Discrete-Time Signals and Systems. The z-transform and Its Application. The Direct z-transform. Region of Convergence. Reference: Sections

Similar documents
Digital Signal Processing

ELEG 305: Digital Signal Processing

Discrete-Time Signals and Systems. Frequency Domain Analysis of LTI Systems. The Frequency Response Function. The Frequency Response Function

y[n] = = h[k]x[n k] h[k]z n k k= 0 h[k]z k ) = H(z)z n h[k]z h (7.1)

Lecture 04: Discrete Frequency Domain Analysis (z-transform)

The Z-Transform. Fall 2012, EE123 Digital Signal Processing. Eigen Functions of LTI System. Eigen Functions of LTI System

6.003: Signals and Systems

Z-Transform. The Z-transform is the Discrete-Time counterpart of the Laplace Transform. Laplace : G(s) = g(t)e st dt. Z : G(z) =

SEL4223 Digital Signal Processing. Inverse Z-Transform. Musa Mohd Mokji

UNIT-II Z-TRANSFORM. This expression is also called a one sided z-transform. This non causal sequence produces positive powers of z in X (z).

Signals and Systems. Problem Set: The z-transform and DT Fourier Transform

Use: Analysis of systems, simple convolution, shorthand for e jw, stability. Motivation easier to write. Or X(z) = Z {x(n)}

ESE 531: Digital Signal Processing

ECE503: Digital Signal Processing Lecture 4

Solutions: Homework Set # 5

Topic 4: The Z Transform

EE 521: Instrumentation and Measurements

ESE 531: Digital Signal Processing

8. z-domain Analysis of Discrete-Time Signals and Systems

Lecture 7 Discrete Systems

Need for transformation?

X (z) = n= 1. Ã! X (z) x [n] X (z) = Z fx [n]g x [n] = Z 1 fx (z)g. r n x [n] ª e jnω

SIGNALS AND SYSTEMS. Unit IV. Analysis of DT signals

EE123 Digital Signal Processing. M. Lustig, EECS UC Berkeley

Your solutions for time-domain waveforms should all be expressed as real-valued functions.

Let H(z) = P(z)/Q(z) be the system function of a rational form. Let us represent both P(z) and Q(z) as polynomials of z (not z -1 )

Chapter 7: The z-transform

Signal Analysis, Systems, Transforms

Lecture 18: Stability

Signals and Systems Lecture 8: Z Transform

Chapter Intended Learning Outcomes: (i) Understanding the relationship between transform and the Fourier transform for discrete-time signals

DIGITAL SIGNAL PROCESSING. Chapter 3 z-transform

(i) Represent discrete-time signals using transform. (ii) Understand the relationship between transform and discrete-time Fourier transform

# FIR. [ ] = b k. # [ ]x[ n " k] [ ] = h k. x[ n] = Ae j" e j# ˆ n Complex exponential input. [ ]Ae j" e j ˆ. ˆ )Ae j# e j ˆ. y n. y n.

Discrete-Time Fourier Transform (DTFT)

! z-transform. " Tie up loose ends. " Regions of convergence properties. ! Inverse z-transform. " Inspection. " Partial fraction

Z Transform (Part - II)

Like bilateral Laplace transforms, ROC must be used to determine a unique inverse z-transform.

Lecture 2 OKAN UNIVERSITY FACULTY OF ENGINEERING AND ARCHITECTURE

Digital Control & Digital Filters. Lectures 1 & 2

The Z transform (2) 1

Digital Signal Processing. Midterm 1 Solution

Signals & Systems Handout #4

Signals and Systems. Spring Room 324, Geology Palace, ,

The Z transform (2) Alexandra Branzan Albu ELEC 310-Spring 2009-Lecture 28 1

If every Bounded Input produces Bounded Output, system is externally stable equivalently, system is BIBO stable

LECTURE NOTES DIGITAL SIGNAL PROCESSING III B.TECH II SEMESTER (JNTUK R 13)

2.161 Signal Processing: Continuous and Discrete Fall 2008

Z-TRANSFORMS. Solution: Using the definition (5.1.2), we find: for case (b). y(n)= h(n) x(n) Y(z)= H(z)X(z) (convolution) (5.1.

z-transform Chapter 6

Transform Analysis of Linear Time-Invariant Systems

University of Illinois at Urbana-Champaign ECE 310: Digital Signal Processing

ECE-S Introduction to Digital Signal Processing Lecture 4 Part A The Z-Transform and LTI Systems

ELEN E4810: Digital Signal Processing Topic 4: The Z Transform. 1. The Z Transform. 2. Inverse Z Transform

Discrete Time Systems

Z - Transform. It offers the techniques for digital filter design and frequency analysis of digital signals.

