Feedback Control of Dynamic Systems 2008

Feedback Control of Dynamic Systems Yves Briere [email protected]

I. Introduction

Introduction • Aim of the course – Give a general overview of classical and modern control theory – Give a general overview of modern control tools

• Prerequisites – Mathematics : complex numbers, linear algebra

9/23/2009

I. Introduction

3

Introduction • Tools – Matlab / Simulink

• Book – « Feedback Control of Dynamics Systems », Franklin, Powell, Amami-Naeini, Addison-Wessley Pub Co – Many many books, websites and free references...

9/23/2009

I. Introduction

4

Introduction 270 BC : the clepsydra and other hydraulically regulated devices for time measurement (Ktesibios)

9/23/2009

I. Introduction

5

Introduction 1136-1206 : Ibn al-Razzaz alJazari “The Book of Knowledge of Ingenious Mechanical Devices” crank mechanism, connecting rod, programmable automaton, humanoid robot, reciprocating piston engine, suction pipe, suction pump, double-acting pump, valve, combination lock, cam, camshaft, segmental gear, the first mechanical clocks driven by water and weights, and especially the crankshaft, which is considered the most important mechanical invention in history after the wheel

9/23/2009

I. Introduction

6

Introduction 1600-1900 : pre-industrial revolution Thermostatic regulators (Cornelius Drebbel 1572 1633)

Water level regulation (flush toilet, steam machine)

Windmill speed regulation. 1588 : mill hoper ; 1745 : fantail by Lee ; 1780 : speed regulation by Mead

9/23/2009

Steam engine pressure regulation (D. Papin 1707)

Centrifugal mechanical governor ( James Watt, 1788) I. Introduction

7

Introduction 1800-1935 : mathematics, basis for control theory Differential equations first analysis and proofs of stability condition for feedback systems (Lagrange, Hamilton, Poncelet, Airy-1840, Hermite-1854, Maxwell-1868, Routh-1877, Vyshnegradsky-1877, Hurwitz-1895, Lyapunov-1892)

Frequency domain approach (Minorsky-1922, Black-1927, Nyquist-1932, Hazen-1934)

1940-1960 : classical period Frequency domain theory : (Hall-1940, Nichols-194, Bode-1938) Stochastic approach (Kolmogorov-1941, Wiener and Bigelow-1942) Information theory (Shannon-1948) and cybernetics (Wiener-1949)

9/23/2009

I. Introduction

8

Introduction 1960-1980 : modern period, aeronautics and spatial industry Non linear and time varying problems (Hamel-1949, Tsypkin-1955, Popov1961, Yakubovich-1962, Sandberg-1964, Narendra-1964, Desoer-1965,Zames1966) Optimal control and Estimation theory (Bellman-1957, Pontryagin-1958,Kalman1960) Control by computer, discrete systems theory : (Shannon-1950, Jury-1960, Ragazzini and Zadeh-1952, Ragazzini and Franklin-1958,(Kuo-1963, Aström-1970) 1980-... : simulation, computers, etc...

9/23/2009

I. Introduction

9

Introduction What is automatic control ? Basic idea is to enhance open loop control with feedback control This seemingly idea is tremendously powerfull Feedback is a key idea in control Perturbation Input reference

Open loop Controler

Input

Process

Output

Perturbation Input reference

Closed loop Controler

Input

Process

Output

Measurement

9/23/2009

I. Introduction

10

Introduction Example : the feedback amplifier Harold Black, 1927 R2 R1 V1

- A +

V2 R2 =− ⋅ V1 R1

Amplifier A has a high gain (say 40dB) V2

1

1+

1  R2  1 +  A R1 

≈−

R2 R1

Resulting gain is determined by passive components ! amplification is linear reduced delay noise reduction 9/23/2009

I. Introduction

11

Introduction Use of block diagrams Capture the essence of behaviour standard drawing abstraction information hiding points similarities between systems

Same tools for : generation and transmission of energy transmission of informaiton transportation (cars, aerospace, etc...) industrial processes, manufacturing mechatronics, instrumentation Biology, medicine, finance, economy... 9/23/2009

I. Introduction

12

Introduction Basic properties of feedback (1) V 2 = A ⋅ (V1 − k ⋅ V 2)

k V1

A

V2

+

V 2 ⋅ (1 + A ⋅ k ) = A ⋅ V1 1 V2 1 1 = ⋅ ≈ k V1 k 1 + 1 A⋅k

Resulting gain is determined by feedback !

9/23/2009

I. Introduction

13

Introduction Basic properties of feedback (2) : static properties kc

e

u

y

kp

r : reference e : error d : disturbance y : output kc : control gain Kp : process gain

d r

e

kc

u

y

kp

d

Open loop control :

y = kp ⋅ kc ⋅e + kp ⋅d

Closed loop control : y =

k p ⋅ kc 1 + kp ⋅ kc

⋅r +

kp 1 + k p ⋅ kc

⋅d

If kc is big enough y tend to r and d is rejected 9/23/2009

I. Introduction

14

Introduction Basic properties of feedback (2) : dynamics properties Closed loop control can : enhance system dynamics stabilize an unstable system make unstable a stable system !

9/23/2009

I. Introduction

15

Introduction The On-Off or bang-bang controller : u = {umax , umin} u

u

e

u

e

e

The proportional controller : u=kc.(r – y)

9/23/2009

I. Introduction

16

Introduction The proportional derivative controller

de(t ) u (t ) = k p ⋅ e(t ) + k d ⋅ dt Gives an idea of future : phase advance The proportional integral controller

u (t ) = k p ⋅ e(t ) + k i ⋅ ∫0 e(τ ) ⋅ dτ t

e(t) tends to zero !

