chapter 6 fourier series
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7/29/2019 Chapter 6 Fourier Series
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Chapter 6
Review of Fourier Series and Its Applications
in Mechanical Engineering Analysis
Tai-Ran Hsu, Professor
Department of Mechanical and Aerospace Engineering
San Jose State University
San Jose, California, USA
ME 130 Applied Engineering Analysis
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Chapter Outline
●
Introduction
● Mathematical Expressions of Fourier Series
● Application in engineering analysis
● Convergence of Fourier Series
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Introduction
Jean Baptiste Joseph Fourier
1749-1829
A French mathematician
Major contributions to engineering analysis:
● Mathematical theory of heat conduction
(Fourier law of heat conduction in Chapter 3)
● Fourier series representing periodical functions
● Fourier transform
Similar to Laplace transform, but for transforming
variables in the range of (-∞ and +∞)
- a powerful tool in solving differential equations
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Periodic Physical Phenomena:
Forces on the needle
Motions of ponies
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Machines with Periodic Physical Phenomena
Mass, M
Elastic
foundation
Sheet metal
x(t)
A stamping
machine involving
cyclic punching
of sheet metals
Cyclic gas pressures
on cylinders,and forces on connecting
rod and crank shaft
In a 4-stroke internal
combustion engine:
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Mathematical expressions for periodical signals from an oscilloscope
by Fourier series:
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The periodic variation of gas pressure in a 4-stoke
internal combustion engine:
1
2
3 4
55
1 - Intake
2 - Compression
3 - Combustion
4 - Expansion5 - Exhaust
Volume, V or Stoke, L
P r e s s u r e ,
PThe P-V Diagram
P = gas pressure
in cylinders
But the stroke l varies with time of the rotating crank shaft, so the
time-varying gas pressure is illustrated as:
Time, t
P r e s s u r e ,
P ( t
)Period Period
1 2 3 4 5
One revolution Next revolution
So, P(t) is a periodic function
with period T
T T
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FOURIER SERIES – The mathematical
representation of periodic physical phenomena
● Mathematical expression for periodic functions:
● If f(x) is a periodic function with variable x in ONE period 2L
● Then f(x) = f(x±2L) = f(x±4L) = f(x± 6L) = f(x±8L)=……….=f(x±2nL)
where n = any integer number
x
f(x)
t
0 π
2π3π
-π
-2π-3π
0L 2L 3L-L-2L-3L
f(t)
(a) Periodic function with period (-π, π)
(b) Periodic function with period (-L, L)
t-2L tt-4L
Period = 2L:
Period: ( -π, π) or (0, 2π)
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Mathematical Expressions of Fourier Series
● Required conditions for Fourier series:
● The mathematical expression of the periodic function f(x) in one period
must be available
● The function in one period is defined in an interval (c < x < c+2L)
in which c = 0 or any arbitrarily chosen value of x, and L = half period
● The function f(x) and its first order derivative f’(x) are either continuous
or piece-wise continuous in c < x < c+2L
● The mathematical expression of Fourier series for periodic function f(x) is:
( ) ( ) .......422
)(1
0 =±=±=⎟ ⎠
⎞⎜⎝
⎛ ++= ∑
∞
=
Lxf Lxf L
xnSinb
L
xnCosa
axf
n
nn
π π (6.1)
where ao, an and bn are Fourier coefficients, to be determined by the following integrals:
..................,3,2,1,0)(1 2
== ∫+
ndxL
xnCosxf L
aLc
cn
π (6.2a)
..................,3,2,1)(1 2
== ∫+
ndxL
xnSinxf
Lb
Lc
cn
π (6.2b)
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Example 6.1
Derive a Fourier series for a periodic function with period (-π, π):
We realize that the period of this function 2L = π – (-π) = 2π
The half period is L = π
If we choose c = -π, we will have c+2L = -π + 2π = π
Thus, by using Equations (6.1) and (6.2), we will have:
∑∞
=
⎟ ⎠
⎞⎜⎝
⎛
=+
=+=
1
0
2)(
n
nnL
xnSinb
L
xnCosa
axf
π
π
π
π
( )∑∞
=
++=1
0 )()(
2
)(
n
nnnxSinbnxCosa
axf
(6.3)
anddx
L
xnCosxf
La
Lc
cn
π
π
π
π π
π === ∫
+−=+
−=)(
1 22
dxL
xnSinxf
Lb
Lc
cn
π
π
π
π π
π === ∫
+−=+
−=)(
1 22
Hence, the Fourier series is:
with..................,3,2,1,0)()(
1== ∫−
ndxnxCosxf an
π
π π
..................,3,2,1)()(1
== ∫−ndxnxSinxf b
n
π
π π
(6.4a)
(6.4b)
We notice the period (-π, π) might not be practical, but it appears to be common in many applied mathtextbooks. Here, we treat it as a special case of Fourier series.
