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<font size= 3> | <font size= 3> | ||
| − | :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. | ||
| − | Ex. 1 Find the Inverse Z transform of the following signal | + | Ex. 1 Find the Inverse Z transform of the following signal |
<math>X(z)=\frac{1}{1-z}, \text{ ROC } |z|<1 </math> | <math>X(z)=\frac{1}{1-z}, \text{ ROC } |z|<1 </math> | ||
| − | :<font size = 2>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.</font size> | + | ::<font size = 2>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.</font size> |
| − | <font size= 2>First we need to manipulate the given ROC | + | :<font size= 2>First we need to manipulate the given ROC inequality to be in the following form, with 'A' being some expression that contains z</font size> |
<center><math>|A| < 1</math></center> | <center><math>|A| < 1</math></center> | ||
| − | <font size= 2>In this case this is already satisfied with </font size> | + | :<font size= 2>In this case this is already satisfied with </font size> |
<center><math>A = z</math></center> | <center><math>A = z</math></center> | ||
| − | <font size=2>The reason for this is so that we can use the formula for an infinite geometric sum and so that we are operating in the correct ROC (this will change your answer) | + | :<font size=2>The reason for this is so that we can use the formula for an infinite geometric sum and so that we are operating in the correct ROC (this will change your answer) |
| − | Using a infinite Geometric sum we can obtain following...</font size> | + | :Using a infinite Geometric sum we can obtain following...</font size> |
| − | <math>X(z) = \sum_{n=0}^{\infty} 1(z^{n}) = \sum_{n=-\infty}^{\infty} z^{n} u[n]</math> | + | <center><math>X(z) = \sum_{n=0}^{\infty} 1(z^{n}) = \sum_{n=-\infty}^{\infty} z^{n} u[n]</math></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><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> |
| − | <font size= 2> By comparison with the Z Transform definition, we can determine <math> x[n] </math></font size> | + | :<font size= 2> By comparison with the Z Transform definition, we can determine <math> x[n] </math></font size> |
| − | <math>x[n] = u[-n]</math> | + | <center><math>x[n] = u[-n]</math></center> |
| + | |||
| + | Ex. 2 Find the Inverse Z transform of the following signal | ||
| + | |||
| + | <math>X(z)=\frac{1}{1-z}, \text{ ROC } |z|>1 </math> | ||
| + | |||
| + | :<font size= 2>First we need to manipulate the given ROC inequality to be in the following form, with 'A' being some expression that contains z</font size> | ||
| + | <center><math>|A| < 1</math></center> | ||
| + | |||
| + | :<font size= 2>In this case</font size> | ||
| + | <center><math>A = \frac{1}{z}</math></center> | ||
| + | |||
| + | :<font size=2>Using a infinite Geometric sum we can obtain</font size> | ||
| + | |||
| + | <center><math>X(z) = \sum_{n=0}^{\infty} 1(\frac{1}{z})^{n}= \sum_{n=-\infty}^{\infty} (\frac{1}{z})^{n} u[n] = \sum_{n=-\infty}^{\infty} u[n] z^{-n}</math></center> | ||
| + | |||
| + | :<font size= 2> By comparison with the Z Transform definition, we can determine <math> x[n] </math></font size> | ||
| + | |||
| + | <center><math>x[n] = u[n]</math></center> | ||
Revision as of 14:38, 13 December 2014
Inverse Z Transform
Overview
- The purpose of this page is to...
- I. Define the Z Transform and Inverse Z Transform
- II. Provide Example Problems of the Inverse Z Transform
I. Definitions
- Z Transform
$ X(z)=\mathcal{L}(x[n])=\sum_{n=-\infty}^{\infty}x[n]z^{-n} $
- Inverse Z Transform
$ x[n]=\mathcal{L}^{-1}(X(z))=\frac{1}{2\pi j}\oint_{c}X(z)z^{n-1}dz $
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
- The reason for this is so that we can use the formula for an infinite geometric sum and so that we are operating in the correct ROC (this will change your answer)
- 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
- Using a infinite Geometric sum we can obtain
- By comparison with the Z Transform definition, we can determine $ x[n] $
