simulink - simon amboise

8/9/2019 Simulink - Simon AMBOISE

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Simon AMBOISE, Weronika KRAKOWIAK

11/06/2008

Velocity control with a PID controller

Simulink project


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Sommaire

1) Introduction..................................................................................................................................... 3

2) Choice of our DC motor................................................................................................................... 3

3) Equations for the motor and for the load : ..................................................................................... 4

4) Motor and load model with simulink .............................................................................................. 6

5) Open-loop........................................................................................................................................ 7

6) Close loop ...................................................................................................................................... 10

7) Digital transformation ................................................................................................................... 12


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1)IntroductionServomechanism is an automatic device for the control of a large power output by means of a

small power input or for maintaining correct operating conditions in a mechanism. It is a type of

feedback control system. The constant speed control system of a DC motor is a servomechanism that

monitors any variations in the motor's speed so that it can quickly and automatically return thespeed to its correct value. Servomechanisms are also used for the control systems of guided missiles,

aircraft, and manufacturing machinery.

In our case, we assume that the direction of the rotational movement of the motor changes in

time. Thats why we have applied huge voltage (170 V) in order to obtain quite a big torque.

2)Choice of our DC motorWe have decided to choose the DC motor AXEM MC23S which is product by Parker SSD Parvex.

This DC motor is used as a part of servomotor.

Fig.1 : F series (on left) and MC series (on right)

The characteristics of this motor are given in the table below:

DC Motor Unit Symbol MC23S

Rated torque Nm Cn 6,1

Rated speed rpm Nn 3000

Rated power output W Pn 1900

Rated voltage V Un 170

Rated current A In 13

EMF at 1000 rpm V Ke 53

Torque constant Nm/A Kt 0,506

Friction coefficient at 1000 rpm Nm Kd 0,013

Armature rsistant R 0,9

Inductance H L 238

Inertia kg.m 10-5 Jm 230


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So, we can calculate the back EMF constant Kb and damping ratio of the mechanical

system b (linear model):

So, Kb=Kt=K=0,506

3)Equations for the motor and for the load :a)The motor

A common actuator in control systems is the DC motor directly provides rotary motion and,

coupled with wheels or drums and cables, can provide transitional motion.

Fig.2 The free body diagram of the rotor showing the direction of rotation.

Fig.3 The electric circuit of the armature.

The motors torque - Tm - is related to the armature current - i - by a constant factor Kt:

The electromotive force - e - is related to the rotational velocity by the following equation:

Tload


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This system will be modeled by summing the torques acting on the rotor inertia and

integrating the acceleration to give the velocity, and integrating velocity to get position.

The first equation is based on the Newtons law, presenting the sum of momentum acting on the

rotor equal to zero:

Basing on the Fig.2, it means on the free body diagram of the motors rotor, we can write the

following equation:

Where:

J Moment of inertia of the rotor

Kt

Integrals of the rotational acceleration

The second equation is based on the Kirchoffs law, presenting the sum of voltages in the electric

circle of the armature:

Electromotive force constant (armature constant)

i Armature current

T Motors torque

b Damping ratio of the mechanical system

Basing on the Fig. 3, it means on the electric circuit of the armature, we can write the following

equation:

Where:

L Electric inductance

R Electric resistance

i Armature current

V Armature voltage

Ke

Rate of change of armature current

Electromotive force constant (motor constant)

b Damping ratio of the mechanical system


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b)The load

At the same way:

The load is coupled with the motor rotor through a somewhat flexible link there will be a spring

action if the positions of the rotor and the load differ. The difference in position (angle) is thus feed

back as a counter torque multiplied by the spring constant. The same is true for the damping. The

difference in velocities between the two bodies will damp the system.

4)Motor and load model with simulink

Fig4 Motor model

Input:

Vin: Voltage input

T_load: load torque

Output:

th_m: motor

dth_m: dmotor/dt

Input:th_m: motor

dth_m: dmotor/dt

Output:

T_load: loads torque

dth_l: dload/dt

th_l: load

Tbrl: spring damping

torque for connection

rotor/load

TKrl: spring constant

torque for connectionrotor/load


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5)Open-loopa) The input voltage

b) Angular Velocity

On this graph we can observe the behavior of the motors angular velocity (yellow line) in

comparison to the loads angular velocity (violet line). We can see certain fluctuations at the

very beginning of the movement, in the close neighbourhood of the point zero, as well as after

1s - when the motor changes its direction of movement. We can see that the value of the

angular velocity approaches 335 rad/s, and that its sign indicates the proper direction of

movement.

