model predictive control: background

Chemical and Biological Engineering

Model Predictive Control: Background

B. Wayne Bequette

• Concise Review of Undergraduate Process Control

• Introduction to Discrete-time Systems• Discrete Internal Model Control (IMC)

Automation and Control

feedforward

feedback

Control Algorithm

Compares measured process output with desired setpoint, and calculates a manipulated input (often a flowrate)

Proportional-integral-derivative (PID)Long history, “workhorse”, lower-level control loops

Model predictive control (MPC)Most widely applied “advanced” control algorithmConstraints, multivariable systems

Common notation

r = setpoint (desired value of process output)

y = measured process output

e = error (setpoint – output, r – y)

u = manipulated input (often a flowrate)

Proportional-Integral-Derivative (PID) Control

( ) ( ) ( )

( ) ( ) ( ) ( )⎥⎦

⎤⎢⎣

⎡+++=

∫ dttdedttetekutu

tytrte

τ 001

ManipulatedInput

Error = setpoint − measured output

Proportionalgain

Integral time Derivative time

Single-input, single-output designProblems with one controller can impact another controller

ConstraintsCan cause “windup” problems

Does not explicitly require a process model

Continuous Linear Models

State Space and Transfer Function

B. Wayne Bequette

• Linearization• State Space Form• Transfer Function• Step Responses• MV Properties

Nonlinear ODE Models

General “Lumped Parameter” Form

( )puxfx ,,=&

y = g x,u, p( )

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

differential state equations

algebraic output equations

⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢

dtdxdtdx

states state derivatives(from accumulationterm)

inputs outputs parameters

Linearization

ssssssss uxj

∂∂

uDxCyuBxAx

′+′=′′+′=′&

( )puxfx ,,=&

y = g x,u, p( )

⎥⎥⎥⎥

⎢⎢⎢⎢

−−

⎥⎥⎥⎥

⎢⎢⎢⎢

−−

⎥⎥⎥⎥

⎢⎢⎢⎢

−−

‘perturbation’ or ‘deviation’ variables

eqn i state j eqn i input j

steady-state solution, us, xs, ys

Where:

Example: Van de vuuse Reaction

AAAAfA

CkCkCVF

CkCkCCVF

−+−=

−−−=

[ ]BsBAfsAf

AsA CCxyCC

x −==⎥⎦

⎤⎢⎣

⎡−−

=⎥⎦

⎤⎢⎣

⎡−−

=⎥⎦

⎤⎢⎣

⎡= 2

[ ] [ ]00 ,10

⎥⎥

⎢⎢

−=⎥⎥⎥

⎢⎢⎢

−−

−−−=

CVFCCB

sAsAfs

etcCAux AsssCf

111 ∂

∂∂∂

“hidden slide” provided for additional background

Continuous State Space Model

DuCxyBuAxx

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

n by n n by m

r by n r by m

Assumes deviation variable form

StabilityEigenvalues of A

n x n A matrix = n eigenvalues = n states If all eigenvalues have a negative real portion – stableIf any eigenvalue has a positive real portion – unstableComplex – generally ‘oscillatory’

( ) 0det =− AIλ

Characteristic Polynomial (n roots): Must have at least 2 states to oscillate (be complex)

eigenvalues are the roots of the equation

Identity matrix

Example: CSTR at 2 Operating PointsOperating condition 1 Operating condition 2

A1 =−1.1680 −0.08862.0030 −0.2443

⎣ ⎢

⎦ ⎥ A2 =

−1.8124 −0.23249.6837 1.4697

⎣ ⎢

⎦ ⎥

( ) ( )( ) ( )( )04628.04123.1

0003.20886.02443.01680.1det2443.00030.2

0886.01680.1

=−−++=−

⎥⎦

⎤⎢⎣

⎡+−

λλλλ

λ = −0.8955 hr-1 and λ = −0.5168 hr-1

λ = −0.8366 hr-1 and λ = 0.4939 hr-1 Operating Point 2 = Unstable

Operating Point 1 = Stable

Can show that the Eigenvalues for Operating Point 2 are:

Laplace Transform

Convert Differential Equations to Algebraic EquationsDesign controllers using algebra rather than differential equations

