Continuous Time. Time-Domain System Analysis. Impulse Response. Impulse Response. Impulse Response. Impulse Response. ( t) + b 0.

Similar documents
( ) ( ) if t = t. It must satisfy the identity. So, bulkiness of the unit impulse (hyper)function is equal to 1. The defining characteristic is

Representing a Signal. Continuous-Time Fourier Methods. Linearity and Superposition. Real and Complex Sinusoids. Jean Baptiste Joseph Fourier

SOLUTIONS TO ECE 3084

Chapter 2. First Order Scalar Equations

Two Coupled Oscillators / Normal Modes

Continuous Time Linear Time Invariant (LTI) Systems. Dr. Ali Hussein Muqaibel. Introduction

Linear Time-invariant systems, Convolution, and Cross-correlation

EECS 2602 Winter Laboratory 3 Fourier series, Fourier transform and Bode Plots in MATLAB

Chapter 1 Fundamental Concepts

Notes 04 largely plagiarized by %khc

Some Basic Information about M-S-D Systems

LAPLACE TRANSFORM AND TRANSFER FUNCTION

Guest Lectures for Dr. MacFarlane s EE3350 Part Deux

DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING EXAMINATIONS 2008

10. State Space Methods

Laplace Transforms. Examples. Is this equation differential? y 2 2y + 1 = 0, y 2 2y + 1 = 0, (y ) 2 2y + 1 = cos x,

The Laplace Transform

Signal and System (Chapter 3. Continuous-Time Systems)

EECE 301 Signals & Systems Prof. Mark Fowler

Basic Circuit Elements Professor J R Lucas November 2001

Section 3.5 Nonhomogeneous Equations; Method of Undetermined Coefficients

Math 333 Problem Set #2 Solution 14 February 2003

Laplace Transform and its Relation to Fourier Transform

d 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3

For example, the comb filter generated from. ( ) has a transfer function. e ) has L notches at ω = (2k+1)π/L and L peaks at ω = 2π k/l,

5. Response of Linear Time-Invariant Systems to Random Inputs

6.003 Homework #13 Solutions

h[n] is the impulse response of the discrete-time system:

EECE 301 Signals & Systems Prof. Mark Fowler

Simulation-Solving Dynamic Models ABE 5646 Week 2, Spring 2010

8. Basic RL and RC Circuits

THE BERNOULLI NUMBERS. t k. = lim. = lim = 1, d t B 1 = lim. 1+e t te t = lim t 0 (e t 1) 2. = lim = 1 2.

Math 334 Fall 2011 Homework 11 Solutions

Answers to Exercises in Chapter 7 - Correlation Functions

Signals and Systems Profs. Byron Yu and Pulkit Grover Fall Midterm 1 Solutions

EE 224 Signals and Systems I Complex numbers sinusodal signals Complex exponentials e jωt phasor addition

INDEX. Transient analysis 1 Initial Conditions 1

KEEE313(03) Signals and Systems. Chang-Su Kim

Signals and Systems Linear Time-Invariant (LTI) Systems

Lecture 13 RC/RL Circuits, Time Dependent Op Amp Circuits

MATH 4330/5330, Fourier Analysis Section 6, Proof of Fourier s Theorem for Pointwise Convergence

CHAPTER 2 Signals And Spectra

Mon Apr 9 EP 7.6 Convolutions and Laplace transforms. Announcements: Warm-up Exercise:

System Processes input signal (excitation) and produces output signal (response)

Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:

e 2t u(t) e 2t u(t) =?

