ELG 4152 :Modern Control

Winter 2007

Printer Belt Drive Design

Presented to :

Prof: Dr.R.Habash

TA: Wei Yang Presented by:

Alaa FarhatMohammed Al-Hashmi

Mubarak Al-Subaie

April , 4. 2007

Outline

Brief Overview Design PD Controller Design PI Controller Design PID Controller Proposed Solution Results References

References “Improved Design of VSS Controller for a Linear Belt-Driven Servomechanism”, by Aleˇs Hace,Karel Jezernik, belt al,

from IEEE/ASME TRANSACTIONS ON MECHATRONICS, VOL. 10, NO. 4, AUGUST 2005

“Adaptive High-Precision Control of Positioning Tables”, by Weiping Li, and Xu Cheng, from IEEE TRANSACTIONS ON CONTROL SYSTEMS TECHNOLOGY,V OL. 2, NO. 3, SEPTEMBER 1994

“Modern Control Systems”, Richard C. Dorf and Robert H. Bishop, Prentice Hall.

“Development of a linear DC motor drive with robust position control”, by Liaw, C.M., Shue, R.Y. et al, from Electric Power Applications, IEEE Proceedings. Volume 148, Issue 2, March 2001 Page(s):111 - 118

“High-precision position control of a belt-driven mechanism”, byPan, J.; Cheung, N.C.; Jinming Yang, fromIndustrial Electronics, IEEE Proceedings. Volume 52, Issue 6, Dec. 2005 Page(s):1644 - 1652

Goal To determine the

effect of the belt spring flexibility using different controllers in

order to improve the system performance

v1

y

T2

T1

Controller Light Sensor

Motor

m

k

k

v2

Shaft

θ1

θ2

A Brief Overview

Controller Light sensor Printing device position

DC motors

Belt

Motor voltage

Printer device

Mass m = 0.2 kg

Light sensorSpring Constant

K1 = 1 V/m

k = 20

Radius of the pulley r = 0.15 m

Motor

Inductance L ≈ 0

Friction b = 0.25N-ms/rad

Resistance R = 2 Ω

Constant Km = 2 N-m/A

Inertia J = Jmotor + Jpulley: J=0.01 kg-m2

Work Distribution

Mubarak Research, Paper Mohammed Research, Paper , Presentation Alaa Research, Simulation

•The motor torque in our case is equivalent to the sum of the torque provided by the DC motor and the undesired load torque caused by the disturbance that will affect the system stability.

The forces present are the tensions T1 and T2 by the following equations:

Mechanical Model

)( 21211 krrkrkT

)( 12122 krrkrkT dt

dk

dt

drmTT 2

22

2

21 2

)()()( sTsTsT dLm

System Model with D Controller

231 xrxx

12

2x

m

kx

J

Tx

J

krx

J

bx

JR

KKKx dm

132

213

2

dt

dvKv 1

22

Td(s)

1/J 1/s r 1/s

X1(s)

2k/mK2K1Km/R

b/J

2Kr /J

1/s

System Model with D Controller (cont.)

qq

kKkkK

L

PPsG

11)(

Mason’s Rule

Td(s)

X1(s) mJRkkrkk

JmbKsJ

krmksJ

bS

sjr

m 21223 22)22(

Design PD Controller

13

122 vKdt

dvKv

2vR

KT mm

13

212 x

R

kKKKx

R

KKKT mmm

Td(s)

1/J 1/s r 1/s

X1(s)

2k/mK2K1Km/R

b/J

RJ

KrRKKKm 213

Design PD Controller

X1(s)

Td(s)

12

213

23 22)2()()2()()( kkkrkkbRskRrskkrkskJRRbsRJs

sRr

mm

Design PID Controller

New State Variables

2

2

44

3

3422

211

,

,

, )(

dt

dx

dt

dx

dt

dyx

xrxxyrx

xyxdttyx

dtvKvK

dt

dvKv 1413

122

System with PID Controller

Td(s)1/J 1/s r 1/s

X1(s)

2k/mK2K1Km/R

b/J

m

m

KKrRKK

RJ

K 213

RJ

KKKm 14

System with PID Controller

rmKKKkbRKKkrKkRmrkRJrKKmkKsmRbsmrJS

sRmr

mmm 14122

133 22)22()(2)(

X1(s)

Td(s)

Open Loop System

231 xrxx

12

2x

m

kx

J

Tx

J

krx

J

bx d 133

2

For the open loop system, the sensor output V1 will be equal to zero. Then our three state space variables equal to the first and second derivative of the displacement and the first derivative of, we get the following state space equations, and the open loop transfer function:

bKmrJkmbsJms

mrs

2)(2 223 Td(s) X1(s)

Range of Stability:

In order to know the range of the gain of our controller, we apply the Routh Test and obtain the appropriate values of the gains:

D controller: ;K2=0.175.035

2 K

System Response using the D Controller

0 0.5 1 1.5 2 2.5 3-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02

Time

Dis

plac

emen

tResponse of the transfer function PD

Implementation of the System with a PD Controller

Similar to the case of the D Controller, in order to know the range of the gain so the system stays stable, we apply the Routh Test. The characteristic equation for our system is:

We then find that and but in order to facilitate our calculation we will assign K2 = 0.1 and K3 =10.

2323 60010001513025)( KksKsssq

35

2K 84.53 K

System Response using thePD controller

0 0.5 1 1.5 2 2.5 3-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02

Time

Dis

plac

emen

tResponse of the transfer function PD

System Stability with thePID Controller Similar to the case of the D and PD

Controller, in order to know the range of the gain so the system is stable, we planned on applying the Routh Test but the coefficients of the characteristic equation for our system are large. Using MATLAB, and fixing K2 to 0.1, we were able to find that and .

In order to facilitate our calculation we will study the case with K2 equal to 0.1, K3 equal to –5 and K4 equal to 10.

System Response using thePID controller

Open-Loop System Resoponse

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02

Time

Dis

plac

emen

tResponse of the OPEN LOOP transfer function

Conclusion

If we are more concerned with the speed of the system, we should go with the PI or PID controller since they are much then the open loop or the PD controller.

They decrease the rise time and settling time and eliminate the steady state error.

Questions