Easy Analysis of “Block Diagrams”

L f t( )[ ]= F s( ) = f t( )e− st dt0

L 1[ ]=1s

( ) ( ) ( )0fssFdt

tdfL −=⎥⎦⎤

⎢⎣⎡

Definition of Laplace Transform

Unit step

Derivative

Laplace Transforms, cont’d

Most Undergraduate CoursesMuch (perhaps too much) time is spent:

Taking Laplace transform of process differential equationLaplace transform of “forcing function” (typically a step)Multiply, then perform partial fraction expansionInvert each term back to the time domain for an analytical expression

In PracticeStep response behavior for process understandingMain use of transfer function is for “controller synthesis”Can easily convert from differential equations (“state space”) to transfer function form using MATLAB, etc.Closed-loop block diagram analysis

Very painful, many nightmares?

Block Diagram AnalysisSetpoint Error Manipulated input Measured output

g c e(s)

controller

(s) u(s)g (s) p

y(s)r(s) +

−process

( ) ( ) ( )( ) ( ) ( )sr

sgsgsgsg

pc ⋅+

Closed-loop characteristic equation(roots determine stability)

Routh Array, etc.

State Space to Transfer Function Form

DuCxyBuAxx

( ) ( ) ( )( ) ( ) ( )sDusCxsy

sBusAxssx+=+=

( ) ( ) ( )( ) ( )[ ] ( )

( ) ( ) ( )susGsy

suDBAsICsy

sBuAsIsx

−=−

Take the Laplace Transform (s = transform variable)

Transfer function matrix

usually, D = 0

( ) ( )[ ]BAsICsG 1−−=

Matrix Transfer Function Form

( ) ( ) ( )susGsy =

y1 s( )M

yr s( )

⎢ ⎢ ⎢

⎥ ⎥ ⎥

=g11 s( ) L g1m s( )

gr1 s( ) L grm s( )

⎢ ⎢ ⎢

⎥ ⎥ ⎥

u1 s( )M

um s( )

⎢ ⎢ ⎢

⎥ ⎥ ⎥

yi s( ) = gij s( )uj s( )

r outputsm inputs

output i input j

r rows by m columns

Transfer Function MatrixFirst subscript = outputSecond subscript = input

Dynamic Behavior: SISO

asasasabsbsbsbsg n

p ++++++++

= −−

−−

( ) ( )( ) ( )( )( ) ( )n

mpzp pspsps

zszszsksg

−−−

−−−=

( ) ( )( ) ( )( )( ) ( )111

ssssssk

sgpnpp

nmnnpp τττ

τττL

Polynomial

Pole-zero

Gain-time constant

Zeros = roots of numeratorPoles = roots of denominator = determine stability

11 1 pp −=τ Large time constant = small, negative, pole

Relative order= n-m

Example of Pole-Zero Cancellation

Number of states = Number of poles (order of numerator of transfer function), except when some poles and zeros ‘cancel’

[ ] [ ]0 018255.35301.1

5640.275.0

9056.00

⎥⎦

⎤⎢⎣

⎡−=⎥

⎤⎢⎣

⎡−−

( ) ( )( )( )2640.23.0

3.05302.16792.0564.2

4590.05302.12 ++

++−−

gp s( ) =−0.6758

0.4417s + 1

Bioreactormodel

Chemical Process Engineers

More familiar with gain-time constant formMost chemical processes are stable

Exceptions: Exothermic or bioreactors, closed-loop systems (mistuned)

Real axis

Imaginary axis

more oscillatory

faster response o

inverse response zeros

xunstable poles

x = poleo = zero

First-Order

( ) ( )susk

τ pdydt

+ y = kpudydt

= −1τ p

time constant

0 2 4 6 8 10 12 14 16 18 200

t, minutes

0 2 4 6 8 10 12 14 16 18 200

t, minutes

Example: gain = 1.43 oC/kWtime constant = 5 minutes

output

First Order + Time-delay

( ) ( )susek

( )θτ −=+ tukydtdy

time constant

time- delay

time constant

long-term change in process output

change in process input

63.2 % of change in process output

time-delay

Second-order Underdamped

period of oscillation

c rise time

decay ratio = b/a

overshoot ratio = a/c

time to first peak

y s( ) =k

τ 2s 2 + 2ζτs +1⋅u s( )