Convolution. Lecture #6 2CT.3 8. BME 333 Biomedical Signals and Systems - J.Schesser

23.2. Representing Periodic Functions by Fourier Series. Introduction. Prerequisites. Learning Outcomes

Chapter 4 The Fourier Series and Fourier Transform

Chapter One Fourier Series and Fourier Transform

ADDITIONAL PROBLEMS (a) Find the Fourier transform of the half-cosine pulse shown in Fig. 2.40(a). Additional Problems 91

Voltage/current relationship Stored Energy. RL / RC circuits Steady State / Transient response Natural / Step response

Solutions to Assignment 1

EE 313 Linear Signals & Systems (Fall 2018) Solution Set for Homework #8 on Continuous-Time Signals & Systems

6.2 Transforms of Derivatives and Integrals.

9/9/99 (T.F. Weiss) Signals and systems This subject deals with mathematical methods used to describe signals and to analyze and synthesize systems.

III-A. Fourier Series Expansion

6.003: Signals and Systems. Fourier Representations

Chapter 8 The Complete Response of RL and RC Circuits

Chapter 6. Systems of First Order Linear Differential Equations

Instructor: Barry McQuarrie Page 1 of 5

Ordinary Differential Equations

Math Week 14 April 16-20: sections first order systems of linear differential equations; 7.4 mass-spring systems.

Laplace transfom: t-translation rule , Haynes Miller and Jeremy Orloff

2.7. Some common engineering functions. Introduction. Prerequisites. Learning Outcomes

ES.1803 Topic 22 Notes Jeremy Orloff

MATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.

Sections 2.2 & 2.3 Limit of a Function and Limit Laws

Communication System Analysis

EE 301 Lab 2 Convolution

Differential Equations

4/9/2012. Signals and Systems KX5BQY EE235. Today s menu. System properties

MAE143A Signals & Systems - Homework 2, Winter 2014 due by the end of class Thursday January 23, 2014.

ENGI 9420 Engineering Analysis Assignment 2 Solutions

CE 395 Special Topics in Machine Learning

Ordinary dierential equations

3.1.3 INTRODUCTION TO DYNAMIC OPTIMIZATION: DISCRETE TIME PROBLEMS. A. The Hamiltonian and First-Order Conditions in a Finite Time Horizon

MATH 128A, SUMMER 2009, FINAL EXAM SOLUTION

R =, C = 1, and f ( t ) = 1 for 1 second from t = 0 to t = 1. The initial charge on the capacitor is q (0) = 0. We have already solved this problem.

EECS20n, Solution to Midterm 2, 11/17/00

Lectures 29 and 30 BIQUADRATICS AND STATE SPACE OP AMP REALIZATIONS. I. Introduction

Homework 4 SOLUTION EE235, Summer 2012

6.302 Feedback Systems Recitation 4: Complex Variables and the s-plane Prof. Joel L. Dawson

CHAPTER 12 DIRECT CURRENT CIRCUITS

ME 452 Fourier Series and Fourier Transform

Lecture 2-1 Kinematics in One Dimension Displacement, Velocity and Acceleration Everything in the world is moving. Nothing stays still.

t is a basis for the solution space to this system, then the matrix having these solutions as columns, t x 1 t, x 2 t,... x n t x 2 t...

u(x) = e x 2 y + 2 ) Integrate and solve for x (1 + x)y + y = cos x Answer: Divide both sides by 1 + x and solve for y. y = x y + cos x

Predator - Prey Model Trajectories and the nonlinear conservation law

Outline Chapter 2: Signals and Systems

University of Cyprus Biomedical Imaging and Applied Optics. Appendix. DC Circuits Capacitors and Inductors AC Circuits Operational Amplifiers

Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k

The fundamental mass balance equation is ( 1 ) where: I = inputs P = production O = outputs L = losses A = accumulation

Chapter #1 EEE8013 EEE3001. Linear Controller Design and State Space Analysis

( ) = Q 0. ( ) R = R dq. ( t) = I t

Signals and Systems Review. 8/25/15 M. J. Roberts - All Rights Reserved 1

From Complex Fourier Series to Fourier Transforms

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.

