progress in fault tolerant - fenix.tecnico.ulisboa.pt

Controlo por Computador 25

J. Miranda Lemos IST-DEEC, Área Científica de Sistemas, Decisão e Controlo

Computer Control

Parts 1, 2 – English version

J. Miranda Lemos

Professor Catedrático do IST

2018

http://en.wikipedia.org/wiki/File:Arduino_Diecimila.jpg



Course syllabus

1. Structure of a computer control system

2. Models in computer control

3. Model identification

4. Controller design



1. Structure of a Computer Control System

Objective: Provide a perspective of the topics addressed in the course.

Franklin, Powell, Workman, Digital Control of Dynamic Systems, 3rd ed., cap 3



Tracking an object with the LEGO NXT robot

r

d

Relação entrea velocidadee a posição(integral)

VelocidadeComandodo motor

Distância

Sensor dedistância

Distânciadesejada

r +

-

Erro

e uK=4

dvMotor ecaixa develocidade

PerurbaçãoSoftware

Sistemafísico



Increasing the gain 𝐾 the motor excitation 𝑢 increases and the robot reacts faster.

Relação entrea velocidadee a posição(integral)

VelocidadeComandodo motor

Distância

Sensor dedistância

Distânciadesejada

r +

-

Erro

e uK=4

dvMotor ecaixa develocidade

PerurbaçãoSoftware

Sistemafísico

Tempo

r

d

e

u=ke



A simple example: Mathematical model of the NXT robot

Objective: Model the NXT moving along a straight line in order to go to a

specified position.

How can we do it?



𝑥 the actual NXT position

𝑅 the desired

“break time in strips” o duration ℎ

Observe the position at instants 0 (initial position), ℎ, 2ℎ, 3ℎ, ..., 𝑛ℎ, ...

𝑛ℎ represensta a generic time instant in which we observe the position of

the robot (𝑛 is called discrete time)

Objective: Relate the position of the robot in different time instants.



Grafic of the position of the robot in different time instants

Present position = Previous position + velocity × time interval

𝑥((𝑛 + 1)ℎ) = 𝑥(𝑛ℎ) + 𝛼𝑢(𝑛ℎ)ℎ

𝑢 is the command of the motor (0 a 100%); 𝛼𝑢 is the robot velocity

Tempo(n-1)h nh0 h 2h 3h 4h ... ...

x(0)

x(nh)x((n-1)h)

x



What is the value of parameter 𝜶?

Do experiments with the robot. We observe that

𝑢 = 100% corresponds to a velocity of 1 𝑚/𝑠

Hecce

100 𝛼 = 1

and

𝛼 = 0,01



Model of the movement of robot:

𝑥((𝑛 + 1)ℎ) = 𝑥(𝑛ℎ) + 𝛼𝑢(𝑛ℎ)ℎ

Control algorithm

Tracjking error (difference between the desired and the actual position):

𝑒(𝑛ℎ) = 𝑅 − 𝑥(𝑛ℎ)

Motor command

𝑢(𝑛ℎ) = 𝐾𝑒(𝑛ℎ)

Model of the controlled system (closed-loop system)

𝑥((𝑛 + 1)ℎ) = 𝑥(𝑛ℎ) + 𝛼𝐾ℎ(𝑅 − 𝑥(𝑛ℎ))



Difference equations

𝑥((𝑛 + 1)ℎ) = 𝑥(𝑛ℎ) + 𝛼𝐾ℎ(𝑅 − 𝑥(𝑛ℎ))

𝑥((𝑛 + 1)ℎ) = (1 − 𝛼𝐾ℎ)𝑥(𝑛ℎ) + 𝛼𝐾ℎ𝑅

A special case: 𝛼 = 0,01 ℎ = 1 𝐾 = 60 then 𝛼𝐾ℎ = 0,6 and 1 − 𝛼𝐾ℎ = 0,4

The difference equation becomes

𝑥(𝑛 + 1) = 0,4𝑥(𝑛) + 0,6𝑅

If we know the initial position 𝑥(0) and the value of the reference 𝑅, we can

compute 𝑥(1) from 𝑥(0), and

𝑥(2) from 𝑥(1),

𝑥(3) from 𝑥(2), ...



𝑥(𝑛 + 1) = 0,4𝑥(𝑛) + 0,6

𝑥(0) = 0

𝑥(1) = 0,4𝑥(0) + 0,6 = 0,4 × 0 + 0,6 = 0,6

𝑥(2) = 0,4𝑥(1) + 0,6 = 0,4 × 0,6 + 0,6 = 0,84

𝑥(3) = 0,4𝑥(2) + 0,6 = 0,4 × 0,84 + 0,6 = 0,936

𝑥(4) = 0,4𝑥(3) + 0,6 = 0,9744

𝑥(5) = 0,4𝑥(4) + 0,6 = 0,98976

The best is to use a computer and a language like MATLAB…

But the important point is the fact that the difference equation represents the

whole sequence of numbers that is its solution



Equilibrium points

The equilibrium corresponds to a situation in which the solution is constant:

𝑥(𝑛 + 1) ≈ 𝑥(𝑛) = ��

For the equation

𝑥(𝑛 + 1) = 0,4𝑥(𝑛) + 0,6𝑅

the equilibrium satisfies

�� = 0,4�� + 0,6𝑅

and is given by

�� =0,6

1 − 0,4𝑅 = 𝑅



Stability

Can we prove that the error goes to zero when the time increases?

Obtain an equation for the erroir and solve it.

𝑥((𝑛 + 1)ℎ) = 𝑥(𝑛ℎ) + 𝛼𝐾ℎ(𝑅 − 𝑥(𝑛ℎ))

Multiply by −1 and add 𝑅 a to both sides:

𝑅 − 𝑥((𝑛 + 1)ℎ) = 𝑅 − 𝑥(𝑛ℎ) − 𝛼𝐾ℎ(𝑅 − 𝑥(𝑛ℎ))

Use the definition of tracking error

𝑒((𝑛 + 1)ℎ) = 𝑒(𝑛ℎ) − 𝛼𝐾ℎ𝑒(𝑛ℎ)

𝑒((𝑛 + 1)ℎ) = [1 − 𝐾ℎ𝛼]𝑒(𝑛ℎ)



Error equation

𝑒((𝑛 + 1)ℎ) = [1 − 𝐾ℎ𝛼]𝑒(𝑛ℎ)

Solution of the error equation

𝑒(𝑛ℎ) = 𝑒(0)(1 − 𝐾ℎ𝛼)𝑛



Error equation

𝑒((𝑛 + 1)ℎ) = [1 − 𝐾ℎ𝛼]𝑒(𝑛ℎ)

Solution of the error equationo

𝑒(𝑛ℎ) = 𝑒(0)(1 − 𝐾ℎ𝛼)𝑛

Stability condition

|1 − 𝐾ℎ𝛼| < 1

0 < 𝐾 <2

ℎ𝛼

Condition so that there are no oscillations

0 < 𝐾 <1

ℎ𝛼



Computer control structure

u yD/A

A/D Sensor

Sistema a Controlar

Porto de Saída

Controlador

Porto de Entrada

Computador de Controlo

Sinal de

comando do

actuador

Variável

Física de

saída

Sinal

proporcional

à variável

Sinal de comando



Data acquisition hardware (simplified)

Relógio

Impulsos do relógioPorto deentrada

Micro-computador

INTCC

Sinal a

amostrar

b0

b7

b1Conversor A/D

Ao receber um impulso de relógio, o conversor

A/D retém uma amostra do sinal e inicia a sua

conversão para um número binário.

Quando os bits b0 a b7 atingem o valor correcto,

o sinal de conversão completa CC é activado e o

pino de nterrupção do microcomputador é

actuado.

Se as interrupções não estiverem inibidas, a

subrotina de interrupção começa a ser executada,

sendo efectuada a leitura do porto de entrada,

onde estão ligados os pinos do A/D.



Software structure for computer control

->Inibe interrupções

->Lê y no porto de entrada, ligado ao A/D

->Cálcula o controlo u

->Escreve u no porto de saída ligado ao D/A

->Desinibe interrupções

->Retorna ao programa principal

Programaprincipal

Salta quando chegauma interrupção do

relógio



Time diagram of computer control

Sinal gerado pelo Relógio

Activa interrupções no flanco ascendente

Interrupção do

relógio

Intervalo de amostragem

Interrupção dorelógio

Lê y no

A/DCalculau(tn)

Escreve

u(tn) no D/A

Espera novainterrupção

tn tn+1

Atraso de cálculo

u(tn)

u(tn-1)

Variável Manipulada

The computer considers the

variables only at sampling times;

There is a delay due to

computation and A/D and D/A

conversion



Remark that:

The manipulated variable 𝑢 is constant between consecutive sampling intervals.

