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Part A (DC motor basic architecture)
A common actuator in control system is the DC motor. It directly provides rotary
motion and, couple with wheels or drums and cables, can provide transitional
motion. The electric circuit of the armature and the free body diagram of the rotor
are shown in Figure 2.
Assume the following values for the physical parameters.
Moment of inertia, Jm = 0.1 kgm2
Damping ratio of the mechanical system, Bm = 0.2275Nms/rad
Electromotive force constant, K=Kb=Kt = 0.3
Electric resistance, Ra = 4 Ω
Input (Va): source voltage
Output (θ0): position of shaft
The rotor and shaft are assumed to be rigid
Solution:
(a) Derive the transfer function that relates output and input.
From the system shown above, the transfer function can be derive which are
Voltage equation:
Va (t) = Raia (t) + eb (t) (1.1)
Motor torque equation:
Tm(t) = Ktia(t) (1.2)
We know that emf is proportional to the angular position,
eb(t) ω(t)
eb(t)
eb(t) = Kbsθo(s) (1.3)
Substitute equation 1.2 and 1.3 into equation 1.1,
Va(s) = Ra + Kbsθo(s) (1.4)
We know that
Tm(s) = (Jms2 + Bms ) θo(s) (1.5)
Substitute equation 3.5 into equation 3.4,
Va(s) = Ra + Kbsθo(s)
Va(s) =
Va(s) Kt = Ra (Jms2 + Bms ) θo(s) + Kb Ktsθo(s)
Factorize the θo(s)
Va(s) Kt = [Ra (Jms2 + Bms ) + Kb Kts] θo(s) (1.6)
Rearrange the equation 1.6 to find the transfer function
=
Substitute the value into the transfer function, where
Moment of inertia, Jm = 0.1 kgm2
Damping ratio of the mechanical system, Bm = 0.2275Nms/rad
Electromotive force constant, K=Kb=Kt = 0.3
Electric resistance, Ra = 4 Ω
=
=
(b) Draw the block diagram of Q3 part A
Kt
BmS
KbS
Figure 5
Va(s)
PART B (DC MOTOR APPLICATION)
The DC motor (in part a) will be used in three-axis articulated arm pick and place
robot for computer integrated manufacturing (CIM) system. The three axes are
simultaneously controlled, and driven by this DC motor. The movement of each axis
must respond with fast acceleration, deceleration and reasonable precision. We have
to design a robot such as this one, and that one of the can be represented by the
block diagram as shown below:
Ө0 (s) +
- Ө0 (s)
Figure 3
Where:
Potentiometer constant (Kp) = 3 v/rad
Input : desire position
Output : actual position
Amplifier (K1): variable gain controller
a) Calculate the range of K1 for the system to remain stable
b) Find the system type and calculate the steady state error for the step input.
Kp K1
Kp
DC motor in part a
c) Calculate the possible value of K1 to acquire the fastest response without overshoot,
and then find Tr and TS. Key in this K1 value in MATLAB SIMULINK or M-FILE
and compare the simulation result value with your calculation.
d) From the range obtained in Q3 Part B (a), find the possible value of K1 to obtained
10% of overshoot, thus find Tr and TS. Key in this K1 value in MATLAB SIMULINK
or M-FILE and compare the simulation result value with your calculation.
e) From the result obtained in Q3 Part B © and Q3 Part (d), conclude the relationship
between variable gain controller (K1) with Tr , TS and %OS.
SOLUTION
a) The range of K1 for the system to remain stable
Ө0 (s) Ө0 (s)
Figure 6
Since, Kp = 3 v/rad
Ө0 (s) Ө0 (s)
Figure 7
G (s) =
Kp K1
Kp
ss 24.0
3.0
Kp K1
Kp
ss 24.0
3.0
1/Kp
=
G (s) =
H (s) =
H (s) = 1
G (s)
Ө0 (s) Ө0 (s)
H (s)
Figure 8
ss
K
21
4.0
9.0
Characteristic equation
R (s) = 0.4s2 + s +0.9K1
s2 0.4 0.9K1
s1 1 0 s0 ai
ai = -1 0.4 0.9K1
1 1 0
= -1 0 – 0.9K1
= 0.9K1
s2 0.4 0.9K1
s1 1 0 s0 0.9K1
0.9K1 > 0
K1 >
K1 > 0
The system is stable if 0 < K1 < ∞
b) The system type and the steady state error for the step input.
The system type
G (s)
Ө0 (s) Ө0 (s) ss
K
21
4.0
9.0
H (s)
Figure 9
The open loop transfer function G (s) H (s) =
=
=
Since the system have one pure integration in the forward path, the system is type 1.
The steady state error for the step input.
Let is static position error constant, Kp
For a type-1 system or higher (N>1)
Therefore, the steady state error,
c) The possible value of K1 to acquire the fastest response without overshoot, then
find Tr and Ts.
G (s)
Ө0 (s) Ө0 (s)
H (s)
Figure 10
The closed loop transfer function is
Hence, the fastest response without overshoot:
ss 24.0
9.0
Transfer function
Compared the transfer function with the closed loop transfer function
The rise time Tr given by:
And substituting for ωr and α, we has
Since,
The settling time Ts for the second order prototype system is given below:
i. For ± 2% band
ii. For ± 5% band
d) The possible value of K1 to obtained 10% of overshoot, thus find Tr and TS.
