Model-based Predictive Control - project MbPC · 2019. 11. 7. · exists a matrix L such that (A+LC ) is stable. Properties testing Let be the set of eigenvalueson or outside the

Model-based Predictive Control -

project

MbPC

S.l. dr. ing. Constantin Florin Caruntu

www.ac.tuiasi.ro/~caruntuc

[email protected]

State-space

model predictive controlOutline

5. Introduction in SS-MPC

6. State-space MPC

7. Stability analysis

8. Robustness analysis

13.01.2014

15:23

State-space

model predictive control

Lecture 5 – Introduction in SS-MPC

� Sample period is T. Sample number is k. Time is t.

� s(t)→ continuous time signal

� s(kT) → discrete time signal (s(k))

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15:23 5.1. State-space models5. Introduction in SS-MPC

Digital control

� A linear continuous time (CT) state-space system:

� → state vector

� → input vector (manipulated variables)

� → output vector (measured variables)

� → controlled variables

� In many cases, H=C, so that y=z

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15:23

c c

c c

c

x A x B u

y C x D u

z H x

= +

= +

=

&

nx∈�mu∈�py∈�qz∈�

5.1. State-space models5. Introduction in SS-MPC

Continuous-time

� With a ZOH at the output of the controller:

u(t) = u(kT) for the interval kT ≤ t ≤ ( k+1)T

� The discrete time (DT) state-space model

is an exact representation of the sampled CT system if

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15:23

( ) ( ) ( )( ) ( ) ( )( ) ( )

x kT T Ax kT Bu kT

y kT Cx kT Du kT

z kT Hx kT

+ = +

= +

=

( )0,c cT

A T A

cA e B e d Bτ τ= = ∫


Discrete-time

� With a ZOH at the output of the controller an exact

representation of the CT system can be obtained if:

• the CT system is linear, or

• the CT system is linear with input saturation

� It is generally not possible to get an exact DT representation

for a nonlinear CT system:

� For a nonlinear model, a DT approximation or numerical

integration can be used to find x(kT)

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15:23

( )( ),

,

x f x u

y h x u

=

=

&


Discrete-time

� Definition (stabilizability): The matrix pair (A, B) is stabilizable if

there exists a matrix K such that (A+BK) is stable.

� Definition (detectability): The matrix pair (C, A) is detectable if there

exists a matrix L such that (A+LC) is stable.

Properties testing

� Let be the set of eigenvalues on or outside the unit circle:

� Proposition (stabilizability): The pair (A, B) is stabilizable if and only if

is a full row rank for all

� Proposition (detectability): The pair (C, A) is detectable if and only if

is full column rank for all

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15:23

Λ

( ) ( ){ }: : 1i iA Aλ λΛ = ≥

( )( )A I Bλ− λ∈Λ

A I

C

λ−

λ∈Λ


Stabilizability and detectability

� Advantages:

� Flexible process modeling

• multivariable

• linear or nonlinear

• deterministic, stochastic or fuzzy

� Incorporates constraints on states and inputs

• physical, safety, environmental, economical

� Optimal performances in closed loop

• horizon dependent, cost function optimization

� Disadvantages:

� Requires online optimization

• nonlinear/uncertain processes → computational power

13.01.2014

15:23 5.2. State-space MPC strategy5. Introduction in SS-MPC

Motivation

� All physical systems have constraints:

• Physical constrains

(actuator limits)

• Performance constraints

(overshoot)

• Safety constraints

(temperature/pressure limits)

� Optimal operating points are often near constraints

� Most control methods address constraints a posteriori:

• Anti-windup methods, “trial and error”

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Constraints

� Classical control:

• Does not take constraints into

account

• Reference far from constaints

• Suboptimal operation

� Predictive control:

� Takes constraints into account

from the design phase

� Reference close to constraints

� Improved operation

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Optimal operation and constraints

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Receding horizon principle

� Predictive control receding horizon principle

� At each sampling period, the predictive controller:

1) Takes a measurement of the system state/output

2) Computes a finite horizon control sequence that:

a) Uses an internal model to predict system behaviour

b) Minimizes some cost function

c) Doesn’t violate any constraint

3) Implements the first part of the optimal sequence

� => This is a feedback control law

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Summary

� Is it a new idea ?

• NO – standard optimal control with finite horizon

• YES – on-line optimization

� Main problems

• The optimization has to be fast enough

• Satisfying constraints for infinite horizon

• The resulting control law may not be stable

� Advantages:

• Systematic method of considering the constraints

• Flexible performance specifications

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Properties


project

MbPC



[email protected]

State-space



6. State-space MPC



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15:23

State-space


Lecture 6 – State-space MPC

� DT system

� Assumptions:

• (A,B) is stabilizable and (C,A) detectable

• C = I => state feedback

• H = C => all outputs/states are controlled variables

o Goal is to regulate the states around the origin

o No delays, disturbances, model errors, noise

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15:23

( ) ( ) ( )( ) ( )( ) ( )

