5

Random Vibration
5.1
5.2

Random Vibration .............................................................
Single Degree of Freedom: The Response to
Random Loads ....................................................................

Formulation
Correlations

†
†

Derivation of Equations † Response
Response Spectral Density

Rutgers University

5-2

5.3
5.4

Response to Two Random Loads ...................................... 5-7
Multi-Degree-of-Freedom Vibration ................................ 5-12

5.5

Multi-Degree-of-Freedom: The Response to
Random Loads .................................................................... 5-17

Deterministic Vibration † Solution by Frequency Response
Function † Modal Analysis

Response due to a Single Random Force † Response to
Multiple Random Forces † Impulse-Response Approach
Modal Analysis Approach

Haym Benaroya

5-1

5.6

†

Continuous System Random Vibration ........................... 5-29
Transverse Vibration of Beams
Vibration

†

Random Transverse

Summary
This chapter summarizes the key ideas of linear random vibration. This discipline focuses on determining the
response statistics of an oscillator or structure to input forces that are definable only in terms of their statistics.
Typical problems include the following: (1) given the power spectrum of the force, find the power spectrum of the

response; (2) given the mean value and variance of the force, find the mean value and variance of the response. The
methodology is built upon the linear theory of vibration for discrete single- and multi-degree-of-freedom (DoF)
systems, and continuous systems. The approaches are essentially the direct method and the modal analysis method.
The direct method may also be called a transfer matrix method (see Chapter 2). Modal analysis (see Chapters 3 and
4) has the same benefit in random vibration as is done in deterministic vibration studies: it can be computationally
more efficient. A number of examples are given, as are a list of representative references.

5.1

Random Vibration

The discipline of random vibration of structures was borne of the need to understand how structures
respond to dynamic loads that are too complex to model deterministically. Examples include aerodynamic
loading on aircraft and earthquake loading of structures. Essentially, the question that must be answered is:
given the statistics (read: uncertainties) of the loading, what are the statistics (read: most likely values with
bounds) of the response? Generally, for engineering applications the statistics of greatest concern are the
mean, or average value, and the variance, or scatter. These concepts are discussed in detail subsequently.
Suppose that we are aircraft designers currently working on the analysis and design of a wing for a new
airplane. As engineers, we are very familiar with the mechanics of solids and can size the wing for static
loads. Also, we have vibration experience and can evaluate the response of the wing to a harmonic
5-1
© 2005 by Taylor & Francis Group, LLC


5-2

Vibration and Shock Handbook

or impulsive forcing. However, this wing provides
Force

lift to an airplane flying through a turbulent
atmosphere. Even though we are not fluid
dynamicists, we know that turbulence is a very
complicated physical process. In fact, the fluid
(air) motion is so complicated that probabilistic
models are required to model the behavior. Here, a
plausibly deterministic but very complicated
Time
dynamic process is taken to be random for the
purposes of modeling. Wing design requires force
FIGURE 5.1 Turbulent force history.
data resulting from the interaction between fluid
and structure. Such data can be shown as the time history in Figure 5.1.
The challenge is to make sense of such intricate fluctuations. The analyst and designer must run scale
model tests. A wing section is set up in the wind tunnel and representative aerodynamic forces are
generated. Data on wind forces and structural response are gathered and analyzed. With additional data
analysis, it is possible to estimate the force magnitudes. Estimates of the mean values of these forces can be
calculated, as well as of the range of possible forces. With these estimates, it is possible to study the behavior
of the wing under a variety of realistic loading scenarios using the tools of probability and statistics to
model this complex physical problem. This text introduces the use of probabilistic information in
mechanical systems, primarily structural and dynamic systems. These tools are applicable to all the sciences
and engineering, even though this text focuses on the mechanical sciences and engineering.

5.2

Single Degree of Freedom: The Response to Random Loads

Consider the second-order differential equation1 governing the linear motion of an oscillator
€ þ 2zvn XðtÞ
_ þ v2n XðtÞ ¼ FðtÞ

XðtÞ

ð5:1Þ

where the input force per unit mass is given by stationary random process FðtÞ and the output
displacement by random process XðtÞ: Note that it is customary to use upper case letters XðtÞ to represent
random processes, and lower case letters xðtÞ to represent the realizations of a random process.
The notation here is standard: 2zvn ¼ c=m where c ¼ viscous damping and m ¼ mass; and v2n ¼ k=m
where k ¼ stiffness.
We assume that the reader understands the concepts of impulse response and convolution. We would
also like to present a priori the results which are derived in the subsequent section; the mean value of the
response mX and the spectral density of the output SXX ðvÞ :

mX ¼ Hð0ÞmF
SXX ðvÞ ¼ lHðvÞl2 SFF ðvÞ
"
#2
1=v2n
2
lHðvÞl ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð1 2 v2 =v2n Þ2 þ ð2zv=vn Þ2
These equations tell us that the mean value of the response mx is proportional to the mean value of the
force mF , and that the response spectral density SXX ðvÞ is proportional to the force spectral density
SFF ðvÞ: For both results, the proportionality constants depend on the structural or system parameters.
1
In this instance, the system parameters z and vn are deterministic and, therefore, so is the governing equation of motion.
If z and vn are either random variables or processes then the governing equation is random. The case of random parameters
is much more complicated because the system itself is random rather than just the forcing. We would have to solve the
problem for the “many randomly prescribed systems” rather than for just one system with randomly prescribed forces.


© 2005 by Taylor & Francis Group, LLC


Random Vibration

5-3

By previewing the end results, the reader will hopefully be able to better follow the mathematical
manipulations which follow.

5.2.1

Formulation

Consider the linear system defined by Equation 5.1 and assume random process input FðtÞ to be
stationary, with mean mF and power spectrum SFF ðvÞ: The stationarity assumption for the forcing
means that transient dynamic behavior cannot be directly considered here.2 Thus, the initial loading
transients of an earthquake, a wind gust, or an extreme ocean wave cannot be considered as stationary.
Assuming that the character of the loading does not change, steady-state behavior can be assumed to be
statistically stationary.

5.2.2

Derivation of Equations

Begin with the convolution equation for the deterministic response of a linear oscillator
ð1
XðtÞ ¼
gðtÞFðt 2 tÞdt
21


ð5:2Þ

where gðtÞ is the impulse response given by
gðtÞ ¼

1 2zvn t
e
sin vd t
vd

and stationary random load per unit mass FðtÞ is applied at t ¼ 21; that is, long before the present
time. This ensures that the impulse response is stationary. Beginning with Equation 5.2, take the
expected value of both sides and use the linear property of mathematical expectation to interchange it
with the integral:
ð1
ð1
gðtÞdt
E{XðtÞ} ¼
gðtÞE{Fðt 2 tÞ}dt ¼ E{FðtÞ}
21

¼ mF

ð1
21

21

gðtÞdt ¼ mF Hð0Þ


where the fact that FðtÞ is stationary was utilized in the second and third equations,3 and HðvÞ is the
frequency response function; Hð0Þ ¼ HðvÞlv¼0 : Therefore, we arrive at the first important result

mX ¼ Hð0ÞmF ¼

1
m
v2n F;

which shows us that since mF is time independent, then so must be E{xðtÞ}:
In order to derive the output spectral density, we must work through the intermediate results involving
the correlation function.

5.2.3

Response Correlations

For a stationary process, the autocorrelation function is given by
ð1
xðtÞxðt þ tÞfX ðxÞdx
RXX ðtÞ ¼
21

where fX ðxÞ is the probability density function of the process. We cannot take the Fourier transform
of this equation to find the response spectral density because we do not know the response density
2

There are, however, clever ways by which stationary solutions can be utilized in nonstationary cases. One possibility is to
multiply the stationary process by a deterministic time function such that the product is an evolutionary, or nonstationary,

process. For example, use AðtÞFðtÞ as the forcing function where AðtÞ is a deterministic transient function and FðtÞ is
stationary.
3
The force is stationary and has a constant mean value.