9/23/2009

I. Introduction

17

II. A first controller design

9/23/2009


18

A first control design • Use of block diagrams • Compare feedback and feedforward control • Insight feedback properties : – Reduce effect of disturbances – Make system insensitive to variations – Stabilize unstable system – Create well defined relationship between output and reference – Risk of unstability de(t ) t u ( t ) = k ⋅ e ( t ) + k ⋅ + k ⋅ • PID controler : p d i ∫0 e(τ ) ⋅ dτ dt

9/23/2009


19

Cruise control A cruise control problem : • Process input : gas pedal u • Process output : velocity v • Reference : desired velocity vr • Disturbance : slope θ

F

θ

mg

Construct a block diagram • Understand how the system works • Identify the major components and the relevant signals • Key questions are : – Where is the essential dynamics ? – What are the appropriate abstractions ?

• Describe the dynamics of the blocks 9/23/2009


20

Cruise control vr

Controller

Throttle

Engine

F

Body v

ext. force

We made the assumptions : • Essential dynamics relates velocity to force • The force respond instantly to a change in the throttle • Relations are linear We can now draw the process equations

9/23/2009


21

Cruise control

Process linear equations :

m⋅

dv(t ) + k ⋅ v = F − m⋅g⋅θ dt

Reasonable parameters according to experience :

dv(t ) + 0.02 ⋅ v = u − 10 ⋅ θ dt

Where : • v in m.s-1 • u : normalized throttle 0 < u < 1 • θ slope in rad 9/23/2009


22

Cruise control Process linear equations :

dv(t ) + 0.02 ⋅ v = u − 10 ⋅ θ dt PI controller :

u (t ) = k ⋅ (v r − v(t )) + k i ⋅ ∫0 (v r − v(τ )) ⋅ dτ t

Combining equations leads to :

d 2 e(t ) de(t ) dθ (t ) + (0.02 + k ) ⋅ + ki ⋅ e(t ) = 10 ⋅ 2 dt dt dt Integral action Steady state and θ = 0 e = 0 ! 9/23/2009


23

Cruise control Now we can tune k and ki in order to achieve a given dynamics

d 2 e(t ) de(t ) dθ(t ) + (0.02 + k ) ⋅ + k i ⋅ e(t ) = 10 ⋅ dt dt dt

d 2 x (t ) dx (t ) + 2 ⋅ σ ⋅ ω0 ⋅ + ω0 2 ⋅ x (t ) = 0 dt dt How to choose ω0 and σ ?

9/23/2009


24

Cruise control Compare open loop and closed loop

Open loop

Closed loop

9/23/2009


25

Cruise control Compare different damping σ (ω0 = 0.1) σ = 0.5 σ=1 σ=2

9/23/2009


26

Cruise control Compare different natural frequencies ω0 (σ= 1) ω0 = 0.05 ω0 = 0.1 ω0 = 0.2

9/23/2009


27

Cruise control Control tools and methods help to : • Derive equations from the system • Manipulate the equations • Understand the equations (standard model) – – – –

Qualitative understanding concepts Insight Standard form Computations

• Find controller parameters • Validate the results by simulation END 1

9/23/2009


28

Standard models Standard models are foundations of the “control language” Important to : Learn to deal with standard models Transform problems to standard model The standard model deals with Linear Time Invariant process (LTI), modelized with Ordinary Differential Equations (ODE) :

d n y(t ) dt

9/23/2009

n

+ a1 ⋅

d n −1 y(t ) dt

n −1

... + a n ⋅ y(t ) = b1 ⋅

d n −1u (t ) dt

n −1


... + b n ⋅ u (t )

29

Standard models Example (fundamental) : the first order equation dy(t ) + a ⋅ y(t ) = 0 dt

⇒ y(t ) = y(0 ) ⋅ e −a ⋅t

dy(t ) + a ⋅ y(t ) = b ⋅ u (t ) dt

⇒ y(t ) = y(0 ) ⋅ e − a ⋅t + b ⋅ ∫0 e − a ⋅(t − τ ) ⋅ u (τ ) ⋅ dτ t

Input signal Initial conditions

9/23/2009


30

Standard models A higher degree model is not so different :

d n y(t ) dt

n

+ a1 ⋅

d n −1 y(t ) dt

n −1

... + a n ⋅ y(t ) = 0

Characteristic polynomial is :

A(s ) = s n + a 1 ⋅ s n −1... + a n If polynomial has n distinct roots αk then the time solution is :

y(t ) =

9/23/2009

∑ C k ⋅ e α k ⋅t n

k =1


31

Standard models Real αk roots gives first order responses :

Complex αk=σ±i.ω roots gives second order responses :

9/23/2009


32

Standard models General case (input u) :

d n y(t ) dt

n

+ a1 ⋅

⇒ y(t ) =

d n −1 y(t ) dt

n −1

... + a n ⋅ y(t ) = b1 ⋅

∑ C k (t ) ⋅ e α k ⋅t + ∫0 g(t − τ) ⋅ dτ n

d n −1u (t ) dt

n −1

... + b n ⋅ u (t )

t

k =1

Where : • Ck(t) are polynomials of t α k ⋅t ' ( ) ( ) g t = C t ⋅ e k ∑ • n

k =1

A system is stable if all poles have negative real parts

9/23/2009


33

Standard models Transfer function without knowing anything about Laplace transform it can be useful to store ak and bk coefficients in a convenient way, the transfer function :

B(s ) s n + a 1 ⋅ s n −1... + a n F(s ) = = A(s ) b1 ⋅ s n −1... + b n

9/23/2009


34

III. The Laplace transform

9/23/2009

III. Laplace transforms

35

Laplace transform (1) : convolution u

System

y

• We assume the system to be LINEAR and TIME INVARIANT The output (y) of the the system is related to the input (u) by the convolution :

y(t) = ∫ u(τ) ⋅ h(t −τ) ⋅ dτ +∞

−∞

• Example : u(t) is an impulsion (0 everywhere except in t = 0)

y(t) = h(t) h(t) is called the impulse response, h(t) describes completely the system • Causality : h(t) = 0 if t < 0 9/23/2009