1x k Ax k Bu k

y k Cx k

z k Hx k

+ = +

=

=

6.1. Unconstrained predictive control6. State-space MPC

Assumptions

� Given the DT system model

the problem is to design a state feedback control law

u(k) = Kx(k) such that the origin of the closed-loop system

is globally asymptotically stable => requires that (A+BK) is stable

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15:23

( ) ( ) ( )1x k Ax k Bu k+ = +

( ) ( ) ( )1x k A BK x k+ = +

6. State-space MPC 6.1. Unconstrained predictive control

State-space control

� Problem: Given an initial state x(0) at time k=0, compute

and implement an input sequence

that minimizes the infinite horizon cost function

• The state weight penalizes non-zero states

• The input weight penalizes non-zero inputs

• Generally Q and R are diagonal and positive definite

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15:23

( ) ( ){ }0 , 1 ,...,u u

( ) ( ) ( ) ( )( )0

T T

k

x k Qx k u k Ru k∞

=

+∑

0Q =f

0R =f


LQR

� The infinite horizon LQR problem has an infinite

number of decision variables

� A simple and closed form solution exists if

• Q is positive semidefinite ( )

• R is positive definite ( )

• The pair is detectable

� A finite horizon version of the LQR problem with the

same assumptions as above will be solved for use in MPC

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15:23

( ) ( ){ }0 , 1 ,...,u u

0Q =f

0R f1

2 ,Q A


LQR

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15:23

1. Obtain measurements of current state x.



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15:23


2. Compute optimal finite horizon input sequence ( ) ( ) ( ){ }* * *0 1 1, ,..., Nu x u x u x−



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15:23


2. Compute optimal finite horizon input sequence

3. Implement first part of optimal input sequence

( ) ( ) ( ){ }* * *0 1 1, ,..., Nu x u x u x−( ) ( )*0:k x u x=



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15:23


2. Compute optimal finite horizon input sequence

3. Implement first part of optimal input sequence

4. Return to step 1.

( ) ( ) ( ){ }* * *0 1 1, ,..., Nu x u x u x−( ) ( )*0:k x u x=



� Problem: Given an initial state x = x(k), compute a finite

horizon input sequence

that minimizes the finite horizon cost function

where

� V(·) is a function of the initial state x and the first N inputs

and not the time index k or the predicted states

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15:23

{ }0 1 1, ,..., Nu u u −

( )( ) ( )1

0 1

0

, ,...,N

T T T

N N N i i i i

i

V x u u x Px x Qx u Ru−

−=

= + +∑

0

1 , 0,1,..., 1i i i

x x

x Ax Bu i N+

=

= + = −

iu

ix


Finite horizon optimal control

� Terminology

• The vector is the prediction of given the current

state and the inputs for all

• is the control horizon

• The matrix is the terminal weight, with

� The stability and performance of a receding horizon

control law based on this problem is determined by

the parameters Q, R, P and N

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15:23

ix ( )x k i+( )x k ( ) iu k i u+ = 0,1,..., 1i N= −

N ∈�n nP ×∈� 0P =f


Finite horizon optimal control

� Note that , and that is

known.

� Can define stacked output and controlled

variables in a similar way.

� Define the stack vectors and as

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15:23

andm ni iu x∈ ∈� �

NmU ∈� NnX ∈�

0

1

2

1

: ,

N

u

u

U u

u −

=

M

1

2

3: ,

N

x

x

X x

x

=

M

( )0x x x k= =

NpY ∈�NqZ ∈�


Notations

� The cost function is defined as

� The value function is defined as

� The optimal input sequence is defined as

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15:23

( ) ( )1

0

,N

T T T

N N i i i i

i

V x U x Px x Qx u Ru−

=

= + +∑

( )* min ( , )U

V x V x U=

( )

( ) ( ) ( ){ }

*

* * *

0 1 1

: argmin ( , )

: , ,...,

U

N

U x V x U

u x u x u x−

=

=


Notations

� Compute prediction matrices and such that

� Rewrite the const function V(·) in terms of x and U

� Compute the gradient

� Set and solve for

� The MPC control law is the first part of the optimal

input sequence

� When there are no constraints, it is possible to do this

analytically.

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15:23

X x U= Φ +Γ

Φ Γ

( ),UV x U∇( ), 0UV x U∇ = ( )*U x

( ) ( ) ( )* *0 0 0mu x I U x= L


Control law design

� Want to find matrices and such that

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15:23

Φ Γ

1 0 0

2 1 1

x Ax Bu

x Ax Bu

= +

= +

M

X x U= Φ +Γ


Construction of prediction matrices

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15:23

1 0 0

2 1 1

x Ax Bu

x Ax Bu

= +

= +

M

( )1 0 0

2

2 0 0 1 0 0 1

1

0 0 2 1

N N

N N N

x Ax Bu

x A Ax Bu Bu A x ABu Bu

x A x A Bx ABu Bu− − −

= +

= + + = + +

= + + + +

M

L

X x U= Φ +Γ

� Want to find matrices and such thatΦ Γ



� Collect terms to get the matrix form

� Recalling that , the prediction matrices and

are:

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15:23

1 0

2

2 1

0

1 2

1

0 0

0

N N N

N N

x uA B

x uA AB Bx

x uA A B A B B− − −

= +

L

L

M MM M M O M

L

0:x x= Φ Γ

2

1 2

0 0

0: , :

N N N

A B

A AB B

A A B A B B− −

Φ = Γ =

L

L

M M M O M

L



� Recall that the cost function is

� Recalling , these can be rewritten in matrix

form as:

� Note that:

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15:23

0:x x=

( ) ( )1

0

1 1 0 0

2 2 1 1

0 0

1 1

, :N

T T T

N N i i i i

i

T T

T

N N N N

V x U x Px x Qx u Ru

x x u uQ R

x x u uQ Rx Qx

Q R

x x u uP R

−

=

− −

= + +

= + +

∑

M M M M

( ), T T TV x U x Qx X X U U= + Ω + Ψ

0 and 0 0

0 0

P Q

R

= = ⇒Ω =

⇒Ψ

f f f

f f


Construction of cost function

� Recall that

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15:23

( ), T T TV x U x Qx X X U UX x U

= + Ω + Ψ

= Φ +Γ

( ) ( ) ( )

( ) ( )

,

2

TT T

T T T T T T

T T T T

T T T T T T

V x U x Qx x U x U U U

x Qx x x U U U U

x U U x

x Q x U U U x

= + Φ +Γ Ω Φ +Γ + Ψ

= + Φ ΩΦ + Γ ΩΓ + Ψ

+ Φ ΩΓ + Γ ΩΦ

= +Φ ΩΦ + Ψ +Γ ΩΓ + Γ ΩΦ


Construction of cost function

� Recall that

where

� Important: this is a convex and quadratic form of U.

