EE4323 Industrial Control Systems
Homework Assignment 1
Dynamics model of a DC motor with gear train
Paulo R. Loma Marconi
May 4, 2017
The objective is to model the dynamics of a DC servo motor with gear train, Fig. 1, and to deduce
two equilibrium points.
Figure 1: DC servo motor with gear train.
1 Free-body diagram analysis
The system can be decomposed in two sections: a rotational mechanical, and an electronmechanical.
The rotational mechanical can be derived as follows,
Figure 2: Rotational mechanical free-body diagram.
where θ is the angular displacement, ω is the angular speed, B is the rotational viscous-damping
coefficient, K is the stiffness coefficient, J is the moment of inertia, f
c
is the contact force between two
gears, and r is the gear radius.
1
The electromechanical section (DC motor) is
Figure 3: Electromechanical free-body diagram.
where R
F
is the field resistance, L
F
is the field inductance, E
F
is the applied constant field voltage,
and i
F
is the input field current. R
A
is the stationary resistance, L
A
is the stationary inductance, and
e
m
is the induced voltage, i
A
is the input stationary current, and e
i
(t) is the applied armature voltage,
and τ
e
is the electromechanical driving torque exerted on the rotor.
If the flux density B is
B =
φ(i
F
)
A
(1)
the torque on the rotor is
τ
e
= Bla i
A
τ
e
=
la
A
φ(i
F
)i
A
(2)
where φ(i
F
) is the flux induced by i
F
, A is the cross-sectional area of the flux path in the air gap
between the rotor and stator, l is the total length of the armature conductors within the magnetic
field, and a is the radius of the armature.
Also, the voltage induced in the armature e
m
can be written as
e
m
=
la
A
φ(i
F
)ω (3)
where both, τ
e
and e
m
, depend on the geometry of the DC motor.
2 Dynamic system
We begin applying D’Alembert’s law (restatement of Newton’s law) to the rotational mechanical
section.
X
τ
all
= 0
J
1
˙ω
1
+ B
1
ω
1
+ r
1
f
c
= τ
e
(t) (4)
J
2
˙ω
2
+ B
2
ω
2
+ K
2
θ − r
2
f
c
= τ
L
(t) (5)
where τ
all
are the torques acting on a body, Kθ is the stiffness torque, Bω is the viscous-frictional
torque, J ˙ω is the inertial torque, τ
e
(t) is the driving torque, τ
L
(t) is the load torque, and rf
c
is the
contact torque.
2
Due to the relation between gears,
θ
1
= Nθ
2
ω
1
= Nω
2
˙ω
1
= N ˙ω
2
N =
r
2
r
1
where N is the gear radius relation. We solve (4) and (5) in terms of ω
2
and θ
2
,
(J
2
+ N
2
J
1
) ˙ω
2
+ (B
2
+ N
2
B
1
)ω
1
+ K
2
θ
2
− Nτ
e
(t) − τ
L
(t) = 0
defining the relations
J
eq
= J
2
+ N
2
J
1
B
eq
= B
2
+ B
2
B
1
it becomes
J
eq
˙ω
2
+ B
eq
ω
2
+ K
2
θ
2
− Nτ
e
(t) − τ
L
(t) = 0 (6)
Now, let us derive the equations of the electromechanical section using Kirchoff’s law.
X
V
all
= 0
e
m
+ V
L
A
+ V
R
A
= e
i
(t) (7)
where V
all
are the induced voltages on the rotor and stator, V
L
A
is the stationary resistance voltage,
V
R
A
is the stationary inductance voltage.
If i
F
is defined as constant, then (2) is
τ
e
(t) =
la
A
φ(i
F
)
i
A
(t)
τ
e
(t) = αi
A
(t) (8)
where α is the internal parameters of the DC motor.
