vibrations chapter 5_2013_mod (1)

7/28/2019 Vibrations Chapter 5_2013_mod (1)

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ENME 361 Spring 2013

ENME 361

Vibrations, Control, and Optimization I

Spring 2013

Chapter 5

Periodic Excitations

Chapter 5 Periodic Excitations

Acknowledgement: Professors B. Balachandran and E. B. Magrab

Assignment 6 due on 3/28/13

Reading assignment: Chapter 4 & 5 and Appendix A and D


In Chapter 5, we shall show how to do the following

Analyze the responses o f single degree-of -freedom systems to harmonicexcitations

Determine the frequency response and phase response of a singledegree-of-freedom system

Interpret the response of a single degree-of-freedom system for anexcitation f requency less than, equal to, and greater than the systems

natural frequency

Determine the system parameters from a measured frequency response

Analyze the responses o f single degree-of -freedom systems wi th rotatingunbalance and with base excitation

Chapter 5 Periodic Excitations

Use accelerometers to measure the responses of sing le degree-of-freedom systems

Isolate vibrations of single degree-of-freedom systems

Analyze the responses of s ingle degree-of-freedom sys tems toexcitations w ith multiple harmonic frequency components


2/48


00

0

0

co

1sin

s sinn

n nd d

t

d

d

d

n

t tV XX e t

e f t

x t

dm

e t

Response

toInitial

Conditions

Response toForcing

21d n

When 0


3/48


Introduction

, ,m k c t x t

Input OutputSystem

x tk

c

m f t


i. Excitation applied from t = 0

CASE 1: Sine Harmonic Excitation

Response to Harmonic Excitation

0sint F t u t

t

1

0

( )u t Unit step function

0 0,

1 0.

tu t

t


4/48


Sine Harmonic Excitation

t

1

sin( ):t

sin( ):u t t

:u t

t

t



0 0,

1 0.

tu t

t

0

0

0

0

0

0sin

sin

s

1sin

sin

sin in

t

nd

d

t

nd

d

tnn

d

d

t

t

t

F ux t e t

F

dm

e t

F

dm

ee t d

m

1 0t

0sinf t F t u t


5/48



: Nondimensional excitation frequency.

: Nondimensional time variable of integration.

n

n

Introduce the nondimensional time ,nt

202

0

sin 1 sin1

F ex e d

k

where,

str sssx x x u

Steady-State PortionTransient Portion

Note that when the excitation frequency is at the natural

frequency; that is, = n, = 1.


Steady-State Portion

2 22

2 22

1

2

1 1: Amplitude Response

1 2

1 2

2tan : Phase Response

1

HD

D

0 sinsssF

x Hk

: Nondimensional excitation frequencyn


6/48


0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

= 0.05

= 0.1

= 0.15

= 0.25

= 0.5

= 1

= /n

H()

Harmonic excitation - Ampli tude Response H()

2 22

1

1 2

H


0 0.5 1 1.5 2 2.5 3

()(rad) = 0.05

= 0.1

= 0.15

= 0.25

= 0.5

= 1

= 0

= 0

0

/2

= /n

Harmonic excitation - Phase Response ()

1 22

tan1


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Transient Portion

0lim sinsssF H

x xk

202

sin 11

str t

H eFx

k

21

2 2

2 1tan

2 1t

After a long period of time many cyclesof forcing:


0 5 10 15 20 25 30 35 40 45 50-1

0

1

xstrans(

)/(F0

/k)

= 0.01, = 0.5

Normalized response of a system to a suddenly applied sine wave forcing function

0 5 10 15 20 25 30 35 40 45 50-2

0

2

xsss

()/(F0

/k)

0 5 10 15 20 25 30 35 40 45 50-2

0

2

x()/(F0/k)

0 5 10 15 20 25 30 35 40 45 50-2

0

2


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0 5 10 15 20 25 30 35 40 45 50-0.5

0

0.5

1

xstrans(

)/(F0/k)

= 0.1, = 0.5


0 5 10 15 20 25 30 35 40 45 50-2

0

2

xsss

()/(F0

/k)

0 5 10 15 20 25 30 35 40 45 50-2

0

2

x()/(F0/k)

0 5 10 15 20 25 30 35 40 45 50-2

0

2


0 5 10 15 20 25 30 35 40 45 50-0.5

0

0.5

xstrans(

)/(F0

/k)

= 0.25, = 0.5


0 5 10 15 20 25 30 35 40 45 50-2

0

2

xsss

()/(F0

/k)