−±−=

step response

Numerator Dynamics

0 10 20 30 40 50 60-0.6

various values of numerator time constant

( ) ( )( )( )11513

ssssg n

nz τ1−=

00 >→< znτ

“inverse response”(right-half-plane zeros)[challenging control problem]

responsefaster large →=nτ

Steady-state Gain

For stable systems, the steady-state gain is found from the long-term response

( ) ( )( ) ( )( )( ) ( ) s

sssksy

nmnnp Δ⋅

111111

ττττττ

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

( ) ( ) uksuksssyty

ssssssk

sssyty

⋅⋅==

+++⋅==

→∞→

limlim

111111

limlimττττττ

p ΔΔ

= ∞→lim

Step input

Final value theorem

Process Gain: Controller Implications

Long-term behavior from steady-state information

Controller really serves as “inverse” of process

uky pΔ=Δ

Δ=Δ1

Process with higher gain is generally easier to control, all else being equal…

Process Zero: Controller Implications

For “tight” control, controller is “inverse” of process

( ) ( )sysg

Inverse of gp(s) is unstable if gp(s) has a right-half-plane zero

( ) ( )susgsy p=)(

Transfer Function to State Space

An infinite number of state space models can yield a given transfer function model

Two different state space “realizations” are normally usedControllable canonical formObservable canonical form

Multivariable Systems: Properties

Matrix-vector Notation

( ) ( )suGsy p=

( )( )

( ) ( )( ) ( )

( )( )⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡susu

sgsgsgsg

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )susgsusgsy

susgsusgsy

2221212

2121111

2 input – 2 output Example

Similar to SISO, controller serves as “process inverse”

( ) ( )syGsu p1−=

Steady-state Implications

⎥⎦

⎤⎢⎣

⎡−

uuyuuy

+−=+=

System 1

Which system will be more difficult to control?

System 2

uuyuuy

⎥⎦

⎤⎢⎣

yKu p1−=

Steady-state Implications

⎥⎦

⎤⎢⎣

⎡−

uuyuuy

+−=+=

System 1

Inverse does not exist (Gain Matrix is singular, rank = 1)Cannot independently control both outputs

System 2

uuyuuy

⎥⎦

⎤⎢⎣

yKu p1−=

Another Example

1195.01

uuyuuy

⎥⎦

⎤⎢⎣

1195.01

Inverse exists – No longer singular (“gut feeling” that there may still be a problem…)

Directional Sensitivity:Singular Value Decomposition (SVD)

G = UΣVT

VUG44 344 2144 344 2144 344 2143421

⎥⎦

⎤⎢⎣

⎡−

−−⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−−=⎥

⎤⎢⎣

72.070.070.072.0

0253.00098.1

70.072.072.070.0

1195.01

Gain matrix Left singular Vector matrix

Right singular Vector matrix

Matrix of singularvalues

Example

Strongest input direction

Strongest output direction

Max singular value

Example, cont’d

VUG44 344 2144 344 2144 344 2143421

⎥⎦

⎤⎢⎣

⎡−

−−⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−−=⎥

⎤⎢⎣

72.070.070.072.0

0253.00098.1

70.072.072.070.0

1195.01

⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡41.138.1

70.072.0

1195.01

⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡70.072.0

⎥⎦

⎤⎢⎣

⎡−=⎥

⎤⎢⎣

⎡72.070.0

⎥⎦

⎤⎢⎣

⎡−=⎥

⎤⎢⎣

⎡−⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡018.0018.0

72.070.0

1195.01

Input in Strongest Direction Output in Strongest Direction

Input in Weakest Direction Output in Weakest Direction

Same magnitude input = very different magnitude output

y = Kp u

Example, cont’d

VUG44 344 2144 344 2144 344 2143421

⎥⎦

⎤⎢⎣

⎡−

−−⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−−=⎥

⎤⎢⎣

72.070.070.072.0

0253.00098.1

70.072.072.070.0

1195.01

High condition number indicates problems

Condition number = σ1/σ2 = 1.98/0.0253 = 78

Note: For SVD analysis it is important to properly scale the inputs and outputs

MV Dynamic Properties: Quadruple Tank

G1 s( ) =

2.662s + 1

1.523s +1( ) 62s +1( )

