2

Frequency-Domain
Analysis
2.1
2.2

Clarence W. de Silva
The University of British Columbia

Introduction ........................................................................... 2-1
Response to Harmonic Excitations ...................................... 2-2
Response Characteristics
(Q-Factor Method)

†

Measurement of Damping Ratio

2.3

Transform Techniques ......................................................... 2-14

2.4

Mechanical Impedance Approach ...................................... 2-25

2.5

Transmissibility Functions .................................................. 2-31

2.6

Receptance Method ............................................................. 2-37

Transfer Function † Frequency-Response Function
(Frequency Transfer Function)
Interconnection Laws

Force Transmissibility † Motion Transmissibility †
General Case † Peak Values of Frequency-Response Functions
Application of Receptance

Appendix 2A Transform Techniques ................................. 2-40

Summary
This chapter presents the frequency-domain analysis of mechanical vibrating systems. In the frequency domain,
the independent variable is frequency. The response of a vibrating system to harmonic excitations under various
levels of damping (overdamped, underdamped, and critically damped) is analyzed. Frequency transfer function
techniques including impedance, mobility, force transmissibility, motion transmissibility, and receptance are
studied. Transform techniques (Fourier and Laplace) are applicable in the frequency-domain analysis. The
Q-factor method of measuring damping is derived. Component interconnection laws are established for frequencydomain analysis. The emphasis is on single-degree-of-freedom (single-DoF) systems.

2.1 Introduction
In many vibration problems, the primary excitation force typically has a repetitive periodic nature, and
in some cases this periodic forcing function may be even purely sinusoidal. Examples are excitations due
to mass eccentricity and misalignments in rotational components, tooth meshing in gears, and
electromagnetic devices excited by AC or periodic electrical signals. In basic terms, the frequencyresponse of a dynamic system is the response to a pure sinusoidal excitation. As the amplitude and
the frequency of the excitation are changed, the response also changes. In this manner, the response
of the system over a range of excitation frequencies can be determined. This represents the frequency
response. In this case, frequency ðvÞ is the independent variable and hence we are dealing with the

frequency domain.
2-1
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Vibration and Shock Handbook

Frequency-domain considerations are applicable even when the signals are not periodic. In fact, a time
signal can be transformed into its frequency spectrum through the Fourier transform. For a given
time signal, an equivalent Fourier spectrum, which contains all the frequency (sinusoidal) components of
the signal, can be determined either analytically or computationally. Hence, a time-domain
representation and analysis has an equivalent frequency-domain representation and analysis, at least
for linear dynamic systems. For this reason, and also because of the periodic nature of typical vibration
signals, frequency-response analysis is extremely useful in the subject of mechanical vibrations. The
response to a particular form of “excitation” is what is considered in the frequency-domain analysis.
Hence, we are specifically dealing with the subject of “forced response” analysis, albeit in the frequency
domain.

2.2 Response to Harmonic Excitations
Consider a simple oscillator with an excitation
force f ðtÞ; as shown in Figure 2.1.
The equation of motion is given by
m€x þ b_x þ kx ¼ f ðtÞ

Spring
k

Mass

m

ð2:1Þ

Suppose that f ðtÞ is sinusoidal (i.e., harmonic).
Pick the time reference such that
f ðtÞ ¼ f0 cos vt

x

Viscous Damper
b

ð2:2Þ

where

FIGURE 2.1

A forced simple oscillator.

v ¼ excitation frequency
f0 ¼ forcing excitation amplitude
For a system subjected to a forcing excitation, we have
Total Response ¼ Homogeneous Response xh þ Particular Response xp
T

H
ðNatural ResponseÞ


¼

Free Response

X
ðDepends only on initial conditionsÞ
does not contain enforced response
and depends entirely on the
natural=homogeneous response

P
ðEnforced ResponseÞ

þ

Forced Response

F
ðDepends only on f0 Þ
but contains a natural=homogeneous component

Using these concepts, we analyze the forced problem, which may be written as
b
k
f
x_ þ x ¼ 0 cos vt ¼ uðtÞ
m
m
m


ð2:3Þ

x€ þ 2zvn x_ þ v2n x ¼ a cos vt ¼ uðtÞ

ð2:4Þ

x€ þ
or

where uðtÞ is the modified excitation. Also,

vn ¼ undamped natural frequency
z ¼ damping ratio
The total response is given by
x ¼ xh þ xp

© 2005 by Taylor & Francis Group, LLC

ð2:5Þ


2-3

Frequency-Domain Analysis

with
xh ¼ C1 el1 þ C2 e l2 t

ð2:6Þ


The particular solution xp ; by definition, is one solution that satisfies Equation 2.4. It should be
intuitively clear that this will be of the form
xp ¼ a1 cos vt þ a2 sin vt {except for the case: z ¼ 0 and v ¼ vn }

ð2:7Þ

where the constants a1 and a2 are determined by substituting Equation 2.7 into the system Equation 2.4
and equating the like coefficient. This is known as the method of undetermined coefficients.
We will consider several important cases.

2.2.1

Response Characteristics

Case 1: Undamped oscillator with excitation frequency

natural frequency

We have
x€ þ v2n x ¼ a cos vt

with v – vn

ð2:8Þ

Homogeneous solution:
xh ¼ A1 cos vn t þ A2 sin vn t

ð2:9Þ


Particular solution:
xp ¼

ðvn2

a
cos vt
2 v2 Þ

ð2:10Þ

It can be easily verified that xp given by Equation 2.10 satisfies the forced system Equation 2.4, with
z ¼ 0: Hence, it is a particular solution.
Complete solution:
x ¼ A1 cos vn t þ A2 sin vn t þ
|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}
H
Satisfies the homogeneous equation

a
cos vt
ðv2n 2 v2 Þ
|fflfflfflfflfflffl{zfflfflfflfflfflffl}

ð2:11Þ

P
Satisfies the equation with input

Now A1 and A2 are determined using the initial conditions:

xð0Þ ¼ x0 and x_ ð0Þ ¼ v0

ð2:12Þ

Specifically, we obtain
x0 ¼ A1 þ

a
v2n 2 v2

v0 ¼ A2 vn

© 2005 by Taylor & Francis Group, LLC

ð2:13aÞ

ð2:13bÞ


2-4

Vibration and Shock Handbook

Hence, the complete response is
v
x ¼ x0 2 2 a 2 cos vn t þ v0 sin vn t þ 2 a 2 cos vt
n
ðvn 2 v Þ
vn 2 v
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

|fflfflfflfflffl
ffl{zfflfflfflfflfflffl}
H
Homogeneous solution

v
¼ x0 cos vn t þ v0 sin vn t
n
|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}

P
Particular solution

a
½cos vt 2 cos vn t
ðv2n 2 v2 Þ |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}

þ

X
Free response
ðdepends only on initial conditionsÞ
comes from xh
pSinusodal at vn

ð2:14aÞ

2 sin

ðvn þvÞ


t sin

ðvn 2vÞ

ð2:14bÞ
t

2
2
|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl
ffl}

F
Forced response ðdepends on inputÞ
comes from both xh and xp
pWill exhibit a beat phenomenon for vn 2v; i:e:;
ðvn þvÞ=2 wave modulated by ðvn 2vÞ=2 wave

This is a stable response in the sense of bounded-input bounded-output (BIBO) stability, as it is
bounded and does not increase steadily.
Note: If there is no forcing excitation, the homogeneous solution H and the free response X will be
identical. With a forcing input, the natural response (the homogeneous solution) will be influenced by it
in general, as is clear from Equation 2.14b.

