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[[Category:signal processing]] | [[Category:signal processing]] | ||
| − | <center | + | <center> |
| − | Inverse Z Transform *under construction* | + | ==Inverse Z Transform *under construction*== |
| − | + | </center> | |
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| − | + | '''Introduction''' | |
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| − | + | ||
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| − | + | ||
| − | < | + | The Z Transform is the generalized version of the DTFT. This is done by replacing <math>e^{j\omega}</math> with <math> re^{j\omega} = z </math>. The DTFT is equal to the Z Transform when <math>|z| =1 </math> |
| − | + | ||
| − | </ | + | |
| − | < | + | <math> |
| − | : | + | \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} | ||
| + | </math> | ||
| + | |||
| + | '''Region of Convergence''' | ||
| + | |||
| + | The R.O.C. 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'. | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | \text{Remember: } | ||
| + | z &=re^{j\omega} | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | ""II. Example Problems of the Inverse Z Transform"" | ||
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:We will find the Inverse Z transform of various signals by manipulation and then using direct Inversion. | :We will find the Inverse Z transform of various signals by manipulation and then using direct Inversion. | ||
:On the first example we will go slowly over each step. | :On the first example we will go slowly over each step. | ||
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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> | ||
| − | :: | + | ::note: It is important to realize that we are not going to try to use the direct formula for an inverse Z transform, Instead our approach will be to manipulate the signal so that we can directly compare it with the Z transform equation and by inspection obtain the Inverse Z transform. |
| − | : | + | :First we need to manipulate the given ROC inequality to be in the following form, with 'A' being some expression that contains z |
<center><math>|A| < 1</math></center> | <center><math>|A| < 1</math></center> | ||
| − | : | + | :In this case this is already satisfied with |
<center><math>A = z</math></center> | <center><math>A = z</math></center> | ||
| − | : | + | :Then we need to manipulate the given signal to be in the following form, B is just some expression that is the result of adjusting the equation (in this case B = 1) |
<center><math>X(z)=B\frac{1}{1-A}</math></center> | <center><math>X(z)=B\frac{1}{1-A}</math></center> | ||
| − | : | + | :Using a infinite Geometric sum we can obtain following... |
<center><math>X(z) = \sum_{n=0}^{\infty} 1(z^{n}) = \sum_{n=-\infty}^{\infty} z^{n} u[n]</math></center> | <center><math>X(z) = \sum_{n=0}^{\infty} 1(z^{n}) = \sum_{n=-\infty}^{\infty} z^{n} u[n]</math></center> | ||
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<center><math>\text{let } k = -n \text{ then, } X(z) = \sum_{k=-\infty}^{\infty} z^{-k} u[-k]= \sum_{k=-\infty}^{\infty}u[-k] z^{-k}</math></center> | <center><math>\text{let } k = -n \text{ then, } X(z) = \sum_{k=-\infty}^{\infty} z^{-k} u[-k]= \sum_{k=-\infty}^{\infty}u[-k] z^{-k}</math></center> | ||
| − | : | + | : By comparison with the Z Transform definition, we can determine <math> x[n] </math> |
<center><math>x[n] = u[-n]</math></center> | <center><math>x[n] = u[-n]</math></center> | ||
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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> | ||
| − | : | + | :First we need to manipulate the given ROC inequality to be in the following form, with 'A' being some expression that contains z |
<center><math>|A| < 1</math></center> | <center><math>|A| < 1</math></center> | ||
| − | : | + | :In this case |
<center><math>A = \frac{1}{z}</math></center> | <center><math>A = \frac{1}{z}</math></center> | ||
| − | : | + | :Manipulate the given signal |
<center><math>X(z)=\frac{1}{1-z} = \frac{1}{z} \frac{1}{1-(-\frac{1}{z})}</math></center> | <center><math>X(z)=\frac{1}{1-z} = \frac{1}{z} \frac{1}{1-(-\frac{1}{z})}</math></center> | ||
| − | : | + | :Using a infinite Geometric sum we can obtain |
