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Control Systems Lab - SC4070

Control techniques

Dr. Manuel Mazo Jr.Delft Center for Systems and Control (TU Delft) [email protected]

Tel.:015-2788131

TU Delft, February 16, 2015(slides modified from the original drafted by Robert Babuska)

M. Mazo Jr. (DCSC/TUD) Dynamics 1 / 36

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Outline

1 Overview of control design methods

2 Continuous vs. discrete time design

3 State-feedback control, observers

4 Control architectures, nonlinear control

5 PID controllers


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Outline





5 PID controllers


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Linear control design methods

P, PD, PI, PID, lead-lag control (classical, in frequency)

state feedback, output feedback (modern, in state-space)

LQR, linear quadratic control (optimal)model predictive control (optimal, finite-horizon, constrained)

robust control (H ∞, µ−synthesis)


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Nonlinear control techniques

feedback linearization

sliding-mode control

nonlinear model predictive control

passivity-based control

knowledge-based control

adaptive control

hybrid control


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Outline





5 PID controllers


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Computer-controlled systems:approach 1

y(t)u(t)Algorithm

Clock

A-D D-A System

1 Design a continuous-time controller,

2 then make sure that the (digital) computer implementationapproximates the continuous-time controller as precisely as possible.


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Computer-controlled systems:approach 2

y(t)u(t)Algorithm

Clock

A-D D-A System

1 Describe the system from the computer’s (digital) viewpoint,

2 and design directly a discrete-time controller.


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System from the computer’s viewpoint

y t ( )u t ( )System

Clock

{ ( )} y t k { ( )}u t k A-DD-A

Describe system (including converters) only over discrete sampling

instants.


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A-D Converter: Zero-Order Hold

Timet t t t t t

0 1 2 3 4 5

u (t ) = u (t k ), t k ≤ t < t k +1

Usually (but not necessarily) t k +1 − t k = const = h.

Higher-order converters are also possible.


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Zero-order hold sampling of systems

Continuous-time system:

dx (t )

dt = Ax (t ) + Bu (t )

y (t ) = Cx (t ) + Du (t )

Discrete-time system:

x (k + 1) = Φx (k ) + Γu (k )

y (k ) = Cx (k ) + Du (k )

with (recall solution of non-autonomous ODE):

Φ = e Ah

, Γ = h

0e As

dsB , where h is the sampling period

Effect on eigenvalues placement.


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Useful basic MATLAB commands

G = ss(A,B,C,D); % LTI continuous-time state-space model

h = 0.1; % sampling period [s]

H = c2d(G,h); % convert to discrete time (ZOH)

H = c2d(G,h,method); % method = ’foh’, ’matched’, ...

G = d2c(H); % convert to continuous time (ZOH)


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Selection of sampling period(1st order)

Number of samples per rise time: N r = T r h ≈ 4− 10

0 5

−1

0

1(a)

0 50

1

0 5

−1

0

1(b)

0 50

1

0 5

−1

0

1(c)

0 50

1

0 5

−1

0

1(d)

Time

0 50

1

Time

(a) N r = 1

(b) N r = 2

(c) N r = 4

(d) N r = 8


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Selection of sampling period(2nd order)

N r = T r h ≈ 4− 10 corresponds to ω0h ≈ 0.2− 0.5

0 50

1

(a)

0 50

1

(b)

0 50

1

(c)

Time

0 50

1

(d)

Time

(a) h = 0.125 (ω0h = 0.23)

(b) h = 0.250 (ω0h = 0.46)

(c) h = 0.500 (ω0h = 0.92)

(d) h = 1.000 (ω0h = 1.83)


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State feedback in DT:problem formulation

Discretize LTI model choosing a sampling interval

Model : x (k + 1) = Φx (k ) + Γu (k )

Linear controller :u (k ) = −Lx (k )

Design parameters : closed-loop poles

Evaluation: compare x (k ) and u (k ) with specifications(trade-off between control magnitude and speed of response)


P l l A k ’ f l

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Poles placement: Ackermann’s formula

Compute L such that (Φ − ΓL) has a desired characteristic polynomialP (z ). Ackermann’s formula:

L = (0 . . . 0 1) C −1P (Φ)

where:

P (Φ) is the desired characteristic polynomial in Φ;

C is the controllability matrix of the pair (Φ, Γ).

Place poles inside unit ball. In Matlab:L = acker(Phi,Gamma,p) (SISO, numerical problems ?)

L = place(Phi,Gamma,p) (MISO, more robust)


Alt ti l h th d i d l

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Alternatively, choose the desired polesin Continuous-Time

Use the continuous-time 2nd order model, study its (continuous-time)characteristic polynomial:

s 2 + 2ζωs + ω2,

which (with zoh sampling) leads to:

z 2 + p 1z + p 2

where:

p 1 = −2e −ζωh cos

ωh

1− ζ 2

p 2 = e −2ζωh

In Matlab use c2d to get the discrete time model (Φ,Γ).