Case 2: Undamped oscillator with v 5 vn (resonant condition)
In this case, the xp that was used before is no longer valid. This is the degenerate case, because otherwise
the particular solution cannot be distinguished from the homogeneous solution and the former will be
completely absorbed into the latter. Instead, in view of the “double-integration” nature of the forced
system equation when v ¼ vn ; we use the particular solution ðPÞ:

xp ¼

at
sin vt
2v

ð2:15Þ

This choice of particular solution is strictly justified by the fact that it satisfies the forced system
equation.
Complete solution:
x ¼ A1 cos vt þ A2 sin vt þ

at
sin vt
2v

ð2:16Þ

Initial conditions:
xð0Þ ¼ x0 and x_ ð0Þ ¼ v0 :
We obtain
x0 ¼ A1

ð2:17aÞ

v0 ¼ vA2

ð2:17bÞ


The total response is
v
x ¼ x0 cos vt þ v0 sin vt
|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}

X
Free response ðdepends on initial conditionsÞ
pSinusodal at v

þ

at sin vt
2vffl{zfflffl
|fflffl
ffl}

ð2:18Þ

F
Forced response ðdepends on inputÞ
pAmplitude increases linearly

Since the forced response increases steadily, this is an unstable response in the BIBO sense.
Furthermore, the homogeneous solution H and the free response X are identical, and the particular
solution P is identical to the forced response F in this case.
© 2005 by Taylor & Francis Group, LLC


2-5


Frequency-Domain Analysis

0

Time t

(a)

0

Time t

(b)

0

Time t

(c)
FIGURE 2.2 Forced response of a harmonically excited undamped simple oscillator: (a) for a large frequency
difference; (b) for a small frequency difference (beat phenomenon); (c) response at resonance.

Note that the same system (undamped oscillator) gives a bounded response for some excitations, while
producing an unstable (steady linear increase) response when the excitation frequency is equal to its
natural frequency. Hence, the system is not quite unstable, but is not quite stable either. In fact, the
undamped oscillator is said to be marginally stable. When the excitation frequency is equal to the natural
frequency it is reasonable for the system to respond in a complementary and steadily increasing manner
because this corresponds to the most “receptive” excitation. Specifically, in this case, the excitation
complements the natural response of the system. In other words, the system is “in resonance” with the
excitation, and the condition is called a resonance. We will address this aspect for the more general case of

a damped oscillator, in the sequel.
Figure 2.2 shows typical forced responses of an undamped oscillator when there is a large difference
between the excitation and the natural frequencies (Case 1); a small difference between the excitation and
the natural frequencies when a beat phenomenon is clearly manifested (also Case 1); and for the resonant
case (Case 2).

Case 3: Damped oscillator
The equation of forced motion is
x€ þ 2zvn x_ þ v2n x ¼ a cos vt

ð2:19Þ

Particular Solution (Method 1):
Since derivatives of both odd order and even order are present in this equation, the
particular solution should have terms corresponding to odd and even derivatives of the
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Vibration and Shock Handbook

forcing function (i.e., sin vt and cos vt). Hence, the appropriate particular solution will be of the
form:
xp ¼ a1 cos vt þ a2 sin vt

ð2:20Þ

Substitute Equation 2.20 into Equation 2.19. We obtain
2v2 a1 cos vt 2 v2 a2 sin vt þ 2zvn ½2va1 sin vt þ va2 cos vt þ v2n ½a1 cos vt þ a2 sin vt ¼ a cos vt

Equate like coefficients:
2v2 a1 þ 2zvn va2 þ v2n a1 ¼ a
2v2 a2 2 2zvn va1 þ v2n a2 ¼ 0
Hence, we have
ðv2n 2 v2 Þa1 þ 2zvn va2 ¼ a

ð2:21aÞ

22zvn va1 þ ðv2n 2 v2 Þa2 ¼ 0

ð2:21bÞ

This can be written in the matrix –vector form:
" 2
#" #
a1
ðvn 2 v2 Þ
2zvn v
ðv2n 2 v2 Þ

22zvn v
The solution is

"

a1
a2

#
¼


1
D

"

a2

¼

ðv2n 2 v2 Þ

22zvn v

2zvn v

ðv2n 2 v2 Þ

" #
a
0
#" #
a
0

ð2:21cÞ

ð2:22Þ

with the determinant

D ¼ ðv2n 2 v2 Þ2 þ ð2zvn vÞ2

ð2:23Þ

On simplification, we obtain
a1 ¼

ðv2n 2 v2 Þ
a
D

ð2:24aÞ

2zvn v
a
D

ð2:24bÞ

a2 ¼

This is the method of undetermined coefficients.

Particular Solution (Method 2): Complex Function Method
Consider
x€ þ 2zvn x_ þ v2n x ¼ ae jvt

ð2:25Þ

where the excitation is complex. (Note: e jvt ¼ cos vt þ j sin vt.)

The resulting complex particular solution is
xp ¼ XðjvÞe jvt
Note that we should take the real part of this solution as the true particular solution.
First substitute Equation 2.26 into Equation 2.25:
X 2 v2 þ 2zvn jv þ v2n e jvt ¼ ae jvt

© 2005 by Taylor & Francis Group, LLC

ð2:26Þ


2-7

Frequency-Domain Analysis

Hence, since e jvt – 0 in general,
X¼

a
½2v2 þ 2zvn jv þ v2n

ð2:27Þ

The characteristic polynomial of the system is
DðlÞ ¼ l2 þ 2zvn l þ v2n

ð2:28aÞ

DðsÞ ¼ s2 þ 2zvn s þ v2n


ð2:28bÞ

DðjvÞ ¼ 2v2 þ 2zvn jv þ v2n

ð2:28cÞ

or, with the Laplace variable s;
If we set s ¼ jv; we have

as

Note that Equation 2.28c is the denominator of Equation 2.27. Hence, Equation 2.27 can be written
X¼

a
DðjvÞ

ð2:29Þ

It follows from Equation 2.26 that the complex particular solution is
a
e jvt
xp ¼
DðjvÞ

ð2:30Þ

Next let
DðjvÞ ¼ lDle jf


ð2:31Þ

Then, by substituting Equation 2.31 in Equation 2.30, we obtain
a jðvt2fÞ
xp ¼
e
lDl

ð2:32Þ

where it is clear from Equation 2.31 that
lDl ¼ magnitude of D ðjvÞ

f ¼ phase angle of DðjvÞ
The actual real particular solution is the real part of Equation 2.32 and is given by
a
cosðvt 2 fÞ
xp ¼
lDl

ð2:33Þ

It can easily be verified that this result is identical to what was obtained previously by Method 1, as
given by Equation 2.20 together with Equations 2.23 and 2.24.
In passing, we will note here that the frequency-domain transfer function (i.e., response and excitation
in the frequency domain) of the system Equation 2.19 is:
GðjvÞ ¼

1
¼

D

1
s2 þ 2zvn s þ v2n

s¼jv

ð2:34Þ

This frequency transfer function (also known as the frequency-response function) is obtained
from the Laplace transfer function GðsÞ by setting s ¼ jv: We will discuss this aspect in more
detail later.
The particular section ðPÞ is equal to the steady-state solution, because the homogeneous solution dies
out due to damping.
The particular solution (Equation 2.33) has the following characteristics:
1. The frequency is the same as the excitation frequency v:
2. The amplitude is amplified by the magnitude 1 lDl ¼ lGðjvÞl:
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Vibration and Shock Handbook

3. The response is “lagged” by the phase angle f of D (or “led” by the phase angle of GðjvÞ; denoted
by ,GðjvÞÞ:
4. Since the homogeneous solution of a stable system decays to zero, the particular solution is also
the steady-state solution.

