estabilidade de sistemas lineares realimentadospalhares/aula7_csl.pdf · estabilidade relativa b...

Estabilidade de Sistemas Lineares Realimentados

1. Estabilidade relativa

2. Estabilidade no espaco de estado

3. Exemplo de projeto: controle de direcao de um veıculo

4. Usando MATLAB c©

5. Projeto sequencial: sistema de leitura de um acionador de disco

c©Reinaldo M. Palharespag.1 Controle de Sistemas Lineares – Aula 7

Estabilidade Relativa

B Routh-Hurwitz indica precisamente a estabilidade absoluta

B A estabilidade relativa de um sistema pode ser definida como a propriedade

que e medida pelo valor da parte real de cada raız ou par de raızes

B Para duas raızes, r1 e r2, se |R{r1}| < |R{r2}|, diz-se que r2 e

relativamente mais estavel

B A estabilidade relativa de um sistema pode tambem ser definida em termos

do coeficiente de amortecimento, ζ, de cada par de raızes complexas e, portanto,

em termos da velocidade de resposta e sobre-elevacao ao inves do tempo de

acomodacao



B O criterio de Routh-Hurwitz pode ser estendido de forma a possibilitar

determinacao da estabilidade relativa atraves de uma simples mudanca de variavel

(deslocamento no eixo real)

Exemplo Pode-se determinar sem calcular as raızes que a parte real de todas as

raızes do polinomio abaixo e menor do que −1 ?

∆(s) = s4 + 14s3 + 71s2 + 154s + 120

Faca s = s′−1, Entao

∆(s′) = s′4 + 10s′3 + 35s′2 + 50s′ + 24



O arranjo e dado por

s′4 1 35 24

s′3 10 50

s′2 30 24

s′1 42

s′0 24

Portanto o polinomio nao tem raızes com R{s′} ≥ 0 ou R{s + 1} ≥ 0 ou

R{s} ≥ −1. Entao todas as raızes tem parte real menor do que −1


Estabilidade no Espaco de Estado

B A EC pode ser obtida diretamente do determinante da regra de Mason, do

diagrama de fluxos de sinais correspondente ao sistema de equacoes

Exemplo Para o sistema de equacoes

x1 = −3x1 + x2

x2 = −Kx1 + x2 + Ku

Diagrama de fluxo de sinais correspondente:

1

1

1

−1

−3

U(s)X1(s)

X1(s)X2(s)X2(s)

1/s1/sK



B Aplicando a regra de Mason:

∆(s) = 1 −∑

L1×1 +∑

L2×2 −∑

L3×3 + · · ·

L1 = 1/s; L2 = −3/s; L3 = −K/s2

∆(s) = 1 − (L1 + L2 + L3) + L1L2 = 1 −

(1

s−

3

s−

K

s2

)

−3

s2= 0

m

∆(s) = s2 + 2s + (K − 3) = 0

Logo, basta K > 3 para que se tenha estabilidade



B Outra forma de verificar estabilidade, para sistemas autonomos (u = 0), e

calcular os autovalores associado a matriz dinamica do sistema

x(t) = Ax(t)

B Calculo dos autovalores ?

λx = Ax, x 6= 0

m

(λI − A)x = 0 ⇒ (λI − A) e singular ⇒ |λI − A| = 0


Exemplo de Projeto

Controle de direcao de um veıculo

Neste caso, o objetivo e realizar o controle de direcao de um veıculo, com

acionamento independente nas duas“rodas” (no caso, veıculos com esteira)

Objetivo Especıfico ? Manter o erro em estado estacionario, para um entrada

em rampa, limitado a um certo patamar – depende de parametros a serem

selecionados...

Por que entrada rampa?

Como funciona? E o modelo? Descritos a seguir


Transportation Technology: Tracked Vehicles

Motion of Vehicle

Sprocket WheelActs as drive wheel,it powers the track.

Idler WheelsCarry the weight of thevehicle. In some cases mayhave sprockets and brakes.

TracksLinks chained together with metal or rubber pads thatcontact and grip the ground. The links mesh with thesproket wheel teeth.

Engine and Steering HousingWaterproof housing for the engine andsteering systems. Exhaust is ported at thehighest point of the vehicle in gasolinepowered vehicles.

Wheelbase

Maximum height of obstaclethat vehicle can climb over

Suspension SystemLets idler wheels move upand down in response tochanges in the terrain.

CG

Tracked vehicle fundamentals

Note: CG = Center of Gravity. In these examples Centerof Gravity is exagerated and does not take into accountany other parts of the vehicle other than the track part.

8" 40"

The weight of the vehicle makes thetire flatten where it contacts theground. This gives a contact area of8" x 12" (the width of the tire) for atotal area of 96 square inches.(Depends on air pressure in the tire)

The weight of the vehicle isdispersed over the entire length oftrack that contacts the ground.This gives a contact area of 40" x12" (the width of the track) for atotal of 480 square inches.

This 4-wheeled vehicle has amaximum of 384 square inches of

contact with the ground. A forth of thetotal weight of the vehicle would be

carried by each wheel.

