INTRODUCTION TO CONTROL SYSTEMSPOSITION CONTROL OF A SPRING-MASS SYSTEM

Goal: Positioning the mass m at the desired positionx. To achieve this goal, a control force Fc is used.

To perform a control task, the transfer function of theconsidered system should be known. The transferfunction of an engineering system can be obtainedusing its governing equation of motion. The Lagrangemethod is a practical way of deriving the equation ofmotion. For this purpose, we first write the kineticand potential energies of the system together withthe virtual work expression.

Lagrange’s Equation:

The equation of motion of the spring-mass system:

Application of the Laplace Transform to the both side of theequation of motion gives the TRANSFER FUNCTION of the system.The TRANSFER FUNCTION relates the Input of the system (Fc ) to theOutput (x) assuming all initial conditions as zero.

s20

kscsm1)s(X 2 ++

=The Laplace transform ofthe output for a step inputwith magnitude 20 N.

step command calculates the response of the system to a unit step input. So, themagnitude of the step input should be specified in the numerator of the output.

The response of the spring-mass system to a control force Fc in the form of a stepfunction with magnitude 20 N can be obtained by using the step command inMatlab.

20

Fc(t)

t (sec)

kFx c

ss =

s20

kscsm1)s(X 2 ++

=

k20

kscsms20slimx 20sss =

++=

→

Magnitude of the step input

clc;clearm=50;c=50;k=2000;num=20;den=[m,c,k];step(num,den)

Spring constant

The response of the system can be obtained by using Matlab Simulink.

>>simulink

0 1 2 3 4 5 6 7 8 9 100

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018

Zaman (s)

x (m

)

Considering that the the spring constant k in the system is constant, the onlyfactor, which determines the magnitude of the response is the amplitude of thecontrol force Fc.

The stedy-state response of the system xss will be different when a disturbanceacts on the system. If the disturbance is shown as Ft

Disturbance

The form of equation of motion with a disturbing (uncontrolled) force

K2=10 Volt/m

We will add an actuator , which converts the electric signal to a force /moment and a sensor, which converts a physical output to an electric signal (voltage)

The relation between the inputs (V1 and Ft) and the output V2 can be written as:

Main input Disturbance

If the desired output of the mass is xss=0.01 m (1 cm), the voltage output ofthe displacement sensor should be V2ss=xss*K2=0.01*10=0.1 V. (The steadystate value of the sensor output). The input voltage, which produces thedesired sensor output can be calculated using the relation given below

V2.010*100

2000*1.0KKkVV

21

ss2ss1 ===

The s term denotes the timederivative and in the steady-state case , all changes(derivatives) are zero.

ss1221ss2 Vkscsm

1KKV++

=

ss121

ss2 VkKKV =

Input voltage in order toobtain mass displacementas 1 cm.

Now, we calculate the response of the system to a step input with magnitude 0.2 V. In this calculation, we take the magnitude of disturbing input Ft as zero.

)s(Vkscsm

1KK)s(V 12212 ++=

s2.0

kscsm1KK)s(V 2212 ++

=

What will the system response be if the disturbing force is different from zero? Let’s assume the step disturbance Ft = 10 N.

Laplace transform of the disturbance

0 1 2 3 4 5 6 7 8 9 100

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

Ft=0

Ft=10 N

V2(t)

Time (s)

DisturbanceInput Voltage output

Displacement output

Δt should be choosen properly !

In a Closed Loop Control system, the output of the system is sent back and compared with the Reference signal.

Controller

The expression for the output of the system can be writen as

For Ft=0, V2ss=0.1 V and for Ft=10 N, V2ss=0.05. This result show that the output ofthe open-loop system is highly influenced from the disturbance. This undesiredeffect may be removed by converting the open loop system to a closed loop controlsystem.

Reference signal(Desired output)

ActuatorSystem

Sensor

Disturbance

EError

Where Hk is the transfer function of the controller used in the closed loopcontrol system. The transfer function of the controller determines the type ofthe operations which will be applied to the error signal.

22

11

VqR00qRV

=−=−

V2

R2

-+V1

R1

11

22 V

RRV −=

q

q

11

22

21

12

1

1

VRRV

VRVR

RVq

−=

=−

=

R2

-+V1

R1

q

q

V2

R

-+

R

V2'

11

21

1

222

11

22

VRRV

RR

RRV

RRV

VRRV

=

−−=′−=

−=′

GAIN CIRCUIT (PROPORTIONAL)

2

1

VqC10

0qRV

=−

=−

C

-+V1 V2

R

∫−= dtVRC1V 12

q

q

∫

∫∫

−=

==

=

=

dtVRC1V

dtVR1dtqq

VR1q

VqR

12

1

1

1

INTEGRAL CIRCUIT

R

-+V1 V2

C

dtdVRCV 1

2 −=

q

q

DERIVATIVE CIRCUIT

2

1

VqR0

0qC1V

=−

=−

dtdVRCV

dtdVCq

CVq

12

1

1

−=

=

=

cVRq0 =−

R

V2

R

R

V1 -+

Vc

R

R-+

21c VVV −=

+-

V1

V2

Vc

2q

3q

-+

V1 Vc

R

R

R

V2

V3

R

1qq

1qq

( )321cc321

c

321

33

22

1111

VVVVVRV

RV

RVR

VqR0qqqq

RVq,

RVq

RVq0qRV

++−=⇒=

++−

=−++=

==

=⇒=−

)VVV(V 321c ++−=

21 qqq += RVq

0qRVRR

11

11

−=

=−−

RVq

0RqV

22

22

=

=−

21c

21c

c

VVVRV

RVRV

VRq0

−=

+−−=

=−

2q

R2

-+

R1

R4

-+

C4

C3

-+

R3 R

R

R

R

-+

V1

V2

1

2P R

RK =33

I CR1K = 44D CRK =

∫ ++=dt

dVKdtVKVKV 1D1I1P2

PID CONTROLLER

P Control (PROPORTIONAL CONTROL)

In this type of control, the error signal is amplified in order to eliminate thenegative effect of the disturbance on the system output.

Choose Hk as 60

Without disturbanceclc;clearm=50;c=50;k=2000;k1=100;k2=10;hk=60;num=[k1*k2*hk];den=[m,c,k+k1*k2*hk];step(num*0.1,den)

With disturbance Ft=10 N

clc;clearm=50;c=50;k=2000;k1=100;k2=10;hk=60;num=[k1*k2*hk];den=[m,c,k+k1*k2*hk];step(num*0.1-k2*10,den)

P-I Control (PROPORTIONAL-INTEGRAL CONTROL)

P-I-D Control (PROPORTIONAL-INTEGRAL –DERIVATIVE CONTROL)