Resonance

The amplification lGðjvÞl ¼ 1 lDl is maximum (i.e., resonance) when lDl is a minimum or lDl2 is a
minimum. As noted earlier, this condition of peak amplification of a system when excited by a sinusoidal
input is called resonance and the associated frequency of excitation is called resonant frequency. We can
determine the resonance of the system (Equation 2.19) as follows:
Equation 2.28c is D ¼ v2n 2 v2 þ 2zvn v j
Hence,
ð2:35Þ
lDl2 ¼ ðv2n 2 v2 Þ2 þ ð2zvn vÞ2 ¼ D
The resonance corresponds to a minimum value of D; or
dD
¼ 2ðv2n 2 v2 Þð22vÞ þ 2ð2zvn Þ2 v ¼ 0
dv

For a minimum:

ð2:36Þ

Hence, with straightforward algebra, the required condition for resonance is
2v2n þ v2 þ 2z2 v2n ¼ 0
or

v2 ¼ ð1 2 2z2 Þv2n

or

qffiffiffiffiffiffiffiffiffiffi
v ¼ 1 2 2z 2 v n
This is the resonant frequency, and is denoted as
qffiffiffiffiffiffiffiffiffiffi
vr ¼ 1 2 2z2 vn


ð2:37Þ
pffiffiffiffiffiffiffiffi
Note that vr # vd # vn , where vd is the damped natural frequency given by vd ¼ 1 2 z2 vn : These
three frequencies (resonant frequency, damped natural frequency, and undamped natural frequency) are
almost equal for small z (i.e., for light damping).
The magnitude and the phase angle plots of GðjvÞ are shown in Figure 2.3. These curves correspond to
the amplification and the phase change of the particular response (the steady-state response) with respect
to the excitation input. This pair, the magnitude and phase angle plots of a transfer function with respect
to frequency, is termed a Bode plot. Usually, logarithmic scales are used for both magnitude (e.g.,
decibels) and frequency (e.g., decades). In summary, the steady-state response of a linear system to a
sinusoidal excitation is completely determined by the frequency transfer function of the system. The total
response is determined by adding H to P and substituting initial conditions, as usual.
For an undamped oscillator ðz ¼ 0Þ; we notice from Equation 2.34 that the magnitude of GðjvÞ
becomes infinity when the excitation frequency is equal to the natural frequency ðvn Þ of the oscillator.
This frequency ðvn Þ is clearly the resonant frequency (as well as natural frequency) of the oscillator. This
fact has been further supported by the nature of the corresponding time response (see Equation 2.18 and
Figure 2.2(c)), which grows (linearly) with time.

2.2.2

Measurement of Damping Ratio (Q-Factor Method)

The frequency transfer function of a simple oscillator (Equation 2.19) may be used to determine the
damping ratio. This frequency-domain method is also termed the half-power point method, for reasons
that should become clear from the following development.
© 2005 by Taylor & Francis Group, LLC


2-9


Frequency-Domain Analysis

Amplification
a
G=
D

Frequency w
Phase Lead
(−f)
0

wr

wn

w

−90

−180
FIGURE 2.3

Magnitude and phase angle curves of a simple oscillator (A Bode plot).

pffiffi
pffiffi
First we assume that z , 1= 2: Strictly speaking, we should assume that z , 1=2 2:
Without loss of generality, consider the normalized (or, non-dimensionalized) transfer function

"
#
v2n
v2n
ð2:38Þ
GðjvÞ ¼ 2
¼ 2
2
2
s þ 2zvn s þ vn s¼jv vn 2 v þ 2zvn vj
As noted before, the transfer function GðsÞ; where s is the Laplace variable, can be converted into the
corresponding frequency transfer function simply by setting s ¼ jv: Its value at the undamped natural
frequency is
GðjvÞlv¼vn ¼

1
2zj

ð2:39aÞ

Hence, the magnitude of GðjvÞ (amplification) at v ¼ vn is
lGðjvÞlv¼vn ¼

1
2z

ð2:39bÞ

For small z we have vr ø vn : Hence, 1=2z is approximately the peak magnitude at resonance (the
resonant peak). The actual peak is slightly larger.

It is clear from Equation 2.39a that the phase angle of GðjvÞ at v ¼ vn is 2p=2:
When power is half of the peak power value (e.g., because the displacement squared is proportional to
potential energy), then the velocity squared is proportional to kinetic energy, and power is the rate of
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Vibration and Shock Handbook

pffiffi
change of energy. Therefore, when the amplification is 1= 2 of the peak value we have half-power points,
given by:
1 1
v2n
pffiffi
¼ 2
¼
vn 2 v2 þ 2jzvn v
2 2z

1
12

2

v
vn

þ 2jz


v
vn

ð2:40Þ

Square Equation 2.40:
1
¼
2 £ 4z 2

1

v
vn

12

2 2

þ 4z 2

v
vn

2

Hence,

v

vn

4

22

v
vn

2

þ 1 þ 4z 2

v
vn

2

¼ 8z 2

or

v
vn

4

2 2ð1 2 2z 2 Þ

v

vn

2

þ ð1 2 8z 2 Þ ¼ 0
|fflfflffl{zfflfflffl}

ð2:41Þ

.0

pffiffi
Now assume that z2 , 1=8 or z , 1=2 2: Otherwise, we will not obtain two positive roots for
ðv=vn Þ2 . Solve for ðv=vn Þ2, which will give two roots v21 and v22 for v2 : Next, assume v22 . v21 :
Compare ðv2 2 v21 Þðv2 2 v22 Þ ¼ 0 with Equation 2.41.
Sum of roots:

v22 þ v21
¼ 2ð1 2 2z 2 Þ
v2n

ð2:42Þ

v22 v21
¼ ð1 2 8z 2 Þ
v4n

ð2:43Þ

Product of roots:


Hence,

v2 2 v1
vn

2

!
qffiffiffiffiffiffiffiffiffiffi
v22 þ v21 2 2v 2 v1
2
z
Þ
2
2
1 2 8z 2
¼
¼
2ð1
2
2
v2n
1
¼ 2 2 4z2 2 2 1 2 £ 8z2 þ Oðz 4 Þ
2
2
2
ø 2 2 4z 2 2 þ 8z {because Oðz 4 Þ ! 0 for small z }
ø 4z 2


or

v2 2 v1
ø 2z
vn
Hence, the damping ratio

zø

ðv2 2 v1 Þ
Dv
v 2 v1
¼
ø 2
2vn
2vn
v2 þ v1

ð2:44Þ

It follows that, once the magnitude of the frequency-response function GðjvÞ is experimentally
determined, the damping ratio can be estimated from Equation 2.44, as illustrated in Figure 2.4.
© 2005 by Taylor & Francis Group, LLC


2-11

Frequency-Domain Analysis


Amplification
1
2z
2 2z

1

0

w1 wnw 2

Frequency w

∆w
FIGURE 2.4

The Q-factor method of damping measurement.

The Q-factor, which measures the sharpness of resonant peak, is defined by
Q-factor ¼

vn
1
¼
2z
Dv

ð2:45Þ

The term originated from the field of electrical tuning circuits where a sharp resonant peak is a

desirable thing (quality factor). Some useful results on the frequency-response of a simple oscillator are
summarized in Box 2.1.

Example 2.1
A dynamic model of a fluid coupling system is shown in Figure 2.5. The fluid coupler is represented by a
rotatory viscous damper with damping constant b: It is connected to a rotatory load of moment of
inertia J; restrained by a torsional spring of stiffness k; as shown. We now obtain the frequency transfer
function of the system that relates the restraining torque t of the spring to the angular displacement
excitation aðtÞ that is applied at the free end of the fluid coupler. If aðtÞ ¼ a0 sin vt; what is the
magnitude (i.e., amplitude) of t at steady state?

Solution
Newton’s Second Law gives
J u€ ¼ bða_ 2 u_Þ 2 ku
Hence,
J u€ þ bu_ þ ku_ ¼ ba_

ðiÞ

u
bs
¼ 2
a
Js þ bs þ k

ðiiÞ

Motion transfer function is

Note that the frequency transfer function is obtained simply by setting s ¼ jv:

The restraining torque of the spring is t ¼ ku: Hence,

t
ku
kbs
¼ GðsÞ
¼
¼ 2
a
a
Js þ bs þ k
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Vibration and Shock Handbook

Box 2.1
HARMONIC RESPONSE
OSCILLATOR

OF A

SIMPLE

Undamped Oscillator:
x€ þ v2n x ¼ a cos vt; xð0Þ ¼ x0 ; x_ ð0Þ ¼ v0

v
x ¼ x0 cos vn t þ v0 sin vn t þ 2 a 2 ½cos vt 2 cos vn t
n
vn 2 v
|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}
|fflfflfflfflfflfflfflfflfflffl
ffl{zfflfflfflfflfflfflfflfflfflfflffl}
X

x ¼ Same X þ

for v – vn

F

at
sin vt
2v

for v ¼ vn
ðresonanceÞ

Damped Oscillator:
x€ þ 2zvn x_ þ v2n x ¼ a cos vt
x¼Hþ
where

a
cosðvt 2 fÞ
lv2n 2 v2 þ 2jzvn vl

|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
P

tan f ¼

2zvn v
;
v2n 2 v2

f ¼ phase lag:

Particular solution P is also the steady-state response.
Homogeneous solution H ¼ A1 el1 t þ A2 el2 t
where l1 and l2 are roots of l2 þ 2zvn l þ v2n ¼ 0
ðcharacteristic equationÞ

A1 and A2 are determined from initial conditions: xð0Þ ¼ x0 ; x_ ð0Þ ¼ v0
pffiffiffiffiffiffiffiffiffiffi
Resonant Frequency: vr ¼ 1 2 2z2 vn
The magnitude of P will peak at resonance.
Damping Ratio:

z¼

Dv
v 2 v1
¼ 2
for low damping
2v n
v2 þ v1


where, Dv ¼ half-power bandwidth ¼ v2 2 v1
v
1
Note: Q-factor ¼ n ¼
for low damping.
2z
Dv
Then, the corresponding frequency-response function (frequency transfer function) is
kbjv
GðjvÞ ¼
ðk 2 J v2 Þ þ bjv

ðivÞ

For a harmonic excitation of

a ¼ a0 e jvt
© 2005 by Taylor & Francis Group, LLC

ðvÞ


2-13

Frequency-Domain Analysis

we have

t ¼ lGðjvÞla0 e


ð jvtþfÞ

Fluid
Coupling
b

ðviÞ

at steady state.
Here, the phase lead of t with respect to a is

a (t)

f ¼ /GðjvÞ ¼ /jv 2 /ðk 2 J v2 þ bjvÞ
¼

p
bv
2 tan21
2
ðk 2 J v2 Þ

t

q
Load
Inertia
J


ðviiÞ

The magnitude of the restraining torque, at
steady state, is

t0 ¼ a0 lGðjvÞl ¼ a0

Restraining
Spring
k

FIGURE 2.5

A fluid coupling system.

kbv
lk 2 J v2 þ bjvl

a0 kbv
ffi
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðk 2 J v2 Þ2 þ b2 v2
Hence,
J a0 v2n v £ 2zvn
ffi
t0 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðv2n 2 v2 Þ2 þ ð2zvn vÞ2

ðviiiÞ


a0 k2zvn v
ffi
t0 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
ðvn 2 v2 Þ2 þ ð2zvn vÞ2

ðixÞ

or

with k=J ¼ v2n ; b=J ¼ 2zvn ; and vn equal to the undamped natural frequency of the load.
Now, define the normalized frequency
r¼

v
vn

ðxÞ

Then, from (ix) we have
2ka0 zr
t0 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð1 2 r2 Þ2 þ ð2zrÞ2

ðxiÞ

For r ¼ 1:

t0 ¼


2ka0 z
¼ ka0
2z

ðxiiÞ

This means, at resonance, the applied twist is directly transmitted to the load spring.
For small r:

t0 ¼

2ka0 zr
1

ðxiiiÞ

which is small, and becomes zero at r ¼ 0: Hence, at low frequencies, the transmitted torque is small.
For larger r:

t0 ¼

2ka0 zr
2ka0 z
¼
2
r
r

ðxivÞ


which is small and goes to zero. Hence, at high frequencies as well, the transmitted torque is small.
The variation of t0 with the frequency ratio r is sketched in Figure 2.6.
© 2005 by Taylor & Francis Group, LLC


2-14

Vibration and Shock Handbook

Magnitude of
Restraining
Torque t 0
ka 0

0
FIGURE 2.6

1

Excitation Frequency Ratio r

Variation of the steady-state transmitted torque with frequency.

2.3 Transform Techniques
Concepts of frequency-response analysis originate from the nature of the response of a dynamic system to
a sinusoidal (i.e., harmonic) excitation. These concepts can be generalized because the time-domain
analysis, where the independent variable is time ðtÞ; and the frequency-domain analysis, where the
independent variable is frequency ðvÞ; are linked through the Fourier transformation. Analytically, it is
more general and versatile to use the Laplace transformation, where the independent variable is the
Laplace variable ðsÞ which is complex (i.e., non-real). This is true because analytical Laplace transforms

may exist even for time functions that do not have “analytical” Fourier transforms. But with compatible
definitions, the Fourier transform results can be obtained from the Laplace transform results simply by
setting s ¼ jv: In the present section, we will formally introduce the Laplace transformation and the
Fourier transformation, and will illustrate how these techniques are useful in the response analysis of
vibrating systems. The preference of one domain over another will depend on such factors as the nature
of the excitation input, the type of the analytical model available, the time duration of interest, and the
quantities that need to be determined.

2.3.1

Transfer Function

The Laplace transform of a piecewise-continuous function f ðtÞ is denoted here by FðsÞ and is given by the
Laplace transformation
ð1
FðsÞ ¼
f ðtÞ expð2stÞdt
ð2:46Þ
0

in which s is a complex independent variable known as the Laplace variable, expressed as
s ¼ s þ jv
ð2:47Þ
pffiffiffiffi
and j ¼ 21: Laplace transform operation is denoted as Lf ðtÞ ¼ FðsÞ: The inverse Laplace transform
operation is denoted by f ðtÞ ¼ L21 FðsÞ and is given by
1 ðsþj1
FðsÞ expðstÞds
ð2:48Þ
f ðtÞ ¼

2p j s2j1
The integration is performed along a vertical line parallel to the imaginary (vertical) axis, located at
s from the origin in the complex Laplace plane (s-plane). For a given piecewise-continuous function
© 2005 by Taylor & Francis Group, LLC


2-15

Frequency-Domain Analysis

f ðtÞ; the Laplace transform exists if the integral in Equation 2.46 converges. A sufficient condition for
this is
ð1
ð2:49Þ
lf ðtÞl expð2stÞdt , 1
0

Convergence is guaranteed by choosing a sufficiently large and positive s: This property is an
advantage of the Laplace transformation over the Fourier transformation. (For a more complete
discussion of the Fourier transformation, see later in this chapter and in Chapter 4.)
By the use of Laplace transformation, the convolution integral equation can be converted into an
algebraic relationship. To illustrate this, consider the convolution integral which gives the response yðtÞ of
a dynamic system to an excitation input uðtÞ; with zero initial conditions. By definition (Equation 2.46),
its Laplace transform is written as
ð1 ð1
YðsÞ ¼
hðtÞuðt 2 tÞdt expð2stÞdt
ð2:50Þ
0


0

Note that hðtÞ is the impulse-response function of the system. Since the integration with respect to t is
performed while keeping t constant, we have dt ¼ dðt 2 tÞ: Consequently,
ð1
ð1
YðsÞ ¼
uðt 2 tÞ exp½2sðt 2 tÞ dðt 2 tÞ
hðtÞ expð2stÞdt
2t

0

The lower limit of the first integration can be made equal to zero, in view of the fact that uðtÞ ¼ 0 for
t , 0: Again using the definition of Laplace transformation, the foregoing relation can be expressed as
YðsÞ ¼ HðsÞUðsÞ

ð2:51Þ

in which
HðsÞ ¼ LhðtÞ ¼

ð1
0

hðtÞ expð2stÞdt

ð2:52Þ

Note that, by definition, the transfer function of a system, denoted by HðsÞ; is given by Equation 2.51.

More specifically, the system transfer function is given by the ratio of the Laplace-transformed output
and the Laplace-transformed input, with zero initial conditions. In view of Equation 2.52, it is clear that
the system transfer function can be expressed as the Laplace transform of the impulse-response function
of the system. The transfer function of a linear and constant-parameter system is a unique function that
completely represents the system. A physically realizable, linear, constant-parameter system possesses a
unique transfer function, even if the Laplace transforms of a particular input and the corresponding
output do not exist. This is clear from the fact that the transfer function is a system model and does not
depend on the system input itself.
Note: The transfer function is also commonly denoted by GðsÞ: However, in the present context we use
HðsÞ in view of its relation to hðtÞ: Some useful Laplace transform relations are given in Table 2.1. Also,
note that the Fourier transform (set s ¼ jvÞ is given by
Forward:
ð1
f ðtÞ expð2jvtÞdt
FðjvÞ ¼
0

Inverse:
f ðtÞ ¼

ðj1
2j1

FðjvÞ expðjvtÞdv

Consider the nth-order linear, constant-parameter dynamic system given by
an

dn y
dn21 y

duðtÞ
dm uðtÞ
þ · · · þ bm
n þ an21
n21 þ · · · þ a0 y ¼ b0 u þ b1
dt
dt
dt
dt m

© 2005 by Taylor & Francis Group, LLC

ð2:53Þ


2-16

Vibration and Shock Handbook

TABLE 2.1

Important Laplace Transform Relations

L21 FðsÞ ¼ f ðtÞ
Ð
1 sþj1
FðsÞ expðstÞds
2pj s2j1

Ð1


k1 f1 ðtÞ þ k2 f2 ðtÞ

k1 F1 ðsÞ þ k2 F2 ðsÞ

expð2atÞf ðtÞ

Fðs þ aÞ

f ðt 2 tÞ

expð2tsÞFðsÞ

f

ðnÞ

Ðt

0

dn f ðtÞ
ðtÞ ¼
dt n

21

Conversion Lf ðtÞ ¼ FðsÞ
f ðtÞ expð2stÞdt


sn FðsÞ 2 sn21 f ð0þ Þ 2 sn22 f 1 ð0þ Þ 2 · · · 2 f n21 ð0þ Þ
Ð0
f ðtÞdt
FðsÞ
þ 21
s
s

f ðtÞdt

Impulse function, dðtÞ

1
1
s
n!
snþ1
1
sþa

Step function, UðtÞ
tn
expð2atÞ
sin vn t

vn
s2 þ v2n

cos vn t


s
s2 þ v2n

for physically realizable systems, m # n: By applying a Laplace transformation and then integrating by
parts, it may be verified that
L

k21
dk f ðtÞ
^ 2 sk21 f ð0Þ 2 sk22 df ð0Þ 2 · · · þ d f ð0Þ
¼ sk FðsÞ
k
dt
dt
dt k21

ð2:54Þ

By definition, the initial conditions are set to zero in obtaining the transfer function. This results in
HðsÞ ¼

b0 þ b1 s þ · · · þ bm sm
a0 þ a1 s þ · · · þ an sn

ð2:55Þ

for m # n: Note that Equation 2.55 contains all the information that is contained in Equation 2.53.
Consequently, the transfer function is an analytical model of a system. The transfer function may be
employed to determine the total response of a system for a given input, even though it is defined in terms
of the response under zero initial conditions. This is quite logical because the analytical model of a system

is independent of the system’s initial conditions.
The denominator polynomial of a transfer function is the system’s characteristic polynomial. Its roots are
the poles or the eigenvalues of the system. If all the eigenvalues have negative real parts, the system is stable.
The response of a stable system is bounded (that is, remains finite) when the input is bounded (which is the
BIBO stability). The zero-input response of an asymptotically stable system approaches zero with time.

2.3.2

Frequency-Response Function (Frequency Transfer Function)

The Fourier integral transform of the impulse-response function is given by
ð1
Hð f Þ ¼
hðtÞ expð2j2p ftÞdt
21

ð2:56Þ

where f is the cyclic frequency (measured in cps or hertz). This is known as the frequency-response function
(or frequency transfer function) of a system. The Fourier transform operation is denoted as F hðtÞ ¼ Hð f Þ:
© 2005 by Taylor & Francis Group, LLC


2-17

Frequency-Domain Analysis

In view of the fact that hðtÞ ¼ 0 for t , 0; the lower limit of integration in Equation 2.56 could be made
zero. Then, from Equation 2.52, it is clear that Hð f Þ is obtained simply by setting s ¼ j2pf in HðsÞ: Hence,
strictly speaking, we should use the notation Hð j2pf Þ and not Hð f Þ. However, for notational simplicity,

we denote Hðj2pf Þ by Hð f Þ: Furthermore, since the angular frequency v ¼ 2pf ; we can express the
frequency-response function by Hð jvÞ; or simply by HðvÞ for notational convenience. It should be noted
that the frequency-response function, like the Laplace transfer function, is a complete representation of a
linear, constant-parameter system. In view of the fact that both uðtÞ ¼ 0 and yðtÞ ¼ 0 for t , 0; we can
write the Fourier transforms of the input and the output of a system directly by setting s ¼ j2pf ¼ jv in
the corresponding Laplace transforms. Specifically, we have, according to the notation used here
Uð f Þ ¼ Uðj2p f Þ ¼ UðjvÞ
and
Yð f Þ ¼ Yð j2p f Þ ¼ Yð jvÞ:
Then, from Equation 2.51, we have
Yð f Þ ¼ Hð f ÞUð f Þ

ð2:57Þ

Note: Sometimes, for notational convenience, the same lowercase letters are used to represent the
Laplace and Fourier transforms as well as the original time-domain variables.
If the Fourier integral transform of a function exists, then its Laplace transform also exists. The
converse is not generally true, however, because of the poor convergence of the Fourier integral in
comparison to the Laplace integral. This arises from the fact that the factor expð2stÞ is not present in the
Fourier integral. For a physically realizable, linear, constant-parameter system, Hð f Þ exists even if Uð f Þ
and Yð f Þ do not exist for a particular input. The experimental determination of Hð f Þ; however, requires
system stability. For the nth-order system given by Equation 2.53, the frequency-response function is
determined by setting s ¼ j2pf in Equation 2.55 as
Hð f Þ ¼

b0 þ b1 j2pf þ · · · þ bm ðj2pf Þm
a0 þ a1 j2pf þ · · · þ an ðj2pf Þn

ð2:58Þ


This, generally, is a complex function of f that has a magnitude denoted by lHð f Þl and a phase angle
denoted by /Hð f Þ:
A further interpretation of the frequency-response function can be given in view of the developments
given in Section 2.2. Consider a harmonic input having cyclic frequency f ; expressed by
uðtÞ ¼ u0 cos 2p ft

ð2:59aÞ

In analysis, it is convenient to use the complex input
uðtÞ ¼ u0 ðcos 2p ft þ j sin 2p ftÞ ¼ u0 expðj2p ftÞ

ð2:59bÞ

and take only the real part of the final result. Note that Equation 2.59b does not implicitly satisfy the
requirement of uðtÞ ¼ 0 for t , 0: Therefore, an appropriate version of the convolution integral, where the
limits of integration automatically account for this requirement, should be used. For instance, we can write
ðt
yðtÞ ¼ Re
hðtÞu0 exp½j2p f ðt 2 tÞ dt
ð2:60aÞ
21

or
yðtÞ ¼ Re u0 expðj2p ftÞ

ðt
21

hðtÞ expð2j2p f tÞdt


ð2:60bÞ

in which Re[·] denotes the real part. As t ! 1; the integral term in Equation 2.60b becomes the frequencyresponse function Hð f Þ; and the response yðtÞ becomes the steady-state response yss : Accordingly,
yss ¼ Re½Hð f Þu0 expðj2p ftÞ

© 2005 by Taylor & Francis Group, LLC

ð2:61aÞ


2-18

Vibration and Shock Handbook

or
yss ¼ u0 lHð f Þl cosð2p ft þ fÞ

ð2:61bÞ

for a harmonic excitation, in which the phase lead angle f ¼ /Hð f Þ: It follows from Equation 2.61b that,
when a harmonic excitation is applied to a stable, linear, constant-parameter dynamic system having
frequency-response function Hð f Þ; its steady-state response will also be harmonic at the same frequency,
but with an amplification factor of lHð f Þl in its amplitude and a phase lead of /Hð f Þ: This result has been
established previously, in Section 2.2. Consequently, the frequency-response function of a stable system
can be experimentally determined using a sine-sweep test or a sine-dwell test. With these methods, a
harmonic excitation is applied as the system input, and the amplification factor and the phase-lead angle in
the corresponding response are determined at steady state. The frequency of excitation is varied
continuously for a sine sweep and in steps for a sine dwell. The sweep rate should be slow enough, and the
dwell times should be long enough, to guarantee steady-state conditions at the output. The pair of plots of
lHð f Þl and /Hð f Þ against f completely represent the complex frequency-response function, and are Bode

plots or Bode diagrams, as noted earlier. In Bode plots, logarithmic scales are normally used for both
frequency f and magnitude lHð f Þl:

Impulse Response
The impulse-response function of a system can be obtained by taking the inverse Laplace transform of the
system transfer function. For example, consider the damped simple oscillator given by the normalized
transfer function:
HðsÞ ¼

v2n
s2 þ 2zvn s þ v2n

ð2:62Þ

The characteristic equation of this system is given by
s2 þ 2zvn s þ v2n ¼ 0

ð2:63Þ

The eigenvalues (poles) are given by its roots. Three possible cases exist, as given below.

Case 1: ðz < 1Þ
This is the case of complex eigenvalues l1 and l2 : Since the coefficients of the characteristic equation are
real, the complex roots should occur in conjugate pairs. Hence,

in which

l1 ; l2 ¼ 2zvn ^ jvd

ð2:64Þ


qffiffiffiffiffiffiffiffi
vd ¼ 1 2 z2 vn

ð2:65Þ

is the damped natural frequency.

Case 2: ðz > 1Þ
This case corresponds to real and unequal eigenvalues,
qffiffiffiffiffiffiffiffi
l1 ; l2 ¼ 2zvn ^ z2 2 1vn ¼ 2a; 2b

ð2:66Þ

with a – b; in which
ab ¼ v2n

© 2005 by Taylor & Francis Group, LLC

ð2:67Þ


2-19

Frequency-Domain Analysis

and
a þ b ¼ 2zvn


ð2:68Þ

Case 3: ðz 5 1Þ
In this case, the eigenvalues are real and equal:
l1 ¼ l2 ¼ 2vn

ð2:69Þ

In all three cases, the real parts of the eigenvalues are negative. Consequently, these second-order
systems of a damped simple oscillator are always stable.
The impulse-response functions hðtÞ corresponding to these three cases are determined by taking
the inverse Laplace transform (Table 2.1) of Equation 2.62 for z , 1; z . 1; and z ¼ 1; respectively. The
following results are obtained:
vn
yimpulse ðtÞ ¼ hðtÞ ¼ pffiffiffiffiffiffiffiffi
expð2zvn tÞ sin vd t for z , 1
ð2:70aÞ
1 2 z2
ab
½expð2atÞ 2 expð2btÞ for z . 1
yimpulse ðtÞ ¼ hðtÞ ¼
ð2:70bÞ
ðb 2 aÞ
yimpulse ðtÞ ¼ hðtÞ ¼ v2n t expð2vn tÞ

for

z¼1

ð2:70cÞ


Step Response
Unit-step function is defined by

(
UðtÞ ¼

1

for t . 0

0

for t # 0

ð2:71Þ

Unit-impulse function dðtÞ may be interpreted as the time derivative of UðtÞ; thus,
dUðtÞ
ð2:72Þ
dt
Note that Equation 2.72 re-establishes the fact that for non-dimensional UðtÞ; the dimension of
dðtÞ is (time)21. Since LUðtÞ ¼ 1=s (see Table 2.1), the unit-step response of the dynamic system
(Equation 2.62) can be obtained by taking the inverse Laplace transform of

dðtÞ ¼

Ystep ðsÞ ¼

1

v2n
2
s ðs þ 2zvn s þ v2n Þ

ð2:73Þ

which follows from Equation 2.73.
To facilitate using Table 2.1, partial fractions of Equation 2.73 are determined in the form
a1
a þ a3 s
þ 2 2
s
ðs þ 2zvn s þ v2n Þ
in which the constants a1 ; a2 ; and a3 are determined by comparing the numerator polynomial; thus,

v2n ¼ a1 ðs2 þ 2dvn s þ v2n Þ þ sða2 þ a3 sÞ
Then a1 ¼ 1; a2 ¼ 22zvn ; and a3 ¼ 1:
The following results are obtained:
1
ystep ðtÞ ¼ 1 2 pffiffiffiffiffiffiffiffi expð2zvn tÞ sinðvd t þ fÞ for z , 1
1 2 z2
1
ystep ¼ 1 2
½b expð2atÞ 2 a expð2btÞ for z . 1
ðb 2 aÞ
ystep ¼ 1 2 ðvn t þ 1Þ expð2vn tÞ

for z ¼ 1

ð2:74aÞ

ð2:74bÞ
ð2:74cÞ

In Equation 2.74c,
cos f ¼ z

© 2005 by Taylor & Francis Group, LLC

ð2:75Þ


2-20

Vibration and Shock Handbook

Transfer Function Matrix
Consider the state-space model of a linear dynamic system as given by
x_ ¼ Ax þ Bu

ð2:76aÞ

y ¼ Cx þ Du

ð2:76bÞ

where, x ¼ the nth order state vector, u ¼ the rth order input vector, y ¼ the mth order output vector,
A ¼ the system matrix, B ¼ the input gain matrix, C ¼ the output (measurement) gain matrix, and
D ¼ the feedforward gain matrix. We can express the input–output relation between u and y; in the
Laplace domain, by a transfer function matrix of the order m £ r:
To obtain this relation, let us Laplace transform the Equations 2.76a and 2.76b and use zero initial

conditions for x; thus,
sXðsÞ ¼ AXðsÞ þ BUðsÞ

ð2:77aÞ

YðsÞ ¼ CXðsÞ þ DUðsÞ

ð2:77bÞ

XðsÞ ¼ ðsI 2 AÞ21 B þ UðsÞ

ð2:78Þ

From Equation 2.77a it follows that,
in which I is the nth order identity matrix. By substituting Equation 2.78 into Equation 2.77b, we obtain
the transfer relation
YðsÞ ¼ ½CðsI 2 AÞ21 B þ D UðsÞ

ð2:79aÞ

YðsÞ ¼ GðsÞUðsÞ

ð2:79bÞ

or
The transfer-function matrix GðsÞ is an m £ r matrix given by
GðsÞ ¼ CðsI 2 AÞ21 B þ D

ð2:80Þ


In practical systems with dynamic delay, the excitation uðtÞ is not fed forward into the response y:
Consequently, D ¼ 0 for systems that we normally encounter. For such systems
GðsÞ ¼ CðsI 2 AÞ21 B

ð2:81Þ

Several examples are given now to illustrate the approaches of obtaining transfer function models
when the time domain differential equation models are given, and to indicate some uses of a transfer
function model. Some useful results in the frequency domain are summarized in Box 2.2.

Example 2.2
Consider the simple oscillator equation given by
m€y þ b_y þ ky ¼ kuðtÞ

ðiÞ

Note that uðtÞ can be interpreted as a displacement input (e.g., support motion) or kuðtÞ can be
interpreted as the input force applied to the mass. Take the Laplace transform of the system equation (i)
with zero initial conditions; thus,
ðms2 þ bs þ kÞYðsÞ ¼ kUðsÞ

ðiiÞ

The corresponding transfer function is
GðsÞ ¼

© 2005 by Taylor & Francis Group, LLC

YðsÞ
k

¼
UðsÞ
ms2 þ bs þ k

ðiiiÞ


2-21

Frequency-Domain Analysis

Box 2.2
USEFUL FREQUENCY-DOMAIN RESULTS
Laplace Transform (L
L):
FðsÞ ¼

ða
0

f ðtÞ expð2stÞdt

Fourier Transform ðF
F ):
FðjvÞ ¼

ða
21

f ðtÞ expð2jvtÞdt


Note: May use FðvÞ to denote FðjvÞ
Note: Set s ¼ jv ¼ j2p f to convert Laplace results into Fourier results.

v ¼ angular frequency (rad/sec)
f ¼ cyclic frequency (cps or Hz)
Transfer function HðsÞ ¼ output input in Laplace domain; with zero initial conditions:
Frequency transfer function (or frequency-response function) ¼ HðjvÞ
Note: Notation ðGðsÞÞ is also used to denote a system transfer function
Note:
HðsÞ ¼ LhðtÞ
hðtÞ ¼ impulse-response function ¼ response to a unit impulse input.
Frequency response:
YðjvÞ ¼ HðjvÞUðjvÞ
where
UðjvÞ ¼ Fourier spectrum of input uðtÞ
YðjvÞ ¼ Fourier spectrum of output yðtÞ
Note:
lHðjvÞl ¼ response amplification for a harmonic excitation of frequency v
/HðjvÞ ¼ response phase “lead” for a harmonic excitation
Multivariable Systems:
State-Space Model:

x_ ¼ Ax þ Bu
y ¼ Cx þ Du

Transfer-Matrix Model:
YðsÞ ¼ GðsÞUðsÞ
where
GðsÞ ¼ CðsI 2 AÞ21 B þ D


© 2005 by Taylor & Francis Group, LLC


2-22

Vibration and Shock Handbook

or, in terms of the undamped natural frequency vn and the damping ratio z; where v2n ¼ k=m and
2zvn ¼ b=m; the transfer function is given by
GðsÞ ¼

v2n
s þ 2zvn s þ v2n

ðivÞ

2

This is the transfer function corresponding to the displacement output. It follows that the output
velocity transfer function is
sYðsÞ
sv2n
¼ sGðsÞ ¼ 2
UðsÞ
s þ 2zvn s þ v2n

ðvÞ

and the output acceleration transfer function is

s2 YðsÞ
s2 v2n
¼ s2 GðsÞ ¼ 2
UðsÞ
s þ 2zvn s þ v2n

ðviÞ

In the output acceleration transfer function, we have m ¼ n ¼ 2: This means that if the acceleration of
the mass that is caused by an applied force is measured, the input (applied force) is instantly felt by the
acceleration. This corresponds to a feedforward action of the input excitation or a lack of dynamic delay.
For example, this is the primary mechanism through which road disturbances are felt inside a vehicle that
has very hard suspension.

Example 2.3
Again let us consider the simple oscillator differential equation
y€ þ 2zvn y_ þ v2n y ¼ v2n uðtÞ

ðiÞ

By defining the state variables as
x ¼ ½x1 ; x2

T

¼ ½y; y_

T

ðiiÞ


a state model for this system can be expressed as
" #
"
#
0
1
0
x_ ¼
uðtÞ
xþ
2
v2n
2vn 22zvn

ðiiiÞ

If we consider both displacement and velocity as outputs, we have
y¼x

ðivÞ

Note that the output gain matrix C is the identity matrix in this case. From Equation 2.79b and
Equation 2.81 it follows that:
"
#21 " #
s
21
0
YðsÞ ¼

UðsÞ
ðvÞ
2
vn s þ 2zvn
v2n
1
YðsÞ ¼ 2
½s þ 2zvn s þ v2n

"

s þ 2zvn

1

2v2n

s

1
YðsÞ ¼ 2
½s þ 2zvn s þ v2n
The transfer function matrix is

"
GðsÞ ¼

© 2005 by Taylor & Francis Group, LLC

"


v2n =DðsÞ
sv2n =DðsÞ

v2n
sv2n

#"

0

v2n

#
UðsÞ

#
UðsÞ

ðviÞ

#
ðviiÞ


2-23

Frequency-Domain Analysis

in which DðsÞ ¼ s2 þ 2zvn s þ v2n is the characteristic polynomial of the system. The first element in

the only column in GðsÞ is the displacement-response transfer function and the second element is
the velocity-response transfer function. These results agree with the expressions obtained in the
previous example.
Now, let us consider the acceleration y€ as an output and denote it by y3 : It is clear from the system
equation (i) that
y3 ¼ y€ ¼ 22zvn y_ 2 v2n y þ v2n uðtÞ

ðviiiÞ

y3 ¼ 22zvn x2 2 v2n x1 þ v2n uðtÞ

ðixÞ

or, in terms of the state variables,

Note that this output explicitly contains the input variable. The feedforward situation implies that the
matrix D is non-zero for the output y3 : Now,
Y3 ðsÞ ¼ 22zvn X2 ðsÞ 2 v2n X1 ðsÞ þ v2n UðsÞ ¼ 22zvn

sv2n
v2
UðsÞ 2 v2n n UðsÞ þ v2n UðsÞ
DðsÞ
DðsÞ

which simplifies to
Y3 ðsÞ ¼

s2 v2n
UðsÞ

DðsÞ

ðxÞ

This again confirms the result for the acceleration output transfer function that was obtained in the
previous example.

Example 2.4
(a) Briefly explain an approach that you could use to measure the resonant frequency of a
mechanical system. Do you expect this measured frequency to depend on whether displacement,
velocity, or acceleration is used as the response variable? Justify your answer.
(b) A vibration test setup is schematically shown in Figure 2.7.
In this experiment, a mechanical load is excited by a linear motor and its acceleration response is
measured by an accelerometer and charge amplifier combination. The force applied to the load by the
linear motor is also measured, using a force sensor (strain-gauge type). The frequency-response function
acceleration/force is determined from the sensor signals, using a spectrum analyzer.
Mechanical
Load

Instrumentation
Charge
Amplifier

a
Spectrum
Analyzer

Force
f(t)


Linear
Motor
Force
Sensor

f

FIGURE 2.7

Acceleration
a

Accelerometer

Mass m

k

b

Measurement of the acceleration spectrum of a mechanical system.

© 2005 by Taylor & Francis Group, LLC


2-24

Vibration and Shock Handbook

Suppose that the mechanical load is approximated by a damped oscillator with mass m; stiffness k; and

damping constant b; as shown in Figure 2.7. If the force applied to this load is f ðtÞ and the displacement
in the same direction is y; show that the equation of motion of the system is given by
m€y þ b_y þ ky ¼ f ðtÞ
Obtain an expression for the acceleration frequency-response function GðjvÞ in the frequency domain,
with excitation frequency v as the independent variable. Note that the applied force f is the excitation
input and the acceleration a of the mass is the response, in this case.
Express GðjvÞ in terms of (normalized) frequency ratio r ¼ v=vn ; where vn is the undamped natural
frequency.
Giving all the necessary steps, determine an expression for r at which the acceleration frequencyresponse function will exhibit a resonant peak. What is the corresponding peak magnitude of lGl?
For what range of values of damping ratio z would such a resonant peak be possible?

Solution
(a) For a single DoF system, apply a sinusoidal forcing excitation at the DoF and measure the
displacement response at the same location. Vary the excitation frequency v in small steps, and
for each frequency at steady state determine the amplitude ratio of the (displacement
response/forcing excitation). The peak amplitude ratio will correspond to the resonance. For a
multi-DoF system, several tests may be needed, with excitations applied at different locations of
freedom and the response measured at various locations as well (see Chapters 10 and 11). In the
frequency domain we have,
lVelocity response spectruml ¼ vxlDisplacement response spectruml
lAcceleration response spectruml ¼ vxlVelocity response spectruml
It follows that the shape of the frequency-response function will depend on whether the
displacement, velocity, or acceleration is used as the response variable. Hence it is likely that the
frequency at which the peak amplification occurs (i.e., resonance) will also depend on the type of
response variable that is used.
(b) A free-body diagram of the mass element is shown in Figure 2.8.
Newton’s Second Law gives
m€y ¼ f ðtÞ 2 b_y 2 ky

ðiÞ


m€y þ b_y þ ky ¼ f ðtÞ

ðiiÞ

y
1
¼
2
f
ðms þ bs þ kÞ

ðiiiÞ

Hence, the equation of motion is
The displacement transfer function is

Note that, for notational convenience, the same
lowercase letters are used to represent the Laplace
transforms as well as the original time-domain
variables (y and f ). The acceleration transfer
function is obtained by multiplying Equation iii
by s2 : (From Table 2.1, the Laplace transform of
d=dt is s; with zero initial conditions). Hence
a
s2
¼
¼ GðsÞ
2
f

ðms þ bs þ kÞ

© 2005 by Taylor & Francis Group, LLC

ðivÞ

y

ky
f(t)

m
by

FIGURE 2.8

Free-body diagram.


2-25

Frequency-Domain Analysis

In the frequency domain, the corresponding frequency-response function is obtained by substituting jv
for s: Hence,
GðjvÞ ¼

2v2
þ bjv þ kÞ


ð2mv2

ðvÞ

Divide throughout by m and use b=m ¼ 2zvn and k=m ¼ v2n ; where vn ¼ undamped natural
frequency and z ¼ damping ratio.
Then,
GðjvÞ ¼

ðv2n

2v2 =m
2r2 =m
¼
2
1 2 r 2 þ 2jzrr
2 v þ 2jzvn vÞ

ðviÞ

where r ¼ v=vn : The magnitude of GðjvÞ gives the amplification of the acceleration signal with respect to
the forcing excitation:
r2 =m
ðviiÞ
lGðjvÞl ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð1 2 r2 Þ2 þ ð2zrÞ2
Its peak value corresponds to the peak value of
pðrÞ ¼

r4

ð1 2 r2 Þ2 þ ð2zrÞ2

ðviiiÞ

and gives the resonance. This occurs when dp=dr ¼ 0: Hence,
½ð1 2 r2 Þ2 þ ð2zrÞ2 4r3 2 r4 ½2ð1 2 r2 Þð22rÞ þ 8z2 r ¼ 0
The solution is
r¼0

or

½ð1 2 r2 Þ2 þ 4z2 r2 þ r2 ½1 2 r 2 2 2z2 ¼ 0

The first result ðr ¼ 0Þ corresponds to static conditions and is ignored. Hence, the resonant peak
occurs when
ð1 2 r2 Þ2 þ 4z2 r2 þ r2 2 r4 2 2z2 r2 ¼ 0
which has the valid root

1
r ¼ pffiffiffiffiffiffiffiffiffiffi
1 2 2z 2

ðixÞ

Note that r has to be real and positive. It follows that, for a resonance to occur we need
1
0 , z , pffiffi
2
Substitute in (vii), the resonant value of r; to obtain
lGlpeak


1
mð1 2 2z2 Þ
¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
1
4z 2
þ
12
2
1 2 2z
1 2 2z 2

or
lGlpeak ¼

1
pffiffiffiffiffiffiffiffi
2mz 1 2 z2

ðxÞ

2.4 Mechanical Impedance Approach
Any type of force or motion variable may be used as input and output variables in defining a system
transfer function. In vibration studies, three particular choices are widely used. The corresponding
© 2005 by Taylor & Francis Group, LLC