1.430s +1( ) 90s +1( )

2.890s +1( )

⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥

G2 s( ) =

1.563s +1

2.539s +1( ) 63s +1( )

2.556s +1( ) 91s +1( )

1.691s +1( )

⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥

z = −0.057 and + 0.013 sec-1

z = −0.060 and − 0.018 sec-1

Operating Point 1 – Minimum Phase

Operating Point 2 – Nonminimum Phase (RHPT zero)

Matrix inverse is unstable

“Transmission zeros” are bothnegative

Multivariable Systems

Can have right-half-plane “transmission zeros” even when no individual transfer function has a RHP zero

Can have individual RHP zeros yet not have a RHPT zero Fine performance when constraints are not activeMay fail when one constraint becomes active or a loop is “opened”

Can exhibit “directional sensitivity” – with some setpointdirections much easier to achieve than others

Some of these MV properties cause challenges independent of control strategy selected

Summary

Nonlinear ModelSolve for steady-state, then linearize (Taylor series expansion)

State Space (linear) ModelDeviation variable form (perturbations from steady-state)

DynamicsEigenvalues of A matrix = poles of transfer function matrix

Pole-zero cancellation may reduce number of polesRight-half-plane zeros = inverse responseMV: Transmission Zeros

Long term behavior from steady-state gainSingular Value Decomposition (SVD)

Discrete Linear Models

B. Wayne Bequette

• Sampling Rules/Assumptions• Continuous to Discrete• Z-transform• Dynamic Properties of Discrete Systems

Discrete Models

Input held constant between sample times (zero-order hold)

Sample time is constant

Rule of thumb for discrete control – select sample time roughly 1/10 of dominant time constant

Discrete Models

DuCxyuxx

+=Γ+Φ=+1 State Space

Input-Output(ARX)mkmkkk

nknkkk

ububububyayayay

−−−

+++++−−−−=

( ) [ ]kyZzy = Z-transform

[ ] ( )zyzyZ k1

− = Backwards shift operator

( ) ( )( ) ( )zuzbzbzbb

zyzazazam

−+−−

=++++1

...1Solve for y(z)

usually b0 = 0

Some texts/papers have different sign conventions

Discrete Models

DiscreteTransfer Function

Multiply by

( ) ( )zuzazazazbzbzbbzy n

−+−−

++++++++

gp z( ) =b0 + b1z

−1 + ... + bm −1z− m +1 + bmz− m

1 + a1z−1 + ... + an−1z

− n+1 + anz− n

gp z( ) =b0z

n + b1zn−1 + ... + bmzn− m

zn + a1z−1 + ... + an −1z

−n+1 + an

zeros = roots of numerator polynomialpoles = roots of denominator polynomial

( ) ( )( ) ( )( )( ) ( )n

mpzp pzpzpz

zzzzzzKzg−−−−−−

Stability

DiscreteContinuous

StableStable

poles in LHP are stable poles inside unit circle are stable

Example

y k( ) = −a1y k − 1( )+ b1u k − 1( ) gp z( ) =b1z

1 + a1z−1 =

z + a1

p = −a1

a1 = 0.5,−0.5,1.5,−1.5 p = −0.5,0.5,−1.5,1.5

Monotonic, unstable

Oscillatory, unstable

Monotonic, stable

Oscillatory, stable

Characteristic behavior

5.06255.06250.06250.0625y(4)

3.375-3.3750.125-0.125y(3)

2.252.250.250.25y(2)

1.5-1.50.5-0.5y(1)

1.5-1.50.5-0.5p

-1.51.5-0.50.5a1

Let y(0) = 1, u(k) = 0

Value of the pole

Simple Example: Results

Poles inside circlestable

Poles outside circleunstable

Negative polesoscillate

First-order discrete systems can oscillateThis cannot happen with continuous systems

Discrete zerosZeros outside unit circle

Inverse is unstable (not necessarily inverse response)

Any continuous system with relative order >2 will have an unstable inverse with a small enough sample time