0 5 10 15 20 25 30 35 40 45 50-2

0

2

x()/(F0/k)

0 5 10 15 20 25 30 35 40 45 50-2

0

2


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CASE 2: Cosine Harmonic Excitation


0cosf t F t u t

t

1

0

( )u t

0 0,

1 0.

tu t

t


Cosine Harmonic Excitation

t

1

cos( ) :t

cos( ) :u t t

:u t

t

t


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0 0,1 0.

tu tt

0

0

0

0

0

0

1sin

sin

si

cos

cos

cosn

t

nd

d

t

nd

d

tn

nd

d

t

t

t

x t e t dm

e t dm

e e t

F t u t

F

dm

t

F t

1 0t



: Nondimensional excitation frequency.

: Nondimensional time variable of integration.

n

n

Introduce the nondimensional time ,nt

202

0

sin 1 cos1

F ex e d

k

where,

ctr cssx x x u

Steady-State PortionTransient Portion


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2 22

1

2

1: Amplitude Response

1 2


1

H

0 coscssF

x Hk


202

cos 11

str t

H eFx

k

21

2 2

2 1tan

2 1t


ii. Excitation Present at All Time


0

0

or

sin

sin

f t F t

f F

0F

0F


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2 22

1

2

1 1: Amplitude Response

1 2


1

HD

0 sinssF

x Hk

Steady-StatePhase

Steady-stateAmplitude



0

0

cos

sin2

ss ss

Fdv x H

d k

FH

k

velocThe s ityteady-state is given by,

acceleratThe steady-state is givenion by,

220

2

2

sinss ss

ss

Fda x H

d k

x


13/48


Undamped Systems - Resonance

2

0

1sinnx t x t F t u tm

a) Response when 1n

0

2

0

2

sin sin1

or

sin sin1

n

Fx

k

Fx t t t

k

Unless the ratio is a rational number, the displacement

resp is not peon rise .odic

n


Undamped Systems - Resonance

b) Response when 1n

0 sin cos2

Fx

k

is nThe ot presponse eriodic.

00

sin sin ,

0, 1.

Fx d

k


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1. For the case where , the response of theundamped system always has a finitemagnitude. Thus, it follows that

where is a positive finite number.

Resonance and Stability

0x A

1


2. When , the term grows linearlyin amplitude with time and, hence, it becomeunbounded after a long time.

This special ratio is called aresonance relation; that is, the linear systemis said to be in resonance when the excitationfrequency is equal to the natural frequency.


1

1 n

cos


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3. Since there is always some amount of dampingin the system, it is clear from

that the response remain bounded whenexcited at the natural frequency .


0lim sin2 2

Fx

k

1 n


0 0.5 1 1.5 2 2.5 30

2

4

6

= 0.1, = 0.7

1.8908

= /n

H()

0 0.5 1 1.5 2 2.5 3

() = 0.1, = 0.7

15.3501

0

/2

= /n

0 2 4 6 8 10 12 14 16 18 20-2

-1

0

1

2

f()/F0,xss(

)/(F0/k)

f()/F0

xss

()/(F0/k)


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0 0.5 1 1.5 2 2.5 30

2

4

6

= 0.1, = 15

= /n

H()

0 0.5 1 1.5 2 2.5 3

() = 0.1, = 1

90

0

/2

= /n

0 2 4 6 8 10 12 14 16 18 20-5

0

5

f()/F0,xss(

)/(F0/k)

f()/F0

xss()/(F0/k)


0 0.5 1 1.5 2 2.5 30

2

4

6

= 0.1, = 1.3

1.3562

= /n

H()

0 0.5 1 1.5 2 2.5 3

() = 0.1, = 1.3

159.353

0

/2

= /n

0 5 10 15-2

-1

0

1

2

f()/F0,xss(

)/(F0/k)

f()/F0

xss

()/(F0/k)


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0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

= /n

H()

Three Regions of SDOF System's Amplitude Response - Amplitude Response H( )

Inertia

dominated

1~xm

Stiffness

dominated

1~xk

Damping

dominated

1~xc


0 0.5 1 1.5 2 2.5 3

()(rad)

0

/2

= /n

Three Regions of SDOF System's Amplitude Response - Phase Response ()

Inertia

dominated

1~xm

Stiffness

dominated

1~xk

Damping

dominated

1~xc


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0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

= 0.05

= 0.1

= 0.15

= 0.25

= 0.5

= 1

= /n

H()

Harmonic excitation - Ampli tude Response H()

2 22

1

1 2

H


0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

= /n

H()

Magnitude and Phase Information - Amplitude Response H()

max2

1 1,

22 1H

maxH

max

11,

2H

12

1

2

2max 1 2

22 1 2

n


19/48


Observations

Stiffness-dominated Region

When the amplitude of the harmonic exciting force isconstant and the excitation frequency is much less thanthe natural frequency of the system, the magnitude of thedisplacement is determined bythe systemstiffness.