Between 2 consecutive sampling instants the manipulated variable is constant. This

fact imposes a constraint on the sampling interval.

There is a delay between the instant tn in which the interrupt arrives (sampling instant)

and the time instant in which 𝑢 is written to the D/A. This delay is mainly due to the

computation time.

The computation may be neglected if it is very small compared with the sampling

interval.

If the computation time is not negligible, it must be considered in control design as an

additional pure delay in the process model.



Control objectves

u yD/A

A/D Sensor

Sistema a Controlar

Porto de Saída

Controlador

Porto de Entrada

Computador de Controlo

Sinal de

comando do

actuador Variável

Física a

controlar

Perturbações

Os objectivos de controlo

dizem respeito à capacidade

de efectuar as manobras

desejadas, tendo em conta a

dinâmica do processo e as

perturbações.

Sinal de

comando



Control objectives (examples)

Keep y at the required value, even in the presence of disturbances

(regulation problem);

Track references for 𝑦, even in the presence of disturbances (servo

problem);

Stabilize the plant to control;

Impose a specified dynamics to the controlled system;

Optimize the system (for instrance minimize the energy) – Optimal Control

Keep a constant behavior even when the plant changes (Adaptive Control).



Exemplo: Controlador Digital com acção integral

Problema: Como actuar no comando do motor do avião para manter o

impulso constante?

Solução: Controlo proporcional

Será que, em regime estacionário, o impulso é igual ao impulso desejado?

Repare-se que não. Se assim for o erro e será nulo e o comando será zero,

ou seja o motor pára.

R yuK+

-

e



Solução para erro estático nulo: Efeito integral

Quando o erro é nulo, a saída do integrador fica constante mas não

necessariamente nula.

R yuK+

-

e

1

sTi

Integrador



A equação que descreve o controlador PI é:

e r y

u t K e tT

e di

t

( ) ( ( ) ( ) )1

0

Quando o erro é nulo o controlo vem dado pelo valor do integrador.

As constantes K e Ti são os ganhos do controlador, podendo ser escolhidas,

por exemplo, de acordo com as regras de Ziegler e Nichols, ou outras mais

adequadas.

Desafio: usando a transformada de Laplace, mostre que se o controlador tiver

um integrador (pólo na origem) o erro estático de seguimento é nulo.



Uma revisão breve: Controlo por realimentação

VKG

KGW

KG

GR

KG

KGYcL

111

𝐾𝐺 é o ganho de malha

R U

W

yK(s) G(s)

v

-



O controlador 𝐾(𝑠) é projectado por forma a “moldar” o ganho de malha:

VKG

KGW

KG

GR

KG

KGYcL

111

|K(jw)G(jw)|

Especificação devida

ao ruído: Atenuação

na alta frequência

Especificação

relativa ao

seguimento da

referência e

rejeição das

perturbações Para que o “quadro” seja completo temos

de considerar a estabilidade e a incerteza

no conhecimento do modelo.



Como implementar em computador as equações do controlador PI?

e r y

u t K e tT

e di

t

( ) ( ( ) ( )1

0

Considere-se a equação do integrador:

u tT

e di

i

t

( ) ( ) 1

0

Derivando ambos os membros da equação:

du

dt Te t

i

i

1

( )

du

dt

u nh u n h

h

i

( ) (( ) )1



Isto resulta nas seguintes equações de diferenças para o

Algoritmo PI digital:

𝑒(𝑛ℎ) = 𝑟 − 𝑦(𝑛ℎ)

u nh u n hh

Te nhi i

i

( ) (( ) ) ( ) 1

u nh K u nh e nhi( ) ( ( ) ( ))

𝑛 é o instante de amostragem (número inteiro)

ℎ é o interval de amostragem (normalmente é omitido)



Pseudocódigo para PI digital

No início de cada intervalo de amostragem, executar recursivamente:

1. Ler no porto de entrada ligado ao A/D a variável y

2.Calcular o erro e r y

3.Calcular a variável manipulada u por

u uh

Te

u K u e

i ianterior

i

i

( )

em que uianterior é a saída do integrador no instante de amostragem anterior

4. Escrever u no porto de saída ligado ao D/A

5. Fazer u uianterior i

6. Esperar nova interrupção



Motivação para o Controlo Adaptativo

Há situações em que a dinâmica do sistema a controlar varia ao longo do

tempo.

Isto pode ser devido, por exemplo, à existência de não linearidades nos

actuadores ou no próprio sistema. Neste caso, a dinâmica linearizada vai

variar com o ponto de trabalho (por exemplo com a velocidade de equilíbrio).

Pode ainda acontecer que a dinâmica varie devido a factores como o

envelhecimento ou alterações do ambiente ou outros factores.

Nesta situação, a afinação do controlador adequada a um ponto de trabalho

pode não o ser para outro.



Exemplo: Válvula não linear

f u u( ) 4 G s

s( )

( )

1

1 3 K 015. Ti 1

A resposta do sistema controlado depende do ponto de funcionamento.

Um controlador bem afinado para um ponto de funcionamento pode não estar

bem afinado para outro.

R yuK(1+ )+

-

e 1Ti s f(.) G(s)

Controlador PI Actuador Sistema



As figuras seguintes mostram a resposta a um escalão na referência de

amplitude 0.1, para duversos pontos de funcionamento, com um PI

sintonizado em torno de r=0.2.

0 10 20 30 40 50 60 70 80 90 1000.1

0.15

0.2

0.25

0.3

0.35

0.4

0 10 20 30 40 50 60 70 80 90 1002.9

2.95

3

3.05

3.1

3.15

3.2

3.25

3.3

0 10 20 30 40 50 60 70 80 90 1004.9

4.95

5

5.05

5.1

5.15

5.2

5.25

5.3



Exemplo de variação da dinâmica numa aeronave

Num avião a dinâmica linearizada altera-se com as condições de voo. A figura

mostra a dependência dos valores próprios do sistema com a velocidade de

equilíbrio

Extraída de Neves da Silva, R. (1994). Controlo de Aeronave não tripulada usando técnicas LQG/LTR de ganho variável Tese de Mestrado, IST -

Departamento de Engenharia Electrotécnica e de Computadores.



Devido à variação da dinâmica com as condições de voo, utilizando um

controlador de ganhos fixos, pode ter-se um bom comportamento numa gama

de funcionamento e um mau funcionamento noutras zonas.

Exemplo do controlo do ângulo de pitch com um controlador fixo quando a

velocidade aumenta progressivamente:

.Extraído de Rato, L. M. (1994). Técnicas de Controlo Adaptativo aplicadas a uma aeronave não tripulada Tese de Mestrado, IST - Departamento de

Engenharia Electrotécnica e de Computadores



Uma possibilidade consiste em adaptar os ganhos para compensar alterações

da dinâmica devidas a variações da velocidade:

Adaptador

Controladoru

y

Ajuste dos

ganhos do controlador



O bloco Adaptador por ser obtido de vários modos:

Tabela que altera os ganhos do controlador de um modo fixo, para cada

valor de velocidade. Esta técnica denomina-se gain-schedulling.

Identificação da dinâmica com o método dos mínimos quadrados, refazendo-

se o cálculo dos ganhos repetidamente, em tempo real. Esta técnica

denomina-se Controlo Adaptativo.



"Gain-scheduling"

x

x

x

+

Controlador

V baixo

Controlador

V médio

Controlador

V alto

u

V

V

V

(V)

(V)

(V)

u

O valor da variável manipulada em

cada instante resulta resulta da

combinação linear das variáveis

calculadas pelos vários

controladores.

Por exemplo, a velocidades baixas, o

controlador respectivo é multiplicado

por um peso próximo de 1 enquanto

o controlador de velocidades altas é

multiplicado por zero.



Controlo Adaptativo vs. "Gain Scheduling"

A solução do controlo de sistemas variáveis no tempo por "gain scheduling" é

útil em muitas situações mas levanta problemas quando:

Não é possível conhecer a priori qual o controlador a utilizar numa dada

situação;

É necessário recorrer a um número muito elevado de controladores.

Uma outra possibilidade, que se estuda neste curso é o Controlo adaptativo.



