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[[Category:inverse z-transform]] | [[Category:inverse z-transform]] | ||
| − | = [[ | + | <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 | ||
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<math>X(Z) = -\frac{\frac{1}{3}}{1-\frac{Z}{3}} - \frac{1}{Z} \frac{1}{1-\frac{2}{Z}}</math> | <math>X(Z) = -\frac{\frac{1}{3}}{1-\frac{Z}{3}} - \frac{1}{Z} \frac{1}{1-\frac{2}{Z}}</math> | ||
| − | Since | + | Since <math>|2|<Z<|3|</math> |
<math>\frac{1}{1-\frac{2}{Z}} = \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n}</math> | <math>\frac{1}{1-\frac{2}{Z}} = \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n}</math> | ||
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Thus, | Thus, | ||
| − | <math>X(Z) = \sum_{n=0}^{+\infty} (\frac{Z}{3})^{n} + \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n}</math> | + | <math>X(Z) = -\frac{1}{3} \sum_{n=0}^{+\infty} (\frac{Z}{3})^{n} + \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n}</math> |
| + | |||
| + | <math>X(Z) = -\frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} + \frac{-1}{Z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{Z})^{n}</math> | ||
| + | |||
| + | <math>X(Z) = -\frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} -\sum_{n=-\infty}^{+\infty} u[n] 2^{n} Z^{-n-1}</math> | ||
| + | |||
| + | In <math>-\frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n}</math>, Let k=-n, then -k=n | ||
| + | |||
| + | In <math>\frac{-1}{Z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{Z})^{n}</math>, Let i=n+1, then n=i-1 | ||
| + | |||
| + | <math>-\sum_{n=-\infty}^{+\infty} u[n] (\frac{1}{3})^{n+1} Z^{n}-\sum_{n=-\infty}^{+\infty} u[n] 2^{n} Z^{-n-1}</math> | ||
| + | |||
| + | <math>-\sum_{n=-\infty}^{+\infty} u[-k] (\frac{1}{3})^{-k+1} Z^{-k}-\sum_{n=-\infty}^{+\infty} u[i-1] 2^{i-1} Z^{-i}</math> | ||
| + | |||
| + | Therefore, <math>x(n) = -u[-n] 3^{n-1} - u[n-1] 3^{n-1}</math> | ||
| + | |||
| + | |||
| + | :<span style="color:blue"> Grader's comment: Made a mistake in the last step . It should be 2 instead of 3 </span> | ||
| + | |||
| + | |||
| + | ===Answer 2=== | ||
| + | |||
| + | Li-Pang Mo | ||
| + | |||
| + | <math>X(z) =\frac{1}{(3-z)(2-z)} </math> | ||
| + | |||
| + | <math>X(z) =\frac{-1}{3-z} + \frac{1}{2-z} </math> | ||
| + | |||
| + | <math>X(z) =(\frac{-1}{3})(\frac{1}{1-\frac{z}{3}}) + (\frac{-1}{z})(\frac{1}{1-\frac{2}{z}}) </math> | ||
| + | |||
| + | <math> |2|<Z<|3| </math>, which makes <math> \frac{z}{3}<1, \frac{2}{z}<1 </math> | ||
| + | |||
| + | |||
| + | |||
| + | Use geometric series: | ||
| + | |||
| + | <math>X(z) =\frac{-1}{3} \sum_{n=0}^{+\infty} (\frac{z}{3})^{n} + \frac{-1}{z} \sum_{n=0}^{+\infty} (\frac{2}{z})^{n} </math> | ||
| + | |||
| + | <math>X(z) =\frac{-1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{z}{3})^{n} + \frac{-1}{z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{z})^{n} </math> | ||
| + | |||
| + | <math>X(z) =\frac{-1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{z}{3})^{n} + \frac{-1}{z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{z})^{n} </math> | ||
| + | |||
| + | <math>X(z) = -\sum_{n=-\infty}^{+\infty} u[n] (z)^{n} (\frac{1}{3})^{n+1} - \sum_{n=-\infty}^{+\infty} u[n] (2)^{n} (z)^{-n-1} </math> | ||