Astrom, KJ, P Hagander & J Sternby “Zeros of Sampled Systems,”Proceedings of the 1984 American Control Conference, 1077-1081

gp s( )=1

s +1( )3 ( ) ( )( )( )33679.0

1238.07990.10803.0−

zzzzg p

Sample time = 1

poles = 0.3679 (multiplicity 3)zeros = -1.7990 & -0.1238

Unstable inverse

Relative order = 1Relative order = 3

poles = -1 (multiplicity 3)

Final & Initial Value Theorems

limn→ ∞

y nΔt( ) = limz→1

1 − z −1( )y z( )

y z( ) = gp z( )u z( ) = gp z( )⋅1

1− z−1

limt→ ∞

y t( )= limz →1

1 − z −1( )y z( ) = limz →1

1 − z −1( )gp z( ) 11− z−1 = lim

z →1 gp z( )

limt → 0

y t( ) = limz →∞

1 − z −1( )y z( )Initial value theorem

Final value theorem

Long-term step response

unit step

So, simply set z = 1 in gp(z) for long-term unit step response

Final Value Theorems

( )( )31

( ) ( )( )( )33679.0

1238.07990.10803.0−

zzzzg p

( ) ( )( )( )

12525.02525.0

3679.011238.017990.110803.01 3 ==

=→zg p

( )( )

10 3 =+

=→sg p

Long-term outputs for unit step inputs

continuous

discrete

State Space Models

DuCxyBuAxx

DuCxyuxx

+=Γ+Φ=+1

Continuous

Discrete

Finite differences approximation for derivative

( )( ) ( )

[ ] kkk

tuBxtAIx

tkxtkxx

Δ+Δ+=

+≈Δ−

Δ−Δ+≈

Φ Γ How good are the approximations?

Exact Discretization

( ) ( ) ( )

( ) ktA

t kAtA

BuAIexex

tBudeetxettx k

−ΔΔ+

Δ+ −ΔΔ

+=Δ+ ∫ σσ

( ) BAIee

1−Δ

−=Γ

tAIΔ=Γ

Δ+=Φ

Approximate

Example Discretization

( ) ⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−−Φ=Γ

⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡−

−==Φ

0157.02592.0

04.004.001.0

8869.00974.007408.0

12.012.003.0

⎥⎦

⎤⎢⎣

⎡=Γ

⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡−

−+⎥

⎤⎢⎣

⎡=Φ

88.012.007.0

12.012.003.0

Approximate

[ ] ⎥⎦

⎤⎢⎣

⎥⎦

⎤⎢⎣

⎡+⎥

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎤⎢⎣

04.004.001.0

( ) ( )( )1251101

sssg p

Example, Continued

Approximate

( ) ( ) DzICzg p +ΓΦ−= −1

( ) 21

2 65702.06277.110136.00157.0

65702.06277.10136.00157.0

−−

zzzzg p

( ) 21

2 616.058.11036.0

616.058.1036.0

−−

zzzg p

Poles/eigenvalues = 0.7408 & 0.8869

Poles/eigenvalues = 0.7 & 0.88

Discrete Transfer Function

-2 0 2 4 6 8

temp, deg C

discrete-time step

-2 0 2 4 6 8

discrete-time step

S = s1 s2 s3 s4 s5 L sN[ ]T

Example Step Response Model

Step Response Model

∞−∞+−−+−−

Δ++Δ+Δ++Δ=

Δ= ∑kNNkNNkNk

usususus

LL 1111

∑−

=−−

∞−−+−−−

Δ++Δ+Δ++Δ=Δ++Δ+Δ++Δ=

iikiNkNk

kNkNNkNk

kNNkNNkNkk

uusususususususy

444 3444 21LL

-2 0 2 4 6 8 10

discrete-time step

-2 0 2 4 6 8 10-0.2

discrete-time step

temp, deg Ch

hi = si − si −1

si = hjj=1

Example Impulse Response Model

Impulse and step responsecoefficients are related

Parameter Estimation, ARX Models

22112211

021102112

120112011

−−−−

−−

++−−=

++−−=++−−=

NNNNN ububyayay

ububyayayububyayay

⎥⎥⎥⎥

⎢⎢⎢⎢

⎡−−

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

−−−−

−−

ˆ..ˆ

ΘΦ= Y

model output measured output known (measured) input

time step index

could changesign on columns1 & 2 of Φ

Optimization/Parameter Solution

yi − ˆ y i( )2

∑ = Y − ˆ Y ( )TY − ˆ Y ( )= Y − ΦΘ( )T Y − ΦΘ( )