Controlo Adaptativo

Numa das famílias de Controlo Adaptativo, o Adaptador é constituído por dois

blocos:

Identificador, que estima continuamente os parâmetros (e. g. a posição dos

pólos, dos zeros e o ganho) de um modelo, a partir dos dados de entrada e

saída medidos.

Projecto do Controlador, que recalcula continuamente os ganhos do

controlador tendo em conta as novas estimativas do modelo.

Deste modo, quando a dinâmica do sistema se altera, o identificador dá conta

desse facto e os ganhos do controlador são alterados para frazer face à nova

situação.



No caso do exemplo do controlo do ângulo de pitch, é possível, recorrendo ao

Controlo Adaptativo, obter uma resposta com características razoavelmente

constantes quando a velocidade varia de 10 a 40 m/s:

Extraído de Rato, L. M. (1994). Técnicas de Controlo Adaptativo aplicadas a uma aeronave não tripulada Tese de Mestrado, IST - Departamento de

Engenharia Electrotécnica e de Computadores



Isto é possível graças ao ajuste dos ganhos do controlador efectuado pelo

adaptador:

Extraído de Rato, L. M. (1994). Técnicas de Controlo Adaptativo aplicadas a uma aeronave não tripulada Tese de Mestrado, IST - Departamento de

Engenharia Electrotécnica e de Computadores.

O adaptador actualiza constantemente o modelo melhor ajustado aos dados e

recalcula os ganhos do controlador de acordo com esta estimativa do modelo.



2 - Models in Computer Control

Objective: Introduce the class of plant models used in this course for control

design

Bibliografy: Astrom e Wittenmark, CCS, Cap. 3, em especial as secções 3.2, 3.4, 3.5, 3.6 e 3.7

Franklin, Powell, Workman, Digital Control of Dynamic Systems, 3rd ed., caps. 3 a 6.



SLITs em tempo discreto

SLITs – Linear Time Invariant Systems in discrete time

Linearity (Superposition Principle holds):

u k y k

u k y k

au k bu k ay k by k

1 1

2 2

1 2 1 2

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

SLITu(k) y(k)



Time Invariance

u k y k

u k k y k k

( ) ( )

( ) ( )

0 0



SLIT description by difference equations

Constant coefficients

y k n a y k n a y kn( ) ( ) ( ) 1 1

b u k m b u k m b u km0 1 1( ) ( ) ( )

Coefficients

Equation

order



n initial conditions specified:

y n

y

( )

( )

1

0

Show that:

The linear difference equation with constant coefficients represents a linear

time invariant system

The solution of the linear difference equations with n initial conditions

specified exists and is unique.



Example of a difference equation

𝑦(𝑘 + 1) = 0,5𝑦(𝑘) + 𝑢(𝑘), initial condition 𝑦(0) = 0

Step response

𝑘 𝑢(𝑘) 𝑦(𝑘)

0 1 0 (initial condition)

1 1 1 + 0,5 × 0 = 1 2 1 1 + 0,5 × 1 = 1,5

3 1 1 + 0,5 × 1,5 = 1,75

4 1 1 + 0,5 × 1,75 = 1,875 5 1 1 + 0,5 × 1,875 = 1,9375

6 1 1 + 0,5 × 1,9375 = 1,96875

What is the number to which the output is approaching (equilibrium)?



Equilibrium

𝑦(𝑘 + 1) = 0,5𝑦(𝑘) + 𝑢(𝑘)

At the equilibrium, 𝑦 is constant, and thus

�� = 0,5�� + 1

�� = 2

End of the example.



Description of SLITs by difference equations

Difference equation with forward shifts:

y k n a y k n a y kn( ) ( ) ( ) 1 1

b u k m b u k m b u km0 1 1( ) ( ) ( )

Difference equation with backwards shifts:

y k a y k a y k nn( ) ( ) ( ) 1 1

b u k n m b u k n m b u k nm0 1 1( ( )) ( ( ) ) ( )

Transform one in the other by shifting time n steps.



Example

𝑦(𝑘 + 2) + 𝑎1𝑦(𝑘 + 1) + 𝑎2𝑦(𝑘) = 𝑢(𝑘 + 1)

In this case, 𝑛 = 2, 𝑚 = 1.

Delaying the time 2 steps (the order of the system):

𝑦(𝑘) + 𝑎1𝑦(𝑘 − 1) + 𝑎2𝑦(𝑘 − 2) = 𝑢(𝑘 − 1)



Causality

A system is causal if y(k) depends only on inputs and outputs up to time k.

System described by a linear difference equation:

y k n a y k n a y kn( ) ( ) ( ) 1 1

b u k m b u k m b u km0 1 1( ) ( ) ( )

This system is causal if

n m



System pure delay

y k a y k a y k nn( ) ( ) ( ) 1 1


An input sample applied at time instant 𝑘 influences the output only after time

𝑘 + (𝑛 − 𝑚)

delay



Hereafter only causal systems are considered.

For causal systems the delay, d, is positive:

d n m 0

In many cases, without loss of generality (why?), we assume

d 1



A detour: The Z transform

Consider the sequence

,2,1,0),( kkf

This sequence is mapped by the Z transform in the function of complex

variable

0

)()(k

kzkfzF



Example

Find the Z transform of the sequence

1)( kf ,2,1,0k

Help ( Z transform definition):

0

)()(k

kzkfzF

Help (Sum of the geometrical series):

rr

i

i

1

1

0 for 1r



Solution

2

00

111)()(

zzzzkfzF

k

k

k

k

11

11

z

z

z



Example

Find the Z transform of the sequence

ahkekf )( ,2,1,0k

Help ( Z transform definition):

0

)()(k

kzkfzF

Help (sum of the geometrical series):

rr

i

i

1

1

0 for 1r



Solution

ahah

k

kah

k

kahk

ez

z

zezezezF

10

1

0 1

1)(



Table Z and Laplace transforms (take kht )



Z transform properties

1.Linearity

)()()()( zGzFkgkfZ

2.Time shift

)()( zFzkfqZ nn

1

0

)()()(n

j

jnn zjfzFzkfqZ

Ex.: :1n )0()()( fzFzkqfZ

:2n 122 )1()0()()( zffzFzkfqZ



3.Final value theorem

)()1(lim)(lim 1

1zFzkf

zk

4.Convolution

The convolution between two sequences )(kf and )(kg is defined by

k

j

jkgjfkgf0

)()()(

Then

)().( zGzFgfZ



Solving difference equations

Solve the difference equation

)()()1( kbukayky com 0)0( y

Approach

Equação dediferenças

Equaçãoalgébrica

Solução daeq. algébrica

Solução daeq. diferenças

? Fácil

TransformadaZ inversa

TransformadaZ

Difícil



Example: Solution of a difference equation with the Z transform

)()()1( kbukayky

Take the Z transform

)()()0()( zbUzaYzyzzY

)0()()( yaz

zzU

az

bzY

With 0)0( y : )()( zU

az

bzY



Step response:

0,1)( kku 1)(

z

zzU

)1)(()()(

zaz

bzzU

az

bzY

Simple fraction decomposition:

1

1

11)( 11

z

z

az

az

za

b

zazzY a

ba

ba

k

a

b aky

1)(1



Discrete transfer function

Assume the model represented by the forward difference equation

y k n a y k n a y k b u k m b u k m b u kn m( ) ( ) ( ) ( ) ( ) ( ) 1 0 11 1

Take the Z transform with zero initial conditions to obtain the discrete transfer

function:

G zY z

U z

b z b z b

z a z a z a

m m

m

n n n

n

( )( )

( )

0 1

1

1

1

2

2

SLITu(k) y(k)



Forward operator

x k qx k( ) ( ) 1

Sucessão

Sucessão avançada

Operador avanço



System operator (forward)

Difference equation:

y k n a y k n a y kn( ) ( ) ( ) 1 1

b u k m b u k m b u km0 1 1( ) ( ) ( )

Replace y k n( ) by q y kn ( )

q y k a q y k a y kn n

n( ) ( ) ( )

1

1 b u k m b u k m b u km0 1 1( ) ( ) ( )

Solving with respect to 𝑦(𝑘)

y kb q b q b q b

q a q a q au k

m m

m m

n n

n n

( ) ( )

0 1

1

1

1

1

1



y kb q b q b q b

q a q a q au k

m m

m m

n n

n n

( ) ( )

0 1

1

1

1

1

1

H qb q b q b q b

q a q a q a

B q

A q

m m

m m

n n

n n

( )( )

( )

0 1

1

1

1

1

1

Operador de transferência

do sistema (avanço)

B(q)

A(q)



Backward operator

x k q x k( ) ( ) 1 1

Sucessão atrasada Sucessão

Operador atraso



System operator (backward)

Difference equation:

y k a y k a y k nn( ) ( ) ( ) 1 1


Replace y k n( ) by q y kn ( ):

y k a q y k a q y kn

n( ) ( ) ( )

1

1


Solve with respect to 𝑦(𝑘):

y k qb b q b q

a q a q a qu kd m

m

n

n( ) ( )

0 1

1

1

1

2

21



Reciprocal polynomial

The recyprocal polynomial of A q( ) is defined as

A q q A qn*( ) ( ) 1

Remark: In general, the reciprocal of the reciprocal is not the original

polynomial!

