InverseZtransform - Rhea

Inverse Z Transform under construction

Introduction

The Z Transform is the generalized version of the DTFT. You can obtain the Z Transform from the DTFT by replacing $e^{j\omega}$ with $re^{j\omega} = z$ . The 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 (ROC)

The ROC 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'. The ROC is always one of three cases;

1. The ROC starts from a circle centered at the origin and extends outward to infinity

2. The ROC starts from a circle centered at the origin and fills in toward the origin

3. The ROC is the space in-between two circles centered at the origin.

If the ROC includes the unit circle then the DTFT exists for that function if it is not included, then it does not exist.

$\begin{align} \text{Remember: } z &=re^{j\omega} \end{align}$

""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

$ |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]

InverseZtransform - Rhea

Inverse Z Transform *under construction*

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Inverse Z Transform under construction