Clase 3 - Control En Matlab

  • Uploaded by: Eric Mosvel
  • 0
  • 0
  • October 2019
  • PDF

This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA


Overview

Download & View Clase 3 - Control En Matlab as PDF for free.

More details

  • Words: 443
  • Pages: 4
1

DISEÑO DE OBSERVADORES DE ESTADO Considere el siguiente sistema:

 x1   0  x  = 20.6  2  x  y = [1 0] 1   x2 

1  x1  0 + u 0  x2  1

b= 0 1 c=[1,0] c= 1

0

co=ctrb(a,b);ob=obsv(a,c);rank(co)

Diseño de un regulador con observador

ans = 2

a=[0,1;20.6,0]

rank(ob)

a=

ans = 2

eig(a)

El sistema es completamente controlable y completamente observable, entonces los polos se pueden ubicar arbitrariamente en cualquier lugar.

0 1.0000 20.6000 0

ans = 4.5387 -4.5387 Se observa que el sistema es inestable, se desea que los polos queden ubicados en

pd = [−1.8 + 2.4i,−1.8 − 2.4i ]

k=place(a,b,pd) place: ndigits= 15 k= 29.6000 3.6000 po=[-8,-8]

pd=[-1.8+2.4i,-1.8-2.4i]

po = -8 -8

pd = -1.8000 + 2.4000i -1.8000 - 2.4000i

h=acker(a',c',po)

b=[0;1]

h= 16.0000 84.6000 h=h' Area de Automática Ing. Mecatrónica. Jimmy Tombé Andrade

2

-1.0000

h= 16.0000 84.6000

0

0

ba=[b;0] ba = 0 1 0

reaconobser .

pda=[pd,-20]

t Clock

To Workspace1

1

K

s Integrator

B

K

y

C

To Workspace

pda = -1.8000 + 2.4000i -1.8000 - 2.4000i 20.0000 kt=acker(aa,ba,pda)

A K Control K

1

K

K

s Integrator1

B1

C1

A1 K

kt = 101.6000 23.6000 -180.0000

H K

kp=kt(1:1,1:2) 1.2

kp = 101.6000 23.6000

1 0.8 Regulador - observador 0.6

ki=kt(1:1,3:3)

0.4 0.2

ki = -180

0 -0.2 -0.4 0

1

2

3

4

5

6

7

8

9

10

reapiconobser

Diseño de un seguidor – observador aa=[a,zeros(2,1);-c,0] aa =

0 1.0000 20.6000 0

0 0 Area de Automática Ing. Mecatrónica. Jimmy Tombé Andrade

3

[num,den]=ss2tf(a,b,c,d)

t Clock 1 s Integrator2

Step

T o Workspace 180

y

In1 Out1

Gain1

To Workspace1 Subsystem

In1 Out1 In2

num = 0 0 den = 1.0000

1 0 -20.6000

Subsystem1 Control K

sysclasico=tf(num,den) 1

1

K

In1

K

s Integrator

B

1 Out1

C

A K

1 Out1 In1

1 2

1

K

In2

K

s Integrator1

B1

Transfer function: 1 ---------s^2 - 20.6 syms s

C1

A1 K

Controlador=k*((s*eye(2)-a+h*c+b*k)^1)*h

H K

Controlador = 2368/5*(5*s+18)/(5*s^2+98*s+756)14436/(5*s^2+98*s+756)+7614/5*(s+16) /(5*s^2+98*s+756)

1.4

1.2

Seguidor - observador

1

0.8

eval(Controlador)

0.6

0.4

0.2

0 0

1

2

3

4

5

6

7

8

9

10

ans = (2368*s+42624/5)/(5*s^2+98*s+756)14436/(5*s^2+98*s+756)+(7614/5*s+121 824/5)/(5*s^2+98*s+756)

Diseño de un controlador clásico

simplify(Controlador)

d=0

ans = 2/5*(9727*s+46134)/(5*s^2+98*s+756)

d= 0

numc=[2*9727,2*46134] Area de Automática Ing. Mecatrónica. Jimmy Tombé Andrade

4

numc = 19454

92268

numc=numc/25 numc = 1.0e+003 * 0.7782 3.6907

denc=[5*5,5*98,5*756] denc = 25

490

3780

denc =denc/25 denc = 1.0000 19.6000 151.2000 control=tf(numc,denc) Transfer function: 778.2 s + 3691 -------------------s^2 + 19.6 s + 151.2

Area de Automática Ing. Mecatrónica. Jimmy Tombé Andrade

Related Documents

Clase 3 - Control En Matlab
October 2019 19
Clase 2 - Control En Matlab
October 2019 24
Clase 1 - Control En Matlab
October 2019 23
Clase 3 Matlab
October 2019 20
Matlab 3
April 2020 14

More Documents from "adnan"