Balakumar Balachandran, Magrab E.B. Vibrations
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where is an (NN) identity matrix, is an (NN) null matrix; that is, a
matrix whose elements are all zero. Equation (8.56) is referred to as the state-
space form
3
of Eq. (8.50).
The initial displacements and the initial velocities of the inertial elements
are given by
(8.59)
The form of Eqs. (8.56) to (8.59) are well suited to numerical evaluation
by standard numerical procedures.
4
Alternatively, a solution for Eq. (8.56)
subject to the initial conditions given by Eqs. (8.59) can be found by using the
Laplace transform method, as discussed in Appendix A. A third choice for
solving Eq. (8.56) is based on the eigenvalues and eigenvectors of the state
matrix and its transpose. This approach parallels the normal-mode approach.
5
It is noted that since the matrix is not symmetric, its eigenvalues are com-
plex valued.
In the vibrations literature, the words “mode of a system” are reserved for
the eigenvectors determined from the eigenvalue problem associated with the
second-order form of the governing equations; that is, eigensystems such as
Eq. (7.79) associated with Eq. (7.73). However, in the broader literature, it is
common to use the words “modes of a system” for the eigenvectors deter-
mined from the eigenvalue problem associated with the state-space form of
the governing equations. This is further explored in Example 8.8.
EXAMPLE 8.6
State-space form of equations for a gyro-sensor
We illustrate how the governing equations obtained for the gyro-sensor in Ex-
ample 7.4 can be written in state-space form. For convenience, the governing
equations determined in Example 7.4 are repeated below.
(a)
where
3M 4e
x
$
y
$
f 3C 4e
x
#
y
#
f 3G 4e
x
#
y
#
f 3K 4e
x
y
f e
f
x
0
f
3A 4
5Y
1
10 26 d
q
1
10 2
q
2
10 2
o
q
N
10 2
t
and
5Y
2
10 26 d
q
#
1
10 2
q
#
2
10 2
o
q
#
N
10 2
t
30 43I 4
460 CHAPTER 8 Multiple Degree-of-Freedom Systems
3
It is noted that the state-space form of a system is not unique; one can obtain different state-space
forms by considering equations different from Eq. (8.54).
4
For example, the MATLAB function ode45 can be used to numerically solve these equations.
5
L. Meirovitch, Fundamentals of Vibrations, McGraw Hill, NY, Chapter 7 (2001) or L. Meirovitch
(1980), ibid.
(b)
Since we have a two degree-of-freedom system, the state vector is a (41)
vector. Noting that N 2 in Eq. (8.56) and identifying q
1
as x and q
2
as y, the
state vector and its time derivative are given by
(c)
In order to construct the (44) state matrix and the (42) matrix
, we first determine the inverse of the inertia matrix . Since the mass
matrix given by Eqs. (b) is a diagonal matrix, the inverse is given by
(d)
Making use of Eqs. (d), (b), and (8.58) and recognizing that the chosen ex-
ample is free of circulatory forces—that is, —we arrive at the state
matrix.
(e)
From Eqs. (8.58) and (d), the matrix is found to be
(f)3B4 D
00
00
1/m 0
0 1/m
T
3B 4
≥
0010
0001
1k
x
mv
2
z
2/m 0 c
x
/m 2v
z
0 1k
y
mv
2
z
2/m 2v
z
c
y
/m
¥
10
01
10
01
0
0
02
⎡
⎣
⎢
⎤
⎦
⎥
−
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
+
−
/
/
m
m
c
c
m
x
y
v
zz
z
m20v
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
⎤
⎦
⎥
⎥
⎥
⎥
⎥
34A
m
m
km
k
xz
y
=
⎡
⎣
⎢
⎤
⎦
⎥
−
⎡
⎣
⎢
⎤
⎦
⎥
−
−
00
00
10
01
0
0
2
/
/
v
mm
z
v
2
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎢
30 43H4
3M 4
1
c
1/m 0
01/m
d
3M 43B 4
3A 4
5Y6 e
Y
1
Y
2
f d
x
y
x
#
y
#
t,
5Y
#
6 e
Y
#
1
Y
#
2
f d
x
#
y
#
x
$
y
$
t,
and
5F6 e
f
x
0
f
3K 4 c
k
x
mv
2
z
0
0 k
y
mv
2
z
d
3M 4 c
m 0
0 m
d,
3C 4 c
c
x
0
0 c
y
d,
3G 4 c
0 2mv
z
2mv
z
0
d
8.3 State-Space Formulation 461
From Eqs. (8.56) and (8.57), and Eqs. (c), (e), and (f), we arrive at the fol-
lowing state-space form of Eq. (a):
(g)
EXAMPLE 8.7
State-space form of equations for a model of a milling system
We revisit the three degree-of-system treated in Example 7.1 for a milling sys-
tem and determine the state-space form of the governing equations. The ap-
proach follows along the lines of Example 8.6. First, from Eqs. (8.57) and
Eqs. (b) of Example 7.1, we identify the state vector as a (61) vector and
construct this vector and its time derivative as follows.
