how to derive and simulate suspension system transfer function

1. Suspension System for Heavy Vehicles
SubmittedTo:
Dr. Yaseer Arfat Durani
SubmittedBy:
Tariq Ullah
17F-MS-ENC-06
M. Fahim Zafar
17F-MS-ENC-13
Subject: Advance Engineering Mathematics
Department of ElectronicEngineering

2. University of Engineering & Technology,
Taxila

3. Suspension System for Heavy Vehicles
Abstract: -
In order to improve handling and comfort performance instead of conventional static
spring and damper system, semi active and active systems are developed. An active suspension
system has been proposed for the purpose of improving the ride comfort. The main purpose of this
work is to illustrate the application of intelligent technique to the control of continuously damping
automotive suspension system. The ride comfort is improved by means of the reduction of the
body acceleration caused by the vehicle body when road disturbances from smooth and real road
roughness. This project also describes the model and controller used in the study and discusses
vehicle response results obtained from a range of road input simulations.
Introduction: -
A suspension system is comprised of some arrangements of springs, shock absorber,
and linkages, which allows a vehicle body to be controlled. If the suspension system of the vehicle
is not robust, and the driver cannot control it, the power produced by the engine is worthless.
Further, to insure the smooth movement, a vehicle needs to be absorb all of the bumps and holes
on the surface of the road using its suspension system. Without this system, even the smallest of
bumps, which may appear negligible to the passenger, will disrupt the smoothness of the journey.
The purpose of the suspension system along with the wheels and the tires to provide greater
handling capabilities as well as increased comfort level.
Suspension System Model: -
In order to design the suspension system for heavy vehicle, one of the four-wheel model is
considered in fig.1
Fig.1: Suspension model for ¼ wheel of vehicle

4. Where:
* body mass (M1) = 2500 kg,
* suspension mass (M2) = 320 kg,
* spring constant of suspension system(k1) = 80,000 N/m,
* spring constant of wheel and tire(k2) = 500,000 N/m,
* damping constant of suspension system(b1) = 350 Ns/m.
* damping constant of wheel and tire(b2) = 15,020 Ns/m.
* control force (u) = force from the controller we are going to design
Design requirements:
A good bus suspension system should have satisfactory road holding ability, while still providing
comfort when riding over bumps and holes in the road. When the bus is experiencing any road
disturbance (i.e. pot holes, cracks, and uneven pavement), the bus body should not have large
oscillations, and the oscillations should dissipate quickly. Since the distance X1-W is very difficult
to measure, and the deformation of the tire (X2-W) is negligible, we will use the distance X1-X2
instead of X1-W as the output in our problem. Keep in mind that this is an estimation.
The road disturbance (W) in this problem will be simulated by a step input. This step could
represent the bus coming out of a pothole. We want to design a feedback controller so that the
output (X1-X2) has an overshoot less than 5% and a settling time shorter than 5 seconds. For
example, when the bus runs onto a 10-cm high step, the bus body will oscillate within a range of
+/- 5 mm and return to a smooth ride within 5 seconds.
Equations of motion:
Our system has two degrees of freedom, since each mass can be moved
in the vertical direction while the other is held still. Thus, we need two simultaneous equations of
motion to describe the given suspension system. The two equations come from free-body diagram
of each mass. Superposition is used to draw the free-body diagram as shown in fig.2
w
Fig.2: Free Body Diagram for suspension system

