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- - 68
II. TWO DEGREE OF FREEDOM SYSTEMS
 Chapter one developed the fundamentals of dynamics
and control of single degree of freedom mechatronic
systems. Since we were dealing with single degrees of
freedom, the relationships that were developed were
temporal in nature. In this chapter, we develop the
fundamentals of dynamics and control of two degree
of freedom systems which are spatial in nature.
A. Modal Analysis and Modal Design
 It’s convenient to transform the coordinates of a two
degree of freedom system into two modal degrees of
freedom. Each modal degree of freedom acts like a
single degree of freedom system. This allows us to
extend the results obtained in the previous chapter to
analyze, design, and control two degree of freedom
systems.
MODAL ANALYSIS AND MODAL DESIGN
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MAE 524 course notes – Spring 2002, Copyrighted by L.Silverberg
- - 69
A.1 Derivation of equations of motion
 Consider the undamped two degree of freedom system
shown. Applying Newton’s second law of motion to
each mass, yields the two equations that govern the
dynamics of the system
(2.1)
m1x1  k2 ( x2  x1 )  k1x1  f1
m2 x2  k2 ( x2  x1 )  k3 x2  f 2
 Equation (2.1) can be rewritten in the matrix-vector
form as
(2.2)
m1 0   x1  k1  k2
 0 m  x     k

2
2  2  
 k2   x1   f1 
  
k2  k3   x2   f 2 
or more compactly as
(2.3)
Mx  Kx  F
where x is called the position vector, F is called the
force vector, M is called the mass matrix, and K is
called the stiffness matrix.
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A.2 The eigenvalue problem
 Consider the freely vibrating two degree of freedom
system, that is let F = 0 in Eq. (2.3). Try solutions in
the form
(2.4)
x  e st
Substituting Eq. (2.4) into Eq. (2.3) and dividing the
result by e st , we get
(2.5)
[s 2M  K ]φ  0
 Equation (2.5) admits a nontrivial solution only if the
matrix above is singular, in which case
(2.6)
det[s 2 M  K]  0
 Substituting the mass and stiffness matrices in Eq.
(2.2) into Eq. (2.6), yields Eq (2.7)
 s 2 m1  k1  k 2