System operator in terms of the reciprocal operator

H q qb b q b q

a q a qq

B q

A q

d m

m

n

n

d*

*

*( )( )

( )

1 0 1

1

1

1

1

11



The representation in terms of the forward operator is better to study stability;

The representation in terms of the backwards operator is better to design

controllers and implement algorithms;



Exercise

)()(5.0)1( kukyky

)(4)1()(3)1(2)2( kukukykyky

For each os these systems:

a) Writte the equation in a form in which the most advanced sample of the

output is )(ky .

b) Writte the transfer function in powers of z and 1z .

c) What is the pure delay in each case?



State model in discrete time

Example: Population model

Divide the population in age ranks (cohorts).

)(kxi is the number of individuals in age range i at time k

Let

Range index: ni ,,2,1,0

Discrete time: ,2,1,0k



If there are no deads, all the individuals of age i in the year k will be of age

1i in the year 1k :

)()1(1 kxkx ii 1,,2,1,0 ni

k=0

k=1

k=2

k=3

x0

x1 x2 x3 x4

k=0 k=1 k=2 k=3

x0

x1

k=0 k=1 k=2 k=3



If some individuals die, only a fraction of individuals with age i in the year k

will have age 1i in the year 1k :

)()1(1 kxkx iii 10 i 1,,2,1,0 ni

The members of the population at age range 0 result from the reproduction of

the individuals in the different ranges:

)()()()1( 11000 kxkxkxkx nn



)()1(1 kxkx iii 10 i 1,,2,1,0 ni

)()()()1( 11000 kxkxkxkx nn

In matrix form:

)(

)(

)(

)(

0000

00

000

000

)1(

)1(

)1(

)1(

2

1

0

1

1

0

210

2

1

0

kx

kx

kx

kx

kx

kx

kx

kx

nn

n

n

)()1( kAxkx



Mckerel captures in the North Sea between 1910 and 1914 (Hjort, 1926).

12

34

56

719101911

19121913

1914

0

20

40

60

80

Idade [ano]Ano

Ca

ptu

ras



Example: State model of the Apach server

J. Hellerstein, X. Diao, S. Parekh, D. Tilbury (2004). Feedback Control of Computing Systems. Wiley Interscience. pp. 229-234.

Inputs: MaxClients (MC), KeepAlive (KA)

Outputs: CPU, MEM

11)()()(1 kKAKAkKAku med 600)()()(2 kMCMCkMCku med

58.0)()()(1 kCPUCPUkCPUky med 55.0)()()(2 kMEMMEMkMEMky med

)(

)(

00028.000025.0

00044.00085.0

)(

)(

63.0026.0

11.054.0

)1(

)1(

2

1

2

1

2

1

ku

ku

kx

kx

kx

kx

)(

)(

10

01

)(

)(

2

1

2

1

kx

kx

ky

ky



Conclusion:

State model for linear systems

)()()1( kBukAxkx

)()()( kDukCxky

State model for nonlinear systems

)(),()1( kukxfkx

))(()( kxhky

In both cases the output equation models sensors.



Conversion between linear models

How to obtain the transfer function given a state model?

Apply the Z transfer with zero initial conditions

𝑥(𝑘 + 1) = 𝐴𝑥(𝑘) + 𝐵𝑢(𝑘)

𝑧𝑋(𝑧) = 𝐴𝑋(𝑧) + 𝐵𝑈(𝑧)

(𝑧𝐼 − 𝐴)𝑋(𝑧) = 𝐵𝑈(𝑧)

𝑋(𝑧) = (𝑧𝐼 − 𝐴)−1𝐵𝑈(𝑧)

𝑌(𝑧) = 𝐶𝑋(𝑧) = 𝐶(𝑧𝐼 − 𝐴)−1𝐵𝑈(𝑧)

Conclusion: The discrete transfer function

𝐺(𝑧) =𝑌(𝑧)

𝑈(𝑧)= 𝐶(𝑧𝐼 − 𝐴)−1𝐵



Model conversion II

Obtain a state model for a given transfer function.

Example: Sistems without zeros

Obtain a state realization for the transfer function

𝐺(𝑧) =𝑏0

𝑧3 + 𝑎1𝑧2 + 𝑎2𝑧 + 𝑎3

Corresponds to the difference equation

𝑦(𝑘 + 3) = −𝑎1𝑦(𝑘 + 2) − 𝑎2𝑦(𝑘 + 1) − 𝑎3𝑦(𝑘) + 𝑏0𝑢(𝑘)



𝑦(𝑘 + 3) = −𝑎1𝑦(𝑘 + 2) − 𝑎2𝑦(𝑘 + 1) − 𝑎3𝑦(𝑘) + 𝑏0𝑢(𝑘)

Take as state variables the output and its first 𝑛 − 1 forward shifts.

(𝑛 is the system order). In this case, 𝑛 = 3, and the state is defined as 𝑥1(𝑘) = 𝑦(𝑘) 𝑥2(𝑘) = 𝑦(𝑘 + 1)

𝑥3(𝑘) = 𝑦(𝑘 + 2) From the definition of the state:

𝑥1(𝑘 + 1) = 𝑦(𝑘 + 1) = 𝑥2(𝑘) 𝑥2(𝑘 + 1) = 𝑦(𝑘 + 2) = 𝑥3(𝑘)

𝑥3(𝑘 + 1) = 𝑦(𝑘 + 3) From the difference equation

𝑥3(𝑘 + 1) = −𝑎3𝑥1(𝑘) − 𝑎2𝑥2(𝑘) − 𝑎1𝑥3(𝑘) + 𝑏0𝑢(𝑘)



𝑥1(𝑘 + 1) = 𝑥2(𝑘)

𝑥2(𝑘 + 1) = 𝑥3(𝑘)

𝑥3(𝑘 + 1) = −𝑎3𝑥1(𝑘) − 𝑎2𝑥2(𝑘) − 𝑎1𝑥3(𝑘) + 𝑏0𝑢(𝑘)

State model in matrix form:

[

𝑥1(𝑘 + 1)𝑥2(𝑘 + 1)𝑥3(𝑘 + 1)

] = [0 1 00 0 1

−𝑎3 −𝑎2 −𝑎1

] [

𝑥1(𝑘)𝑥2(𝑘)𝑥3(𝑘)

] + [00𝑏0

] 𝑢(𝑘)

𝑦(𝑘) = [1 0 0] [

𝑥1(𝑘)𝑥2(𝑘)𝑥3(𝑘)

]



Systems with zeros

𝐺(𝑧) =𝑏0𝑧 + 𝑏1

𝑧3 + 𝑎1𝑧2 + 𝑎2𝑧 + 𝑎3

Break the system in a part without zeros and a part without poles.

The state is defined as before, but taking 𝑥1 instead of 𝑦.

Only the output equation is affected:

𝑦(𝑘) = 𝑏1𝑥1(𝑘) + 𝑏0𝑥2(𝑘)

u yx1b0z+b1

Sistema semzeros



Using MATLAB for linear model conversion

MATLAB (Control Systems Toolbox) can be used to convert models:

ss2tf converts the state model to the transfer function;

tf2ss converts the transfer function to the state model (remark that the

state realization is not the one defined above))

tf define a system by its transfer function.

series transfer function of a series of 2 transfer functions.

Use the help or doc of these functions to know how they work and to know

more functions.



Model of a sampled system

What is the discrete transfer function “seen” by the computer?

Relógio

D/A A/DG(s)

Sistema

u(kh) y(kh)u(t) y(t)



To compute a transfer function:

Apply a signal to the input, with zero initial conditions

Observe the output

Compute the Z transforms of the input and of the corresponding output

Make their quotient

What is the most convenient test signal to apply?



