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Chapter 6
Review of Fourier Series and Its Applications in Mechanical Engineering Analysis
Tai-Ran Hsu, ProfessorDepartment of Mechanical and Aerospace Engineering
San Jose State UniversitySan Jose, California, USA
ME 130 Applied Engineering Analysis
Chapter Outline
● Introduction
● Mathematical Expressions of Fourier Series
● Application in engineering analysis
● Convergence of Fourier Series
Introduction
Jean Baptiste Joseph Fourier1749-1829
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 transformSimilar to Laplace transform, but for transformingvariables in the range of (-∞ and +∞)- a powerful tool in solving differential equations
Periodic Physical Phenomena:
Forces on the needle
Motions of ponies
Machines with Periodic Physical Phenomena
Mass, M
Elastic foundation
Sheet metal
x(t)
A stampingmachine involving
cyclic punchingof sheet metals
Cyclic gas pressureson cylinders,
and forces on connecting rod and crank shaft
In a 4-stroke internalcombustion engine:
Mathematical expressions for periodical signals from an oscilloscopeby Fourier series:
The periodic variation of gas pressure in a 4-stoke internal combustion engine:
12
3 4
55
1 - Intake
2 - Compression
3 - Combustion
4 - Expansion
5 - Exhaust
Volume, V or Stoke, L
Pre
ssur
e, PThe P-V Diagram
P = gas pressurein cylinders
But the stroke l varies with time of the rotating crank shaft, so the time-varying gas pressure is illustrated as:
Time, t
Pres
sure
, P(t)
Period Period
1 2 3 4 5
One revolution Next revolution
So, P(t) is a periodic functionwith period T
T T
FOURIER SERIES – The mathematicalrepresentation 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 2π 3π-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)
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 continuousor piece-wise continuous in c < x < c+2L
● The mathematical expression of Fourier series for periodic function f(x) is:
( ) ( ) .......422
)(1
0 =±=±=⎟⎠⎞
⎜⎝⎛ ++= ∑
∞
=
LxfLxfLxnSinb
LxnCosa
axf
nnn
ππ(6.1)
where ao, an and bn are Fourier coefficients, to be determined by the following integrals:
..................,3,2,1,0)(1 2== ∫
+ndx
LxnCosxf
La
Lc
cnπ (6.2a)
..................,3,2,1)(1 2== ∫
+ndx
LxnSinxf
Lb
Lc
cnπ (6.2b)
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) = 2πThe half period is L = πIf 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)(
nnn L
xnSinbL
xnCosaaxfπ
ππ
π
( )∑∞
=
++=1
0 )()(2
)(n
nn nxSinbnxCosaa
xf (6.3)
anddx
LxnCosxf
La
Lc
cn ππ
πππ
π === ∫
+−=+
−=)(1 22
dxL
xnSinxfL
bLc
cn ππ
πππ
π === ∫
+−=+
−=)(1 22
Hence, the Fourier series is:
with ..................,3,2,1,0)()(1== ∫− ndxnxCosxfan
π
ππ
..................,3,2,1)()(1== ∫− ndxnxSinxfbn
π
ππ
(6.4a)
(6.4b)
We notice the period (-π, π) might not be practical, but it appears to be common in many applied math textbooks. Here, we treat it as a special case of Fourier series.
Example 6.2
Derive a Fourier series for a periodic function f(x) with a period (-ℓ, ℓ)
∑∞
=⎟⎠⎞
⎜⎝⎛
=+
=+=
1
0
2)(
nnn L
xnSinbL
xnCosaaxfll
ππLet us choose c = -ℓ, and the period 2L = ℓ - (-ℓ) = 2ℓ, and the half period L = ℓ
dxL
xnCosxfL
aLc
cnll
ll
l === ∫
+−=+
−=
π)(1 22
dxL
xnSinxfL
bLc
cnll
ll
l === ∫
+−=+
−=
π)(1 22
Hence the Fourier series of the periodic function f(x) becomes:
∑∞
=⎟⎠⎞
⎜⎝⎛ ++=
1
0
2)(
nnn
xnSinbxnCosaa
xfll
ππ (6.5)
with
..................,3,2,1,0)(1== ∫− ndxxnCosxfan
ll
l
l
π
..................,3,2,1)(1== ∫− ndxxnSinxfbn
ll
l
l
π
(6.6a)
(6.6b)
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)(
nnn LL
xnSinbLLxnCosaaxf ππ
dxLLxnCosxf
La
LLc
cn === ∫
+−=+
=
π)(1 202
0l
dxLLxnSinxf
LLb
LLc
cn === ∫
+=+
=
π)(1 202
0
The corresponding Fourier series thus has the following form:
∑∞
=⎟⎠⎞
⎜⎝⎛ ++=
1
0
2)(
nnn L
xnSinbL
xnCosaa
xf ππ (6.7)
dxxfL
aL
)(1 2
00 ∫=
..................,3,2,1)(1 2
0== ∫ ndx
LxnCosxf
La
L
nπ
..................,3,2,1)(1 2
0== ∫ ndx
LxnSinxf
Lb
L
nπ
(6.8b)
(6.8c)
(6.8a)
Periodic functions with periods (0, 2L) are more realistic. Equations (6.7) and (6.8) areThus more practical in engineering analysis.