The displacement response is in phase with the

excitation force.


Observations

Damping-dominated Region

When the amplitude of the harmonic exciting force isconstant and the excitation frequency equals the naturalfrequency of the system, the magnitude of thedisplacement is magnified for , and theamount of magnification is determined by the dampingcoefficient.

The displacement response lags the excitation force

by 90o.

0 1/ 2


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Observations

Inertia-dominated Region

When the amplitude of the harmonic exciting force isconstant and the excitation frequency is much greaterthan the natural frequency of the system, themagnitude of the displacement response isdetermined by the systems inertia. When greaterthan the magnitude of the amplituderesponse is always less than1.

The displacement response is almost 180o

out ofphase with the excitation force.

22 1 2


Frequency-Response Function

1. G() provides a relationship between a systems inputand a systems output.

2. G() is a complex-valued function of frequency, and thisfunction provides information about:

i. the magnitude, and

ii. the phase

of the steady-state response of a linear vibratory systemas a function of the excitation frequency.

Frequency-response function

of a systemG()

ForceInput

Displacementoutput

( )f t ( )x t


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3. For the cases treated, where the force isapplied to the mass directly, this function hasthe following form:

21 ( )( ) , 1( .) ,n

jG H je

k

( ) : provides the magnitude, and

( ): provides the phase lag associated with the response.

H



4. The magnitude of the frequency-responsefunction is given by,

1 ( )( ) ( )

1 ( )( )

1

( ) cos ( ) sin ( )

jG H e

k

jH e

k

H jk

1( ) ( )G H

k


22/482


The Relationship of the Frequency-Response Function to

the Transfer Function

When the initi al conditions are zero, it was shown in

Appendix D that in the Laplace trans form domain

( )( )

( )

F s mX s

D s

The transfer function of the vibratory system is

defined as

( ) Displacement Output( )( ) Force Input

X sG sF s

where

2 2( ) 2 n nD s s s


2 2

2

1 1( )( ) 2

1

1 2

n n

n n

G smD s m s s

k s s

Thus,

We now use this result to obtain the frequency-response

function of the vib ratory system, denoted G(j), which isobtained by setting the complex variable s =j.Hence,

2

2

( )

1( )

1 2

1

1 2

1( ) ( )

n n

j

G jk j

k j

G j H ek


23/482


and, therefore,

( ) ( )H kG j k G j

Thus, the amplitude response is equal to the product o f

the magnitude of the frequency-response function andthe stiffness of the system.

2

2

( )

1( )

1 21

1 2

1( ) ( )

n n

j

G j

k j

k j

G j H ek


Al ternat ives fo rms of the frequency-response funct ion

Assume a forcing function of the form f(t) = Foejt

and a solut ion of the formx(t) =Xo()ejt. Upon

substituting these expressions into

( )( ) jooF

X H ek

2

2 2

1 2 ( ) j to

n n

Fd x dx f t x e

dt dt k k

we find that

The velocity and acceleration are, respectively,

/2

2 2

( )

( )

j tj t j t

o o o

j tj t j t

o o o

x t V e j X e X e

x t A e X e X e


24/482


Sensitivity to System Parameters and Filter Characteristics

Filters

A band pass filteris a system that lets frequency

components in a signal that are within its pass band pass

relatively unattenuated or amplified, while frequency

components in a signal that are outside the pass band are

attenuated.

The pass band is determined by the cutoff frequencies,

which are those frequencies at which

max( )2

HH


max( ) 0dH

d

2max 1 2

max 2

1

2 1H

max( )2

HH

3dB


25/482


Cutoff Frequencies of a Filter

The cutoff frequencies are determinedby solving

max

2 2 22

1 1

2 2 2(1 ) 1 2

H

Solving this equation for the upper and lower cutof f

frequency ratios, we obtain, respectively,

2 2

2 2

1 2 2 1

1 2 2 1

cu

cl

whereandcu cl cu cl

n n


2 2

2 2

1 2 2 1

1 2 2 1

cu

cl

whereandcu cl cu cl

n n

The lower cutoff frequency exists only for those values

of for which01221 22

Upon solving for, we find that1

2 2=0.38272


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The non dimensional filter bandwidth is

for 0.3827cu cl w cu cl n

B

The center frequency c is definedby its geometric mean

c cu cl

The quality factor (Q factor) is defined as

c

w

QB

We note that for < 0.1,1

2c

w

QB

When < 0.1, the error in us ing this approximation is < 3%.


