Diseño de Controladores en El Dominio de La Frecuencia

Diseño de sistemas de control

Control por Computadora

ESCUELA DE POSGRADO

Curso: Control por Computadora

Tema: Diseño de controladores en el dominio de la frecuencia

Presentado por: CONTRERAS MARTINEZ, DIMEL ARTURO

Docente: Dr. Juan Javier Sotomayor Moriano Msc. Luis Enciso Salas

2016

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Diseño de sistemas de control

Control por Computadora

1.-Una línea de llenado de botellas utiliza un mecanismo de tornillo de alimentación, tal como se muestra en la figura. Se requiere un sistema de control para mantener una velocidad deseada durante durante el traslado de botellas. botell as.

El diagrama de bloques del motor-tornillo (velocidad lineal / variable de control amplif.), viene dada por:

Luego del llenado simultaneo de 5 botellas (que dura de 2 seg. a 4 seg. dependiendo del volumen de la botella) se produce el traslado y posicionamiento del siguiente grupo de botellas (ciclo de llenado). Se requiere:

 Que la velocidad deseada sea controlada en un rango de 0 a 1 m/seg., con una señal de referencia de 0 a 10 VDC,

 Que el error estacionario = 0,  Alcanzar una velocidad deseada en aprox. 2 seg. y  Con mínimo sobreimpulso (mp=0).  Realizar el llenado de aprox. de 2500 botellas por hora.

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Diseño de sistemas de control

Control por Computadora

Sistema de control digital a) Trazar el diagrama diagrama de bloques del sistema sistema de control control digital de la planta dada. Asumir elementos adicionales necesarios para el lazo de control digital: sensor, controlador digital, etc. Seleccionar el periodo de muestreo muestreo T’. fundamentar fundamentar la elección del d el controlador a emplear.

Para seleccionar el periodo de muestreo obtenemos las constantes de tiempo de la planta, y de ellas elegimos la menor, el periodo de muestreo será igual o menor a la quinta parte de esta constante de tiempo elegida.

 () = (.+)(+)  = 0.1  = 1 min = 0.1  = 5 = 05.1 = 0.02 Diseño del controlador digital en el Dominio de la Frecuencia b) Obtener la respuesta en frecuencia de lazo abierto. Obtener las especificaciones en el dominio de la frecuen f recuencia. cia.

Función de transferencia en lazo abierto:

() = {ℎ () () ()} − 1  10  10 

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Diseño de sistemas de control

Control por Computadora

Diagrama de Bode:

MG

′ ′ 

MF

21.2 dB 7.83 54.8° 31.2

Graficar la respuesta en el tiempo del sistema con entrada escalón y sin controlador. Comprobar si el sistema exhibe una respuesta en el tiempo de acuerdo a las l as especificaciones deseadas.

Utilizando la función de transferencia en lazo abierto:

  0.01729 () = 0. 0186 1.799   0.8025

 Aplicamos un escalón a la planta, la l a lectura de la variable de salida es la que se obtiene con el

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Diseño de sistemas de control

Control por Computadora

[y, t] = step(GHz,10) ref = ones(1,length(y)); plot(t,y,'-r' plot(t,y,'-r',t,ref, ,t,ref,'-b' '-b') ) grid on legend('salida' legend('salida', ,'referencia' 'referencia') )

Observamos

que

el

sistema

sin

controlador

no

cumple

con

las

espcificaciones deseadas. Para una entrada escalón unitario la respuesta deberia ser 1VDC y podemos ver que el sistema no llega a ese valor, el error estacionario no es cero.

c) Partiendo de la respuesta en frecuencia del sistema, calcular los parámetros del controlador digital D(z), que permita alcanzar las especificaciones deseadas. Para las condiciones deseadas, se elige un control PI debido a que se quiere que el error estacionario sea “0”.

Controlador PI:

() = (1 (1  1)

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Diseño de sistemas de control

Control por Computadora

Metodología: a. Se construye el diagrama de Bode de la FT de lazo abierto GH(jw). Determinar MF y MG.



Esto ya se calculo calculo anteriormente: anteriormente:

b. Obtenemos la frecuencia de cruce de ganancia(

′

) para el

’

deseado.