• The unique and global minimum occurs at the point where

• The optimal input sequence is therefore

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15:23

( ) ( )1,2

T T T TV x U U GU U Fx x Q x= + + +Φ ΩΦ

( ) ( ): 2 0, 0 si 0: 2

T

T

G

F

= Ψ +Γ ΩΓ Ω = Ψ

= Γ ΩΦ

f f f

( ), 0UV x U GU Fx∇ = + =

( )* 1U x G Fx−= −


Finding the solution

� The optimal input sequence is

� The MPC control law is defined by the first part of

� Define

such that .

o This is a time invariant linear control law

o It approximates the optimal infinite horizon control law

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15:23

( )* 1U x G Fx−= −

( )*U x

( ) ( ) ( )* *0 0 0mu x I U x= L

( ) 10 0MPC mK I G F−= − LMPCu K x=


Predictive control law

� The solution to the infinite horizon LQR problem is

where

and P is the solution to the Algebraic Riccati Equation

� If the terminal weight P in the finite cost function V(·)

is a solution to the Algebraic Riccati Equation above,

then:

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15:23

( ) ( )LQRu k K x k=

( ) 1T TLQRK B PB R B PA−

= − +

( ) 1T T T TP A PA A PB B PB R B PA Q−= − + +

MPC LQRK K=


Equivalence between LQR and MPC

� In practice, system variables are always constrained by:

• Physical limitations

� Input constraints (actuator limits)

� typically active during transients

� State constraints (reservoir capacities)

� can be active during transients or in steady state

• Safety considerations (critical temperatures/pressures)

• Performance specifications (limit overshoot)

13.01.2014

15:23 6. State-space MPC 6.2. Constrained predictive control

Variables constraints

13.01.2014

15:23

1

2

3

4

5

1 4

1.5 4

1.5 3

2 3

2 3

y

y

y

y

y

≤ ≤

≤ ≤

≤ ≤

≤ ≤

≤ ≤

� Clasify constraints (hard or soft):

� Hard constraints must be satisfied at all the times,

otherwise the problem is infeasible

� Soft constraints can be violated to avoid infeasibility

6. State-space MPC 6.2. Constrained predictive control

Variables constraints

� A common system nonlinearity is input saturation.

� Easily transformed into a constraint on a linear system:

The component of is:

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15:23

( ) ( ) ( )( ) ( )( ) ( )

1 sat nonlinearx k Ax k B u k

y k Cx k

+ = +

=

( )th 1,...,i i m∈ ( )sat u

( ){ }{ } { } { }

{ } { } { } { }

{ } { } { }

if

sat : if

if

i i i

i i i ii

i i i

u u u

u u u u u

u u u

<

= ≤ ≤

>


Systems with saturated input

� Sub-optimal methods to consider saturation:

� Saturate the unconstrained control law

(the constraints are ignored in the design phase of the

controller)

� Re-design an unconstrained control law

(increasing the weighting factor for u in the performance

criterion for optimal control)

�Anti-windup strategies

(dynamically limit the controller state -> I component)

13.01.2014


Constraint handling

� Problem: Given an initial state x(0) at time k=0, compute

and implement an input sequence

that minimizes the infinite horizon cost function

while guaranteeing that constraints are satisfied for all time.

� it is usually impossible to solve this problem

� predictive control provides and approximate solution

� the MPC laws with constraints will be nonlinear

13.01.2014

15:23

( ) ( ){ }0 , 1 ,...,u u

( ) ( ) ( ) ( )( )0

T T

k

x k Qx k u k Ru k∞

=

+∑


LQR problem with constraints

� Problem: Given an initial state x = x(k), compute a finite



where

while guaranteeing all constraints are satisfied over the

prediction horizon

13.01.2014

15:23

{ }0 1 1, ,..., Nu u u −

( )1

0

NT T T

N N i i i i

i

x Px x Qx u Ru−

=

+ +∑

0

1 , 0,1,..., 1i i i

x x

x Ax Bu i N+

=

= + = −

0,1,...,i N=


Optimal control with finite horizon and constraints

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15:23

1. Obtain measurement of current output/state.



13.01.2014

15:23


2. Compute optimal finite horizon input sequence subject to constraints.



13.01.2014

15:23



3. Implement first part of optimal input sequence.



13.01.2014

15:23



3. Implement first part of optimal input sequence.

4. Return to step 1.



� Recall that the sequence of predicted states X was

solved in terms of the stacked inputs U

or, defining ,

� The matrices and are prediction matrices.

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15:23

1 0

2

2 1

0

1 2

1

0 0

0

N N N

N N

x uA B

x uA AB Bx

x uA A B A B B− − −

= +

L

L

M MM M M O M

L

0:x x=

Φ Γ

.X x U= Φ +Γ


Prediction matrices

� Now incorporate a set of linear inequality constraints

on the predicted states and inputs

� Many constraints take this form:

• input constraints only

• state constraints only

• Can include constraints on outputs or controlled variables.