( )∑=

iiibbaa

,,,ˆmin

Θ = ΦTΦ( )−1ΦTY

Objective Function

Solution

parameterestimatevector

measuredoutputvector

Example

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-0.2

time, min

PRBS estimation example

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

time, min

PRBSinput

Result

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

−−

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

−−−−−

−−−−

−−−

−−−−

−−−

−−−−−

−−

0502.01134.01162.01513.00172.03846.05007.02003.00463.01123.01879.00485.01558.02171.03446.01771.04618.01564.00137.0

111162.01134.0111513.01162.0110172.01513.0113846.00172.0115007.03846.0112003.05007.0110463.02003.0111123.00463.0111879.01123.0110485.01879.0111558.00485.0112171.01558.0113446.02171.0111771.03446.0114618.01771.0111564.04618.0110137.01564.0110889.00137.01100889.0

Θ = ΦTΦ( )−1ΦTY

1.1196−0.3133−0.0889

0.2021

⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥

⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥

( )( )( )5481.05716.0

274.20889.0

3133.01196.112021.00889.0

3133.01196.12021.00889.0

−−−−

+−+−

−−

azazbzbzg p

Identified Model

( ) ( )

3133.01196.112021.00889.0

3133.01196.12021.00889.0

zuzzzzzy

azazbzbzg p

⋅+−+−

+−+−

−−

2021.00889.03133.01196.1 2121 −−−− +−−= kkkkk uuyyy

Subspace Identification

Subspace ID can be used to develop discrete state space models from input-output data

SummaryDiscrete models

State space, ARX, discrete transfer function

Zeros & polesPoles inside unit circle are stable (negative = oscillate)Zeros inside unit circle have stable inverses

Parameter estimationExample with PRBS input

Step and impulse response models

Discrete (Digital) Control

B. Wayne Bequette

• Review of Digital Control Techniques• Discrete Internal Model Control• Examples• Summary

Manfred Morari

Digital Feedback Control

r(z) e(z) u(z)gc(z) gp(z)

Digital Controller DesignResponse specification-based

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

( ) ( ) ( )zrzgzy

zrzgzg

zgzgzy

Desired response

Controller Design

( ) ( ) ( )( ) ( )

( ) ( )( )

( )zgzg

zgzgzgzg

−⋅=

1Desired responsesolve forcontroller

Internal Model Form

u(z)q(z)r(z)

y(z) ++

+ -~g (z) p

g (z) p

y(z) ~

Process

Process model

IMC controller

For controller design, consider perfect model and no disturbances

IMC Design

r(z) u(z)q(z) gp(z)

( ) ( ) ( ) ( )( ) ( ) ( )zrzgzy

zrzqzgzy

Desired response ( ) ( )( )zg

solve forcontroller

Really based on the model

( ) ( )( ) ( ) ( )zgzgzgzgzq pCL

CL 1~~

Can implement IMC in Classic Feedback Form

( ) ( )( ) ( )zgzq

c ~1−=

Digital Controller Design

Deadbeat

Dahlin’s Controller

State Deadbeat

State Deadbeat with Filter (Vogel-Edgar)

Modified Dahlin’s Controller

NEW (IMC) DESIGN

John Ragazzini

Deadbeat

Achieve setpoint in the minimum time (if no time-delay, then one time step)

kk ry =+1

( ) ( )zrzzy 1−=Backwards shift operator

( ) ( )zgzzq p11~−−=

( ) 1−= zzgCL

IMC Form

( ) ( ) 1

−−

−⋅=

zzzgzg pc Classic Feedback Form

Example 0, FO Process

( )110

sg p ( ) 1

9048.010952.0~

zzzg pΔt = 1

( ) ( )00952.0

9048.00952.0

9048.011

−−

( ) ( ) ( )

