| (42 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
| − | Inverse Z-transform | + | '''Inverse Z-transform''' |
| − | <math>x[n]= \frac{1}{2\ | + | |
| + | By formula we have, | ||
| + | |||
| + | <math>x[n]= \frac{1}{2\pi j} \oint_C X(z) z^{n-1} dz \ </math> | ||
<math>= \sum_ {poles a_i of X(z) z^{n-1}} \ </math> Residue <math>X(z) z^{n-1} \ </math> | <math>= \sum_ {poles a_i of X(z) z^{n-1}} \ </math> Residue <math>X(z) z^{n-1} \ </math> | ||
| Line 9: | Line 12: | ||
So inverting X(z) involves power series | So inverting X(z) involves power series | ||
| − | <math> f(x)= \sum_ | + | <math> f(x)= \sum_{n =0}^{\infty} \frac {f^{n} x_0 (x-x_0)^{n}} {n!} \ </math> |
| − | <math>\frac{1}{1-x} = \sum_{n = | + | <math>\frac{1}{1-x} = \sum_{n =0}^{\infty} x^{n} \ </math> , geometric series when |x|< 1 |
| + | |||
| + | |||
| + | Shortcut equivalent to complex integration formula's | ||
| + | |||
| + | 1) Write x(z) as a power series. | ||
| + | |||
| + | <math>x(z)= \sum_{n =-\infty}^{\infty} C_n z^n \ </math> ,series must converge for all z's in the ROC of x(z) | ||
| + | |||
| + | 2) Observe that | ||
| + | |||
| + | <math>x(z) = \sum_{n =-\infty}^{\infty} x[n] z^{-n} \ </math> | ||
| + | |||
| + | i.e., | ||
| + | |||
| + | <math>x(z) = \sum_{n =-\infty}^{\infty} x[-n] z^n \ </math> | ||
| + | |||
| + | 3) By comparison | ||
| + | |||
| + | <math>x[-n]= C_n \ </math> | ||
| + | |||
| + | or | ||
| + | |||
| + | <math>x[n]= C_{-n} \ </math> | ||
| + | |||
| + | Example 1: <math>x(z) = \frac{1}{1-z} \ </math> | ||
| + | |||
| + | Two possible ROC's | ||
| + | Case 1: |z|< 1 | ||
| + | |||
| + | <math>x(z)= \sum_{n =0}^{\infty} z^n \ </math> , by the formula of goemetric series | ||
| + | |||
| + | <math>= \sum_{k =-\infty}^{0} z^{-k} \ </math> | ||
| + | |||
| + | <math>= \sum_{k =-\infty}^{\infty} u[-k] z^{-k} \ </math> | ||
| + | |||
| + | So x[n]=u[-n] | ||
| + | |||
| + | Consistent of having inside of 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 | ||
| + | |||
| + | <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 =-\infty}^{\infty} -u[k-1] z^{-k} \ </math> | ||
| + | |||
| + | By comparison with the Z- transform formula we have, | ||
| + | |||
| + | x[n]= -u[n-1] | ||
| + | |||
| + | Consistent with having outside of circle as ROC. | ||
Latest revision as of 08:50, 23 September 2009
Inverse Z-transform
By formula we have,
$ x[n]= \frac{1}{2\pi j} \oint_C X(z) z^{n-1} dz \ $
$ = \sum_ {poles a_i of X(z) z^{n-1}} \ $ Residue $ X(z) z^{n-1} \ $ $ = \sum_ {poles a_i of X(z) z^{n-1}} \ $ Coefficient of degree(-1) term in the power 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!} \ $
$ \frac{1}{1-x} = \sum_{n =0}^{\infty} x^{n} \ $ , geometric series when |x|< 1
Shortcut 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 in 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's Case 1: |z|< 1
$ x(z)= \sum_{n =0}^{\infty} z^n \ $ , by the formula of goemetric series
$ = \sum_{k =-\infty}^{0} z^{-k} \ $
$ = \sum_{k =-\infty}^{\infty} u[-k] z^{-k} \ $
So x[n]=u[-n]
Consistent of having inside of 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
$ = \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 =-\infty}^{\infty} -u[k-1] z^{-k} \ $
By comparison with the Z- transform formula we have,
x[n]= -u[n-1]
Consistent with having outside of circle as ROC.