� For simplicity, assume that

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15:23

ix iu

, for all 0,1,..., 1

.

i i i i i

N N N

M x E u b i N

M x b

+ ≤ = −

≤

0sM = ⇒

0sE = ⇒

, si , for all 0,1,..., 1i i iE E M M b b i N= = = = −


Incorporating constraints

� Suppose that there are the following input and output

constraints:

� Recalling that this is equivalent to:

� Similar expression for terminal constraint (in terms of only).

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15:23

, 0,1,..., 1

, 0,1,...,

low i high

low i high

u u u i N

y y y i N

≤ ≤ = −

≤ ≤ =

,i iy Cx=

0

0, 0,1,..., 1

0

0

low

high

i i

low

high

uI

uIx u i N

yC

yC

−− + ≤ = − −−

Nx



� From the previous example, the constraints can be

written in the form:

by defining:

and

13.01.2014

15:23

0

0: , : , : , 0,1,..., 1

0

0

low

high

i i i

low

high

uI

uIM E b i N

yC

yC

−− = = = = − −−

, for all 0,1,..., 1

.

i i i i i

N N N

M x E u b i N

M x b

+ ≤ = −

≤

: , :low

N N

high

yCM b

yC

−− = =



� Taking all of the constraints together:

� By appropriately defining yields:

� Next, X will be eliminated by using the prediction

matrices.

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15:23

00 0

1 0

1 1

0

1

1

0 0 0

00

0

00 0 0

N

N N

N N

bM Ex u

M bx

Ex u

M b

−−

+ + ≤

L L

L M O MM M

M O M MM L

L L

( )0, , and :c x xε∆ ϒ =

.x X U cε∆ + ϒ + ≤



� Substitute in

and collect terms. The constraints can be rewritten as:

where

and

� The constraints are now in terms of the input

sequence U and the initial state

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15:23

X x U= Φ +Γ

x X U cε∆ + ϒ + ≤

JU c Wx≤ +

:J ε= ϒΓ +

:W = −∆ − ϒΦ

( )0 .x x x k= =



� In summary, the basic procedure is:

� Define linear inequalities in

� Write the constraints in the form:

� Stack the constraints to get them in the form:

� Substitute and rearrange to the form:

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15:23

X x U= Φ +Γ

.x X U cε∆ + ϒ + ≤

.JU c Wx≤ +

, , and i i i iu x y z

, for all 0,1,..., 1

.

i i i i i

N N N

M x E u b i N

M x b

+ ≤ = −

≤



� Recall that the cost function

can be rewritten (with ) as

for some matrices defined previously.

� Remember that

13.01.2014

15:23

( )1

0

( , ) :N

T T T

N N i i i i

i


=

= + +∑

0:x x=

( ) ( )1,2

T T T TV x U U GU U Fx x Q x= + + +Φ ΩΦ

, and F G Ω

0 if 0, 0 and 0.G P Q R= =f f f f


Cost function

� Definition (quadratic problem - QP):

Given the matrices Q and A and the vectors c and b, the

optimization problem

is called a quadratic problem.

� Propozition If in the above quadratic problem, then:

1. The optimization problem is strictly convex.

2. A global minimum can be always found.

3. The global minimum is unique.

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15:23

1min

2

subject to

T TQ c

A b

θθ θ θ

θ

+

≤

0Q f


Quadratic problems (QP)

13.01.2014


Solution of quadratic problems

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15:23

� Many problems can be recasted as quadratic problems

� Example: Nonlinear constraints.

� Example: Nonlinear cost function.

{ }

1min

2

subject to max ,

T T

T T

Q c

e f b

θθ θ θ

θ θ

+ ≤

1min

2

subject to ,

T T

T T

Q c

e b f b

θθ θ θ

θ θ

+⇒

≤ ≤

min

subject to

Tc

A b

θθ

θ

≤

,min

subect to

0

0

T

T

A b

c

c

θ δδ

θ

θ δ

θ δ

⇒

≤

− ≤

− − ≤


Writing problems as QP

� The constrained optimal control problem is

� This is a quadratic problem in its standard form.

� The optimal solution is:

1. A global minimum (when ).

2. Unique (when ).

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15:23

1min

2

subject to

T T

UU GU U Fx

JU c Wx

+

≤ +

( )

U Q G c Fx

A J b c Wx

θ → → →→ → +

0G =f

0G f


Constrained optimal control as QP

� Some QP parameters are dependent on the current state x

• Without constraints, is a linear function of x.

• With constraints, is a nonlinear function.

� must be calculated by solving an online quadratic

problem for every x.

� In Matlab, the quadratic problem can be solved by using

13.01.2014

15:23

( )*U x( )*U x

( )*U x

( )U=quadprog G,F* x,J,C+Wx


Solution using QP

� The MPC law is the first part of the optimal input

sequence

� Since is no longer linear, is a

nonlinear control law.

� The dynamics of the closed loop system are nonlinear

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15:23

( ) ( ) ( )* *0: 0 0MPC mK u x I U x= = L

( )*U x : n mMPCK →� �

( ) ( ) ( )1 MPCx k Ax k BK x+ = +


MPC implementation

� Example: Duble integrator

Constraints:

Prediction horizon=12.