( ) ( )( ) 1

1504.9

1504.10

1504.9504.10

10952.09048.0

stepunit 1

−−

zzzzzu

zzrlet

zrzqzu

Large control action up,then down

Control action

setpoint change

Classic Feedback Form for Deadbeat Design

19048.015.10

19048.01

0952.01

19048.0

0952.01

0952.09048.0

0952.09048.01~

9048.00952.0

9048.010952.0~

−−

⋅=−

−⋅⎟

⎠⎞

⎜⎝⎛=

−−

⋅⎟⎠⎞

⎜⎝⎛=

−⋅

( ) ( ) ( )1

−⋅=

−⋅= −

−−

zzzgzg ppc

Classic Feedback Implementation

( ) ( ) ( )

( ) ( ) ( ) ( )

9048.05.10

9048.015.101

19048.015.10

−−

−⋅+=

−⋅=−

−−

zezzuz

zezgzu

IMC Implementation

u(z)q(z)r(z)

y(z) ++

~g (z)p

r(z)~ g (z) p

y(z) ~

Process

Process model

IMC controller

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )zrzqzu

zdzrzr

zyzyzd

zuzgzy p

=( ) ( ) ( )

( ) ( ) ( )

~9048.0~0952.01

~9048.010952.01

~9048.010952.0

~~0952.0~9048.0~

0952.0~9048.01

−−−

−−

−⎟⎠⎞

⎜⎝⎛=

−⎟⎠⎞

⎜⎝⎛=

+==−

zrzzzuz

uyyzuzzyz

Discussion

~9048.0~0952.01

~~0952.0~9048.0~

−−

−⎟⎠⎞

⎜⎝⎛=

( )11 9048.05.10 −− −⋅+=−=

eeuuyre

Standard feedback control clearly has the current control actionas a function of the previous control action. Why doesn’t IMC?

Standard Feedback

Example 0 (First-order), Deadbeat Design

-2 0 2 4 6 8

deadbeat, FO example

-2 0 2 4 6 8

Example 1, Second-Order Process

( ) ( )( )1251101~

sssg p ( ) 21

65702.06277.110136.00157.0~

−−

zzzg pΔt = 3

( )00136.00157.0

65702.06277.10136.00157.0

65702.06277.12

+++−

+−= −−

−−−

zzzzzzq

“ringing” behavior as shown next

zero = -0.0136/0.0157 = -0.8694

controller has pole = -0.8694, which is stable, but oscillates

Example 1 (Second-order): Deadbeat Design

-5 0 5 10 15 20 25 30

ydeadbeat, SO example

-5 0 5 10 15 20 25 30-100

Perfect control, at sample times

“ringing” behavior (intersample)

Dahlin’s Controller

Desired first-order + time-delay response to setpointchange

( ) ( )( )

( ) ( ) ( )( ) ( )zgz

zgzgzq pp

CL 111

~−−

⋅−−

=⋅−−

==αα

( ) ( )1

−−

=zzzgCL α

αwhere α = exp(-Δt/λ)For no time-delay:

( ) ( )

( )( ) 0136.00157.0

65702.06277.11

0136.00157.065702.06277.11

⋅−−

= −−

−−

αααα

For the second-order example:

Second-order Process: Dahlin’s Controller

Still “ringing”, but more damped than deadbeat

For λ = 3, α= 0.3679

-5 0 5 10 15 20 25 30

Dahlin, SO example

-5 0 5 10 15 20 25 30-100

First-order response, at sample times

Dahlin’s Modified Controller

Substitute zeros at origin for unstable (|zero|>1) or ringing (|zero|<1 but negative) zeros. Also, keep the gain the same

Problem: Dahlin applied this to gc(z), but should have applied it to the IMC form, q(z)!

