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:(c) H(z) must have more poles than zeros. | :(c) H(z) must have more poles than zeros. | ||

: One possible reasoning: Assuming that a causal system has the ROC defined in (a), then assuming H(z) has more zeros than poles would be a contradiction, since H(z) can could then be decomposed by long division or partial fractions into a function of z that diverges in the region |z| > z<sub>0</sub> and <math>\infty</math>. Simply taking the limit of H(z) as z approaches <math>\infty</math> would show a contradiction since the ROC would not include z = <math>\infty</math>. | : One possible reasoning: Assuming that a causal system has the ROC defined in (a), then assuming H(z) has more zeros than poles would be a contradiction, since H(z) can could then be decomposed by long division or partial fractions into a function of z that diverges in the region |z| > z<sub>0</sub> and <math>\infty</math>. Simply taking the limit of H(z) as z approaches <math>\infty</math> would show a contradiction since the ROC would not include z = <math>\infty</math>. | ||

+ | |||

+ | ==Z Transform and Digital Filters== | ||

+ | The z-transform properties above can also give insight into characteristics of FIR and IIR filters | ||

+ | :(a) FIR filters: | ||

+ | ::(i) since h[n] is of finite duration, it has no poles except possibly z = 0 or z = <math>\infty</math> | ||

+ | ::(ii) The ROC of H(z) includes the entire z-plane besides possibly z = 0 or z = <math>\infty</math> | ||

+ | (from (b) above) | ||

+ | :(b) IIR filters: | ||

+ | ::(i) H(z) has non-zero poles in the finite z-plane | ||

+ | ::(ii) The denominator of H(z) must have a degree > 1 | ||

+ | ::Reasoning: the ROC of H(z) for a neither sided signal is a ring, implying that there must be two non-zero poles to bound the ROC. |

## Latest revision as of 00:57, 2 December 2019

## Contents

# Z Transform and LTI System Properties Study Guide

## Introduction

This page will go over some of my conclusions about properties of the z-transform and discuss some examples of how they may be used to draw conclusions about LTI systems and digital filters. This topic assumes some basic knowledge of the z-transform and signal processing, however some definitions are provided below as a refresher or for reference. The information provided is intended to clarify or augment some of the z-transform properties presented in class. Some of the information includes statements or arguments to validate or clarify claims; these are intended to aid understanding but not substitute for a proof.

## definitions

z-transform: $ X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} $ z ∈ ℂ

DTFT: $ X(\omega) = \sum_{n=-\infty}^{\infty} x[n]e^{-j\omega n} $

right sided: $ x[n] $ such that $ x[n] = 0 $ ∀ n < n_{0}

left sided: $ x[n] $ such that $ x[n] = 0 $ ∀ n > n_{0}

both sided: $ x[n] $ such that $ x[n] = 0 $ ∀ n > n_{1} and n < n_{2} ,n_{1} < n_{2}

neither sided: $ x[n] $ has finite duration, meaning $ x[n] = 0 $ ∀ n < n_{1} and n > n_{2}, n_{1} < n_{2}

Absolutely Summable: $ x[n] $ is absolutely summable if: $ \sum_{n=-\infty}^{\infty} |x[n]| = c $, where $ c $ is a constant.

z-plane: the complex plane.

## Relationship Between Fourier Transform and Z Transform

In the z-transform definition above, z is any complex number, which can be represented in polar coordinates by $ z = re^{j\omega} $ Then the z-transform can be written as: $ X(z) = X(re^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n](re)^{-j \omega n} $. Thus if r = 1, meaning that z is constrained to z = $ e^{j\omega} $, then the z-transform is equivalent to the DTFT, and the DTFT is the z-transform with z constrained to the unit circle in the z-plane.

Since the z-transform exists outside the unit circle, it is useful for analyzing signal that don't have a DTFT, analyzing LTI system stability, and looking at the transfer function characteristics among other uses.

## Summary of Z Transform Properties

- (a) The DTFT(x[n]) converges if the ROC of X(z) contains the unit circle on the z-plane.
- (b) No poles of X(z) are included in the ROC of X(z), poles of X(z) shape the ROC. E.g. if X(z) has a pole at 0, then z = 0 is not included in the ROC.
- (c) For Right Sided x[n], the ROC of X(z):
- (i) is generally defined by |z| > z
_{0}(outside a circle in the z-plane) - (ii) includes $ \infty $ if n
_{0}≥ 0 - (iii) z = 0 could possibly be included in the ROC
- (iv) the farthest pole from the origin lies just inside |z
_{0}|

- (i) is generally defined by |z| > z
- (d) For Left Sided x[n], the ROC of X(z):
- (i) is generally defined by |z| < z
_{0}(inside a circle in the z-plane) - (ii) includes z = 0 if n
_{0}≤ 0 - (iii) z = $ \infty $ could possibly be included
- (iv) the nearest pole to the origin lies just outside of |z
_{0}|

- (i) is generally defined by |z| < z
- (e) For Both Sided x[n], the ROC of X(z):
- (i) is generally a ring in the z-plane such as z
_{1}< |z| < z_{2} - (ii) does not include z = 0 or z = $ \infty $

- (i) is generally a ring in the z-plane such as z
- (f) For finite duration x[n] (neither sided), the ROC of X(z):
- (i) includes the entire complex plane
- (ii) x[n] has no poles except maybe z = 0 or z = $ \infty $
- (iii) if for x[n], n
_{1}≥ 0, z = $ \infty $ is included - (iv) if for x[n], n
_{2}≤ 0, z = 0 is included - (v) if n
_{1}< 0 and n_{2}> 0, then neither z = 0 or z = $ \infty $ is included

## LTI Systems and Z Transform Properties

(1) System Stability

An LTI System with impulse response h[n] is BIBO stable if h[n] is absolutely summable.

- (a) This is a condition for the DTFT to exist, so it must also be true that the ROC of H(z) contains the unit circle in the z-plane.
- (b) This implies that the poles of a right sided h[n] must be inside the unit circle and poles of a left sided h[n] must be outside of the unit circle in the z-plane for h[n] to be BIBO stable.

(2) Causal LTI Systems From statements in (1) and (c) above, a few conclusions about causal system can be made:

- (a) The ROC of H(z) includes z = $ \infty $ and |z| > z
_{0} - (b) For H(z) to be BIBO stable, it must have poles inside the unit circle in the z-plane.
- (c) H(z) must have more poles than zeros.
- One possible reasoning: Assuming that a causal system has the ROC defined in (a), then assuming H(z) has more zeros than poles would be a contradiction, since H(z) can could then be decomposed by long division or partial fractions into a function of z that diverges in the region |z| > z
_{0}and $ \infty $. Simply taking the limit of H(z) as z approaches $ \infty $ would show a contradiction since the ROC would not include z = $ \infty $.

## Z Transform and Digital Filters

The z-transform properties above can also give insight into characteristics of FIR and IIR filters

- (a) FIR filters:
- (i) since h[n] is of finite duration, it has no poles except possibly z = 0 or z = $ \infty $
- (ii) The ROC of H(z) includes the entire z-plane besides possibly z = 0 or z = $ \infty $

(from (b) above)

- (b) IIR filters:
- (i) H(z) has non-zero poles in the finite z-plane
- (ii) The denominator of H(z) must have a degree > 1
- Reasoning: the ROC of H(z) for a neither sided signal is a ring, implying that there must be two non-zero poles to bound the ROC.