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<center> | <center> | ||
| − | ==Inverse Z Transform *under construction*== | + | ==Inverse Z Transform *under construction last updated 4:11 am 12/20/14*== |
</center> | </center> | ||
'''Introduction''' | '''Introduction''' | ||
| − | + | The Z Transform is the generalized version of the DTFT. You can obtain the Z Transform from the DTFT by replacing <math>e^{j\omega}</math> with <math> re^{j\omega} </math> which is equivalent to z. The The DTFT is equal to the Z Transform when <math>|z| =1 </math> | |
| − | The Z Transform is the generalized version of the DTFT. You can obtain the Z Transform from the DTFT by replacing <math>e^{j\omega}</math> with <math> re^{j\omega} | + | |
<math> | <math> | ||
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'''Region of Convergence (ROC)''' | '''Region of Convergence (ROC)''' | ||
| − | |||
The ROC determines the region on the Z Plane where the Z Transform converges. The ROC depends solely on the 'r' value that is contained in 'z'. The ROC is always one of three cases; | The ROC determines the region on the Z Plane where the Z Transform converges. The ROC depends solely on the 'r' value that is contained in 'z'. The ROC is always one of three cases; | ||
:1. The ROC starts from a circle centered at the origin and extends outward to infinity | :1. The ROC starts from a circle centered at the origin and extends outward to infinity | ||
:2. The ROC starts from a circle centered at the origin and fills in toward the origin | :2. The ROC starts from a circle centered at the origin and fills in toward the origin | ||
:3. The ROC is the space in-between two circles centered at the origin. | :3. The ROC is the space in-between two circles centered at the origin. | ||
| − | If the ROC includes the unit circle then the DTFT | + | If the ROC includes the unit circle then the DTFT converges for that function if it is not included, then it does not. |
<math> | <math> | ||
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</math> | </math> | ||
| − | + | The ROC is determined when preforming Z transforms and is given when preforming inverse Z transforms. | |
| + | '''Solving an inverse Z Transform''' | ||
| + | To find the Inverse Z transform of signals use manipulation then direct Inversion. Do not use formula directly! | ||
| − | + | The Infinite Geometric Series formula is used in most problems involving Inv. Z transform. | |
| − | + | ||
| − | + | <math> | |
| + | \begin{align} | ||
| + | \text{Infinite Geometric Series: } | ||
| + | X(z) &= \frac{a}{1-r}\\ | ||
| + | &= \sum_{n=0}^{\infty} (a)r^{n} , \text{if } |z| < 1\\ | ||
| + | &= \sum_{n=-\infty}^{\infty} (a)r^{n} u[n]\\ | ||
| + | \end{align} | ||
| + | </math> | ||
| − | + | it can be seen that this general form is already starting to look like that of the Z Transform, with some change of variables we can manipulate this equation to be that of a Z transform and then by comparison find the inverse z transform. | |
| − | + | '''Examples''' | |
| − | + | Ex. 1 Find the Inverse Z transform of the following signal | |
| − | + | ||
| − | + | <math>X(z)=\frac{1}{1-z}, \text{ ROC } |z|<1 </math> | |
| − | + | ||
| − | + | Solution | |
| − | + | ||
| − | + | <math> | |
| − | + | \begin{align} | |
| − | + | X(x) &= \frac{1}{1-z}\\ | |
| − | + | &= \sum_{n=0}^{\infty} 1(z^{n}) \text{ if } |z| < 1\\ | |
| − | + | &= \sum_{n=-\infty}^{\infty} z^{n} u[n]\\ | |
