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Which if we choose the variable <math>\omega</math> becomes:<br /> | Which if we choose the variable <math>\omega</math> becomes:<br /> | ||
<big><math>\,\mathcal{X}(\omega)=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}\,</math></big><br /> | <big><math>\,\mathcal{X}(\omega)=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}\,</math></big><br /> | ||
+ | |||
+ | Taking r=1 for z which gave us <math>z=e^{j\omega}</math> means that we are dealing with a circle which has a radius 1 and is centered around the origin on the z-plane. In other words it is a unit circle on the z-plane. Thus we can conclude that the z-transform of the signal can exist anywhere in the z-plane but the DTFT of the signal can only exist on the unit circle.<br /> | ||
+ | <br /> | ||
+ | |||
+ | ==== 3. Region of Convergence for a z-transform ==== |
Revision as of 09:19, 25 November 2016
Contents
The Z-transform, Shilton Saha
1. Introduction
The z-transform in very simple terms is a mathematical process of going from the discrete time domain to the z domain also known as the complex frequency domain. In the discrete time domain, a signal is usually defined as a sequence of real or complex numbers which is then converted to the z-domain by the process of z-transform. The z-transform is a very useful and important technique, used in areas of signal processing, system design and analysis and control theory.
The formula used to convert a discrete time signal x[n] to X[z] is as follows:
$ X(z) = \sum_{n=-\infty}^{\infty}x[n]z^{-n} \ $
Where x[n] is the discrete time signal and X[z] is the z-transform of the discrete time signal. Now the z-transform comes in two parts. The first part is the formula as shown above and the second part is to define a region of convergence for the z-transform. Both parts are needed for a complete z-transform as a z-transform without a ROC would not be of much help in signal processing. More on the region of convergence will be discussed below. Although the z-transform achieved by directly applying this formula, the inverse z-transform requires some mathematical manipulations that is related to the power series and geometric series. More on this will be discussed in the next sections.
2. Relation to DTFT
There is a very close relation between DTFT and z-transform. Even each of their respective formulas are also quite similar, which is often overlooked. So before going any further, let’s look at the formulas for both the DTFT and the z-transform for a signal x[n]. First let’s look at the DTFT formula:
$ \,\mathcal{X}(\omega)=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}\, $
Which we can also write as:
$ \,\mathcal{X}(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}\, $
Now we write down the z-transform formula:
$ X(z) = \sum_{n=-\infty}^{\infty}x[n]z^{-n} \ $
Now if we denote $ z=re^{j\omega} $, the z-transform formula, with some change of variable can be written as:
$ \,\mathcal{X}(re^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]({{re}^{j\omega}})^{-n} \, $
$ \,\mathcal{X}(re^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]{r^{-n}e^{-j\omega n} }\, $
Now this formula is quite similar to the z-transform formula except for the ‘r’ term. So, if we take r=1, then $ \, \mathcal z=e^{j\omega}\, $. Thus the final DTFT formula can be written as:
$ \,\mathcal{X}(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}\, $
Which if we choose the variable $ \omega $ becomes:
$ \,\mathcal{X}(\omega)=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}\, $
Taking r=1 for z which gave us $ z=e^{j\omega} $ means that we are dealing with a circle which has a radius 1 and is centered around the origin on the z-plane. In other words it is a unit circle on the z-plane. Thus we can conclude that the z-transform of the signal can exist anywhere in the z-plane but the DTFT of the signal can only exist on the unit circle.