| (40 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
| − | + | *<span style="color:green"> Hmm... This looks a lot like my course notes... Perhaps you want to write this somewhere, otherwise one might think that you are pretending that you wrote this yourself. --[[User:Mboutin|Mboutin]] 12:26, 23 September 2009 (UTC) </span> | |
| − | <math> x[n] = \oint_C {X( | + | '''Inverse Z-transform |
| + | ''' | ||
| + | <math> x[n] = \frac{1}{2 \prod j} \oint_C {X(z)} {z ^ {n-1}} dz \ </math> | ||
where C is a closed counterwise countour inside the ROC of the Z- transform and around the origin. | where C is a closed counterwise countour inside the ROC of the Z- transform and around the origin. | ||
| − | <math> = \sum_{poles a_i ( X( | + | <math> = \sum_{poles a_i ( X(z) z ^ {n-1})} Residue ( X(z) z ^ {n-1}) \ </math> |
| − | <math> = \sum_{poles a_i ( X( | + | <math> = \sum_{poles a_i ( X(z) z ^ {n-1})} \ </math> Coefficient of degree (-1) term on the power series expansion of <math> ( X(z) z ^ {n-1}) \ </math> <math> about a_i \ </math> |
| − | So inverting X( | + | So inverting X(z) involves power series. |
| − | |||
| − | <math>\frac{1}{(1-x)} = \sum_{n=0}^\infty X^n \ </math> , geometric series where |X| | + | <math>f(X)= \sum_{n=0}^\infty \frac{f^n (X_0) (X-X_0)^{n}}{n!} \ </math> , near <math>X_0</math> |
| + | |||
| + | <math>\frac{1}{(1-x)} = \sum_{n=0}^\infty X^n \ </math> , geometric series where |X|< 1 | ||
| Line 18: | Line 21: | ||
Computing equivalent to complex integration formula's | Computing equivalent to complex integration formula's | ||
| − | 1) Write X( | + | 1) Write X(z) as a power series. |
| − | <math>X( | + | <math>X(z) = \sum_{n=-\infty}^{\infty} \ C_n z^n \ </math> , series must converge for all z's on the ROC of X(z) |
2) Observe that | 2) Observe that | ||
| − | <math>X( | + | <math>X(z) = \sum_{n=-\infty}^{\infty} \ x[n] z^{-n} \ </math> |
i.e., | i.e., | ||
| − | <math>X( | + | <math>X(z) = \sum_{n=-\infty}^{\infty} \ x[-n] z^n \ </math> |
3) By comparison | 3) By comparison | ||
| Line 36: | Line 39: | ||
or | or | ||
| − | <math>X[n] = \ C_ -n</math> | + | <math>X[n] = \ C_ {-n}</math> |
| + | |||
| + | |||
| + | |||
Example 1: | Example 1: | ||
| + | |||
| + | <math>X(z) = \frac{1}{(1-z)} \ </math> | ||
| + | |||
| + | Two possible ROC | ||
| + | |||
| + | Case 1: |z|<1 | ||
| + | |||
| + | <math>X(z) = \sum_{n=0}^\infty z^n \ </math> | ||
| + | |||
| + | <math> = \sum_{k=-\infty}^{0} \ z^{-k} \ </math> | ||
| + | <math> = \sum_{n=-\infty}^{\infty} \ u(-k) z^{-k} \ </math> | ||
| + | |||
| + | so, x[n]=u[-n] | ||
| + | |||
| + | Consistent as having inside a circle as ROC. | ||
| + | |||
| + | Case 2: |z|>1 | ||
| + | |||
| + | <math>X(z) = \frac{1}{(1-z)} \ </math> | ||