36

Laplace transform (1) : definition y(t)

u(t)

U(s)

h(t)

Y(s) H(s)

Time space

Laplace space

t : real (time)

s : complex (frequency)

X(s) = ∫ x ( t ) ⋅ e −s⋅t ⋅ dt +∞

x(t)

c − j⋅∞ 1 x(t ) = ⋅ ∫ X(s) ⋅ e s⋅t ⋅ ds 2 ⋅ π ⋅ j c − j⋅∞

y(t) = ∫ u(τ) ⋅ h(t −τ) ⋅ dτ +∞

0

X (s) Y(s) = H(s) . U(s)

−∞

☺ Mathematical formulas are never used ! 9/23/2009


37

Laplace transform (2) : properties Impulse fonction t=0 : x(t) = infinite x(t)=0

X(s) = 1

Step fonction : t<0 : x(t) = 0 t>0 : x(t) = 1

X(s) = 1/s

Derivation :

d y(t) = x ( t ) dt

Y(s) = s.X(s) − x (0 + )

Sinusoïdal fonction :

y(t) = sin(ω ⋅ t )

9/23/2009

Y(s) =


1 s 2 + ω2 38

Laplace transform (3) : properties

Delay :

y(t) = x ( t − t d ) Initial value theorem :

Y(s) = X(s) ⋅ e − t d ⋅s y(0 + ) = lim(s ⋅ Y(s )) s →∞

Final value theorem (if limit exists) :

y( + ∞) = lim(s ⋅ Y(s )) s →0

9/23/2009


39

Laplace transform (4) : tables

From t to s 9/23/2009


40

Laplace transform (4) : tables

From s to t 9/23/2009


41

Laplace transform and differential equations a o x (t ) + a 1x (t ) + a 2 x(t ) = b 0 u (t ) + b1u (t ) + Theorem of differentiation

a o X(s ) + a1 ⋅ (s ⋅ X(s ) − x (0 + )) + a 2 (s ⋅ (s ⋅ X(s ) − x (0 + )) − x (0 + )) = b 0 U(s ) + b1 (s ⋅ U(s ) − u (0 + ))

9/23/2009


42

Laplace transform and differential equations a o X(s ) + a1 ⋅ (s ⋅ X(s ) − x (0 + )) + a 2 (s ⋅ (s ⋅ X(s ) − x (0 + )) − x (0 + )) = b 0 U(s ) + b1 (s ⋅ U(s ) − u (0 + ))

(a

)

2 + a ⋅ s + a ⋅ s ⋅ X(s ) − (a1 + a 2 ⋅ s ) ⋅ x (0 + ) − a 2 ⋅ x (0 + ) o 1 2

= (b 0 + b1 ⋅ s ) ⋅ U(s ) − b1 ⋅ u (0 + )

X(s ) =

(a1 + a 2 ⋅ s ) ⋅ x(0+ ) + a 2 ⋅ x (0+ ) − b1 ⋅ u (0+ ) b 0 + b1 ⋅ s ( ) U s + a o + a1 ⋅ s + a 2 ⋅ s 2 a o + a1 ⋅ s + a 2 ⋅ s 2 Initial conditions Transfer function

9/23/2009


43

Laplace transform and differential equations

( b 0 + b1 ⋅ s a 1 + a 2 ⋅ s ) ⋅ x (0 + ) + a 2 ⋅ x (0 + ) − b1 ⋅ u (0 + ) X(s ) = U(s ) + 2 a o + a1 ⋅ s + a 2 ⋅ s a o + a1 ⋅ s + a 2 ⋅ s 2

X(s ) = H(s ) ⋅ U(s ) + I(s )

x (t ) = h (t ) ∗ u (t ) + i (t ) ☺ Initial conditions are generally assumed to be null ! 9/23/2009


44

Finding output response with Laplace transform

u(t)

System

y(t)

What is the output y(t) from a given input u(t) ?

y(t)

u(t)

Table of transform

Table of transform Y(s) = H(s) . U(s)

Y(s)

U(s)

9/23/2009


45

Finding final value with Laplace transform

u(t)

y(t) System

What is the final output y(inf) from a given input u(t) ? y( + ∞) = lim(s ⋅ Y (s ))

u(t) Table of transform

s →0

Theorem of final value Y(s) = H(s) . U(s) Y(s)

U(s)

9/23/2009


46

Poles and zeros U

Y

H(s)

Transfer function is a ratio of polynomials : b 0 + b1s + b 2 s 2 + ... Y(s) N(s) = H(s) = = U(s) D(s) a 0 + a 1s + a 2 s 2 + a 3 s 3 + ...

Poles and zeros : b (s − z1 ) ⋅ (s − z 2 ) ⋅ ⋅ ⋅ Y(s) = H(s) = 0 ⋅ U(s) a 0 (s − p1 ) ⋅ (s − p 2 ) ⋅ (s − p 3 ) ⋅ ⋅ ⋅

poles : p1, p2, p3…

Im p1

z1 p3

p2 9/23/2009

zero : z1, z2, …

Re z2

• Poles and zeros are either into the left plane ore into the right plane • Complex poles and zeros have a conjugate


47

• Poles are the roots of Transfer function denominator – Real values or conjugate complex pairs

• Poles are also the eigenvalues of matrix A • Poles = modes

9/23/2009


48

Poles and zeros : decomposition U

H(s)

Y

Transfer function can be expansed into a sum of elementary terms : b 0 + b1s + b 2 s 2 + ... Y(s) = H(s) = U(s) a 0 + a 1s + a 2 s 2 + a 3 s 3 + ...

