| Line 1: | Line 1: | ||
| − | + | = Question about computing the inverse z-transform= | |
| − | + | I'm not sure if I've done this correctly in the first place. I would like your input before I actually attempt to inverse it. 0 | |
| + | '''I tried the LaTeX but it failed miserably. ''' . Maybe it would be easier if you write it out while you are reading my LaTeX fail trying to help me, otherwise I can come in and see you on Monday. | ||
| + | :I fixed it. Now it works. For future references, [http://meta.wikimedia.org/wiki/Help:Displaying_a_formula here] is a good page to bookmark: it explains how to use latex in mediawiki, and gives many examples that you can cut and paste. -pm | ||
| + | ---- | ||
| + | == Computation of inverse z-transform by a student (with question about how to obtain ROC)= | ||
| + | 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)},\\ | ||
| + | &= \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 | |
| + | ---- | ||
| + | ==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, \\ | ||
| + | \Rightarrow X(z) &= \sum_{k=-2}^\infty (a/z)^{\color{red}n} + \sum_{k=0}^\infty (a/z)^{\color{red}n} ,{\color{red}\text{Oops! The terms inside the summation contain n, but the summation is over k.}} \\ | ||
| + | &=\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> | ||
| − | + | To answer 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?"), | |
| − | + | ---- | |
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
Answer from Prof. MImi | Answer from Prof. MImi | ||
Revision as of 11:43, 12 September 2010
Question about computing the inverse z-transform
I'm not sure if I've done this correctly in the first place. I would like your input before I actually attempt to inverse it. 0 I tried the LaTeX but it failed miserably. . Maybe it would be easier if you write it out while you are reading my LaTeX fail trying to help me, otherwise I can come in and see you on Monday.
- I fixed it. Now it works. For future references, here is a good page to bookmark: it explains how to use latex in mediawiki, and gives many examples that you can cut and paste. -pm
= Computation of inverse z-transform by a student (with question about how to obtain ROC)
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)},\\ &= \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, \\ \Rightarrow X(z) &= \sum_{k=-2}^\infty (a/z)^{\color{red}n} + \sum_{k=0}^\infty (a/z)^{\color{red}n} ,{\color{red}\text{Oops! The terms inside the summation contain n, but the summation is over k.}} \\ &=\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} $
To answer 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?"),
Answer from Prof. MImi
- In the step where you replaced -n by k, you forgot to replace the n inside the summation. Also, the first sum should then go from -2 to -infinity, instead of infinity.
- Actually, I do not see why you replaced -n by k in both sums. IN the first sum, you should have set k=n-2. In the second sum, you did not need to make any change of variable.
- The arrow in the middle of your computations, and the one towards the end should both be replaced by equal signs.
- The simplification of the first summation following the arrow is incorrect: you would need to add two terms instead of just one.
- The equality following the arrow is only valid when |z|>|a|.You must write this next to the equality!
- This explanation would be much clearer if you had typed in your answer: this way I could make notes directly inside the computations and cross-out and replace stuff using different colors.
Anybody sees anything else? Do you have more questions?
