AV243 lab 1: Compensator design for a typical
electromechanical engine gimbal control
(EGC) system and Linear system performance
assessment using MATLAB/SIMULINK
Karun Mathews Manoj (SC19B120)
March 12, 2021
Objective:
of (proportional
+ rate
feedback),
(Proportional
+pensators
rateand
feedback
+
Integral)
andDesign
(Proportional
+ Integral
+
Derivative)
type
ofand
comfor
typical
engine
gimbal
systemmodel
of launch
vehicles
linear
system
performance
assessment
using control
SIMULINK
SIMULINK
LTIa
Viewer.electromechanical
Theory:
The
block
diagram
representation
of
a
typical
electromechanical
engine
gimbal control system using proportional plus rate feedback controller is shown in
Fig.1.
Figure
1: Block
diagram
representation
proportional
+ rate
feedback
controller of a typical electromechanical EGC system with
The parameter values are given below:
1
J = Jm
B=B
+ K .K /R
m
t
b
a
+(
Nm
Nl
+(
Nm
)2B
Nl
l
(1)
)2 B K = K /R
l
t
a
(2)
(motor torque constant) Kt = 0.181Nm/A
(3)
(motor back emf constant) Kb = 0.181 V/(rad/sec)
(4)
(MI of motor rotating assembly) Jm = 1.1694 10−4 kg.m2
∗
(5)
(MI of engine gimbal) Jl = 12.753 kg.m2
Nm θl 1 (6)
(Ball screw gear ratio)Nl = θm= 398
(Viscous damping coefficient of motor shaft) Bm = 2.943 10−4Nm/(rad/sec)
(7)
∗
(8)
(Viscous damping coefficient of the engine gimbal) Bl = 58.86Nm/(rad/sec)
(9)
(Armature resistance) Ra = 8.6 ohms
(10)
(Position sensor scale factor) Kp = 0.36V/rad
(11)
(Tacho generator scale factor) Ktg = 0.1 V/(rad/sec)
(12)
(13)
Controller gain design equations In order to get a closed loop system with transfer
equation:
w2n
Gd(s) =
;
(14)
n
s2 + 2ζ.wn.s + w2
2
the design equations for K1 and K2 are given by:
wn2.J
K1 =
=
2ζ.wn.J−K.K
B K2
K
tgp.K
(15)
(16)
(17)
Specifications for closed system
(-3 dB bandwidth) wb = 5 ∗ 2 ∗ πrad/sec; (Damping factor) ζ = 0.6;
(18)
Use the following formula for finding wn, thenaturalfrequencyoftheclosedloopsystem :
√
.
wb = wn (1 − 2ζ 2 ) +
4ζ 4 − 4ζ 2 + 2
Procedure for controller design and simulation:
(19)
1. Find
outwhich
the (k1)
values
natural
frequency
(wn), sencontroller
andof
rateundamped
feedback
controller
gain (k2)
using
a proportional
MATprogram
usespecifications.
the
parameter
values
of the
dynamics,
sor LAB
scale
factors
andgain
closed
Write
down
theplant
values
below:
2.
3.
4.
wn = 27.36 rad/sec;
(20)
k1 = 19.5077;
(21)
k2 = 0.9538;
(22)
Draw
the Simulink
fortime
the LTI
closed
system
and
find out M-peak,
the following
performances
SIMULINK
-3fordBstep
bandwidth,
peak
overshoot,
riseusing
time,model
settling
(the Viewer:
lastloop
3 are
response)
Add
an integral
controller
Ki/s, and
in parallel
withmeasurement
the pro- portional
controller
gain
‘K1’,
with(2).
awith
gainG(s)
Ki = =K1/10
repeat the
of the
parameters
given
in step
Add
a
derivative
controller
with
G(s)
=
Kd.s/(1+s/300),
where
Kd
=
K2*KTG/Kp
in parallel
with the
PI controller,
and remove the rate feed- back
controller.
Repeat
the exercise
in step
(2).
5. Make a performance comparison table as given below:
3
Performance parameter
P + Rate comparison
feedback table P + I + rate feedback
Table 1: Performance
-3 dB Bandwidth (Hz)
5.9046 (@ -3.03 dB)
5.9046 (@ -3.02 dB)
M peak (dB)
2.38
2.43
Peak Overshoot (%)
23.5%
24%
Rise time (ms) (%)
54.6
54.5
6. Settling
Give your
performance comparison.
timecomments
(ms) (%) on the 307
304
The
PID
controller
outperforms
the
other
two
controllers
in
the
two
as-the
pects
of
peak
overshoot
and settling
significantly
other
two.
This implies
that thetime
PID- having
controlled
system lower
wouldvalues
havethan
much
better
stability.
However,
the PID
controller
provides
it
will provide
a weaker
output
signal.much lesser gain than the other two, hence
The P + Rate feedback and P+I+Rate feedback controllers have very similar
performances.
the
system
haswould
aθslightly
smaller
as compared
to the
other sive
two. In
my PID
opinion,
make
the bandwidth
PID control
system less
responto
varying
inputsthis
(like
c = A.sin(wt)).
4
P+I+D
5.5704 (@ -3.02 dB)
0.592
10.7%
58.3
202