If we use a discrete step, a continuous step is issued, rendering easier the

computations

Relógio

D/A A/DG(s)

Sistema




y t TLs

G s( ) ( )

1

1

y kh TL

sG s

t kh( ) ( )

11

The equivalent discrete transfer function is

G zZ y kh

Z u khd ( )

( )

( )

Relógio

D/A A/DG(s)

Sistema




Let u(kh) be a step. Its Z transform is

Z u khz

( )

1

1 1

kht

d sGs

TLZzzG )(1

)1()( 11



Model of a SLIT with zero order hold sampling (Conclusion)

kht

d sGs

TLZzzG )(1

)1()( 11

Relógio

D/A A/DG(s)

Sistema




MATLAB functions

MATLAB functions to:

Compute the discrete equivalent to a given continuous system: c2d

Inverse operation (the solution may not be unique): d2c

They only work for linear models.



Example – Model of a sampled system

Obtain the discrete equivalent to

G sa

s a( )

?

Solution:

G z z Z TLa

s s ad t kh( ) ( )

( )

1 1 1



Simple fraction decomposition

a

s s a s s a( )

1 1

11 0

sf t t ( )

f kh k

F zz

( )

( )

1 0

1

1 1

10

0

1

1 1

s af t e t

f kh e k

F ze z

at

ahk

ah

( )

( )

( )



G z zz e zd ah( ) ( )

11

1

1

1

1

1 1

G ze z

e zd

ah

ah( )( )

1

1

1

1

The convergence region must be chosen so that the system is causal.



The continuous system response to the step coincides, at the sampling

instants, with the response of the discrete system

This does not happen with other type of insputs, for instance a sinusoid.

For this reason the method is also called step invariant method.

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7 8-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1



Pole transform

A pole at si is transformed in a pole zi of

the discrete model, given by

z ei

s hi h = sampling interval



2nd

order system with conjugate poles

w

w w0

2

2

0 0

22s s

The poles are transformed in the

roots of the polynomial

z a z a2

1 2

a e hh

1

2

02 10 w wcos

a e h

2

2 0 w

These formulas are useful to establish specifications.



Zero mapping

There is no simple rule to mapp the zeros.

A continuous time minimum phase system can be mapped into a non-

minimum-phase system (in which there are zeros outside the unit circle) in

discrete time. This fact happens when decreasing the sampling interval.

Decreasing the sampling rate is not always a good decision (opposite to signal

processing).

See exemples AW pp. 73-75



Sampling the state model

Given the continuous state model:

)()(

)()()(

tCxty

tbutAxtx

What is the model “seen” in discrete time?



Obtaining the sampled state model

Definition of the sampling instants ( h is the sampling period):

htt kk 1

Using the formula of variation of constants1:

1

11 )()()()()(

1

k

k

kkk

t

t

stA

k

ttA

k dssbuetxetx

If u is constant during each sampling period:

)()()(1

1 )(

1 k

t

t

stA

k

Ah

k tbudsetxetxk

k

k

1 Ver Controlo em Espaço de Estados



)()()(1

1 )(

1 k

t

t

stA

k

Ah

k tbudsetxetxk

k

k

To compute the integral, make the change of variable stk 1:

h

A

h

A

t

t

stAded

d

dsedse

k

k

k

0

0

(1

1 )

Therefore

)()()(0

1 k

h

A

k

Ah

k tbudetxetx



)()()(0

1 k

h

A

k

Ah

k tbudetxetx

Defining

Ahe

h

A bde0

The sampled state model is

)()(

)()()1(

kCxky

kukxkx



Approximation for fast sampling (small h ):

AhIeAh bhbAIbde h

h

A

0

2

02

This corresponds to approxima the derivative by finite differences:

)()()()1(

kbukAxh

kxkx



Example: 1st

order system

)()()( tutxtx R ,0

he

10

h

h

ede

)(1)()1( kuekxekx hh



Example: double integrator

uxdt

dx

1

0

00

10

xy 01

ht

Ah AsITLe

11

s

sAsI

0

1

2)det( sAsI (as expected!)



s

s

s

sAsIadj

T

0

1

1

0)(

s

ss

AsI

AsIadjAsI

10

11

)det(

)( 21

10

111t

AsITL



10

1 heAh

h

hdd

hh

211

0

10

12

00

End of example



3.Parametric identification.

Parameter estimation.

Recursive and non-recursive methods.

Least squares and maximum likelihood.

Referência: AW cap. 13, FP



Fitting a curve to experimental data

An experimental situation:

We want to relate the current I with the

tension V in the circuit..

Apply different values of the tension and

register data

Tension [volt] Current [mA]

V1=1 I1=2.1

V2=2 I2=3.9

V3=3 I3=6.2

V4=4 I4=7.9

+

-V g=?

I



The model assumed:

I gV

g is a parameter to be estimated from data.

Due to experimental measuring errors, the data

Does not satisfy the model.

How to decide which is straight line that

best fits data?

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

1

2

3

4

5

6

7

8

9

10

V [volt]

I [m

A]



According to the Principle of Least Squares, the parameter is estimated such

as to estimate the sum of the squares of the deviations.

Accordingly, the estimate of g minimizes

J g g g g g( ) . . . . 21 1 39 2 6 2 3 7 9 42 2 2 2

What we expect to observe

given the model (depends on

the estimate g)

The actual

observation



Least squares cost J g( )

J(g)

g g^



Since J g( ) is a quadratic function of g , the least squares estimate verifies

1

20

dJ

dgg g

or

1 21 1 2 39 2 3 62 3 4 7 9 4 0. . . . g g g g

This equation simplifies to 60 1201 0 .g and the LS estimate is given by

.g 2 00



Ajuste de uma recta a dados experimentais (Caso geral)

Suponhamos que a relação teórica entre duas grandezas X e Y é do tipo

Y X

em que é um parâmetro desconhecido que se pretende estimar.

Repare-se que, conhecendo uma estimativa de podemos responder a

perguntas do tipo "Se X valer … quanto se espera que valha Y ? "



Suponhamos que são observados n pares X Y i ni i, , , , 1 correspondentes a

outros tantos ensaios experimentais. Dispõe-se da tabela

Y

X1 Y1

X2 Y2

X3 Y3

X4 Y4

X5 Y5

X



Pretende-se estimar a recta melhor ajustada aos dados experimentais, de

acordo com o critério de mínimos quadrados.

De acordo com esta critério, a estimativa é

tal que minimiza a soma dos quadrados dos

desvios:

J Y Xi i

i

n

( )

2

1

O que estamos à

espera que seja Yi

O que efectivamente

observamos

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

1

2

3

4

5

6

7

8

9

10

V [volt]

I [m

A]

Desvio i

Xi

Yi



A estimativa de mínimos quadrados verifica a equação (eq. "normal"):

1

20

dJ

d

ou seja

X Y Xi i i

i

n

1

0

Esta equação tem por solução

X Y

X

i ii

n

ii

n

1

2

1



Minimum or not?

A condição 1

20

dJ

d

não garante necessariamente que J( ) seja mínimo

para . É necessário impor uma condição na segunda derivada:

d J

dX i

i

n2

2

2

1

0

Neste exemplo, esta condição é verificada se for feita pelo menos uma

medida com X 0 (o que tem uma interpretação geométrica imediata).

Veremos a seguir que se estimarmos mais do que um parâmetro a segunda

derivada deixa de ser um escalar. A condição de mínimo é então a de que os

dados sejam tais que a matriz de segundas derivadas seja definida positiva.



Bom, ou apenas óptimo?

A estimativa de mínimos quadrados é "óptima" no sentido em que minimiza

um funcional de custo. No entanto, o funcional de custo pode não ser o mais

adequado.

Como caricatura, pode dizer-se que os bons relógios são os que estão

parados pois dão horas absolutamente certas duas vezes por dia.

Um outro exemplo é o de um caçador que vê dois pombos. Se disparar para o

ponto que minimiza a distância média quadrática aos dois pombos…

Isto sugere que por vezes são necessários outros critérios.



Outros critérios de Estimação

Ops exemplos anterios sugerem a utilidade de utilizar critérios que

ultrapassem as limitações dos Mínimos Quadrados. Um dos mais utilizados

em Estimação é o critério de Máxima Verosimilhança.

No entanto, quando a motivação é o Controlo Adaptativo, os Mínimos

Quadrados gozam (quando integrados num sistema de controlo em cadeia

fechada) de propriedades que os tornam suficientes para muitas aplicações.