Then,

1 1 2w cu cl B and

21 1 1 1c cu cl

and, therefore,

1

2c

w

QB

2n n

1 2 1

1 2 1

cu

cl

2 2

2 2

1 2 2 1

1 2 2 1

cu

cl


27/482


0 0.5 1 1.5 20

1

2

3

4

5

6

Hmax

Hmax

/2

cl

cu

max

H()

Bw 21

2

For < 0.1 [Error < 3%]

c 1

3 dB re Hmax

The quality factor is a measure of the maximum

magnification of the system.


5.4 Systems with Rotating Unbalanced Mass

tm

Ft

m

mx

dt

dx

dt

xd aonn

sinsin2

22

2

2

where Fa = mo2, m = mo + M, andn

k

m

Setting = nt, we have2

2

22 sin( )

d x dxx M

d d

where

m

mM o

We recall that


28/482


Displacement Response

Noting that Fa/k= M2, we find that

Displacementmagnitude

( ) ( )sin( ( ))ub

Phase

x M H

22

2 22

1

2

( ) ( )

1 2

2( ) tan1

o

ub

mM

m

H H

where ( ) ( )sin ( )oF

x Hk


0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

= 0.05

= 0.1

= 0.15

= 0.25

= 0.5

= 1

= /n

Hub

()

Harmonic excitation due to rotating unbalance -Ampl itude Respons


29/482


0 0.5 1 1.5 2 2.5 3

()

= 0.05

= 0.1

= 0.15

= 0.25

= 0.5

= 1

= 0

= 0

0

/2

= /n

Harmonic excitation due to rotating unbalance - Phase Response


Design Guideline for Vibration Attenuation

In order to reduce the displacement of the massof a single degree-of-freedom system when themass is subjected to a harmonic unbalancedforce, the natural frequency of the systemshouldbe at least twice the excitation frequency or thenatural frequency should be at least 50% lower

than the excitation frequency. These ranges holdirrespective of the systems damping for .0 1


30/483


5.5 Systems with Base Excitation

yd

dyx

d

dx

d

xd

22

2

2

If ( ) sin( )oy y

2

22 2 cos( ) sin( )o o

d x dxx y y

d d

then,

2

( ) ( ) 1 2 sin ( ) ( )ox y H

The governing equation is

where 1

1

2

tan 2

2( ) tan

1


By using the appropriate trigonometric identities, we

obtain

PhaseDisplacementmagnitude

( ) ( )sin( ( ))o mbx y H

1 1 1tan tan tan1

x yx y

xy

where

2

2 22

31

2 2

1 2( )

1 22

( ) tan1 4 1

mbH


31/483


0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

= 0.05

= 0.1

= 0.15

= 0.25

= 0.5

= 1

= /n

Hmb

()

Excitation due to moving base -Ampl itude Response

Notice that the curvesobtained for thedifferent dampingfactors all have thesame amplitude valueat =2.


0 0.5 1 1.5 2 2.5 3

= 0.02

= 0.1

= 0.3

= 0.5

= 0.7

= 1

= 0

= 0

0

/2

= /n

()

Excitation due to moving base - Phase Response


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Design Guideline for Vibration Attenuation

In order to reduce the displacement of the massof a single degree-of-freedom system when thebase is subjected to a harmonic excitation, thenatural frequency of the system should be eitherfive times higher than the excitation frequency orthe natural frequency should be at least 30%lower than the excitation frequency.


5.7 Vibration IsolationSystem with Direct Excitation of Inertial Element

The displacement response is

( ) ( )sin ( )oF

x Hk

The force transmitted to the base (ground)

( ) ( ) 2Tdx

F k xd

Then

( ) ( ) sin ( ) 2 cos ( )

( )sin ( )

T o

o mb

F F H

F H

23

1

2 22 22

1 2 2( ) , ( ) tan

1 4 11 2mbH

where


33/483


Then, the magnitude of the ratio of the force transmitted to

the ground to that applied to the mass is

( )( )T mb

o

FH

F

Recall that for System Subjected to Base Excitation

We obtained that

( ) ( )sin( ( ))o mbx y H

Then, the magnitude of the ratio of the displacement

transmitted to the mass to that applied to the base is

( )( )mb

o

xHy


Transmissibility Ratio

To minimize the force transmitted to the base from the

vibrations of a directly excited system or to minimize

the magnitude of the base motion on a system, we

require that

( ) 1mbH

We define the transmissibility ratio TRas

2

2 22

1 2( )