Web Appendix N - Derivations of the Properties of the LaplaceTransform

Transcription:

Time-Domain Sysem Analysis Coninuous Time. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 1. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 2 Le a sysem be described by a 2 y ( ) + a 1 y ( ) + a 0 y( ) = x( ) and le he exciaion be a uni impulse a ime = 0. Then he zero-sae response y is he impulse response h. a 2 h ( ) + a 1 h ( ) + a 0 h( ) = δ ( ) Since he impulse occurs a ime = 0 and nohing has excied he sysem before ha ime, he impulse response before ime = 0 is zero (because his is a causal sysem). Afer ime = 0 he impulse has occurred and gone away. Therefore here is no longer an exciaion and he impulse response is he homogeneous soluion of he differenial equaion. a 2 h ( ) + a 1 h ( ) + a 0 h( ) = δ ( ) Wha happens a ime, = 0? The equaion mus be saisfied a all imes. So he lef side of he equaion mus be a uni impulse. We already know ha he lef side is zero before ime = 0 because he sysem has never been excied. We know ha he lef side is zero afer ime = 0 because i is he soluion of he homogeneous equaion whose righ side is zero. These wo facs are boh consisen wih an impulse. The impulse response migh have in i an impulse or derivaives of an impulse since all of hese occur only a ime, = 0. Wha he impulse response does have in i depends on he form of he differenial equaion.. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 3. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 4 Coninuous-ime LTI sysems are described by differenial equaions of he general form, a n y ( n) ( ) + a n 1 y ( n 1) ( ) + + a 1 y ( ) + a 0 y( ) ( ) + b m 1 x ( m 1) ( ) + + b 1 x ( ) + b 0 x( ) = b m x ( m) For all imes, < 0: If he exciaion x i is zero. The response y ( ) is an impulse, hen for all ime < 0 ( ) is zero before ime = 0 because here has never been an exciaion before ha ime. For all ime > 0: The exciaion is zero. The response is he homogeneous soluion of he differenial equaion. A ime = 0: The exciaion is an impulse. In general i would be possible for he response o conain an impulse plus derivaives of an impulse because hese all occur a ime = 0 and are zero before and afer ha ime. Wheher or no he response conains an impulse or derivaives of an impulse a ime = 0 depends on he form of he differenial equaion a n y ( n) ( ) + a n 1 y ( n 1) ( ) + + a 1 y ( ) + a 0 y( ) = b m x ( m) ( ) + b m 1 x ( m 1) ( ) + + b 1 x ( ) + b 0 x( ). J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 5. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 6 1

a n y ( n) ( ) + a n 1 y ( n 1) ( ) + + a 1 y ( ) + a 0 y( ) ( ) + b m 1 x ( m 1) ( ) + + b 1 x ( ) + b 0 x( ) = b m x ( m) Case 1: m < n If he response conained an impulse a ime = 0 hen he nh derivaive of he response would conain he nh derivaive of an impulse. Since he exciaion conains only he mh derivaive of an impulse and m < n, he differenial equaion canno be saisfied a ime = 0. Therefore he response canno conain an impulse or any derivaives of an impulse. a n y ( n) ( ) + a n 1 y ( n 1) ( ) + + a 1 y ( ) + a 0 y( ) ( ) + b m 1 x ( m 1) ( ) + + b 1 x ( ) + b 0 x( ) = b m x ( m) Case 2: m = n In his case he highes derivaive of he exciaion and response are he same and he response could conain an impulse a ime = 0 bu no derivaives of an impulse. Case 3: m > n In his case, he response could conain an impulse a ime = 0 plus derivaives of an impulse up o he (m n)h derivaive. Case 3 is rare in he analysis of pracical sysems.. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 7. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 8 Example ( ) + 3y( ) = x ( ) + 3h( ) = δ ( ). We know ha ( ) = 0 for < 0 and ha h( ) is he homogeneous soluion for Le a sysem be described by y x is an impulse we have h ( ). If he exciaion h > 0 which is h( )=Ke 3. There are more derivaives of y han of x. Therefore he impulse response canno conain an impulse. So he impulse response is h( ) = Ke 3 u( ).. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 9. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 10. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 11. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 12 2

