Exercises for lectures 23 Discrete systems

Similar documents
1985 AP Calculus BC: Section I

Answer: 1(A); 2(C); 3(A); 4(D); 5(B); 6(A); 7(C); 8(C); 9(A); 10(A); 11(A); 12(C); 13(C)

+ x. x 2x. 12. dx. 24. dx + 1)

CDS 101: Lecture 5.1 Reachability and State Space Feedback

Lectures 9 IIR Systems: First Order System

Probability & Statistics,

CDS 101: Lecture 5.1 Reachability and State Space Feedback

PURE MATHEMATICS A-LEVEL PAPER 1

Time : 1 hr. Test Paper 08 Date 04/01/15 Batch - R Marks : 120

Part B: Transform Methods. Professor E. Ambikairajah UNSW, Australia

Control Systems. Transient and Steady State Response.

Periodic Structures. Filter Design by the Image Parameter Method

Problem Value Score Earned No/Wrong Rec -3 Total

Review Exercises. 1. Evaluate using the definition of the definite integral as a Riemann Sum. Does the answer represent an area? 2

UNIT 2: MATHEMATICAL ENVIRONMENT

CYLINDER THRUST CALCULATION

Session : Plasmas in Equilibrium

Discrete Fourier Transform (DFT)

ELG3150 Assignment 3

Multiple Short Term Infusion Homework # 5 PHA 5127

MONTGOMERY COLLEGE Department of Mathematics Rockville Campus. 6x dx a. b. cos 2x dx ( ) 7. arctan x dx e. cos 2x dx. 2 cos3x dx

Chapter 2 Feedback Control Theory Continued

15/03/1439. Lectures on Signals & systems Engineering

8(4 m0) ( θ ) ( ) Solutions for HW 8. Chapter 25. Conceptual Questions

Worksheet: Taylor Series, Lagrange Error Bound ilearnmath.net

Introduction to Quantum Information Processing. Overview. A classical randomised algorithm. q 3,3 00 0,0. p 0,0. Lecture 10.

Course Outline. Designing Control Systems. Proportional Controller. Amme 3500 : System Dynamics and Control. Root Locus. Dr. Stefan B.

Option 3. b) xe dx = and therefore the series is convergent. 12 a) Divergent b) Convergent Proof 15 For. p = 1 1so the series diverges.

Scattering Parameters. Scattering Parameters

Windowing in FIR Filter Design. Design Summary and Examples

APPENDIX: STATISTICAL TOOLS

INTRODUCTION TO SAMPLING DISTRIBUTIONS

ECE594I Notes set 6: Thermal Noise

Blackbody Radiation. All bodies at a temperature T emit and absorb thermal electromagnetic radiation. How is blackbody radiation absorbed and emitted?

Chapter Taylor Theorem Revisited

Theory of Control: I. Overview. Types of Structural Control Active Control Systems. Types of Structural Control

National Quali cations

Systems in Transform Domain Frequency Response Transfer Function Introduction to Filters

Frequency Response & Digital Filters

GUC (Dr. Hany Hammad) 4/20/2016

2. Finite Impulse Response Filters (FIR)

ln x = n e = 20 (nearest integer)

1973 AP Calculus BC: Section I

ECE 599/692 Deep Learning

Rational Function. To Find the Domain. ( x) ( ) q( x) ( ) ( ) ( ) , 0. where p x and are polynomial functions. The degree of q x

EC1305 SIGNALS & SYSTEMS

COMPLEX NUMBERS AND ELEMENTARY FUNCTIONS OF COMPLEX VARIABLES

Improved Three-Step Input Shaping Control of Crane System

LECTURE 13 Filling the bands. Occupancy of Available Energy Levels

Statistics 3858 : Likelihood Ratio for Exponential Distribution

Lecture 37 (Schrödinger Equation) Physics Spring 2018 Douglas Fields

CDS 101: Lecture 8.2 Tools for PID & Loop Shaping

Chapter 2 Infinite Series Page 1 of 11. Chapter 2 : Infinite Series

z 1+ 3 z = Π n =1 z f() z = n e - z = ( 1-z) e z e n z z 1- n = ( 1-z/2) 1+ 2n z e 2n e n -1 ( 1-z )/2 e 2n-1 1-2n -1 1 () z

NAME: SOLUTIONS EEE 203 HW 1

Physics 302 Exam Find the curve that passes through endpoints (0,0) and (1,1) and minimizes 1