α3 α1 α2 Y(s) = H(s) = + + + ... U(s) s − p1 s − p 2 s − p 3 j⋅θ − j⋅θ p1 and p2 are conjugate :p1 = −ω0 ⋅ e , p2 = −ω0 ⋅ e

α 1, 2 α3 Y(s) = H(s) = 2 + + ... 2 U(s) s − p3 s + 2 ⋅ ω0 ⋅ cos θ ⋅ s + ω0

First orders

Second orders

☺ Complex system response is the sum of first order and second order systems responses 9/23/2009


49

Dynamic response of first order systems U

H(s) =

Y(s ) =

1 s + p1

Y

1 U(s ) s + p1

Example 1 : impulse response u(t) is an impulsion (0 everywhere, except in 0 : ∞)

y( t ) = e − p1 ⋅t

u(t)

Table of transform

Table of transform

Y(s ) = H(s ) ⋅ U(s ) U(s)=1 9/23/2009


Y(s ) =

1 s + p1 50

Dynamic response of first order systems U

H(s) =

Y(s ) =

1 s + p1

1 U(s ) s + p1

Example 2 : step response u(t) is a step

(

1 y( t ) = 1 − e − p1 ⋅t p1

u(t)

9/23/2009

)

Table of transform

Table of transform 1 U(s ) = s

Y

Y(s ) = H(s ) ⋅ U(s )


Y(s ) =

1 1 ⋅ s + p1 s 51

Properties of first order systems U

H(s) =

Y(s) =

1 s + p1

Y

1 U(s) s + p1

Step response t1 = 1/p1 is the time constant of the system : after t = t1, 63% of the final value is obtained

9/23/2009


52

Dynamic response of second order systems U

H(s) =

Y

1 s 2 + 2 ⋅ σ ⋅ ω0 ⋅ s + ω0

2


u(t)

y( t ) =

1 ω0 1 − σ

(

)

Table of transform

Table of transform

Y(s ) = H(s ) ⋅ U(s ) U(s)=1 9/23/2009

2

⋅ e −σ⋅ωà ⋅t ⋅ sin ω0 1 − σ 2 ⋅ t


Y(s) =

1 s 2 + 2 ⋅ σ ⋅ ω0 ⋅ s + ω0

2

53


H(s) =

Y

1 s 2 + 2 ⋅ σ ⋅ ω0 ⋅ s + ω0

2

Example 2 : step response u(t) is an step

u(t)

y( t ) = 1 −

1 ω0 1 − σ

2

)

Table of transform

Table of transform

Y(s ) = H(s ) ⋅ U(s ) U(s)=1/s 9/23/2009

(

⋅ e −σ⋅ωà ⋅t ⋅ sin ω0 1 − σ 2 ⋅ t + ar cos(σ )


Y(s) =

1 s 2 + 2 ⋅ σ ⋅ ω0 ⋅ s + ω0

2

⋅

1 s

54

Properties of second order systems U

H(s) =

Step response : y( t ) = 1 −

Y

1 s 2 + 2 ⋅ σ ⋅ ω0 ⋅ s + ω0

1 ω0 1 − σ

2

2

(

)


σ is the damping factor ω0 is the natural frequency

overshoot

ω 0 1 − σ 2 is the pseudo-frequency 5% of the final value is obtained after : t 5% ≈

3 σ ⋅ ω0

Overshoot increases as σ decreases

9/23/2009


55


H(s) =

Y

1 s 2 + 2 ⋅ σ ⋅ ω0 ⋅ s + ω0

2

Step response (continued) :

±θ poles : p1, 2 = −ω0 ⋅ e cos(θ)=σ

Im

ω

0

θ

9/23/2009

Re


56

Im

Unstable

Stable

Stability

Re unstable pole, deverges like exp(t) stable pole, decays like exp(-4.t) Any pole with positive real part is unstable Any input (even small) will lead to instability See animation 9/23/2009


57

fast slow

« fast poles » vs « slow poles »

Im

Re

slow pole, decays like exp(-t) constant time : t1 = 1s fast pole, decays like exp(-4.t) constant time : t1 = 4s

Fast poles can be neglected See animation 9/23/2009


58

Effect of zeros See animation Im • Fast zero : neglected

Re

• Slow zero : transient response affected • Positive zero : non minimal phase system, step response start out in the wrong direction

Zeros modify the transient response 9/23/2009


59

Ex. analysis of a feedback system Process model :

dv(t ) + 0.02 ⋅ v = u − 10 ⋅ θ dt Transfer functions :

s ⋅ V(s ) + 0.02 ⋅ V(s ) = U(s )  s ⋅ V(s ) + 0.02 ⋅ V(s ) = −10 ⋅ θ(s )

1  V(s ) = F ( s ) =  U(s ) 0.02 + s  ⇒  V(s ) = − 10  θ(s ) 0.02 + s

9/23/2009


60

Ex. analysis of a feedback system Transfer function of the controller (PID) :

de(t ) t u (t ) = k ⋅ e(t ) + k d ⋅ + k i ⋅ ∫0 e(t ) ⋅ dτ dt U (t ) 1 ⇒ = k + kd ⋅s + ki ⋅ E(t ) s

9/23/2009


61

Ex. analysis of a feedback system We can now combine transfer functions : vr

u PID e

F

v

-10 θ

V(s ) =

9/23/2009

1 −10 ⋅ Vr (s ) + ⋅ E (s ) 1 + F(s ) ⋅ PID(s ) 1 + F(s ) ⋅ PID(s )


62

IV. Design of simple feedback

9/23/2009

IV. Design of simple feedbacks

63

Introduction Standard problems are often first orders or second orders • Standard problem standard solution d r

u C(s)

y P(s)

-

b P(s ) = s+a

9/23/2009

P(s ) =

b1s + b 2 s 2 + a1 ⋅ s + a 2


64

Control of a first order system Most physical problems can be modeled as first order systems Step 1 : transform your problem in a first order problem :