Then, simplifying and using (6) and (7) the dynamic system is,
J
eq
˙ω
2
+ B
eq
ω
2
+ K
2
θ
2
− Nτ
e
− τ
L
= 0 (9)
L
A
˙
i
A
+ R
A
i
A
+ αω
1
− e
i
= 0 (10)
3 State-space equations
Let us define the state-space equations for x =
h
θ
2
˙
θ
2
i
A
i
|
. From the dynamic system,
J
eq
¨
θ
2
+ B
eq
˙
θ
2
+ K
2
θ
2
− Nαi
A
− τ
L
= 0
L
A
˙
i
A
+ R
A
i
A
+ αω
1
− e
i
= 0
reordering,
¨
θ
2
= −
B
eq
J
eq
˙
θ
2
−
K
2
J
eq
θ
2
+
Nα
J
eq
i
A
−
1
J
eq
τ
L
˙
i
A
= −
R
A
L
A
i
A
−
Nα
L
A
˙
θ
2
+
1
L
A
e
i
3
defining the states as
x
1
= θ
2
, ˙x
1
=
˙
θ
2
= x
2
x
2
=
˙
θ
2
, ˙x
2
=
¨
θ
2
= −
B
eq
J
eq
x
2
−
K
2
J
eq
x
1
+
Nα
J
eq
x
3
−
1
J
eq
τ
L
x
3
= i
A
, ˙x
3
=
˙
i
A
= −
R
A
L
A
x
3
−
Nα
L
A
x
2
+
1
L
A
e
i
then
˙x
1
˙x
2
˙x
3
=
0 1 0
−
K
2
J
eq
−
B
eq
J
eq
Nα
J
eq
0 −
Nα
L
A
−
R
A
L
A
| {z }
A
x
1
x
2
x
3
|{z}
x
+
0 0 0
0 −
1
J
e
q
0
0 0
1
L
A
| {z }
B
0
τ
L
e
i
|{z}
u
(11)
˙x = Ax + Bu (12)
The output y = ˙ω
2
can be defined as
y =
0 1 0
| {z }
C
x
1
x
2
x
3
+
0 0 0
| {z }
D
e
i
(13)
y = C ˙x (14)
4 Equilibrium point x
0
Using ˙x = 0 in (12), the equilibrium point x
0
can be calculated as
0 = Ax
0
+ Bu (15)
x
0
= −A
−1
Bu (16)
x
1
0
x
2
0
x
3
0
= −
0 1 0
−
K
2
J
eq
−
B
eq
J
eq
Nα
J
eq
0 −
Nα
L
A
−
R
A
L
A
−1
0 0 0
0 −
1
J
e
q
0
0 0
1
L
A
0
τ
L
e
i
(17)
Solving for no external torque τ
L
= 0, constant applied armatrue voltage e
i
= E
0
, and K
2
6= 0,
0 = x
2
0
0 = −
K
2
J
eq
x
1
0
−
B
eq
J
eq
x
2
0
+
Nα
J
eq
x
3
0
0 = −
Nα
L
A
x
2
0
−
R
A
L
A
x
3
0
+
1
L
A
E
0
due to x
2
0
= 0, we have
0 = −
K
2
J
eq
x
1
0
+
Nα
J
eq
x
3
0
0 = −
R
A
L
A
x
3
0
+
1
L
A
E
0
then
x
1
0
=
Nα
K
2
R
A
E
0
x
3
0
=
1
R
A
E
0
4
therefore the equilibrium point is
x
0
=
x
1
0
x
2
0
x
3
0
=
Nα
K
2
R
A
0
1
R
A
E
0
(18)
This equilibrium point indicates that a constant angular displacement (twist) produced by
x
1
0
= θ
2
0
is sufficient to balance the constant applied armature voltage e
i
= E
0
.
On the other hand, if we solve for no external torque τ
L
= 0, constant applied armature voltage
e
i
= E
0
, and no stiffness K
2
= 0. The problem is,
x
1
0
x
2
0
x
3
0
= −
0 1 0
0 −
B
eq
J
eq
Nα
J
eq
0 −
Nα
L
A
−
R
A
L
A
−1
0 0 0
0 −
1
J
e
q
0
0 0
1
L
A
0
0
E
0
if we eliminate x
1
0
because the first column of A
−1
has zeros, the problem reduces to
x
2
0
x
3
0
= −
"
−
B
eq
J
eq
Nα
J
eq
−
Nα
L
A
−
R
A
L
A
#
−1
−
1
J
e
q
0
0
1
L
A
0
E
0
(19)
solving, we have
x
2
0
x
3
0
=
"
Nα
B
eq
R
A
+(Nα)
2
−B
eq
B
eq
R
A
+(Nα)
2
#
E
0
(20)
which indicates that a constant angular speed produced by x
2
0
=
˙
θ
2
0
is needed to balance the
constant applied armature voltage e
i
= E
0
.
References
[1] Close, Charles M. and Frederick, Dean K. and Newell, Jonathan C., Modeling and Analysis of
Dynamic Systems, 2001, ISBN 0471394424.
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