Control Engineering
Problems - Solutions
VI.
July, 2001
Department of Automation and Applied Control
Ruth Bars - Jenõ Hetthéssy
Problem #1
Model of a DC motor
A rough structural diagram of a DC (direct current) motor is seen on the Fig. below. The
output of the motor is the angular velocity , the input signals are the armature voltage Ua
and the torque mt acting at the shaft of the motor (disturbance). The excitation voltage Ue
is constant.
Ra
La
mt
ia
Ua

Ui
Ue= const
The resistance of the rotating part of the motor (armature, rotor) is denoted by Ra, its
inductance is La. The inertia torque (reduced to its shaft) is denoted by , the armature
current is ia.
The motor converts electrical energy to mechanical energy.
The constant excitation voltage produces constant excitation current in the stator
(standing part of the motor) coil. The excitation current creates a constant flux  (it is
supposed that the flux is proportional to the excitation current). Switching on the
armature voltage Ua the armature current ia is developing in the armature circuit creating
the armature flux. As an interaction between the excitation flux and the armature flux a
driving moment is created which - working against the disturbance torque - rotates the
rotor. Then - according to the principle of Lenz - a voltage Ui is induced in the rotor coil.
Tasks:
a./ Give the voltage equation of the armature circuit. Determine the mechanical equation
of the motor.
b./ The state variables of the system are the angular velocity  and the armature current ia.
The output signal is the angular velocity . Give the state equation of the system.
c./ Determine the transfer function of the motor between the angular velocity  as output
signal and the armature voltage Ua as input signal. Give the transfer function between
the angular velocity  as output signal and the disturbance torque mt as input signal.
Solution:
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a./ Write the Kirchhoff equation for the armature circuit. The armature voltage is the sum
of the ohmic, the inductive and the induced voltage. The induced voltage is proportional
to the product of the constant excitation flux and the angular velocity.
The difference of the driving moment (torque) of the motor and the disturbance torque
produces the accelerating moment, which can be expressed with the product of the inertia
torque and the angular acceleration. The driving moment of the motor is proportional to
the product of the excitation flux and the armature current.
The equations describing the behaviour of the motor:
U a  ia Ra  La
dia
Ui
dt
U i  k1
m  mt  
d
dt
m  k 2 ia
Give the Laplace transform of the equations above:
U a ( s)  ia ( s)( Ra  sLa )  k1 ( s)
k 2 ia ( s)  mt ( s)  s ( s)
On the basis of the equations above - following the causal relations - the following blockdiagram can be derived. The difference of the armature voltage and the induced voltage
creates the armature current. The interaction between the armature current and the
excitation flux produces the driving moment of the motor. The difference between the
driving moment of the motor and the disturbance torque gives the accelerating moment,
which determine the angular velocity.
Ua
mt
1
Ra  sLa
ia
k 2
k1
b./ The general form of the state equation:
3
m
1
s

  AX  BU
X
Y  CX  DU
where X is the vector of the state variables, U is the vector of the input signals and Y
denotes the vector of the output signals.
Here
U 
 x  i 
X   1    a  ; U   a  and Y = y1= .
 x2   
 mt 
The state equation is obtained by rearranging the differential equation.
dia
R
1
1

U a  a ia 
k1
dt
La
La
La
d 1
1
 k 2 i a  m t
dt 

In matrix-vector form:
A 
B 


 dia 
1

 dt   Ra  k1 
0 



i
U a 
 x1   


a
L
L
La
    a
 
 x       k a
1 m

 2   d   2
0    0   t 

 

 dt   
C
D

ia   U a 
Y    0 1   0 0 
 
 mt 
c./ Supposing zero initial conditions and applying the Laplace transform to the state
equation, the following relationship is obtained between the output and input vectors:
Y(s)  [C (sI  A) 1 B  D] U( s)  W (s)U( s)
W(s) is the transfer matrix, its elements are the transfer functions, giving the relationship
between the particular outputs and inputs.
In our case
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Ra

s  L
a
W ( s )  0 1
k

2

 
k1 
La 

s 

-1
1
L
 a
0


0 

1

 
 k 2

 La
R 
1
( s  a )