EE 4343 Lab#4 PID Control Design of Rigid Bodies

Analytic Continuation

A Review of Complex Arithmetic

CATAVASII LA NAȘTEREA DOMNULUI DUMNEZEU ȘI MÂNTUITORULUI NOSTRU, IISUS HRISTOS. CÂNTAREA I-A. Ήχος Πα. to os se e e na aș te e e slă ă ă vi i i i i

TABLES AND FORMULAS FOR MOORE Basic Practice of Statistics

Chapter (8) Estimation and Confedence Intervals Examples

EE 570: Location and Navigation: Theory & Practice

( A) ( B) ( C) ( D) ( E)

Lecture 13. Graphical representation of the frequency response. Luca Ferrarini - Basic Automatic Control 1

Chapter 2: Numerical Methods

EE Control Systems LECTURE 11

Discrete Fourier Transform. Nuno Vasconcelos UCSD

H2 Mathematics Arithmetic & Geometric Series ( )

ME 375 FINAL EXAM Friday, May 6, 2005

Course: INS335 R - DEMOCRATIZATION Responsible Faculty: Kanet, Roger E --- Survey Comparisons --- Responses Individual INS All. Med.

IV Design of Discrete Time Control System by Conventional Methods

DTFT Properties. Example - Determine the DTFT Y ( e ) of n. Let. We can therefore write. From Table 3.1, the DTFT of x[n] is given by 1

Statistical Fundamentals and Control Charts

(Reference: sections in Silberberg 5 th ed.)

Module 5: IIR and FIR Filter Design Prof. Eliathamby Ambikairajah Dr. Tharmarajah Thiruvaran School of Electrical Engineering & Telecommunications

I M P O R T A N T S A F E T Y I N S T R U C T I O N S W h e n u s i n g t h i s e l e c t r o n i c d e v i c e, b a s i c p r e c a u t i o n s s h o

Slide 1. Slide 2. Slide 3 DIGITAL SIGNAL PROCESSING CLASSIFICATION OF SIGNALS

Announce. ECE 2026 Summer LECTURE OBJECTIVES READING. LECTURE #3 Complex View of Sinusoids May 21, Complex Number Review

Additional Math (4047) Paper 2 (100 marks) y x. 2 d. d d

Finite-length Discrete Transforms. Chapter 5, Sections

UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN MECHANICAL ENGINEERING SEMESTER 1 EXAMINATION 2016/2017

[ ] Review. For a discrete-time periodic signal xn with period N, the Fourier series representation is

MTH 142 Exam 3 Spr 2011 Practice Problem Solutions 1

SOLVED EXAMPLES. Ex.1 If f(x) = , then. is equal to- Ex.5. f(x) equals - (A) 2 (B) 1/2 (C) 0 (D) 1 (A) 1 (B) 2. (D) Does not exist = [2(1 h)+1]= 3

PHYS ,Fall 05, Term Exam #1, Oct., 12, 2005

Chapter 3 Fourier Series Representation of Periodic Signals

Sample Midterm This midterm consists of 10 questions. The rst seven questions are multiple choice; the remaining three

NET/JRF, GATE, IIT JAM, JEST, TIFR

امتحانات الشهادة الثانوية العامة فرع: العلوم العامة

Consider serial transmission. In Proakis notation, we receive

10. The Discrete-Time Fourier Transform (DTFT)

DFT: Discrete Fourier Transform

Bifurcation Theory. , a stationary point, depends on the value of α. At certain values

Thomas Whitham Sixth Form

Continuous Functions

Discrete Fourier Series and Transforms

EEO 401 Digital Signal Processing Prof. Mark Fowler

The second condition says that a node α of the tree has exactly n children if the arity of its label is n.

ELEC9721: Digital Signal Processing Theory and Applications

Transcription:

Exrciss for lcturs 3 Discrt systms Michal Šbk Automatické říí 06 30-4-7

Stat-Spac a Iput-Output scriptios Automatické říí - Kybrtika a robotika Mols a trasfrs i CSTbx >> F=[ ; 3 4]; G=[ ;]; H=[ ]; J=0; >> Pss = ss(f,g,h,j,- a = x x x x 3 4 b = u x x c = x x y = u y 0 Samplig tim: uspcifi Discrt-tim mol. >> Ptf=tf(Fss Trasfr fuctio: 4 + ------------- ^ - 5 - Samplig tim: uspcifi >> Psf = sf(pss Psf = + 4 ------------- - - 5 + ^ Michal Šbk ARI-Pr-3-0