Li d ti t l LQR

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Linear quadratic control: LQR

J =N

k =1

x (k )T Qx (k ) + u (k )T Ru (k ),

where (matrices, weights) Q

, R

are design parameters.A state feedback matrix L that gives a minimal J can be found bysolving an associated algebraic Riccati equation.

Similar to pole placement, but no need to define poles!

In Matlab: dlqr(Phi,Gamma,Q,R) % state weighting

In Matlab: dlqry(Phi,Gamma,C,D,Q,R) % output weighting


O tli

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Outline





5 PID controllers


St t ti ti b

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State estimation: observers

x (k + 1) = Φx (k ) + Γu (k )

y (k ) = Cx (k )

Assume input and output are available, reconstruct the state:

Direct calculation;Luenberger observer (model-based);

Kalman filter (optimal in presence of Gaussian noise).

Note: the terms “observer”, “estimator”, “filter” are in this context usedsynonymously.


M del b ed e ti ti

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Model-based estimation

1 Consider the model:

x (k + 1) = Φx (k ) + Γu (k )

2 Introduce ”feedback” from measured y (k )

x (k + 1) = Φx (k ) + Γu (k ) + K [y (k )− C x (k )]

3 Define the estimation error e = x − x

e (k + 1) = Φe (k )− KCe (k ) = [Φ− KC ]e (k )


Observer block diagram

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Observer – block diagram

K

1z

u k ( ) x k+( 1)^ y k ( )^

y k ( )

C x k ( )^

x k ( )^


Output feedback (observer + state

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Output feedback (observer + statefeedback)

ˆx

(k

+ 1) = Φˆx

(k

) + Γu

(k

) + K y

(k

)− C

ˆx

(k

)

u (k ) = −Lx (k )


Poles of the closed loop system

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Poles of the closed-loop system

x (k + 1) = Φx (k ) + Γu (k )

e (k + 1) = (Φ− KC )e (k )

u (k ) = −L(x (k )− e (k ))

x (k + 1)

e (k + 1)

=

Φ− ΓL ΓL

0 Φ− KC

x (k )

e (k )

Separation principle:

(Closed-loop) Process poles: Ar (z ) = det(zI − Φ + ΓL)

Observer poles: Ao (z ) = det(zI − Φ + KC )


Outline

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Outline





5 PID controllers


Feed-forward (two-degree-of-freedom)

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Goal: respond to a reference signal with desired specs.

Approach: Replace u (k ) = −Lx (k ) by: u (k ) = −Lx (k ) + Lc u c (k )

u x^

Lc

Observer

Process-L

uc

y



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Closed-loop system:

x (k + 1) = (Φ− ΓL)x (k ) + ΓLe (k ) + ΓLc u c (k )

e (k + 1) = (Φ− KC )e (k )

y (k ) = Cx (k )

Transfer function from u c to y (for impulse response):

H cl (z ) = C (zI − Φ + ΓL)−1ΓLc = Lc

B (z )

Ar (z )


Application to a nonlinear system

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Application to a nonlinear system

Feedback controller

Reference Nonlinear

system

y0

y

u0

y0

Feedforward controller

δ

δ


Control by local linear controller

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Control by local linear controller

u

u0

Linear

controller

Nonlinear system

y0

y

Linearized

model

δ

δ


Model-based adaptive control

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Model based adaptive control

u

Adaptation

Design parameters

Controller

-

Process

Linear model

yr

ym


Outline

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Outline





5 PID controllers


Continuous-time PID controller

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Continuous time PID controller

The “textbook” version of a PID controller:

u (t ) = K

e (t ) +

1

T i

t e (s )ds + T d

de (t )

dt

A more realistic PID controller:

U (s ) = K

(U c (s )− Y (s )) + 1sT i

(U c (s )− Y (s ))− sT d 1 + sT d /N Y (s )

Note: u c (t ) is the reference that we want the output to follow.


Discrete-time PID controller

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P-term: P (k ) = K (u c (k )− y (k ))

I-term: I (k + 1) = I (k ) + K

T i e (k )

D-term: D (k ) = T d T d +Nh

D (k − 1)− KT d N

T d +Nh (y (k )− y (k − 1))

u (k ) = P (k ) + I (k ) + D (k )

Note: Backward-difference used to approximate the D-term, i.e. s → z −1hz


PID tuning

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g

Pole placement

Root locus

Bode diagram

(Heuristic) Tuning rules (Ziegler-Nichols, λ−tuning)

G (s ) = e −t 0s

K p

(τ s + 1) ⇒ K c =

τ

K p (λ + t 0), T i = τ , T d =

t 0

2


Example of a complete controller:

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p pCascaded control on an InvertedPendulum

Reference

Position controller Inverted pendulum

Angle controller


The End

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Thanks for your attention!Questions?


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Hyu 01213