InverseZtransform - Rhea

Inverse Z Transform under construction

Introduction

The Z Transform is the generalized version of the DTFT. This is done by replacing $e^{j\omega}$ with $re^{j\omega} = z$ . 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

The R.O.C. 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'.

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

""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