| − | + | &\text{let }k = -n\\ | |
| − | + | &= \sum_{k=-\infty}^{\infty} z^{-k} u[-k]\\ | |
| − | + | &= \sum_{k=-\infty}^{\infty}u[-k] z^{-k}\\ | |
| − | + | &\text{By comparison with the Z Transform definition..}\\ | |
| + | x[n] &= u[-n]\\ | ||
| + | \end{align} | ||
| + | </math> | ||
Ex. 2 Find the Inverse Z transform of the following signal | Ex. 2 Find the Inverse Z transform of the following signal | ||
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<math>X(z)=\frac{1}{1-z}, \text{ ROC } |z|>1 </math> | <math>X(z)=\frac{1}{1-z}, \text{ ROC } |z|>1 </math> | ||
| − | + | Solution | |
| − | + | ||
| − | + | <math> | |
| − | + | \begin{align} | |
| − | + | X(z) &= \frac{1}{1-z} \\ | |
| − | + | &= \frac{1}{z} \frac{1}{1-(-\frac{1}{z})}\\ | |
| − | + | & \text{Using a infinite Geometric series...}\\ | |
| − | + | &= \sum_{n=0}^{\infty} (\frac{-1}{z})^{n}\frac{1}{z} \text{ if } |-\frac{1}{z}| < 1\\ | |
| − | + | &= \sum_{n=0}^{\infty}(-1)^{n} z^{-n-1} \\ | |
| − | + | &= \sum_{n=-\infty}^{\infty} (-1)^{n} z^{-n-1} u[n]\\ | |
| − | + | &\text{ let } k=n+1 \\ | |
| − | + | &= \sum_{k=-\infty}^{\infty}(-1)^{k-1} z^{-k} u[k-1] \\ | |
| − | + | &= \sum_{k=-\infty}^{\infty} (-1)^{k-1} u[k-1] z^{-k}\\ | |
| − | + | &\text{By comparison with the Z Transform definition...}\\ | |
| − | + | x[n] &=(-1)^{n-1} u[n-1]\\ | |
| + | \end{align} | ||
| + | </math> | ||
| − | |||
Ex. 4 Find the Inverse Z transform of the following signal | Ex. 4 Find the Inverse Z transform of the following signal | ||
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<math>X(z)=\frac{1}{1-2z}, \text{ ROC } |2z|>1 </math> | <math>X(z)=\frac{1}{1-2z}, \text{ ROC } |2z|>1 </math> | ||
| − | + | Solution | |
| − | + | <math> | |
| − | + | \begin{align} | |
| − | + | X(z) &= \frac{1}{1-2z}\\ | |
| − | + | &= \frac{1}{2z} \frac{1}{1-(-\frac{1}{2z})}\\ | |
| − | + | &= \sum_{n=0}^{\infty} (\frac{-1}{2z})^{n}\frac{1}{2z} \text{ if } |-\frac{1}{2z}| < 1\\ | |
| − | + | &= \sum_{n=0}^{\infty} (-2z)^{-n}(2z)^{-1}\\ | |
| − | + | &= \sum_{n=-\infty}^{\infty} \frac{1}{2}(-2)^{-n}z^{-n-1} u[n]\\ | |
| − | + | &\text{ let } k=n+1\\ | |
| − | + | &= \sum_{k=-\infty}^{\infty} \frac{1}{2}(-2)^{-k+1}u[k-1] z^{-k}\\ | |
| − | + | &\text{By comparison with the Z Transform definition...}\\ | |
| − | + | x[n] &= \frac{1}{2}(-2)^{-k+1}u[n-1]\\ | |
| + | \end{align} | ||
| + | </math> | ||
Revision as of 22:50, 19 December 2014
Inverse Z Transform *under construction last updated 4:11 am 12/20/14*
Introduction
The Z Transform is the generalized version of the DTFT. You can obtain the Z Transform from the DTFT by replacing $ e^{j\omega} $ with $ re^{j\omega} $ which is equivalent to z. The The DTFT is equal to the Z Transform when $ |z| =1 $
$ \begin{align} \text{DTFT: } X(w) &= \sum_{n=-\infty}^\infty x[n]e^{-j\omega n}\\ \text{Z-Transform: } X(z) &= \sum_{n=-\infty}^\infty x[n]z^{-n}\\ \text{Inv. Z-Transform: } x[n] &= \frac{1}{2\pi j}\oint_{c}X(z)z^{n-1}dz \end{align} $
Region of Convergence (ROC) The ROC determines the region on the Z Plane where the Z Transform converges. The ROC depends solely on the 'r' value that is contained in 'z'. The ROC is always one of three cases;
- 1. The ROC starts from a circle centered at the origin and extends outward to infinity
- 2. The ROC starts from a circle centered at the origin and fills in toward the origin
- 3. The ROC is the space in-between two circles centered at the origin.