(a)
Then, the inverse of the inertia matrix is obtained. Since the mass matrix
given in Eqs. (b) of Example 7.1 is a diagonal matrix, the inverse of this
matrix is
(b)
Making use of Eqs. (8.58) and Eqs. (b) of Example 7.1 and noting that this
system is free of gyroscopic and circulatory forces—that is, and
—we find the following state matrix:3H 4 304
3G 4 304
3M 4
1
C
1/m
1
00
01/m
2
0
001/m
3
S
3M 4
5Y6 e
Y
1
Y
2
f f
x
1
x
2
x
3
x
#
1
x
#
2
x
#
3
v
and
5Y
#
6 e
Y
#
1
Y
#
2
f f
x
#
1
x
#
2
x
#
3
x
$
1
x
$
2
x
$
3
v
D
00
00
1/m 0
0 1/m
Te
f
x
0
f
d
x
#
y
#
x
$
y
$
t D
0010
0001
1k
x
mv
2
z
2/m 0 c
x
/m 2v
z
0 1k
y
mv
2
z
2/m 2v
z
c
y
/m
T d
x
y
x
#
y
#
t
462 CHAPTER 8 Multiple Degree-of-Freedom Systems
(c)
Carrying out the matrix multiplication operations, the matrix in Eq. (c) is re-
duced to the form
(d)
In this case, from Eqs. (8.58) and Eqs. (b), we find that the (63) matrix
is given by
(e)
Hence, the state-space form of Eqs. (b) of Example 7.1 is given by
(f)
where the matrices and are given by Eqs. (d) and (e), respectively.3B 43A4
f
x
#
1
x
#
2
x
#
3
x
$
1
x
$
2
x
$
3
v 3A4f
x
1
x
2
x
3
x
#
1
x
#
2
x
#
3
v 3B4c
f 1t2
0
0
s
3B 4 F
000
000
000
1/m
1
00
0 1/m
2
0
001/m
3
V
3B 4
3A 4 F
00 010 0
00 001 0
00 000 1
k
1
/m
1
k
1
/m
1
0 c
1
/m
1
c
1
/m
1
0
k
1
/m
2
1k
1
k
2
2/m
2
k
2
/m
2
c
1
/m
2
1c
1
c
2
2/m
2
c
2
/m
2
0 k
2
/m
3
1k
2
k
3
2/m
3
0 c
2
/m
3
1c
2
c
3
2/m
3
V
100
010
001
100
01 0
001
1
2
3
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
−
⎡
⎣
⎢
⎢
⎢
/
/
/
m
m
m
⎤⎤
⎦
⎥
⎥
⎥
−
−+−
−+
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎤
cc
ccc c
ccc
11
112 2
223
0
0
⎦⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
34A
m
m
m
=
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
−
⎡
000
000
000
100
01 0
001
1
2
3
/
/
/
⎣⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
−
−+−
−+
⎡
⎣
⎢
⎢
⎢
⎤
⎦
kk
kkk k
kkk
11
112 2
223
0
0
⎥⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
8.3 State-Space Formulation 463
EXAMPLE 8.8 Eigenvalues and eigenvectors of a proportionally damped system
from the state matrix
In this example, we revisit Examples 7.21 and 7.22 and illustrate by using the
state-space form of equations for the undamped system and the proportion-
ally damped system that the eigenvectors have a similar structure in both
cases. This was pointed out in Section 7.2.3. Let us suppose that the matrices
given in Example 7.21 correspond to the system shown in Figure 7.1. The
governing equations of motion of this two degree-of-freedom system are re-
peated below after setting the forces to zero.