5. From the picture above and Newton's law, we can obtain the dynamic equations as the following:
𝑀1 𝑋̈1+𝑏1 𝑋̇1+𝐾1 𝑋1 - 𝑏1 𝑋̇2-𝐾1 𝑋2= U (1)
𝑀2 𝑋̈2+ (𝑏1 + 𝑏2)𝑋̇2+ (𝐾1 + 𝐾2)𝑋2- 𝑏1 𝑋̇1-𝐾1 𝑋1= U +𝑏2 𝑊̇ + 𝐾2 𝑊 (2)
Transfer Function Equation:
Assume that all of the initial condition are zeroes, so these equations represent the situation when
the bus wheel goes up a bump. The dynamic equations above can be expressed in a form of transfer
functions by taking Laplace Transform of the above equations. The derivation from above
equations of the Transfer Functions G1(s) and G2(s) of output, X1-X2, and two inputs U and W,
is as follows. Taking the Laplace transform of equation (1) and (2) we get the following equation
(3) and (4).
(𝑀1 𝑠2
+𝑏1 𝑠 + 𝑘1)𝑋1( 𝑠) – (𝑏1 𝑠 + 𝐾1) 𝑋2( 𝑠) = U(s) (3)
– (𝑏1 𝑠 + 𝐾1) 𝑋1( 𝑠) + (𝑀2 𝑠2
+(𝑏1 + 𝑏2 )𝑠 + (𝑘1 + 𝑘2))𝑋2( 𝑠) = (𝑏2 𝑠 + 𝐾2)𝑊(𝑠) -U(s) (4)
Equation (3) and (4) in matrix form as:
[
( 𝑀1 𝑠2
+ 𝑏1 𝑠 + 𝑘1) – ( 𝑏1 𝑠 + 𝐾1)
– ( 𝑏1 𝑠 + 𝐾1) (𝑀2 𝑠2
+ ( 𝑏1 + 𝑏2) 𝑠 + ( 𝑘1 + 𝑘2))
][
𝑋1( 𝑠)
𝑋2( 𝑠)
] =[
𝑈(𝑆)
(𝑏2 𝑠 + 𝐾2)𝑊(𝑠) − U(s)
]
Let A = [
( 𝑀1 𝑠2
+ 𝑏1 𝑠 + 𝑘1) – ( 𝑏1 𝑠 + 𝐾1)
– ( 𝑏1 𝑠 + 𝐾1) (𝑀2 𝑠2
+ ( 𝑏1 + 𝑏2) 𝑠 + ( 𝑘1 + 𝑘2))
]
∆ = det [
( 𝑀1 𝑠2
+ 𝑏1 𝑠 + 𝑘1) – ( 𝑏1 𝑠 + 𝐾1)
– ( 𝑏1 𝑠 + 𝐾1) (𝑀2 𝑠2
+ ( 𝑏1 + 𝑏2) 𝑠 + ( 𝑘1 + 𝑘2))
]
Or, ∆ = ( 𝑀1 𝑠2
+ 𝑏1 𝑠 + 𝑘1).(𝑀2 𝑠2
+ ( 𝑏1 + 𝑏2) 𝑠 + ( 𝑘1 + 𝑘2)) - ( 𝑏1 𝑠 + 𝐾1). ( 𝑏1 𝑠 + 𝐾1)
To find the value of X1(s) and X2(s), we used inverse matrix method (X=A-1B). Find the inverse
of A and then multiply with input U(s) and W(s) on the right-hand side as the following:
[
𝑋1( 𝑠)
𝑋2( 𝑠)
] =
1
∆
[
( 𝑀1 𝑠2 + 𝑏1 𝑠 + 𝑘1) – ( 𝑏1 𝑠 + 𝐾1)
– ( 𝑏1 𝑠 + 𝐾1) (𝑀2 𝑠2 + ( 𝑏1 + 𝑏2) 𝑠+ ( 𝑘1 + 𝑘2))
] [
𝑈(𝑆)
(𝑏2 𝑠 + 𝐾2 𝑊(𝑠) − U(s)
]
Multiply right-side matrix and Rearrange we will get the following matrix form.
[
𝑋1( 𝑠)
𝑋2( 𝑠)
] =
1
∆
[
( 𝑀1 𝑠2
+ 𝑏1 𝑠 + 𝑘1) – ( 𝑏1 𝑠 + 𝐾1)
– ( 𝑏1 𝑠 + 𝐾1) (𝑀2 𝑠2
+ ( 𝑏1 + 𝑏2) 𝑠 + ( 𝑘1 + 𝑘2))
][
𝑈(𝑆)
W(s)
]

6. 𝑋1( 𝑠) =
M2s2U (s) + (b1+b2)sU(s) +(k1+k2)U(s) + {(b1s+k1)(b2s+k2)}W(s) – (b1s+k1)U(s)
∆
X2(s)=
(b1s+k1)U(s)+M1s2(b2s+k2)W(s)−M1s2U(s)+b1s(b2s+k2)W(s)+k1(b2s+k2)W(s)−(b1s+k1)U(s)
∆
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 the following:
G1(s) = [
X1(s) −X2(s)
𝑈(𝑠)
] =
( 𝑀1+𝑀2 ) 𝑠2
−𝑏2 𝑠+𝐾2
∆
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 the following:
G2(s) =
X1(s) −X2(s)
𝑊(𝑠)
=
−𝑀1 𝑏2 𝑠3
−𝑀1 𝐾2 𝑠2
∆
Matlab Code:
To implement and simulate the above transfer function we have used Matlab and Matlab Simulink.
Matlab code for above equation is:
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);
figure(1)
step(G1)
title(' Response of unit step actuated force input, U(s)');
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);
figure(2)
step(G2)
title('response for a step disturbance input W(s)');
numf=num1;
denf=nump;