P(s ) =

b s+a

Step 2 : choose a PI controller

ki C(s ) = k + s Step 3 : combine equations and tune k and in ki in order to achieve the desired closed loop behavior (mass-spring damper analogy)

ki  b  ⋅k +  P(s ) ⋅ C(s ) 1 + b'⋅s s+a  s  = CL(s ) = = K⋅ ki  b  1 + P(s ) ⋅ C(s ) 2⋅σ s2 1+ ⋅k +  1+ ⋅s + 2 s+a  s  ω0 ω0

9/23/2009


65


H(s) =

Y

1 s 2 + 2 ⋅ σ ⋅ ω0 ⋅ s + ω0

2


u(t)

y( t ) =

1 ω0 1 − σ

Y(s ) = H(s ) ⋅ U(s )

9/23/2009

(

)

Table of transform

Table of transform U(s)=1

2

⋅ e −σ⋅ωà ⋅t ⋅ sin ω0 1 − σ 2 ⋅ t

Y(s) =


1 s 2 + 2 ⋅ σ ⋅ ω0 ⋅ s + ω0

2

66


H(s) =

Y

1 s 2 + 2 ⋅ σ ⋅ ω0 ⋅ s + ω0

2

Example 2 : step response u(t) is an step

u(t)

y( t ) = 1 −

1 ω0 1 − σ

Table of transform

9/23/2009

(

)

Table of transform

Y(s ) = H(s ) ⋅ U(s ) U(s)=1/s

2


Y(s) =


1 s 2 + 2 ⋅ σ ⋅ ω0 ⋅ s + ω0

2

⋅

1 s

67


H(s) =

Step response : y( t ) = 1 −

Y

1 s 2 + 2 ⋅ σ ⋅ ω0 ⋅ s + ω0

1 ω0 1 − σ

σ is the damping factor ω0 is the natural frequency

2

2

(

)


overshoot

ω 0 1 − σ 2 is the pseudo-frequency 5% of the final value is obtained after : t 5% ≈

3 σ ⋅ ω0

Overshoot increases as σ decreases

9/23/2009


68


H(s) =

Y

1 s 2 + 2 ⋅ σ ⋅ ω0 ⋅ s + ω0

2

Step response (continued) :

±θ poles : p1, 2 = −ω0 ⋅ e cos(θ)=σ

Im

ω

0

θ

9/23/2009

Re


69

Control of a second order system

Step 1 , step 2 : idem (PI controller) Step 3 : Transfer function is now third order

CL(s ) =

P(s ) ⋅ C(s ) = K⋅ 1 + P(s ) ⋅ C(s )

1 + b'⋅s

2  2⋅σ s (1 + a ⋅ s )1 + ⋅s + 2 ω0 ω0 

   

2 dof (k and ki) : the full dynamics (order 3) cannot be totally chosen

9/23/2009


70

Simulation tools Matlab or Scilab Transfer function is a Matlab object Adapted to transfer function algebra (addition, multiplication…) Simulation, time domain analysis

9/23/2009


71

Conclusion

Laplace Transform + Simulation tools Design of simple feedbacks

9/23/2009

VI. Design of simple feedbacks (Ctd)

72

V. Frequency response

9/23/2009


73

Introduction Frequency response : • One way to view dynamics • Heritage of electrical engineering (Bode) • Fits well block diagrams • Deals with systems having large order – electronic feedback amplifier have order 50-100 ! • input output dynamics, black box models, external description • Adapted to experimental determination of dynamics

9/23/2009


74

The idea of black box u

System

y

The system is a black box : forget about the internal details and focus only on the input-output behavior Frequency response makes a “giant table” of possible inputs-outputs pairs Test entries are enough to fully describe LTI systems ☺ - Step response - Impulse response - sinusoids

9/23/2009


75

What is a LTI system A Linear Time Invariant System is : • Linear

If (u1,y1) and (u2,y2) are input-output pairs then (a.u1+ b.u2 , a.y1+ b.y2) is an input-output pair : Theorem of superposition

• Time Invariant

(u1(t),y1(t)) is an input-output pair then (u1(t-T),y1(t-T)) is an inputoutput pair

The “giant table” is drastically simplified : y(t) = ∫− ∞ h (t − τ ) ⋅ u (τ ) ⋅ dτ +∞

⇒ Y (s ) = H (s ) ⋅ U(s )

9/23/2009


76

What is the Fourier Transform Fourier’s idea : an LTI system is completely determined by its response to sinusoidal signals • Transmission of sinusoid is given by G(jω) • The transfer function G(s) is uniquely given by its values on the imaginary axes • Frequency response can be experimentally determined The complex number G(jω) tells how a sinusoid propagates through the system in steady states : u(t) = sin (ω ⋅ t )

⇒ y(t) = G ( j ⋅ ω) ⋅ sin (ω ⋅ t + arg(G ( j ⋅ ω)))

9/23/2009


77

Steady state response Fourier transform deals with Steady State Response : u(t) = cos(ω0 ⋅ t ) + i ⋅ sin (ω0 ⋅ t ) = e i⋅ω0 ⋅t 1 ⇒ U(s ) = s − i ⋅ ω0 ⇒ Y(s ) = G (s ) ⋅

G (i ⋅ ω0 ) Rk 1 = +∑ s − i ⋅ ω0 s − i ⋅ ω0 s − αk

⇒ y(t ) = G (i ⋅ ω0 ) ⋅ e i⋅ω0 ⋅t + ∑ R k ⋅ e α k ⋅t

(System has distinct poles αk)

Decays if all αk are negatives 9/23/2009


78

Steady state response

9/23/2009


79

Nyquist stability theorem Nyquist stability theorem tells if a system WILL BE stable (or not) with a simple feedback u(t) C(s).