 k2
0  det 

2
 k2
s m2  k 2  k 3 

( s 2 m1  k1  k 2 )( s 2 m2  k 2  k 3 )  k 22 
m1 m2 s 4  [m1 (k 2  k 3 )  m2 (k1  k 2 )]s 2  (k1 k 2  k 2 k 3  k 3 k1 )
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 Equation (2.7) is a quadratic equation in terms of s2,
admitting the two solutions. Each solution is
substituted back into Eq. (2.5) in order to obtain the
associated vectors φ1 and φ 2 . Equation (2.5) is called
the eigenvalue problem associated with the equations
of motion, Eq. (2.3). The complex numbers s are
called eigenvalues and the associated vectors,
φ1 and φ2 , are called eigenvectors. The eigenvectors
are also called natural modes of vibration.
 Since the eigenvalue problem is a homogeneous
equation, the eigenvectors are unique up to a
multiplicative constant. Therefore, without loss of
generality, we shall normalize the eigenvectors by
letting the first entry in each eigenvector equal one,
that is
(2.8)
1
1
φ 1   , φ 2   
1 
 2 
The eigenvectors are found by substituting Eq. (2.8)
back into Eq. (2.5).
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Properties of eigenvalues and eigenvectors
 In Eq. (2.2) the mass matrix and the stiffness matrix
were both symmetric matrices. In all two degree of
freedom systems composed of mass and stiffness, the
equations can be derived so that the mass and stiffness
matrices are symmetric.
 Furthermore, the mass and stiffness matrices have
another property. The property is analogous to the sign
property of numbers. The mass matrix is positive and
the stiffness matrix is nonnegative. Of course, M and
K are matrices, so we’ll need to define what we
precisely mean by positive and nonnegative within the
context of matrices.
 A symmetric matrix A is defined to be positive if the
number yT Ay  0 for all nonzero y. Similarly, a
symmetric matrix A is defined to be nonnegative if the
number yT Ay  0 for all nonzero y.
 The mass and stiffness matrices satisfy
M0
(2.9)
M  MT
K0
K  KT
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 The eigenvalues and eigenvectors of systems that
satisfy Eqs. (2.9) are said to be normal mode systems.
The first normal mode property states that the
eigenvalues are pure imaginary. Pre-multiplying Eq.
(2.5) by φT , yields
(2.10)
0  φT [s 2M  K ]φ  s 2 (φTMφ)  (φTKφ)
From Eq. (2.9)
(2.11)
φTMφ
s  T
 0,
φ Kφ
2
so s  i0 is pure imaginary, in which  0 is called a
natural frequency of oscillation. The eigenvalue
problem can now be written as
2
(2.12)  0 Mφ  Kφ
 The second normal mode property states that the
eigenvectors are real up to a multiplicative constant.
To show this, write the eigenvector in terms of its real
and imaginary components as
(2.13)
φ  φ R  φ Ii
 Substitute Eq. (2.13) into Eq. (2.12) and separate the
real and imaginary parts to get
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(2.14)
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 02 Mφ R  Kφ R ,  02 Mφ I  Kφ I
 Notice in Eq. (2.14) that the real part of the
eigenvector and the imaginary part of the eigenvector
satisfy the same eigenvalue problem. Therefore, the
eigenvectors must be real – up to a multiplicative
constant.
 The next normal mode property is the orthogonality
property. To show this property, pre-multiply the
eigenvalue problem, Eq. (2.12), by the transpose of the
eigenvectors, as shown below
2
φ T2 :  01
Mφ 1  Kφ 1
(2.15)
2
φ 1T :  02
Mφ 2  Kφ 2
Next, subtract the two equations from each other,
recognizing the properties given in Eq. (2.9), to get
(2.16)
2
2
( 01
  02
)φ T2 Mφ 1  0
 Equation (2.16) reveals that eigenvectors having
distinct eigenvalues are orthogonal with respect to the
mass matrix. Substituting Eq. (2.15) into Eq. (2.16)
further reveals that eigenvectors having distinct
eigenvalues are orthogonal with respect to the stiffness
matrix. Indeed,
(2.17)
φ 1T Mφ 2  0, φ 1T Kφ 2  0
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 Next, let’s normalize the eigenvectors, not in the
manner given in Eq. (2.8), but rather according to the
following
(2.18)
φ 1T Mφ 1  1, φ T2 Mφ 2  1
Substituting Eq. (2.15) into Eq. (2.18) further reveals
that
(2.19)
2
2
φ 1T Kφ 1   01
, φ 1T Kφ 2   02
 Finally, the orthogonality properties and the
normalization given in Eqs. (2.16) through Eq. (2.19)
are written compactly as
(2.20)
φ Tr Mφ s   rs , φ Tr Kφ s   02r  rs
where  rs is the Kronecker-delta symbol in which
 rs = 0 when r and s are different and  rs = 1 when r
and s are equal.
 In summary, the eigenvalues and eigenvectors of two
degree freedom systems composed of mass and
stiffness are normal in the sense that the eigenvalues
are pure imaginary, the eigenvectors are real, and they
satisfy orthogonality properties with respect to the
mass and stiffness matrices.
MODAL ANALYSIS AND MODAL DESIGN
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A.3 Modal equations of motion
 Since the eigenvectors form a basis set, any vector x
can be expressed as a linear combination of the
eigenvectors, whether x is associated with the freely
vibrating system or not.
 We begin by representing x as
(2.21)
x(t )  φ1q1 (t )  φ2q2 (t )
where q1 (t ) and q2 (t ) are called modal
displacements. From Eq. (2.21), the r-th modal
displacement can be regarded as the level of
participation of the r-th mode of vibration in the
overall response. Pre-multiplying Eq. (2.21) by
1T M and  2T M , we get
(2.22)
q1 (t )  φ 1T Mx(t ), q2 (t )  φ T2 Mx(t )
 Equations (2.21) and (2.22) are transformations
between the physical displacements x and the modal
displacements q1 (t ) and q2 (t ) .
 Let’s now replace the physical coordinates in Eq. (2.4)
with modal coordinates. Substituting Eq. (2.21) into
Eq. (2.4) yields Eq (2.23):
M[φ 1q1 (t )  φ 2 q2 (t )]  K[φ 1q1 (t )  φ 2 q2 (t )]  F
MODAL ANALYSIS AND MODAL DESIGN
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 Pre-multiplying Eq. (2.23) by 1T and 2T , and invoking
the orthogonality properties, Eq. (2.20), we get the
equations that govern the modal motions of the two
degree of freedom system
2
q1 (t )   01
q1 (t )  Q1 (t )
(2.23)
2
q2 (t )   02
q2 (t )  Q2 (t )
where
(2.24)
Q1 (t )  φ 1T F, Q2 (t )  φ T2 F
are called modal forces.
 What are Q1 (t ) and Q2 (t ) , and how are they related to
the physical forces? The physical forces are expressed
in terms of the modal forces as follows
(2.25)
F  Mφ 1Q1 (t )  Mφ 2 Q2 (t )
 Equation (2.26) can be verified by substituting Eq.
(2.25) into Eq. (2.26) to render an identity. Thus, just
as the modal displacements qr represents the degree to
which a mode of vibration φr participates in the
overall response, so too does a modal force Qr
represent the degree to which a mode of force Mφ r
participates in the overall force.
MODAL ANALYSIS AND MODAL DESIGN
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MAE 524 course notes – Spring 2002, Copyrighted by L.Silverberg
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 In summary, the modal transformations of the
displacements and of the force are Eq (2.27)
x(t )  φ 1q1 (t )  φ 2 q2 (t ), q1 (t )  φ 1T Mx(t ), q2 (t )  φ T2 Mx(t )
F  Mφ 1Q1 (t )  Mφ 2 Q2 (t ), Q1 (t )  φ 1T F, Q2 (t )  φ T2 F
 The modal displacements act as single degree of
freedom system coordinates governed by independent
modal equations of motion, Eq. (2.24). Modal
displacements are excited by modal forces. The
spatial relationship between the modal components of
force and motion are as follows.
x(t )  φ 1 q1 (t )  F  Mφ 1Q1 (t )
(2.28)
x(t )  φ 2 q 2 (t )  F  Mφ 2 Q2 (t )
MODAL ANALYSIS AND MODAL DESIGN
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A.4 Dynamics of a double pendulum
 As an illustration, consider the double pendulum
shown. Summing moments about each of the pivot
points yields the two equations governing the motion
of the double pendulum
(2.29)
m ga
I 11  ka 2 ( 2   1 )  1  1  M 1
2
m ga
I 22  ka 2 ( 2   1 )  2  2  M 2
2
 Equation (2.29) can be written in the compact matrix1 
 F1 
vector form of Eq. (2.4) letting x   , F   ,
 2 
F2 
(2.30)
0  m1 a 2
0 