Quadratic cost function:

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15:23

( ) ( ) ( )1 1 0

10 1 1

x k x k u k

+ = +

1, 12u x∞

≤ ≤

1 0, 1, 1

0 1Q R P

= = =

� Solution: Controller with

57 regions. Each region i

has ( )*0 i iu x v K x= +


Solutions complexity


project

MbPC



[email protected]

State-space



6. State-space MPC



13.01.2014

15:23

State-space


Lecture 7 – Stability analysis

� Problem: Given the initial state x = x(k), compute a finite



where

� Assume that are positive definite.

13.01.2014

15:23

{ }0 1 1, ,..., Nu u u −

( ) ( )1

0

, :N

T T T

N N i i i i

i


=

= + +∑

0

1 , 0,1,..., 1i i i

x x

x Ax Bu i N+

=

= + = −

, and Q R P

7. Stability analysis 7.1. Introduction

Optimal control with finite horizon with constraints

� The value function is defined as

� The optimal input sequence is defined as

� Define the predicted state trajectory

where:

� The first part of defines a MPC law

13.01.2014

15:23

( )* min ( , )U

V x V x U=

( ) ( ) ( ) ( ){ }* * * *0 1 1: argmin ( , ) : , ,..., NU

U x V x U u x u x u x−= =

( ) ( ) ( )* * *0 1, ,..., Nx x x x x x( )( ) ( ) ( )

*

0

* * *

1 , 0,..., 1i i i

x x x

x x Ax x Bu x i N+

=

= + = −

( )*U x ( ) ( )*0MPCK x u x=

7. Stability analysis 7.1. Introduction

Notations

� Consider a discrete time system

with continuous and .

o Definition (Lyapunov function): A continuous function

defined on a region containing the

origin in its interior is called a Lyapunov function if

1.

2.

3.

13.01.2014

15:23

( ) ( )( )1x k f x k+ =: n nf →� � ( )0 0f =

:V S R→ nS ⊂ �

( )0 0V =

( ) 0, , with 0V x x S x> ∀ ∈ ≠( )( ) ( ) 0,V f x V x x S− ≤ ∀ ∈

7. Stability analysis 7.2. Lyapunov functions

…for discrete-time systems

o Proposition (Asymptotic stability): If there exists a

Lyapunov function

and

then the origin is an asymptotically stable equilibrium

point for

with region of attraction S.

13.01.2014

15:23

( )( ) ( ) 0, , with 0V f x V x x S x− < ∀ ∈ ≠

( ) as V x x→∞ →∞

( ) ( )( )1x k f x k+ =


…for discrete-time systems

13.01.2014

15:23

� Consider the linear DT system

and the candidate Lyapunov function

•

•

•

( ) ( )1x k Ax k+ =

( )0 0V =( ) ( )if P 0, 0 and as V x V x x> →∞ →∞f

( ) ( ) ( ) ( ) ( )T T T TV Ax V x Ax P Ax x Px x A PA P x− = − = −

( ) TV x x Px=


Lyapunov stability for discrete-time linear systems

13.01.2014

15:23

� Problem: To ensure asymptotic stability of the

system

chose such that

� Choosing P to satisfy the above is possible if, for

some , the discrete Lyapunov equation

has a positive definite solution

� System is stable if and only if can be found for

any

( ) ( )1x k Ax k+ =0P f 0TA PA P− p

0Z fTA PA P Z− = −

0P f

0P f

0Z f



� Consider a discrete time system

with continuous and

o Definition (Control Lyapunov function): A continuous

function defined on a region containing

the origin in its interior is called a Control Lyapunov

function if

1.

2.

3. There exists a continuous control law such that

13.01.2014

15:23

( ) ( ) ( )( )1 ,x k f x k u k+ =: n m nf × →� � � ( )0,0 0f =

:V S R→ nS ⊂ �

( )0 0V =

( ) 0, , if 0V x x S x> ∀ ∈ ≠

( )( )( ) ( ), 0,V f x K x V x x S− ≤ ∀ ∈( )u K x=


Control Lyapunov functions

o Proposition (Asymptotic Stability): If there exists a

Control Lyapunov function and a continuous control law

such that

and

then the origin is an asymptotically stable equilibrium point

for

with region of attraction S.

13.01.2014

15:23

( )( ) ( ) 0, , cu 0V f x V x x S x− < ∀ ∈ ≠

( ) as V x x→∞ →∞

( ) ( ) ( )( )1 ,x k f x k K x+ =

( )u K x=


Control Lyapunov functions

13.01.2014

15:23

� Consider the DT linear system

and the candidate Control Lyapunov function

and control law

•

•

•

( ) ( ) ( )1 ,x k Ax k Bu k+ = +

( )0 0V =( ) ( )if P 0, 0 si as V x V x x> →∞ →∞f

( )( ) ( ) ( )( ) ( )( )( ) ( )( )

T T

TT

V A BK x V x A BK x P A BK x x Px

x A BK P A BK P x

+ − = + + −

= + + −

( ) TV x x Px=( )u K x=



13.01.2014

15:23

� Problem: To ensure asymptotic stability for the system

cchoose such that

� Choosing P to satisfy the above is possible if, for some

the discrete Lyapunov equation

has a positive definite solution

� is stable if and only if can be found for any

( ) ( ) ( )1x k Ax k Bu k+ = +0P f ( ) ( ) 0TA BK P A BK P+ + − p

0Z f

( ) ( )TA BK P A BK P Z+ + − = −0P f

0P f0Z f

( )A BK+



13.01.2014

15:23

� Theorem (MPC stability): The origin is an asympotically

stable equlibrium point of the closed loop system

if the following conditions are satisfied:

1) Q and R are positive definite.