Automatické říí - Kybrtika a robotika Rspos by log ivisio Log ivisio ca b us for primag calculatio from -imag It os ot by calculatig th rst, but cotius to "gativ powrs This ca b us to calculat rspos from or - TF 3 4 3 4 3 4 3 4 3 4 Michal Šbk ARI-Pr-3-0 3

Trasfr fuctio i a - = Automatické říí - Kybrtika a robotika Exampls: Ivstigat orr, is it propr? b bˆ a a ( ( ( ˆ( b bˆ a a ( ( ( ˆ( b bˆ a a ( ( ( ˆ( b( ˆ b( a( aˆ ( Michal Šbk ARI-Pr-3-0 4

Pols a ros i a - = Automatické říí - Kybrtika a robotika Oprator chag (complx varioabls f( = - = It is circular ivrsio plus a rflctio accorig to th ral axis 0 ½ j j j ½j j ½ ½ j Aras of stability a istability ar ovrtur Michal Šbk ARI-Pr-3-0 5

Pols a ros i a s Automatické říí - Kybrtika a robotika To sig a iscrt cotrol for a iscrt systm by th mtho of th pol placmt with giv spcificatios i th tim omai, o w to kow whr to plac thm? W ca us formulas for cotius systms with formulas for sampl systms For st orr systm hs For orr systm, hs, h( j h( j h( j Michal Šbk ARI-Pr-3-0 6

Sttlig tim T s Automatické říí - Kybrtika a robotika Sam sttlig tim Ts I th s-pla: pols lyig o th vrtical lis σ = cost. It th -pla: coctric circls with a ctr at th origi h kost cost. k hs h( j h( j h( j s >> T=;sigma=;omga=0:.0:pi/T; >> xplus=xp(-sigma+j*omga; >> xmius=xp(-sigma-j*omga; >> x=[xplus xmius]; >> plot(ral(x,imag(x,'.' >> hol Currt plot hl >> T=;sigma=;omga=0:.0:pi/T; >> xplus=xp(-sigma+j*omga; >> xmius=xp(-sigma-j*omga; >> x=[xplus xmius]; 3 >> plot(ral(x,imag(x,'.' >> T=;sigma=3;omga=0:.0:pi/T; >> xplus=xp(-sigma+j*omga; >> xmius=xp(-sigma-j*omga; >> x=[xplus T xmius]; >> plot(ral(x,imag(x,'.' 3 Michal Šbk ARI-Pr-3-0 7

First maximum timt p Automatické říí - Kybrtika a robotika Sam first max. tim Tp I th s-pla: horiotal lis ω = cost. I th -pla: raial rays from th origi j h kost cost. hs h( j h( j h( j 0 s >> T=;sigma=;omga=0:.0:pi/T; xplus=xp(-sigma+j*omga; xmius=xp(-sigma-j*omga; x=[xplus xmius]; plot(ral(x,imag(x,'.' >> hol Currt plot hl >> T=;sigma=;omga=0:.0:pi/T; xplus=xp(-sigma+j*omga; [ra] xmius=xp(-sigma-j*omga; x=[xplus xmius]; plot(ral(x,imag(x,'.' 0 >> T=;sigma=3;omga=0:.0:pi/T; xplus=xp(-sigma+j*omga; xmius=xp(-sigma-j*omga; x=[xplus xmius]; plot(ral(x,imag(x,'.' Michal Šbk ARI-Pr-3-0 8

Th sam ris timt r Automatické říí - Kybrtika a robotika Th sam ris tim.8 Tr I th s-pla: pols o coctric circls ω = cost. I th -pla: curvs hs h( j h( j h( j s cos Michal Šbk ARI-Pr-3-0 9

Th sam ovrshoot a ampig Automatické říí - Kybrtika a robotika Th sam ovrshoot % OS 00 ( a ampig I th s-pla: lis passig through th origi I th -pla: part of a spiral hs h( j h( j h( j s Michal Šbk ARI-Pr-3-05 0

Rspos rquirmts usig pol positio: orr Automatické říí - Kybrtika a robotika Cotius Rquir ris tim 4 Im.8 s, 3 r Discrt hs, h( j,, s j j Im h( j 0 - - -3 R R -4-5 -4.5-4 -3.5-3 -.5 - -.5 - -0.5 0 Rquir srrlig tim k% Ts s R s, s 4 Im 4 Im 3 0 - - -3 R s, k% s R 3 0 - -, k% s R -4-5 -4.5-4 -3.5-3 -.5 - -.5 - -0.5 0-3 -4-5 -4.5-4 -3.5-3 -.5 - -.5 - -0.5 0 Michal Šbk Pr-ARI-03-0