São além disso simples e de convergência robusta.



Carl Frederich Gauss (1777-1855) utilizou pela primeira vez

o critério de mínimos quadrados para a estimação de

parâmetros em equações.

Em 1801, o astronomo italiano Piazzi observou pela primeira

vez um pequeno planeta denominado Ceres. Infelizmente, a

duração das observações era muito curta devido a Ceres se

ser escondido atrás do Sol, pelo que estas eram insuficientes para estimar os

parâmetros da sua órbita pelos métodos tradicionais. Recorrendo ao critértio

dos mínimos quadrados, Gauss efectuou uma estimativa (bastante diferente

das obtidas pelos métodos clássicos) que foi brilhantemente confirmada pelas

observações experimentais.



What is the least squares estimate of the

Gravity acceleration g ?

Model: h gt

e 2

2

t [s] h [m]

1 8.49

2 20.05

3 50.65

4 72.19

5 129.85

6 171.56

h(t)

0



Parameter estimates in difference equations

Model:

y t a y t a y t n b u t b u t mn m( ) ( ) ( ) ( ) ( ) 1 01 1 1

Problem: From samples of u and y , estimate parameters a bi j,

u y

ESTIMADOR

Sistema a

Identificar



Example

Consider the model

)1()()()1( kekbukayky

Data

y kk

2

1

1000

30( )

u kk

2

1

1000

50( )

y k y kk

( ) ( )

1 11

1000

y k u kk

( ) ( )

1 361

1000

y k u kk

( ) ( )

201

1000

Find the LS estimate of parameters a and b

Write the LS functional

Compute the partial derivatives of the cost with respect to the parameters and equate to zero

White noise; corresponds

To model deviations



J y k ay k bu kk

N

1

21

2

1

( ) ( ) ( )

J

ay k y k ay k bu k

k

N

( ) ( ) ( ) ( )1 01

J

bu k y k ay k bu k

k

N

( ) ( ) ( ) ( )1 01

( ) ( ) ( ) ( ) ( )a y k b y k u k y k y kk

N

k

N

k

N2

1 1 1

1

30 20 1 a b

( ) ( ) ( ) ( ) ( )a u k y k b u k u k y kk

N

k

N

k

N

1

2

1 1

1 20 50 36 a b

. .a b 0 61 0 964



Matrix notation

y t a y t a y t n b u t b u t m e tn m( ) ( ) ( ) ( ) ( ) ( ) 1 01 1 1

Define the regressor , ,by

( ) ( ) ( ) ( ) ( )t y t y t n u t u t m1 1 1 1

And the vector of parameters to estimate, , by

a a b bn m1 1

The model is written (linear regression model)

y t t e t( ) ( ) ( ) 1



Least Squares Criteria

Given N observations, estimate the vector of parameters o by a vector by

minimizing:

JN

y t tt

N

( ) ( ) ( )

1

21

2

1



y

y

y N N

e

e

e N

( )

( )

( )

( )

( )

( )

1

2

0

1

1

1

2

y

y

y N N

e

e

e N

( )

( )

( )

( )

( )

( )

1

2

0

1

1

1

2



y

y

y N N

e

e

e N

( )

( )

( )

( )

( )

( )

1

2

0

1

1

1

2

The set of N observations verifies

y

y N 1

N np np 1

N 1



Another way to write the LS functional

JN N

1

2

1

2

2

Since:

y

It is:

JN

y y 1

2

JN

y y y 1

22



The least squares estimate satisfies

J 0

Gradient of a quadratic form

x x Ax x A2

Recall that

JN

y y y 1

22

Therefore

JN

y1

22 2



The least squares estimate satisfies the equation

0222

1 y

NJ

or

y

Computing the transpose

y



Normal equation

The least squares estimate of the parameter vector in the linear regression

model

y t t e t( ) ( ) ( ) 1

Satisfies the normal equation

y

If the inverse of exists, the least squares estimate is unique and is given

by

1

y



Example - Invertibility of the matrix

Consider the system

y k b u k b u k e k( ) ( ) ( ) ( ) 0 11 2

a) is it possible to find estimates of b0 and b1 when the input is always

u k k( ) 1 ?

b) What about if we know that b1 0 ?

c) And if u( )0 0 and u k k( ) 1 1?

Suggestion: Write the matrix when N=4 and then consider what happens

with higher values of N.

Zero mean

noise



Condições de Excitação Persistente

Para que a estimativa de mínimos quadrados exista e seja única é necessário

que a matriz seja definida positiva. Caso contrário o funcional de

mínimos quadrados não tem um mínimo.

A matriz é definida positiva se os dados forem suficientemenbte ricos, o

que depende da entrada do sistema.

As condições na entrada do sistema que levam a que 0 dizem-se

condições de excitação persistente.



Estimation error covariance matrix and information matrix

Definition of the error covariance matrix

P E

When the residuals forma white sequence, it is true that:

i)The least squares estimate is unbiased:

E 0

ii)When the variance of the residuals is 1:

P

1



This fact motivates the definition of the information matrix by

P 1



Recursive Estimator

Objective: Obtain the current estimate by combining the previous estimate with

the most recent data, without the need to writte the normal equation

RLS = Recursive Least Squares

RLS

New data

y t t( ), ( ) 1

Previous estimate ( )t 1

and auxiliary variables

P t( )1

New estimate ( )t and

auxiliary variables P t( )



Non-recursive estimator given t observations:

( ) t y

1

Can be written as:

( ) ( ) ( ) ( ) t t y k kk

t

1

1

1

Where the information matrix is

( ) ( ) ( )t k kk

t

1 11



The information matrix verifies

( ) ( ) ( ) ( )t t t t 1 1 1 (1)

Furthermore

( ) ( ) ( ) ( ) t t y k kk

t

1

1

1 (2) e

( ) ( ) ( ) ( )t t y k kk

t

11

(3)

We want: Write ( ) t as a function of ( ) t 1 , ( )t , y t( ) and ( )t 1

Sugesttion: Isolate the last term of the sum in (2).

Writte (3) for t 1;

use (1);



( ) ( ) ( ) ( ) t t y k kk

t

1

1

1

( ) ( ) ( ) ( ) ( ) ( ) t t y k k y t tk

t

1

1

1

1 1

( ) ( ) ( ) ( )t t y k kk

t

1 1 11

1

( ) ( ) ( ) ( ) ( ) ( ) t t t t y t t 1 1 1 1



( ) ( ) ( ) ( ) ( ) ( ) t t t t y t t 1 1 1 1

( ) ( ) ( ) ( )t t t t 1 1 1

Therefore:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) t t t t t t t y t t 1 1 1 1 1 1

( ) ( ) ( ) ( ) ( ) ( ) ( ) t t t t y t t t 1 1 1 11



The equations of the recursive estimator are:

( ) ( ) ( ) ( ) ( ) ( ) ( ) t t t t y t t t 1 1 1 11

( ) ( ) ( ) ( )t t t t 1 1 1

Drawback: We need to invert a matrix in each iteration.

Difference between what we expect to be the

output y(t) given the estimate and what we

really observe.

New

estimate Previous

estimate Vector gain

We need to invert a matrix in

each iteration



Is it possible to avoid matrix inversion?

( ) ( ) ( ) ( ) ( ) ( ) ( ) t t t t y t t t 1 1 1 11

( ) ( ) ( ) ( )t t t t 1 1 1

In terms of the covariance the equations become:

( ) ( ) ( ) ( ) ( ) ( ) ( ) t t P t t y t t t 1 1 1 1

P t t t t( ) ( ) ( ) ( )

1 1 11



P t t t t( ) ( ) ( ) ( )

1 1 11

Matrix inversion lemma

A BCD A A B DA B C DA

1 1 1 1 1

11

Apply this lemma with

A t P t 1 1 1 1 ( ) ( ) , B t ( )1 , C 1, D t ( )1

We get:

)1()1(1)1()1()1()1()1()1()(1

tPtttPtttPtPtP

Scalar



Recursive Least Squares algorithm (RLS)

Model:

y t t e to( ) ( ) ( ) 1

Estimator:

( ) ( ) ( ) ( ) ( ) ( ) t t K t y t t t 1 1 1

K t P t t( ) ( ) ( ) 1 "Kalman gain"

P t P tP t t t P t

t P t t( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

1

1 1 1 1

1 1 1 1

"Riccati equation"



Using algebraic manipulations, the Kalman gain and the Riccati equation can

be written as:

K tP t t

t P t t( )

( ) ( )

( ) ( ) ( )

1 1

1 1 1 1

P t I K t t P t( ) ( ) ( ) ( ) 1 1

The RLS parameter estimation algorithm is a special case of the Kalman filter.