2) The terminal cost P is chosen such that and

where K is any matrix with

( ) ( ) ( )( )*01x k Ax k Bu x k+ = +

0P f

( ) ( )T TA BK P A BK P Q K RK+ + − = − −p

( ) 1.A BKρ + <

7. Stability analysis 7.3. MPC stability

13.01.2014

15:23

� For the closed loop MPC system

the value function will be used as a Lyapunov

function.

� Since Q and R are positive definite, it follows that:

1)

2)

3)

� Need to ensure in addition that

( ) ( ) ( )( )*01x k Ax k Bu x k+ = +( )*V ⋅

( )* 0 0V =( )* 0, 0TV x x Px x≥ > ∀ ≠( )* as V x x→∞ →∞

( )( ) ( )( )* *1 0, 0V x k V x k x+ − < ∀ ≠


Proof

13.01.2014

15:23

� Consider the optimal input sequence

� At the next time, consider a shifted input sequence

where K is such that

( ) ( ) ( ) ( ){ }* * * *0 1 1, ,..., NU x u x u x u x−=( )U x%

( ) ( ) ( ) ( ) ( ){ }* * * * *0 1 2 1, , ..., ,N NU x u x u x u x u x− −=

( ) ( ) ( ) ( ) ( ){ }* * * *1 2 1, , ..., ,N NU x u x u x u x Kx x−=%applied

new tail

( ) 1.A BKρ + <


Proof

13.01.2014

15:23

� Define the stage cost as

� Define the terminal cost as

� Recall

� Find the cost associated with the shifted sequence

( ), : .T Tl x u x Qx u Ru= +( ) : .TfV x x Px=

( ) ( ) ( )( )*01 .x k Ax k Bu x k+ = +

( )( ) ( ) ( )( )( )( )( )

( ) ( )( )( )( )( )

( )( ) ( )( )( )( ) ( )( )( )

*

*

0

*

* *

*

: 1 ,

old optimal cost

, old first stage cost

old terminal cost

, new ( -1) stage cost

new terminal cost

f N

th

N N

f N

U x k V x k U x k

V x k

l x k u k

V x x k

l x x k Kx x k N

V A BK x x k

+ =

+

−

−

+

+ +

% %


Proof

13.01.2014

15:23

� Problem: Find conditions that the last hree terms are

negative:

� Chose the terminal cost to be a control Lyapunov

function with the terminal control law

� Remember that the stage cost

for all when Q and R are positive definite

( )( )( ) ( )( ) ( )( )( ) ( ) ( )( )( )* * * *, ( ), 0x u ≠


Proof

13.01.2014

15:23

� For quadratic cost functions, it can be ensured that:

by requiring that

� Can choose if and only if

� Define and solve the inequation

� Note that

( ) ( ) ( )- , 0, 0f fV Ax BKx V x l x Kx x+ ≤ − < ∀ ≠

( ) ( )T TA BK P A BK P Q K RK+ + − = − −p

0P f ( ) 1A BKρ + <: TZ Q K RK= +

0 and 0 0Q R Z⇒f f f


Proof

13.01.2014

15:23

� Conditions were established to ensure that:

� Since the stage cost , then

� Therefore

� This proves that the value function is a control

Lyapunov function with the MPC control law

( ), 0l x u ≥

( ) ( )( )( ) ( )( ) ( ) ( )( )* *01 , , , 0V x k U x k V x k l x k u k x+ < − ∀ ≠%

( ) ( )( )( ) ( )( )*1 , , 0V x k U x k V x k x+ < ∀ ≠%

( )( ) ( )( )* *1 , 0V x k V x k x+ < ∀ ≠( )*V ⋅

( ) ( )( )*u k u x k=


Proof


horizon input sequence that minimizes the finite horizon

cost function

where

subject to

13.01.2014

15:23

( )1

0

NT T T

N N i i i i

i

x Px x Qx u Ru−

=

+ +∑

0

1 , 0,1,..., 1i i i

x x

x Ax Bu i N+

=

= + = −

, 0,1,..., 1i iMx Eu b i N+ ≤ = −

7. Stability analysis 7.4. MPC feasibility


� Definition (Invariant set): The setul is called an

invariant set for the system

if

� Definition (Constraint admissible set): Given a control

law a set of states and a set of

constraints the set S is constraint admissible

if

� For the previously defined problem

13.01.2014

15:23

( ) ( )( )1x k f x k+ =

nS ⊂ �

( ) ( )( )0 , 0.x S f x k S k∈ ⇒ ∈ ∀ ≥

( ) ,u K x= ,nS ⊂ �,n mZ ⊂ ×� �

( )( ), ,x K x Z x S∈ ∀ ∈

( ){ }: , :Z x u Mx Eu b= + ≤


Invariant sets for discrete-time systems

� Given K such that a matrix and a vector

can be chosen such that

is invariant for the closed loop system

and constraint admissible for the control law and

constraint set Z.

� For every , it is required that:

13.01.2014

15:23

( ) 1,A BKρ + < NMNb

{ }: :n N NS x M x b= ∈ ≤�

( ) ( ) ( )1x k A BK x k+ = +u Kx=

x S∈

( )( )

(invariant)

(constraint admissible)

N NM A BK x b

M EK x b

+ ≤

+ ≤


Invariant terminal constraints




where

13.01.2014

15:23

{ }0 1 1, ,..., Nu u u −

( ) ( )1

0

, :N

T T T

N N i i i i

i


=

= + +∑

0

1

,

, 0,1,..., 1

, 0,1,..., 1

. (invariant and constraint admissible)

i i i

i i i i i

N N N

x x

x Ax Bu i N

M x E u b i N

M x b

+

=

= + = −

+ ≤ = −

≤



13.01.2014

15:23

� Recall the shifted input sequence

� If is feasible at time k, then the shifted sequence

is feasible at time k+1 if is implemented.