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Example 6.2
Derive a Fourier series for a periodic function f(x) with a period (-ℓ, ℓ)
∑∞
=
⎟ ⎠
⎞⎜⎝
⎛
=+
=+=
1
0
2)(
n
nnL
xnSinb
L
xnCosa
axf
ll
π π
Let us choose c = -ℓ, and the period 2L = ℓ - (-ℓ) = 2ℓ, and the half period L = ℓ
dxL
xnCosxf
La
Lc
cn
ll
ll
l === ∫+−=+
−=
π )(1
22
dxL
xnSinxf
Lb
Lc
cn
ll
ll
l === ∫
+−=+
−=
π )(
1 22
Hence the Fourier series of the periodic function f(x) becomes:
∑∞
=
⎟ ⎠
⎞⎜⎝
⎛ ++=
1
0
2)(
n
nn
xnSinb
xnCosa
axf
ll
π π (6.5)
with
..................,3,2,1,0)(1
== ∫− ndx
xn
Cosxf anll
l
l
π
..................,3,2,1)(1
== ∫−ndx
xnSinxf b
nll
l
l
π
(6.6a)
(6.6b)
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Example 6.3
Derive a Fourier series for a periodic function f(x) with a period (0, 2L).
As in the previous examples, we choose c = 0, and half period to be L. We will have
the Fourier series in the following form:
∑∞
=
⎟ ⎠
⎞⎜⎝
⎛
=+
=+=
1
0
2)(
n
nnLL
xnSinb
LL
xnCosa
axf
π π
dxLLxnCosxf
La LLc
cn
=== ∫
+−=+
=π )(1 202
0l
dxLL
xnSinxf
LLb
LLc
cn
=== ∫
+=+
=
π )(
1 202
0
The corresponding Fourier series thus has the following form:
∑∞
=
⎟ ⎠
⎞⎜⎝
⎛ ++=
1
0
2)(
n
nnL
xnSinb
L
xnCosa
axf
π π (6.7)
dxxf
L
aL
)(1 2
00 ∫=
..................,3,2,1)(1 2
0== ∫ ndx
L
xnCosxf
La
L
n
π
..................,3,2,1)(1 2
0== ∫ ndx
L
xnSinxf
Lb
L
n
π
(6.8b)
(6.8c)
(6.8a)
Periodic functions with periods (0, 2L) are more realistic. Equations (6.7) and (6.8) are
Thus more practical in engineering analysis.
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Example (Problem 6.4 and Problem (3) of Final exam S09)
Derive a function describing the position of the sliding block M in one period in a slide
mechanism as illustrated below. If the crank rotates at a constant velocity of 5 rpm.
