GOVERNMENT COLLEGE OF ENGINEERING, AURANGABAD��AN AUTONOMOUS INSTITUTE OF GOVERNMENT OF MAHARASHTRA��ELECTRICAL ENGINEERING DEPARTMNET ��BHANUDAS SABHAHGRAH RAILWAY STATION RD, RACHANAKAR COLONY, NEW USMANPURA, AURANGABAD, MAHARASHTRA 431005
ONLINE PRESENTATION ON
“Design Of Automotive Suspension Control Of A Vehicle”
By
Lokesh Kishor Chaudhari
(BE20S03F001)
Rahul Hanumant Gade
(BE20S03F003)
THIRD YEAR ELECTRICAL ENGINEERING
ACADEMIC YEAR (2021-2022)
Under the guidence of
Prof. Mrs. S. S. Kulkarni
OUTLINE
1. Physical setup
2. System parameters
3. Equations of motion
4. Transfer function models
5. Entering equations in MATLAB
1. System model
2. System parameters
3. Design requirements
4. Open-loop step response
1. Plotting the root locus
2. adding a notch filter
3. Finding the gain from the root locus
4. plotting the closed-loop response
INTRODUCTION
SUSPENSION: SYSTEM MODELING�
1. Physical setup:
2. System parameters:
(M1) 1/4 bus body mass 2500 kg
(M2) suspension mass 320 kg
(K1) spring constant of suspension system 80,000 N/m
(K2) spring constant of wheel and tire 500,000 N/m
(b1) damping constant of suspension system 350 N.s/m
(b2) damping constant of wheel and tire 15,020 N.s/m
(U) control force
3. Equations of motion:
From the picture above and Newton's law, we can obtain the dynamic equations as the following:
TRANSFER FUNCTION MODEL
Solving above Newton’s equations for getting transfer function of vehicle, we get
Find the inverse of matrix A and then multiply with inputs U(s)and W(s) on the righthand side as follows:
When we want to consider the control input U(s) only, we set W(s) = 0. Thus we get the transfer function G1(s) as in the following:
When we want to consider the disturbance input W(s) only, we set U(s) = 0. Thus we get the transfer function G2(s) as in the following:
The final transfer function is:
Where:
5. Entering equations in MATLAB:
We can generate the above transfer function models in MATLAB by entering the following commands in the MATLAB command window.
M1 = 2500;
M2 = 320;
K1 = 80000;
K2 = 500000;
b1 = 350;
b2 = 15020;
s = tf('s');
G1 = ((M1+M2)*s^2+b2*s+K2)/((M1*s^2+b1*s+K1)*(M2*s^2+(b1+b2)*s+(K1+K2))-(b1*s+K1)*(b1*s+K1));
G2 = (-M1*b2*s^3-M1*K2*s^2)/((M1*s^2+b1*s+K1)*(M2*s^2+(b1+b2)*s+(K1+K2))-(b1*s+K1)*(b1*s+K1));
SUSPENSION: SYSTEM ANALYSIS�
1. System model
2. System parameters
3. Design requirements
4. Open-loop step response:
M1 = 2500;
M2 = 320;
K1 = 80000;
K2 = 500000;
b1 = 350;
b2 = 15020;
s = tf('s');
G1 = ((M1+M2)*s^2+b2*s+K2)/((M1*s^2+b1*s+K1)*(M2*s^2+(b1+b2)*s+(K1+K2))-(b1*s+K1)*(b1*s+K1));
step(G1)
STEP RESPONSE FOR OPEN LOOP SYSTEM
IMPULSE RESPONSE FOR OPEN LOOP SYSTEM
STEP RESPONSE FOR WHEEL DEFORMATION W(S)
By considering wheel deformation of 0.1 magnitude
G2 = (-M1*b2*s^3-M1*K2*s^2)/((M1*s^2+b1*s+K1)*(M2*s^2+(b1+b2)*s+(K1+K2))-(b1*s+K1)*(b1*s+K1));
step(0.1*G2)
SUSPENSION: ROOT LOCUS CONTROLLER DESIGN
From the main problem, the dynamic equations in transfer function form are the following:
Eq. 1:
Eq. 2:
Where,
And the system schematic is the following where F(s)G1(s) = G2(s).
m1 = 2500;
m2 = 320;
k1 = 80000;
k2 = 500000;
b1 = 350;
b2 = 15020;
nump=[(m1+m2) b2 k2];
denp=[(m1*m2) (m1*(b1+b2))+(m2*b1) (m1*(k1+k2))+(m2*k1)+(b1*b2) (b1*k2)+(b2*k1) k1*k2];
G1=tf(nump,denp);
num1=[-(m1*b2) -(m1*k2) 0 0];
den1=[(m1*m2) (m1*(b1+b2))+(m2*b1) (m1*(k1+k2))+(m2*k1)+(b1*b2) (b1*k2)+(b2*k1) k1*k2];
G2=tf(num1,den1);
numf=num1;
denf=nump;
F=tf(numf,denf);
R = roots(denp)
R =
-23.9758 +35.1869i
-23.9758 -35.1869i
-0.1098 + 5.2504i
-0.1098 - 5.2504i
1. Plotting the root locus:
rlocus(G1)
z=-log(0.05)/sqrt(pi^2+(log(0.05)^2))
sgrid(z,0)
z =
0.6901
2. Adding a notch filter:
z1=3+3.5i;
z2=3-3.5i;
p1=30; p2=60;
numc=conv([1 z1],[1 z2]);
denc=conv([1 p1],[1 p2]);
C=tf(numc,denc);
rlocus(C*G1)
Changing the axis for pole zero plot:
axis([-40 10 -30 30])
z=-log(0.05)/sqrt(pi^2+(log(0.05)^2))
sgrid(z,0)
z =
0.6901
PLOTTING THE CLOSED LOOP RESPONSE
IMPULSE RESPONSE FOR CLOSED LOOP SYSTEM
ADVANTAGES
CONCLUSION
REFERENCE
THANK YOU!