,

2
I 2   0
m2 a 
 2 m1 ga

2
ka


ka


2
K
m2 ga 
2
2
  ka

ka 
2


I
M 1
0
MODAL ANALYSIS AND MODAL DESIGN
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 Substituting Eq. (2.30) into Eq. (2.6) yields Eq. (2.31)
m1 ga
 2

2
2

I

ka

ka
 0 1

2
2
0  det[ 0 M  K ]  det 
m2 ga 
2
2
2


ka
 0 I 2  ka 
2 

2
3k 3g   3g  3k 3k   3g  
4
2  3k
  
 0  0 



 
   
m
m
a
2
a
m
m
2
a

 
2
2 
 1
   1
 Equation (2.31) is a quadratic equation in  02 . Solving
Eq. (2.31),
1/ 2
2

 3k 3k 3 g 
 

 

 


a 
 
1  3k 3k 3 g   m1 m2
2
  
 01,02   

 
2  m1 m2
a    3 g  3k 3k   3g  2   

4  2a  m  m    2a    
2 
  

   1
3g

1  3k 3k 3 g   3k 3k  
2a
  
  
 



2  m1 m2
a   m1 m2   3 g  3k  3k
 2a m1 m2
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 Assuming eigenvectors in the form of Eq. (2.5) and
substituting Eqs. (2.32) back into Eq. (2.5), we obtain
the eigenvectors of the double pendulum
(2.30)
 3k
 m
1

 3k
 m2
 3k
m
 2
 3k
 m2
3k 
1
m1   1  0
      φ1   
3k  1  0
1

m2 
3k 
m 
m1   1  0
      φ 2   2 
3k   2  0
 m1 
m1 
 Finally, let’s normalize the modes according to Eq.
(2.20). Doing this yields the normalized modes of
vibration, Eq. (2.31)

3
φ 1  
2
 (m1  m2 )a



1/ 2

1
3
 ,φ 2  
2
1
 m1 m2 (m1  m2 )a
MODAL ANALYSIS AND MODAL DESIGN
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