| + | <math> = \frac{1}{z(\frac{1}{z}-1)} \ </math> | ||
| + | <math>= \frac {-1}{z} {\frac{1}{1-\frac{1}{z}}} \ </math> ,observe <math>\ |\frac {1}{z}|< 1 \ </math> | ||
| + | |||
| + | Now by using the geometric series formula, the series can be formed as | ||
| + | |||
| + | <math>= \frac {-1}{z} \sum_{n=0}^\infty (\frac {1}{z})^{n} \ </math> | ||
| + | |||
| + | <math>= - \sum_{n=0}^\infty z^{-n-1} \ </math> | ||
| + | |||
| + | Let k=(n+1) | ||
| + | |||
| + | <math>= - \sum_{k=1}^\infty z^{-k} \ </math> | ||
| + | |||
| + | <math>= \sum_{k=1}^\infty -u(k-1) z^{-k} \ </math> | ||
| + | |||
| + | By coparison with the Z- transform formula | ||
| + | |||
| + | <math>x[n]= -u[n-1] \ </math> | ||
| + | |||
| + | Consistent as having outside of circle as the ROC. | ||
Latest revision as of 07:27, 23 September 2009
- Hmm... This looks a lot like my course notes... Perhaps you want to write this somewhere, otherwise one might think that you are pretending that you wrote this yourself. --Mboutin 12:26, 23 September 2009 (UTC)
Inverse Z-transform
$ x[n] = \frac{1}{2 \prod j} \oint_C {X(z)} {z ^ {n-1}} dz \ $
where C is a closed counterwise countour inside the ROC of the Z- transform and around the origin.
$ = \sum_{poles a_i ( X(z) z ^ {n-1})} Residue ( X(z) z ^ {n-1}) \ $ $ = \sum_{poles a_i ( X(z) z ^ {n-1})} \ $ Coefficient of degree (-1) term on the power series expansion of $ ( X(z) z ^ {n-1}) \ $ $ about a_i \ $
So inverting X(z) involves power series.
$ f(X)= \sum_{n=0}^\infty \frac{f^n (X_0) (X-X_0)^{n}}{n!} \ $ , near $ X_0 $
$ \frac{1}{(1-x)} = \sum_{n=0}^\infty X^n \ $ , geometric series where |X|< 1
Computing equivalent to complex integration formula's
1) Write X(z) as a power series.
$ X(z) = \sum_{n=-\infty}^{\infty} \ C_n z^n \ $ , series must converge for all z's on the ROC of X(z)
2) Observe that
$ X(z) = \sum_{n=-\infty}^{\infty} \ x[n] z^{-n} \ $
i.e.,
$ X(z) = \sum_{n=-\infty}^{\infty} \ x[-n] z^n \ $
3) By comparison
$ X[-n] = \ C_n \ $
or
$ X[n] = \ C_ {-n} $
Example 1:
$ X(z) = \frac{1}{(1-z)} \ $
Two possible ROC
Case 1: |z|<1
$ X(z) = \sum_{n=0}^\infty z^n \ $
$ = \sum_{k=-\infty}^{0} \ z^{-k} \ $ $ = \sum_{n=-\infty}^{\infty} \ u(-k) z^{-k} \ $
so, x[n]=u[-n]
Consistent as having inside a circle as ROC.
Case 2: |z|>1
$ X(z) = \frac{1}{(1-z)} \ $
$ = \frac{1}{z(\frac{1}{z}-1)} \ $ $ = \frac {-1}{z} {\frac{1}{1-\frac{1}{z}}} \ $ ,observe $ \ |\frac {1}{z}|< 1 \ $
Now by using the geometric series formula, the series can be formed as
$ = \frac {-1}{z} \sum_{n=0}^\infty (\frac {1}{z})^{n} \ $
$ = - \sum_{n=0}^\infty z^{-n-1} \ $
Let k=(n+1)
$ = - \sum_{k=1}^\infty z^{-k} \ $
$ = \sum_{k=1}^\infty -u(k-1) z^{-k} \ $
By coparison with the Z- transform formula
$ x[n]= -u[n-1] \ $
Consistent as having outside of circle as the ROC.