© 2005 by Taylor & Francis Group, LLC


5-4

Vibration and Shock Handbook

function fX ðxÞ: There are two other equivalent ways to proceed: (i) utilize the ergodic definition of the
autocorrelation,4 or (ii) use Equation 5.2, utilizing available information on FðtÞ; as follows.
First, derive the cross-correlation between FðtÞ and XðtÞ: Multiply both sides of Equation 5.2 by
Fðt 2 a1 Þ and take the expected values of both sides:
ð1
E{XðtÞFðt 2 a1 Þ} ¼
gðt1 ÞE{Fðt 2 t1 ÞFðt 2 a1 Þ}dt1
21

where E{Fðt 2 t1 ÞFðt 2 a1 Þ} ¼ RFF ðt1 2 a1 Þ and E{XðtÞFðt 2 a1 Þ} ¼ RXF ða1 Þ; the cross-correlation
between loading FðTÞ and response XðtÞ: Thus,
ð1
gðt1 ÞRFF ðt1 2 a1 Þdt1
ð5:3Þ
RXF ða1 Þ ¼
21

and RFF is known from experimental data. Next, multiply both sides of Equation 5.2 by Xðt þ a2 Þ and

take expected values of both sides:
ð1
E{Xðt þ a2 ÞXðtÞ} ¼
gðt2 ÞE{Xðt þ a2 ÞFðt 2 t2 Þ}dt2
21

RXX ða2 Þ ¼

ð1

21

gðt2 ÞRXF ðt2 þ a2 Þdt2

ð5:4Þ

Substitute Equation 5.3 into Equation 5.4 to find
ð1 ð1
gðaÞgðbÞRFF ðt þ b 2 aÞda db
RXX ðtÞ ¼
21

ð5:5Þ

21

which is a double convolution. To evaluate the variance, we use

s2X ¼ E{XðtÞ2 } 2 E2 {XðtÞ} ¼ RXX ð0Þ 2 ½Hð0ÞE{FðtÞ}


2

While this information on the correlation is of interest, a more important result is the response spectral
density that we derive in the next section.

5.2.4

Response Spectral Density

Begin with the Fourier transform relation between power spectrum and correlation function
1 ð1
SXX ðvÞ ¼
R ðtÞe2ivt dt;
2p 21 XX
and substitute Equation 5.5 for RXX ðtÞ; with l ¼ t þ b 2 a :
1 ð1 2ivt ð1 ð1
e
gðaÞgðbÞRFF ðlÞda db dt
2p 21
21 21
ð1
ð1
1 ð1
¼
gðaÞe2iva da
R ðlÞe2ivl dl
gðbÞeþivb db £
2p 21 FF
21
21

¼ HðvÞH p ðvÞSFF ðvÞ

SXX ðvÞ ¼

by definition, where p denotes complex conjugate and therefore
SXX ðvÞ ¼ lHðvÞl2 SFF ðvÞ

ð5:6Þ

This is the fundamental result for random vibration and linear systems theory that allows us to
evaluate the output spectral density, given the input spectral density and the system frequency response.
4

Rxx ðtÞ ¼ limT!1

1 ðþT
xðtÞxðt þ tÞdt:
2T 2T

© 2005 by Taylor & Francis Group, LLC


Random Vibration

5-5

It is emphasized here that the derivation of Equation 5.6 made use of the convolution equation that is
valid for linear systems and structures. Any generalization for nonlinear behavior requires problemspecific approaches.

Example 5.1


Oscillator Response to White Noise

Consider a simple application of the above ideas to an oscillator. What is the response of a damped
oscillator to a force with white noise density?

Solution
The governing equation of motion is
€ þ 2zvn XðtÞ
_ þ v2n XðtÞ ¼ FðtÞ
XðtÞ
where FðtÞ is the external force per unit mass, the system transfer function is given by
HðvÞ ¼

1
v2n þ i2zvn v þ ðivÞ2

and the squared magnitude of HðvÞ is given by
lHðvÞl2 ¼

ðv2n

1
2 v Þ þ ð2zvn vÞ2
2 2

Therefore, given any input spectral density SFF ðvÞ; the response spectral density is
SXX ðvÞ ¼ lHðvÞl2 SFF ðvÞ ¼

ðv2n


2

SFF ðvÞ
2
v Þ2 þ ð2zv

n vÞ

2

Suppose, for mathematical simplicity, that the forcing is white noise, SFF ðvÞ ¼ S0 : Then,
SXX ðvÞ ¼

S0
ðv2n 2 v2 Þ2 þ ð2zvn vÞ2

ð5:7Þ

and the mean-square response is given by
ð1
pm2 S0
pS0
E{XðtÞ2 } ¼
SXX ðvÞdv ¼
¼
kc
2v3n z
21
The mean-square response can also be written in terms of a one-sided spectrum

pm2 S0
m2 W0
E{XðtÞ2 } ¼
¼
kc
4kc
where the one-sided density W0 is related to the two-sided density by S0 ¼ pW0 =4:
This integral is not standard, but can be found in texts on random vibration.5 Even though infinite
mean-square energy is input to the system,6 it responds with finite mean-square energy. See Figure 5.2 for
plots of the components of Equation 5.7. Only the positive frequency ranges are plotted as they are
5

For example, the integral of this example problem is a specialized version of
2
ð1
B0 þ ivB1
pðA0 B21 þ A2 B20 Þ
dv ¼
2
A0 A1 A2
21 A0 þ ivA1 2 v A2

where
A0 ¼ v2n ;
6

A1 ¼ 2zvn ;

A2 ¼ 2;


B0 ¼ 1;

B1 ¼ 0:

The energy input equals to the area under the spectral density, which for white noise is
ð1
S0 dv ¼ 1:
21

© 2005 by Taylor & Francis Group, LLC


5-6

Vibration and Shock Handbook

symmetric about the abscissae. White noise is
useful and frequently used, even though it is
nonphysical, because it leads to good approximate
results.

SFF

Input Spectrum
So

ω

Example 5.2 Response to Colored
Noise

Suppose the same system as in the last
example is subjected to more complex loading,
where the spectral density of the forcing is not a
constant, but a function of v: How would the above
analysis change?

|H(iω)|2
System function

ω
Solution
ωn
The output spectral density becomes a more
complex function of frequency, for example, if
SXX
the loading density is similar to those found for
Response spectrum
wind loads. Then, the mean-square response must
be evaluated numerically.
The applied problems are always solved numerically, although hopefully after some significant
ω
analytical exposition. There are numerical
ωn
methods that are specifically geared to handling
uncertainties. Of particular note is the group
FIGURE 5.2 SFF ðvÞ; lHðivÞl2 ; SXX ðvÞ:
known as Monte Carlo methods. These methods
utilize the massive computational power available
today to account for uncertainties. This is
accomplished by generating random numbers that are used to represent random parameters. For each

of these generated random values, the program recalculates the problem. After running enough cycles to
ensure the convergence of the statistics, these numerical realizations are averaged to find the mean value
and variance of the relevant parameters.
We summarize the key results for the random vibration of a single-DoF linear oscillator in Table 5.1 to
Table 5.3. Figure 5.3 shows some of the most important correlation and spectral density pairs.

TABLE 5.1

Mean-Value Response

mX ¼ Hð0ÞmF
2
6
lHðvÞl2 ¼ 4

1=v2n
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 2 2
ð1 2 v =vn Þ þ ð2zv=vn Þ2

32
7
5

mF is known from force data

TABLE 5.2
RXX ðtÞ ¼

s2X


Output Correlation Function/Variance

Ð1 Ð1
21

21

gðaÞgðbÞRFF ðt þ b 2 aÞda db

¼ RXX ð0Þ 2 ½Hð0ÞE{FðtÞ}

2

RFF ðt þ b 2 aÞ is known from force data

© 2005 by Taylor & Francis Group, LLC


Random Vibration

5-7
TABLE 5.3

Output Spectral Density

SXX ðvÞ ¼ lHðvÞl2 SFF ðvÞ
lHðvÞl2 from structural parameters
SFF ðvÞ is the input force power spectrum


FIGURE 5.3

5.3

Correlation functions and corresponding power spectral densities.

