3.10 State-Space Approach

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The state-space approach to modal analysis may be used for any linear dynamic system. The starting

point is to formulate a state-space model of the system, which is a set of coupled first-order differential

equations:

x_ ¼ Ax þ Bu ð3:65Þ

where x ¼ state vector; u ¼ input vector; A ¼ system matrix; and B ¼ input gain matrix.

m

m

k

c1

k

c2

k

y2

y1

⇒ f(t)

FIGURE 3.12 A system with linear viscous damping.

m

m

k

c

k

c

k

c y2

y1

⇒ f(t)

FIGURE 3.13 A system with proportional damping in

proportion to stiffness.

m

m

k

c

k

k

c y2

y1

f1(t) f2(t)

FIGURE 3.14 A system with proportional damping in

proportion to inertia.

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There are many approaches to formulating a vibration problem as a state-space model 3.65. One simple

method is to first obtain the conventional, coupled, second-order differential equations:

My€ þ Cy_ þ Ky ¼ f ðtÞ

Next, the state vector and the input vector are defined as

x ¼

y

y_

" #

and u ¼ f ðtÞ ð3:66Þ

Note that Equation 3.59 may be written as

y€ ¼ 2M21Ky 2 M21Cy_ þ M21f ðtÞ ð3:67Þ

which is identical to

y€ ¼ ½2M21K 2 M21C􀀉

y

y_

" #

þ f ðtÞ ð3:67aÞ

This, together with the identity y_ ¼ y_; can be expressed in the form

y_

y€

" #

¼

0 I

2M21K 2M21C

" #

y

y_

" #

þ

0

M21

" #

fðtÞ ð3:68Þ

which is in the state-space form 3.65 where

A ¼

0 I

2M21K 2M21C

" #

and B ¼

0

M21

" #

ð3:69Þ

Note that, in Equation 3.68 and Equation 3.69, I denotes an identity matrix of an appropriate size.

3.10.1 Modal Analysis

Consider the free motion ðu ¼ 0Þ of the nth order system given by Equation 3.65. Its solution is given by

x ¼ FðtÞxð0Þ ð3:70Þ

It is known that the state transition matrix FðtÞ is given by the matrix-exponential expansion equation.

FðtÞ ¼ expðAtÞ ¼ I þ At þ

1

2!

A2t2 þ · · · ð3:71Þ

To discuss the rationale for this exponential response further, we begin by assuming a homogeneous

solution of the form

x ¼ X expðltÞ ð3:72Þ

By substituting Equation 3.72 in the homogeneous equation of motion (that is, Equation 3.65 with

u ¼ 0), the following matrix-eigenvalue problem results:

ðA 2 sIÞX ¼ 0 ð3:73Þ

We shall assume that the n eigenvalues ðl1; l2; …; lnÞ of A are distinct. Then, the corresponding

eigenvectors X1; X2; …; Xn are linearly independent vectors; that is, any one eigenvector cannot be

expressed as a linear combination of the rest of the eigenvectors in the set. Thus, the general solution for

free dynamics is

xðtÞ ¼ X1 expðl1tÞ þ X2 expðl2tÞ þ · · · þ Xn expðlntÞ ð3:74Þ

Modal Analysis 3-37

© 2005 by Taylor & Francis Group, LLC

Each of the n eigenvectors has an unknown parameter. The total of n unknowns is determined using n

initial conditions:

xð0Þ ¼ x0 ð3:75Þ

3.10.2 Mode Shapes of Nonoscillatory Systems

Since the eigenvectors are independent, if the initial state is set at x0 ¼ Xi; then the subsequent motion

should not have any Xj terms with j – i in Equation 3.74. Otherwise, when we set t ¼ 0; Xi becomes a

linear combination of the remaining eigenvectors, which contradicts the linear independence. Hence, the

motion due to this eigenvector initial condition is given by xðtÞ ¼ Xi expðlitÞ; which vector is parallel to Xi

throughout the motion. Thus, Xi gives the mode shape of the system corresponding to the eigenvalue li:

3.10.3 Mode Shapes of Oscillatory Systems

The analysis in the preceding section is valid for real eigenvalues and eigenvectors. In vibratory systems,

li and Xi generally are complex. Let

li ¼ si þ jvi ð3:76Þ

Xi ¼ Ri þ jIi ð3:77Þ

For real systems, corresponding complex conjugates exist:

l􀀊i ¼ si 2 jvi ð3:78Þ

X􀀊 j ¼ Ri 2 jIi ð3:79Þ

Equation 3.76 to Equation 3.79 represent the ith mode of the system. The corresponding damped natural

frequency is vi and the damping parameter is si: The net contribution of the ith mode to the solution