State Deadbeat Controller Design

Brings outputs & inputs to new steady-state in the minimum number of time stepsDoes not invert zeros at all

( )000293.0

65702.06277.10293.0

65702.06277.11~2

+++−

==−−

−− zz

zzzzgzq p

0136.00157.00136.00157.0

65702.06277.110136.00157.0

65702.06277.110136.00157.0~

⋅+−+

−−

zzzzzg p

( )zg p−~ ( )zg p+

second-order example

Second-order Example: State Deadbeat Design

-5 0 5 10 15 20 25 30

State Deadbeat, SO example

-5 0 5 10 15 20 25 30-100

Controller Forms

( ) ( ) ( ) ( )( ) ( )( ) ( )n

nkkpzp pzpz

azazazazKzGzg

−−

−−−−==

−−

Zafiriou & Morari notation Assumes stable poles

+ia−

iazero inside unit circle = = zero outside unit circle

( ) ( ) ( )( ) ( ) n

nSD zaaK

pzpzzq −− −−−−

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

11 −

−− −−

⋅−−−−

zaaKpzpzzFqzq n

nSDVE α

State Deadbeat:

State Deadbeat with Filter (Vogel-Edgar):

Notice that neither the state deadbeat nor Vogel-Edgar controllers try to invert zeros (even good ones!)

IMC Design SummaryModel Factorization (“good stuff” and “bad stuff”)

( ) ( ) ( )zgzgzg ppp +−= ~~~

• zeros outside unit circle• zeros inside unit circle that are negative• “all-pass” by including pole at 1/zi for each positive zi outside the unit circle (but not the negative ones)

• numerator one order less than denominator

( ) ( ) ( )zFzgzq p1~−−=

Controller: Invert “good stuff” part of model

• “Filter” for desired response, often first-order

Example 2: Third-Order (3-tank)

( )( )31

sg p ( ) ( )( )( )33679.0

1238.07990.10803.0~−

zzzzg p

( ) ( )( )( )

( )( )( )( ) 23

1238.17990.21238.07990.1

3679.01238.17990.20803.0~

zzzg p

( ) ( ) ( )zgzgzg ppp +−= ~ ~ ~

Sample time = 1

( ) ( ) ( ) ( ) ( )

( ) ( )αααα

−−

⋅−

−−

⋅−

== −

−−

zzzFzgzq p

12526.0

3679.0

2526.03679.0~

Note gain is 1 (value at z = 1)

where α = exp(-Δt/λ)

zeros at -1.7990 (outside unit circle)-0.1238 (inside, but negative)

Example 2 (Third-Order) IMC Design

-2 0 2 4 6 8 10

Discrete IMC, TO example

-2 0 2 4 6 8 10-1

For λ = 1, α= 0.135

Third-Order, Inverse Response (Ex. 3, Z&M, 1985)

( ) ( )( )( )( )5.221

5.1333.3~+++

sssssg p ( ) ( )( )

( )( )( )779.0819.0905.0792.0162.101316.0~

−−−+−−

=zzzzzzg p

( )( )( )( )

( )( )( )

( )( )

( )( )792.1162.0162.11

162.111792.0162.1

779.0819.0905.0162.111

162.11792.1162.001316.0

−⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −+−

⋅−−−⎟

⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −−−

zzzg p

Δt = 0.1

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

( )αα

−−

⋅−

−−−== −

− zzzzzzzFzgzq p

18606.002741.0

779.0819.0905.0~ 1

where α = exp(-Δt/λ)

zeros at 1.162 (outside unit circle)-0.792 (inside, but negative)

( ) ( )( )( )( )779.0819.0905.0

8606.002741.0~−−−

−=− zzz

zzzg p

Response (Ex. 3, Z&M ’85)

-1 0 1 2 3 4 5-1

Discrete IMC, RHPZ - Example 3 Z&M

-1 0 1 2 3 4 5-2

For λ = 0.5, α= 0.8187

“Real-World” Discussion

Have assumed a “perfect model” for these simulations. In practice, the real-world “plant” is not perfectly modeled (indeed, there are usually nonlinearities involved).

Can approximate real-world challenges by having the “plant” be different than the model used for controller design. Also, it is important to incorporate constraints and noise in the simulations.

Summary

Digital Control TechniquesDeadbeatDahlin’sState Deadbeat, State Deadbeat w/filter (Vogel-Edgar)Modified Dahlin’s (mis-developed by Dahlin!)

Internal Model ControlFactorization of zeros outside unit circle and negative zeros inside the circleForm “all-pass”(reflection of positive zeros outside the unit circle)Invert “good stuff” and use first-order filter

model predictive control: background

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