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 =