Example h( ) = ( 3 /16)e 3 /4 u( ) + ( 1 / 4)δ ( ) The original differenial equaion is 4 h ( ) + 3h( ) = δ ( ). Subsiuing he soluion we ge 4 d ( 3 /16)e 3 /4 u( ) + ( 1 / 4)δ ( ) d = δ ( ) +3 ( 3 /16)e 3 /4 u( ) + ( 1 / 4)δ ( ) 4 ( 3 /16)e 3 /4 δ ( ) + ( 9 / 64)e 3 /4 u( ) + ( 1 / 4) δ ( ) = δ ( ) +3 ( 3 /16)e 3 /4 u( ) + ( 1 / 4)δ ( ) ( 3 / 4)e 3 /4 δ ( ) + ( 9 /16)e 3 /4 u( ) + δ ( ) ( 9 /16)e 3 /4 u( ) + ( 3 / 4)δ ( ) = δ ( ) δ ( ) = δ ( ) Check.. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 13. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 14 The Convoluion Inegral If a coninuous-ime LTI sysem is excied by an arbirary exciaion, he response could be found approximaely by approximaing he exciaion as a sequence of coniguous recangular pulses of widh T p. The Convoluion Inegral Exac Exciaion Approximae Exciaion. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 15. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 16 The Convoluion Inegral Le he response o an unshifed pulse of uni area and widh T p be he uni pulse response h p ( ). Then, invoking lineariy, he response o he overall exciaion is (approximaely) a sum of shifed and scaled uni pulse responses of he form ( ) T p x nt p y n= ( )h p nt p ( ) As T p approaches zero, he uni pulses become uni impulses, he uni pulse response becomes he uni impulse response h() and he exciaion and response become exac. The Convoluion Inegral Le he uni pulse response be ha of he RC lowpass filer h p ( )/RC ( ) = 1 +T e p /2 T p ( )/RC u + T p 2 1 +Tp /2 e T p u T p 2. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 17. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 18 3

The Convoluion Inegral The Convoluion Inegral Le x( ) be his smooh waveform and le i be approximaed by a sequence of recangular pulses. The approximae exciaion is a sum of recangular pulses.. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 19. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 20 The Convoluion Inegral The Convoluion Inegral The approximae response is a sum of pulse responses. T p = 0.1 s. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 21. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 22 The Convoluion Inegral The Convoluion Inegral As T p approaches zero, he expressions for he approximae exciaion and response approach he limiing exac forms T p = 0.05 s Superposiion Inegral Convoluion Inegral ( ) = x τ x ( )δ ( τ )dτ y ( ) = x τ ( )h( τ )dτ. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 23. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 24 4

The Convoluion Inegral Anoher (quicker) way o develop he convoluion inegral is o sar wih x ( ) = x τ ( )δ ( τ )dτ which follows direcly from he sampling propery of he impulse. If h ( ) is he impulse response of he sysem, and if he sysem is LTI, hen he response o x( τ )δ ( τ ) mus be x( τ )h( τ ). Then, invoking addiiviy, if x ( ) = x τ ( )δ ( τ )dτ, hen y ( ) = x τ ( )h( τ )dτ. A Graphical Illusraion of he Convoluion Inegral The convoluion inegral is defined by x( ) h ( ) = x τ ( )h( τ )dτ ( ) and he ( ) be he wo funcions below. For illusraion purposes le he exciaion x impulse response h. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 25. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 26 A Graphical Illusraion of he Convoluion Inegral In he convoluion inegral here is a facor h( τ ). We can begin o visualize his quaniy in he graphs below. A Graphical Illusraion of he Convoluion Inegral ( ) o h( τ ) is ( ( )) = h( τ ) The funcional ransformaion in going from h ( ) τ τ ( ) τ τ h τ h τ h τ. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 27. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 28 A Graphical Illusraion of he Convoluion Inegral The convoluion value is he area under he produc of x and h( τ ). This area depends on wha is. Firs, as an example, le = 5. ( ) A Graphical Illusraion of he Convoluion Inegral ow le = 0. For his choice of he area under he produc is zero. Therefore if y() = x( ) h( ) hen y(5) = 0. Therefore y( 0) = 2, he area under he produc.. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 29. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 30 5