Example: Recursive estimation of a parameter

0 20 40 60 80 100 120 140 160 180 2000

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Ganho fixo=0.03

Ganho fixo=0.5

Ganho de Kalman

y t t e to

o

( ) ( ) ( )

1

2

Estimator gain:

High Fast initial convergence but high

fluctuations in steady-state.

Low slow convergence.

The Kalman gain is initially high and

then decreases.



MATLAB code for the example

thrls=0;

thsm=0;

thbig=0;

p=10;

theta0=2;

tfinal=200;

for t=1:tfinal

pp(t)=p;

fi=1+0.1*rand;

y=theta0*fi+0.2*randn;

p=p-p*fi*fi*p/(1+fi*p*fi);

kalm=p*fi;

thrls=thrls+kalm*(y-thrls*fi);

thsm=thsm+0.03*(y-thsm*fi);

thbig=thbig+0.5*(y-thbig*fi);

sth(t,1)=thrls;

sth(t,2)=thsm;

sth(t,3)=thbig;

end;

axis([0 tfinal 0 5])

hold on

plot(sth)

hold off



0 20 40 60 80 100 120 140 160 180 20010

-3

10-2

10-1

100

101

This example illustrates the convergence of the RLS algorithm

Initially, since our uncertainty about our true value of the parameter to estimate

is high, we select a high value for the

covariance, in this case P 10.

The Kalman gain is high and the convergence

is fast.

As time goes on and we receive more data, P

decreases and the kalman gain also, causing

the convergence of the estimates.



The estimator may be envisaged as a

Mechanism that reduces our

uncertainty about the true value of the

parameter through the observations.

The uncertainty is measured by the

Probability density function of the

Estimation error.

In this example the uncertainty is

described by a gaussian pdf with

variance proportional to P .

0 0.5 1 1.5 2 2.5 3 3.5 40

1

2

3

4

5

6

Parameter probability density

given 200 observations

Initial parameter probability

density function



An example with unidentifiable parameters

Increasing the number of observations, we reduce the uncertainty about the

sum, but there is a high uncertainty in the perpendicular direction.

-5 -4 -3 -2 -1 0 1 2 3 4 5-5

-4

-3

-2

-1

0

1

2

3

4

5

y k b u k b u k e k( ) ( ) ( ) ( ) 0 11 2

Only the sum can be identified

Parameter

probability

density



An example of unidentifiable parameters in closed loop

Process model:

y t ay t bu t e t( ) ( ) ( ) ( ) 1 1 (*)

This system is coupled with the controller:

u t Ky t( ) ( )

Let be an arbitrary constant. From the controller definition

u t Ky t( ) ( ) 0

Since u t Ky t( ) ( ) 0 , we can add this quantity to (*) and obtain:

y t a K y t b u t e t( ) ( ) ( ) ( ) ( ) ( ) 1 1



Therefore, the controlled system is described by any model of the form

y t a K y t b u t e t( ) ( ) ( ) ( ) ( ) ( ) 1 1

This fact shows that any parameters �� and �� that satisfy

a a K b b

Lead to the same input/output relation. Eliminating , we obtain the following

relation between the parameters to estimate

b bK

a a 1

Any estimate ( , )a b that verifies this condition descries the data equally well.



The estimates belong to the straight line, but they are not equal to the true

process values. In to obtain estimates close to the true ones, we can either

Change the controller gain K

Add an external signal to the control action

a

b1/K

a

^

b



Sleeping in RLS

If the data is adequate, the entries of the Kalman gain decrease and may

become very small.

From this point on, the estimates become approximately constant. If the

system to identify changes, it takes a long time for the estimates to converge

to a new value.

We say that the algorithm is “sleeping”.



Exponential forgetting

Sleeping in RLS happens because the algorithm weights equaly recent data

and remote past data.

To avoid this effect, the least squares functional is modified to include an

exponential weight that weights less the observations in the past:

J y t kt k

k

t

1

21

2

1

( ) ( )

is called the forgetting factor. It satisfies 0 1

Exponential weight, smaller for oldest data



Assimptotic memory

The asymptotic memory N provides an idea of the number of data points that

influence the current estimate.

N

1

1

samll -> Small # data retained; “agile” to track parameter changes

big -> Many data points retained; the algorithm is more slow to track

changes but is more precise

N

1

0.99 100

0.98 50

0.95 20



RLS with exponential forgetting

Minimizes the cost functional with exponential forgetting.

( ) ( ) ( ) ( ) ( ) ( ) t t K t y t t t 1 1 1

K tP t t

t P t t( )

( ) ( )

( ) ( ) ( )

1 1

1 1 1

P t I K t t P t( ) ( ) ( ) ( ) / 1 1

Show that these equations minimize the weighted functional.

Observe that the information matrix is updated according to

( ) ( ) ( ) ( )t t t t 1 1 1



Covariance windup

We want to estimate parameters a and b in the model

y t ay t bu t e t( ) ( ) ( ) . ( ) 1 01 1

The true value of the parameters is

a 06. b 01.

(In practice, you never know the true value; it is used here for comparison!)

Use RLS with exponential forgetting and 095.

Consider two different situations for the input 𝑢:

Constant input

Constant input added with a white noise signal (dither noise)



Results with a constant input

Initially, the trace of 𝑃 decreases and the estimate approaches its true value

of a 06. . However, since the system is not excited and a forgetting factor is

used, 𝑃 starts to increase, leading to an increase of the Kalman gain and

strong fluctuations of the estimate.

0 50 100 150 200 250 3000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 50 100 150 200 250 30010

-1

100

101

102

103

104

tr(P)

a



Results with a variable input

When the input 𝑢 is sufficiently exciting, the trace of 𝑃 does go up again due to

forgetting (because there is new information arriving) and the estimate is kept

close to the true value, although with fluctuations.

0 50 100 150 200 250 3000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 50 100 150 200 250 30010

-2

10-1

100

101

102

103

104

tr(P)

a



Time varying systems

0 50 100 150 200 250 3000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Lbd=0.99

Lbd=0.95

The parameter a changes at t=100 from

0.6 to 0.8.

The figure shows the results with two

different values of the forgetting factor.

When 𝜆 is smaller, the adaptation of the

estimate to the new value is faster, but

the steady-state estimate has more

fluctuations.



Other types of forgetting algorithms

Variable forgetting factor. The forgetting factor is close to 1 when the

power of the residues is low and is decreased when the power increases.

Directional forgetting. The forgetting factor acts only proportionally to the

direction of the incoming information to avoid forgetting about parameters for

which no new information is available.



Numerically robust algorithms

The equations for the RLS that have been presented are not numerically

robust. One way to solve this problem is to use Bierman’s UD algorithm in

which what is propagated are matrices 𝑈 (an upper triangular matrix with 1’s in

the diagonal) and 𝐷 (a diagonal matrix), such that P UDU T .

The positive definite and symmetric characters of 𝑃 are ensured in this way.

Furthermore, since 𝑈 is a kind of square-root of 𝑃, the range of numbers used

by the algorithm is smaller, a fact that increases the relative precision and

reduces numerical problems.

Chapter 13 of AW includes a PASCAL procedure for this algorithm.



The effect of colored noise

0 100 200 300 400 500 600 700 800 900 10000.2

0.3

0.4

0.5

0.6

0.7

0.8

Ruído colorido

Ruído branco

Valor verdadeiro

When the noise is white the estimate is

unbiased, i. e., the mean of the error is

zero.

This is no longer true when the noise is

colored. The figure shows the estimate

of parameter a in

y t ay t bu t e t ce t( ) ( ) ( ) ( ) ( ) 1 1

when c 0 and when c 095.



Asymptotic behavior of the least squares estimator

Question: Does the LS estimate converge to the true value of the parameters?

True model:

)()()( 0 tvtty

The least squares estimate is

N

t

N

t

tytttN1

1

1

)()()(')()(ˆ



)()()( 0 tvtty

N

t

N

t

tytttN1

1

1

)()()(')()(ˆ

N

t

N

t

tvttttN1

0

1

1

)()(')()(')()(ˆ

And hence

N

t

N

t

tvtN

ttN

N1

1

1

0 )()(1

)(')(1

)(ˆ

The LS estimate is given by the true value added to a bias.