La 
R
k k 2
s2  s a  1 2
La
La

The components of the transfer matrix are:
k 2
1
La
Am
 ( s)
k1


 2
2
La
Ra
R kk
U a ( s)
s2
s
 1 s TmTe  sTm  1
s2  s a  1 2
2
2
k1k 2
k1k 2
La
La
1
is the gain factor of the motor. The steady value of the angular
k1
velocity is obtained multiplying the steady value of the armature voltage by the gain.
Ra
Tm 
is the electromechanical time constant. Its value depends both on
k1k 2 2
L
electrical and mechanical parameters. Te  a is the electrical time constant.
Ra
where Am 
The transfer function between the angular velocity and the disturbance torque:
Ra
L
R
1

(1  s a )
(s  a )
2
Ra

La
A (1  sTe )
k1 k 2
 (s)


 2 t
2
La
Ra
mt ( s )
R
kk
s TmTe  sTm  1
s2
s
1
s2  s a  1 2
2
2
k1 k 2
k1 k 2
La
La

Here gain factor At gives the effect of the disturbance torque to the steady value of the
angular velocity. The negative sign indicates that the increase of the torque will cause
a decrease in the steady value of the angular velocity.
Similar result is obtained applying the superposition theorem to our linear system.
Starting from the Laplace transform of the original differential equations first
determine the ratio of the Laplace transform of the angular velocity and that of the
armature voltage supposing zero disturbance torque. Then calculate the relationship
between the angular velocity and the disturbance torque supposing zero armature
voltage.
We can also use the block-diagram above showing the relationships between the
individual signals in the motor. Calculating the resulting transfer functions in a
negative feedback circuit we get the previous results.
Considering the previous results the model of the motor can be given also according
to the block-diagram below.
5
mt
At (1  sTe )
1  sTm  s 2TeTm
Ua

Am
1  sTm  s 2TeTm
Problem #2
Model of a DC motor
Let us consider the case when the excitation voltage is also a changing variable. The
excitation flux is proportional to the excitation current ie. The resistance of the excitation
coil is Re, its inductance is Le.
The inputs of the motor are: the armature voltage Ua, the excitation voltage Ue and the
disturbance torque mt (disturbance). Its outputs are the angular velocity  and the
armature current ia.
Tasks:
a./ Give the voltage equation for the armature circuit and the mechanical equation of
motion.
b./ The state variables of the system are the angular  and the armature current ia. The
output signal is . Determine the state equation of the system.
c./ The working points of the individual variables are: U a 0 , ia 0 , ie 0 , mt 0 , 0 .
In the small vicinity of the working points give the variables as the sum of their
working points and the small variation around it.
U a  U a 0  U a
ia  ia 0  ia
ie  ie 0  ie
mt  mt 0  mt
   0  
Determine the linearised state equation of the system.
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d./ Draw the block-diagram of the linearised system.
On the basis of the block-diagram give the transfer function of the system between the
change of the angular velocity  , as output signal and the change of the armature
voltage U a as input signal. Determine the transfer function between the change of
the angular velocity  , as output signal and the change of the excitation current
ie as input signal. Give the transfer function between the change of the armature
current ia as output signal and the change of the disturbance torque mt as input
signal.
Solution:
a./ The behaviour of the motor is described by the following equations:
di
U a  i a R a  La a  U i
dt
U i  k1
  k 2 ie
d
dt
m  k 3 ia  k 2 k 3ie ia
m  mt  
The system is nonlinear. The induced voltage and the torque of the motor can be given as
a product of two changing variables.
b./ Let us express the derivatives of the state variables ia (armature current) and 
(angular velocity) from the equations above.
The state space representation of the system:
dia
R
kk
1
  a ia  1 2 ie  U a  f1 (ia , ie ,  ,U a )
dt
La
La
La
d 1
1
 k 2 k3ieia  mt  f 2 (ia , ie , mt )
dt 

