m |
|||
| (14 intermediate revisions by 4 users not shown) | |||
| Line 1: | Line 1: | ||
| − | + | [[Category:problem solving]] | |
| − | + | [[Category: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 from a student== | ||
| + | Take <math>x[n] = a^n\left( u[n-2]+u[n]\right) </math>. We then have | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | X(z) &= \sum_{n=-\infty}^\infty x[n]z^{-n} \text{ (by definition of the z-transform)},\\ | ||
| + | &= \sum_{n=-\infty}^\infty a^n(u[n-2]+u[n])z^{-n}, \\ | ||
| + | &= \sum_{n=2}^\infty a^n(z^{-n}) + \sum_{n=0}^\infty a^n(z^{-n}). \\ | ||
| + | \text{Now let }k=-n, \\ | ||
| + | \Rightarrow X(z) &= \sum_{k=-2}^\infty (a/z)^n + \sum_{k=0}^\infty (a/z)^n ,\\ | ||
| + | &=\sum_{k=0}^\infty \left( (a/z)^n + 2)\right) + \sum_{k=0}^\infty \left( \frac{a}{z}\right)^n \\ | ||
| + | & = \left(\frac{1}{1-a/z}+2\right) + \left(\frac{1}{1-a/z}\right), \\ | ||
| + | & = \frac{z}{z-a}+2 + \frac{z}{z-a}, \\ | ||
| + | & = \frac{z}{z-a}+2\frac{z-a}{z-a} + \frac{z}{z-a} , \\ | ||
| + | & = \frac{4z-2a}{z-a}, \\ | ||
| + | & = \frac{4-2a/z}{1-a/z}, \text{ for } |z|<a ??? | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | So if I end up with something that says 1/1-(1/z), I am confused. does it converge when |z|>a or when |z|<a? | ||
:~ksoong | :~ksoong | ||
| + | ---- | ||
| − | + | ==Comments/corrections from [[user:mboutin|Prof. Mimi]]== | |
| − | + | Take <math>x[n] = a^n(u[n-2]+u[n]) </math>. We then have | |
| − | + | <math> | |
| − | + | \begin{align} | |
| − | + | X(z) &= \sum_{n=-\infty}^\infty x[n]z^{-n} \text{ (by definition of the z-transform)}, {\color{OliveGreen}\surd}\\ | |
| − | + | &= \sum_{n=-\infty}^\infty a^n(u[n-2]+u[n])z^{-n}, {\color{OliveGreen}\surd} \\ | |
| − | + | &= \sum_{n=2}^\infty a^n(z^{-n}) + \sum_{n=0}^\infty a^n(z^{-n}). {\color{OliveGreen}\surd} \\ | |
| + | \text{Now let }k=-n,& {\color{red}\text{This change of variable is not useful, unfortunately.}} \\ | ||
| + | \Rightarrow X(z) &= \sum_{k=-2}^{\color{red}-\infty} (a/z)^{\color{red}n} + \sum_{k=0}^\infty (a/z)^{\color{red}n} ,{\color{red}\text{The terms inside the summation contain n, but the summation is over k.}} \\ | ||
| + | &=\sum_{k=0}^{\color{red}-\infty} \left( (a/z)^n {\color{red} -(a/z)^{-2}-(a/z)^{-1}} )\right) + \sum_{k=0}^\infty \left( \frac{a}{z}\right)^n \\ | ||
| + | & = \left(\frac{1}{1-a/z}+{\color{red} -(a/z)^{-2}-(a/z)^{-1}} )\right) + \left(\frac{1}{1-a/z}\right), {\color{red}\text{For this last step, you need to assume } \left| \frac{a}{z}\right|<1, \text{ else both sums diverge.}} | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | <span style="color:green"> The answer to your initial question ("if I end up with something that says 1/1-(1/z), I am confused. does it converge when |z|>a or when |z|<a?") is in the last step. As you can see from this step, X(z) only converges if | ||
| + | |a|<|z|. Note that, since a could be a complex number, it is important not to say a< |z|</span> | ||
---- | ---- | ||
| − | + | ||
| + | A good way to check your answer is to use the Z-transform table. You can use the time shift property on the first term (a^n *u[n-2]) and the second term (a^n *u[n]) can be directly converted using the table. Your final answer should match up with what you ended with above. -[[User:sbiddand|Sbiddand]] | ||
| + | |||
---- | ---- | ||
| + | |||
| + | Anybody see anything else? Do you have more questions? Comments? Please feel free to add below. | ||
| + | |||
| + | ---- | ||
| + | What if we find a mistake in our computation of the Z transform in the last homework? Do we work with the wrong Z transform and end up with something that isn't the x[n] we chose for that question on homework 2 or do we take the correct Z transform for that x[n] that we chose in homework 2 and then calculate the inverse Z transform? | ||