1 2

mbTR H


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2 3 4 5 6 7 8 9 100

5

10

15

20

25

30

=0

=0.1

=0.2

=0.4

TR(%)

0 0.5 1 1.5 2 2.50

1

2

3

4

5

6

Hmb

()

= 0.02

= 0.1

= 0.3

= 0.5

= 0.7

TR(%) = 100TR

2

2 22

1 2

1 2

TR

We examine TR for > 2 and < 0.4.


5.6 Acceleration Measurement: Accelerometer

Piezoelectricelement

m

kc

y(t)

x(t)

Voltageoutput

y(t)

Piezoelectricelement

m

k c

x(t)

Voltageoutput


35/483


2

2

2

2

2

d

ydz

d

dz

d

zd

The governing equation is

If the base is subjected to a harmonically varying displacement,

then ( ) sin( )oy y

22

22 sin( ) sin( )o o

d z dz z y a

d d

Thus,

where aon2 = yo2. Then, the solu tion is ( ) ( )sin ( )oz a H

Therefore, if the acceleration amplitude response and the

phase response are to be relatively constant over a wide

frequency range, then the parameters of the accelerometer

have to be chosen so that H() varies by less than d,where |d|


36/483


Since H(0) =1 and the damping factor is very low, the

frequency range is determined from

22 22

1 11

11 2

d

or

11

1a

d

0 0.5 1 1.5 2 2.50

1

2

3

4

5

6

H()

= 0.02

= 0.1

= 0.3

= 0.5

= 0.7

d

a


Example 5.10: Design of an accelerometer

We shall determine the working range of an accelerometer

with a natural frequency of 60 kHz and whose variation in

the amplitude response is less than 2%.

1 11 1 0.14

1 1 0.02a

d

where a = a/n = fa/fn and fa is the frequency below whichthe amplitude response varies by less than d.

Thus, fa = (0.14)(60 kHz ) = 8.4 kHz.

In addition, the phase response () 0 for 0 awhen is small. Hence, the working range of theaccelerometer is 0 < fa 8.4 kHz.

For d= 0.02


37/483


5.9 Response to Excitation with Harmonic Components

1. Excitation with two harmonic components:

or, 1 sint tBf

1 sinx Hk

B

1 sinBf

1 22 221 2

, tan 11 2

H

,n

n

t


Response to Excitation with Harmonic Components

or,

2 cost tAf

2 cosx Hk

A

2 cosAf

1 22 221 2

, tan1

1 2

H

,n

n

t


38/483



Let,

1 2

sin cos

cos sin

x x x

H Hk k

H

kA

B

B

A

1 2 sin cosBf f Af

1 2x x x LINEAR SYSTEM



Then,

where,

sinH

xk

C

2

1

2

tan

B

B

C A


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,H 1f 1x

,H 2f 2x

,H 1 2f f 1 2x x

Linear System



2. Excitation with multiple harmonic components

or,

where the are distinct.

1

cos sinN

j j j j

j

t A t B t

, andj

n j

n

t

1

cos sinN

j j j j

j

f A B

k

j

j j j

n n

t


40/484



where,

The corresponding displacement response to isgiven by

1

N

j

j

f f

cos sinj j j j jf A B

jf

cos

sin

j

j j j j

j j j

Hx A

k

B



where,

sinj

j jj j

Hx C

k

2

1

2

tanj

j j j

j

j

j

j

n

C A B

B


41/484



Let,

LINEAR SYSTEM

11

sin

N

j j j j jj

x H Ck

1

N

j

j

x x

1

N

j

j

f f



,j jH jf jx

1, ,j N

,j jH 1

N

j

j

f

1

N

j

j

x

1, ,N

Linear System


42/484


3. Periodic excitation

Fundamental Frequency:


T

f t

t

t f t n T

t f t n T

00 22

TT

Period


T

t

f(t)

noo 2o 3o 4o

an a1a2

a3a4

Notice that spectrum isdiscrete

The Fourier series decomposes a periodic signal in the time

domain into its discrete frequency components; that is, the

time-varying signal has been transformed into the frequency

domain.