A Graphical Illusraion of he Convoluion Inegral A Graphical Illusraion of he Convoluion Inegral The process of convolving o find y() is illusraed below. v ou ( ) = x( τ )h( τ )dτ = u τ ( ( ) e τ )/ RC RC u( τ )dτ. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 31. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 32 A Graphical Illusraion of he Convoluion Inegral Convoluion Example < 0 : v ou > 0 : v ou ( ) = 0 ( ) = u τ ( ( ) e τ )/ RC RC u( τ )dτ ( )/ RC v ou ( ) = 1 ( RC e τ )/ RC dτ = 1 e τ = RC 1 / RC e τ 0 0 For all ime, : / RC v ou ( ) = ( 1 e )u( ) ( )/ RC 0 / RC = 1 e. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 33. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 34 Convoluion Example Convoluion Inegral Properies ( ) = Ax( 0 ) x( ) Aδ 0 If g( ) = g 0 ( ) δ ( ) hen g( 0 ) = g 0 ( 0 ) δ ( ) = g 0 ( ) δ 0 If y( ) = x( ) h( ) hen y ( ) = x ( ) h( ) = x( ) h ( ) and y( a) = a x( a) h( a) Commuaiviy Associaiviy x( ) y( ) z Disribuiviy ( ) + y( ) x x( ) y( ) = y( ) x( ) z ( ) = x ( ) = x ( ) y( ) z( ) ( ) z ( ) + y ( ) z ( ) ( ). J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 35. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 36 6

The Uni Triangle Funcion ( ) = 1, < 1 ri 0, 1 = rec ( ) rec ( ) Sysem Inerconnecions If he oupu signal from a sysem is he inpu signal o a second sysem he sysems are said o be cascade conneced. I follows from he associaive propery of convoluion ha he impulse response of a cascade connecion of LTI sysems is he convoluion of he individual impulse responses of hose sysems. The uni riangle, is he convoluion of a uni recangle wih Iself. Cascade Connecion. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 37. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 38 Sysem Inerconnecions If wo sysems are excied by he same signal and heir responses are added hey are said o be parallel conneced. I follows from he disribuive propery of convoluion ha he impulse response of a parallel connecion of LTI sysems is he sum of he individual impulse responses. Parallel Connecion Uni and Uni Sep Response In any LTI sysem le an exciaion x( ) produce he response ( x( ) ) will produce he response y( ). Then he exciaion d d d ( y( ) ). I follows hen ha he uni impulse response h( )is d he firs derivaive of he uni sep response h 1 ha he uni sep response h 1 impulse response h( ). ( ) and, conversely ( ) is he inegral of he uni. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 39. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 40 Sabiliy and A sysem is BIBO sable if is impulse response is absoluely inegrable. Tha is if ( ) d h is finie. Sysems Described by Differenial Equaions The mos general form of a differenial equaion describing an LTI sysem is a k y (k ) ( ) = b k x (k ) ( ). Le x le y ( ) = Xe s and ( ) = Ye s. Then x ( k) ( ) = s k Xe s and y ( k) ( ) = s k Ye s and a k s k Ye s = b k s k Xe s. The differenial equaion has become an algebraic equaion. Ye s a k s k = Xe s b k s k Y X = b k s k a k s k. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 41. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 42 7