� Note that:

( ).U x%

( ) ( ) ( ) ( ) ( ){ }* * * * *0 1 2 1, , ..., ,N NU x u x u x u x u x− −=

( ) ( ) ( ) ( ) ( ){ }* * * *1 2 1, , ..., ,N NU x u x u x u x Kx x−=%applied

new tail

( )*U x ( )U x%( ) ( )*0u k u x=

( ) ( )( ) ( )

* *

*

*

N N N

N N N

N N N

M x x EKx x bM x b

M A BK x x b

+ ≤≤ ⇒

+ ≤


Proof

13.01.2014

15:23

� Theorem (MPC feasibility) Consider the system

with state and input constraints

� If the terminal constraint is chosen to be a

constraint admissible invariant set for the closed loop

system

for some K and the constrained finite horizon control

problem is feasible at time k=0, then it is feasible for all

if the control input is given by the MPC law1k ≥

( ) ( ) ( )1x k Ax k Bu k+ = +.Mx Eu b+ ≤

N N NM x E u b+ ≤

( ) ( ) ( )1x k A BK x k+ = +

( ) ( )( )*0 .u k u x k=


13.01.2014

15:23

� In order to guarantee infinite horizon constraint satisfaction

and stability:

• Chose Q and R positive definite.

• Compute a stabilizing terminal control law

� Stability

• Compute a positive definite P such that the terminal cost

is a Control Lyapunov function with

� Feasibility

• Compute a and such that the terminal constraint

is constraint admissible and invariant for the closed loop system

.u Kx=

T

fV x Px=.u Kx=

NM Nb

N N NM x E u b+ ≤

( ) ( ) ( )1 .x k A BK x k+ = +

7. Stability analysis 7.5. Summary


project

MbPC



[email protected]

State-space



6. State-space MPC



13.01.2014

15:23

State-space


Lecture 8 – Robustness analysis

� So far, only the problem of regulating the states and

inputs around the origin was considered.

� In practice, some nonzero, time-varying setpoint or

reference signal has to be folloed.

• Aircraft landing by autopilot.

• Robot arm with pre-specified trajectory.

• Radar tracking problems.

� In what follows, the tracking of piecewise constant

reference signals will be considered.

13.01.2014

15:23 8. Robustness analysis 8.1. Reference tracking

13.01.2014

15:23

� A linear discrete time state-space system

� → state vector

� → input vector (manipulated variables))

� → output vector (measured variables)

� → controlled variables

( ) ( ) ( )( ) ( ) ( )( ) ( )

x kT T Ax kT Bu kT

y kT Cx kT Du kT

z kT Hx kT

+ = +

= +

=nx∈�mu∈�py∈�qz∈�

8. Robustness analysis 8.1. Reference tracking

Discrete-time linear systems

13.01.2014

15:23

� Control the system so that as if the reference

signal tends to a constant value (assume future values of r are known)

� At each time, given the current reference value

� Target Calculator computes target input and target state

� Regulator controls the system around the target

( ) ( )z k r k→ k →∞

( )r k( )u k∞ ( )x k∞

( ) ( )( ),x k u k∞ ∞


General problem

13.01.2014

15:23

� Given a linear state feedback control gain (either from

uncosntrained MPC or LQR) with implement the

control law:

� Closed loop system is stable if and are constant

� Want to design the target calculator such that as

K

( ) 1A BKρ + <( ) ( ) ( ) ( )( )ˆ |u k u k K x k k x k∞ ∞= + −

( )u k∞ ( )x k∞( ) ( )z k r k→ k →∞


General problem

� All of the outputs are not always controlled variables

(i.e., ).

� It is generally not possible to control all of the outputs

or states to an arbitrary setpoint

� Example: it is not possible to maintain a car in a

fixed position while also maintaining a nonzero

velocity.

� Generally, the controlled variables are a linear

combination of the states or outputs

� Problem:Which choices of H are allowed ?

13.01.2014

15:23

( ) ( ) so that H C z k y k≠ ≠

( )z k( )x k ( )y k


Choice of controlled variables

o Definition (Target equilibrium pair): Given a reference

input r for a linear discrete time system, the pair

is called an (offset free) target equilibrium pair if

� Rearranging the above gives

� Depending on the values of H and r target equilibrium pair

may exist or may not exist.

13.01.2014

15:23

x Ax Bu

Hx r

∞ ∞ ∞

∞

= +

=

( ) ( ),x k u k∞ ∞

( ) 00

xI A B

u rH

∞

∞

− − =


Target equilibrium pairs

o Proposition: A sufficient condition for guaranteeing the

existence of a target equilibrium pair for any reference r is

that

is full row rank.

� The above condition implies that:

• H has to be full row rank

• Number of inputs >= number of controlled variables

13.01.2014

15:23

( )0

I A B

H

− −

m q≥


Target equilibrium pairs existence

� At each time instant, the target calculator solves the

following set of linear inequalities

� Note that if a solution exists, it may not be unique.

