Chapter 03 Mathematical Methods - University of...

1

Slides to accompany lectures in

Vibro-Acoustic Design in Mechanical Systems © 2012 by D. W. Herrin

Department of Mechanical Engineering University of Kentucky

Lexington, KY 40506-0503 Tel: 859-218-0609

[email protected]

Chapter 3 – Mathematical Methods

ME 510 Vibro-Acoustic Design

Euler’s Formula

θθθ sincos je j +=

The Power of Complex Numbers

( ) ( ) ( ) ( ) ( ) ( ) ( )0!

0!4

0!3

0!2

44

33

22

1 nn

fnxfxfxfxaxfafxf +⋅⋅⋅+++++=

Taylor’s Series Proof

⋅⋅⋅+−+−=!6!4!2

1cos642 θθθ

θ

⋅⋅⋅+−+−=!7!5!3

sin753 θθθ

θθ

⋅⋅⋅+++−−+= jjje j!5!4!3!2

15432 θθθθ

θθ

Dept. of Mech. Engineering University of Kentucky

2


Complex Numbers - Example

jz 321 +=

Addition and Subtraction

jzz 2721 +=+

jz 152 −=

jzz 4321 +−=−


3 ME 510 Vibro-Acoustic Design

Complex Numbers - Example

jz 321 +=

Multiplication and Division (Rationalization)

jjjzz 131332151021 +=+−+=

jz 152 −=

( )( )( )( ) 26

17755532

*22

*21

2

1 jjjjj

zzzz

zz +

=+−

++==


4


Complex Numbers - Review Close and Frederick, 2002



Complex Numbers - Review Close and Frederick, 2002


6

2


Polar Form - Review

θcoszx =θsinzy =

22 yxz +=

⎟⎠

⎞⎜⎝

⎛=xyarctanθ

θjezz =

Close and Frederick, 2002



Complex Number Notations - Review

jyxz +=

( )θθ sincos jzz +=

θ∠= zzθjezz =

Rectangular Form

Polar Form

Close and Frederick, 2002


8


Polar Form - Review

jjezj

=⎟⎠

⎞⎜⎝

⎛+⎟⎠

⎞⎜⎝

⎛==⎟⎠

⎞⎜⎝

⎛

2sin

2cos2 ππ

π

jjezj

−=⎟⎠

⎞⎜⎝

⎛+⎟⎠

⎞⎜⎝

⎛==⎟⎠

⎞⎜⎝

⎛

23sin

23cos2

3ππ

π

( ) ( ) ( ) 1sincos −=+== πππ jez j

( ) ( ) ( ) 12sin2cos2 =+== πππ jez j



Polar Form - Examples

jz 321 += jz 152 −=

jez 983.01 13= jez 197.0

2 26 −=

jezzj

1313338 421 +==

⎟⎠

⎞⎜⎝

⎛ π

26177

21

2613 18.1

197.0

983.0

*22

*21

2

1 jeee

zzzz

zz j

j

j +====

−


10


Polar Form - Examples

jz 321 += jz 152 −=

°∠= 3.56131z °∠=°−∠= 7.348263.11262z

( ) °∠=°−°∠= 453383.113.5633821zz

( ) °∠=°−−°∠=°−∠

°∠= 6.67

213.113.56

21

3.11263.5613

2

1

zz



Polar Form – Complex Conjugate

jz 321 +=

jez 983.0*1 13 −=

26177

21

2626

2613 18.1

197.0

197.0

197.0

983.0

*22

*21

2

1 jeee

ee

zzzz

zz j

j

j

j

j +====

+

+

−

jz 32*1 −=

jez 983.01 13=


12

3


MATLAB Commands

r = abs(z)

theta = angle(z)

x = real(z)

y = imag(z)


13 ME 510 Vibro-Acoustic Design Dept. of Mech. Engineering University of Kentucky

14

Example 1 Write the harmonically varying force as a complex force.