If the ROC includes the unit circle then the DTFT converges for that function if it is not included, then it does not.
$ \begin{align} \text{Remember: } z &=re^{j\omega} \end{align} $
The ROC is determined when preforming Z transforms and is given when preforming inverse Z transforms.
Solving an inverse Z Transform To find the Inverse Z transform of signals use manipulation then direct Inversion. Do not use formula directly!
The Infinite Geometric Series formula is used in most problems involving Inv. Z transform.
$ \begin{align} \text{Infinite Geometric Series: } X(z) &= \frac{a}{1-r}\\ &= \sum_{n=0}^{\infty} (a)r^{n} , \text{if } |z| < 1\\ &= \sum_{n=-\infty}^{\infty} (a)r^{n} u[n]\\ \end{align} $
it can be seen that this general form is already starting to look like that of the Z Transform, with some change of variables we can manipulate this equation to be that of a Z transform and then by comparison find the inverse z transform.
Examples
Ex. 1 Find the Inverse Z transform of the following signal
$ X(z)=\frac{1}{1-z}, \text{ ROC } |z|<1 $
Solution
$ \begin{align} X(x) &= \frac{1}{1-z}\\ &= \sum_{n=0}^{\infty} 1(z^{n}) \text{ if } |z| < 1\\ &= \sum_{n=-\infty}^{\infty} z^{n} u[n]\\ &\text{let }k = -n\\ &= \sum_{k=-\infty}^{\infty} z^{-k} u[-k]\\ &= \sum_{k=-\infty}^{\infty}u[-k] z^{-k}\\ &\text{By comparison with the Z Transform definition..}\\ x[n] &= u[-n]\\ \end{align} $
Ex. 2 Find the Inverse Z transform of the following signal
$ X(z)=\frac{1}{1-z}, \text{ ROC } |z|>1 $
Solution
$ \begin{align} X(z) &= \frac{1}{1-z} \\ &= \frac{1}{z} \frac{1}{1-(-\frac{1}{z})}\\ & \text{Using a infinite Geometric series...}\\ &= \sum_{n=0}^{\infty} (\frac{-1}{z})^{n}\frac{1}{z} \text{ if } |-\frac{1}{z}| < 1\\ &= \sum_{n=0}^{\infty}(-1)^{n} z^{-n-1} \\ &= \sum_{n=-\infty}^{\infty} (-1)^{n} z^{-n-1} u[n]\\ &\text{ let } k=n+1 \\ &= \sum_{k=-\infty}^{\infty}(-1)^{k-1} z^{-k} u[k-1] \\ &= \sum_{k=-\infty}^{\infty} (-1)^{k-1} u[k-1] z^{-k}\\ &\text{By comparison with the Z Transform definition...}\\ x[n] &=(-1)^{n-1} u[n-1]\\ \end{align} $
Ex. 4 Find the Inverse Z transform of the following signal
$ X(z)=\frac{1}{1-2z}, \text{ ROC } |2z|>1 $
Solution
$ \begin{align} X(z) &= \frac{1}{1-2z}\\ &= \frac{1}{2z} \frac{1}{1-(-\frac{1}{2z})}\\ &= \sum_{n=0}^{\infty} (\frac{-1}{2z})^{n}\frac{1}{2z} \text{ if } |-\frac{1}{2z}| < 1\\ &= \sum_{n=0}^{\infty} (-2z)^{-n}(2z)^{-1}\\ &= \sum_{n=-\infty}^{\infty} \frac{1}{2}(-2)^{-n}z^{-n-1} u[n]\\ &\text{ let } k=n+1\\ &= \sum_{k=-\infty}^{\infty} \frac{1}{2}(-2)^{-k+1}u[k-1] z^{-k}\\ &\text{By comparison with the Z Transform definition...}\\ x[n] &= \frac{1}{2}(-2)^{-k+1}u[n-1]\\ \end{align} $