DSP-I DSP-I DSP-I DSP-I

Review of Fundamentals of Digital Signal Processing

EEL3135: Homework #4

How to manipulate Frequencies in Discrete-time Domain? Two Main Approaches

UNIVERSITY OF OSLO. Please make sure that your copy of the problem set is complete before you attempt to answer anything.

Lecture 19 IIR Filters

Generalizing the DTFT!

Discrete-time signals and systems

ECSE 512 Digital Signal Processing I Fall 2010 FINAL EXAMINATION

Digital Filters Ying Sun

Digital Signal Processing:

1. Z-transform: Initial value theorem for causal signal. = u(0) + u(1)z 1 + u(2)z 2 +

Z-Transform. 清大電機系林嘉文 Original PowerPoint slides prepared by S. K. Mitra 4-1-1

Module 4 : Laplace and Z Transform Problem Set 4

Very useful for designing and analyzing signal processing systems


2.161 Signal Processing: Continuous and Discrete Fall 2008

EC Signals and Systems

Review of Fundamentals of Digital Signal Processing

Review of Discrete-Time System

Z-Transform. x (n) Sampler

Digital Signal Processing, Homework 2, Spring 2013, Prof. C.D. Chung. n; 0 n N 1, x [n] = N; N n. ) (n N) u [n N], z N 1. x [n] = u [ n 1] + Y (z) =

Lecture 11 FIR Filters

Digital Signal Processing Lecture 4

EE-210. Signals and Systems Homework 7 Solutions

Digital Signal Processing. Outline. Hani Mehrpouyan, The Z-Transform. The Inverse Z-Transform

Digital Signal Processing Lecture 3 - Discrete-Time Systems

Chapter 13 Z Transform

Digital Signal Processing

Transform analysis of LTI systems Oppenheim and Schafer, Second edition pp For LTI systems we can write

Discrete Time Systems

2. Typical Discrete-Time Systems All-Pass Systems (5.5) 2.2. Minimum-Phase Systems (5.6) 2.3. Generalized Linear-Phase Systems (5.

Analog vs. discrete signals

7.17. Determine the z-transform and ROC for the following time signals: Sketch the ROC, poles, and zeros in the z-plane. X(z) = x[n]z n.

Digital Signal Processing Lecture 9 - Design of Digital Filters - FIR

Discrete-Time David Johns and Ken Martin University of Toronto

The z-transform Part 2

Stability Condition in Terms of the Pole Locations

ENT 315 Medical Signal Processing CHAPTER 2 DISCRETE FOURIER TRANSFORM. Dr. Lim Chee Chin

Introduction to the z-transform

GATE EE Topic wise Questions SIGNALS & SYSTEMS

Digital Signal Processing. Lecture Notes and Exam Questions DRAFT

VI. Z Transform and DT System Analysis

III. Time Domain Analysis of systems

Transcription:

Discrete-Time Signals and Systems The z-transform and Its Application Dr. Deepa Kundur University of Toronto Reference: Sections 3. - 3.4 of John G. Proakis and Dimitris G. Manolakis, Digital Signal Processing: Principles, Algorithms, and Applications, 4th edition, 2007. Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 2 / 36 The Direct z-transform Region of Convergence Direct z-transform: Notation: X (z) x(n)z n n X (z) {x(n)} the region of convergence (ROC) of X (z) is the set of all values of z for which X (z) attains a finite value The z-transform is, therefore, uniquely characterized by:. expression for X (z) 2. ROC of X (z) x(n) X (z) Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 3 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 4 / 36

Power Series Convergence Power Series Convergence For a power series, For a power series, f (z) a n (z c) n a 0 + a (z c) + a 2 (z c) 2 + n0 f (z) n0 a n (z c) n a 0 + a (z c) + a 2 (z c) 2 + there exists a number 0 r such that the series convergences for z c < r, and diverges for z c > r may or may not converge for values on z c r. there exists a number 0 r such that the series convergences for z c >r, and diverges for z c <r may or may not converge for values on z c r. Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 5 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 6 / 36 Region of Convergence Region of Convergence: r > r 2 Consider X (z) n n x(n)z n x(n)z n + x( n )z n + n 0 }{{} ROC: z < r x(n)z n n0 x(n)z n n0 }{{} ROC: z > r 2 x(0) }{{} ROC: all z Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 7 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 8 / 36