F t( ) = F cos ωt( )F t( ) = Fe jωt = F cos ωt( )+ j sin ωt( )( )F t( ) = Re Fe jωt( )

ME 510 Vibro-Acoustic Design Dept. of Mech. Engineering University of Kentucky

15


F t( ) = F1 cos ωt( )− F2 sin ωt( )F t( ) = F1 + jF2( )e jωt = F1 + jF2( ) cos ωt( )+ j sin ωt( )( )F t( ) = F1 cos ωt( )− F2 sin ωt( )+ j F1 sin ωt( )− F2 cos ωt( )( )


16


F t( ) = F1 cos ωt( )− F2 sin ωt( ) = F cos ωt +ϕ( )

F = F 21 + F

22

ϕ = arctan F2 F1( )F t( ) = Fe jωt+ jϕ = Fe jωt

F t( ) = Re Fe jωt( )


17

Displacement, Velocity and Acceleration

v t( ) = ve j ωt+ϕv( )

a t( ) = ddtv t( ) = jωve j ωt+ϕv( ) =ωve j ωt+ϕv+π /2( )

x t( ) = v t( )∫ dt = 1jωve j ωt+ϕv( ) =

1ωve j ωt+ϕv−π /2( )


18

Complex Conjugate

F = FF* = Fe j ωt+ϕ( )Fe− j ωt+ϕ( ) = FF = F

Re F( ) = 12F+F*( )

=12F cos ωt( )+ j sin ωt( )( )+ F cos ωt( )− j sin ωt( )( )( )

= F cos ωt( )

4


19

Derivation Mechanical Power F t( ) = F cos ωt +ϕF( ) = Re Fe j ωt+ϕF( )( )v t( ) = vcos ωt +ϕv( ) = Re ve j ωt+ϕv( )( )W t( ) = F t( )v t( ) = Re F t( )( )Re v t( )( )

W t( ) = 12F t( )+F t( )*( ) 12 v t( )+ v t( )

*( )W t( ) = 1

4F t( )v t( )+F t( )* v t( )* +F t( )* v t( )+F t( )v t( )*( )

W t( ) = 12Re F t( )v t( )( )+Re F t( )* v t( )( )( )

W t( ) = Re Fve j 2ωt+ϕF+ϕv( )( ) 2+Re Fve j ϕF−ϕv( )( ) 2W t( ) = Fv cos 2ωt +ϕF +ϕv( )+ cos ϕF −ϕv( )( )


20

Derivation Mechanical Power

W t( ) = Fv cos 2ωt +ϕF +ϕv( )+ cos ϕF −ϕv( )( )

W =1T

W t( )dt0

T

∫ =12Fvcos ϕF −ϕv( )

W =12Re Fv*( ) = 12 Re F

*v( )


The Transfer Function

Hi1 =aiF1=outputinput

•  Measure using impact testing or shaker.

•  Simulate by using unit force and determining acceleration.

F1 ai



Transfer Functions

•  Measure using impact testing or shaker.

•  Simulate by using unit force and determining acceleration.

Unit Force


22


Assumption Superposition

F1 F2 F2

F1

+ =

aiI ai

II aiI + ai

II

aiI + ai

II = Hi1F1 +Hi2F2



Assumption Linearity

F1 ai 2×F1 2×ai

Doubling the force will double the response.


24

5


Linear Superposition

+

=

+ +

In Operation

F1 F2

F3 F4Dept. of Mech. Engineering University of Kentucky


M B

K

Total response = Transient Response + Steady State Response •  Transient response depends on (a) the system parameters (M, B,

and K) and (b) the I.C’s

•  Steady state response depends on (a) the system parameters and (b) the amplitude and frequency of the forcing function

Equation of Motion:

Single DOF Systems

( )tf

( )tx

( ) ( )tFtfKxxBxM ωcos==++


26


0 2 4 6 8 10

Time (s)

f(t) a

nd x

(t)

Transient Steady StateTotal Response Forcing Function

( ) ( ) ( )φωθωζω −++= − tXtAetx Dtn coscos

Transient and Steady State Response



Complex Exponential Form

M B

K ( )tf

( )txMx +Bx +Kx = f t( ) = F cos ωt( )x t( ) = Xe jωt f t( ) = Fe jωt

−Mω 2 + jωB+K( ) Xe jωt = Fe jωt

X = FK −Mω 2( )+ jωB

=F

K −Mω 2( )2+ ωB( )2

e− jϕ

x t( ) = Re Xe jωt( )= F

K −Mω 2( )2+ ωB( )2

cos ωt −ϕ( )

ϕ = arctan ωBK −Mω 2

"

#$

%

&'


28


29

The Amplification Factor ω0 =

KM

δ =B2M

g t( ) =F t( )M

X = g

ω02 −ω 2( )

2+ 2δω( )2

ϕ = arctan 2δωω 2 −ω0

2

"

#$

%

&'+ nπ , n = 0,1, 2,...