43/484



Fourier Series:

where,

0 0 01

cos sin2

i i

i

bt i t i t a

a

0

0

0

0 , 0,1,2,

, 12

2cos

,n ,2siT

i

T

ia f t i t d

b f t i t d T

i

T

t

it

Fourier

coefficients


Fundamental frequency:

for are called the higher harmonics of

0 0 01

cos sin2

i i

i

af t a i t b i t

0i

Fourier-series expansion off

1i 0

0



44/484



LINEAR SYSTEMS:

0

1

1cos

2

sin

i i i i

i i i

i

ax H a

k k

b

01

1sini i i i i

ix c c

k


01

1sin i i ii i

i

cxk

c

1

0

2 2

0

2

tan ii

i i i i i

i

a

H a b

c

b

c

a

0i

n

i



45/484


,i iH 1,2,i Linear System

01

1sini i i i i

i

x c ck

0 0 01

cos sin2

i i

i

af t a i t b i t



Glossary Chapter 5

Accelerancethe rat io of the Laplace transform of a

vibratory systems acceleration output to the Laplace

transform of the force input; the corresponding frequency-

response function is obtained by substitut ing s =j.Accelerometer a dev ice whose output is propor tional to

acceleration.

Admi ttance the ratio o f the Laplace transfo rm of a

vibratory systems output displacement to the Laplace

transform of the force input; the corresponding frequency-

response function is obtained by substitut ing s =j; usedsynonymously w ith receptance and compliance functions.


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Ampl itude response the non-dimensional frequency-

response function used to relate the output response ofa linear vibratory system to the input.

Band pass filter a system that allows frequency

components in the input that are within the region

defined by it s lower and upper cutoff frequencies to

pass relatively unattenuated while those frequency

components outside th is region are attenuated.

Bandwidth the frequency range for a system; fo r a

band pass filter, this is the difference between the upper

and lower cutoff frequencies; for a low pass f ilter, it is

the upper cutoff frequency; for a high pass filter, it is

the lower cutof f f requency.


Cutoff frequency fo r a vibrato ry system, the frequency at

which the magnitude of the displacement response has

decreased to 0.7071 (or 1/2) of its maximum value; that is,the system, has the half the power it has at the maximum

value.

Damping-dominated response the excitation frequency

range, in which the system response is heavily influenced

by the viscous damping of the system; in this range, the

response is inversely proportional to the damping

coefficient.

Frequency-response function the ratio of the output of asystem to the input of the system as a function of

frequency; can be obtained from the transfer function by

substituting s =j.


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Half-power poin ts exc itation frequencies fo r which the

displacement response has half the power that it has at its

maximum value; the frequencies cor responding to the half-

power points are used to define the cutoff frequencies.

Harmonic any frequency that is an in teger mu lt ip le of a

basic frequency.

Harmonic excitation in mechanical vibrations it refers to

either a sine function of given amplitude and frequency, a

cosine function of g iven amplitude and frequency, or the

combination of a sine function of given amplitude and

cosine function of g iven amplitude each with the same

frequency.


High pass filter a system that allows frequency

components in the input that are greater than its cutoff

frequency to pass relatively unattenuated while those

frequency components below it are attenuated.

Inertia-dominated response the exci tat ion frequency

range in which the system response is heavily influenced

by the inertia of the sys tem; in th is range, the response is

inversely proportional to the system mass.

Low pass filter a system that allows f requency

components in the input that are less than its cuto ff

frequency to pass relatively unattenuated while thosefrequency components above it are attenuated.


48/48


Periodic excitation a time-varying exc itation whose

waveform repeats itself every constant time interval calledthe period; the simplest type of periodic excitation is a

waveform that varies with time as a sine or cosine

function.

Phase response the phase lead or lag of the ampl itude

response with respect to the input as a function of

frequency.

Resonance the condi tion at wh ich the exci tat ion

frequency i s equal to one of the natural frequencies of a

system; also refers to the response of an undamped or

lightly damped linear system at frequencies close to or at

one of the systems natural frequencies; for a nonlinearsystem, resonances can occur at rational multiples of a

natural frequency

Stiffness-dominated response the exci tat ion frequency

range in which the system response is heavily influenced

by the stiffness of the system; in this range, the

response is inversely proportional to the system

stiffness

Transmissibility ratio a measure of the amount of the

applied force to the mass that is t ransmitted to the

ground or the amount of displacement applied to the

base that is transmitted to the mass.

Vibration i solationmeans used to reduce a systems

transmissibility ratio.

vibrations chapter 5_2013_mod (1)

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