7. F=tf(numf,denf);
Kd = 208025;
Kp = 832100;
Ki = 624075;
C = pid(Kp,Ki,Kd);
sys_cl=F*feedback(G1,C);
t=0:0.05:5;
figure(3)
step(0.1*sys_cl,t)
title('Step response under PID Control')
Kd=5*Kd;
Kp=5*Kp;
Ki=3*Ki;
C=pid(Kp,Ki,Kd);
sys_cl=F*feedback(G1,C);
figure(4)
step(0.1*sys_cl,t)
title('Step Response under High-Gain PID')
axis([0 5 -.01 .01])
Results:
Figure 3 show the response of unit step actuated input U(s) only, when W(s) = 0.
Fig 3. Open loop response when W(s)=0.
This is the open loop response when no feedback and controller are used. From the graph of the
open-loop response for a unit step actuated force, we can see that the system is under-damped.
People sitting in the bus will feel very small amount of oscillation. Moreover, the bus takes an
unacceptably long time to reach the steady state (the settling time is very large).

8. Figure 4 show the response for unit step disturbance input W(s) response only, when U(s) = 0.
Fig 4. Open loop response when U(s)=0.
This is the open loop response when no feedback and controller are used. We can see from the
graph when the bus passes a bump or hole on the road, the bus body will oscillate for an
unacceptably long time (~50 seconds) with an initial amplitude of 0.8 m. People sitting in the bus
will not be comfortable with such an oscillation due to the large overshoot and long settling time.
Adding a PID controller:
In open loop output we cannot obtain our required system specifications. The solution to these
problems is to add a feedback controller into the system to improve the performance. After adding
the controller, the whole suspension design model is shown given below.
Fig 5 suspension design with controller

9. There is different type of controllers, which we can used to improve ours result. But here we used
Proportional-Integral-Derivative (PID) controller. PID control is the most common control
algorithm used in industry and has been universally accepted in industrial control. The popularity
of PID controllers can be attributed partly to their robust performance in a wide range of operating
conditions and partly to their functional simplicity, which allows engineers to operate them in a
simple, straightforward manner. The basic idea behind a PID controller is to read the data, then
compute the desired actuator output by calculating proportional, integral, and derivative responses
and summing those three components to compute the output. The transfer function of PID
controller is.
𝐶( 𝑠) = 𝐾𝑝 +
𝐾𝑖
𝑠
+ 𝐾𝑑 =
𝐾 𝑑 𝑠2
+𝐾 𝑝 𝑠+𝐾 𝑝
𝑠
where 𝐾𝑝 is the proportional gain, 𝐾𝑖 is the integral gain, and 𝐾𝑑 is the derivative gain. Here we
will need all three of these gains in our controller. After adding the code of PID transfer function
in Matlab code we get the following result.
Results:
when the road disturbance (W) is simulated by a unit step input and put 𝐾𝑑=624075, 𝐾𝑖=832100
and 𝐾𝑝=208025, then the result shown given below in figure 6.
Fig 6 output when using PID controller
From the graph, the percent overshoot is 9mm, which is larger than the 5mm requirement, but the
settling time is satisfied, less than 5 seconds. To get the require percent overshoot we choose other
value for PID.
when the road disturbance (W) is simulated by a unit step input and put 𝐾𝑑=4*624075,
𝐾𝑖=4*832100 and 𝐾𝑝=3*208025, then the result shown given below in figure 7.

10. Fig 7 output when using PID controller
Now we see that the percent overshoot and settling time meet the requirements of the system. The
percent overshoot is about 3% of the input's amplitude and settling time is 1.8 seconds which is
less than the 5 second requirement. This mean that when we change the value of parameter of PID
we get more precise result.
Simulink Modeling:
The transfer function of our suspension system and feedback controller in Simulink shown below
in figure 9.
Fig 9 close loop with PID controller
The output of the close system is given below.

11. Fig 10 close loop output using PID
Now we see that the percent overshoot and settling time meet the requirements of the system. The
percent overshoot is about 3% of the input's amplitude and settling time is 1.8 seconds which is
less than the 5 second requirement. This mean that when we change the value of parameter of PID
we get more precise result.
Conclusion:
This paper describes about the designing and analyzing the suspension system using PID
controller. The suspension has been designed and analyzed on the basis of vehicle dynamics. It
also helps to understand and overcome the theocratical difficulties of the vehicle. We have made
this project for the purpose of stability of heavy vehicle on brumby or irregular surfaces by using
the PID controller