φ =X ω( )

X ω = 0( )=

1

1− ωω0( )

2"

#$

%

&'

2

+ 4 δω0( )

2ωω0( )

2

The Amplification Factor


-1.5

-1

-0.5

0

0.5

1

1.5

0 0.005 0.01 0.015 0.02Time (s)In

put,

Out

put

f (t)

x(t)

Vibrating System with transfer function H(s)

( ) ( )tFtf ωcos= x t( ) = X cos ωt −ϕ( )

H s( )→s= jω

H jω( )∠ϕ

X∠ϕ = F H jω( )∠ϕ

Frequency Response Function

Sinusoidal Steady State Response


30

6


31

Frequency Response Functions

H ω( ) =Y ω( )X ω( )

H ω( ) = x ω( ) F ω( )Dynamic Flexibility or Receptance

Mobility or Mechanical Admittance

Accelerance

Dynamic Stiffness

Mechanical Impedance

Acoustic Impedance

Specific Impedance

H ω( ) = v ω( ) F ω( )A ω( ) = a ω( ) F ω( )

κ ω( ) = F ω( ) x ω( )

Z ω( ) = F ω( ) v ω( )

Z ω( ) = p ω( ) Q ω( )

Z ω( ) = p ω( ) u ω( )


32

Frequency Response Functions

H ω( ) =X ω( )F ω( )

=1

−ω 2M + jω2Mδ +K=

1 K1− ω ω0( )2 + j2 ωδ ω0

2( )

δ =ω0

δ >ω0

δ <ω0

δ = 0

Critically Damped

Overdamped

Underdamped

Undamped

ξ =δω0

Often a damping ratio is defined


33

Dynamic Flexibility Magnitude


34

Dynamic Flexibility Phase


35

Nyquist Diagram


Derivation of Loss Factor

Ekin =M2ω 2x2 cos2 ωt +ϕ( )

Epot =K2x2 sin2 ωt +ϕ( )

Edis = Fd dx∫ = B∫ dxdtdx = B∫ dx

dt"

#$

%

&'2

dt

= Bx2ω 2 cos2 ωt +ϕ( )dt0

T

∫ = Bx2ω 2 T2= Bx2ω 2 2π

2ω= Bx2ωπ

η =Edis 2πmax Epot( )

=ωBK

= 2ωδω02


36

7


Material Damping

Creep Test

σ0 σ

t

ε0

ε

t

σ0 J(t) ε∞

Relaxation Test

ε0 ε

t

σ0 σ

t

ε0 E(t) σ∞



The Kelvin-Voigt 2-Parameter Model

E R

Creep Compliance ( ) ( )

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

=

− tRE

eE

tJ

tJt

110σε

Stress Relaxation

( ) ( )( ) EtE

tEt=

= 0εσ

Material Damping


38


The 3-Parameter Model

E0 R

Creep Compliance ( ) ( )

( )⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

=

− tqq

eqqp

qtJ

tJt

1

0

1

01

0

0

111

σε

E1

Constants

1

101

00

11

EEERq

EqERp

+=

=

=

Stress Relaxation ( ) ( )

( )⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

=

− tpe

pqqqtE

tEt

1

1

1

100

0εσ

Material Damping



For Low Frequency Harmonic Processes the Kelvin-Voigt Model is Applicable

E ω( ) ≈ E0 + jωR = E0 1+ jη( )

( )02 ER

UWD ωπ

ωη ≈=

The Material Loss Factor (Steel ~ .0001)

Energy Dissipated per Cycle

Maximum Potential Energy

Material Damping


40


Component and Structural Damping

•  Component loss factor differs because stress and distortion conditions are position-dependent.

•  Unless particular measures are taken to increase component damping, damping at the contact surfaces of joints (bolted, riveted, clamped) is the predominant damping mechanism.

•  Damping is due to relative motion of the mating surfaces. This can happen in the form of macro-slip (relative motion across the entire face of the joint), micro-slip (interfacial slip of small areas), and gas pumping losses.

η ~ 0.01



Structural Damping

•  Damping in structures is normally not viscous. It is due to mechanisms such as hysteresis in the material and slip in connections.

•  In reality, due to computation easiness, damping in structures are approximated by viscous damping.

•  Phenomenological damping methods physical dissipative mechanisms are modeled.

•  Spectral damping methods viscous damping is introduced by means of specified fractions of critical damping.

