THE UNIVERSITY OF TEXAS AT
ARLINGTON
EE5329
CONTROL SYSTEMS DESIGN PROJECT
LAB 1 REPORT
Identification of Static Motor Module Parameters
Compiled by:
Ognjen Kuljača
09/21/2002
2
CONTENTS
1. INTRODUCTION .....................................................................3
2. GENERAL GAIN G ..................................................................4
3. TIME CONSTANT  .................................................................7
V
4. TRANSFER FUNCTION tach ................................................8
Vin
5. CONCLUSION ..........................................................................9
3
1. INTRODUCTION
The goal of the lab was to find the simplified transfer function of the DC motor
module. The simplified transfer function does not deal with the nonlinearities present in
the DC motor system. The simplified transfer function has the following form:
H(s) 
Vtach
G

Vin
s  1
( 1-1 )
where: G – general gain
 - time constant in [s]
Vin – input voltage [V]
Vtach – tachometer output voltage [V]
For the measurement and data manipulation Real Time Windows Target for
Matlab and NI PCI1200 data acquisition cards were used.. The system was regarded to be
linear. The simplified motor module block for finding transfer function H(s) from ( 1-1 )
is shown in Error! Reference source not found..
B0
V in
A
Tm
+
Va
1/R a
Kt
_
_
w
1 /J
1/s
-K t ach
+
Kb
Figure 1: Simplified motor module block diagram
- Kb
V tach
4
2. GENERAL GAIN G
Vtach
. Static
Vin
characteristic is measured 1 time and than gain G is taken as mean value of the measured
slopes on the left and on the right of deadband. Deadband is excluded. The results are
shown in Table 1.
General gain G is measured as the slope of the static characteristic
Vi
n
Vt
G
ach
-5
-4
-3
-2
-1
4.8
3.8
-
0
-
0
0.4
-
0
0.2
0
0
0.
0
2
0.
0
4
0.
0
6
0.
0
8
5
0.
70
0
-
4
0.
85
0.7
0.6
3
0.
90
1.7
-
2
0.
95
2.7
0.8
1
0.
96
0.
00
0.
00
0.
00
0.
00
0.
00
0.
00
0.
00
0.
00
0.
00
0.
7
0.
70
1.
7
0.
85
2.
8
0.
93
3.
8
0.
95
4.
8
0.
96
5
Table 1: General gain G – measured values
From Figure 2 and Table 1 it can be seen that deadzone value is between –1 and 1
V input. There is no enough data to determine possible saturation limit.
6
Data from Table 1 is shown in Figure 2.
Vin vs. V tach
5
4
3
2
V
tach
1
0
-1
-2
-3
-4
-5
-5
-4
-3
-2
-1
0
Vin
1
2
3
4
5
Figure 2: Static characteristic Vin vs. Vtach
The values for the deadzone are excluded when gain is calculated. Data used for
the gain calculation with overall gain computed as mean value of particular
measurements is shown in:
Vi
n
Vt
G
ach
-5
-4
-3
-2
-1
4.8
0.
96
3.8
0.
95
2.7
0.
90
1.7
0.
85
0.7
0.
70
0.
1
7
0.
70
1.
2
7
0.
85
2.
3
8
0.
93
3.
4
8
0.
95
4.
5
8
m
ean
0.
96
0.
88
Figure 3: Data used for the gain calculation
7
Gain is calculated as K = 0.88. however, the values for 1 and –1 look a little bit
off, so without them gain is 0.92.
3. TIME CONSTANT 
Time constant is measured from step response of the motor module.
Measurements are taken for 3 V, 4 V and 5 V inputs. Then, a time point at which
response reached 63.2 % of its steady state value is read from the response plots.
Measurement No.
1.
2.
3.
Step input
[V]
5
4
3
Steady state
value
4.8
3.8
2.8
63.2% of steady
state value
3.0
2.4
1.8
Mean value:
Time constants
[s]
0.33
0.30
0.30
0.31
Table 2: Time constant  – measured values and mean value
From Table 2, time constant is calculated as  = 0.31 s.
Plots for the step response for inputs 3 V, 4 V and 5 V are given in Figure 4,
Figure 5 and Figure 6:
Step response for input 3 V
3
2.5
V
tach
2
1.5
1
0.5
0
0
1
2
3
4
5
6
7
t [s]
Figure 4: Step response for input 3 V
8
9
8
Step response for input 4 V
4
3.5
3
V
tach
2.5
2
1.5
1
0.5
0
0
1
2
3
4
5
6
7
8
9
t [s]
Figure 5: Step response for input 4 V
Step response for input 5 V
5
4.5
4
3.5
V
tach
3
2.5
2
1.5
1
0.5
0
0
1
2
3
4
5
6
7
8
9
t [s]
Figure 6: Step response for input 5 V
4. TRANSFER FUNCTION
Vtach
Vin
Transfer function is derived from Figure 1. The velocity m is marked as w in
Figure 1.
The transfer function G1 can be found as:
G1 
m
1

Tm Js  B j
The transfer function G2 can be found as:
( 4-1 )
9
1
K t G1
Ra
G2 
1
1 Kb
K t G1
Ra
m
is then:
Vin
The transfer function G =
G   AG1 
The transfer function G =
( 4-2 )
AK t
R a Js  R a B j  K b K t
( 4-3 )
Vtach
is then:
Vin
G  AG1K b K tach 
AK t K b K tach
R a Js  R a B j  K b K t
( 4-4 )
With the given values in lab handout:
G
1.0341
0.2629s  1
( 4-5 )
0.88
0.31s + 1
( 4-6 )
From measured values:
G=
5. ANALYSIS OF TRANSFER FUNCTIONS
Tables and plots required are given in previous sections. It can be seen that there
is a deadzone present in the system. That might yield problems if precise control at small
speed is required since in that case a small reference voltage or control signal will not be
able to go through the deadzone.
From equations ( 4-5 ) and ( 4-6 ) it easy to see that poles are –0.2629 and –0.31
for theoretical and measured transfer functions respectively.
There are no zeros (theoretically there is one zero in infinity for each transfer
function).
6. CONCLUSION
10
The measured and calculated values are relatively well matched. The differences
are due to neglecting the nonlinearities in the system, especially the saturation. That can
be seen from difference in gain in particular.
Since poles are negative it is clear that system is stable. However, there is steady
state error present. For the step input R, it is easy to see that steady state error will be:
Ess = Rss-Yss=R(1-K).