| + | |||
| + | <math>let p = -n , n = -p, q = n+1 , n = q-1 </math> | ||
| − | |||
| − | + | <math>X(z) = -\sum_{n=-\infty}^{+\infty} u[-p] (z)^{-p} (\frac{1}{3})^{-p+1} - \sum_{n=-\infty}^{+\infty} u[q-1] (2)^{q-1} (z)^{q-1} </math> | |
| − | + | By observation: | |
| − | + | <math>x(n) = -u[-n] (\frac{1}{3})^{-n+1} - u[n-1](2)^{n-1} </math> | |
| + | :<span style="color:blue"> Grader's comment: Answer is Correct </span> | ||
| − | |||
| − | |||
===Answer 3=== | ===Answer 3=== | ||
Write it here. | Write it here. | ||
Latest revision as of 12:55, 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 2<|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
Ruofei
$ X(Z) = \frac{1}{(3-Z) (2-Z)} $
$ X(Z) = -\frac{1}{3-Z} + \frac{1}{2-Z} $
$ X(Z) = -\frac{\frac{1}{3}}{1-\frac{Z}{3}} + \frac{1}{Z} \frac{1}{\frac{2}{Z}-1} $
$ X(Z) = -\frac{\frac{1}{3}}{1-\frac{Z}{3}} - \frac{1}{Z} \frac{1}{1-\frac{2}{Z}} $
Since $ |2|<Z<|3| $
$ \frac{1}{1-\frac{2}{Z}} = \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n} $
$ \frac{1}{1-\frac{Z}{3}} = \sum_{n=0}^{+\infty} (\frac{Z}{3})^{n} $
Thus,
$ X(Z) = -\frac{1}{3} \sum_{n=0}^{+\infty} (\frac{Z}{3})^{n} + \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n} $
$ X(Z) = -\frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} + \frac{-1}{Z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{Z})^{n} $
$ X(Z) = -\frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} -\sum_{n=-\infty}^{+\infty} u[n] 2^{n} Z^{-n-1} $
In $ -\frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} $, Let k=-n, then -k=n
In $ \frac{-1}{Z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{Z})^{n} $, Let i=n+1, then n=i-1
$ -\sum_{n=-\infty}^{+\infty} u[n] (\frac{1}{3})^{n+1} Z^{n}-\sum_{n=-\infty}^{+\infty} u[n] 2^{n} Z^{-n-1} $
$ -\sum_{n=-\infty}^{+\infty} u[-k] (\frac{1}{3})^{-k+1} Z^{-k}-\sum_{n=-\infty}^{+\infty} u[i-1] 2^{i-1} Z^{-i} $
Therefore, $ x(n) = -u[-n] 3^{n-1} - u[n-1] 3^{n-1} $
- Grader's comment: Made a mistake in the last step . It should be 2 instead of 3
Answer 2
Li-Pang Mo
$ X(z) =\frac{1}{(3-z)(2-z)} $
$ X(z) =\frac{-1}{3-z} + \frac{1}{2-z} $
$ X(z) =(\frac{-1}{3})(\frac{1}{1-\frac{z}{3}}) + (\frac{-1}{z})(\frac{1}{1-\frac{2}{z}}) $
$ |2|<Z<|3| $, which makes $ \frac{z}{3}<1, \frac{2}{z}<1 $
Use geometric series:
$ X(z) =\frac{-1}{3} \sum_{n=0}^{+\infty} (\frac{z}{3})^{n} + \frac{-1}{z} \sum_{n=0}^{+\infty} (\frac{2}{z})^{n} $
$ X(z) =\frac{-1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{z}{3})^{n} + \frac{-1}{z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{z})^{n} $
$ X(z) =\frac{-1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{z}{3})^{n} + \frac{-1}{z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{z})^{n} $
$ X(z) = -\sum_{n=-\infty}^{+\infty} u[n] (z)^{n} (\frac{1}{3})^{n+1} - \sum_{n=-\infty}^{+\infty} u[n] (2)^{n} (z)^{-n-1} $
$ let p = -n , n = -p, q = n+1 , n = q-1 $
$ X(z) = -\sum_{n=-\infty}^{+\infty} u[-p] (z)^{-p} (\frac{1}{3})^{-p+1} - \sum_{n=-\infty}^{+\infty} u[q-1] (2)^{q-1} (z)^{q-1} $
By observation:
$ x(n) = -u[-n] (\frac{1}{3})^{-n+1} - u[n-1](2)^{n-1} $
- Grader's comment: Answer is Correct
Answer 3
Write it here.
Answer 4
Write it here.