(a)
To find the eigenvalues and eigenvectors associated with the two degree-
of-freedom system, we write Eq. (a) in the following state-space form:
(b)
Following the procedure outlined in Example 8.5, we find that the state ma-
trix is given by
(c)
Since Eq. (b) is a set of linear ordinary differential equations with constant
coefficients, one can assume a solution of the form
(d)
On substituting Eq. (d) into Eq. (b) and canceling the common factor of e
lt
on
both sides of the equations, we arrive at the following system
(e)
The algebraic system of equations given by Eq. (e) constitutes an eigen-
value problem, since we are seeking those special values of l for which the
vector {X} will be nontrivial. For the chosen physical system, we have a
system of four algebraic equations, which will mean that we will have four
eigenvalues and a set of four corresponding eigenvectors. These eigenvalues
are determined from
3A 45X6 l5X6
5x6 5X6e
lt
3A 4 D
0010
0001
1k
1
k
2
2/m
1
k
2
/m
1
1c
1
c
2
2/m
1
c
2
/m
1
k
2
/m
2
1k
2
k
3
2/m
2
c
2
/m
2
1c
2
c
3
2/m
2
T
d
x
#
1
x
#
2
x
$
1
x
$
2
t 3A4d
x
1
x
2
x
#
1
x
#
2
t 3A45x6
c
k
1
k
2
k
2
k
2
k
2
k
3
de
x
1
x
2
f e
0
0
f
c
m
1
0
0 m
2
de
x
$
1
x
$
2
f c
c
1
c
2
c
2
c
2
c
2
c
3
de
x
#
1
x
#
2
f
464 CHAPTER 8 Multiple Degree-of-Freedom Systems
(f)
Since matrix is not a symmetric matrix, the eigenvalues of this matrix can
no longer be expected to be real. Furthermore, from Eqs. (b) and (d), it is seen
that the first two entries in each eigenvector are associated with the displace-
ment states, which are x
1
and x
2
, and that the next two entries in each eigenvec-
tor are associated with the velocity states, which are and . From Eq. (d), we
see that and ; hence, the third entry of the eigenvector asso-
ciated with the eigenvalue l is l times the first entry of this eigenvector and the
fourth entry of this eigenvector is l times the second entry of this eigenvector.
We numerically determine the eigenvalues and eigenvectors associated
with the state matrix for the parameter values given in Example 7.21 and
compare the results obtained in the undamped and damped cases presented in
Example 7.22.
Eigenvalues and Eigenvectors in the Undamped Case
In the undamped case, we set the damping coefficients c
i
0 in Eq. (c). From
Eq. (a) of Example 7.21, we then make the substitutions k
1
k
2
2, k
2
1,
k
2
k
3
1, m
1
2, and m
2
1 and find that the state-space matrix is given by
(g)
Upon substituting Eq. (g) into Eq. (f), we find that the eigenvalues
6
are
(h)
and the corresponding eigenvectors are
(i)
The imaginary eigenvalues are identical to those given in Eqs. (c) and (d)
of Example 7.22. Upon comparing the results of Example 7.22 and this ex-
ample, it is clear that in the undamped case
(j)l
11,12
jv
1
and
l
21,22
jv
2
5X6
21
d
0.351
0.496
j0.459
j0.648
t,
5X6
22
d
0.351
0.496
j0.459
j0.648
t,
5X6
11
d
0.508
0.718
j0.275
j0.389
t,
5X6
12
d
0.508
0.718
j0.275
j0.389
t,
l
21,22
j1.307
l
11,12
j0.541
3A 4 D
0010
0001
11/200
1 100
T
3A 4
x
#
2
lx
2
x
#
1
lx
1
x
#
2
x
#
1
3A 4
det 33A4 l3I44 0
8.3 State-Space Formulation 465
6
The MATLAB function eig was used.
or
where the v
i
are the natural frequencies of the system. Equations ( j) are con-
sistent with Eqs. (7.35). The four eigenvalues occur in the form of two com-
plex conjugate pairs. The eigenvectors associated with a complex conjugate
pair of eigenvalues are complex conjugates of each other; for example, the
eigenvector {X}
11
associated with the eigenvalue l
11
is the complex conjugate
of the eigenvector {X}
12
associated with the eigenvalue l
12
. In addition, it can
be verified that the third entry of each eigenvector is the corresponding eigen-
value times the first entry of this eigenvector and that the fourth entry of each
eigenvector is the corresponding eigenvalue times the second entry of this
eigenvector.