Cook et al., 2001


42

8


K

C

M fa(t)

( )

( )tfM

xxx

tfKxxCxM

nn12 2 =++

=++

ωξω

x

Models for Damped Structures

( )( ) tj

tj

FetfXetx

ω

ω

=

=



( ) FM

XXjX

FM

xxx

nn

nn

12

12

22

2

=++−

=++

ωωξωω

ωξω


( ) ( )

( ) ( )

FM

XXjX

tfM

xjMKx

tfxjKxM

nn1

11

1

222 =++−

=++

=++

ωηωω

η

η

Viscous Damping Representation Loss Factor Representation

At Resonance ξη 2=

Used in SEA Codes Used in FEM Codes


44



[ ] [ ] [ ] ( ){ }tfxKxCxM a=++

Assuming special damping matrices (which simplify calculation) with viscous damping or with structural damping, the equations of motion can be decoupled by modal transformation. This yields equations of motion for simple damped systems which can be solved in the familiar manner. If this assumption is not justified by the physical damping behavior of a structure, then the coupling effects should be incorporated in the model.

[ ] [ ] [ ]MKC βα +=

VDI 3830




( ) ( )

( ) ( )

( )tfM

xxx

tfM

xxx

tfKxxMKxM

nn

nn

12

1

2

22

=++

=+++

=+++

ωξω

ωβαω

βα

( )

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=

nn

nn

ωβαωξ

βαωξω

121

2 2

Proportional Damping Representation Relation to Viscous Damping Ratio


46


Rayleigh or Proportional Damping

1ω

⎟⎠

⎞⎜⎝

⎛ +=ωβ

αωξ21

2ω

1ξ

2ξ

Design Spectrum

( )21

22

11222ωωωξωξ

α−

−=

( )21

22

1221212ωω

ωξωξωωβ

−

−=

Estimating α and β

Cook et al., 2001



Structural Vibration at a Single DOF

At a single degree of freedom

u =ϕ cos ωt( )

u =displacement of a nodal DOFϕ =amplitude


48

9


Structural Vibration for the Entire Structure

u{ }= ϕ{ }cos ωt( )

u{ }=Vector of nodal displacements

ϕ{ }=vector of amplitudes for each DOF



Mode Shapes

The vector of amplitudes is a mode shape

2φ 3φ 4φ 5φ 6φ7φ

8φ

Use subscript i to differentiate mode shapes and natural frequencies

u{ }= ϕ{ }i cos ωit( )


50


Appropriate Initial Conditions

If the I.C.s are a scalar (a) multiple of a specific mode shape then the structure will vibrate in the corresponding mode and natural frequency

ICs = a ϕ{ }i



Determining Natural Frequencies

Consider a multi DOF system

M[ ] u{ }+ K[ ] u{ }= 0{ }

Note that the system is ü  Undamped ü  Not excited by any external forces


52


Modal Analysis

u{ }= ϕ{ }i cos ωit( )

If the system vibrates according to a particular mode shape and frequency

ϕ{ }i =mode shape i

ωi =natural frequency i



Modal Analysis

u{ }= −ωi ϕ{ }i sin ωit( )

g  First derivative (Velocity)

g  Second derivative (Acceleration)

u{ }= −ωi2 φ{ }i cos ωit( )


54

10


Modal Analysis

− M"# $%ωi2 + K"# $%( ) φ{ }i = 0{ }

Plug velocity and acceleration into the equation of motion

Two possible solutions

φ{ }i = 0{ }

det − M"# $%ωi2 + K"# $%( ) = 0



The Eigen Problem

The Eigen Problem

A!" #$ x{ }= λ x{ }

det − M"# $%ωi2 + K"# $%( ) = 0

λEigenvalue

x{ }Eigenvector


56


Modal Analysis – An Eigen Problem

− M"# $%ωi2 + K"# $%( ) φ{ }= 0"# $%

Natural frequencies are eigenvalues

The mode shapes are eigenvectors

K!" #$ φ{ }=ωi2 M!" #$ φ{ }

M!" #$−1K!" #$ φ{ }=ωi2 φ{ }

ωi2

φ{ }Dept. of Mech. Engineering University of Kentucky


Modal Analysis – An Eigen Problem

Solving the eigen problem

M!" #$−1K!" #$ φ{ }=ωi2 φ{ }

A!" #$ x{ }= λ x{ }

A!" #$ x{ }−λ x{ }= 0A!" #$− λ I!" #$( ) x{ }= 0det A!" #$− λ I!" #$( ) = 0


58


Example Eigen Problem

2 33 2

!

