Difference between revisions of "InverseZtransform" - Rhea

Revision as of 15:44, 13 December 2014

Inverse Z Transform *under construction*

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} (-2z)^{-n}(2z)^{-1}= \sum_{n=-\infty}^{\infty} \frac{1}{2}(-2)^{-n}z^{-n-1} u[n]

\text{ let } k=n+1, \text{ then } X(z) = \sum_{k=-\infty}^{\infty} \frac{1}{2}(-2)^{-k+1}u[k-1] z^{-k}

By comparison with the Z Transform definition, we can determine

$ x[n] $

x[n] = \frac{1}{2}(-2)^{-k+1}u[n-1]

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Difference between revisions of "InverseZtransform" - Rhea

Revision as of 15:44, 13 December 2014

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