Eigenvalues and Eigenvectors in the Damped Case
In the damped case, we find from Eq. (a) of Example 7.21 that c
1
c
2
0.8,
c
2
0.2, and c
2
c
3
0.4. Upon substituting these values and the values for
k
i
and m
i
used to obtain Eq. (g), we find that the state-space matrix is given by
(k)
After substituting Eq. (k) into Eq. (f), we find that the eigenvalues are
7
(l)
and the associated eigenvectors are
(m)
On comparing the eigenvalues determined for the damped system with
those determined in Example 7.22, we see that the eigenvalues determined in
both cases are equal and that they are in the form
(n)l
d1
1,2
z
1
v
1
jv
d1
and
l
d2
1,2
z
2
v
2
jv
d2
5X6
d2
1
d
0.233 j0.263
0.329 j0.371
0.399 j0.227
0.564 j0.321
t,
5X6
d2
2
d
0.233 j0.263
0.329 j0.371
0.399 j0.227
0.564 j0.321
t
5X6
d1
1
d
0.141 j0.488
0.200 j0.690
0.238 j0.137
0.337 j0.194
t,
5X6
d1
2
d
0.141 j0.488
0.200 j0.690
0.238 j0.137
0.337 j0.194
t,
l
d2
1,2
0.271 j1.278
l
d1
1,2
0.129 j0.526
3A 4 D
00 1 0
00 0 1
11/2 0.4 0.1
1 1 0.2 0.4
T
1l
11,12
2
2
v
2
1
and
1l
21,22
2
2
v
2
2
466 CHAPTER 8 Multiple Degree-of-Freedom Systems
7
The MATLAB function eig was used. However, the eigenvalues and associated eigenvectors
presented here are not in the same order as that determined by MATLAB.
where z
i
are the damping factors and v
di
are the damped natural frequencies.
At first glance, the eigenvectors determined for the undamped case and pre-
sented in Eqs. (i) appear to have a different form from the eigenvectors deter-
mined for the damped case and presented in Eqs. (m). To compare them, we
normalize each eigenvector by dividing throughout with the first entry of that
eigenvector; this means that the first entry of each eigenvector will be 1 in
both the undamped and damped cases. The normalized eigenvectors for the
undamped and damped cases are shown below.
Normalized eigenvectors in the undamped case
(o)
Normalized eigenvectors in the damped case
(p)
Comparing the eigenvectors given in Eqs. (o) and (p), we see that the first
two entries of an eigenvector determined for the undamped case are identical
to the first two entries of the corresponding eigenvector in the proportionally
damped case; this means that the ratio of the displacement states for modes of
the undamped system and the ratio of the displacement states in the corre-
sponding modes of the proportionally damped system are the same. For this
reason, it is often loosely stated in the literature that the modes of the un-
damped system are “identical” to the corresponding modes of the proportion-
ally damped system. From Eqs. (o) and (p), it is clear that since the first entry
of each eigenvector is one, the third entry of each eigenvector is the corre-
sponding eigenvalue. Recall the discussion following Eq. (f ).
5X6
d21
d
1.000
1.414
0.270 j1.278
1.748 j0.598
t,
5X6
d22
d
1.000
1.414
0.270 j1.278
1.748 j0.598
t
5X6
d11
d
1.000
1.414
0.129 j0.525
0.183 j0.743
t,
5X6
d12
d
1.000
1.414
0.129 j0.525
0.183 j0.743
t,
5X6
21
d
1.000
1.413
j1.307
j1.846
t,
5X6
22
d
1.000
1.413
j1.307
j1.846
t
5X6
11
d
1.000
1.413
j0.541
j0.766
t,
5X6
12
d
1.000
1.413
j0.541
j0.766
t,
8.3 State-Space Formulation 467
EXAMPLE 8.9 Free oscillation comparison for a system with an arbitrary damping
model and a system with a constant modal damping model
We compare the free oscillations of two systems with identical mass and stiff-
ness matrices, but with different damping matrices. In one case, the damping
matrix product term
(a)
is a diagonal matrix while in the other case, this matrix product is not diago-
nal. For the diagonal matrix case, the damping factors are the same for each
of the two modes and, hence, this case is referred to as a constant modal
damping case. In this case, as illustrated in Example 8.2, one can determine
the solution for free oscillations by using the normal-mode approach. How-
ever, this is not possible to do when the damping matrix is arbitrary or the
matrix product is not a diagonal matrix. Here, we consider the constant
modal damping case of Example 8.2 as a baseline case and make changes to
the damping matrix determined for this baseline case so that the matrix
product is not a diagonal matrix. In both cases, the solutions for free os-
cillations of the displacement states are determined by making use of the
state-space formulation; that is, Eq. (8.56). The initial conditions for both
cases are given by Eqs. (a) of Example 8.2.