c./ In the vicinity of its working point a multivariable function can be approximated by
the sum of its working point and the small variations around it.
f
f  f 0  f  f 0  
xi
i xi x1 0 ,
x2 0 .
Applying this approximation to the nonlinear expressions in our state equation:
ie  ie 0 0  (ie )  ie 0 0  ie 0    0 ie
ieia  ie 0ia 0  ie 0 ia  ia 0 ie
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Let us substitute the variables in state equation given in point b./ by their working points
plus the small variation around it.
d (ia 0  ia )
R
1
kk
kk
kk
  a (ia 0  ia )  (U a 0  U a )  1 2 ie 0 0  1 2 ie 0   1 2  0 ie
dt
La
La
La
La
La
d ( 0   ) k 2 k3
kk
kk
1
1

ie 0ia 0  2 3 ie 0 ia  2 3 ia 0 ie  mt 0  mt
dt





From the equations above the following relationships are obtained between the working
points:
dia 0
R
1
kk
 0   a ia 0  U a 0  1 2 ie 0 0
dt
La
La
La
d 0
kk
1
 0  2 3 ie 0ia 0  mt 0
dt


It is seen that the working points are not independent of each other. The values of
working points of the disturbance torque and of the excitation current determine
nominal value of the armature current. The nominal values of the armature current,
armature voltage and that of the excitation current provide the nominal value of
angular velocity.
For small variations around the working points the following equations are described:
dia
R
kk
1
kk
  a ia  1 2 ie 0   U a  1 2  0 ie
dt
La
La
La
La
kk
d k 2 k3
1

ie 0 ia  2 3 ia 0 ie  mt
dt



Let us give the linearised state equations in matrix-vector form:
A 
B 



R
k
k
k
k

1

 dia    a
 1 2 ie 0 
 2 3 0
0  U a 

 dt   L

i


a


La
L
La
    a
  ie 
 d    k k a
k
k
1


2 3
   0

  2 3i
0
ia 0    mt 


 dt    e 0

 
C
D  U a 

 ia  
  0 1   0 0 0 ie 
 
 mt 
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the
the
the
the
d./ On the basis of the Laplace transform of the equations above the block-diagram of the
linearised system is given in the Fig. below.
In a system with a unity negative feedback in the resulting transfer function between a
given output and input signal the numerator contains the transfer function of the forward
path between the input and the output signal. The denominator consists of the transfer
function of the open loop increased by 1.
ie
k1k 2 0
k 2 k 3ia 0
mt
ia
U a
1
Ra  sLa
k 2 k 3 ie 0
1
s
k1 k 2 ie 0
The resulting transfer functions are as follows:

U a

ie
ia
mt
ie  0
mt  0
U a  0
mt  0
U a  0
ie  0
k 2 k 3 ie 0
1
s( Ra  sLa )
k1 k 2 i e 0


2
2
Ra
La
k1 k 2 k 3 i e 0
1 s
 s2
1
2
2
k1 k 2 k 3 i e 0
k1 k 22 k 3 ie20
s( Ra  sLa )
ia 0

k1 k 22 k 3 0 ie 0
1
( Ra  sLa )  0
k 2 k 3ia 0

2
ie 0
s s( Ra  sLa )
k1 k 2 i e 0


2
2
Ra
La
k1 k 2 k 3 i e 0
1 s
 s2
1
2
2
k1 k 2 k 3 i e 0
k1 k 22 k 3ie20
s( Ra  sLa )
k1 k 2 i e 0
1
s( Ra  sLa )
k 2 k 3 ie 0


2
2
Ra
La
k1 k 2 k 3 i e 0
1 s
 s2
1
2
2
k1 k 2 k 3 i e 0
k1 k 22 k 3 ie20
s( Ra  sLa )
9

The resulting transfer functions are proportional elements with two time lags. It is seen
that the denominator of the individual transfer functions is the same. The values of the
parameters (gains and time constants) depend also on the values of the working points.
Increasing of the excitation current - depending on the values of the working points - may
increase or decrease the static value of the angular velocity.
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