Difference between revisions of "InverseZtransform" - Rhea

Revision as of 16:26, 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

$ |A| < 1 $

In this case this is already satisfied with

$ A = z $

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)

X(z)=B\frac{1}{1-A}

Using a infinite Geometric sum we can obtain following...

X(z) = \sum_{n=0}^{\infty} 1(z^{n}) = \sum_{n=-\infty}^{\infty} z^{n} u[n]

\text{let } k = -n \text{ then, } X(z) = \sum_{k=-\infty}^{\infty} z^{-k} u[-k]= \sum_{k=-\infty}^{\infty}u[-k] z^{-k}

By comparison with the Z Transform definition, we can determine

$ x[n] $

$ x[n] = u[-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

$ |A| < 1 $

In this case

A = \frac{1}{z}

Manipulate the given signal

X(z)=\frac{1}{1-z} = \frac{1}{z} \frac{1}{1-(-\frac{1}{z})}

Using a infinite Geometric sum we can obtain

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]

\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}

By comparison with the Z Transform definition, we can determine

$ x[n] $

x[n] = (-1)^{n-1} u[n-1]

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

X(z)=\frac{1}{1-2z} = \frac{-1}{2z} \frac{1}{1-\frac{1}{2z}}

Using a infinite Geometric sum we can obtain

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]

\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}

By comparison with the Z Transform definition, we can determine

$ x[n] $

$ x[n] = -u[n-1] $

@@ Line 52: / Line 52: @@
 <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>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>
@@ Line 72: / Line 74: @@
 :<font size= 2>In this case</font size>
 <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>
-<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>
+<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>
-<center><math>x[n] = u[n]</math></center>
+<center><math>x[n] = -u[n-1]</math></center>

Difference between revisions of "InverseZtransform" - Rhea

Revision as of 16:26, 13 December 2014

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