| Line 52: | Line 52: | ||
<center><math>A = z</math></center> | <center><math>A = z</math></center> | ||
| − | :<font size=2> | + | :<font size= 2>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)</font size> |
| − | :Using a infinite Geometric sum we can obtain following...</font size> | + | <center><math>X(z)=B\frac{1}{1-A}</math></center> |
| + | |||
| + | :<font size=2>Using a infinite Geometric sum we can obtain following...</font size> | ||
<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> | ||
| Line 72: | Line 74: | ||
:<font size= 2>In this case</font size> | :<font size= 2>In this case</font size> | ||
<center><math>A = \frac{1}{z}</math></center> | <center><math>A = \frac{1}{z}</math></center> | ||
| + | |||
| + | :<font size= 2>Manipulate the given signal</font size> | ||
| + | <center><math>X(z)=\frac{1}{1-z} = \frac{1}{z} \frac{1}{1-(-\frac{1}{z})}</math></center> | ||
:<font size=2>Using a infinite Geometric sum we can obtain</font size> | :<font size=2>Using a infinite Geometric sum we can obtain</font size> | ||
| − | <center><math>X(z) = \sum_{n=0}^{\infty} | + | <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> \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> | ||
| + | |||
| + | :<font size= 2> By comparison with the Z Transform definition, we can determine <math> x[n] </math></font size> | ||
| + | |||
| + | <center><math>x[n] = (-1)^{n-1} u[n-1]</math></center> | ||
| + | |||
| + | Ex. 4 Find the Inverse Z transform of the following signal | ||
| + | |||
| + | <math>X(z)=\frac{1}{1-2z}, \text{ ROC } |2z|>1 </math> | ||
| + | |||
| + | :<font size= 2>Manipulate the given signal</font size> | ||
| + | <center><math>X(z)=\frac{1}{1-2z} = \frac{-1}{2z} \frac{1}{1-\frac{1}{2z}}</math></center> | ||
| + | |||
| + | :<font size=2>Using a infinite Geometric sum we can obtain</font size> | ||
| + | |||
| + | <center><math>X(z) = - \sum_{n=0}^{\infty} (\frac{1}{2z})^{n}\frac{1}{2z} = - \sum_{n=0}^{\infty} z^{-n-1}2^{-n-1} = - \sum_{n=-\infty}^{\infty} z^{-n-1}2^{-n-1} u[n]</math></center> | ||
| + | |||
| + | <center><math> \text{ let } k=n+1, \text{ then } X(z) = - \sum_{k=-\infty}^{\infty} z^{-k} 2^{} u[k-1] = - \sum_{k=-\infty}^{\infty} u[k-1] 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> | ||
| − | <center><math>x[n] = u[n]</math></center> | + | <center><math>x[n] = -u[n-1]</math></center> |
Revision as of 15:26, 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
- 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] $