Response to Two Random Loads

Previously, the system responses were due to a single randomly varying force. In general, the situation is
more complicated, because more than one load may act on a system and the resulting response depends
not only on the properties of each force, but also on the correlation between the two forces.
© 2005 by Taylor & Francis Group, LLC


5-8

Vibration and Shock Handbook

Consider the response of the system to two
Q(t)
P(t)
random loadings, PðtÞ and QðtÞ; acting simultaneously but at different points on the system, as
X(t)
shown in Figure 5.4. We are interested in
calculating the response statistics of the displacement XðtÞ at an arbitrary point on the system.
Assume that E{PðtÞ} ¼ 0 and E{QðtÞ} ¼ 0: Also,
by utilizing available data, we are able to estimate
FIGURE 5.4 Response to two random loads.
RPP ðtÞ and RQQ ðtÞ: Our interest is in evaluating
RXX ðtÞ and its Fourier transform SXX ðvÞ: Using linear superposition and the convolution integral, the

response due to both forces is given by
ð1
gXP ðt1 ÞPðt 2 t1 Þ þ gXQ ðt1 ÞQðt 2 t1 Þ dt1
XðtÞ ¼
21

Similarly, for Xðt þ tÞ :
Xðt þ tÞ ¼

ð1
21

gXP ðt2 ÞPðt þ t 2 t2 Þ þ gXQ ðt2 ÞQðt þ t 2 t2 Þ dt2

where gXP is the impulse response function at coordinate X due to force P and gXQ is the impulse response
function at X due to Q: Then,
ð1
gXP ðt1 ÞPðt 2 t1 Þ þ gXQ ðt1 ÞQðt 2 t1 Þ dt1
RXX ðtÞ ¼ E{XðtÞXðt þ tÞ} ¼ E
21
ð1

21

gXP ðt2 ÞPðt þ t 2 t2 Þ þ gXQ ðt2 ÞQðt þ t 2 t2 Þ dt2

Now expand the product, and then move the expectation operator to the random processes, as follows:
ð1
ð1
gXP ðt2 ÞE{Pðt 2 t1 ÞPðt þ t 2 t2 Þ}dt2 dt1

gXP ðt1 Þ
RXX ðtÞ ¼
21

21

þ
þ
þ

ð1

gXP ðt1 Þ

21

ð1

gXQ ðt1 Þ

21

ð1

gXQ ðt1 Þ

21

ð1


gXQ ðt2 ÞE{Pðt 2 t1 ÞQðt þ t 2 t2 Þ}dt2 dt1

21

ð1

gXP ðt2 ÞE{Qðt 2 t1 ÞPðt þ t 2 t2 Þ}dt2 dt1

21

ð1

gXQ ðt2 ÞE{Qðt 2 t1 ÞQðt þ t 2 t2 Þ}dt2 dt1

21

In this expression
E{Pðt 2 t1 ÞPðt þ t 2 t2 Þ} ¼ RPP ðt þ t1 2 t2 Þ
E{Qðt 2 t1 ÞQðt þ t 2 t2 Þ} ¼ RQQ ðt þ t1 2 t2 Þ
The expectations in the second and third terms are cross-correlations of the form RPQ ðtÞ ¼
E{PðtÞQðt þ tÞ}: Therefore, the autocorrelation of the response becomes
RXX ðtÞ ¼

ð1

gXP ðt1 Þ

21
ð1


þ
þ
þ

© 2005 by Taylor & Francis Group, LLC

21

ð1

21

ð1

21

ð1

gXP ðt1 Þ
gXQ ðt1 Þ
gXQ ðt1 Þ

gXP ðt2 ÞRPP ðt þ t1 2 t2 Þdt2 dt1

21
ð1

21

ð1


21

ð1

21

gXQ ðt2 ÞRPQ ðt þ t1 2 t2 Þdt2 dt1
gXP ðt2 ÞRQP ðt þ t1 2 t2 Þdt2 dt1
gXQ ðt2 ÞRQQ ðt þ t1 2 t2 Þdt2 dt1


Random Vibration

5-9

The importance of this result is primarily in the observation that we cannot derive RXX ðtÞ unless the
cross-correlations RPQ ðtÞ and RQP ðtÞ are also known. Using the Fourier transform relation between
RXX ðtÞ and SXX ðvÞ; that is
1 ð1
R ðtÞe2ivt dt
SXX ðvÞ ¼
2p 21 XX
we obtain
p
p
ðvÞHXP ðvÞSPP ðvÞ þ HXP
ðvÞHXQ ðvÞSPQ ðvÞ
SXX ðvÞ ¼ HXP


p
p
þ HXQ
ðvÞHXP ðvÞSQP ðvÞ þ HXQ
ðvÞHXQ ðvÞSQQ ðvÞ

ð5:8Þ

where
p
HXP
ðvÞHXP ðvÞ ¼ lHXP ðvÞl2
p
HXQ
ðvÞHXQ ðvÞ ¼ lHXQ ðvÞl2

and
SPQ ðvÞ ¼

1 ð1
R ðtÞe2ivt dt
2p 21 PQ

As expected, the evaluation of the output spectral density requires knowledge about the cross-spectra
SPQ ðvÞ and SQP ðvÞ: If there are more than two forces, then we will have additional cross-spectra between
each pair of forces.
Examining Equation 5.8 closely, we find that we can write SXX ðvÞ as
"
#"
#

h
i SPP ðvÞ SPQ ðvÞ
HXP ðvÞ
p
p
SXX ðvÞ ¼ HXP ðvÞ HXQ ðvÞ
ð5:9Þ
SQP ðvÞ SQQ ðvÞ
HXQ ðvÞ

Example 5.3 Conjugates of Cross Spectra
It was briefly mentioned that RPQ ðtÞ ¼ RQP ð2tÞ: How are SPQ ðvÞ and SQP ðvÞ related?
Solution
By the definition of spectral density
SPQ ðvÞ ¼
Replacing RPQ ðtÞ with RQP ð2tÞ
SPQ ðvÞ ¼

1 ð1
R ðtÞexpð2ivtÞdt
2p 21 PQ
1 ð1
R ð2tÞexpð2ivtÞdt
2p 21 QP

Letting 2t ¼ t
SPQ ðvÞ ¼

1 ð1
R ðtÞexpðivtÞdt

2p 21 PQ

Then,
SPQ ðvÞ ¼ SQP ð2vÞ ¼ SpQP ðvÞ

Example 5.4

Response Spectrum due to Two Random Loads

Consider a mass– spring–damper system in Figure 5.5 subject to two random forces PðtÞ and QðtÞ: Find
the response spectrum SXX ðvÞ assuming that
SPP ðvÞ ¼ SP ;

© 2005 by Taylor & Francis Group, LLC

SPQ ðvÞ ¼ 0;

SQQ ðvÞ ¼ SQ


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Vibration and Shock Handbook

Solution
The equation of motion for this system is given by
mX€ þ cX_ þ kX ¼ PðtÞ þ QðtÞ

X(t)
k


First, assume that QðtÞ ¼ 0 in order to first obtain
HXP ðvÞ: Taking the Fourier transform, the equation
of motion is given by

Q(t)

m
c

P(t)

ð2mv2 þ civ þ kÞXðvÞ ¼ PðvÞ
Then, the frequency response function HXP ðvÞ is
1
HXP ðvÞ ¼
2
ð2mv þ civ þ kÞ

FIGURE 5.5 Single-DoF system subjected to two
random loads.

or
1
mð2v2 þ 2vn ziv þ v2n Þ

HXP ðvÞ ¼

Similarly, HXQ ðvÞ is obtained by setting PðtÞ ¼ 0 and is also given by
1

HXQ ðvÞ ¼
2
mð2v þ 2vn ziv þ v2n Þ
Then, the spectral density of the response is given by
SXX ðvÞ ¼

h

m
2
4

v2n

2

2 v 2 2vn ziv

21

ðmðv2n 2 v2 þ 2vn zivÞÞ21

m

v2n

2

2 v 2 2vn ziv


3

21

2
i SPP ðvÞ
4
SQP ðvÞ

SPQ ðvÞ

3
5

SQQ ðvÞ

5 ¼ SPP ðvÞ þ SPQ ðvÞ þ SQP ðvÞ þ SQQ ðvÞ
m2 ðv2n 2 v2 Þ2 þ ð2vn zÞ2
ðmðv2n 2 v2 þ 2vn zivÞÞ21

In our case, the spectral density is reduced to
SXX ðvÞ ¼

SP þ SQ
m2

v2n

2


2 v2 þ 2vn z

2

Spectral densities are rather complicated expressions to evaluate in general. A number of special cases
will help better understand the effects of the cross-terms:
1. PðtÞ and QðtÞ arise from independent sources and are, therefore, uncorrelated.7 Then, RPQ ðtÞ ¼ 0;
RQP ðtÞ ¼ 0 and SPQ ðvÞ ¼ 0; SQP ðvÞ ¼ 0:
2. PðtÞ and QðtÞ are directly correlated; that is, QðtÞ ¼ kPðtÞ where k is a constant.
3. PðtÞ and QðtÞ are exponentially correlated, E{PðtÞQðt þ tÞ} ¼ kPQ exp{2 at} where kPQ is a
constant.
4. PðtÞ and QðtÞ are correlated in a “simplified” exponential; that is, with a triangular correlation
defined by E{PðtÞQðt þ tÞ} ¼ kPQ ð1 2 t=t1 Þ; 2t1 # t # t1 :

7

Independence implies that
E{Pðt1 ÞQðt2 Þ} ¼ E{Pðt1 Þ}E{Qðt2 Þ}

They are uncorrelated if
CovðPðt1 ÞQðt2 ÞÞ ¼ E{Pðt1 ÞQðt2 Þ} 2 E{Pðt1 Þ}E{Qðt2 Þ} ¼ 0
Independent processes are always uncorrelated whereas uncorrelated processes may not be independent.