"#

$

%&x1x2

!"#

$#

%&#

'#= λi

x1x2

!"#

$#

%&#

'#

det 2 33 2

(

)*

+

,-− λi

1 00 1

!

"#

$

%&

'

())

*

+,,= 0



Find Eigenvalues

det 2−λ 33 2−λ

#

$%

&

'(

)

*++

,

-..= 0

2−λ( ) 2−λ( )−9 = λ2 − 4λ −5Eigen values

λ1 = −1λ2 = 5


60

11


For Eigen Vector 1

λ1 = −1

2 33 2

#

$%

&

'(− −1( ) 1 0

0 1

#

$%

&

'(

)

*++

,

-..

x1x2

/01

21

341

511

= 00

/01

21

341

51

2+1 33 2+1

#

$%

&

'(

)

*++

,

-..

x1x2

/01

21

341

511

= 00

/01

21

341

51



For Eigen Vector 1

3 33 3

!

"#

$

%&

'

())

*

+,,

x1x2

-./

0/

12/

3/1

= 00

-./

0/

12/

3/

x1 can be any value

x1x2

!"#

$#

%&#

'#1

= 1−1

!"#

$#

%&#

'#= .7071

−.7071

!"#

$#

%&#

'#


62


For Eigen Vector 2

λ2 = 5

2 33 2

"

#$

%

&'− 5( ) 1 0

0 1

"

#$

%

&'

)

*++

,

-..

x1x2

/01

21

341

512

= 00

/01

21

341

51

2−5 33 2−5

"

#$

%

&'

)

*++

,

-..

x1x2

/01

21

341

512

= 00

/01

21

341

51



For Eigen Vector 2

−3 33 −3

"

#$

%

&'

(

)**

+

,--

x1x2

./0

10

230

402

= 00

./0

10

230

40

x2 = x1

x2 can be any value

x1x2

!"#

$#

%&#

'#2

= 11

!"#

$#

%&#

'#= .7071

.7071

!"#

$#

%&#

'#


64


Example

Equations of motions

M1 M2 K1 K2

x1 x2

M1x1 + K1x1 − K1x2 = 0M 2x2 + K2x2 + K1x2 − K1x1 = 0

M1 = 5 kgM 2 =10 kgK1 = 800 N/mK2 = 400 N/m



Matrix Form

M1 0

0 M 2

!

"

##

$

%

&&

x1x2

'()

*)

+,)

-)+

K1 −K1−K1 K1 + K2

!

"

##

$

%

&&

x1x2

'()

*)

+,)

-)= 0

0

'()

*)

+,)

-)

5 00 10

!

"#

$

%&x1x2

'()

*)

+,)

-)+ 800 −800

−800 1200

!

"#

$

%&x1x2

'()

*)

+,)

-)= 0

0

'()

*)

+,)

-)


66

12


Set Up Eigen Problem

M!" #$−1K!" #$ φ{ }=ωi2 φ{ }

M!" #$−1K!" #$=

160 −160−80 120

!

"(

#

$)

det 160 −160−80 120

!

"(

#

$)− λ 1 0

0 1

!

"(

#

$)

+

,--

.

/00= 0



Solve Eigen Problem

det 160−λ −160−80 120−λ

#

$%

&

'(

)

*++

,

-..= 0

ω1 = λ1 = 5.011 rad/s

f1 =5.011

2π= .7975 Hz

ω2 = λ2 =15.965 rad/s

f2 =15.965

2π= 2.5409 Hz

φ1φ2

"#$

%$

&'$

($1

= 0.76450.6446

"#$

%$

&'$

($

φ1φ2

"#$

%$

&'$

($2

= −0.86010.5101

"#$

%$

&'$

($


68


Mode 1

M1 M2 K1 K2

0.7645 0.6446



Mode 2

M1 M2 K1 K2

-0.8601 0.5101


70


Weighted Orthogonality

ϕ1ϕ2

!"#

$#

%&#

'#1

T

M[ ]ϕ1ϕ2

!"#

$#

%&#

'#2

= 0

ϕ1ϕ2

!"#

$#

%&#

'#1

T

K[ ]ϕ1ϕ2

!"#

$#

%&#

'#2

= 0



Modal Vector Scaling

ϕ1ϕ2

!"#

$#

%&#

'#1

T

M[ ]ϕ1ϕ2

!"#

$#

%&#

'#1

=1

g  Unity Modal Mass


72

13


Modal Vector Scaling

.7645X

.6446X

!"#

$%&1

T5 00 10

'