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 velocityω = 5 RPM
ω
X(t)
Dead EndA
Dead EndB
(X = 0) (X = 2R)
Sliding Block, M
The class is encouraged to study Examples 6.4 and 6.5
Solution:
(a) Illustrate the periodic function in three periods:
Determine the angular displacement of the crank:
A B
Rωθ
Dead-end A:x = 0t = 0
Dead-end B:x = 2R
t = 1/5 min
We realize the relationship: rpm N = ω/(2π), and θ = ωt, where ω = angular velocityand θ = angular displacement relating to the position of the sliding block
One revolution
For N = 5 rpm, we have: tt 5
512==
πθ Based on one revolution (θ=2π) corresponds
to 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
A B
Rωθ
Dead-end A:x = 0t = 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:
2RR
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)]
(b) Formulation of Fourier Series:We have the periodic function: x(t) = R[1 – Cos(10πt)] with a period: 0 < t < 1/5 min
If 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
nnn
o
nnn
o
tnSinbtnCosaaL
tnSinbL
tnCosaatx
ππ
ππ
(b)
with ( ) ( ) ( )⎥⎦⎤
⎢⎣⎡
++
+−−
−== ∫ nnSin
nnSinRdttnCostxan 1
121
122
1010
11 5
1
0
πππ
π (c)
We may obtain coefficient ao from Equation (c) to be ao = 0:
( ) ( )
( ) ( )[ ] ( ) ( )[ ]11212
11212
101011010100
51
0
−++
+−−−
=
−== ∫∫∞
ππ
ππ
πππ
nCosn
RnCosn
R
dttnSintCosRdttnSintxbn
(d)
The other coefficient bn can be obtained by:
Convergence of Fourier Series
We have learned the mathematical representation of periodic functions by Fourier series In Equation (6.1):
( ) ( ) .......422
)(1
0 =±=±=⎟⎠⎞
⎜⎝⎛ ++= ∑
∞
=
LxfLxfLxnSinb
LxnCosa
axf
nnn
ππ (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 the Fourier series” and the “accurate representation of the periodic function”:
Example 6.6
Derive the Fourier series for the following periodic function:
( ) 000int
≤≤−≤≤= t
tStf ππ
( ) 000int
≤≤−≤≤= t
tStf ππ
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
nn nxSinbnxCosaa
xf (a)
with
( ) 11
11)0(1)()(120
≠−
+=+== ∫∫∫ −−
nforn
nCosdtntCostSindtntCosdtntCostfan ππ
ππππ
π
π
π
and.....,3,2,11)0(1)()(1 0
0=+== ∫ ∫∫ −−
ndtntSintSindtntSindtntSintfbn π
ππ
π πππor
101
)1(1
)1(211
0
≠=⎭⎬⎫
⎩⎨⎧
⎥⎦⎤
⎢⎣⎡
++
−−−
= nforn
tnSinn
tnSinbn
π
πFor the case n =1, the two coefficients become:
02
1 2
01 === ∫π
π
ππo
tSindttCostSina 211
01 == ∫ dttSintSinbπ
πand
( ) 000int
≤≤−≤≤= t
tStf ππ
0 π 2π 3π-π-2π-3πt
f(t)
The Fourier series for the periodic function with the coefficients become:
( )∑∞
=
+++=22
1)(n
nn ntSinbntCosatSintfπ
(b)
The Fourier series in Equation (b) can be expanded into the following infinite series:
⎟⎠⎞
⎜⎝⎛ ++++−+= ................
638
356
154
322
21)( tCostCostCostCostSintf
ππ(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)
π1
1 =f
0 π-π
( )π1
1 == fxf
Graphically it will look like
Observation: Not even closely resemble- The Fourier series with one term does not converge to the function!
( ) 000int
≤≤−≤≤= t
tStf ππ
0 π 2π 3π-π-2π-3πt
f(t)
Case 2: Include 2 terms in Expression (b):
( )2
1)(2tSintftf +==
πt
f(t)
0 π-π
21)(2
tSintf +=π
Observation: A Fourier series with 2 termshas shown improvement in representing the function
Case 3: Include 3 terms in Expression (b):
( )ππ 3
222
1)(3tCostSintftf −+==
t
f(t)
0 π-π
ππ 322
21)(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 wouldbe sufficiently accurate in representing most periodic functions
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 x3x4
f(x)
x
Period, 2L
f1(x)
f2(x)
f3(x)
(1)
(2)
(3)
= f1(x) 0 < x < x1= f2(x) x1 < x <x2= f3(x) x2 < x < x4
f(x) = <
The periodic function f(x) hasdiscontinuities at: xo, x1 , x2 and x4
The Fourier series for this piece-wisecontinuous periodic function willNEVER 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 thefunction except these discontinuities, at which the series will converge HALF-WAY ofthe function values at these discontinuities.
Convergence of Fourier Series at Discontinuities of Periodic Functions
0 x1 x2 x3x4
f(x)
x
Period, 2L
f1(x)
f2(x)
f3(x)
(1)
(2)
(3)
Convergence of Fourier series at HALF-WAY points:
[ ])()(21)( 12111 xfxfxf +=
[ ])()(21)( 23222 xfxfxf +=
)0(21)()( 1434 fxfxf ==
at Point (1)
at Point (2)
at Point (3)
same value as Point (1)
)0(21)0( 1ff =