This 2-tracked vehicle has amaximum of 960 square inches ofcontact with the ground. A sixth of

the total weight of the vehicle iscarried by each idler wheel.

Wheel's and Tires Vs. Tracks

Wheel Tracks

Width of wheel and track = 12"

Note: not to scaleWheel's contact

area exagerated forthis illustration.

VariableDifferential

Engine

SteeringControls

Fas

ter

Slo

wer

Direction of Travel(Turn Right)

Steering a Tracked Vehicle

Tracked vehicles are steered byadjusting the speed of the track onone side of the vehicle inrelation to the speed of the trackon the other side. This creates atorque on the vehicle and causesit to pivot around the slowertrack. The steering controlschange individual track speedsthrough a variable speeddifferential. The differential isusually controlled by levers.

By reversing one track entirelythe vehicle can pivot and spin inplace.

Most of the track is slidingsideways compared to theforward motion of the vehiclewhen turning. This can causedamage to the ground or torodadways.

Grip Area

Tensioning SpringKeeps tension on the trackand allows for smallchanges in terrain.

Support Base(while climbing)

Support Base Length

CG

A line drawn straight down from the center of gravity does notpass outside of the support base so vehicle does not tip over

Greater support base lengthmakes flat tracks more stable.

Tank-style tracks are used forvery rough terrain or higherspeeds. Flatter tracks are usedon earth-moving equipmentbecause more track can contactthe ground and climbing overobsticals is less important. Byhaving a larger wheelbase theflat track is more stable. Anobject is stable as long as a linepointing straight down from itscenter of gravity does not gooutside of its support base.

Obstruction

Different Kinds of Tracks: All Terrain VS. Flat

Grip AreaCG

Note: Vehicles are more likely to tip overfrom side to side because the supportbase's width is narrower than its length

SupportBase

Width

Front View

Side Views

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2

1R(s)

Desireddirectionof turning

(s 1 a)(s 1 1)

Ks (s 1 2)(s 1 5)

ThrottleSteering

Tracktorque

Power trainand controller

Vehicle

Right

Left

Difference in track speed

Y(s)Direction of

travel

(b)

(a)

Power train andvehicle G(s)

ControllerGc(s)

Y(s)

Figure 6.8 (a) Turning control system for a two-track vehicle (b) Block diagram

Exemplo: Controle de Direcao de um Veıculo

B EC do sistema realimentado

1 + Gc(s)G(s) = 1 +K(s + a)

s(s + 1)(s + 2)(s + 5)= 0

ou

s(s+1)(s+2)(s+5)+K(s+a) = s4+8s3+17s2+(K+10)s+Ka = 0

Determinar a regiao de estabilidade para valores de K e a. De que forma?

Routh-Hurwitz...



Arranjo de Routh associado:

s4 1 17 Ka

s3 8 (K + 10) 0

s2 b3 Ka

s1 c3

s0 Ka

⇒

b3 =126 − K

8> 0

c3 =b3(K + 10) − 8Ka

b3

> 0

K < 126

Ka > 0

(K + 10)(126 − K) − 64Ka > 0


MASTER 77

Copyright © 1998 by Addison Wesley Longman. All rights reserved.

0

1.0

0.6

2.0

3.0

500 70 100 126 150

K

a

Stableregion

Selected K and a

Figure 6.9 The stable region


B Do objetivo especıfico, o erro em estado estacionario para entrada em rampa,

R(s) = A/s2 e obtido de

ess = lims→0

s

E(s)︷︸︸︷

1

1 + Gc(s)G(s)

= lims→0

ss(s + 1)(s + 2)(s + 5)

s(s + 1)(s + 2)(s + 5) + K(s + a)

A

s2

=10A

Ka



B Da figura, nota-se que pode-se selecionar uma infinidade de valores para o par

K e a a fim de atender o desempenho especificado

B Eg, para ess = 23.8% do valor de A e necessario que

ess =10A

Ka= 0.238A ⇒ Ka =

10

0.238= 42

O que pode ser conseguido, selecionando K = 70 e a = 0.6, ou K = 50 e

a = 0.84, ou...


Usando MATLAB c©

B roots calcula as raızes de um polinomio

B eig calcula os autovalores de uma matriz, particularmente

eig(A) % devolve todos os autovaleres da matriz A


MASTER 78


K

a Stable region

(a 5 0.6, K 5 70)

Characteristicpolynomial

For a given value of K: determinefirst value of a for instability.

twotrackstable.m

% the stability region for the two track vehicle% control problem%a=[0.1:0.01:3.0]; K=[20:1:120];x=0*K; y=0*K;n=length(K); m=length(a);for i=1:n for j=1:m q=[1, 8, 17, K(i)+10, K(i)*a(j)]; p=roots(q); if max(real(p)) > 0, x(i)=K(i); y(i)=a(j-1); break; end endendplot(x,y), grid, xlabel('K'), ylabel('a')

Range of a and K

Initialize plot vectors as zerovectors of appropriate lengths.