On the two small graphs we can observe exactly the behavior of the motors (yellow line) and

loads (violet line) angular velocities in the places where the fluctuations appear. On the first

graph we can see that at the very beginning the motor starts its normal movement and then -

as the movement of the load appears - the angular velocity of the motor stabilizes and

afterwards both velocities become almost equal, which we can observe on the graph (~0.065s)

by the two lines almost covering each other perfectly. On the second graph we observe the

behavior of the system after 1s. We can see that as the direction of the movement changes

(~1s), the behavior of the system is similar to the one at the beginning. The motor starts its

On this graph we can see the behavior of the

motor with the load as it starts to rotate, then weobserve that after 1s its direction of movement

changes. These are the initial condition that we

have applied. We observe how the voltage

changes during this process at the beginning its

value is equal to the nominal value of the

motors voltage (170V) and afterwards, after 1s,

as the voltage value changes sign to the opposite

the direction of the motors movement changes.


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normal movement in the opposite direction as previously and then as the movement of the

load also changes, the angular velocity of the motor stabilizes and both velocities become

almost equal it is visible on the graph graph (~1.065s) by the two lines almost covering each

other perfectly. We can see that the time after which the both velocities become almost equal

measuring it as the movement of the motor starts or as the direction of the movement

changes - in each case, is very similar.

c) Angular position

d) Torque

On this graph we can observe the motors

(yellow line) and loads position (violet line)

behavior presented in radians. We can see that

the lines cover each other almost perfectly.

However, on the two graphs presented below,

we may notice some fluctuations in the

neighbourhood of the point zero as the motor,and then the load, starts its movement and after

1s as the motors, and than the loads,

direction of movement changes. We may

observe that the movement after one second

doesnt start immediately.


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On this graph we may observe the phenomenon of damping. On the vertical line we have

placed the torque appearing on the rod due to the movement of the load with respect to the

motor. Damping is any effect, either deliberately engendered or inherent to a system, that

tends to reduce the amplitude of oscillations of an oscillatory system. It depends on the

angular velocity and the damping coefficient. We can observe this phenomenon at the

beginning of the movement of the system and after 1s when there is a change in the

direction. We can see that the values of the torque are very small, we can even say negligable.
http://en.wikipedia.org/wiki/Oscillationhttp://en.wikipedia.org/wiki/Oscillation


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On this graph we may observe the behavior of the torque developed by the spring constant.

On the vertical line we have placed the torque. It depends on the spring constant and the

position of the system. We can observe this phenomenon as before - at the beginning of the

movement of the system and after 1s when there is a change in the direction. We can see

that the values of the torque are very big in comparison to the previous values (of the torque

appearing due to the relative moevement of the components).

The total torque is equal to the sum of two torques that we have mentioned before: the one

due to the movement and the second due to the spring constant. As the first torque is

negligable we may say that the total torque of the system is equal to the one which is

developed due to the movement.

6)Close loopa) Motor+Load transfert function

First, we have transformed the Motor+load system by one transfert function. We used the matlab

functions :

[A,B,C,D]=linmod('positionanalogicTrFun')

[num,den]=ss2tf(A,B,C,D)

So we obtained as transfert function :


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b) Close loop presentation

At the beginning, we assume that the input angular velocity is equal 50 rad/sec, and after one

second, we change its value to -50 rad/sec. We have added a PID controller and we have searched

the parameters with the Ziegler-Nichols method.

c) The Ziegler-Nichols methodThe "P" gain is increased until it reaches the "critical gain" Kc

at which the output of the loop

starts to oscillate.

We have found Kc = 256 and the oscillation period P c = 0,005 s.

Kc and the oscillation period Pc are used to set the gains as shown:

Kp = 0,6 * Kc = 153,6Ki = 0,5 *Pc = 0,0025

Ki = 0,125 *Pc

d) The results= 0,000625


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On this graph we can observe the behavior of the loads angular velocity. At the beginning we

assume that the value of this velocity is equal to 50 rad/s. We may see that after 0.03s the angular

velocity of the load stabilises and there are no fluctuations visible. The similar behavior we may

observe when the direction of the rotational movement of the whole system changes (after 1s).

7)Digital transformationa) Open loop discret function


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First, we have included Zero-order Hold blocks with sample time equal to 0.001s. The

terminator blocks are needed to block the parameters which are not used. Then we have used the

matlab functions :

[A,B,C,D]=dlinmod('VelocitydigitalTrFun',.001)

[num,den]=ss2tf(A,B,C,D)

With sample time equal to 0.001s.

We obtain the discret transfer function :

The result for the open loop is :

b) Close loop discret functionIn the same way, we obtain the discret transfer function:

For a sample time of 0.001 s.

The shape of this curve is nearly the same as

the shape of the curve presented above

(analog one)

We may see the sample time (0.001s) and the digital profile of this

digital representation.


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As a result, wa have :

Conclusion

Our DC motor needs an high voltage because we need a great torque at the input. With open-loop, we may observe the influence of the load and its inertia on the angular

velocity, angular position. We have considered the influence of the connection rod too.

We used a PID controller in order to optimize, stabilize and improve the output systemsreaction. With the Ziegler-Nichols method, we have found the parameters for the PID

controller. As a result, for our input demand, the response time is small (< 0,05s).

We have digitized this system using Matlabs function : dlinmod. Its very important to takean appropriate sample time in order to have an acceptable systems response.

simulink - simon amboise

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