P(s).

y(t) L(s)=C(s).P(s)

y(t)

(1) Standard system with negative unitary feedback

-1

(2) Nyquist standard form

(2) : if L(i.ω0) = -1 then oscillation will be maintained

9/23/2009


80

Nyquist stability theorem Step 1 : draw Nyquist curve Imag ω=+∞

Real ω=0

ω=-∞

L(i.ω)

Step 2 : where is (-1,0) ?

9/23/2009


81

Nyquist theorem

When the transfer loop function L does not have poles in the right half plane the closed loop system is stable if the complete Nyquist curve does not encircle the critical (-1,0) point. When the transfer loop function L has N poles in the right half plane the closed loop system is stable if the complete Nyquist curve encircle the critical (-1,0) point N times.

9/23/2009


82

Nyquist stability theorem Nyquist stability theorem compares L(i.ω) with (-1,0) Imag

Real (-1,0)

ω=0 L(i.ω)

(-1,0)

(-1,0) STABLE 9/23/2009

UNSTABLE V. Frequency response

83

Nyquist theorem

• Focus on the characteristic equation • Difficult to see how the characteristic equation L is influenced by the controller C Question is : how to change C ?

• Strong practical applications • Possibility to introduce stability margin : how close to instability are we ?

9/23/2009


84

Stability margin Stability margin definitions : ϕM Phase margin 1/gM

45°- 60° gM Gain margin 2-6

(-1,0)

d

d Shortest distance to critical point

ϕM

0.5 - 0.8

9/23/2009


85

The Bode plot Nyquist theorem is spectacular but not very efficient… Impossible to distinguish C(s) and P(s) Bode plots two curves : one for gain, one for phase :

dB scale for gain Logarithmic frequency axis Linear scale for phase 9/23/2009


86

The Bode plot Bode’s plot main properties : Asymptotic curves (gain multiple of 20dB/dec) are ok Simple interpretation of C(s) and P(s) in cascade : GaindB(C(s).P(s) = GaindB(C(s)) + GaindB(P(s)) Phase(C(s).P(s) = Phase(C(s)) + Phase(P(s))

9/23/2009


87

The Bode stability criteria

Gain Margin > 0 : closed loop system will be stable

Phase Margin > 0 : closed loop system will be stable

One criteria is sufficient in most cases because gain and margin are closely related 9/23/2009


88

Close loop frequency response Typical property of Hbo(s) are :

H bo ( j ⋅ ω) >> 1 for ω << ωc H bo ( j ⋅ ω) << 1 for ω >> ωc

H bf ( j ⋅ ω) =

H bo ( j ⋅ ω) ≈ 1 for ω << ωc 1 + H bo ( j ⋅ ω)

H bo ( j ⋅ ω) H bf ( j ⋅ ω) = ≈ H bo ( j ⋅ ω) for ω >> ωc 1 + H bo ( j ⋅ ω)

9/23/2009


89

Close loop frequency response AdB

ωc Log10ω

Open loop Bode plot Close loop Bode plot

9/23/2009


90

Close loop frequency response Phase margin effect AdB

ωc Log10ω

Open loop Bode plot Close loop Bode plot (phase margin 90°) Close loop Bode plot (phase margin 30°)

9/23/2009


91

Static gain H bo ( j ⋅ ω) H bf ( j ⋅ ω) = 1 + H bo ( j ⋅ ω) For small value of ω (low frequency) :

9/23/2009

If Hbo(ω) << 1

then

Hbf(ω) ≅ Hbo(ω)

If Hbo(ω) >> 1

then

Hbf(ω) ≅ 1

If Hbo(ω) → ∞

then

Hbf(ω) → 1


92

Controller specifications ε

e(s)

C(s)

u(s)

G(s)

y(s)

-

Open loop transfer function : OLTF = C.G Close loop transfer function : CLTF = C.G / (1 + C.G)

9/23/2009


93

Controller specifications • Static gain close to 1 → Low frequency : high gain

• Perturbation rejection → High frequency : low gain |C(jω).G(jω)|

• Stability → phase margin > 0

• Bandwith

ωc

→ Cross over frequency ωc

• Overshoot ≈ 25% → phase margin ≈ 45° → Gentle slope in transition region 9/23/2009


94

Controller design • Proportionnal feedback

C (s ) = K

– Effect : lifts gain with no change in phase – Bode : shift gain by factor of K

9/23/2009


95

Controller design • Lead compensation

1+τ ⋅ s C (s ) = K 1 + A ⋅τ ⋅ s

– Effect : lifts phase by increasing gain at high frequency – Very usefull controller : increase phase margin – Bode : add phase between zero and pole

9/23/2009


96

Modern loop shaping • Use of rltool (Matlab Control Toolboxe)

9/23/2009


97

VI. Design of simple feedback (Ctd)

9/23/2009


98

Introduction More complete standard problem :

d e

r F(s)

n x

u C(s)

y

P(s)

-

Controller : feedback C(s) and feedforward F(s) Load disturbance d : drives the system from its desired state x Measurement disturbance n : corrupts information about x Main requirement is that process variable x should follow reference r

9/23/2009


99

Introduction Controller’s specifications : A. Reduce effects of load disturbance B. Does not inject too much measurement noise into the system C. Makes the closed loop insensitive to variations in the process D. Makes output follow reference signal

Classical approach : deal with A,B and C with controller C(s) and deal with D with feedforward F(s)

9/23/2009


100

Introduction Controller’s specifications : A. Reduce effects of load disturbance B. Does not inject too much measurement noise into the system C. Makes the closed loop insensitive to variations in the process D. Makes output follow reference signal Classical approach : deal with A,B and C with controller C(s) and deal with D with feedforward F(s) : Design procedure • Design the feedback C(s) too achieve – – –

•

Small sensitivity to load disturbance d Low injection of measurement noise n High robustness to process variations

Then design F(s) to achieve desired response to reference signal r 9/23/2009


101

Relations between signals Three interesting signals (x, y, u) Three possible inputs (r, d, n) Nine possible transfer functions !