1/ 2
 m2 



m
 1
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 The normalized modes of force are then Eq. (2.35)


a2

Mφ 1  
 3(m1  m2 ) 
1/ 2
 a 2 m1 m2 
 m1 

 , M φ 2  
m 2 
 3(m1  m2 ) 
1/ 2
1
 
 1
 Equation (2.32) provides us with temporal information
concerning the double pendulum and Eq. (2.34) and
(2.35) provide us with spatial information concerning
the double pendulum. We see that the first mode is a
“pendulum” mode wherein the system moves as a
rigid body pendulum. The second mode is a “pseudoelastic” mode associated with the spring and gravity.
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A.5 Actuator dynamics
 As mentioned earlier, the dynamics of an actuator can
interact with a system, producing a coupled system
comprised of the mechanical part and the actuator. Of
course, such interactions are undesirable in the sense
that they complicate the design process – violating the
goal to separate the components of a system into
independent parts. The method of mounting an
actuator to a system effects the coupling of the
actuator to the system.
 Consider the two degree of freedom system given in
Eq. (2.2) and Eq. (2.3). Assume that the actuator
harmonically excites the mechanical part through the
applied force, in which case
(2.36)
 0  0
X 
F   it    eit  F0eit , x p   1 eit  Xeit
X 2 
F0e  F0 
in which  denotes the excitation frequency.
Substituting Eq. (2.36) into Eq. (2.3) and dividing by
e it yields
(2.37)
[ 2M  K ] X  F0
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 Let’s now recall the explicit form of the mass and
stiffness matrices,
(2.38)
0
m
k1  k2
M 1
,
K


 k
 0 m2 

2
 k2 
k2 
and let’s express the physical parameters in these
matrices in terms of the quantities
m2
k
k
, 1  1 ,  2  2
m1
m1
m2
where r is the mass ratio,  1 is the frequency of the
mechanical part,  2 is the frequency of the mount, and
 is the frequency of the excitation. Substituting Eq.
(2.39) into Eq. (2.37), dividing the first equations by
m1 , and dividing the second equations by m 2 , yields
the mass and stiffness matrices
(2.39)
r
(2.40)
12  r 22
1 0
M
, K  
2
0
1


  r 2
 r 22 

r 22 
and from Eq. (2.38), we get
(2.41)
  2  12  r 22

  22

 r 22 
1
X

F0
2
2
m
   2 
2
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 Solving Eq. (2.41) yields the response of the two
degree of freedom system
(2.42)

r 22
F0 
X

,
m2    2  12  r 22 
   4   2 [12   22  r 22 ]  12 22
 Also notice that the mount force between the masses
is
F0 22
(2.43)
FM  k2 ( X 2  X1 ) 
( 2  12 )

 The actuator and the system are uncoupled only if the
excitation frequency has no affect on X1. From Eq.
(2.42), it follows that the actuator and the sensor can
be uncoupled only if the excitation frequency is
significantly lower than the frequency of the
mechanical part; we impose the necessary condition
(2.44)
  1
 The actuator dynamics and the dynamics of the
mechanical part are uncoupled in the following two
situations. In the first situation, the actuator prescribes
a force and in the second situation the actuator
prescribes a displacement.
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Prescribed force (hard mount)
 Let the actuator be significantly less massive than the
mechanical part and let it be “hard mounted,” that is
let r << 1 and 2 >> From Eq. (2.43)
(2.45)
FM = F0.
 The mount force is equal to the excitation force and
independent of the excitation frequency.
Prescribed displacement (soft mount)
 Let the actuator be significantly less massive than the
mechanical part and Let it be “soft mounted,” that is
let r << 1 and 2 << From Eq. (2.42)
F0
.
k2
 The displacement of the actuator is proportional to the
excitation force and independent of the excitation
frequency.
(2.46)
X2 
 In summary, an actuator that acts on a single degree
of freedom system, if it produces a force according to
Eq. (2.36), can be uncoupled from the system if the
excitation frequency is much lower than the frequency
of the mechanical part. Under these conditions a hard
mounted actuator prescribes a force and a soft
mounted actuator prescribes a displacement. These
decoupling conditions are used in design to reduce
complexity.
MODAL ANALYSIS AND MODAL DESIGN
Lecture notes prepared by L.Silverberg and J.Morton