Region of Convergence: r < r 2 ROC Families: Finite Duration Signals Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 9 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 0 / 36 ROC Families: Infinite Duration Signals z-transform Properties Property Time Domain z-domain ROC Notation: x(n) X (z) ROC: r 2 < z < r x (n) X (z) ROC x 2 (n) X (z) ROC 2 Linearity: a x (n) + a 2 x 2 (n) a X (z) + a 2 X 2 (z) At least ROC ROC 2 Time shifting: x(n k) z k X (z) ROC, except z 0 (if k > 0) and z (if k < 0) z-scaling: a n x(n) X (a z) a r 2 < z < a r Time reversal x( n) X (z ) r < z < r 2 Conjugation: x (n) X (z ) ROC dx (z) z-differentiation: n x(n) z r d < z < r Convolution: x (n) x 2 (n) X (z)x 2 (z) At least ROC ROC 2 among others... Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 2 / 36

Convolution Property x(n) x (n) x 2 (n) X (z) X (z) X 2 (z) Convolution using the z-transform Basic Steps:. Compute z-transform of each of the signals to convolve (time domain z-domain): X (z) {x (n)} X 2 (z) {x 2 (n)} 2. Multiply the two z-transforms (in z-domain): X (z) X (z)x 2 (z) 3. Find the inverse z-transformof the product (z-domain time domain): x(n) {X (z)} Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 3 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 4 / 36 Common Transform Pairs Common Transform Pairs Signal, x(n) z-transform, X (z) ROC δ(n) All z 2 u(n) z z > 3 a n u(n) az 4 na n u(n) az ( az ) 2 5 a n u( n ) az 6 na n az u( n ) 7 cos(ω 0 n)u(n) 8 sin(ω 0 n)u(n) 9 a n cos(ω 0 n)u(n) 0 a n sin(ω 0 n)u(n) ( az ) 2 z cos ω 0 z < a z < a 2z cos ω 0 +z 2 z > z sin ω 0 2z cos ω 0 +z 2 z > az cos ω 0 2az cos ω 0 +a 2 z 2 az sin ω 0 2az cos ω 0 +a 2 z 2 Signal, x(n) z-transform, X (z) ROC δ(n) All z 2 u(n) z z > 3 a n u(n) az 4 na n u(n) az ( az ) 2 5 a n u( n ) az 6 na n az u( n ) 7 cos(ω 0 n)u(n) 8 sin(ω 0 n)u(n) 9 a n cos(ω 0 n)u(n) 0 a n sin(ω 0 n)u(n) ( az ) 2 z cos ω 0 z < a z < a 2z cos ω 0 +z 2 z > z sin ω 0 2z cos ω 0 +z 2 z > az cos ω 0 2az cos ω 0 +a 2 z 2 az sin ω 0 2az cos ω 0 +a 2 z 2 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 5 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 6 / 36

Why Rational? Poles and eros X (z) is a rational function iff it can be represented as the ratio of two polynomials in z (or z): X (z) b 0 + b z + b 2 z 2 + + b M z M a 0 + a z + a 2 z 2 + + a N z N zeros of X (z): values of z for which X (z) 0 poles of X (z): values of z for which X (z) For LTI systems that are represented by LCCDEs, the z-transform of the unit sample response h(n), denoted H(z) {h(n)}, is rational Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 7 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 8 / 36 Poles and eros of the Rational z-transform Poles and eros of the Rational z-transform Let a 0, b 0 0: X (z) B(z) A(z) b 0 + b z + b 2 z 2 + + b M z M a 0 + a z + a 2 z 2 + + a N z ( ) N b0 z M z M + (b /b 0 )z M + + b M /b 0 a 0 z N z N + (a /a 0 )z N + + a N /a 0 b 0 z M+N (z z )(z ) (z z M ) a 0 (z p )(z p 2 ) (z p N ) Gz N M M k (z z k) N k (z p k) X (z) Gz N M M k (z z k) N k (z p k) where G b 0 a 0 Note: finite does not include zero or. X (z) has M finite zeros at z z,,..., z M X (z) has N finite poles at z p, p 2,..., p N For N M 0 if N M > 0, there are N M zero at origin, z 0 if N M < 0, there are N M poles at origin, z 0 Total number of zeros Total number of poles Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 9 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 20 / 36