| + | |||
| + | VG | ||
| + | |||
| + | :No, use the correct answer instead. (I recommend checking all your answers using a table of z-transforms). --[[User:Mboutin|Mboutin]] 19:26, 14 September 2010 (UTC) | ||
| + | ---- | ||
| + | [[2010_Fall_ECE_438_Boutin|Back to ECE438, Fall 2010, Prof. Boutin]] | ||
| + | |||
| + | [[ECE438|Back to ECE438]] | ||
Latest revision as of 11:44, 26 November 2013
Practice Question on "Digital Signal Processing"
Topic: Computing an inverse z-transform
Question from a student
Take $ x[n] = a^n\left( u[n-2]+u[n]\right) $. We then have
$ \begin{align} X(z) &= \sum_{n=-\infty}^\infty x[n]z^{-n} \text{ (by definition of the z-transform)},\\ &= \sum_{n=-\infty}^\infty a^n(u[n-2]+u[n])z^{-n}, \\ &= \sum_{n=2}^\infty a^n(z^{-n}) + \sum_{n=0}^\infty a^n(z^{-n}). \\ \text{Now let }k=-n, \\ \Rightarrow X(z) &= \sum_{k=-2}^\infty (a/z)^n + \sum_{k=0}^\infty (a/z)^n ,\\ &=\sum_{k=0}^\infty \left( (a/z)^n + 2)\right) + \sum_{k=0}^\infty \left( \frac{a}{z}\right)^n \\ & = \left(\frac{1}{1-a/z}+2\right) + \left(\frac{1}{1-a/z}\right), \\ & = \frac{z}{z-a}+2 + \frac{z}{z-a}, \\ & = \frac{z}{z-a}+2\frac{z-a}{z-a} + \frac{z}{z-a} , \\ & = \frac{4z-2a}{z-a}, \\ & = \frac{4-2a/z}{1-a/z}, \text{ for } |z|<a ??? \end{align} $
So if I end up with something that says 1/1-(1/z), I am confused. does it converge when |z|>a or when |z|<a?
- ~ksoong
Comments/corrections from Prof. Mimi
Take $ x[n] = a^n(u[n-2]+u[n]) $. We then have $ \begin{align} X(z) &= \sum_{n=-\infty}^\infty x[n]z^{-n} \text{ (by definition of the z-transform)}, {\color{OliveGreen}\surd}\\ &= \sum_{n=-\infty}^\infty a^n(u[n-2]+u[n])z^{-n}, {\color{OliveGreen}\surd} \\ &= \sum_{n=2}^\infty a^n(z^{-n}) + \sum_{n=0}^\infty a^n(z^{-n}). {\color{OliveGreen}\surd} \\ \text{Now let }k=-n,& {\color{red}\text{This change of variable is not useful, unfortunately.}} \\ \Rightarrow X(z) &= \sum_{k=-2}^{\color{red}-\infty} (a/z)^{\color{red}n} + \sum_{k=0}^\infty (a/z)^{\color{red}n} ,{\color{red}\text{The terms inside the summation contain n, but the summation is over k.}} \\ &=\sum_{k=0}^{\color{red}-\infty} \left( (a/z)^n {\color{red} -(a/z)^{-2}-(a/z)^{-1}} )\right) + \sum_{k=0}^\infty \left( \frac{a}{z}\right)^n \\ & = \left(\frac{1}{1-a/z}+{\color{red} -(a/z)^{-2}-(a/z)^{-1}} )\right) + \left(\frac{1}{1-a/z}\right), {\color{red}\text{For this last step, you need to assume } \left| \frac{a}{z}\right|<1, \text{ else both sums diverge.}} \end{align} $
The answer to your initial question ("if I end up with something that says 1/1-(1/z), I am confused. does it converge when |z|>a or when |z|<a?") is in the last step. As you can see from this step, X(z) only converges if |a|<|z|. Note that, since a could be a complex number, it is important not to say a< |z|
A good way to check your answer is to use the Z-transform table. You can use the time shift property on the first term (a^n *u[n-2]) and the second term (a^n *u[n]) can be directly converted using the table. Your final answer should match up with what you ended with above. -Sbiddand
Anybody see anything else? Do you have more questions? Comments? Please feel free to add below.
What if we find a mistake in our computation of the Z transform in the last homework? Do we work with the wrong Z transform and end up with something that isn't the x[n] we chose for that question on homework 2 or do we take the correct Z transform for that x[n] that we chose in homework 2 and then calculate the inverse Z transform?
VG
- No, use the correct answer instead. (I recommend checking all your answers using a table of z-transforms). --Mboutin 19:26, 14 September 2010 (UTC)
