Problem Set 4 Solutions 1

Similar documents
Problem Set 5 Solutions 1

Problem Set 4 Solution 1

Chapter 5. Standard LTI Feedback Optimization Setup. 5.1 The Canonical Setup

6.245: MULTIVARIABLE CONTROL SYSTEMS by A. Megretski. Solutions to Problem Set 1 1. Massachusetts Institute of Technology

Problem Set 8 Solutions 1

10 Transfer Matrix Models

Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!

Problem Set 2 Solutions 1

Time Response Analysis (Part II)

6.241 Dynamic Systems and Control

Massachusetts Institute of Technology. Department of Electrical Engineering and Computer Science : MULTIVARIABLE CONTROL SYSTEMS by A.

Problem Set 3: Solution Due on Mon. 7 th Oct. in class. Fall 2013

Control Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control D-MAVT ETH Zürich

Dr Ian R. Manchester

Problem set 5 solutions 1

Lecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design.

Introduction & Laplace Transforms Lectures 1 & 2

Control Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control D-MAVT ETH Zürich

6.302 Feedback Systems Recitation 16: Compensation Prof. Joel L. Dawson

CONTROL DESIGN FOR SET POINT TRACKING

Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!

ECE 388 Automatic Control

Prüfung Regelungstechnik I (Control Systems I) Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!

Stability. X(s) Y(s) = (s + 2) 2 (s 2) System has 2 poles: points where Y(s) -> at s = +2 and s = -2. Y(s) 8X(s) G 1 G 2

Raktim Bhattacharya. . AERO 632: Design of Advance Flight Control System. Preliminaries

Exercise 1 (A Non-minimum Phase System)

( ) ( = ) = ( ) ( ) ( )

Zeros and zero dynamics

Topic # Feedback Control Systems

Introduction. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31

Outline. Control systems. Lecture-4 Stability. V. Sankaranarayanan. V. Sankaranarayanan Control system

Controls Problems for Qualifying Exam - Spring 2014

Exam. 135 minutes + 15 minutes reading time

Analysis of SISO Control Loops

Exercise 1 (A Non-minimum Phase System)

Control Systems. Root Locus & Pole Assignment. L. Lanari

Linear State Feedback Controller Design

Outline. Classical Control. Lecture 1

Hankel Optimal Model Reduction 1

2.161 Signal Processing: Continuous and Discrete Fall 2008

Transform Solutions to LTI Systems Part 3

Balanced Truncation 1

6.241 Dynamic Systems and Control

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

Raktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Dynamic Response

L2 gains and system approximation quality 1

An Internal Stability Example

Exam. 135 minutes, 15 minutes reading time

Chapter 6 - Solved Problems

Pole placement control: state space and polynomial approaches Lecture 2

Module 4. Related web links and videos. 1. FT and ZT

Fall 線性系統 Linear Systems. Chapter 08 State Feedback & State Estimators (SISO) Feng-Li Lian. NTU-EE Sep07 Jan08

EEE 184: Introduction to feedback systems

LTI Systems (Continuous & Discrete) - Basics

Classify a transfer function to see which order or ramp it can follow and with which expected error.

Control Systems. Laplace domain analysis

Grades will be determined by the correctness of your answers (explanations are not required).

ẋ n = f n (x 1,...,x n,u 1,...,u m ) (5) y 1 = g 1 (x 1,...,x n,u 1,...,u m ) (6) y p = g p (x 1,...,x n,u 1,...,u m ) (7)

4.1 Fourier Transforms and the Parseval Identity

Time Response of Systems

Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!

DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD

Massachusetts Institute of Technology

A DESIGN METHOD FOR SIMPLE REPETITIVE CONTROLLERS WITH SPECIFIED INPUT-OUTPUT CHARACTERISTIC

9.5 The Transfer Function

Control Systems 2. Lecture 4: Sensitivity function limits. Roy Smith

Topic # Feedback Control

ME 375 Final Examination Thursday, May 7, 2015 SOLUTION

Control Systems. Frequency domain analysis. L. Lanari

ECE 3620: Laplace Transforms: Chapter 3:

Notes for ECE-320. Winter by R. Throne

Robust fixed-order H Controller Design for Spectral Models by Convex Optimization

Chapter 2. Classical Control System Design. Dutch Institute of Systems and Control

Raktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency Response-Design Method

EL2520 Control Theory and Practice

Automatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year

Laplace Transform Analysis of Signals and Systems

Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!