© 2005 by Taylor & Francis Group, LLC


Random Vibration

5-11


We consider in more detail the first two cases listed above. Where the loads are independent, because
the cross-correlations are identically zero, the output spectral density is just the sum of the two respective
spectral densities obtained with the forces acting separately:
p
p
SXX ðvÞ ¼ HXP ðvÞHXP
ðvÞSPP ðvÞ þ HXQ ðvÞHXQ
ðvÞSQQ ðvÞ

¼ lHXP ðvÞl2 SPP ðvÞ þ lHXQ ðvÞl2 SQQ ðvÞ

ð5:10Þ

We note that the output spectral density for a linear system follows a strict interpretation of the principle
of linear superposition only when the forces are uncorrelated.
For the case where QðtÞ ¼ kPðtÞ, we have
RPQ ðtÞ ¼ E{PðtÞkPðt þ tÞ} ¼ kRPP ðtÞ
RQP ðtÞ ¼ E{kPðtÞPðt þ tÞ} ¼ kRPP ðtÞ
RQQ ðtÞ ¼ E{kPðtÞkPðt þ tÞ} ¼ k2 RPP ðtÞ
Then, we can obtain the respective spectral densities
SPQ ðvÞ ¼ SQP ðvÞ ¼ kSPP ðvÞ
SQQ ðvÞ ¼ k2 SPP ðvÞ
leading to the spectral density of the response
p
p
p
ðvÞHXP ðvÞSPP ðvÞ þ HXP
ðvÞHXQ ðvÞkSPP ðvÞ þ HXQ
ðvÞHXQ ðvÞkSPP ðvÞ
SXX ðvÞ ¼ HXP

p
p
p
þ HXQ
ðvÞHXQ ðvÞk2 SPP ðvÞ ¼ ðHXP þ kHXQ ÞðHXP
þ kHXQ
ÞSPP ðvÞ

ð5:11Þ

2

¼ lHXP ðvÞ þ kHXQ ðvÞl SPP ðvÞ
This last expression is related to the relative phase between the two functions HXP ðvÞ and HXQ ðvÞ: The
addition of two frequency response functions HXP þ kHXQ is shown graphically in Figure 5.6.
Suppose SPP ðvÞ ¼ SQQ ðvÞ ¼ SðvÞ Then, from Equation 5.10, for independent loadings
SXX ðvÞ ¼ ½lHXP l2 þ lHXQ l2 SðvÞ

ð5:12Þ

If the forces are directly correlated with parameter k ¼ 1; Equation 5.11 yields
SXX ðvÞ ¼ lHXP þ HXQ l2 SðvÞ ¼ ½lHXP l2 þ lHXQ l2 þ 2lHXP llHXQ lcos f SðvÞ

ð5:13Þ

where f is the phase difference between HXP and HXQ as shown in Figure 5.6. The law of cosines is
used for the second relation. Therefore, a comparison of Equation 5.12 with Equation 5.13
shows that the results of an uncorrelated loading
will be identical to those that are correlated where
Imaginary

cos f ¼ 0; that is, f ¼ ^p=2: This is when the two
vectors in Figure 5.6 are perpendicular to each
other. For other values of f, the spectral density in
the correlated case may have any value in the
HXP +kHXQ
range defined by ½lHXP l2 ^ lHXQ l2 SðvÞ depending
kHXQ
on the value of f:
If, at some frequency, HXP ¼ 2HXQ , the spectral
HXP
density at that frequency for the correlated case
φ
with k ¼ 1 will be zero. For any case where
HXP ¼ HXQ , the spectral density with correlation
Real
will be twice that obtained without correlation.
Another specialized result is where QðtÞ follows FIGURE 5.6 Addition of frequency-response
PðtÞ after a lag of t0 so that QðtÞ ¼ Pðt þ t0 Þ: functions.
© 2005 by Taylor & Francis Group, LLC


5-12

Vibration and Shock Handbook

Then,
RPQ ðtÞ ¼ E{PðtÞPðt þ t0 þ tÞ} ¼ RPP ðt0 þ tÞ
with the respective spectral density
1 ð1
1 ð1

SPQ ðvÞ ¼
RPQ ðtÞe2ivt dt ¼
R ðt þ tÞe2ivt dt
2p 21
2p 21 PP 0
1 ivt0 ð1
RPP ðt0 þ tÞe2ivðt0 þtÞ dðt0 þ tÞ ¼ eivt0 SPP ðvÞ
¼
e
2p
21
Since SPQ ðvÞ ¼ SQP ðvÞ are complex conjugates
SQP ðvÞ ¼ e2ivt0 SPP ðvÞ
Also, SPP ðvÞ ¼ SQQ ðvÞ: Then,
p
p
p
p
SXX ðvÞ ¼ HXP ðvÞHXP
ðvÞ þ eivt0 HXP ðvÞHXQ
ðvÞ þ e2ivt0 HXQ ðvÞHXP
ðvÞ þ HXQ ðvÞHXQ
ðvÞ SPP ðvÞ

5.4

Multi-Degree-of-Freedom Vibration

In this section, a review of deterministic vibration is provided. These results are used in the next section,
where the forcing is taken to be a random process.


5.4.1

Deterministic Vibration

Consider a case where there is more than one DoF.
The equations of motion can be written as

F1(t)
c1

½m {€xðtÞ} þ ½c {_xðtÞ} þ ½k {xðtÞ} ¼ {FðtÞ}

F2(t)

x1(t)

x2(t)

c2

m2
For an N-DoF system, the matrices ½m ; ½c and ½k
k1
k2
are of dimension N £ N; and the response {xðtÞ}
and force {FðtÞ} vectors are of dimension N £ 1:
For purposes of demonstration and discussion,
FIGURE 5.7 Two-DoF mass –spring –damper system.
the necessary concepts will be introduced primarily by working through the solution of a two-DoF

system. All these ideas transfer to larger systems, but with the two-DoF model we can demonstrate the
key ideas without the complications of the major algebraic and numerical demands made by the larger
systems.
For the system shown in Figure 5.7, we can derive the coupled equations of motion using either
Newton’s Second Law of motion applied to a free body diagram for each mass, as shown in Figure 5.8, or
by Lagrange’s equation. In either case, we find the governing equations to be
m1

m1 x€ 1 ðtÞþðc1 þ c2 Þ_x1 ðtÞ þ ðk1 þ k2 Þx1 ðtÞ 2 c2 x_ 2 ðtÞ 2 k2 x2 ðtÞ ¼ F1 ðtÞ;
m2 x€ 2 ðtÞ þ c2 x_ 2 ðtÞ 2 c2 x_ 1 ðtÞ 2 k2 x1 ðtÞ þ k2 x2 ðtÞ ¼ F2 ðtÞ
or in matrix form
"

m1

0

0

m2

#(

x€ 1 ðtÞ
x€ 2 ðtÞ

"

)
þ


c1 þ c2

2c2

2c2

c2

© 2005 by Taylor & Francis Group, LLC

#(

x_ 1 ðtÞ
x_ 2 ðtÞ

"

)
þ

k1 þ k2

2k2

2k2

k2

#(


x1 ðtÞ
x2 ðtÞ

)

(
¼

F1 ðtÞ
F2 ðtÞ

)
ð5:14Þ


Random Vibration

5-13

F1(t)

k1 x1
c1 x1

m1

FIGURE 5.8

F2(t)


x1(t)
k2(x2−x1)

x2(t)

m2

c2(x2−x1)

Free body diagram of a two-DoF system.