Sysems Described by Differenial Equaions The ransfer funcion for sysems of his ype is ( ) = H s b k s k a k s k = b s + b 1 s 1 + + b 2 s 2 + b 1 s + b 0 a s + a 1 s 1 + + a 2 s 2 + a 1 s + a 0 This ype of funcion is called a raional funcion because i is a raio of polynomials in s. The ransfer funcion encapsulaes all he sysem characerisics and is of grea imporance in signal and sysem analysis. ow le x Sysems Described by Differenial Equaions ( ) = Xe jω and le y( ) = Ye jω. This change of variable s jω changes he ransfer funcion o he frequency response. ( ) = b ( jω ) + b 1 ( jω ) 1 + + b 2 ( jω ) 2 + b 1 ( jω ) + b 0 a ( jω ) + a 1 ( jω ) 1 + + a 2 ( jω ) 2 + a 1 ( jω ) + a 0 H jω Frequency response describes how a sysem responds o a sinusoidal exciaion, as a funcion of he frequency of ha exciaion.. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 43. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 44 Sysems Described by Differenial Equaions ATLAB Sysem Objecs I is shown in he ex ha if an LTI sysem is excied by a sinusoid x( ) = A x cos( ω 0 +θ x ) ha he response is y( ) = A y cos( ω 0 +θ y ) where A y = H( jω 0 ) A x and ( ) +θ x. θ y = H jω 0 A ATLAB sysem objec is a special kind of variable in ATLAB ha conains all he informaion abou a sysem. I can be creaed wih he f command whose synax is sys = f(num,den) where num is a vecor of numeraor coefficiens of powers of s, den is a vecor of denominaor coefficiens of powers of s, boh in descending order and sys is he sysem objec.. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 45. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 46 ATLAB Sysem Objecs For example, he ransfer funcion s H 1 ( s) 2 + 4 = s 5 + 4s 4 + 7s 3 +15s 2 + 31s + 75 can be creaed by he commands»num = [1 0 4] ; den = [1 4 7 15 31 75] ;»H1 = f(num,den) ; Discree Time»H1 Transfer funcion: s ^ 2 + 4 ---------------------------------------- s ^ 5 + 4 s ^ 4 + 7 s ^ 3 + 15 s ^ 2 + 31 s + 75. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 47. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 48 8

Discree-ime LTI sysems are described mahemaically by difference equaions of he form [ ] + a 1 y[ n 1] + + a y[ n ] = b 0 x[ n] + b 1 x[ n 1] + + b x[ n ] [ ] he response y[ n] can be found by [ ] as he only forcing funcion on he a 0 y n For any exciaion x n finding he response o x n righ-hand side and hen adding scaled and ime-shifed versions of ha response o form y[ n]. If x n [ ] is a uni impulse, he response o i as he only forcing funcion is simply he homogeneous soluion of he difference equaion wih iniial condiions applied. The impulse response is convenionally designaed by he symbol h[ n]. Since he impulse is applied o he sysem a ime n = 0, ha is he only exciaion of he sysem and he sysem is causal he impulse response is zero before ime n = 0. [ ] = 0, n < 0 h n Afer ime n = 0, he impulse has come and gone and he exciaion is again zero. Therefore for n > 0, he soluion of he difference equaion describing he sysem is he homogeneous soluion. [ ] = y h [ n], n > 0 h n Therefore, he impulse response is of he form, h[ n] = y h [ n]u[ n]. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 49. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 50 Example Example Le a sysem be described by 4 y n if he exciaion is a uni impulse, 4 h n [ ] 3y[ n 1] = x n [ ] 3h[ n 1] = δ n [ ]. Then, [ ]. The eigenfuncion is he complex exponenial z n. Subsiuing ino he homogeneous difference equaion, 4z n 3z n 1 = 0. Dividing hrough by z n-1, 4z 3 = 0. Solving, z = 3 / 4. The homogeneous soluion is hen of he form h n [ ] = K ( 3 / 4) n.. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 51. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 52 Example Example Le a sysem be described by 3y n Then, if he exciaion is a uni impulse, 3h n [ ] + 2y[ n 1] + y[ n 2] = x n [ ] + 2h[ n 1] + h[ n 2] = δ [ n] The eigenfuncion is he complex exponenial z n. Subsiuing ino he homogeneous difference equaion, 3z n + 2z n 1 + z n 2 = 0. Dividing hrough by z n 2, 3z 2 + 2z +1 = 0. Solving, z = 0.333 ± j0.4714. The homogeneous soluion is hen of he form [ ] = K 1 ( 0.333+ j0.4714) n + K 2 ( 0.333 j0.4714) n h n [ ].. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 53. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 54 9