(Equation 3.74) is

ðRi cos vit 2 Ii sin vitÞ2 expðsitÞ

It should be clear, for instance from Equation 3.66, that only some of the state variables in xðtÞ

correspond to displacements of the masses (or spring forces). These can be extracted through an output

relationship of the form

y ¼ Cx ð3:80Þ

The contribution of the ith mode to displacement variables is

Yi ¼ C½Ri cos vit 2 Ii sin vit􀀉2 expðsitÞ ð3:81Þ

If Equation 3.81 can be expressed in the form

Yi ¼ Si sinðvit þ fiÞexpðsitÞ ð3:82Þ

in which Si is a constant vector that is defined up to one unknown, then it is possible to excite the system

so that every independent mass element undergoes oscillations in phase (hence, passing through the

equilibrium state simultaneously) at a specific frequency vi: We have noted that this type of motion is

known as normal mode motion. The vector Si gives the mode shape corresponding to the (damped)

natural frequency vi: A normal mode motion is possible for undamped systems and for certain classes of

damped systems. The initial state that is required to excite the ith mode is x0 ¼ Ri: The corresponding

displacement and velocity initial conditions are obtained from Equation 3.81; thus

Yið0Þ ¼ CRi ð3:83Þ

Y_ ið0Þ ¼ CðRisi 2 IiviÞ ð3:84Þ

Note that the constant factor 2 has been ignored, because Xi is known up to one unknown complex

parameter.

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© 2005 by Taylor & Francis Group, LLC

Example 3.9

A torsional dynamic model of a pipeline segment

is shown in Figure 3.15(a). Free-body diagrams in

Figure 3.15(b) show internal torques acting at

sectioned inertia junctions for free motion. A state

model is obtained using the generalized velocities

(angular velocities Vi) of the inertia elements and

the generalized forces (torques Ti) as state

variables. A minimum set that is required for

complete representation determines the system

order. There are two inertia elements and three

spring elements — a total of five energy-storage

elements. The three springs are not independent,

however. The motion of two springs completely

determines the motion of the third. This indicates

that the system is a fourth-order system. We obtain

the model as follows.

Newton’s Second Law gives

I1

V_ 1 ¼ 2T1 þ T2 ðiÞ

I2

V_ 2 ¼ 2T2 2 T3 ðiiÞ

and Hooke’s Law gives

T_ 1 ¼ k1V1 ðiiiÞ

T_ 2 ¼ k2ðV2 2 V1Þ ðivÞ

Torque T3 is determined in terms of T1 and T2; using the displacement relation for the inertia I2:

T1

k1 þ

T2

k2 ¼

T3

k3 ðvÞ

The state vector is chosen as

x ¼ ½V1; V2; T1; T2􀀉T ðviÞ

The corresponding system matrix is

A ¼

0 0 2

1

I1

1

I1

0 0 2

1

I2

k3

k1

􀀏 􀀐

2

1

I2

1 þ

k3

k2

􀀏 􀀐

k1 0 0 0

2k2 k1 0 0

2

66666666664 3 77777777775

ðviiÞ

The output-displacement vector is

y ¼

T1

k1

;

T1

k1 þ

T2

k2

􀀒 􀀓

ðviiiÞ

(a)

I1

I2

k1 k2 k3

T1 T2 T3

(b)

T1

T1

T2

T2

T2

T2

T3

T3

W1 W2

FIGURE 3.15 (a) Dynamic model of a pipeline

segment; (b) free-body diagrams.

Modal Analysis 3-39

© 2005 by Taylor & Francis Group, LLC

which corresponds to the output-gain matrix

C ¼

0 0

1

k1

0

0 0

1

k1

1

k2

2

6664

3

7775

ðixÞ

For the special case given by I1 ¼ I2 ¼ I and k1 ¼

k3 ¼ k; the system eigenvalues are

l1;l􀀊1 ¼ ^jv1 ¼ ^j

ffiffiffi

k

I

s

ðxÞ

l2;l􀀊2 ¼ ^jv2 ¼ ^j

ffiffiffiffiffiffiffiffiffiffiffi

k þ 2k2

I

s

ðxiÞ

and the corresponding eigenvectors are

X1; X􀀊 1 ¼ R1 ^ j I1 ¼

a1

2 ½v1; v1; 7jk1; 0􀀉T ðxiiÞ

X2; X􀀊 2 ¼ R2 ^ j I2 ¼

a2

2 ½v2; 2v2; 7jk1; ^2jk2􀀉T ðxiiiÞ

In view of Equation 3.81, the modal contributions to the displacement vector are

Y1 ¼

1

1

" #

a1 sin v1t and Y2 ¼

1

21

" #

a2 sin v2t ðxivÞ

This Equation xiv is of the form given by Equation 3.82. The mode shapes are given by the vectors

S1 ¼ ½1; 1􀀉T and S2 ¼ ½1; 21􀀉T; which are illustrated in Figure 3.16. In general, each modal contribution

introduces two unknown parameters, ai and fi; into the free response (homogeneous solution) where fi

are the phase angles associated with the sinusoidal terms. For an n-DoF (order-2n) system, this results

in 2n unknowns, which require the 2n initial conditions x(0).

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1st Mode

Node 2nd Mode

FIGURE 3.16 Mode shapes of the pipeline segment.

3-40 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

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