In this case, the mass, damping, and stiffness matrices are given by
"
"
#
"
#
k1 þ k2
c1 þ c2 2c2
m1 0
;
½k ¼
;
½c ¼
½m ¼
2k2
2c2
c2
0 m2


5.4.2

2k2

#

k2

Solution by Frequency Response Function

Begin by taking the Fourier transform of the equations of motion, obtaining
ð2v2 ½m þ iv½c þ ½k Þ{XðvÞ} ¼ {FðvÞ}
where {XðvÞ} and {FðvÞ} are Fourier transforms of {xðtÞ} and {FðtÞ}; respectively. The matrix
ð2v2 ½m þ iv½c þ ½k Þ is denoted as ½ZðvÞ : Then,
{XðvÞ} ¼ ½ZðvÞ
The matrix ½ZðvÞ

21

21

{FðvÞ}

is identical to the frequency response matrix denoted as ½HðvÞ :
{XðvÞ} ¼ ½HðvÞ {FðvÞ}

For the two-DoF system in Figure 5.7
"
2m2 v2 þ ivc2 þ k2

1
½HðvÞ ¼
det½Z
ivc2 þ k2

ð5:15Þ
ivc2 þ k2

#

2m1 v2 þ ivðc1 þ c2 Þ þ ðk1 þ k2 Þ

det½Z ¼ v4 m1 m2 2 iv3 ðm1 c2 þ c1 m2 þ c2 m2 Þ þ ivðc1 k2 þ k1 c2 Þ
2 v2 ðm1 k2 þ c1 c2 þ k1 m2 þ k2 m2 Þ þ k1 k2
Note that the frequency response function of a two-DoF system is given as a matrix with dimension of
2 £ 2. To clarify the meaning of each element, let us expand Equation 5.15:
X1 ðvÞ ¼ H11 ðvÞF1 ðvÞ þ H12 ðvÞF2 ðvÞ
X2 ðvÞ ¼ H21 ðvÞF1 ðvÞ þ H22 ðvÞF2 ðvÞ
Now, the meaning of each element is clear. Each element Hij ðvÞ is the frequency response function for
coordinate i due to a force at j: In general, ½HðvÞ is a symmetric matrix since ½m ; ½c and ½k are
symmetric.
For a deterministic response, recall that the impulse response function is related to the frequency
response function by
1 ð1
H ðvÞexpðivtÞdv
gij ðtÞ ¼
2p 21 ij

© 2005 by Taylor & Francis Group, LLC



5-14

Vibration and Shock Handbook

which is written for each element in the matrix. For the two-DoF system
1 ð1 2m2 v2 þ ivc2 þ k2
g11 ðtÞ ¼
expðivtÞdv
2p 21
det½Z
1 ð1 ivc2 þ k2
expðivtÞdv
g12 ðtÞ ¼
2p 21 det½Z
1 ð1 ivc2 þ k2
expðivtÞdv ¼ g12 ðtÞ
g21 ðtÞ ¼
2p 21 det½Z
1 ð1 2m1 v2 þ ivðc1 þ c2 Þ þ ðk1 þ k2 Þ
g22 ðtÞ ¼
expðivtÞdv
2p 21
det½Z
Again, ½gðtÞ is a symmetric matrix.
Using the convolution integral, the response xi ðtÞ due to Fj is
ð1
gij ðtÞFj ðt 2 tÞdt
xi ðtÞ ¼
21


The total response xi ðtÞ of mass mi is then equal to the sum of the individual responses to each of the
forces. In our case, the responses are
ð1
x1 ðtÞ ¼
½g11 ðtÞF1 ðt 2 tÞ þ g12 ðtÞF2 ðt 2 tÞ dt
21

x2 ðtÞ ¼

ð1

21

½g21 ðtÞF1 ðt 2 tÞ þ g22 ðtÞF2 ðt 2 tÞ dt

We can extend our analysis to a system with N DoF. The equations of motion are written in matrix form
as
½m {€xðtÞ} þ ½c {_xðtÞ} þ ½k {xðtÞ} ¼ {FðtÞ}
where {xðtÞ} and {FðtÞ} are response and force vectors and the frequency response matrix is given by
½HðvÞ ¼ ð2v2 ½m þ iv½c þ ½k Þ21
The impulse response matrix ½gðtÞ is given by
½gðtÞ ¼

1 ð1
½HðvÞ expðivtÞdv
2p 21

and the response matrix is given by
{xðtÞ} ¼


ð1
21

½gðtÞ {Fðt 2 tÞ}dt

Note that the impulse response vector is not trivial to evaluate, much less the convolution integrals. More
importantly, this method becomes much harder for each additional DoF. Therefore, this method is rarely
used, and we often rely on modal analysis.

5.4.3

Modal Analysis

Let us first consider an undamped system in free vibration with equations of motion
½m {€xðtÞ} þ ½k {xðtÞ} ¼ 0
If we assume that {xðtÞ} is harmonic with frequency v;
{xðtÞ} ¼ A0 {u}eivt
the equations of motion are reduced to the eigenvalue problem given by
ð2v2 ½m þ ½k Þ{u} ¼ {0}

© 2005 by Taylor & Francis Group, LLC

ð5:16Þ


Random Vibration

5-15


The goal is to find the nonzero vector {u}: Therefore, equations must be linearly dependent,8 or there
should be no inverse for ð2v2 ½m þ ½k Þ: This can be satisfied if the determinant of ð2v2 ½m þ ½k Þ is
zero:

l 2 v2 ½m þ ½k l ¼ 0
There are several values of v2 which can make the determinant equal zero, and these values of v2 are
called eigenvalues. For each nonrepeating eigenvalue, there is a corresponding vector {u} that satisfies
Equation 5.16. This vector is called the eigenvector. Physically, the eigenvalues of this problem are the
natural frequencies squared, and the eigenvectors are the mode shapes, since they represent the respective
motion of each modal coordinate.
By solving this eigenvalue problem, we can obtain N sets of eigenvalues v2i with corresponding
eigenvectors {u}i for an N-DoF system.9 It is customary that the eigenvalues are arranged in ascending
order

v21 , v22 , · · · , v2N
The eigenvectors can be normalized with respect to the mass matrix,
{u}Ti ½m {u}i ¼ 1;
where

i ¼ 1; 2; …; N

9
8
u1i >
>
>
>
>
>
=

<
..
{u}i ¼
.
>
>
>
>
>
>
;
:
uNi

For the ith and jth set of eigenvalues and eigenvectors, we can write
2v2i ½m {u}i þ ½k {u}i ¼ {0}
2v2j ½m {u}j þ ½k {u}j ¼ {0}
Multiplying the first equation by {u}Tj and the second equation by {u}Ti ; we obtain
2v2i {u}Tj ½m {u}i þ {u}Tj ½k {u}i ¼ {0}

ð5:17Þ

2v2j {u}Ti ½m {u}j þ {u}Ti ½k {u}j ¼ {0}

ð5:18Þ

Take the transpose of Equation 5.18,
2v2j {u}Tj ½m T {u}i þ {u}Tj ½k T {u}i ¼ {0}
Since ½m and ½k are symmetric, ½m ¼ ½m


T

and ½k ¼ ½k T ; and

2v2j {u}Tj ½m {u}i þ {u}Tj ½k {u}i ¼ {0}
8

An example of a linear dependent equation is

or

"

1

2

2

4

#(

u1
u2

)
¼

ð5:19Þ


( )
0
0

u1 þ 2u2 ¼ 0
2u1 þ 4u2 ¼ 0
In this case, u1 and u2 need not be zeros. Instead, the equations are satisfied as long as u1 ¼ 2u2 : Therefore, there is an infinite
number of solutions.
9
An eigenvalue problem will be demonstrated in the next example.

© 2005 by Taylor & Francis Group, LLC


5-16

Vibration and Shock Handbook

Subtracting Equation 5.19 from Equation 5.17, we obtain
ðv2j 2 v2i Þ{u}Tj ½m {u}i ¼ 0
If the eigenvalues are unique ðv2j – v2i Þ; then
{u}Tj ½m {u}i ¼ 0;

i–j

{u}Tj ½k {u}i ¼ 0;

i–j


From Equation 5.19, similarly

That is, the eigenvectors are orthogonal (or normal) to each other with respect to the mass and the stiffness
matrices. If the eigenvectors are normalized with respect to the mass matrix, Equation 5.19 with i ¼ j can
be written as
2v2i {u}Ti ½m {u}i þ {u}Ti ½k {u}i ¼ {0}
and therefore
{u}Ti ½k {u}i ¼ v2i
In summary, we just found that the eigenvectors have the following properties:
(
1; i ¼ j
{u}Ti ½m {u}j ¼
0; i – j
and

