jn8t9o5ezcxxos0%2fbh%2f9s0texaioq71vbfrix1z4jw4%3d

1Lecture 12- SDOF- Periodic Input, Fourier Series

MEEM 3700 1

MEEM 3700MEEM 3700Mechanical VibrationsMechanical Vibrations

Mohan D. Rao Chuck Van Karsen

Mechanical Engineering-Engineering MechanicsMichigan Technological University

Copyright 2003

Lecture 12- SDOF- Periodic Input, Fourier Series

MEEM 3700 2

Response Under a General Periodic ForceResponse Under a General Periodic Force

( )mx cx kx F t+ + =&& &( ) ( )F t F t = +Where for any value of Where for any value of tt

F(t) = General Periodic ForceF(t) = General Periodic Force

This force can be expressed in terms of itsThis force can be expressed in terms of itsFourier Series CoefficientsFourier Series Coefficients

1) Harmonic (sin, cos)

tt

F(t)

4) Random

tt

F(t))

2) Transient

tt

F(t))

3) Periodic

tt

F(t)


MEEM 3700 3

Time to Frequency Domain Conversion:Time to Frequency Domain Conversion:

01 2 3

1 2 3

( ) cos( ) cos(2 ) cos(3 )2

sin( ) sin(2 ) sin(3 )

ax t a t a t a t

b t b t b t

= + + + ++ + + +

L

K

)t(x)t(x +=

Time

Am

plitu

de

Periodic Function:Periodic Function: ( ) ( )F t F t T= +

Trigonometric series representation: Trigonometric series representation: (where )(where )2T =

Conditions for application:Conditions for application:1.) Function is periodic1.) Function is periodic2.) Discontinuities are finite within any period2.) Discontinuities are finite within any period3.) Function has a finite number of maxima and minima duri3.) Function has a finite number of maxima and minima during any periodng any period4.) Function is absolutely 4.) Function is absolutely integrableintegrable over any periodover any period ( )

0

Tx t dt <

( ) ( )F t F t T= +


MEEM 3700 4

Time to Frequency Domain Conversion:Time to Frequency Domain Conversion: Fourier SeriesFourier SeriesSeries coefficients are evaluated as:Series coefficients are evaluated as:

0

2 ( ) cos( )T

na x t n t dt= 0

2 ( ) sin( )T

nb x t n t dt= Evaluation is simplified by the Evaluation is simplified by the OrthogonalityOrthogonality conditions of Sine & Cosineconditions of Sine & Cosine

0

sin( ) sin( )T

m t n t =

0

sin( ) cos( ) 0T

m t n t =

0

cos( ) cos( )T

m t n t =

0 for

for 02

m n

m n

=

0 for

for 02

m n

m n

=


MEEM 3700 5

Time to Frequency Domain Conversion:Time to Frequency Domain Conversion: Fourier SeriesFourier SeriesX (t)

4 4 4 4( ) s in ( ) s in (3 ) s in (5 ) s in (7 )3 5 7

x t t t t t = + + + + + L

an bn

2T =


MEEM 3700 6

Time to Frequency Domain Conversion:Time to Frequency Domain Conversion: Fourier SeriesFourier Series

Example of Example of 44 termsterms

Example of Example of 88 termsterms


MEEM 3700 7

Harmonic Forcing Function: ExampleHarmonic Forcing Function: Example

( ) 0 j tF t F e =( )x t

k c

m

( )mx cx kx F t+ + =&& &EOM:EOM:0

j tF e =Steady State Response:Steady State Response: j t

ssX Xe=

( )2 0j t j tm j c k Xe F e + + =BecomesBecomes

( )201X

F k m j c = +

( )22 2 201X

F k m c =

+1

2tanc

k m

=


MEEM 3700 8

To obtain the nonTo obtain the non--dimensionalizeddimensionalized forms of the previous forms of the previous equatiosnsequatiosns use:use:

c

cc

=n

r = nkm

=

Harmonic Forcing Function: ExampleHarmonic Forcing Function: Example

( ) ( )0

2 221 2ss

Fk

Xr r

= +

12

2tan1

rr

=


MEEM 3700 9

Response Under a General Periodic ForceResponse Under a General Periodic ForceExpand the periodic force in a Fourier SeriesExpand the periodic force in a Fourier Series

( ) ( ) ( )01 1

cos sin2 n nn naF t a n t b n t

= == + +

wherewhere ( ) ( )0

2 0,1, 2,c .os 3 ..T

n F t n t da t nT= =

( ) ( )0

2 0,1,2,s .in 3 ..T

n F t n t db t nT= =

Therefore, the equation of motion of the system can be expressedTherefore, the equation of motion of the system can be expressed as:as:

( ) ( )01 1

cos sin2 n nn namx cx kx a n t b n t

= =+ + = + + && &

* The total solution is the sum of these responses ** The total solution is the sum of these responses *