Luego de las pruebas realizadas, el valor elegido para el Margen de fase deseado es:

’ = 60°

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Diseño de sistemas de control

Control por Computadora

 A partir del diagrama de bode, se se encuentra el cruce de ganancia:

′ ′ |( )| = 20() 1.4 = 20()  = 0.8511 1 = ′  (/)  10  = ′10  = 1.4472  = 6.91 rad/s

c. Se calcula la atenuación que el controlador deberá proveer para que el sistema compensado tenga

 :

La ganancia proporcional resulta:

d. Se determina el valor de Ti:

e. Se obtiene la FT del controlador D(w) y se pasa al plano – plano  –z z : D(z)

() = 0.8511(1 511(1 0.691) Para pasar de w -> z , hacemos uso de la transformación bilineal o Tustin:

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Diseño de sistemas de control

Control por Computadora

Resumen comparando: También se obtuvo para el controlador para otros valores de Margen de fase deseado MF’ :

MF’ 

′   () ()

60°

70°

90°

6.91 rad/s

5.21 rad/s

3 rad/s

0.8511

0.6026

0.32

1.447

1.919

3.33

0.8511(1 0.6026(1 0.521) 0.32(1 0.3 )   0. 6 91  ) 124122.3 116. 116.2 115  115 106.9 106.2 144.7144.7 191. 333 333 333 191.9 191  191..9 333

d) Obtener la respuesta en frecuencia de lazo abierto del sistema con el controlador diseñado y comparar con la respuesta en frecuencia del sistema sin controlador. De sus conclusiones. >> Script en Matlab:

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Diseño de sistemas de control

Control por Computadora

Respuesta de lazo abierto con controlador D(z)

124 124  122. 122.3 144.7 144.7 Inicialmente diseñado con MF’=60° MF’=60°

Respuesta de lazo abierto sin controlador

Se puede observar que los márgenes de ganancia MG y de fase MF son

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Diseño de sistemas de control

Control por Computadora

Graficar la respuesta en el tiempo del sistema con entrada escalón y con controlador, comprobar si el sistema exhibe una respuesta en el tiempo

de

acuerdo

a

las

especificaciones

deseadas.

De

sus

conclusiones.

La función de transferencia en lazo cerrado con el controlador D(z) es:

()  =   ()() () ()()()

R(z) es un escalón unitario. >> Script en Matlab:

figure(5) FTz = feedback(Dz*Gz,Hz); feedback(Dz*Gz,Hz);%FT %FT lazo cerrado [y, t] = step(FTz,10) step(FTz,10) ref = ones(1,length(y)); ones(1,length(y)); plot(t,y,'-r' plot(t,y,'-r',t,ref/Hz, ,t,ref/Hz,'-b' '-b') ) grid on legend('salida legend('salida lazo cerrado + D(z)', D(z)','referencia' 'referencia') )

Se analizará que la respuesta en el tiempo se acerque más a los requerimientos:

 error estacionario = ess = 0  Tiempo de establecimiento establecimiento = tss aprox. 2 seg.  Mínimo sobreimpulso (min Mp)

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Diseño de sistemas de control

Control por Computadora

Los casos analizados son los siguientes: MF’ 

60°

   

Car ac te rí s ti c as

 =   = .

 ,

para un error 5%.

124 124  122. 122.3 144.7 144.7

70°

 = %

 =   = .. 

 , para

un error 5%.

116. 116.2   115 115 191.9 191.9

 = .. %

90°

 = 

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Diseño de sistemas de control

Control por Computadora

Tras realizar varias pruebas, con diferentes MF deseados, no se eliminó por completo el sobreimpulso ya que la respuesta del del sistema se hace muy lenta, y en este caso el tiempo

es muy i mportante porque es un proceso

industrial rápido. La elección del controlador más adecuado se encuentra entre elegir el controlador diseñado para MF’=60° para tener una respuesta más rápida y el controlador controlador diseñado para MF’=70° para tener un mínimo sobre impulso. Verificaremos para el control de posición cual es el más adecuado. Funcionamiento Funcionamiento de la línea de llenado e) Elabore la señal de referencia apropiada para que el traslado de las botellas cubra la distancia prevista, sin sobreimpulsos significativos y con suficiente velocidad. Lo anterior, a fin de cumplir el requerimiento de botellas llenadas por hora.

Distancia prevista por ciclo: Para el llenado de 5 botellas a la vez: X = 5 * 0.1m = 0.5m

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Diseño de sistemas de control

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Señal de referencia de velocidad:

Se va a considerar los siguientes tiempos, luego de haber hecho pruebas de tiempos para lograr alcanzar x = 0.5m cada ciclo:

Tiempo de llenado mínimo t =2s De acuerdo a las pruebas realizadas se obtuvo: Proceso

Tiempo

Tiempo de

4 seg, total

posicionamiento

t1 = 1s t2 = 2s

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Diseño de sistemas de control

Control por Computadora

Velocidad y recorrido:

La velocidad de referencia y la controlada se muestran muestran a continuación:

La posición de avance de la faja es:

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Diseño de sistemas de control  Entonces

Control por Computadora

en 1hora:

1hora = 3600s

# # = 3600 6 ∗5 # = 3000 Se genera una señal periódica de referencia de la siguiente manera: function y function  y = fcn(u) n = floor(u/7); tt = 4*n; if tt>0 if  tt>0 tll = 3*n; %Tiempo de llenado else tll = 0; end y = 2.5*(u-tt-tll)*(u=(tt+tll+1))*(u=(tt+tll+2))*(u