(
{u}Ti ½k

{u}j ¼

v2i ; i ¼ j
0;

i–j

If we construct a matrix composed of eigenvectors
½P ¼ ½ {u}1

···


{u}N

ð5:20Þ

we can write
½p T ½m ½p ¼ ½I
2 2
3
0
v1 0
6
7
6
7
..
7 ¼ diagðv2 Þ
½p T ½k ½p ¼ 6
6 0
7
.
0
4
5
2
0
0 vN
where ½I is the identity matrix and [P] is called the modal matrix.
Now, let us return to our original equations of motion given by
½m {€xðtÞ} þ ½c {_xðtÞ} þ ½k {xðtÞ} ¼ {FðtÞ}
We restrict our problem to the proportional damping case, where ½c is a linear combination of ½m

and ½k :
½c ¼ a½m þ b½k
with constant a and b:
Let us define a new set of coordinates {zðtÞ}; so that
{xðtÞ} ¼ ½P {zðtÞ}
where ½p is the modal matrix defined in Equation 5.20. The equations of motion then become
½m ½P {€zðtÞ} þ ½c ½P {_zðtÞ} þ ½k ½P {zðtÞ} ¼ {FðtÞ}

© 2005 by Taylor & Francis Group, LLC

ð5:21Þ


Random Vibration

5-17

Multiplying by ½P T ;
½P T ½m ½P {€zðtÞ} þ ½P T ½c ½P {_zðtÞ} þ ½P T ½k ½P {zðtÞ} ¼ ½P T {FðtÞ}
which reduces to
{€zðtÞ} þ ða½I þ b £ diagðv2 ÞÞ{_zðtÞ} þ diagðv2 Þ{zðtÞ} ¼ ½P T {FðtÞ}
or
z€1 ðtÞ þ a þ bv21 z_1 ðtÞ þ v21 zðtÞ ¼ {u}T1 {FðtÞ}
..
.
z€N ðtÞ þ a þ bv2N z_N ðtÞ þ v2N zðtÞ ¼ {u}TN {FðtÞ}
It is clear now why proportional damping is considered. The damping matrix ½c becomes diagonalized
so that the equations of motion are decoupled. The coordinates zi ðtÞ are called the modal coordinates, and
{u}Ti {FðtÞ} are called the modal forces. We often let


a þ bv2i ¼ 2vi zi
{u}Ti {FðtÞ} ¼ qi ðtÞ
to simplify the equations to
z€1 ðtÞ þ 2v1 z1 z_1 ðtÞ þ v21 z1 ðtÞ ¼ q1 ðtÞ
..
.

ð5:22Þ

z€N ðtÞ þ 2vN zN z_N ðtÞ þ v2N zðtÞ ¼ qN ðtÞ
The solution to these equations can be obtained using the impulse response function and the convolution
integral for each coordinate
ð1
zi ðtÞ ¼
gi ðtÞgi ðt 2 tÞdt; i ¼ 1; 2; …; N
21

where

8
1 2zvi t
>
<
e
sin vdi t; t $ 0
gi ðtÞ ¼ vdi
>
:
0;
t,0


and

qffiffiffiffiffiffiffiffi
vdi ¼ vi 1 2 z2i

Once we find all the zi ðtÞ; we recover the physical coordinates by using the transformation in Equation
5.21 or
{xðtÞ} ¼ ½P {zðtÞ}

5.5

Multi-Degree-of-Freedom: The Response to Random Loads

From the analysis of the Single-DoF system, we know that knowledge of the frequency response function
HðvÞ is crucial in obtaining the response statistics to random loads. For a system with more than one
DoF, there is more than one frequency response function HðvÞ: For an N-DoF system, the frequency
response functions will be given as an N £ N symmetric matrix HðtÞ:
© 2005 by Taylor & Francis Group, LLC


5-18

Vibration and Shock Handbook

In the single-DoF system, the response statistics were given in terms of the mean response mX and
the spectral density of the response SXX ðvÞ: We will find here that the response statistics of an N-DoF
system are given in terms of the mean response for each DoF, mX1 ; …; mXN ; and a matrix of spectral
densities of the response SðvÞ; which is also of dimension N £ N:
We present a priori the results which are derived in the subsequent section; the mean response vector

mX and the matrix spectral density of the output SXX ðvÞ:
mX ¼ HXX ð0ÞmF
SXX ðvÞ ¼ HpXX ðvÞSFF ðvÞHTXX ðvÞ
Here, we will first obtain the frequency response function of a typical two-DoF system. We will then
consider a two-DoF system subjected to a single random load. Finally, we will expand our analysis to an
N-DoF system subjected to N random loads.

5.5.1

Response due to a Single Random Force

When a body vibrates, points on the body move relative to each other. To describe this relative
motion of two points on a body subjected to random loading, we must know both their crossspectral densities and cross-correlations, and also the spectral densities and the autocorrelation
function of the individual motions of each point. This requirement is true even though there is
only one force acting on the system. This is similar to the problem of the single mass loaded by
two random loads.
Consider a body excited by a single randomly varying force PðtÞ with autocorrelation RTT ðtÞ and
spectral density SPP ðvÞ. Owing to this loading, two points on the body have displacements XðtÞ and YðtÞ
with means mX and mY ; autocorrelations RXX ðtÞ and RYY ðtÞ; and spectral densities SXX ðvÞ and SYY ðvÞ;
respectively. Note again that upper case letters are used to signify random processes.
We know the following from our earlier studies
E{XðtÞ} ¼ HXP ð0ÞmP
E{YðtÞ} ¼ HYP ð0ÞmP
and
SXX ðvÞ ¼ lHXP ðvÞl2 SPP ðvÞ

ð5:23Þ

SYY ðvÞ ¼ lHYP ðvÞl2 SPP ðvÞ


ð5:24Þ

where
HXP ðvÞ ¼
HYP ðvÞ ¼

ð1
21

ð1
21

gXP ðtÞe2ivt dt
gYP ðtÞe2ivt dt

and
XðtÞ ¼
YðtÞ ¼

© 2005 by Taylor & Francis Group, LLC

ð1
21

ð1
21

gXP ðtÞPðt 2 tÞdt
gYP ðtÞPðt 2 tÞdt



Random Vibration

5-19

With these relations, we can now find the expressions for the cross-correlation RXY ðtÞ and cross-spectral
density SXY ðvÞ: Using the definition of a cross-correlation function, we obtain
RXY ðtÞ ¼ E{XðtÞYðt þ tÞ} ¼ E
¼
¼

ð1
21

ð1
21

gXP ðt1 Þ
gXP ðt1 Þ

ð1
21

ð1
21

ð1
21

gXP ðt1 ÞPðt 2 t1 Þdt1


ð1
21

gYP ðt2 ÞPðt þ t 2 t2 Þdt2

gYP ðt2 ÞE{Pðt 2 t1 ÞPðt þ t 2 t2 Þ}dt2 dt1
gYP ðt2 ÞRPP ðt þ t1 2 t2 Þdt2 dt1

Even though we have an explicit expression, it is generally not easy to interpret.
The cross-spectral density can be derived by using the Fourier transform relation between RXY ðtÞ and
SXY ðvÞ: Thus,
ð1
1 ð1
1 ð1 ð1
gYP ðt2 ÞRPP ðt þ t1 2 t2 Þdt2 dt1 e2ivt dt
gXP ðt1 Þ
RXY ðtÞe2ivt dt ¼
2p 21
2p 21 21
21
ð1
ð1
1 ð1
gXP ðt1 Þeþivt1
¼
gYP ðt2 Þe2ivt2
RPP ðt þ t1 2 t2 Þe2ivðtþt1 2t2 Þ dt dt2 dt1
ð5:25Þ
2p 21

21
21

SXY ðvÞ¼

In order to integrate this last expression, it is easier to proceed by transforming the variables in the
innermost integral according to

l ; t þ t1 2 t2 and dl ¼dt
where t1 and t2 are dummy variables. Then, Equation 5.25 becomes
SXY ðvÞ¼

ð1
ð1
1 ð1
gXP ðt1 Þeþivt1 dt1
gYP ðt2 Þe2ivt2 dt2
RPP ðlÞe2ivl dl
2p 21
21
21

p
ðvÞHYP ðvÞSPP ðvÞ
¼HXP

ð5:26Þ

Where points X and Y coincide, this result reduces to the classical fundamental relation
SXX ðvÞ¼lHXP l2 SPP ðvÞ:

Equation 5.23, Equation 5.24 and Equation 5.26 can be written in matrix form
"

or

"

SXX ðvÞ

SXY ðvÞ

SYX ðvÞ

SYY ðvÞ

SXX ðvÞ SXY ðvÞ
SYX ðvÞ

SYY ðvÞ

#

"
¼

"

#
¼


p
ðvÞ
HXP
p
ðvÞ
HYP

#
SPP ðvÞ HXP ðvÞ

HYP ðvÞ

p
p
ðvÞSXX ðvÞHXP ðvÞ HXP
ðvÞSXY ðvÞHYP ðvÞ
HXP
p
ðvÞSYX ðvÞHXP ðvÞ
HYP

ð5:27Þ

#

p
ðvÞSYY ðvÞHYP ðvÞ
HYP

Example 5.5 Response Spectra for a Two-Degree-of-Freedom System Subjected

to a Single Random Force
Consider the mass–spring –damper system shown in Figure 5.9. Assume that the random force F1 ðtÞ is a
white noise random process with SF1 F1 ðvÞ ¼ S0 and F2 ðtÞ ¼ 0: (i) Find the frequency response functions
H11 ðvÞ and H21 ðvÞ; and (ii) find the spectral densities SX1 X1 ðvÞ; SX1 X2 ðvÞ and SX2 X2 ðvÞ:
© 2005 by Taylor & Francis Group, LLC


5-20

Vibration and Shock Handbook

F1(t)
X1(t)
c1

X2(t)

c2
m2

m1
k1

FIGURE 5.9

k2

Two-DoF system excited by a single random force.