Poles and eros of the Rational z-transform Example: X (z) 2z + 6z 3 + 6z + 5 (z ( 2 (z 0) + j ))(z ( j )) 2 2 2 (z ( + j 3))(z ( j 3))(z ( )) 4 4 4 4 2 Pole-ero Plot Example: poles: z 4 ± j 3 4, 2, zeros: z 0, 2 ± j 2 unit circle poles: z 4 ± j 3 4, 2 zeros: z 0, 2 ± j 2 POLE ERO Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 2 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 22 / 36 Pole-ero Plot Graphical interpretation of characteristics of X (z) on the complex plane ROC cannot include poles; assuming causality... Pole-ero Plot For real time-domain signals, the coefficients of X (z) are necessarily real complex poles and zeros must occur in conjugate pairs note: real poles and zeros do not have to be paired up X (z) z2 2z + 6z 3 + 6z + 5 ROC ROC Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 23 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 24 / 36

Pole-ero Plot Insights For causal systems, the ROC will be the outer region of the smallest (origin-centered) circle encompassing all the poles. For stable systems, the ROC will include the unit circle. ROC Causal? Yes. Stable? Yes. For stability of a causal system, the poles will lie inside the unit circle. The System Function Therefore, h(n) time-domain impulse response y(n) x(n) h(n) H(z) H(z) Y (z) X (z) z-domain system function Y (z) X (z) H(z) Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 25 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 26 / 36 The System Function of LCCDEs The System Function of LCCDEs y(n) a k y(n k) + b k x(n k) k {y(n)} { {y(n)} Y (z) k0 a k y(n k) + k a k {y(n k)} + k a k z k Y (z) + k k0 b k x(n k)} k0 b k {x(n k)} k0 b k z k X (z) Y (z) + Y (z) [ a k z k Y (z) k + ] a k z k k H(z) Y (z) X (z) b k z k X (z) k0 X (z) b k z k k0 M [ k0 b kz k + ] N k a kz k LCCDE Rational System Function Many signals of practical interest have a rational z-transform. Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 27 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 28 / 36

Inversion of the z-transform Expansion into Power Series Three popular methods: Contour integration: x(n) X (z)z n dz 2πj C Expansion into a power series in z or z : X (z) x(k)z k k and obtaining x(k) for all k by inspection Partial-fraction expansion and table lookup Example: By inspection: X (z) log( + az ), ( ) n+ a n z n ( ) n+ a n n n n x(n) n { ( ) n+ a n n n 0 n 0 z n Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 29 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 30 / 36. Find the distinct poles of X (z): p, p 2,..., p K and their corresponding multiplicities m, m 2,..., m K. 2. The partial-fraction expansion is of the form: X (z) K k ( Ak + A ) 2k z p k (z p k ) + + A mk 2 (z p k ) m k where p k is an m k th order pole (i.e., has multiplicity m k ). 3. Use an appropriate approach to compute {A ik } Example: Find x(n) given poles of X (z) at p 2 and a double pole at p 2 p 3 ; specifically, X (z) X (z) z ( + 2z )( z ) 2 (z + 2)(z ) 2 (z + 2)(z ) 2 A z + 2 + A 2 z + A 3 (z ) 2 Note: we need a strictly proper rational function. DO NOT FORGET TO MULTIPLY BY z IN THE END. Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 3 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 32 / 36

(z + 2) A (z + 2) + A 2(z + 2) (z + 2)(z ) 2 z + 2 z A (z ) 2 + A 2(z + 2) z A 4 9 + A 3(z + 2) (z ) 2 + A 3(z + 2) (z ) 2 z 2 (z ) 2 A (z ) 2 (z + 2)(z ) 2 z + 2 (z + 2) A (z ) 2 z + 2 A 3 3 + A 2(z ) 2 z + A 3(z ) 2 (z ) 2 + A 2 (z ) + A 3 z Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 33 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 34 / 36 (z ) 2 A (z ) 2 (z + 2)(z ) 2 z + 2 d dz (z + 2) [ ] (z + 2) + A 2(z ) 2 z + A 3(z ) 2 (z ) 2 A (z ) 2 + A 2 (z ) + A 3 z + 2 d [ ] A (z ) 2 + A 2 (z ) + A 3 dz z + 2 z A 2 5 9 Therefore, assuming causality, and using the following pairs: X (z) 4 9 a n u(n) na n u(n) + 2z + 5 9 az az ( az ) 2 z + 3 x(n) 4 9 ( 2)n u(n) + 5 9 u(n) + 3 nu(n) [ ( 2) n+2 + 5 9 9 + n ] u(n) 3 z ( z ) 2 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 35 / 36 Dr. Deepa Kundur (University of Toronto) The z-transform and Its Application 36 / 36