STABILITY ANALYSIS. Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated using cones: Stable Neutral Unstable

Data Based Design of 3Term Controllers. Data Based Design of 3 Term Controllers p. 1/10

EE Experiment 11 The Laplace Transform and Control System Characteristics

Homework 7 - Solutions

EECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators.

H 2 Optimal State Feedback Control Synthesis. Raktim Bhattacharya Aerospace Engineering, Texas A&M University

Control Systems I. Lecture 9: The Nyquist condition

Software Engineering/Mechatronics 3DX4. Slides 6: Stability

D(s) G(s) A control system design definition

Feedback Stabilization over Signal-to-Noise Ratio Constrained Channels

LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593

Control Systems Design

Laplace Transforms and use in Automatic Control

GATE EE Topic wise Questions SIGNALS & SYSTEMS

ECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27

EL 625 Lecture 10. Pole Placement and Observer Design. ẋ = Ax (1)

1. Find the solution of the following uncontrolled linear system. 2 α 1 1

EE C128 / ME C134 Final Exam Fall 2014

Lecture 15: H Control Synthesis

MAE143 B - Linear Control - Spring 2017 Final, June 15th

Frequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability

e st f (t) dt = e st tf(t) dt = L {t f(t)} s

Transcription:

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6245: MULTIVARIABLE CONTROL SYSTEMS by A Megretski Problem Set 4 Solutions Problem 4T For the following transfer matrices G, find co-prime factorization G = D N, the natural state space for the associated transfer matrix model, and the corresponding natural state space model (a) discrete time transfer matrix G(z) = [ z z z z Answer: one possible co-prime factorization G = D N is given by [ z [ D(z) = z, N(z) = z z The corresponding canonical state space X H2(T) 2 consists of all constant functions [ a x(z), a R, with update/output equations [ x + = x + Version of December 9, 2 [ w, y = x( ),

2 which is equivalent to a(t + ) = a(t) + [ w(t), y(t) = [ a(t) Reasoning: the identity DG = N can be verified by a direct multiplication To see that the factorization G = D N is co-prime, ie that there exists ɛ > such that D(z) D(z) + N(z) N(z) ɛi z D + = {z C : z > }, note that D(z) is continuous on D + { }, and not singular for all z D + { } except z =, at which point D(z) D(z) + N(z) N(z) = D() D() + N() N() = 2I > Note that there is no co-prime factorization with D(z) = ( /z)i (the easy choice) because this would yield a double pole of D N at z = The canonical state space X for G is the subset of all functions V H 2 2(T) which can be approximated arbitrarily well by functions V H 2 2(T) of the form V = NW DY, where W, Y are Fourier transforms of square summable sequences w, y : Z R 2 such that w (t) = and y (t) = for t Using the fact that zn(z) and zd(z) are polynomials in z while comparing the coefficients at the z-expansions in the equality V = NW DY, one can see that V has to be a constant function of the form [ a V (z) = lim z{n(z)w (z) D(z)Y (z)} z To see that a can be an arbitrary real number, use [ [ az W (z) =, Y (z) = The update/output laws for the canonical state x(t) H 2 2(T) defined by x(t)(z) = z t [N(z) W t (z) D(z)Ỹt(z) + x()(z), where t t W t (z) = w(τ)z τ, Y t (z) = y(τ)z τ, τ= τ=

3 are given by where Substituting x + (t)(z) = z[x(t)(z) D(z)y(t) + N(z)w(t), x(t)(z) y(t) = D( ) [x(t)( ) + N( )w(t) [ a(t) [, D( ) =, N( ) = yields the answer (b) continuous time transfer matrix [ /s /s G(s) = /s /s Answer: one possible co-prime factorization G = D N is given by [ s [ D(s) = s+, N(s) = s+ s+ The corresponding canonical state space X H2(jR) 2 consists of all functions [ a x(s) s+, a R, with update/output equations dx(t)(s) dt = [ [ w(t), y(t) = 2x(t)(), which is equivalent to ȧ(t) = [ w(t), y(t) = [ a(t) Reasoning: the identity DG = N can be verified by a direct multiplication To see that the factorization G = D N is co-prime, ie that there exists ɛ > such that D(s) D(s) + N(s) N(s) ɛi s C + = {s C : Re{s} > },

