\frac{3}{Z}} $
$ X(z) = \frac{1}{-Z} \sum_{n=0}^{+\infty} (\frac{3}{3})^n $
$ X(z) = \sum_{n=0}^{+\infty} 3^{n} Z^{-n-1} $
$ X(z) = \sum_{-\infty}^{+\infty} u[n] 3^{n} Z^{-n-1} $
Let -k = -n-1, so n = k-1
$ X(z) = \sum_{-\infty}^{+\infty} u[k-1] 3^{k-1} Z^{-k} $
By comparison with the x-transform formula
$ x[n] = 3^{n-1} u[n-1] $
- Grader's comment: Forgot the negative sign
Answer 6
$ X(z) = \frac{1}{3-z} $
we have |z| > 3,
$ X(z) = \frac{1}{3} \frac{1}{1-\frac{z}{3}} $
$ x[n] = (\frac{1}{3})^{-n+1} u[-n] $
- Grader's comment: Wrong approach
Answer 7
$ X(z) = \frac{1}{3-z} $
$ X(z) = \frac{1}{\frac{3}{z} z -z} $
$ X(z) = \frac{1}{z} \frac{1}{\frac{3}{z} - 1} $
$ X(z) = \frac{-1}{z} \frac{1}{1-\frac{3}{z}} $
$ X(z) = \frac{-1}{z} \frac{1-\left( \frac{3}{z} \right) ^{n}}{1-\frac{3}{z}}, \quad \text{As n goes to} \quad \infty \text{ since ROC} \quad |z|>3 $
$ X(z) = \frac{-1}{z} \sum_{n=0}^{\infty} \left( \frac{3}{z} \right)^{n} $
$ X(z) = \frac{-1}{z} \sum_{n=0}^{\infty} 3^n z^{-n} $
$ X(z) = -\sum_{n=0}^{\infty} 3^n z^{-\left(n+1 \right)}, \quad \text{Replace n+1 with k} $
$ X(z) = -\sum_{k=1}^{\infty} 3^{k-1} z^{-k} $
$ X(z) = -\sum_{-\infty}^{\infty} 3^{k-1} u[k-1] z^{-k}, \quad \text{By comparing with DTFT equation we get} $
$ x[n] = -3^{\left( n-1 \right) } u[n-1] $
- Grader's comment: Correct Answer
Answer 5
$ X(z) = \frac{1}{\frac{3z}{z}-z} $
$ = -(\frac{1}{z})(\frac{1}{1-\frac{3}{z}}) $
Using the geometric series formula,
$ = -(\frac{1}{z})\sum_{n=0}^{\infty}(\frac{3}{z})^n $
$ = -(\frac{1}{z})\sum_{n=0}^{\infty}3^nz^{-n} $
$ = \sum_{n=-\infty}^{\infty}-u[n]3^nz^{-n-1} $
Let k = n+1,
$ = \sum_{k=-\infty}^{\infty}-u[k-1]3^{k-1}z^{-k} $
By comparison with the Z-transform formula,
x[n] = -u[n-1]3n-1
- Grader's comment: Correct Answer
Answer 9
$ X(z)=\frac{1}{3-Z} $
$ = \frac{-1}{z}*\frac{1}{1-\frac{3}{z}} $
$ = \frac{-1}{z}\sum_{n=0}^{\infty}(\frac{3}{z})^n $
$ = \sum_{n=0}^{\infty}(-z^{-1})(3z^{-1})^n $
$ = \sum_{n=0}^{\infty}-3^n z^{-n-1} $
$ = \sum_{n=-\infty}^{\infty}-3^n u[n] z^{-n-1} $
substituting n+1 with k:
$ = \sum_{n=-\infty}^{\infty}-u[k-1]3^{k-1}z^{-k} $
Comparing to common Z-transform tables:
x[n] = (−3^(n−1))u[n−1]
- Grader's comment: Correct Answer