Transfer function
Compared the transfer function with the closed loop transfer function
The rise time Tr given by:
Since,
The settling time Ts for the second order prototype system is given below:
iii. For ± 2% band
iv. For ± 5% band
e) From the result from the calculation above. The changes of variable gain controller, K 1
whether it is increased or decreased, it does not effect to the value of Ts in 2% band
and in 5% band. But the value of variable gain controller, K1 will affect the value of
Tr. The value calculated show that when the increasing of the value variable gain
controller, it will in decrease the value of Tr and when the value of variable gain
controller, K1 is decreased, the value of Tr is increased. For the percentage of
overshoot, %OS, the result shows that the there is a relationship between the value of
variable gain controller and percentages of overshoot, %OS. When the percentages of
overshoot, %OS is in the low percentage or 0%, the value of variable gain controller,
K1 also is in the low value compared to the value of variable gain controller, K1 in the
high value of percentages of overshoot, %OS. When the percentage of overshoot,
%OS is in the high value of percentages, the value of variable gain controller, K 1 also
in the high value compared to the value of variable gain controller, K1 in the low value
of percentages of overshoot, %OS.
From the calculation above, it is shows that the value of variable gain controller, K1 is
inversely proportional to the value of rise time, Tr. Then, the variable gain controller
does not have any relationship to the settling time, Ts. Besides, the value of variable
gain controller, K1 is proportional to the percentage of overshoot, OS%.
MATLAB SIMULINK or M-FILE and the simulation result question 3 part B ©
Figure 11: Command using MATLAB
Transfer function:
1.562
-------------------
s^2 + 2.5 s + 1.562
poles =
-1.2500
-1.2500
omegan =
1.2500
zeta =
1.0000
Tr =
-8.0000e-001 -9.9346e+004i
Ts =
3.2000
Figure 12: Step response waveform
MATLAB SIMULINK or M-FILE and the simulation result question 3 part B (d)
Figure 13: Command using MATLAB
Transfer function:
4.471
-------------------
s^2 + 2.5 s + 4.471
poles =
-1.2500 + 1.7055i
-1.2500 - 1.7055i
omegan =
2.1145
zeta =
0.5912
Tr =
1.2919
Ts = 3.2000
Figure 14: Step response waveform
PART C (REDUCING SYSTEM OVERSHOOT)
The variable gain value in Q3 Part B (d) will be use in this section. Normally by
increase gain value the overshoot will present. To overcome this problem there are
two possible solutions, one is by using magnetic brake and another one is by using
system velocity feedback. We interested to reduce overshoot by using the second
method.
(a) Design the system to reduce overshoot in Figure 3 using velocity feedback if
the given tachometer has a constant value, Kg = 10 m v/rads-1
(b) Determine the transfer function of Q3 Part C (a).
(c) Velocity feedback can reduce overshoot for a positional control system,
explain how it is work.
(d) Calculate the %OS, Tr and Ts for this system and compare the result obtain
using MATLAB SIULINK or M-FILE.
(e) Calculate the percent of %OS reduction obtained by using this method.
(f) Give your opinion on how to obtain more percent of overshoot by using this
method.
SOLUTION
(a)
Figure 15
Figure 16
Figure 17
Kp
1/s
Kg
Kp 1/s
Kg
Kp
Θi
(s)
Θi
Kp
Kp 1/sΘi
(s)
Θo
(s)
Θo
(s)
Θo
(s)
Kp
1/sK1
Kg
Kp
Figure 18
=
=
=
=
3
3Θi
(s)Θo
(s)
b)
c) Velocity feedback can reduce overshoot for a positional control system. In this task, if
we compare the previous value of overshoot (10%) with the present value (9.79%), it’s
certainly has decreasing in terms of the overshoot. As we all know, the overshoot
represent a distortion of the signal. The overshoot may result from circuit design
parameters that are intended to decrease the response time of the circuit. The velocity
feedback will be function to feed the signal desired back to the input.
d)
= 0.5947 (
µs =
=
=
= 0.0979
%µs = 9.79%
Tr =
=
=
= 1.2989s
Ts =
= 3.1812s
Matlab simulation result for part c:
Figure 19: Command using MATLAB
Figure 20: Step response waveform from M-FILE
Figure 21: Settling time waveform from M-FILE
Figure 22: Rise time waveform from M-FILE
Figure 23: Overshoot (%) and peak amplitude of part c step response
Comparison between MATLAB simulation and calculation result:
MATLAB Manual calculation
Overshoot (%) 9.96% 9.79%
Rise time (Tr) 0.87s 1.30s
Settling time (Ts) 2.8s 3.18s
e) Percent of %Over S reduction,
=
= 2.1%
f) We may observe that to ensure good accuracy it would appear that we should
make ζ as small as possible by increasing the gain Kp as large as possible or by
reducing the viscous friction. However, by making ζ small will make the system
intolerably oscillatory.
