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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)
-
2Lecture 12- SDOF- Periodic Input, Fourier Series
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= +
Lecture 12- SDOF- Periodic Input, Fourier Series
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
=
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3Lecture 12- SDOF- Periodic Input, Fourier Series
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 =
Lecture 12- SDOF- Periodic Input, Fourier Series
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
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4Lecture 12- SDOF- Periodic Input, Fourier Series
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
=
Lecture 12- SDOF- Periodic Input, Fourier Series
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
=
-
5Lecture 12- SDOF- Periodic Input, Fourier Series
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 *
Lecture 12- SDOF- Periodic Input, Fourier Series
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
=
= +
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6Lecture 12- SDOF- Periodic Input, Fourier Series
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
=
Lecture 12- SDOF- Periodic Input, Fourier Series
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:
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7Lecture 12- SDOF- Periodic Input, Fourier Series
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)
Lecture 12- SDOF- Periodic Input, Fourier Series
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
+ + =
+ + =
&& &
&& &
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8Lecture 12- SDOF- Periodic Input, Fourier Series
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= =
Lecture 12- SDOF- Periodic Input, Fourier Series
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 =
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9Lecture 12- SDOF- Periodic Input, Fourier Series
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
Lecture 12- SDOF- Periodic Input, Fourier Series
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