()

*

+,.7645X.6446X

!"#

$%&1=1

g  Unity Modal Mass

X = .376

ϕ1ϕ2

!"#

$#

%&#

'#1

= .287.242

!"$

%&'



Example Vibrating Machine Isolation

m1 =100 kgm2 = 500 kgκ1 = 5E6 N/mκ2 =1E6 N/mdv1 =100kg/sdv2 = 200kg/s


74


Eigenvalues and Eigenvectors

ω1 = 40.7 rad/sf1 = 6.48 Hzψ1 =

15.8

!"#

$%&

ω2 = 245.6 rad/sf2 = 39.1 Hzψ2 =

1−0.034

"#$

%&'



Forces With / Without Excitation

F1 t( ) = F1e jωt

F2 t( ) = F2e jωt

x1p t( ) = x1pe jωt

x2 p t( ) = x2 pe jωt

Fwithout = FexcFwith =κ1x1e

jωt + jωdv1x1ejωt


76


Simultaneous Differential Equations

−m1ω2x p1 + jω dv1 + dv2( ) x p1 + κ1 +κ2( ) x p1 − jωdv2x p2 −κ2x p2 = Fexc

− jωdv2x p1 −κ2x p1 −m2ω2x p2 + jωdv2x p2 +κ2x p2 = 0



Simultaneous Differential Equations

γ ω( ) = FwithoutFwith

=κ2 + jωdv2 −ω

2m2( ) κ1 +κ2( )+ jω dv1 + dv2( )−ω 2m1( )− κ2 + jωdv2( )2

κ1 + jωdv1( ) κ2 + jωdv2( )


78

14


The Frequency Spectrum

Dynamic signals can be represented in the frequency domain by a frequency spectrum

f (Hz)

10

1 2

amplitude vs. time

time domain frequency domain

amplitude vs. frequency

Fundamental (first harmonic)

Second harmonic



Fourier Analysis Used to determine the frequency spectrum of dynamic signals – spectrum analyzer hardware

T = 2π/ω Is the period of the signal, ω is the fundamental frequency (first harmonic), 2ω is the second harmonic, etc.

y t( ) = Ao2+ An cos nωt( )+Bn sin nωt( )( )

n=1

∞

∑

=Ao2+ Cn cos nωt −ϕn( )( )

n=1

∞

∑

A0 =2T

y t( )dt0

T

∫

An =2T

y t( )cos nωt( )dt0

T

∫ Bn =2T

y t( )sin nωt( )dt0

T

∫

Cn = An2 +Bn

2 ϕn = tan−1 Bn An( )


80


Example

t

A

-A

T

y(t)

Phase between An & Bn

An =2T

y t( )0

T

∫ cos nωt( )dt = 2T

Acos nωt( )dt +0

T 2

∫ −A( )cos nωt( )dtT 2

T

∫#

$%%

&

'((= 0

i.e., Acos nωt( )dt0

T 2

∫ =Anωsin nωt( )

0

T 2

=Anω

sin n 2πTT2

)

*+

,

-.− 0

#

$%

&

'(= 0

Bn =2T

y t( )0

T

∫ sin nωt( )dt =4A nπ , n odd0, n even

/01

21

Cn = An2 +Bn

2 = Bn ϕn = tan−1 Bn An( ) = π 2



Example – Fourier Series Expansion

( ) ( ) ( ) ( ) ( ) ⎟⎠

⎞⎜⎝

⎛ +++++= tnn

tttAty ωωωωπ

sin15sin513sin

31sin4

πA4

π34A


82


Example – Fourier Series Expansion



amplitude vs. time

time domain

t

A

-A

T

y(t)

ω 3ω

frequency domain

amplitude vs.frequency

5ω

etc.

Example – Fourier Line Spectrum

( ) ( ) ( ) ( ) ( ) ⎟⎠

⎞⎜⎝

⎛ +++++= tnn

tttAty ωωωωπ

sin15sin513sin

31sin4

πA4

π34Aπ54A


84

15


“How much is the vibration?”