(a) Illustrate the periodic function in three periods, and(b) Derive the appropriate Fourier series describing the position of
the sliding block x(t) in which t is the time in minutes
A B
CrankRadius
R
Rotational velocity
ω = 5 RPM
ω
X(t)
Dead End A
Dead EndB
(X = 0) (X = 2R)
Sliding Block, M
The class is encouraged to study Examples 6.4 and 6.5
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Solution:
(a) Illustrate the periodic function in three periods:
Determine the angular displacement of the crank:
A B
R ω
θ
Dead-end A:
x = 0
t = 0
Dead-end B:
x = 2R
t = 1/5 min
We realize the relationship: rpm N = ω/(2π), and θ = ωt, where ω = angular velocity
and θ = angular displacement relating to the position of
the sliding block
One revolution
For N = 5 rpm, we have: t
t
5
5
12 ==π
θ
Based on one revolution (θ=2π
) correspondsto 1/5 min. We thus have θ = 10πt
Position of the sliding block along the x-direction can be determined by:
x = R – RCosθ
or x(t) = R – RCos(10πt) = R[1 – Cos(10πt)] 0 < t < 1/5 min
x
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A B
R ω
θ
Dead-end A:
x = 0
t = 0
Dead-end B:
x = 2R
t = 1/5 min
One revolutionx
x(t) = R[1 – Cos(10πt)]
We have now derived the periodic function describing the instantaneous position of the
sliding block as:
0 < t < 1/5 min (a)
Graphical representation of function in Equation (a) can be produced as:
2R
R
0π/2 π 3π/2 2π
0 1/10 1/5 min
Time, t (min)
x(t)
One revolution
(one period)
2nd period 3rd period
Θ =
Time t =
x(t) = R[1 – Cos(10πt)]
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(b) Formulation of Fourier Series:
We have the periodic function: x(t) = R[1 – Cos(10π
t)] with a period: 0 < t < 1/5 minIf we choose c = 0 and period 2L = 1/5, we will have the Fourier series expressed in
the following form by using Equations (6.7) and (6.8):
( )
[ ]∑
∑∞
=
∞
=
++=
⎥
⎦
⎤⎢
⎣
⎡
=
+
=
+=
1
1
10102
10/110/12
n
nn
o
n
nn
o
tnSinbtnCosaa
L
tnSinb
L
tnCosa
atx
π π
π π
(b)
with ( )( ) ( )
⎥
⎦
⎤⎢
⎣
⎡
+
++
−
−−== ∫
n
nSin
n
nSinRdttnCostxa
n
1
12
1
12
2
10
101
15
1
0
π π
π π (c)
We may obtain coefficient ao from Equation (c) to be ao = 0:
( ) ( )
( )( )[ ]
( )( )[ ]112
12112
12
10101101010 0
5
1
0
−++
+−−−
=
−== ∫∫
∞
π π
π π
π π π
nCosn
RnCos
n
R
dttnSintCosRdttnSintxbn
(d)
The other coefficient bn can be obtained by:
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Convergence of Fourier Series
We have learned the mathematical representation of periodic functions by Fourier series
In Equation (6.1):
( ) ( ) .......422
)(1
0 =±=±=⎟ ⎠
⎞⎜⎝
⎛ ++= ∑
∞
=
Lxf Lxf L
xnSinb
L
xnCosa
axf
n
nn
π π (6.1)
This form requires the summation of “INFINITE” number of terms, which is UNREALISTIC.
The question is “HOW MANY” terms one needs to include in the summation in order to
reach an accurate representation of the periodic function?