Constant Modal Damping Case
In Example 8.2, we considered damped oscillations of a two degree-of-
freedom system. In this example, which is a continuation of Examples 7.14 and
7.20, it is assumed that the damping matrix product term can be approximated
by the following equation, which is obtained by assuming constant modal
damping and dropping the off-diagonal terms because they are “small.”
(b)
To determine the damping matrix , which will lead to the diagonal matrix
shown in Eq. (b), we carry out a series of matrix multiplications and find from
Eq. (b) that the damping matrix needs to take the form
(c)
Equation (c) can be used to determine the damping coefficients c
1
, c
2
, and c
3
that will provide constant modal damping and a matrix that will be a di-
agonal matrix.
Previously, in Example 7.20, the modal matrix and the modal masses were
determined along with the system natural frequencies. On using Eq. (7.68a),
Eqs. (a), (b), and (g) of Example 7.20, and the value of z 0.05 used in Exam-
ple 8.2, Eq. (c) results in
3C
t
4
3C 4 c
c
1
c
2
c
2
c
2
c
2
c
3
d 33£ 4
T
4
1
3M
D
432zv 43£ 4
1
3C 4
3C 4
3M
D
4
1
3£ 4
T
3C 43£ 4 312zv24 c
2zv
1
0
02zv
2
d
3C
t
4
3C 4
3C
t
4
3C 4
3C
t
4 3M
D
4
1
3£ 4
T
3C43£ 4
468 CHAPTER 8 Multiple Degree-of-Freedom Systems
(d)
from which it is found that
(e)
For these values, it can be verified that
(f)
which, as expected, is a diagonal matrix.
In order to determine the associated free oscillations, we first used Eq. (d)
and the matrices and that are given by Eqs. (a) of Example 7.20 to
form the matrices and given by Eqs. (8.58). These matrices are used,
in turn, in Eq. (8.56) to numerically obtain the state vector {Y}.
8
The results
obtained for free oscillations of the displacement states, which are in agree-
ment with those obtained analytically in Example 8.2 and shown in Figure 8.1,
are presented in Figure 8.7.
Arbitrary Damping Case
If we increase the values of c
1
, c
2
, and c
3
, from those given by Eqs. (e), then
the resulting damping matrix will not necessarily result in a diagonal
matrix on carrying out the operations given in Eq. (a). This can help assess
how the free oscillation characteristics change when there are non-zero off-
diagonal terms in the matrix .
It is found that doubling the values of either c
1
or c
3
has “little” effect on
the responses x
j
(t) compared to those obtained with the damping matrix prod-
uct given in Eq. (f). However, doubling the value of c
2
from 0.246 N s/m to
0.49 N s/m does result in noticeable differences in the responses of the iner-
tia elements. In this case, the damping matrix takes the form
(g)
and evaluating the damping matrix product we obtain
(h)3M
D
4
1
3£ 4
T
3C43£ 4 c
0.253 0.025
0.009 0.855
d
3C 4 c
0.821 0.246
0.246 1.144
d N
#
s/m
#
#
3C
t
4
3C 4
3B 43A 4
3K 43M4
3M
D
4
1
3£ 4
T
3C43£ 4 c
0.252 0
0 0.562
d
c
3
0.654 N
#
s/m
c
2
0.246 N
#
s/m
c
1
0.332 N
#
s/m
c
0.577 0.246
0.246 0.900
dNs/m
c
0.1 2.519 0
0 0.1 5.623
dc
0.893 2518
11
d
1
c
c
1
c
2
c
2
c
2
c
2
c
3
d c
0.893 1
2.518 1
d
1
c
3.658 0
0 10.31
d
8.3 State-Space Formulation 469
8
The MATLAB function ode45 was used.