MEEM 3700 10

Response Under a General Periodic Force: SolutionsResponse Under a General Periodic Force: Solutions0

2amx cx kx+ + =&& & 0

2ssaXk

=steady state solution:steady state solution:

( )1

cosnn

mx cx kx a n t=

+ + = && &steady state solution:steady state solution:

( ) ( )2 22 21 1 2n

ssn

ak

Xn r nr

=

= +

( )

1sinn

nmx cx kx b n t

=+ + = && &

steady state solution:steady state solution:

( ) ( )2 22 21 1 2n

ssn

bk

Xn r nr

=

= +


MEEM 3700 11

Response Under a General Periodic Force: SolutionsResponse Under a General Periodic Force: Solutions

The complete solution is the summation from the The complete solution is the summation from the principle of superpositionprinciple of superposition

( ) ( ) ( ) ( )

( ) ( ) ( )

02 22 21

2 22 21

cos ...2 1 2

sin1 2

n

ss nn

n

nn

akax t n t

k n r nr

bk

n tn r nr

=

=

= + + +

+ +

12 2

2tan1n

nrn r

=


MEEM 3700 12

Response Under a General Periodic Force:Response Under a General Periodic Force:

1.) If 1.) If nn = = nn for any n, the amplitude of the correspondingfor any n, the amplitude of the correspondingharmonic will be very large.harmonic will be very large.

2.) As n becomes larger, the amplitude becomes smaller. Thus2.) As n becomes larger, the amplitude becomes smaller. Thusthe first few terms are generally sufficient to obtain ththe first few terms are generally sufficient to obtain the e response with response with resonableresonable accuracy.accuracy.

Notes:Notes:


MEEM 3700 13

Fourier Series ExampleFourier Series Example

Jo ( )x t

c

k

( )F to

l

T

( )F t

oF

t

Given:Given:210 1

sec40 4000

oJ kg m l mN Nc km m

= == =

Initial Conditions:Initial Conditions:

( )( )0 0 10

0 0 1 secox F

x T

= == =&

Find x(t)Find x(t)


MEEM 3700 14

Fourier Series Example: FBD & EOMFourier Series Example: FBD & EOM

( )x t2lc &

2lk

( )F t

x l= for small anglesfor small angles

( )( )

0 0

2 2

0

2 2

0

4 4

4 4

M J

kl cl F t l J

cl klJ F t l

= + =

+ + =

&&& &&

&& & xl

=

( )( )

0

02

4 4

4 4

J cl klx x x F t llJ c kx x x F tl

+ + =

+ + =

&& &

&& &


MEEM 3700 15

( ) ( ) ( )0 cos sin2 n nn naF t a t b t = + +

Fourier Series Example: Determining Fourier Fourier Series Example: Determining Fourier CoeffCoeff..

20 00 0 020

0 020 0

02 2

0

22 |2

22 2 2cos cos

22 sin2cos

2

T T

T T

n

T

n

F Fa t dt t FT T T

F Fn na t t dt t t dtT T T T T

ntntF nt Tan T T

= = =

= = = +

2T =

0 for 1, 2,3, 4...na n= =


MEEM 3700 16

Fourier Series Example: Determining Fourier Fourier Series Example: Determining Fourier CoeffCoeff..

00

02

0 0 01 2 3

2 2sin

2 2sin

, , , ...2 3

T

n

n

F ntb t dtT T TF t ntbT TF F Fb b b

= =

= = =

( ) ( ) ( ) ( ) ( )0 0 0 0 0 sin sin 2 sin 3 ... sin2 2 3F F F F FF t t t t n t

n =


MEEM 3700 17

Fourier Series Example: Solving with NumbersFourier Series Example: Solving with Numbers

( ) ( ) ( )10 10 1000 5 10sin 2 5sin 4 3.33sin 6 ...x x x t t t + + = +&& &

1000 1010 0.05 sec10 2 10000n d nrad = = = =

( ) ( ) ( )( ) ( ) ( ) ( )0.5 0 1 2 3 cos 10 sin 10 sin 2 sin 4 sin 6 ...tx t e A t B t X X t X t X t = + + + + + +

XkF


MEEM 3700 18

-0.0013-3.330.390.056/103-0.0084-5.001.680.054/1020-0.016-10.001.640.05/5100.015751.000.0500XFrn

( ) ( ) ( )( ) ( ) ( ) ( )0.5 cos 10 sin 10 0.0157 0.016sin 2 0.0084sin 4 0.0013sin 6 ...tx t e t t t t tBA = + + +

Fourier Series Example: Solving with NumbersFourier Series Example: Solving with NumbersXkF

Apply initial conditions to solve for Apply initial conditions to solve for AA and and BB

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