4 note that D(s) is continuous on C + { }, and not singular for all s C + { } except s =, at which point D(s) D(s) + N(s) N(s) = D() D() + N() N() = 2I > Note that there is no co-prime factorization with D(s) = because this would yield a double pole of D N at s = s I (the easy choice) s+ The canonical state space X for G is the subset of all functions V H 2 2(jR) which can be approximated arbitrarily well by functions V H 2 2(jR) of the form V = NW DY, where W, Y are Fourier transforms of square integrable functions w, y : R R 2 such that w (t) = and y (t) = for t Using the fact that (s+)n(s) and (s+)d(s) are polynomials in s while comparing the inverse Fourier transforms of both sides of V = NW DY, one can see that V has to be a rational function of the form V (s) = [ a s+ To see that a can be an arbitrary real number, use [ 2a [ W (s) = s, Y (s) = The canonical state x(t) H 2 2(jR) is defined by where Substitution x(t)(s) = e ts [N(s)W t (s) D(s)Y t (s) + x()(s), W t (s) = t e τs w(τ)dτ, Y t (s) = x(t)(s) = [ a(t) s+, t e τs y(τ)dτ followed by differentiation with respect to t yields [ [ ȧ(t)/(s + ) a(t)/(s + ) = s + N(s)w(t) D(s)y(t), which is equivalent to the state space equations given above

5 r e K(s) u P (s) v Figure : Design objectives of Problem 2 Problem 42P Consider the feedback design setup from Figure Let us define closed loop bandwidth of the feedback system as the largest ω > such that S(jω) for all ω [, ω, where S = + P K is the closed loop sensitivity function (in this definition, the threshold is a bit arbitrary) It is frequently claimed that location of unstable zeros of P limits the maximal achievable closed loop bandwidth While mathematically this is not exactly true, the only way to achieve a substantially larger bandwidth is by making S(jω) extremely large at other frequencies You are asked to verify this for P (s) = s a s(s + 2a), where a > is a real parameter (determining location of the open loop zero), using hinfsynm to estimate the maximal bandwidth achievable by a stabilizing LTI controller K = K(s) of order not larger than 5, satisfying the closed loop sensitivity magnitude bound S(jω) < 2 at all frequencies ω, as a function of a Generate a plot of your estimate, as a function of a > Hint: write an algorithm which attempts to achieve a given closed loop bandwidth, using pre-designed low-pass filter, like the Butterworth filter, to incorporate the bandwidth constraint into H-Infinity optimization Once this is accomplished, use binary search to find (approximately) the maximal bandwidth

6 Conclusion: maximal achievable closed loop bandwidth grows linearly with a, at a rate of at least 43a The rate can be made slightly larger at the expense of using much higher controller gains See the plot on Figure 2 Approach: since Figure 2: Bandwidth vs a for Problem 42 ap (as) = P (s) = s s(s + 2) for every a >, controller K (s) stabilizes P and achieves bandwidth ab if and only if controller K (s) = K (as)/a stabilizes P and achieves bandwidth b Hence it is sufficient to find the best achievable bandwidth for a = Figure 3: Open loop model for Problem 42

7 We use SIMULINK models ps42desmdl (see Figure 3) and ps42testmdl (see Figure 4) to define, respectively, the open and closed loop systems It introduces two distur- Figure 4: Closed loop model for Problem 42 bance inputs w, w 2 and three cost outputs e, e 2, e 3 : w is the main input (reference/sensor noise); w 2 is artificial small disturbance, to prevent sensor singularity due to the imaginary axis pole of P ; e enforces the closed loop sensitivity bound S 2, frovided the overall closed loop L2 gain in not larger than ; e 2 is defined using a low-pass filter W such that W (jω) for ω b, where b is the desired closed loop bandwidth, and therefore enforces the bandwidth constraint frovided the overall closed loop L2 gain in not larger than ; e 3 is artificial small cost to prevent control singularity due to the fact that P is strictly proper MATLAB function ps42m controls design optimization and testing The essential part of its code is shown below: function [K,S=ps42(a,b,d) s=tf( s ); % useful constant P=(s-a)/(s*(s+2*a)); % the plant [A,B,C,D=butter(3,b, s ); % Butterworth filter W=ss(A,B,C,D); W=(/abs(squeeze(freqresp(W,i*b))))*W; assignin( base, d,d) % export variables assignin( base, P,P) assignin( base, W,W) p=linmod( ps42des ); % extract open loop