Example The impulse response is hen ( 0.1665 + j0.1181) 0.333+ j0.4714 h[ n] ( = )n + ( 0.1665 j0.1181) ( 0.333 j0.4714) n which can also be wrien in he forms, ( 0.1665 + j0.1181)e h[ n] = ( 0.5722) j 2.1858n n + ( 0.1665 j0.1181)e j 2.1858n u n ( ) u[ n] [ ] 0.1665 e j 2.1858n + e j 2.1858n h[ n] = ( 0.5722) n + j0.1181( e j 2.1858n e j 2.1858n ) u[ n] [ ] = ( 0.5722) n 0.333cos( 2.1858n) 0.2362sin( 2.1858n) h[ n] = 0.4083( 0.5722) n cos( 2.1858n + 0.6169) h n [ ] u n Example h[ n] = 0.4083( 0.5722) n cos( 2.1858n + 0.6169). J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 55. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 56 Sysem Response Simple Sysem Response Example Once he response o a uni impulse is known, he response of any LTI sysem o any arbirary exciaion can be found Any arbirary exciaion is simply a sequence of ampliude-scaled and ime-shifed impulses Therefore he response is simply a sequence of ampliude-scaled and ime-shifed impulse responses. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 57. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 58 ore Complicaed Sysem Response Example Sysem Exciaion Sysem Impulse Response Sysem Response The Convoluion Sum The response y[ n] o an arbirary exciaion x[ n] is of he form y[ n] = x[ 1]h[ n +1] + x[ 0]h[ n] + x[ 1]h[ n 1] + where h[ n] is he impulse response. This can be wrien in a more compac form [ ] = x m y n m= called he convoluion sum. [ ]h n m [ ]. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 59. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 60 10

A Convoluion Sum Example A Convoluion Sum Example. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 61. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 62 A Convoluion Sum Example A Convoluion Sum Example. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 63. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 64 Convoluion Sum Properies Convoluion is defined mahemaically by y[ n] = x[ n] h[ n] = x m m= [ ]h n m [ ] The following properies can be proven from he definiion. x[ n] Aδ [ n n 0 ] = Ax[ n n 0 ] Le y[ n] = x[ n] h[ n] hen y[ n n 0 ] = x[ n] h[ n n 0 ] = x[ n n 0 ] h[ n] y[ n] y[ n 1] = x[ n] ( h[ n] h[ n 1] ) = ( x[ n] x[ n 1] ) h n and he sum of he impulse srenghs in y is he produc of he sum of he impulse srenghs in x and he sum of he impulse srenghs in h. [ ] Convoluion Sum Properies (coninued) Commuaiviy Associaiviy ( x[ n] y[ n] ) z n Disribuiviy ( x[ n] + y[ n] ) z n x[ n] y[ n] = y[ n] x[ n] [ ] = x n [ ] = x n ( [ ]) [ ] y[ n] z n [ ] z n [ ] + y n [ ] z n [ ]. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 65. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 66 11