<center><math>X(z) = \sum_{n=0}^{\infty} (\frac{-1}{z})^{n}\frac{1}{z} = \sum_{n=0}^{\infty}(-1)^{n} z^{-n-1} = \sum_{n=-\infty}^{\infty} (-1)^{n} z^{-n-1} u[n]</math></center> | <center><math>X(z) = \sum_{n=0}^{\infty} (\frac{-1}{z})^{n}\frac{1}{z} = \sum_{n=0}^{\infty}(-1)^{n} z^{-n-1} = \sum_{n=-\infty}^{\infty} (-1)^{n} z^{-n-1} u[n]</math></center> | ||
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<center><math> \text{ let } k=n+1, \text{ then } X(z) = \sum_{k=-\infty}^{\infty}(-1)^{k-1} z^{-k} u[k-1] = \sum_{k=-\infty}^{\infty} (-1)^{k-1} u[k-1] z^{-k}</math></center> | <center><math> \text{ let } k=n+1, \text{ then } X(z) = \sum_{k=-\infty}^{\infty}(-1)^{k-1} z^{-k} u[k-1] = \sum_{k=-\infty}^{\infty} (-1)^{k-1} u[k-1] z^{-k}</math></center> | ||
| − | : | + | :By comparison with the Z Transform definition, we can determine <math> x[n] </math> |
<center><math>x[n] = (-1)^{n-1} u[n-1]</math></center> | <center><math>x[n] = (-1)^{n-1} u[n-1]</math></center> | ||
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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> | ||
| − | : | + | :Manipulate the given signal |
<center><math>X(z)=\frac{1}{1-2z} = \frac{1}{2z} \frac{1}{1-(-\frac{1}{2z})}</math></center> | <center><math>X(z)=\frac{1}{1-2z} = \frac{1}{2z} \frac{1}{1-(-\frac{1}{2z})}</math></center> | ||
| − | : | + | :Using a infinite Geometric sum we can obtain |
<center><math>X(z) = \sum_{n=0}^{\infty} (\frac{-1}{2z})^{n}\frac{1}{2z} = \sum_{n=0}^{\infty} (-2z)^{-n}(2z)^{-1}= \sum_{n=-\infty}^{\infty} \frac{1}{2}(-2)^{-n}z^{-n-1} u[n]</math></center> | <center><math>X(z) = \sum_{n=0}^{\infty} (\frac{-1}{2z})^{n}\frac{1}{2z} = \sum_{n=0}^{\infty} (-2z)^{-n}(2z)^{-1}= \sum_{n=-\infty}^{\infty} \frac{1}{2}(-2)^{-n}z^{-n-1} u[n]</math></center> | ||
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<center><math> \text{ let } k=n+1, \text{ then } X(z) = \sum_{k=-\infty}^{\infty} \frac{1}{2}(-2)^{-k+1}u[k-1] z^{-k}</math></center> | <center><math> \text{ let } k=n+1, \text{ then } X(z) = \sum_{k=-\infty}^{\infty} \frac{1}{2}(-2)^{-k+1}u[k-1] z^{-k}</math></center> | ||
| − | : | + | : By comparison with the Z Transform definition, we can determine <math> x[n] </math> |
<center><math>x[n] = \frac{1}{2}(-2)^{-k+1}u[n-1]</math></center> | <center><math>x[n] = \frac{1}{2}(-2)^{-k+1}u[n-1]</math></center> | ||
Revision as of 14:04, 19 December 2014
Inverse Z Transform *under construction*
Introduction
The Z Transform is the generalized version of the DTFT. This is done by replacing $ e^{j\omega} $ with $ re^{j\omega} = z $. 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
The R.O.C. 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'.
$ \begin{align} \text{Remember: } z &=re^{j\omega} \end{align} $
""II. Example Problems of the Inverse Z Transform""
- We will find the Inverse Z transform of various signals by manipulation and then using direct Inversion.
- On the first example we will go slowly over each step.
Ex. 1 Find the Inverse Z transform of the following signal
$ X(z)=\frac{1}{1-z}, \text{ ROC } |z|<1 $
- note: It is important to realize that we are not going to try to use the direct formula for an inverse Z transform, Instead our approach will be to manipulate the signal so that we can directly compare it with the Z transform equation and by inspection obtain the Inverse Z transform.
- First we need to manipulate the given ROC inequality to be in the following form, with 'A' being some expression that contains z
- In this case this is already satisfied with
- Then we need to manipulate the given signal to be in the following form, B is just some expression that is the result of adjusting the equation (in this case B = 1)
- Using a infinite Geometric sum we can obtain following...
- By comparison with the Z Transform definition, we can determine $ x[n] $
Ex. 2 Find the Inverse Z transform of the following signal
$ X(z)=\frac{1}{1-z}, \text{ ROC } |z|>1 $
- First we need to manipulate the given ROC inequality to be in the following form, with 'A' being some expression that contains z
- In this case
- Manipulate the given signal
- Using a infinite Geometric sum we can obtain
- By comparison with the Z Transform definition, we can determine $ x[n] $
Ex. 4 Find the Inverse Z transform of the following signal
$ X(z)=\frac{1}{1-2z}, \text{ ROC } |2z|>1 $
- Manipulate the given signal
- Using a infinite Geometric sum we can obtain
- By comparison with the Z Transform definition, we can determine $ x[n] $