8 p=ss(pa,pb,pc,pd); [K,G=hinfsyn(p,,); % optimize controller fprintf( success flag for w=%f: %f<\n,b,norm(g,inf)) assignin( base, K,K) % export controller S=linmod( ps42test ); % extract closed loop fprintf( max(real(eig(a)))=%f\n,max(real(eig(sa)))) S=ss(Sa,Sb,Sc,Sd); fprintf( norm(s,inf)=%f\n,norm(s,inf)) ww=logspace(-2,5,); Sw=abs(squeeze(freqresp(S,i*ww))); k=find(sw>,); % find actual bandwidth w=ww(k-); fprintf( actual bandwidth: %f (vs %f)\n,w,b) At a =, b = 43, d = the code generates a stabilizing controller K of order 5 which yields closed loop bandwidth of slightly more than 43 Logarithmic plots of the closed loop sensitivity function and controller are provided on Figure 5 Problem 43P Continuous time scalar signal q = q(t) models the (one dimensional) position of an oscillator driven by random forces in the absence of friction, according to q(t) + ω 2 q(t) = f (t), where f is a noise signal, and ω > is a parameter The result g(t) of measuring q(t) in real time is modeled according to g(t) = q(t) + f 2 (t), where f 2 is another noise signal Assuming that f = [f ; f 2 is a normalized vector-valued white noise, use h2synm to find an LTI system (a filter ) which takes g = g(t) as an input and outputs an estimate ˆq = ˆq(t) of q = q(t), minimizing the steady state value J = lim t E[ e(t) 2 of the variance of the estimation error e = q ˆq Plot the minimal J as the function of ω > Hint: the system, as described, is not stabilizable (there is no provision for a feedback loop from g back to f ) Therefore, before applying h2synm, one has to massage the setup into a stabilizable format One

9 Figure 5: Closed loop Bode plots for Problem 42 way to do this is by starting with a particular filter F producing an estimate ˆq which achieves J <, and then using h2synm to design an LTI filter which takes g ˆq as measurement, and outputs the optimal estimate ˆδ of δ = q ˆq, so that ˆq = ˆq + ˆδ is the optimal estimate of q Conclusion: the minimal mean square estimation error decreases with as ω increases, as shown on Figure 6 The design uses H2 optimization according to the setup shown on Figure 7, where W (s) =, L(s) = s 2 + ω 2 s 2 + ω 2 s 2 + ω 2 + L s + L 2, F (s) = L s + L 2 s 2 + ω 2 + L s + L 2,

Figure 6: Minimal mean square estimation error in Problem 43 where L > and L 2 + ω > Here W represents the original oscillator, F is the classical output estimator for W, and L specifies the mandatory zeros of F Figure 7: Design setup for Problem 43 Indeed, for a state space model for W, eg ẋ = x 2, ẋ 2 = ωx 2 + f, y = x + f 2,

the classical observer for q = x is ˆx = ˆx 2 + L (y ˆx ), ˆx 2 = ωx 2 + L 2 (y ˆx ), ˆq = ˆx, which has transfer function F (from y to ˆq) Also, an estimator F achieves a finite mean square error if and only if both F and W F belong to class H 2, which is equivalent to F having representation of the form F = F + L for some H 2, which justifies the use of the setup from Figure 7 This SIMULINK diagram, as well as the testing model Figure 8: Test setup for Problem 43 shown on Figure 8, are controlled by MATLAB code ps43m, the essential parts of which are shown below: function [J,F=ps43(w,L) s=tf( s ); % useful constant F=(s*L()+L(2))/(s^2+w^2+s*L()+L(2)); W=/(s^2+w^2); L=(s^2+w^2)/(s+w+)^2; assignin( base, F,F) % export variables assignin( base, W,W) assignin( base, L,L) p=linmod( ps43des ); % extract open loop p=minreal(ss(pa,pb,pc,pd)); % remove uncontrollable modes K=h2syn(p,,); % optimize F=minreal(F+K*L); % complete filter assignin( base, F,F) % export variables p=linmod( ps43test ); % extract error system G=minreal(ss(pa,pb,pc,pd)); % remove uncontrollable modes J=norm(G)^2;