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| − | [[ | + | = [[:Category:Problem solving|Practice Question]], [[ECE438]] Fall 2013, [[User:Mboutin|Prof. Boutin]] = |
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| − | + | On computing the inverse z-transform of a discrete-time signal. | |
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---- | ---- | ||
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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.) | ||
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---- | ---- | ||
| − | ==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! | ||
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---- | ---- | ||
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| − | + | === 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> |
| + | |||
| + | <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 |
| − | < | + | <span class="texhtml">''x''[''n''] = ''u''[ − ''n'']( − 3<sup>''n'' − 1</sup> − 2<sup>''n'' − 1</sup>)</span> |
| − | === Answer 2=== | + | |
| − | Kyungjun Kim | + | === Answer 2 === |
| + | |||
| + | Kyungjun Kim | ||
Using a partial fraction expansion, we can change the original equation to | Using a partial fraction expansion, we can change the original equation to | ||
| − | <math>X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math> Where A = 1, B = -1, so we get | + | <math>X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math> Where A = 1, B = -1, so we get |
| − | <math>= -\frac{1}{3-z}-\frac{1}{2-z}</math> | + | <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 | By factoring out 1/3 for the first term, and 1/2 for the second term, we can have both terms in form of | ||
| − | <math> \frac{1}{1-r} </math>, which is equal to <math> \sum_{n=0}^{+\infty} (\frac{1}{r})^n </math> | + | <math> \frac{1}{1-r} </math>, which is equal to <math> \sum_{n=0}^{+\infty} (\frac{1}{r})^n </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}(\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> | ||
| − | Then let k=-n | + | Then 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> | ||
| − | Comparing it with z-transform formula, we can get | + | 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> |
| − | ===Answer 3=== | + | === Answer 3 === |
| − | + | ||
| − | + | By Yeong Ho Lee | |
| − | + | First, using partial fraction we get.. | |
| − | A | + | <math> X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math> |
| − | + | A(2-z) + B(3-z) = 1 | |
| − | let z= | + | let z=2, then B=1 |
| − | + | let z=3, then A=-1 | |
| − | <math>= -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})+\frac{1}{2}(\frac{1}{1-\frac{z}{2}})</math> | + | <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> | ||
| − | <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> | + | <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 | + | 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> | + | <math>= -\frac{1}{3}\sum_{n=0}^{+\infty} 3^{k} z^{-k} +\frac{1}{2}\sum_{n=0}^{+\infty} 2^{k}z^{-k}</math> |
| − | by comparison with z-transfrom formula | + | 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> | ||
| − | ===Answer 4=== | + | === Answer 4 === |
| − | <math>X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math> | + | <math>X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math> |
| − | <math>= -\frac{1}{3-z} - \frac{1}{2-z}</math> | + | <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}(\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>= -\frac{1}{3}\sum_{k=0}^{+\infty} (\frac{z}{3})^k -\frac{1}{2}\sum_{k=0}^{+\infty} (\frac{z}{2})^k</math> | ||
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<math>= \sum_{k=0}^{+\infty}[(-\frac{1}{3})(\frac{1}{3})^k + (-\frac{1}{2})(\frac{1}{2})^k]u[k] * z^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 | + | Substitute k with -n |
<math>= \sum_{n=-\infty}^{+\infty}[(-\frac{1}{3})3^{-n} + (-\frac{1}{2})2^{-n}]u[-n] * z^{-n}</math> | <math>= \sum_{n=-\infty}^{+\infty}[(-\frac{1}{3})3^{-n} + (-\frac{1}{2})2^{-n}]u[-n] * z^{-n}</math> | ||
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Look up Z transform equation on RHEA table and see that X(z) becomes... | 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> | ||
| + | |||
| + | === Answer 5 === | ||
| + | |||
| + | by partical fraction, we get, | ||
| + | |||
| + | <math>X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | [[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]] |
Revision as of 22:32, 19 September 2013
Contents
Practice Question, ECE438 Fall 2013, Prof. Boutin
On computing the inverse z-transform of a discrete-time signal.
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.)
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} $
$ = -\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)
Answer 2
Kyungjun Kim
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
$ = -\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
$ \frac{1}{1-r} $, which is equal to $ \sum_{n=0}^{+\infty} (\frac{1}{r})^n $
$ = -\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)
Answer 3
By Yeong Ho Lee
First, using partial fraction we get..
$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $
A(2-z) + B(3-z) = 1
let z=2, then B=1
let z=3, then A=-1
$ = -\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} $
by comparison with z-transfrom formula
x[n] = − 3n − 1u[ − n] + 2n − 1u[ − n]
x[n] = ( − 3n − 1 + 2n − 1)u[ − n]
Answer 4
$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $
$ = -\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]
Answer 5
by partical fraction, we get,
$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $
$ = -\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]