umerical Convoluion umerical Convoluion nx = -2:8 ; nh = 0:12 ; % Se ime vecors for x and h x = usd(nx-1) - usd(nx-6) ; % Compue values of x h = ri((nh-6) / 4) ; % Compue values of h y = conv(x,h) ; % Compue he convoluion of x wih h % % Generae a discree- ime vecor for y % ny = (nx(1) + nh(1)) + (0:(lengh(nx) + lengh(nh) - 2)) ; % % Graph he resuls % subplo(3,1,1) ; sem(nx,x,'k','filled') ; xlabel('n') ; ylabel('x') ; axis([-2,20,0,4]) ; subplo(3,1,2) ; sem(nh,h,'k','filled') ; xlabel('n') ; ylabel('h') ; axis([-2,20,0,4]) ; subplo(3,1,3) ; sem(ny,y,'k','filled') ; xlabel('n') ; ylabel('y') ; axis([-2,20,0,4]) ;. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 67. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 68 umerical Convoluion umerical Convoluion Coninuous-ime convoluion can be approximaed using he conv funcion in ATLAB. y( ) = x( ) h( ) = x τ ( )h( τ )dτ Approximae x( ) and h( ) each as a sequence of recangles of widh T s. ( ) x nt s x h n= ( ) h nt s n= nt ( s T s / 2 )rec T s nt ( s T s / 2 )rec T s. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 69. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 70 umerical Convoluion The inegral can be approximaed a discree poins in ime as y( nt s ) x( mt s )h(( n m)t s )T s m= This can be expressed in erms of a convoluion sum as ( ) T s x m m= ( ) and h n y nt s where x n = x nt s h n m = h ( nt s ). = T s x n h n Sabiliy and I can be shown ha a discree-ime BIBO-sable sysem has an impulse response ha is absoluely summable. Tha is, h[ n] is finie. n=. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 71. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 72 12

Sysem Inerconnecions Sysem Inerconnecions The cascade connecion of wo sysems can be viewed as a single sysem whose impulse response is he convoluion of he wo individual sysem impulse responses. This is a direc consequence of he associaiviy propery of convoluion. The parallel connecion of wo sysems can be viewed as a single sysem whose impulse response is he sum of he wo individual sysem impulse responses. This is a direc consequence of he disribuiviy propery of convoluion.. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 73. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 74 Uni and Uni Sequence Response In any LTI sysem le an exciaion x n y[ n]. Then he exciaion x n response y n [ ] y[ n 1]. [ ] produce he response [ ] x[ n 1] will produce he I follows hen ha he uni impulse response is he firs backward difference of he uni sequence response and, conversely ha he uni sequence response is he accumulaion of he uni impulse response. The mos common descripion of a discree-ime sysem is a difference equaion of he general form If x n a k y[ n k] = b k x[ n k]. [ ] = Xz n, y[ n] has he form y[ n] = Yz n where X and Y are [ ] = z k Xz n and complex consans. Then x n k [ ] = z k Yz n and a k z k Yz n y n k Sysems Described by Difference Equaions = b k z k Xz n. Rearranging Yz n a k z k = Xz n b k z k Y X = b k z k a k z k. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 75. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 76 The ransfer funcion is ( ) = H z or, alernaely, ( ) = H z k =0 k =0 k =0 k =0 b k z k a k z k b k z k a k z k = b 0 + b 1 z 1 + b 2 z 2 + + b z a 0 + a 1 z 1 + a 2 z 2 + + a z = z b 0 z + b 1 z 1 + + b 1 z + b a 0 z + a 1 z 1 + + a 1 z + a The ransform can be wrien direcly from he difference equaion and vice versa. Sysems Described by Difference Equaions Le x n Frequency Response [ ] = Xe jωn. Then y[ n] = Ye jωn and x[ n k] = e jωk Xe jωn and [ ] = e jωk Ye jωn. Then he general difference equaion y n k descripion of a discree-ime sysem becomes a k y[ n k] = b k x n k Ye jωn [ ] a k e jωk = Xe jωn b k e jωk. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 77. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 78 13

Frequency Response Frequency Response Example. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 79. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 80 Frequency Response Example. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 81 14