Peak = 0.51 mm Peak-to-peak = 1.04 mm Average = - 8.1x10-4 mm rms = 0.15 mm

Typical Situation



Spectrum reveals several harmonics under 200 Hz, probably related to engine rpm and firing frequency.

Typical Situation


86


Digitization of Analog Signals Time signals are sampled using an analog-to-digital converter (a digital logic hardware device) to yield a digitized version of the waveform for further processing.

ADC

Analog voltage signal from transducer

Time (s) Voltage (V)

0.0 0.0

0.001 0.15

0.002 0.18

0.003 0.28

etc.

Voltage values stored in computer memory



samplegiven any of timeN)3 2, (1,number index sample

period sample totalsamples ofnumber total

Hz)or ec(samples/s /1 rate sample interval sample

=Δ=

⋅⋅⋅=

Δ==

=

Δ==

=Δ

trtr

tNTN

tft

s

Sampling Parameters

y(t) y(rΔt) y(t)


88


Discrete Fourier Transform •  Continuous (analog) form:

•  Discrete (digital) form:

The discrete form will yield frequency values up to N/2 only.

( ) ( ) ( )( )

( ) ( ) ( ) ( )∫∫

∑

==

++=∞

=

T

n

T

n

nnn

o

dttntyT

BdttntyT

A

tnBtnAAty

00

1

sin2cos2

sincos2

ωω

ωω

( ) ( )

( ) ⎟⎠

⎞⎜⎝

⎛⋅Δ=

=⎟⎠

⎞⎜⎝

⎛⋅Δ=Δ⎟⎠

⎞⎜⎝

⎛ Δ⋅⋅Δ

⋅Δ

=

∑

∑∑

=

==

Nrntry

NB

NnNrntry

Nttrn

tNty

tNA

N

rn

N

r

N

rrn

π

ππ

2sin22

,2,12cos22cos2

1

11



Frequencies in Discretely Sampled Signals 1.  Like its analog counterpart, the discrete Fourier series

contains frequency components spaced 1/T apart. This is called the frequency resolution Δf :

Δf = 1/T (Hz)

Δf = 1/T f

Fourier Coefficients


90

16


Frequencies in Discretely Sampled Signals

2s

Nyqff =

2.  Signals must be sampled at a minimum rate fs twice the highest frequency of interest (the Nyquist frequency fNyq). The highest frequency resolved in a DFT is:

If this is not done, aliasing error occurs.



The Aliasing Phenomenon

fs = f

(a)

(b)

(c)

2f > fs > f

fs = 2f (minimum sample rate)

The signal f will appear as 2f/3 in the spectrum


92


The Aliasing Phenomenon

True spectrum

fNyq

f Measured spectrum

fNyq

f

•  Aliasing results in harmonics above the Nyquist frequency being “folded” back onto their neighbors.

•  Commercial spectrum analyzers set the sampling frequency at least 2.5 times the highest frequency of interest.

•  Signal is low-pass filtered before sampling.



Non–Periodic (transient, random) Signals

The Fourier Integral (Transform):

F(ω) = f (t)e− jωt dt−∞

∞

∫

f (t) = 12π

F(ω)e jωt dω−∞

∞

∫

t

f(t)

ω

F(ω)

Time Domain Frequency Domain

•  F(ω) is continuous, complex

•  limt→∞f t( ) = 0

94


The Fourier Integral

Example:

t

A

-A

T

f(t)

F(ω) = Ae− jωt0

T /2

∫ dt + (−A)e− jωtT /2

T

∫ dt = jAω

2e− jωT /2 − e− jωT −1( )

F(ω) = F(ω)F *(ω) = Aω

6−8cos πω ωo( )+ cos 2πω ωo( )

where ωo = 2π /T


The Fourier Integral

Example:

0

1

2

3

4

5

0 1 2 3 4 5 6 7 8 9 10

ω/ωo

F(ω)A /ωo

Compare with Fourier series results

96

17


Frequency Response Functions (FRF’s)

An FRF is the ratio of two Fourier transforms:

H12 (ω) =F2 (ω)F1(ω)

What is the FRF between the sound pressures at two points in a duct carrying a +x wave?

A

x1 x2


Frequency Response Functions (FRF’s)

H12 (ω) = P2 (ω)P1(ω)

=Ae− jkx2

Ae− jkx1= e− jk (x2−x1 )

Note: k(x2 − x1) =ωΔt where Δt is the propagation time

A

x1 x2

98

Chapter 03 Mathematical Methods - University of...

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