homework 5 solutions
DESCRIPTION
control systemsTRANSCRIPT
ME 50400 and ECE 59500
AUTOMOTIVE CONTROL
Assignment 4
Question 1
The linearized vehicle model is –
10
1
22
2
y
J
lC
vm
C
vJ
lClC
J
lClC
vm
lClC
vm
CC
W
Z
FF
COGCOG
F
COGZ
FFRR
Z
FFRR
COGCOG
FFRR
COGCOG
RF
Now applying the given values,
10
69.8
79.0
52.368.24
87.066.2
y
W
From the model –
69.8
79.0
52.368.24
87.066.2
B
A
As the output of the model is Yaw rate, ; 10C
and assuming rational and proper transfer function, 0D
Now the Transfer function,
84.3018.6
61.4269.8)(
2
ss
ssG
The poles of the transfer function are in the left side of the S-plane. So the system is
Stable.
Root Locus Plot:
SIMULINK Model:
Observation and Stability limit:
As shown on the root locus, the gain values can vary from zero to infinity without
passing the imaginary axis thus the system is stable. Besides since the poles of the open
loop are not close the imaginary axis the system should not be oscillatory.
-20 -15 -10 -5 0 5-6
-4
-2
0
2
4
6
0.996
0.350.60.760.870.9250.965
0.984
0.996
2.557.51012.51517.520
0.350.60.760.870.9250.965
0.984
Root Locus
Real Axis (seconds-1)
Imagin
ary
Axis
(seconds-1
)
Question 2
The transfer function is,
)()(
)(1
)(
)()(
211222112211
2
11
aaaasaas
asJ
sM
sRsG Z
Z
where,
52.3
68.24
8743.0
66.2
22
21
12
11
a
a
a
a
[Reference: Problem - 1]
So, )8.3018.6(3545
66.2
)(
)()(
2
ss
s
sM
sRsG
Z
Now, PI controller, sKsKsG IPc /)()(
So, Transfer function, )()(1
)()()(
sGsG
sGsGsT
c
c
)66.2)(()8.3018.6(3545
3545/)66.2)((
)(*
)(2
sKsKsss
sKsK
NUMKsKDENs
NUMKsK
IP
IP
IP
IP
So the Characteristics equation is-
066.2)66.28.30()18.6(3 2 II KsKKpskps ---------
------(i)
Now as, )/(10
7.0
sradn
Desired Characteristics equation, 0100142 ss
We need an additional pole (s+a) hence,
0100)10014()14()( 23 asasassD --------------- (ii)
From (i) and (ii) equating the coefficients —Gains are,
248648
34528
I
P
K
K and a=1.92
O.L step response
C.L step response:
O.L ramp response:
C.L ramp response:
Observation and Stability limit:
Since the additional pole that we found from equating the desired characteristic
equitation and the C.L.T.F is also on the L.H.P (s= -1.92) thus the system is stable.
Question 3
Vehicle parameter values are given.
Desired slip ratio 0.15
The peak friction coefficient is 0.3 at slip 0.15. 60% static weight is distributed on front
axle and 40% on rear axle.
Neglect the dynamic weight transfer due to deceleration and steering.
We are given, )/(5
9.0
sradn
,
Now, to design PD Anti Lock Brake (ABS) control system –
Close loop characteristics equation is given by –
0))((1
0)()(1
as
bsKK
sGsPD
DP
0)( bsKKas DP --------------------- (i)
215.0
3.0)(
i
iisi
k
k ,
,
)(
i
sizisi
i
I
Rb
gFI
Ra
Now compute normal force at contact patch,
Normal force at contact patch of frontal wheel, gFzfi 17402
6.0 kg-m/s
2
= 5120.82 kg-m/s2
Normal force at contact patch of rear wheel, gFzri 17402
4.0 kg-m/s
2
= 3413.88 kg-m/s2
PI controller for Front Wheel
Transfer function, 22.1274
35.0)(
sas
bsG
Where, a = 1274.22 & b = 0.35
Since the C.L characteristic equation order is just one, we have to use the desired
equation from the same order we know the for the second order system
, s+4.5= 0 (ii)
When we equate (i) and (ii) we get one equation and two unknowns thus we will have
infinite set of . One of them would be
2.0
1.3627
D
P
K
K
The characteristic equation we know that the only pole of the system is -4.5 thus the
system is stable. But since the gains are found negative, the desired performance is not
achievable. We need to either change the desired pole or tune them to find reasonable
gains.
2.0
106
D
P
K
K
O.L step response:
Response
C.L step response:
PD controller for Rear Wheel
Transfer function, 02.856
35.0)(
sas
bsG
Where, a = 856.02 & b = 0.35
Now comparing the coefficients of equations (i) & (ii), we get similar gain values for the
rear wheel controller as well and again we need to tune the gains.
Observations:
Since the plant (for both front and rear wheel ) has a pole very far in the left half plane,
both O.L and C.L response have super fat response while O.L has huge steady state error,
this error is removed by very large proportional gain.
Question 4
Vehicle parameter values are given.
Desired acceleration slip at 0.2
Response
The road surface is covered with packed snow. The peak friction coefficient is 0.3 at slip
0.2. 60% static weight is distributed on front axle and 40% on rear axle.
Neglect the dynamic weight transfer due to deceleration & steering and no engine
intervention.
Assuming, Rising time, 1Tr = 70 ms and Max Overshoot < 10 %.
So we get, )/(2
8.0
sradn
Desired Characteristics equation, 042.3)( 2 sssD ---------------- (i)
Now,
5.12.0
3.0)(
th
thisi
k
k
Now PID controller was preferred for the Brake based Traction Control System. [you can
use any other controller]
Close loop characteristics equation is given by –
0))((1
0)()(1
as
b
s
KK
sGsPID
IP
0)(2^ IP bKsbKas --------------------- (ii)
where,
i
sizisi
i
I
Rb
gFI
Ra
)(
Now compute normal force at contact patch,
Normal force at contact patch of frontal wheel, gFzfi 17402
6.0 kg-m/s
2
= 5120.82 kg-m/s2
Normal force at contact patch of rear wheel, gFzri 17402
4.0 kg-m/s
2
= 3413.88 kg-m/s2
PI controller for Front Wheel
Transfer function, 67.955
35.0)(
sas
bsG
Where, a = 955.67 & b = 0.35
Now comparing the coefficients of equations (i) & (ii) –
43.11
4.2721
I
P
K
K
Since the Kp is negative, the desired performance is not reachable and tuning the gains
which yields reasonable performance are:
19650
1220
I
P
K
K
O.L step response:
C.L step response:
PID controller for Rear Wheel
Transfer function, 02.642
35.0)(
sas
bsG
Where, a = 642.02 & b = 0.35
Now comparing the coefficients of equations (i) & (ii) –43.11
1825
I
P
K
K
Again since Kp is negative , the desired performance of the system is not reachable so we
need to tuned the gains to just find a reasonable performance close to what we firstly
wished to.
The tuned gains which give us similar step response for this section as well.
16450
960
I
P
K
K
O.L step response:
C.L step response: