18.2 Frequency-Domain Formulation

Back

Frequency-domain analysis of vibrating systems is very useful in a wide variety of applications.

The analytical convenience of frequency-domain methods results from the fact that differential

equations in the time domain become algebraic equations in the frequency domain. Once the

18-2 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

necessary analysis is performed in the frequency domain, it is often possible to interpret the results

without having to transform them back to the time domain through inverse Fourier transformation.

In the context of the present chapter, frequency-domain representation is particularly important

because it is the frequency transfer functions that are used for extracting the necessary modal

parameters.

For the convenience of notation, we shall develop the frequency-domain results using the Laplace

variable, s: As usual, the straightforward substitution of s ¼ jv or s ¼ j2pf gives the corresponding

frequency-domain results.

18.2.1 Transfer-Function Matrix

Let us consider a linear mechanical system that is represented by

My€ þ Cy_ þ Ky ¼ f ðtÞ ð18:1Þ

where

f(t) ¼ forcing excitation vector (nth order column)

y ¼ displacement response vector (nth order column)

M ¼ mass (inertia) matrix ðn £ nÞ

C ¼ damping (linear viscous) matrix ðn £ nÞ

K ¼ stiffness matrix ðn £ nÞ

If the assumption of proportional damping is made, the coordinate transformation

y ¼ Cq ð18:2Þ

decouples Equations 18.1 into the canonical form of modal equations

M􀀊 q€ þ C􀀊 q_ þ Kq ¼ CTf ðtÞ ð18:3Þ

where

C ¼ modal matrix ðn £ nÞ of n independent modal vector vectors ½c1; c2; …; cn􀀉

M¯ ¼ diagonal matrix of modal masses Mi

C¯ ¼ diagonal matrix of modal damping constants Ci

K¯ ¼ diagonal matrix of modal stiffnesses Ki

Specifically, we have

M􀀊 ¼ CTMC ð18:4Þ

C􀀊 ¼ CTCC ð18:5Þ

K􀀊 ¼ CTKC ð18:6Þ

If the modal vectors are assumed to be M-normal, then we have

Mi ¼ 1

Ki ¼ v2i

and furthermore, we can express Ci in the convenient form

Ci ¼ 2zivi

where

vi ¼ undamped natural frequency

zi ¼ modal damping ratio

Experimental Modal Analysis 18-3

© 2005 by Taylor & Francis Group, LLC

By Laplace transformation of the response canonical equations of modal motion (Equation 18.3),

assuming zero initial conditions, we obtain

s2 þ 2zv1s þ v21

0

s2 þ 2zv2s þ v22

. .

.

0 s2 þ 2zvns þ v2

n

2

66666664

3

77777775

QðsÞ ¼ CTFðsÞ ð18:7Þ

Laplace transforms of the modal response (or generalized coordinate) vector, qðtÞ; and the forcing

excitation vector, f ðtÞ; are denoted by the column vectors, QðsÞ and FðsÞ; respectively. The square matrix

on the left-hand side of Equation 18.7 is a diagonal matrix. Its inverse is obtained by inverting the

diagonal elements. Consequently, the following modal transfer relation results:

QðsÞ ¼

G1 0

G2

. .

.

0 Gn

2

66666664

3

77777775

CTFðsÞ ð18:8Þ

in which the diagonal elements are the damped simple-oscillator transfer functions

GiðsÞ ¼

1 􀀑

s2 þ 2zivis þ v2i

􀀜 for i ¼ 1; 2; …; n ð18:9Þ

Note that vi; the ith undamped natural frequency (in the time domain), is only approximately equal to

the frequency of the ith resonance of the transfer function (in the frequency domain), as given by

vri ¼

ffiffiffiffiffiffiffiffiffiffi

1 2 2z2i

q

vi ð18:10Þ

As we have discussed before, and as is clear from Equation 18.10, the approximation improves for

decreasing modal damping. Consequently, in most applications of EMA, the resonant frequency is taken

to be equal (approximately) to the natural frequency for a given mode.

From the time-domain coordinate transformation (Equation 18.2), the Laplace domain coordinate

transformation relation is obtained as

YðsÞ ¼ CQðsÞ ð18:11Þ

Substitute Equation 18.8 into Equation 18.11; thus

YðsÞ ¼ C

G1 0

G2

. .

.

0 Gn

2

66666664 3 77777775

CTFðsÞ ð18:12Þ

Equation 18.12 is the excitation – response (input – output) transfer relation. It is clear that the n £ n

transfer function matrix, G, for the n-DoF system is given by

GðsÞ ¼ C

G1 0

G2

. .

.

0 Gn

2

66666664

3

77777775

CT ð18:13Þ

18-4 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

Notice in particular that GðsÞ is a symmetric matrix; specifically

GTðsÞ ¼ GðsÞ ð18:14Þ

which should be clear from the matrix transposition property, ðABCÞT ¼ CTBTAT:

An alternative version of Equation 18.13 that is extensively used in EMA can be obtained by using the

partitioned form (or assembled form) of the modal matrix in Equation 18.13. Specifically, we have

GðsÞ ¼ ½c1; c2; …; cn􀀉

G1 0

G2

. .

.

0 Gn

2

66666664

3

77777775

cT

1

cT

2

.. .

cTn

2

66666664

3

77777775

ð18:15Þ

On multiplying out the last two matrices on the right-hand side of Equation 18.15 term by term, the

following intermediate result is obtained:

GðsÞ ¼ ½c1; c2; …; cn􀀉

G1cT

1

G2cT

2

.. .

GncTn

2

66666664

3

77777775

Note that Gi are scalars while ci are column vectors. The two matrices in this product can be multiplied

out now to obtain the matrix sum

GðsÞ ¼ G1c1cT

1 þ G2c2cT

2 þ · · · þ GncncTn

¼

Xn

r¼1

Gr cr cT

r ð18:16Þ

in which cr is the rth modal vector that is normalized with respect to the mass matrix. Notice that each

term cr cT

r in the summation (Equation 18.16) is an n £ n matrix with the element corresponding to its

ith row and kth column being ðcickÞr : The ikth element of the transfer matrix GðsÞ is the transfer function

GikðsÞ; which determines the transfer characteristics between the response location, i; and the excitation

location, k: From Equation 18.16, this is given by

GikðsÞ ¼

Xn

r¼1

Gr ðcickÞr ¼

Xn

r¼1

􀀑 ðcickÞr

s2 þ 2zrvr s þ v2r

􀀜 ð18:17Þ

with s ¼ jv ¼ j2pf in the frequency domain. Note that ðciÞr is the ith element of the rth modal vector,

and is a scalar quantity. Similarly, ðcickÞr is the product of the ith element and the kth element of the rth

modal vector, and is also a scalar quantity. This is the numerator of each modal transfer function within

the right-hand side summation of Equation 18.17, and is the residue of the pole (eigenvalue) of that

mode.

Equation 18.17 is useful in EMA. Essentially, we start by determining the residues ðcickÞr of the poles

in an admissible set of measured transfer functions. We can determine the modal vectors in this manner.

In addition, by analyzing the measured transfer functions, the modal damping ratios, zi; and the natural

frequencies, vi; can be estimated. From these results, an estimate for the time-domain model (i.e., the

matrices M, K, and C) can be determined.

18.2.2 Principle of Reciprocity

By the symmetry of transfer matrix, as given by Equation 18.14, it follows that

GikðsÞ ¼ GkiðsÞ ð18:18Þ

Experimental Modal Analysis 18-5

© 2005 by Taylor & Francis Group, LLC

This fact is further supported by Equation 18.17. This symmetry can be interpreted as Maxwell’s principle

of reciprocity. To understand this further, consider the complete set of transfer relations given by Equation

18.12 and Equation 18.13:

Y1ðsÞ ¼ G11ðsÞF1ðsÞ þ G12ðsÞF2ðsÞ þ · · · þ G1nðsÞFnðsÞ

Y2ðsÞ ¼ G21ðsÞF1ðsÞ þ G22ðsÞF2ðsÞ þ · · · þ G2nðsÞFnðsÞ

.. .

YnðsÞ ¼ Gn1ðsÞF1ðsÞ þ Gn2ðsÞF2ðsÞ þ · · · þ GnnðsÞFnðsÞ

ð18:19Þ

Note that the diagonal elements, G11; G22; …; Gnn; are driving-point transfer functions (or autotransfer

functions) and the rest are cross-transfer functions. Suppose that a single excitation, FkðsÞ; is applied at the

kth DoF with all the other excitations set to zero. The resulting response at the ith DoF is given by

YiðsÞ ¼ GikðsÞFk ðsÞ ð18:20Þ

Similarly, when a single excitation, FiðsÞ; is applied at the ith DoF, the resulting response at the kth DoF is

given by

YkðsÞ ¼ GkiðsÞFiðsÞ ð18:21Þ

In view of the symmetry that is indicated by Equation 18.18, it follows from Equation 18.20 and

Equation 18.2.1 that if the two separate excitations, FkðsÞ and FiðsÞ; are identical then the corresponding

responses, YiðsÞ and YkðsÞ; are also identical. In other words, the response at the ith DoF due to a single

force at the kth DoF is equal to the response at the kth DoF when the same single force is applied at the ith

DoF. This is the frequency-domain version of the principle of reciprocity.