P P⋅C P⋅C⋅F ⋅d + ⋅n + ⋅r 1+ P ⋅ C 1+ P ⋅ C 1+ P ⋅ C P 1 P⋅C⋅F y= ⋅d + ⋅n + ⋅r 1+ P ⋅ C 1+ P ⋅ C 1+ P ⋅ C P⋅C C C⋅F u= ⋅d + ⋅n + ⋅r 1+ P ⋅ C 1+ P ⋅ C 1+ P ⋅ C x=

Six distinct transfer functions…

9/23/2009


102

Relations between signals Nine frequency responses… Bode Diagram From: n

From: d

From: r

To: x

0 -20

Magnitude (dB) To: y

-40

0 -20 -40

To: u

0 -20 -40 10

-1

10

0

1 -1

1010

10

0

1 -1

1010

10

0

10

1

Frequency (rad/sec) 9/23/2009


103

Relations between signals Nine step responses… Step Response From: n

From: d

From: r

To: x

2 0

Amplitude To: y

-2 2 1 0

To: u

-1 2 0 -2 0 9/23/2009

10

20

300

10 20 Time (sec)

300

10

20


30 104

Relations between signals A correct design means that each transfer has to be evaluated… Need to be a little bit organized ! Need less criteria

9/23/2009

Concept of sensibility functions


105

Sensibility functions P P⋅C P⋅C⋅F ⋅d + ⋅n + ⋅r 1+ P ⋅C 1+ P ⋅ C 1+ P ⋅C P 1 P⋅C⋅F ⋅d + ⋅n + ⋅r y= 1+ P ⋅C 1+ P ⋅ C 1+ P ⋅C P⋅C C C⋅F ⋅d + ⋅n + ⋅r u= 1+ P ⋅C 1+ P ⋅ C 1+ P ⋅C x=

L = P⋅C 1 1 S= = 1+ L 1+ P ⋅ C L P⋅C T= = 1+ L 1+ P ⋅ C

Loop sensitivity function Sensibility function Complementary sensibility function

L tells everything about stability : common denominator of each transfer functions

9/23/2009


106

Sensibility functions L=PC tells everything about stability : common denominator of each transfer functions

Magnitude (dB)

Bode Diagram 0

-20

-40

Phase (deg)

-90 -180 -270 -360 -1 10

9/23/2009

0

10 Frequency (rad/sec)

10


107

Sensibility functions S=1/(1+L) tells about noise reduction d e

r=0

n

u C

x

y

P

-

Without feedback : y ol = n + P ⋅ d

With feedback control : 1 P y cl = n+ ⋅ d = S ⋅ y ol 1+ P ⋅ C 1+ P ⋅ C

Disturbances with |S(iω)| < 1 are reduced by feedback Disturbances with |S(iω)| > 1 are amplified by feedback

9/23/2009


108

Sensibility functions It would be nice to have |S(iω)| < 1 for all frequencies !

Cauchy Integral Theorem : for stable open loop system :

∫

(i ⋅ ω) = 0

∞ log S 0

For unstable or time delayed systems :

∫

(i ⋅ ω) > 0

∞ log S 0

Conclusion : water bed effect… log|S(iω)|

ω

9/23/2009


109

Sensibility functions Nyquist stability criteria :

∀ω, C∆P < 1 + L ⇔ ∆P P

<

1 T

(-1,0)

1/T tells how much P is allowed to vary until system becomes unstable

d = 1+ L C∆P

9/23/2009


110

Sensibility functions Nyquist stability criteria : Minimum value of d tells how close of instability is the system dmin is a measure of robustness : the bigger is M=1/d the more robust is the system (-1,0) log|S(iω)| d = 1+ L C∆P

M ω

9/23/2009


111

VII. Feedforward design

9/23/2009


112

Introduction Feedforward is a useful complement to feedback. Basic properties are: + Reduce effects of disturbance that can be measured + Improve response to reference signal + No risk for instability - Design of feedforward is simple but requires good model and/or measurements + Beneficial when combined with feedback

9/23/2009


113

Attenuation of measured disturbance d F u

y P1

P2

Y = P2 ⋅ (1 − P1 ⋅ F ) D

Disturbance is eliminated if F is chosen such as: F = P1-1 Need to measure d P1 needs to be inversible 9/23/2009


114

Combined Feedback and Feedforward d F r

u C

P1

P2

y

Disturbance d is attenuated both by F and C : Y P2 ⋅ (1 − P1 ⋅ F ) = D 1+ P ⋅ C

9/23/2009


115

System inverse

The ideal feedforward needs to compute the inverse of P1. That’s might be tricky… Examples: 1 1+ s

F ( s ) = P −1 ( s ) = 1 + s

Differentiation

e −s P (s ) = 1+ s

F ( s ) = P −1 ( s ) = (1 + s ) ⋅ es

Prediction

F ( s ) = P −1 ( s ) =

P (s ) =

P (s ) =

9/23/2009

1− s 1+ s

1+ s 1− s


Unstable

116

Approximate system inverse

The ideal feedforward needs to compute the inverse of P1. That’s might be tricky… Examples: 1 1+ s

F ( s ) = P −1 ( s ) = 1 + s

Differentiation

e −s P (s ) = 1+ s

F ( s ) = P −1 ( s ) = (1 + s ) ⋅ es

Prediction

F ( s ) = P −1 ( s ) =

P (s ) =

P (s ) =

9/23/2009

1− s 1+ s

1+ s 1− s


Unstable

117

Approximate system inverse Since it is difficult to obtain an exact inverse we have to approximate. One possibility is to find the transfer function which minimizes : J = ∫0 ( u ( t ) − v ( t ) ) ⋅ dt ∞

