Vehicle Suspension System Analysis: Mass-Spring-Damper & MATLAB
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This report provides a comprehensive analysis of a vehicle suspension system modeled as a mass-spring-damper system. The analysis includes a mathematical derivation of the system's equations of motion, focusing on damping ratio, natural frequency, and the characteristic equation. Diff...

Running Head: Mass-Spring-Damper System 1
Vehicle Suspension System
Institutional Affiliation
Assignment
Date of Submission
Vehicle Suspension System
Institutional Affiliation
Assignment
Date of Submission
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Mass-Spring-Damper System 2
SECTION 1: MATHEMATICAL ANALYSIS OF SYSTEM (25 marks)
The vehicle suspension system is modeled as illustrated in the figure below,
The suspension system comprises of two mass-spring-damper systems stacked on each other. A
simple mass spring damper is analyzed for each section,
X= input body or surface position, x=0
M=mass
Y= mass height with reference to the ground surface, y=0
(1) Free body diagram
(2) Motion of the system
ms ¨y (t ) +Cs ˙y ( t ) +k s y ( t )=Cs ˙x ( t ) + ks x ( t )
¨y ( t ) + Cs
ms
˙y ( t ) + k s
ms
y ( t ) = Cs
ms
˙x ( t ) + ks
ms
x ( t )
(3) Laplace transform and resultant transfer function
s2 Y ( s )+ Cs
ms
sY ( s ) + ks
ms
Y ( s )= Cs
ms
sX ( s )+ k s
ms
X (s)
SECTION 1: MATHEMATICAL ANALYSIS OF SYSTEM (25 marks)
The vehicle suspension system is modeled as illustrated in the figure below,
The suspension system comprises of two mass-spring-damper systems stacked on each other. A
simple mass spring damper is analyzed for each section,
X= input body or surface position, x=0
M=mass
Y= mass height with reference to the ground surface, y=0
(1) Free body diagram
(2) Motion of the system
ms ¨y (t ) +Cs ˙y ( t ) +k s y ( t )=Cs ˙x ( t ) + ks x ( t )
¨y ( t ) + Cs
ms
˙y ( t ) + k s
ms
y ( t ) = Cs
ms
˙x ( t ) + ks
ms
x ( t )
(3) Laplace transform and resultant transfer function
s2 Y ( s )+ Cs
ms
sY ( s ) + ks
ms
Y ( s )= Cs
ms
sX ( s )+ k s
ms
X (s)

Mass-Spring-Damper System 3
Forming a transfer function,
H ( s ) =
( Cs
ms
s + ks
ms )
( s2+ Cs
ms
s+ ks
ms )
The denominator forms the characteristic equation as,
s2 + Cs
ms
s + ks
ms
=0
(4) Damping ratio and the natural frequency
2 ζ ωn =Cs
ms
ωn
2= ks
ms
→ ωn = √ ks
ms
ζ =
1
2 ωn
∗Cs
ms
= Cs
2 ms ωn
(5) Characteristic equation and Eigen values
s2 + Cs
ms
s + ks
ms
… .∈the S−domain
(6) Natural response based on different cases of the damping ratio
Cs=2 m √ k
m =2m ωn
ξ= c
cs
= c
2 mωn
s1,2= −c
2 m ± √ ( c
2m )
2
− k
m
s1,2= (−ξ ± √ξ2−1 ) ωn
Forming a transfer function,
H ( s ) =
( Cs
ms
s + ks
ms )
( s2+ Cs
ms
s+ ks
ms )
The denominator forms the characteristic equation as,
s2 + Cs
ms
s + ks
ms
=0
(4) Damping ratio and the natural frequency
2 ζ ωn =Cs
ms
ωn
2= ks
ms
→ ωn = √ ks
ms
ζ =
1
2 ωn
∗Cs
ms
= Cs
2 ms ωn
(5) Characteristic equation and Eigen values
s2 + Cs
ms
s + ks
ms
… .∈the S−domain
(6) Natural response based on different cases of the damping ratio
Cs=2 m √ k
m =2m ωn
ξ= c
cs
= c
2 mωn
s1,2= −c
2 m ± √ ( c
2m )
2
− k
m
s1,2= (−ξ ± √ξ2−1 ) ωn
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Mass-Spring-Damper System 4
x ( t ) = A e ( −ζ + √ ζ2 −1 ) ωn t + B e ( −ζ − √ ζ 2−1 ) ωn t
Damping cases:
C> Cc …. Heavy damping
The roots are real.
x (t )= A es1 t +B es2 t
A , B … .. constants of integration
α >0 , β>0 , β=α2 − k
m < α2
s1=−α+ β
s2=−α −β
α = c
2 m β= 1
2 m √c2−4 mk
C=Cc … . critical damping
β= 1
2 m √c2−4 mk
s1=s2=−α−¿ Cc
2m =−ωn
x ( t ) = ( A +Bt ) e−ωn t
x (t )= [ x0 + ( ˙x0 +ωn x0 ) t ] e−ωn t
The motion is not vibratory and it decays to the desired equilibrium position.