The following example will give some idea on the relationship of the “number of terms in theFourier series” and the “accurate representation of the periodic function”:
Example 6.6
Derive the Fourier series for the following periodic function:
( ) 00
0int
≤≤−≤≤= t
tStf
π
π
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( ) 00
0int
≤≤−≤≤= t
tStf
π
π
This function can be graphically represented as:
0 π 2π 3π-π-2π-3πt
f(t)
We identified the period to be: 2L = π- (-π) = 2π, and from Equation (6.3), we have:
( )∑∞
=
++=1
0 )()(2
)(n
nnnxSinbnxCosa
axf (a)
with
( )1
1
11)0(
1)()(
120
≠−
+=+== ∫∫∫ −−
nforn
nCosdtntCostSindtntCosdtntCostf a
nπ
π
π π π
π
π
π
π
and
.....,3,2,11
)0(1
)()(1 0
0=+== ∫ ∫∫ −−
ndtntSintSindtntSindtntSintf bn
π
π π
π π π π
or 10
1
)1(
1
)1(
2
11
0
≠=⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡
+
+−
−
−= nfor
n
tnSin
n
tnSinbn
π
π
For the case n =1, the two coefficients become:
0
2
12
0
1 ===
∫
π
π
π π o
tSindttCostSina
2
11
0
1 == ∫ dttSintSinbπ
π
and
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( ) 00
0int
≤≤−≤≤= t
tStf
π
π
0 π 2π 3π-π-2π-3π
t
f(t)
The Fourier series for the periodic function with the coefficients become:
( )∑∞
=
+++=22
1)(
n
nnntSinbntCosa
tSintf
π (b)
The Fourier series in Equation (b) can be expanded into the following infinite series:
⎟ ⎠
⎞⎜⎝
⎛ ++++−+= ................63
8
35
6
15
4
3
22
2
1)(
tCostCostCostCostSin
tf π π
(c)
Let us now examine what the function would look like by including different number of
terms in expression (c):
Case 1: Include only one term:
t
f(t)
π
11 =f
0π-π
( )π
11 == f xf
Graphically it will look like
Observation: Not even closely resemble- The Fourier series with one term does not converge to the function!
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( ) 00
0int
≤≤−≤≤= t
tStf
π
π
0 π 2π 3π-π-2π-3π
t
f(t)
Case 2: Include 2 terms in Expression (b):
( ) 2
1)(
2
tSintf tf +==
π
t
f(t)
0π-π
2
1)(2
tSintf +=
π
Observation: A Fourier series with 2 terms
has shown improvement in representing the function
Case 3: Include 3 terms in Expression (b):
( )π π 3
22
2
1)(3
tCostSintf tf −+==
t
f(t)
0π-π
π π 3
22
2
1)(3
tCostSintf −+=
Observation: A Fourier series with 3 terms
represent the function much better than the two previous cases with 1 and 2 terms.
Conclusion: Fourier series converges better to the periodic function with more terms
included in the series.
Practical consideration: It is not realistic to include infinite number of terms in the
Fourier series for complete convergence. Normally an approach with 20 terms would
be sufficiently accurate in representing most periodic functions
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Convergence of Fourier Series at Discontinuities of Periodic Functions
Fourier series in Equations (6.1) to (6.3) converges to periodic functions everywhereexcept at discontinuities of piece-wise continuous function such as:
0 x1 x2 x3
x4
f(x)
x
Period, 2L
f 1(x)
f 2(x)
f 3
(x)
(1)
(2)
(3)
= f 1(x) 0 < x < x1
= f 2(x) x1 < x <x2
= f 3(x) x2 < x < x4
f(x) = <
The periodic function f(x) has
discontinuities at: xo, x1 , x2 and x4
The Fourier series for this piece-wisecontinuous periodic function will
NEVER converge at these discontinuous points even with ∞ number of terms
● The Fourier series in Equations (6.1), (6.2) and (6.3) will converge every where to the
function except these discontinuities, at which the series will converge HALF-WAY of the function values at these discontinuities.
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Convergence of Fourier Series at Discontinuities of Periodic Functions
0 x1 x2 x3x4
f(x)
x
Period, 2L
f 1(x)
f 2(x)
f 3(x)
(1)
(2)
(3)
Convergence of Fourier series at HALF-WAY points:
[ ])()(2
1)( 12111 xf xf xf +=
[ ])()(2
1)( 23222 xf xf xf +=
)0(2
1)()( 1434 f xf xf ==
at Point (1)
at Point (2)
at Point (3)
same value as Point (1)
)0(2
1)0( 1f f =