Where: V = P⋅X⋅U

And where U is a particular input (ex: a step signal). This gives for instance: P (s ) =

1 1+ s

P ( s ) = e −s P (s ) = 9/23/2009

1− s 1+ s

1+ s 1+ T ⋅s

P −1 ( s ) ≈

P −1 ( s ) ≈ 1

F ( s ) = P −1 ( s ) = 1 VII. Feedforward design

118

Improved response to reference signal The reference signal can be injected after the controller: Mu r

My

ym

C

u

P2

y

ym is the desired trajectory. Choose Mu = My / P Design concerns: Mu approximated My adapted such that My/P feasible

9/23/2009


119

Combining feedback and feedforward

Feedback Closed loop Acts only when there are deviations Market driven Robust to model errors Risk for instability

Feedforward Open loop Acts before deviation shows up Planning Not robust to model errors No risk for instability

Feedforward must be used as a complement to feedback. Requires good modeling.

9/23/2009


120

VIII. State feedback

9/23/2009


121

Introduction -

Simple design becomes difficult for high order systems What is the State concept ? - State are the variables that fully summarize the actual state of the system - Future can be fully predicted from the current state - State is the ideal basis for control

9/23/2009


122

State feedback Let us suppose the system is described by the following equation (x is a vector, A, B and C are matrixes) :  dx  = A ⋅ x + B⋅u  dt y = C ⋅ x  The general linear controller is : u = −K ⋅ x + L ⋅ u

The closed loop system then becomes :  dx  = A ⋅ x + B ⋅ ( −K ⋅ x + L ⋅ u ) = ( A − B ⋅ K ) ⋅ x + B ⋅ L ⋅ u  dt y = C ⋅ x  The closed loop system has the characteristic equation:

P ( s ) = det ( s ⋅ I − ( A − B ⋅ K ) )

9/23/2009


123

State feedback Let us suppose the system is described by the following equation (x is a vector, A, B and C are matrixes) :  dx  = A ⋅ x + B⋅u  dt y = C ⋅ x  The general linear controller is :

Main mathematical tool is linear algebra and matrixes !

u = −K ⋅ x + L ⋅ u

The closed loop system then becomes :  dx  = A ⋅ x + B ⋅ ( −K ⋅ x + L ⋅ u ) = ( A − B ⋅ K ) ⋅ x + B ⋅ L ⋅ u  dt y = C ⋅ x  The closed loop system has the characteristic equation:

P ( s ) = det ( s ⋅ I − ( A − B ⋅ K ) )

9/23/2009


124

Pole placement Original (open loop) system behavior depends on its poles, solution of the characteristic equation:

POL ( s ) = det ( s ⋅ I − A ) Closed loop system behavior depends on its poles, solution of the characteristic equation:

PCL ( s ) = det ( s ⋅ I − ( A − B ⋅ K ) ) Needs to tune N parameters (N : dimension of x and K) Appropriate choice of K allow to place the poles anywhere ! (Needs simple mathematical skills (not detailed here ☺) Two problems : observability, controllability 9/23/2009


125

Pole placement

Poles of the OL system Poles of the CL system

9/23/2009


126

First problem : observability In the control feedback equation x is supposed to be known. If one can access (measure) x, there is no problem. Sometimes, x cannot be measured but can be observed. System described by:

 dx  = A ⋅ x + B⋅u  dt y = C ⋅ x 

Only u and y accessible, A and B known. Solution is to estimate internal state x with a “state observer” of gain Ko :  dx obs = A ⋅ x obs + B ⋅ u + K obs ⋅ ( y − yobs )   dt y = C ⋅ x obs  obs

Appropriate choice of Kobs minimizes yobs – y : xobs tends to x Poles of the observer are the poles of: 9/23/2009

P ( s ) = det ( s ⋅ I − ( A − K obs ⋅ C ) )


127

First problem : observability In the control feedback equation x is supposed to be known. If one can access (measure) x, there is no problem. Sometimes, x cannot be measured but can be observed. System described by:

 dx  = A ⋅ x + B⋅u  dt y = C ⋅ x 

Only u and y accessible, A and B known. Solution is to estimate internal state x with a “state observer” of gain Ko :  dx obs = A ⋅ x obs + B ⋅ u + K obs ⋅ ( y − yobs )   dt y = C ⋅ x obs  obs

Appropriate choice of Kobs minimizes yobs – y : xobs tends to x Poles of the observer are the poles of: 9/23/2009

P ( s ) = det ( s ⋅ I − ( A − K obs ⋅ C ) )


128

First problem : observability Poles of the observer are the poles those of:

P ( s ) = det ( s ⋅ I − ( A − K obs ⋅ C ) ) Poles of the system Poles of the observer

9/23/2009


129

First problem : observability Problem : is the system observable ?

In most cases : yes Sometimes, the state is not observable : The observer does not converge to the true state, whatever Kobs is. Can be derived from a mathematical analyses of (A,C):

rank(A,AC,AAC,AAAC…) = N

9/23/2009


130

Combining an observer and a state feedback

True (with x) state feedback can be replaced by an observed (xobs) state feedback:

u = −K ⋅ x obs + L ⋅ u

Poles of the OL system Poles of the observer Poles of the CL system

9/23/2009


131

Second problem : controllability Sometimes a state is not controllable : means that whatever the command u is, some parts of the state are not controllable Can be derived from a mathematical analyses of (A,B):

rank(A,AB,AAB,AAAB…) = N Problem if : - A state is not controllable and unstable - A state is not controllable and slow No problem if : - A state is not controllable and fast (decays rapidly)

9/23/2009


132

Feedback Control of Dynamic Systems 2008

Recommend Documents