0<C <Cc … . light damping
c2 <4 mk
β=i ω¿ whereω¿= 1
2 m √4 mk−c2= √ ( k
m )− c2
4 m2
s1=−α +i ω¿
s2=−α −iω¿
α= c
2 m
x=e−αt ( a cos ω¿ t+ B sin ω¿ t )
x ( t ) = A e ( −ζ + √ ζ2 −1 ) ωn t + B e ( −ζ − √ ζ 2−1 ) ωn t
Damping cases:
C> Cc …. Heavy damping
The roots are real.
x (t )= A es1 t +B es2 t
A , B … .. constants of integration
α >0 , β>0 , β=α2 − k
m < α2
s1=−α+ β
s2=−α −β
α = c
2 m β= 1
2 m √c2−4 mk
C=Cc … . critical damping
β= 1
2 m √c2−4 mk
s1=s2=−α−¿ Cc
2m =−ωn
x ( t ) = ( A +Bt ) e−ωn t
x (t )= [ x0 + ( ˙x0 +ωn x0 ) t ] e−ωn t
The motion is not vibratory and it decays to the desired equilibrium position.
0<C <Cc … . light damping
c2 <4 mk
β=i ω¿ whereω¿= 1
2 m √4 mk−c2= √ ( k
m )− c2
4 m2
s1=−α +i ω¿
s2=−α −iω¿
α= c
2 m
x=e−αt ( a cos ω¿ t+ B sin ω¿ t )
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Mass-Spring-Damper System 5
x=C e−αt cos (ω¿ t−φ0 )
x ( t ) =e−ζ ωn t ( A cos ( √ 1−ζ2 ωn t ) +B sin ( √ 1−ζ2 ωn t ) )
x (t )= [ C cos ( ωd t −φ0 ) ] e−ξωn t
ωd= √ 1−ξ2
(7) Expressing natural resonance and resonant frequency
Replacing with actual values where the top mass is given as m=340kg and ks=21000N/m
ωn= √ k s
ms
= √ 21000
340 =7.859 rad /sec
¨y +15.72 ζ ˙y+ 61.765 y =0
To find the resonance frequency,
ωn=2 π f n
f n= ωn
2 π = 7.859
2 π =11.921 Hz
(8) Value of Cs with unity damping ratio
Cs
2meq ωn
=1
Cs=2∗( 340 kg )∗(7.859 rad
s )
Cs=5344.12 N .m/ s
x=C e−αt cos (ω¿ t−φ0 )
x ( t ) =e−ζ ωn t ( A cos ( √ 1−ζ2 ωn t ) +B sin ( √ 1−ζ2 ωn t ) )
x (t )= [ C cos ( ωd t −φ0 ) ] e−ξωn t
ωd= √ 1−ξ2
(7) Expressing natural resonance and resonant frequency
Replacing with actual values where the top mass is given as m=340kg and ks=21000N/m
ωn= √ k s
ms
= √ 21000
340 =7.859 rad /sec
¨y +15.72 ζ ˙y+ 61.765 y =0
To find the resonance frequency,
ωn=2 π f n
f n= ωn
2 π = 7.859
2 π =11.921 Hz
(8) Value of Cs with unity damping ratio
Cs
2meq ωn
=1
Cs=2∗( 340 kg )∗(7.859 rad
s )
Cs=5344.12 N .m/ s

Mass-Spring-Damper System 6
SECTION II: MATLAB IMPLEMENTATION (25 marks)
SECTION I
...Vehicle suspension system
ms=340;
ks=21000;
cs=5344.12;
num=[cs/ms ks/ms];
den=[1 cs/ms ks/ms];
Hs=tf(num,den)
%converting the state space from transfer function
[A,B,C,D]=tf2ss(num,den);
Hs =
15.72 s + 61.76
---------------------
s^2 + 15.72 s + 61.76
Continuous-time transfer function.
SECTION II
figure(1)
subplot(2,1,1)
step(Hs)
grid on
subplot(2,1,2)
impulse(Hs)
grid on
figure(2)
bode (Hs)
grid on
% to determine system stability & control
figure(3)
subplot(2,1,1)
nyquist(Hs)
grid on
subplot(2,1,2)
SECTION II: MATLAB IMPLEMENTATION (25 marks)
SECTION I
...Vehicle suspension system
ms=340;
ks=21000;
cs=5344.12;
num=[cs/ms ks/ms];
den=[1 cs/ms ks/ms];
Hs=tf(num,den)
%converting the state space from transfer function
[A,B,C,D]=tf2ss(num,den);
Hs =
15.72 s + 61.76
---------------------
s^2 + 15.72 s + 61.76
Continuous-time transfer function.