20 40 60 80 100 120

2.5

2.0

1.5

1.0

0.5

0

(b)

(a)

Figure 6.16 (a) Stability region for a and K for two-track vehicle turning control (b) MATLAB script

MASTER 79


0 2 4 6 8 10 12 14 16

Ramp input

y(t) Steady-state error

u 5 unit ramp input

a 5 0.6 and K 5 70

Linear simulation

aKramp.m

% This script computes the ramp response% for the two-track vehicle turning control% problem with a=0.6 and K=70%t=[0:0.01:10]; u=t;numgc=[1 0.6]; dengc=[1 1];numg=[70]; deng=[1 7 10 0];[numa,dena]=series(numgc,dengc,numg,deng);[num,den]=cloop(numa,dena);[y,x]=lsim(num,den,u,t);plot(t,y,t,u), gridxlabel('Time (sec)'), ylabel('y(t)')

(b)

(a)

Time (sec)

0

2

4

6

8

10

12

14

16

y(t)

Figure 6.17 (a) Ramp response for a = 0.6 and K = 70 for two-trackvehicle turning control (b) MATLAB script

Exemplo de Projeto Sequencial

Sistema de Leitura de um Drive

B No bloco anterior – desempenho de sistemas realimentados – considerou-se

apenas um ganho estatico, Ka, associado ao amplificador no projeto sequencial

B Nesta bloco, a questao do ajuste do ganho Ka sera reavaliado quando se

considera um sensor de realimentacao de velocidade, como apresentado a seguir


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22

1 2

1R(s) Y(s)

Position

1s

1s + 20

Amplifier Motor coil

Ka G1(s)

H(s) = 1

D(s)

Velocity

Position sensor

K1

Velocity sensor

Switch

2

1 2

1R(s) Y(s)Ka G1(s)

1 + K1s

G2(s)

D(s)

Figure 6.22 The closed-loop disk drive head system with an optional velocity feedback

Figure 6.23 Equivalent system with the velocity feedback switch closed


B Inicialmente deixando a chave aberta, a FT malha fechada e

Y (s)

R(s)=

KaG1(s)G2(s)

1 + KaG1(s)G2(s)

sendo

G1(s) =5000

s + 1000e G2(s) =

1

s(s + 20)

Logo a EC e

s(s + 20)(s + 1000) + 5000Ka = s3 + 1020s2 + 20000s + 5000Ka = 0



Arranjo de Routh

s3 1 20000

s2 1020 5000Ka

s1 b1

s0 5000Ka

⇒

b1 =(20000)1020 − 5000Ka

1020> 0

para estabilidade: Ka < 4080

B Se b1 = 0 obtem-se estabilidade marginal. Isto e obtido para Ka = 4080.

Portanto a equacao auxiliar e dada por

1020s2 + 5000(4080) = 0 ⇒ s1,2 = ±j141.4

B Esta analise permite responder uma questao relevante quando abordou-se

anteriormente o projeto sequencial: o ganho Ka nao pode assumir qualquer valor

neste caso com modelo completo - sem reducao...



B Considerando a realimentacao de velocidade (ie, fechando a chave), a FT

malha fechada passa a ser:

Y (s)

R(s)=

KaG1(s)G2(s)

1 + [KaG1(s)G2(s)] (1 + K1s)

e a EC e

s(s + 20)(s + 1000) + 5000Ka(1 + K1s) =

s3 + 1020s2 + [20000 + 5000KaK1] s + 5000Ka = 0



com arranjo de Routh

s3 1 [20000 + 5000KaK1]

s2 1020 5000Ka

s1 b1

s0 5000Ka

b1 =1020 [20000 + 5000KaK1] − 5000Ka

1020> 0

B Selecionar o par (Ka, K1) tal que b1 > 0, com Ka > 0...

B Eg, K1 = 0.05 e Ka = 100 ...


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TABLE 6.2 Performance of the Disk Drive System Compared to theSpecifications

Performance Measure Desired Value Actual Response

Percent overshoot Less than 5% 0%

Settling time Less than 250 ms 260 ms

Maximum responseto a unit disturbance Less than 5 2 10 13 2 2 1013

Table 6.2 Performance of the disk drive system compared to the specifications


B Usando MATLABc© para avaliar: (i) a resposta de r(t) para y(t); (ii) a maxima

resposta ao disturbio unitario d(t) para y(t).

ka = 100; k1 = 0.05; % Ganhos selecionados

s=tf(’s’) % variavel "s"

g1=tf([5000],[1 1000]) % G1(s)

g2=tf([1],[conv([1 0],[1 20])]) % G2(s)

FTmf=(ka*g1*g2)/(1+(ka*g1*g2)*(1+k1*s)) % FT malha fechada R->Y

FTdisturbio=(-g2)/(1+(ka*g1*g2)*(1+k1*s)) % FT malha fechada D->Y

subplot(2,1,1)

step(FTmf)

subplot(2,1,2)

step(FTdisturbio)



0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.2

0.4

0.6

0.8

1

Time (sec)

Am

plitu

deFTmf

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5−2

−1.5

−1

−0.5

0x 10

−3

Time (sec)

Am

plitu

de

FTdisturbio


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