Пресс-релиз популярных книг

Авторы: 111 А Б В Г Д Е Ж З И Й К Л М Н О П Р С Т У Ф Х Ц Ч Ш Щ Э Ю Я

Книги: 164 А Б В Г Д Е Ж З И Й К Л М Н О П Р С Т У Ф Х Ц Ч Ш Щ Э Ю Я

На сайте 111 авторов, 92 книг, 72 статей, 5913 глав.

2.3 Transform Techniques

Back

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

FðsÞ ¼

ð1

f ðtÞ expð2stÞdt ð2:46Þ

in which s is a complex independent variable known as the Laplace variable, expressed as

s ¼ s þ jv ð2:47Þ

and j ¼

ffiffiffiffi

21 p : Laplace transform operation is denoted as Lf ðtÞ ¼ FðsÞ: The inverse Laplace transform

operation is denoted by f ðtÞ ¼ L21FðsÞ and is given by

f ðtÞ ¼

2p j

ðsþj1

s2j1

FðsÞ expðstÞds ð2:48Þ

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

0 1 Excitation Frequency Ratio r

Magnitude of

Restraining

Torque t0

ka0

FIGURE 2.6 Variation of the steady-state transmitted torque with frequency.

2-14 Vibration and Shock Handbook

f ðtÞ; the Laplace transform exists if the integral in Equation 2.46 converges. A sufficient condition for

this is

ð1

lf ðtÞl expð2stÞdt , 1 ð2:49Þ

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

Y ðsÞ ¼

ð1

hðtÞuðt 2 tÞdt expð2stÞdt ð2:50Þ

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,

Y ðsÞ ¼

ð1

uðt 2 tÞ exp½2sðt 2 tÞ􀀉dðt 2 tÞ

ð1

hðtÞ expð2stÞdt

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

Y ðsÞ ¼ HðsÞU ðsÞ ð2:51Þ

in which

HðsÞ ¼ LhðtÞ ¼

ð1

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:

FðjvÞ ¼

ð1

f ðtÞ expð2jvtÞdt

Inverse:

f ðtÞ ¼

ðj1

2j1

FðjvÞ expðjvtÞdv

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

dny

dtn þ an21

dn21y

dtn21 þ · · · þ a0y ¼ b0u þ b1

duðtÞ

dt þ · · · þ bm

dmuðtÞ

dtm ð2:53Þ

Frequency-Domain Analysis 2-15

for physically realizable systems, m # n: By applying a Laplace transformation and then integrating by

parts, it may be verified that

dkf ðtÞ

dtk ¼ skF^ðsÞ 2 sk21f ð0Þ 2 sk22 df ð0Þ

2 · · · þ

dk21f ð0Þ

dtk21 ð2:54Þ

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

HðsÞ ¼

b0 þ b1s þ · · · þ bmsm

a0 þ a1s þ · · · þ ansn 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

Hð f Þ ¼

ð1

hðtÞ expð2j2p ftÞdt 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 FhðtÞ ¼ Hð f Þ:

TABLE 2.1 Important Laplace Transform Relations

L21 FðsÞ ¼ f ðtÞ Conversion Lf ðtÞ ¼ FðsÞ

2pj

Ðsþj1 s2j1 FðsÞ expðstÞds

f ðtÞ expð2stÞdt

k1 f1 ðtÞ þ k2 f2ðtÞ k1 F1 ðsÞ þ k2 F2 ðsÞ

expð2atÞf ðtÞ Fðs þ aÞ

f ðt 2 tÞ expð2tsÞFðsÞ

f ðnÞðtÞ ¼

dn f ðtÞ

dtn sn FðsÞ 2 sn21 f ð0þÞ 2 sn22 f 1 ð0þÞ 2 · · · 2 f n21ð0þÞ

Ðt2

1 f ðtÞdt

FðsÞ

s þ

Ð02

1 f ðtÞdt

Impulse function, dðtÞ 1

Step function, UðtÞ

tn n!

snþ1

expð2atÞ

s þ a

sin vn t

s2 þ v2

cos vn t

s2 þ v2

2-16 Vibration and Shock Handbook

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

U ð f Þ ¼ Uðj2p f Þ ¼ UðjvÞ

and

Y ð f Þ ¼ Y ð j2p f Þ ¼ Y ð jvÞ:

Then, from Equation 2.51, we have

Y ð f Þ ¼ Hð f ÞUð fÞ 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 Y ð 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 b1j2pf þ · · · þ bmðj2pf Þm

a0 þ a1j2pf þ · · · þ anðj2pf 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 u0ðcos 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

yðtÞ ¼ Re

ðt

hðtÞu0 exp½j2p f ðt 2 tÞ􀀉dt

􀀒 􀀓

ð2:60aÞ

yðtÞ ¼ Re u0 expðj2p ftÞ

ðt

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􀀉 ð2:61aÞ

Frequency-Domain Analysis 2-17

yss ¼ u0lHð 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

s2 þ 2zvns þ v2

n ð2:62Þ

The characteristic equation of this system is given by

s2 þ 2zvns þ v2

n 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,

l1; l2 ¼ 2zvn ^ jvd ð2:64Þ

in which

vd ¼

ffiffiffiffiffiffiffiffi

1 2 z2

vn ð2:65Þ

is the damped natural frequency.

Case 2: ðz > 1Þ

This case corresponds to real and unequal eigenvalues,

l1; l2 ¼ 2zvn ^

ffiffiffiffiffiffiffiffi

z2 2 1

vn ¼ 2a; 2b ð2:66Þ

with a – b; in which

ab ¼ v2

n ð2:67Þ

2-18 Vibration and Shock Handbook

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:

yimpulseðtÞ ¼ hðtÞ ¼

ffiffivffiffinffiffiffiffi

1 2 z2

p expð2zvntÞ sin vdt for z , 1 ð2:70aÞ

yimpulseðtÞ ¼ hðtÞ ¼

ðb 2 aÞ ½expð2atÞ 2 expð2btÞ􀀉 for z . 1 ð2:70bÞ

yimpulseðtÞ ¼ hðtÞ ¼ v2

nt expð2vntÞ for z ¼ 1 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,

dðtÞ ¼

dUðtÞ

dt ð2:72Þ

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

YstepðsÞ ¼

ðs2 þ 2zvns þ v2

nÞ 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

s þ

a2 þ a3s

ðs2 þ 2zvns þ v2

nÞ

in which the constants a1; a2; and a3 are determined by comparing the numerator polynomial; thus,

n ¼ a1ðs2 þ 2dvns þ v2

nÞ þ sða2 þ a3sÞ

Then a1 ¼ 1; a2 ¼ 22zvn; and a3 ¼ 1:

The following results are obtained:

ystepðtÞ ¼ 1 2

1 ffiffiffiffiffiffiffiffi

1 2 z2

p expð2zvntÞ sinðvdt þ fÞ for z , 1 ð2:74aÞ

ystep ¼ 1 2

ðb 2 aÞ ½b expð2atÞ 2 a expð2btÞ􀀉 for z . 1 ð2:74bÞ

ystep ¼ 1 2 ðvnt þ 1Þ expð2vntÞ for z ¼ 1 ð2:74cÞ

In Equation 2.74c,

cos f ¼ z ð2:75Þ

Frequency-Domain Analysis 2-19

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

From Equation 2.77a it follows that,

XðsÞ ¼ ðsI 2 AÞ21B þ UðsÞ ð2:78Þ

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 AÞ21B þ D􀀉UðsÞ ð2:79aÞ

YðsÞ ¼ GðsÞUðsÞ ð2:79bÞ

The transfer-function matrix GðsÞ is an m £ r matrix given by

GðsÞ ¼ CðsI 2 AÞ21B 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 AÞ21B 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

my€ þ by_ 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 kÞY ðsÞ ¼ kUðsÞ ðiiÞ

The corresponding transfer function is

GðsÞ ¼

Y ðsÞ

U ðsÞ ¼

ms2 þ bs þ k ðiiiÞ

2-20 Vibration and Shock Handbook

Box 2.2

USEFUL FREQUENCY-DOMAIN RESULTS

Laplace Transform (L):

FðsÞ ¼

ða

f ðtÞ expð2stÞdt

Fourier Transform ðF):

FðjvÞ ¼

ða

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:

Y ðjvÞ ¼ HðjvÞU ðjvÞ

where

UðjvÞ ¼ Fourier spectrum of input uðtÞ

Y ðjvÞ 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 AÞ21B þ D

Frequency-Domain Analysis 2-21

or, in terms of the undamped natural frequency vn and the damping ratio z; where v2

n ¼ k=m and

2zvn ¼ b=m; the transfer function is given by

GðsÞ ¼

s2 þ 2zvns þ v2

n ðivÞ

This is the transfer function corresponding to the displacement output. It follows that the output

velocity transfer function is

sY ðsÞ

U ðsÞ ¼ sGðsÞ ¼

sv2

s2 þ 2zvns þ v2

n ðvÞ

and the output acceleration transfer function is

s2Y ðsÞ

UðsÞ ¼ s2GðsÞ ¼

s2v2

s2 þ 2zvns þ v2

n 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 2zvny_ þv2

ny ¼ v2

nuðtÞ 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

x_ ¼

0 1

2v2

n 22zvn

" #

x þ

" #

uðtÞ ðiiiÞ

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

y ¼ x 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:

YðsÞ ¼

s 21

n s þ 2zvn

" #21 0

" #

UðsÞ ðvÞ

YðsÞ ¼

½s2 þ 2zvns þ v2

n􀀉

s þ 2zvn 1

2v2

n s

" #

U ðsÞ

YðsÞ ¼

½s2 þ 2zvns þ v2

n􀀉

sv2

" #

UðsÞ ðviÞ

The transfer function matrix is

GðsÞ ¼

n=DðsÞ

sv2

n=DðsÞ

" #

ðviiÞ

2-22 Vibration and Shock Handbook

in which DðsÞ ¼ s2 þ 2zvns þ v2

n 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 22zvny_ 2v2

ny þ v2

n uðtÞ ðviiiÞ

or, in terms of the state variables,

y3 ¼ 22zvnx2 2 v2

nx1 þ v2

nuðtÞ ðixÞ

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,

Y3ðsÞ ¼ 22zvnX2ðsÞ 2 v2

nX1ðsÞ þ v2

nU ðsÞ ¼ 22zvn

sv2

DðsÞ

UðsÞ 2 v2

DðsÞ

UðsÞ þ v2

nU ðsÞ

which simplifies to

Y3ðsÞ ¼

s2v2

DðsÞ

U ðsÞ 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.

Acceleration

Force

f(t)

Mass m

Mechanical

Load

Linear

Motor

Instrumentation

Charge

Amplifier

Spectrum

Analyzer

a Accelerometer

Force

Sensor

FIGURE 2.7 Measurement of the acceleration spectrum of a mechanical system.

Frequency-Domain Analysis 2-23

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

my€ þ by_ 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

my€ ¼ f ðtÞ 2 by_ 2 ky ðiÞ

Hence, the equation of motion is

my€ þ by_ þ ky ¼ f ðtÞ ðiiÞ

The displacement transfer function is

f ¼

ðms2 þ bs þ kÞ ðiiiÞ

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

f ¼

ðms2 þ bs þ kÞ ¼ GðsÞ ðivÞ

b y

k y

f (t)

FIGURE 2.8 Free-body diagram.

2-24 Vibration and Shock Handbook

In the frequency domain, the corresponding frequency-response function is obtained by substituting jv

for s: Hence,

GðjvÞ ¼

2v2

ð2mv2 þ bjv þ kÞ ðvÞ

Divide throughout by m and use b=m ¼ 2zvn and k=m ¼ v2

n; where vn ¼ undamped natural

frequency and z ¼ damping ratio.

Then,

GðjvÞ ¼

2v2=m

ðv2

n 2 v2 þ 2jzvnvÞ ¼

2r2=m

1 2 r2 þ 2jzrr ðviÞ

where r ¼ v=vn: The magnitude of GðjvÞ gives the amplification of the acceleration signal with respect to

the forcing excitation:

lGðjvÞl ¼

r2=m ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ð1 2 r2Þ2 þ ð2zrÞ2

p ðviiÞ

Its peak value corresponds to the peak value of

pðrÞ ¼

ð1 2 r2Þ2 þ ð2zrÞ2 ðviiiÞ

and gives the resonance. This occurs when dp=dr ¼ 0: Hence,

½ð1 2 r2Þ2 þ ð2zrÞ2􀀉4r3 2 r4½2ð1 2 r2Þð22rÞ þ 8z2r􀀉 ¼ 0

The solution is

r ¼0 or ½ð1 2 r2Þ2 þ 4z2r2􀀉 þ r2½1 2 r2 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 r2Þ2 þ 4z2r2 þ r2 2 r4 2 2z2r2 ¼ 0

which has the valid root

r ¼

1 ffiffiffiffiffiffiffiffiffiffi

1 2 2z2

p ðixÞ

Note that r has to be real and positive. It follows that, for a resonance to occur we need

0 , z ,

1ffiffi

2 p

Substitute in (vii), the resonant value of r; to obtain

lGlpeak ¼

mð1 2 2z2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÞffiffiffiffiffiffiffiffiffiffiffiffiffi

1 2

1 2 2z2

􀀏 􀀐2

4z2

1 2 2z2

lGlpeak ¼

2mz

ffiffiffiffiffiffiffiffi

1 2 z2

p ðxÞ

Пресс-релиз популярных книг

2.3 Transform Techniques

Популярные книги

Популярные статьи

Навигация