SECTION II
figure(1)
subplot(2,1,1)
step(Hs)
grid on
subplot(2,1,2)
impulse(Hs)
grid on
figure(2)
bode (Hs)
grid on
% to determine system stability & control
figure(3)
subplot(2,1,1)
nyquist(Hs)
grid on
subplot(2,1,2)
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Mass-Spring-Damper System 7
nichols(Hs)
grid on
nichols(Hs)
grid on
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Mass-Spring-Damper System 8
SECTION III: STATE SPACE ANALYSIS (25 marks)
(15) State space analysis and types of problems state space analysis is applied to
The dynamic equations that are obtained from lumped-parameter continuous system is
represented using the state equation and output equation. The model representation is obtained
using the linearized technique as it illustrates the system state matrix, input matrix, and output
matrix respectively.
˙x=f ( x ) ( t , u ( t ) , t ) … state equation
y=g ( x ( t ) , u ( t ) , t ) … output equation
The state space equation is obtained from the transfer function that demonstrates the relationship
between the system inputs and outputs. The state space form provides the structural information
of the system using system states which are the internal description of the system. The transfer
functions expressed as state space models can be represented using the Laplace transform to
determine the eigenvalues of the system. The eigenvalues determine the natural modes of the
SECTION III: STATE SPACE ANALYSIS (25 marks)
(15) State space analysis and types of problems state space analysis is applied to
The dynamic equations that are obtained from lumped-parameter continuous system is
represented using the state equation and output equation. The model representation is obtained
using the linearized technique as it illustrates the system state matrix, input matrix, and output
matrix respectively.
˙x=f ( x ) ( t , u ( t ) , t ) … state equation
y=g ( x ( t ) , u ( t ) , t ) … output equation
The state space equation is obtained from the transfer function that demonstrates the relationship
between the system inputs and outputs. The state space form provides the structural information
of the system using system states which are the internal description of the system. The transfer
functions expressed as state space models can be represented using the Laplace transform to
determine the eigenvalues of the system. The eigenvalues determine the natural modes of the

Mass-Spring-Damper System 9
system (Kisi.deu, 2016). The state variables determine the future behavior of the system and the
system excitation signals are given as the set of the state variables such as,
[ x1 ( t ) x2 ( t ) … xn ( t ) ]
It avoids redundancy in the system analysis by using the least number of state variables. For the
linear systems, the equations are obtained in matrix forms (Piconepress, 2012). The state space
models can be used to monitor the internal behavior of the system, the model can easily
incorporate the very complex output variables. It is quite useful in determining the cost of the
operators and solution. The linear algebra uses the Cayley-Hamilton Theorem.
(16) Example system that uses the state space analysis
The state space variables and model representation is widely used in the engineering
problem and especially in the field of mechanical engineering. The representation uses the matrix
transformation to change from one state variable to another. It is implemented in MIMO case in
the canonical model formats.
The state space for the robot manipulator,
system (Kisi.deu, 2016). The state variables determine the future behavior of the system and the
system excitation signals are given as the set of the state variables such as,
[ x1 ( t ) x2 ( t ) … xn ( t ) ]
It avoids redundancy in the system analysis by using the least number of state variables. For the
linear systems, the equations are obtained in matrix forms (Piconepress, 2012). The state space
models can be used to monitor the internal behavior of the system, the model can easily
incorporate the very complex output variables. It is quite useful in determining the cost of the
operators and solution. The linear algebra uses the Cayley-Hamilton Theorem.
(16) Example system that uses the state space analysis
The state space variables and model representation is widely used in the engineering
problem and especially in the field of mechanical engineering. The representation uses the matrix
transformation to change from one state variable to another. It is implemented in MIMO case in
the canonical model formats.
The state space for the robot manipulator,
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Mass-Spring-Damper System 10
(17) Advantages of using state space techniques to analyze the complete suspension system
including both tire and wheel base.
The state representation has similarity transformations and the state space model focuses on the
controllability, reachability, and system stability. It has the observability, ability of a system to
reconstruct, and detectability.
(17) Advantages of using state space techniques to analyze the complete suspension system
including both tire and wheel base.
The state representation has similarity transformations and the state space model focuses on the
controllability, reachability, and system stability. It has the observability, ability of a system to
reconstruct, and detectability.
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Mass-Spring-Damper System 11
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