� If are calculated as above, then it is possible

to show that as if:

• The sequence converges to a constant value

• The controller is

• The matrix and the observer are stable

13.01.2014

15:23

( ) 00

xI A B

u rH

∞

∞

− − =

( ) ( )( ),x k u k∞ ∞( ) ( )z k r k→ k →∞

( )r ⋅

( ) ( ) ( ) ( )( )ˆ |u k u k K x k k x k∞ ∞= + −A BK+


Target equilibrium pairs computation – without constraints

13.01.2014

15:23

� Future values of and are assumed unknown

� The observer estimates the current state and disturabnce

o Problem: Control the system so that as

reference signal and disturbance tend to constant values

( ) ( )z k r k→ k →∞

( )r ⋅ ( )d ⋅( )ˆ |x k k

( )ˆ |d k k

8. Robustness analysis 8.2. Disturbance rejection

General problem

13.01.2014

15:23

� Consider a linear model with a constant disturbance acting

on the states and outputs

� Disturbance inputs , with

� Estimate by forming an augmented system for the

states and disturbances

( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( ) ( )

1

1

d

d

d

x k Ax k Bu k B d k

d k d k

y k Cx k C d k

z k Hx k H d k

+ = + +

+ =

= +

= +

( ) ld k ∈�, ,n l n l n ld d dB C H

× × ×∈ ∈ ∈� � �

( )d k( )x k ( )d k


Constant disturbance model

13.01.2014

15:23

� Augment the state of the system with the disturbance

� An observer will be used to provide state and

disturbance estimates

( )( )

( )( ) ( )

( ) ( ) ( )( )

( ) ( ) ( )( )

1

1 0 0

d

d

d

x k x kA B Bu k

d k d kI

x ky k C C

d k

x ky k H H

d k

+ = + +

=

=

( )ˆ |x k k( )ˆ |d k k


Extended model with disturbances

� The augmented system is not guaranteed to be detectable

for every

o Proposition (Detectability): The augmented system is

detectable if and only if is detectable and the matrix

is full column rank.

� The above implies that:

• The number of disturbances has to be smaller or equal

than the number of outputs (for detectability)

13.01.2014

15:23

( )0

I A B

H

− −

and d dB C

( ),C A


Estimating states and disturbances

� At each time instant, target calculator solves the following

set of linear equalities

or, in matrix form

� As in the undisturbed case, this is solvable given an

appropriate rank condition.

13.01.2014

15:23

( ) ˆˆ0

d

d

B dxI A B

uH r H d

∞

∞

− − = −

ˆ

ˆ

d

d

x Ax Bu B d

r Hx H d

∞ ∞ ∞

∞

= + +

= +


Target equilibrium pairs computation – with constraints

� Let the following conditions hold:

• The sequence and tend to constant values.

• The rank conditions are satisfied.

• The state/disturbance observer is stable.

• The control input is given by

where are chosen as on previous slide.

• The matrix K is such that is stable.

• Number of disturbances = number of outputs.

� If all of the above hold, then

13.01.2014

15:23

( )r ⋅ ( )d ⋅

( ) ( ) ( ) ( )( )ˆ |u k u k K x k k x k∞ ∞= + −( ) ( )( ),x k u k∞ ∞

( )A BK+

( ) ( ) as z k r k k→ →∞


Results

� Suppose there are constraints on the states and inputs

� As before, assume that:

• The rank conditions are satisfied.

• The state/disturbance observer is stable.

• Number of disturbances = number of outputs.

� Aditional condition: are such that target

equilibrium pairs satisfying the above constraint always

exist.

13.01.2014

15:23

, 0,1,..., 1

, 1,...,

low i high

low i high

u u u i N

y y y i N

≤ ≤ = −

≤ ≤ =

( ) ( )ˆ and |r k d k k



13.01.2014

15:23

� At each time instant, the target calculator is given the

current reference r and the current disturbance estimate

� The target calculator computes the target equilibrium pair

by solving the following quadratic program:

subject to:

� The ideal steady state for the inputs is

( ) ˆˆ0

d

d

B dxI A B

uH r H d

∞

∞

− − = −

d̂

( ),x u∞ ∞

( ) ( ),

1min

2

T

x uu u u u

∞ ∞∞ ∞− −

ˆ

low high

low d high

u u u

y Cx C d y

∞

∞

≤ ≤

≤ + ≤.u



� Problem: Given compute an optimal finite

horizon sequence of inputs that minimize

subject to

� Can be written as a quadratic program.

13.01.2014

15:23

{ }0 1 1, ,..., Nu u u −

( ) ( ) ( ) ( )

( ) ( )

1

0

NT T

i i i i

i

T

N N

x x Q x x u u R u u

x x P x x

−

∞ ∞ ∞ ∞=

∞ ∞

− − + − − +

+ − −

∑

0

1

ˆ

ˆ, 0,1,..., 1

, 0,1,..., 1

ˆ , 0,1,...,

i i i d

low high

low d high

x x

x Ax Bu B d i N

u u u i N

y Cx C d y i N

+

∞

∞

=

= + + = −

≤ ≤ = −

≤ + ≤ =

ˆˆ, , and x u x d∞ ∞


Predictive control with disturbance rejection

13.01.2014

15:23

� At each time instant k:

• Observer takes measurement and computes extimates

and

• Target calculator computes

• Regulator computes an input by solving the finite horizon problem

as a quadratic problem and by implementing the first input

sequence

• If quadratic problems are always feasible and the system is stable,

then

( )y k ( )ˆ |x k k( )ˆ |d k k

( ) ( ) and x k u k∞ ∞

( ) ( )( )*0u k u x k=

( ) ( ) as z k r k k→ →∞


Predictive control with disturbance rejection

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