Under what conditions is the bias zero?



In order for the LS estimate to be unbiased, i. e.

0)(ˆ ENE

It must be

0)()(1

N

t

tvtE

In general, this happens if )(tv is uncorrelated with )(,),1( ntyty , which

is equivalent to:

)(tv must be uncorrelated with ),2(),1( tvtv

For the LS estimate to be unbiased the noise must be white.



Examplo

Consider the process described by the ARMAX model

𝑦(𝑡) = 𝑎𝑦(𝑡 − 1) + 𝑏𝑢(𝑡 − 1) + 𝑒(𝑡) + 𝑐𝑒(𝑡 − 1)

where u and e are white and independent, with zero mean and unit variance.

Find the LS estimates a and b as a function of cba ,, .

Sugestion: )()()( 0 tvtty

N

t

N

t

tytN

ttN

N1

1

1

)()(1

)(')(1

)(ˆ

Assume valid the approximations

)(')()(')(1

1

ttEttN

N

t

)()()()(

1

1

tytEtytN

N

t



)1()()1()1()( tcetetbutayty

)1()()1(

)1()(

tcete

tu

tybaty

0' )1( t

)(tv



𝜑(𝑡 − 1) = [𝑦(𝑡 − 1) 𝑢(𝑡 − 1)]𝑇 𝜃 = [𝑎 𝑏]𝑇

Leastquares estimate

�� = 𝐸[𝜑(𝑡 − 1)𝜑𝑇(𝑡 − 1)]−1. 𝐸[𝜑(𝑡 − 1)𝑦(𝑡)]

Calculus of the information matrix

Λ = 𝐸 {[𝑦(𝑡 − 1)

𝑢(𝑡 − 1)] [𝑦(𝑡 − 1) 𝑢(𝑡 − 1)]} =

= [𝐸[𝑦2(𝑡 − 1)] 𝐸[𝑦(𝑡 − 1)𝑢(𝑡 − 1)]

𝐸[𝑦(𝑡 − 1)𝑢(𝑡 − 1)] 𝐸[𝑢2(𝑡 − 1)]] = [

𝜎𝑦2 0

0 1]

Λ−1 = [

1

𝜎𝑦2 0

0 1

]



𝐸[𝜑(𝑡 − 1)𝑦(𝑡)] = [𝐸[𝑦(𝑡)𝑦(𝑡 − 1)]

𝐸[𝑦(𝑡)𝑢(𝑡 − 1)]]

Use the model to observe that

𝐸[𝑦(𝑡)𝑢(𝑡 − 1)] = 𝐸[(𝑎𝑦(𝑡 − 1) + 𝑏𝑢(𝑡 − 1) + 𝑒(𝑡) + 𝑐𝑒(𝑡 − 1))𝑢(𝑡 − 1)] = 𝑏

Furthermore

𝐸[𝑦(𝑡)𝑦(𝑡 − 1)]

= 𝑎𝜎𝑦2 + 𝑏𝐸[𝑦(𝑡 − 1)𝑢(𝑡 − 1)] + 𝐸[𝑒(𝑡)𝑦(𝑡 − 1)] + 𝑐𝐸[𝑒(𝑡 − 1)𝑦(𝑡 − 1)]

= 𝑎𝜎𝑦2 + 𝑐

Finaly �� = [

1

𝜎𝑦2 0

0 1] [

𝑎𝜎𝑦2 + 𝑐

𝑏] = [

𝑎 +𝑐

𝜎𝑦2

𝑏]



Estimation in the presence of colored noise

In the presence of colored noise, least squares yield a biased estimate. This

fact means that, after many observations, the estimate does not approximate

the true parameter value.

Methods to solve the problem:

Instrumental Variables

Extended Least Squares

Maximum Likelihood

Of these, ML is the most general and powerful, although the most heavy from

the computational point of view.



Extended least squares

Heuristic generalization of least squares for ARMAX models:

)()(')( tvtty o )()1()()( 1 ntectectetv n )(te white

Assume that, at time 𝑡 the following estimates are available

)(,),1( ntt of )(,),1( ntete

The estimate )(t of )(te can be computed from

nc

cnttttyt

ˆ

ˆ

ˆ

)()1()(')()( 1



)()1()()(')( 1 ntectectetty no

Let

)()1()(':)(' nttttes nes cc ˆˆ'ˆ:'ˆ 1

The extended LS estimate is computed according to

)(ˆ)(')()()()1(ˆ tttytKtt eseseseses

)(tKes is the Kalman gain that corresponds to the regressor )(tes .

Finally:

)(ˆ)(')()( tttyt es



Maximum Likelihood method

Let y be a random variable (r.v.) whose probability density function ),( yp

depends on an unknown parameter

We want to estimate as a function of the observations of y and assuming

that we know the function ),( yp .

It is remarked that ),( yp becomes a function of only when an

observation y is made.



Maximum likelihood estimator

Given an observation y choose the value of that maximizes the likelihood

function defined as

),(:),( ypyL

Or, equivalently

,log yL

Thus

,logmaxˆ yLML or 0,log ˆ

ML

yL



Example

)()( tety

is an unknown constant that we want to estimate from observations of y

(corrupted by the noise )(te ).

At each time t the pdf of e is

2

)(2

1exp

2

1)( eep te

)(),( 2121tetett independent r.v. (white noise)

We want to estimate by the method of maximum likelihood, as a function of

')()1( tyyY .



Sugestion

Let 1x and 2x be independent r.v. Then

)()(),(2121 xxxx ppp

Considering this fact and the form of the pdf of )(ty , compute the joint pdf of

the vector of observations Yp (that depends on ).

Take the logarithm and equate the derivative to zero.



Considering the observation model, for each observation:

2

)( ))((2

1exp

2

1)),((

tytyp ty

Since the )(te are independent for different 𝑡

t

i

iyY iypYp1

)( )),((),(

Therefore,

t

itY iyYp

1

2

2/))((

2

1exp

2

1),(



t

itY iyYp

1

2

2/))((

2

1exp

2

1),(

t

i

Y iyt

Yp1

2)(

2

12log

2,(log

The maximum likelohoiod estimate satisfies

0ˆ,(log

MLY Yp

or

0ˆ)(1

t

i

MLiy and the ML estimate is

t

i

ML iyt 1

)(1



Properties of the ML estimator

Consistency

-20 -10 0 10 20 30 400

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

N muito grande

N grande

N pequeno

o

p(

ML)

ML



Cramer-Rao bound

The precision of an unbiased estimator is bounded by the Cramer-Rao

inequality: 1 JP

T

ooEP )ˆ)(ˆ( is the estimation error covariance matrix

LEJ log

2

2

is Fisher information matrix

The ML estimator is asymptotically efficient, i. e., when the number of

observations grows unbound, the error covariance matrix approaches the

Cramer-Rao bound.



Estimation with MATLAB

z=[y u];

na=4;

nb=1;

nc=4;

nk=1;

nn=[na nb nc nk];

th=armax(z,nn);

yh=idsim(u,th);

plot(t,[y yh]);

[Phi,Gamma,C,D,K,X0]=th2ss(th);

T=0.05;

[A,B]=d2c(Phi,Gamma,T);

[num,den]=ss2tf(A,B,C,D);

The function armax approximates ML estimates. It is suitable for colored noise

Least squares estimates are obtained with the function arx



A simple example with colored noise (simulation)

0 20 40 60 80 100 120 140 160 180 200-2

-1

0

1

2Mínimos Quadrados

0 20 40 60 80 100 120 140 160 180 200-2

-1

0

1

2Máxima Versomilhança



Example: Identification of a flexible bar

We want to identify the model from experimental data.

M otor

B raçoflexível

V eiodo m otor

D/A

A m pl i ficadorde potência

S ensor A /D

u

y

Medida do ângulo doveio do motorC

Controlador doângulo doveio do motor

Dinâmica domotor controlado

Dinâmica da barraflexível

Comandodo motor

Posição doveio do motor

Posição daponta da barra



Open loop step response

2490 2500 2510 2520 2530 2540 2550 2560 2570 2580 2590 2600

-8

-6

-4

-2

0

2

4

6

8

Tempo [número da amostra]

y

Output data

What features can we see in

the response?



Response features

Oscillatory with small damping. (conjugate poles)

High frequency oscillation. (add conjugate poles in high frequency)

Nonminimum phase zero

progress in fault tolerant - fenix.tecnico.ulisboa.pt

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