Solution

The equation of motion for this system has been obtained as Equation 5.14 with F2 ðtÞ ¼ 0: We found
previously that the frequency response function for this system is given by
"
#
2m2 v2 þ ivc2 þ k2
ivc2 þ k2
1
½HðvÞ ¼
det½Z
ivc2 þ k2
2m1 v2 þ ivðc1 þ c2 Þ þ ðk1 þ k2 Þ
where
det½Z ¼ v4 m1 m2 2 iv3 ðm1 c2 þ c1 m2 þ c2 m2 Þ þ ivðc1 k2 þ k1 c2 Þ
2 v2 ðm1 k2 þ c1 c2 þ k1 m2 þ k2 m2 Þ þ k1 k2
Since F2 ¼ 0; the frequency response functions due to F2 ; H12 ðvÞ and H22 ðvÞ; are zero. The frequency
response functions due to F1 are given by
H11 ðvÞ ¼

2m2 v2 þ ivc2 þ k2
det½Z

H21 ðvÞ ¼

ivc2 þ k2
det½Z

Using Equation 5.27, the response spectra are given by
p
ðvÞH11 ðvÞSF1 F1 ðvÞ ¼ S0
SX1 X1 ðvÞ ¼ H11


ðk2 2 m2 v2 Þ2 þ ðvc2 Þ2
ðdet½Z Þ2

p
ðvÞH22 ðvÞSF1 F1 ðvÞ ¼ S0
SX2 X2 ðvÞ ¼ H22

k22 þ ðvc2 Þ2
ðdet½Z Þ2

p
ðvÞH21 ðvÞSF1 F1 ðvÞ ¼ S0
SX1 X2 ðvÞ ¼ H11

ð2m2 v2 2 ivc2 þ k2 Þðivc2 þ k2 Þ
ðdet½Z Þ2

p
ðvÞH11 ðvÞSF1 F1 ðvÞ ¼ S0
SX2 X1 ðvÞ ¼ H21

ð2m2 v2 þ ivc2 þ k2 Þð2ivc2 þ k2 Þ
ðdet½Z Þ2

Note that SX2 X1 ðvÞ ¼ SX1 X2 ð2vÞ ¼ SpX1 X2 ðvÞ:

5.5.2

Response to Multiple Random Forces


Consider a multi-DoF system subjected to multiple random forces. The goal is to obtain the relation
equivalent to Equation 5.27 for the general response for N DoF by first considering the response of a
Single-DoF to N forces, and then generalizing to N DoF. This will be done in two ways. First, by extending
the impulse-response method leading to the convolution integral, and then by utilizing the modal
© 2005 by Taylor & Francis Group, LLC


Random Vibration

5-21

analysis approach where the coupled physical equations of motion are decoupled in a new modal
coordinate system.

5.5.3

Impulse-Response Approach

Let us start by expressing the response Xk ðtÞ using the impulse-response functions gki ðtÞ: Recall that gki ðtÞ
is the impulse response of mass mk to force Fi ðtÞ: By linear superposition, the total response Xk ðtÞ of mass
mk is equal to the sum of the individual responses to each of the N forces
Xk ðtÞ ¼

N
X
i¼1

Xki ðtÞ ¼


N ð1
X
i¼1

21

gki ðtÞFi ðt 2 tÞdt

Assuming that the forces are stationary with respective means mFi and cross-correlations RFi Fj ðtÞ; the
mean and correlation of the response can be found as follows:
N ð1
X

E{Xk ðtÞ} ¼

21

i¼1
N
X

mXk ¼

i¼1

mFi

gki ðtÞE{Fi ðt 2 tÞ}dt

ð1

21

gki ðtÞdt ¼

N
X
i¼1

mFi Hki ð0Þ

In matrix form
mX ¼ HXX ð0ÞmF
This is a static response and, therefore, can be ignored here and added on at the end of the computations.
We next evaluate the expressions for the response correlations and spectral densities, from which we can
evaluate the mean-square response. By definition, the correlations are given by
( N
)
N
X
X
Xkm ðtÞ
Xjn ðt þ tÞ
RXk Xj ðtÞ ¼ E{Xk ðtÞXj ðt þ tÞ} ¼ E
(
¼E

m¼1

N ð1
X


m¼1

¼

gkm ðz ÞFm ðt 2 z Þdz

N X
N ð1 ð1
X
m¼1 n¼1

¼

21

21

21

N X
N ð1 ð1
X
m¼1 n¼1

21

21

n¼1


N ð1
X
n¼1

21

)
gjn ðj ÞFn ðt þ t 2 j Þdj

gkm ðz Þgjn ðj ÞE{Fm ðt 2 z ÞFn ðt þ t 2 j Þ}dz dj
gkm ðz Þgjn ðj ÞRFm Fn ðt 2 j þ z Þdz dj

The response spectral density is by definition
SXk Xj ðvÞ ¼

1 ð1
R ðtÞexpð2ivtÞdt
2p 21 Xk Xj

Before substituting the correlation function into this equation, multiply it by expð2iv½z 2 j Þ
£ expðiv½z 2 j Þ: Also, define n ¼ t 2 j þ z with dn ¼ dt: All of these manipulations allow us to put
the spectral density in the following form:
SXk Xj ðvÞ ¼
¼

N X
N ð1
X
m¼1 n¼1

N X
N
X
m¼1 n¼1

21

gkm ðzÞexpðivzÞdz

p
ðvÞHjn ðvÞSFm Fn ðvÞ
Hkm

© 2005 by Taylor & Francis Group, LLC

ð1
21

gjn ðjÞexpð2ivjÞdj £

1 ð1
ðnÞexpð2ivnÞdn
R
2p 21 Fm Fn


5-22

Vibration and Shock Handbook


where the star denotes complex conjugate. In matrix form
SXk Xj ðvÞ ¼ Hpk ðvÞSFF ðvÞHTj ðvÞ

ð5:28Þ

In this notation, Hpk ðvÞ is a row vector of dimension 1 £ N
p
Hpk ðvÞ ¼ Hk1
ðvÞ · · ·

HjT ðvÞ is a column vector of dimension N £ 1

2

6
6
HTj ðvÞ ¼ 6
6
4
and SFF ðvÞ is a matrix of dimension N £ N
2
6
6
SFF ðvÞ ¼ 6
6
4

p
ðvÞ
HkN


Hj1 ðvÞ
..
.
HjN ðvÞ

3
7
7
7
7
5

SF1 F1 ðvÞ

···

SF1 FN ðvÞ

..
.

..

..
.

SFN F1 ðvÞ

.


· · · SFN FN ðvÞ

3
7
7
7
7
5

Equation 5.28 can now be generalized for any Xk and Xj
SXX ðvÞ ¼ Hp ðvÞSFF ðvÞHT ðvÞ
where

2
6
6
HðvÞ ¼ 6
4

and

2
6
6
SXX ðvÞ ¼ 6
6
4

H11 ðvÞ


···

H1N ðvÞ

..
.

..

..
.

HN1 ðvÞ

···

.

ð5:29Þ
3
7
7
7
5

HNN ðvÞ

SX1 X1 ðvÞ


···

SX1 XN ðvÞ

..
.

..

..
.

SXN X1 ðvÞ

.

