Lecture7-ModellingElectro-MechanicalSystems

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Page : 1 EE406 Control Systems Lecture 7 : Modeling Electro-Mechanical Systems

UCSI University Faculty of Engineering Kuala Lumpur, Malaysia Department of Mechatronics

Lecture 7

Modeling Electro-Mechanical System

Mohd Sulhi bin Azman Lecturer

Department of Mechatronics UCSI University

[email protected]

1 August 2011

Contents

• Typical electro-mechanical system

• Motors : DC and AC

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Introduction

• It is a hybrid of mechanical and

electromechanical system.

• Basically, a motor is an electromechanical

component that yields a displacement output for

a voltage input, that is, a mechanical output

generated by an electrical input.

• We shall consider the application of DC motor.

DC Motor

• In motor control applications, DC control is much easier than AC control strategies.

• DC motor controls are mostly used in open and closed loop control systems to control variables like speed, torque, displacement etc.

(3)

DC Motor Construction

• General construction:

DC Motor Construction

(4)

Principle of Operation of DC Motor

(5)

Typical DC Motor Circuit

Nomenclatures

• I_a = armature current • T_m = motor torque

• J_m= inertia of load with motor shaft

• D_m= coefficient of viscous friction at the armature • L_a = armature inductance

• e_a = applied armature voltage • e_b = back EMF

• θ_m= displacement

• ω_m= angular velocity of the motor • K_b= back EMF constant

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Modeling DC Motors

• In field controlled motor, the field current

controls the motor output, whereas in armature

controlled motor, the field current is maintained

constant by keeping voltage to a constant value.

• The armature current I

_a

, is varied to change the

torque T

m

, of the load connected to the motor

shaft. Here, the input of the motor is the

armature voltage e

_a

, at the armature terminals

and the output is the load torque, T

_m

.

Modeling DC Motors

• Let us evaluate the transfer function of motor. The torque developed by the motor is proportional to armature current:

• Now, the current carrying armature is rotating in a magnetic field, therefore, its voltage is proportional to the motor speed, ω_m. Hence:

( )

m t a

T t

K I t

T s

K I s

=

( )

m

b b m b

b b m

d

t

e t

K

t

K

dt

e s

sK

s

θ

ω

θ

=

…(i)

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Modeling DC Motors

• Now, apply the KVL to the armature circuit:

• Now, re-arrange equations (i) and (ii) and substitute them in equation (iii):

( )

a a

a L R b

a

a a a a b

a a a a a b

e t

e

t

e

t

e t

dI

e

L

I R

e

dt

e s

sL I

I R

e s

=

+

=

+

=

+

…(iii)

( )

m m

a a a b m

t t

T

e

sL

R

sK

s

K

θ



 



=



 

+



+



 



Modeling DC Motors

• Re-arranging gives:

• Now, our transfer function must be in terms of

e_a(s)/θ(s). Now, how can we eliminate the T_m? Easy, what we have to do is that we must write a separate equation for the rotational mechanical system. Now consider the following figure:

(

)

m

a a b m a

t

T

sL

R

sK

e

K

θ

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Modeling DC Motors

• Referring to the figure from the previous slide, we observe that the torque developed in the motor for displacing the load is obtained from the torque equation of mechanical elements:

(

)

2

2 in out

mass m motor

m m

m m m

m m m m m

m m m m

T

D

T

d

J

D

T

dt

J

sD

T

s J

sD

T

θ

=

+

=

+

=

+

=

+

=

…(v)

Modeling DC Motors

• Now, substitute equation (v) into equation (iv):

• Factorizing:

• Assume L_a<< R_a, hence L_a ~ 0:

(

)

(

2 m m

)

m

a a b m a

t

s J

sD

sL

R

sK

e

K

θ

+

=

(

) (

m m

)

a a b m a

t

sJ

sD

sL

R

K

s

e

K

θ

+





+

=









Take out the variable “s” and “θm”.

(

)

a

m m b m a

t

R

sJ

sD

K

s

e

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Modeling DC Motors

• Now, we can write our transfer function in the

form θ

m

(s)/e

a

(s).

(

)

1

m

a a

m m b

t

e

R

s

sJ

sD

K

θ

=





+









…(vi)

Quiz

• Show that equation (vi), as shown in page 17, can

be reduced to:

(

)

(

)

1

t a m m

a _t _b

m

m a

K

R J

K

e

K K

s s

D

J

R

θ

α

=

+









+













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Finding Motor Constants

• For derivation, read Nise, chapter 2, subsection 2.8.

• To find electrical constants K_t/R_a and K_b, we use the torque-speed curve and the following relationships:

t stall a a

K

T

R

=

e

a b

no load

e

K

ω

−

=

Recap

• To model a DC motor, we:

1. Find the expression for the torque. 2. Find the expression for the back EMF.

3. Use KVL to write equation for armature circuit

4. Re-arrange and substitute (1) and (2) into (3) to give the torque-voltage equation.

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Example 1

• Derive the modeling equation for the system

shown below:

Example 1

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Solution to Example 1

• Begin by finding mechanical constants: First, we

calculate the total inertia at the armature of the motor (this is our destination).

2

1

2

1

5 700

12

10

m a L

m

N

J

N

J





=

+













= +





=





Recall the concept of destination-over-source.

Solution to Example 1

• Next, we find the total damping at the armature motor:

• Now, from the torque-speed characteristic, we find electrical constants K_t/R_a and K_b. The parameters that we are looking for is T_stall, ω_no-loadand e_a.

2 1

2 2

1

2 800

10

m a L

m

N

D

N

D





=

+













= +





=





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Solution to Example 1

• And we obtain the following values:

• Now, we proceed to find the transfer function,

by using the equation obtained in page 18.

500

5

100

t stall a a

K

T

R

=

e

=

100

2

50

a b no load

e

K

ω

−

=

Solution to Example 1

• From page 18, we have:

• On substitution, one obtains:

(

)

(

)

1

t a m m

a _t _b

m

m a

K

R J

K

e

K K

s s

D

J

R

θ

α

=

+









+

















[

]

(

)

5 / 12

0.417

1

1.667

10 (5)(2)

12

m a

e

s s

θ

=

+





+





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Solution to Example 1

• Now if you notice, the transfer function in the previous slide is of the form θ_m/e_a, which is in another word, the output (displacement of the motor) – over – the input (input voltage).

• The real transfer function (output) is not the motor displacement, but the load displacement, θ_L. Thus, to find θ_Lwe need to use the gear ratio formula. We note that θ_Lis the destination and θ_m is the source. Thus:

1

2

1

2

destination source

100

(0.417) 0.0417 1000

L m

N N

θ θ

= =

= = =

Solution to Example 1

• And lastly, the transfer function in page (26)

can re-arrange to be:

• And this is our “true” transfer function. That is,

the displacement of the motor, represented by

the load, is our real output.

(

0.0417

1.667

)

L

a

e

s s

θ

=

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Next Step

• Textbook reference : Chapter 2.

• Homework 7 has been posted on the course

website. Attempt them. You do not have to

submit Homework 7 as it will not be graded.

• Thank You.

Wise Word

“Persistence is the twin sister of

excellence. One is a matter of quality; the

Lecture7-ModellingElectro-MechanicalSystems

Lecture 7

Introduction

DC Motor Construction

Modeling DC Motors

shaft. Here, the input of the motor is the

Modeling DC Motors

Modeling DC Motors

Modeling DC Motors

Recap

Example 1

Solution to Example 1

Solution to Example 1

Solution to Example 1

Solution to Example 1

Next Step

other, a matter of time.”