Rspos rquirmts usig pol positio: orr Automatické říí - Kybrtika a robotika Cotius Rquir ovrshoot 4 3 s j j, Im Discrt,, hs h( j h( j 0 - - arccos mi R -3-4 -5-4.5-4 -3.5-3 -.5 - -.5 - -0.5 0 mi Rquir ovrshoot a 4 3 l p 00 max l p 00 Im max 0 - - arccos k% s R s, mi R -3-4 -5-4.5-4 -3.5-3 -.5 - -.5 - -0.5 0 Michal Šbk Pr-ARI-03-0

Discrt Root Locus Automatické říí - Kybrtika a robotika CL pols graph pig o K, i.. KL( 0 Th graph is raw accorig to th sam ruls as i cotiuous cas Hovwr th itrprtatio of its positio is of cours iffrt >> Ls=(s+3*(s+4/(s+/(s+ Ls = +7s+s^ / +3s+s^ >> rlocus(ls,sgri >> L=(+3*(+4/(+/(+ Ls = +7+^ / +3+^ >> rlocus(l,gri stabl vrywhr ustabl vrywhr Michal Šbk ARI-Pr-3-0 3

Discrt Nyquist stability critrio Automatické říí - Kybrtika a robotika Discrt Cotiuous CL systém has Z P N ustabl pols, whr Z N P N umbr of poit - Nyquist graph L(s but th N is P umbr of ustabl OL pols. Nyquist stability critrio CL systém is stabl P N PN N umbr of Nyquist graph L(s but P umbr of ustabl OL pols is N, So it is th sam A spcial cas: Nyquist stability critrio for stabl OL systm If th OL systém is stabl, th th CL systém is stabl. Niquist graph L(s os ot circl th critical poit -. Michal Šbk ARI-Pr-3-0 4

Automatické říí - Kybrtika a robotika Paralll rivatio of both for compariso umbr of ros H( (= OL pols = umbr of pols H( (= CL pols Z umbr of ustabl CL pols = umbr of ustabl ros P umbr of ustabl OL pols = umbr of ustabl ros N umbr of circlig poit - by Nyquist graph I th sam irctio i which w surrou th ara ur cosiratio Cotiuous Discrt Circlig th istability ara clockwis Circlig th istability ara coutrclockwis From argumt pricipl N Z P That s why Z P N CL stabl if Z 0 i.. PN Thus coutr-clockwis From argumt pricipl N Z P P Z That s why Z P N CL stabl if Z 0 i.. P N Thus coutr-clockwis Michal Šbk ARI-Pr-3-0 5

Exampl Automatické říí - Kybrtika a robotika OL trasfr fuctio is ustabl -> P = Nyquist graph is L ( a N = Accorig to th critrio, th clos loop will b stabl It is rally a stabl, CL charactristic polyomial is >> a=-,b= a = - + b = >> yquist(b/a >> a+b as = c( Michal Šbk ARI-Pr-3-0 6

Automatické říí - Kybrtika a robotika Exampl Dtrmi th CL stability of th iscrt systm with G( s s( s Samplig at frqucy 0.5 H (i.. samplig prio T = s With ro orr shapig (ZOH A with iscrt proportioal cotrollr L( KG( >> G=/(+s/s G = / s(s+ >> G3=c(tf(G, Trasfr fuctio:.35 + 0.594 --------------------- ^ -.35 + 0.353 Samplig tim: >> pk(g3 Zro/pol/gai:.353 (+0.53 ----------------- (- (-0.353 Samplig tim: K=;L=K*G3; yquist(l N 0, P 0 Z 0 >> pformat rootc >> Gp=sf(G3; >> K=;L=K*Gp; >> cl_char=l.um+l. cl_char = (+0.8540i(-0.8540i >> isstabl(cl_har_pol as = Michal Šbk ARI-Pr-3-0 7

Automatické říí - Kybrtika a robotika Exampl: Discrt PM a GM For systém G( s s( s with samplig frq. 5 H, ZOH a iscrt P cotrollr with K = Fi iscrt PM a GM >> G=c(tf(/(+s^/s,/5,'oh'; >> pk(g Zro/pol/gai: 0.00077(+3.38(+0.4 ---------------------------- (-(-0.887^ Samplig tim: 0. >> L=G;yquist(L GM.7 5B, PM 7.5º Cotius valus ar almost th sam: GM 6B PM º Corrctio: PM is PM spoj PM spoj 9T s 9 0.60. 3.5 7.5 Michal Šbk ARI-Pr-3-0 8