Example 18.1

Consider the two-DoF system shown in Figure 18.1. Assume that the excitation forces, f1ðtÞ and f2ðtÞ; act

at the y1 and y2 DoFs, respectively. The equations of motion are given by

m 0

0 m

" #

y€

c 0

0 c

" #

y_ þ

2k 2k

2k 2k

" #

y ¼ f ðtÞ ðiÞ

This system has proportional damping (specifically, it is clear that C is proportional to M) and hence

possesses the same real modal vectors as does the undamped system. Let us first obtain the transfer

matrix in the direct manner. By taking the Laplace transform (with zero initial conditions) of the

equations of motion (i), we have

ms2 þ cs þ 2k 2k

2k ms2 þ cs þ 2k

" #

YðsÞ ¼ FðsÞ ðiiÞ

m

m

k

c

k

k

c y2

y1

f f2(t) 1(t)

FIGURE 18.1 A vibrating system with proportional damping.

18-6 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

Hence, in the relation YðsÞ ¼ GðsÞFðsÞ; the transfer matrix G is given by

GðsÞ ¼

ms2 þ cs þ 2k 2k

2k ms2 þ cs þ 2k

" #21

¼

1 􀀑

ðms2 þ cs þ 2kÞ2 2 k2

􀀜

ms2 þ cs þ 2k k

k ms2 þ cs þ 2k

" #

ðiiiÞ

The characteristic polynomial DðsÞ of the system is

DðsÞ ¼ ðms2 þ cs þ 2kÞ2 2 k2 ¼ ðms2 þ cs þ kÞðms2 þ cs þ 3kÞ ðivÞ

and is common to the denominator of all four transfer functions in the matrix. Specifically,

G11ðsÞ ¼ G22ðsÞ ¼

ms2 þ cs þ 2k

DðsÞ ðvÞ

G12ðsÞ ¼ G21ðsÞ ¼

k

DðsÞ ðviÞ

This result implies that the characteristic equation characterizes the entire system (particularly,

the natural frequencies and damping ratios) and, no matter what transfer function is measured

(or analyzed), the same natural frequencies and modal damping are obtained.

We can put these transfer functions into the partial fraction form. For example,

ms2 þ cs þ 2k

ðms2 þ cs þ kÞðms2 þ cs þ 3kÞ ¼

A1s þ A2

ðms2 þ cs þ kÞ þ

A3s þ A4

ðms2 þ cs þ 3kÞ ðviiÞ

By comparing the numerator coefficients, we find that A1 ¼ A3 ¼ 0 (this is the case when the modes are

real; with complex modes, A1 – 0 and A3 – 0 in general) and A2 ¼ A4 ¼ 1=2: These results are

summarized below:

G11ðsÞ ¼ G22ðsÞ ¼

􀀑 1=ð2mÞ

s2 þ 2z1v1s þ v21

􀀜 þ

􀀑 1=ð2mÞ

s2 þ 2z2v2s þ v22

􀀜 ðviiiÞ

G12ðsÞ ¼ G21ðsÞ ¼

􀀑 1=ð2mÞ

s2 þ 2z1v1s þ v21

􀀜 2

􀀑 1=ð2mÞ

s2 þ 2z2v2s þ v22

􀀜 ðixÞ

with

v1 ¼

ffiffiffiffiffi

k=m p ; v2 ¼

ffiffiffiffiffiffi

3k=m p ; z1 ¼ c

􀀋 ffiffiffiffiffiffi

4mk p ; and z2 ¼ c

􀀋 ffiffiffiffiffiffiffi

12mk p

By comparing the residues (numerators) of these expressions with the relation expressed in Equation

18.17, we can determine the M-normal modal vectors. Specifically, by examining G11:

ðc 21

Þ1 ¼

1

2m

; ðc 21

Þ2 ¼

1

2m

and by examining at G12:

ðc1c2Þ1 ¼

1

2m

; ðc1c2Þ2 ¼ 2

1

2m

We need consider only two admissible transfer functions (e.g., G11 and G12; or G11 and G21; or G12 and

G22; or G21 and G22) in order to completely determine the modal vectors. Specifically, we obtain

c1

c2

" #

1¼

1=

ffiffiffiffi

2m p

1=

ffiffiffiffi

2m p

2

4

3

5 and

c1

c2

" #

2¼

1=

ffiffiffiffi

2m p

21=

ffiffiffiffi

2m p

2

4

3

5

Note that the modal masses are unity for these modal vectors. Also, there is an arbitrariness in the sign.

As usual, we have overcome this problem by making the first element of each modal vector positive.

Experimental Modal Analysis 18-7

© 2005 by Taylor & Francis Group, LLC

Навигация