· · · SXN XN ðvÞ

3
7
7
7
7
5

Let us recall Equation 5.6, Equation 5.9, and Equation 5.27. Equation 5.6 is the response spectral density
of a single-DoF system to a single random force, Equation 5.9 is the response spectral density of a singleDoF system to two random forces, and Equation 5.27 is the response spectral densities of a two-DoF
system due to a single force. Comparing them with Equation 5.29, we realize that they are all special cases
of this equation.
For a zero mean, the mean-square response of the jth coordinate is given by the relation

ð1
SXj Xj ðvÞdv
s2Xj ¼
ð5:30Þ
21

Example 5.6

Spacecraft Design Problem (based on Wirsching et al., p. 228)

A rocket mounting is to be designed. The rocket of mass m1 ¼ 181 kg is mounted on a supporting
structure of mass m2 ¼ 362 kg using structural brackets that can be modeled as springs of constant
k ¼ 1:40 MN/m. There is also damping with coefficient c ¼ 3:50 kNsec/m. When the engine fires,
the thrust can be interpreted as being composed of a mean value thrust and a random fluctuating thrust.
© 2005 by Taylor & Francis Group, LLC


Random Vibration

5-23

The random variations are due to factors such as fluctuations in fuel mix and eccentricities in thrust
direction. The mean thrust is given by mT ¼ 16 kN and the random component is given by T, which is
a random function of time. T is characterized as white noise of intensity 1980 N2/Hz over the frequency
range 0 –50 Hz. The mean thrust can be viewed as a static load while the random thrust as
the dynamic load.
The design must consider that components in the supporting structure will be damaged if the
acceleration is too large. Also, the structural brackets will fail if the relative displacement between the
rocket and the supporting structure is too large. Both these design issues are difficult because the input is
random. Therefore, there is a trade-off between the cost of making a component stronger and making it

more reliable. The probabilistic framework allows the analyst to quantitatively understand the trade-off.

Solution
The goal of the analysis is to determine the acceleration of the supporting mass X€ 2 and the relative
displacement Z ¼ X2 2 X1 :
First we solve the mean value problem. The mean value acceleration is
a¼

F
16;000
29:46
¼
¼ 29:46 m=sec2 ¼
¼ 3:00g
m1 þ m2
181 þ 362
9:81

The mean force in the bracket is found by applying the equation of motion to m2

mFb ¼ m2 a2 ¼ 362 £ 29:46 ¼ 10665 N ¼ 10:66 kN
The mean bracket deflection is

mZ ¼

mFb
10:66 £ 103
¼
¼ 7:61 £ 1023 m ¼ 7:61 mm
k

1:40 £ 106

While the mean value is the single most important descriptor of randomness, the variance is important
because it is a measure of the scatter of possible results. In order to calculate the variance, we need to
derive the response spectral density and the frequency response function (see Equation 5.30). This we
compute next. The equations of motion are given by
m1 X€ 1 2 kðX2 2 X1 Þ 2 cðX_ 2 2 X_ 1 Þ ¼ FðtÞ
m2 X€ 2 þ kðX2 2 X1 Þ þ cðX_ 2 2 X_ 1 Þ ¼ 0
Given the problem statement, the variables of interest are ðZ; X2 Þ: Transform variables using
Z ¼ X2 2 X1 : The equations of motion become
m1 X€ 2 2 m1 Z€ 2 kZ 2 cZ_ ¼ FðtÞ
m2 X€ 2 þ kZ þ cZ_ ¼ 0
To derive the frequency response function, assume harmonic excitation
" #
F0 i vt
FðtÞ ¼
e
0
and harmonic response of the form

"

ZðtÞ
X€ 2

#

"
¼


Z0
X€ 0

#
eivt

Substitute these equations into the matrix equation of motion to find
m1 X€ 0 þ m1 v2 Z0 2 kZ0 2 icvZ0 ¼ F0
m2 X€ 0 þ kZ0 þ icvZ0 ¼ 0

© 2005 by Taylor & Francis Group, LLC


5-24

Vibration and Shock Handbook

The frequency response function is then found to be
"
#
m2
2m1
1
HðvÞ ¼
DðvÞ 2k 2 icv m1 v2 2 k 2 icv

DðvÞ ¼ m1 m2 v2 2 kðm1 þ m2 Þ 2 icvðm1 þ m2 Þ
There is only one applied force, and the spectral density function matrix is
"
#

S0 0
SF ðvÞ ¼
0 0
Apply Equation 5.29 to find
"
SXX ðvÞ ¼

SZZ

SZ X€

SXZ
€

SX€ X€

#

S0
¼
lDðvÞl2

"

m22

2km2 þ icm2 v

2km2 þ icm2 v


k2 þ ðcvÞ2

#

To determine the root mean square of relative displacement ZðtÞ; note that the spectral density function
of ZðtÞ is the top term on the diagonal of SXX ðvÞ:
SZZ ðvÞ ¼ S0

m22
lDðvÞl2

The mean-square value of ZðtÞ is then

s2Z ¼

ð1
21

SZZ ðvÞ ¼

pS0 m22
ckðm1 þ m2 Þ2

The above are for a two-sided spectrum. The given spectrum is one-sided, therefore, S0 ¼ W0 =4p: Then,
with appropriate substitutions, we find sZ . 0:212 mm: In summary, the relative displacement has
mean and root mean-square

mZ ¼ 7:61 mm;

sZ ¼ 0:212 mm


The force in the spring has mean and root mean-square

mFb ¼ kmZ ¼ 1:40 £ 106 £ 7:61 ¼ 10:65 kN
sFb ¼ ksZ ¼ 1:40 £ 106 £ 0:212 ¼ 0:297 kN
Finally, the spectral density of the acceleration is the second diagonal term of SX ðvÞ:
SX€ X€ ¼ S0

s2X€ ¼

k2 þ ðcvÞ2

ð1

lDðvÞl2

21

SX€ X€ ðvÞdv

and

s2X€ ¼ pS0

k
c
þ
cðm1 þ m2 Þ2
ðm1 þ m2 Þm1 m2


Upon making the appropriate substitutions, we find

sX€ ¼ 0:852 m=sec2 ¼ 0:087g
In summary, the mean and standard deviation of X€ 2 are
mX€ 2 ¼ 3g;

sX€ 2 ¼ 0:087g

See Example 5.1 for a discussion on the above nonstandard integrals.
A designer would now use these mean and standard deviation numbers to decide how reliable to make
the components. For example, if we assume that Fb and X€ 2 are normal density, we can specify a design based
© 2005 by Taylor & Francis Group, LLC


Random Vibration

5-25

fX (x)

on m ^ 3s, which tells us the probability of
exceeding this 6s range is less than 0.1%. If this
design is too expensive, then the designer will try a
design based on m ^ 3s; one that will have a higher
probability of failure. In this way, a trade-off study is
performed weighing cost against probability of
failure. Figure 5.10 demonstrates a design based on
5% failure probability.

0.95


mX −1.96σX mX
FIGURE 5.10
probability.

5.5.4

mX + 1.96σX

Normal density with 5% failure

Modal Analysis Approach

The response of a multi-DoF discrete system loaded by random forces is generally a very complicated
problem to formulate and solve. Here, to demonstrate one possible approach, we study the modal
analysis of such a structure and work through some of the intricacies.
We will start the analysis at the point where the modal equations of motion have been formulated and
the assumption of proportional damping has been made. From Equation 5.22, for a random vibration,
the modal equations in indicial notation are
Z€ i þ 2zi vi Z_ i þ v2i Zi ¼ Qi ðtÞ;

i ¼ 1; 2; …; N

ð5:31Þ

where the parameters are familiar. We further assume that the modal forces Qi ðtÞ are ergodic random
excitations. The transformation between physical and modal spaces is
Xj ðtÞ ¼

N

X
i¼1

uji Zi ðtÞ

and the transformation between physical and modal forces is
Qj ðtÞ ¼

N
X
i¼1

where ui is the modal vector for the ith DoF and

uij Fi ðtÞ ¼ uTj F

2

u1i

ð5:32Þ

3

6
7
6
7
ui ¼ 6 ... 7
4

5
uNi
In matrix form, we can write
X ¼ PZ
Q ¼ PT F
where P is the modal matrix given by
P ¼ ½ u1

···

uN

For a two-DoF system, we can be specific
Xj ðtÞ ¼
Qj ðtÞ ¼

© 2005 by Taylor & Francis Group, LLC

2
X
i¼1
2
X
i¼1

uji Zi ðtÞ ¼ uj1 Z1 ðtÞ þ uj2 Z2 ðtÞ;

j ¼ 1; 2
ð5:33Þ


uij Fi ðtÞ ¼ u1j F1 ðtÞ þ u2j F2 ðtÞ;

j ¼ 1; 2