chapter 6 fourier series

7/29/2019 Chapter 6 Fourier Series

http://slidepdf.com/reader/full/chapter-6-fourier-series 1/22

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



Chapter Outline

●

Introduction

● Mathematical Expressions of Fourier Series

● Application in engineering analysis

● Convergence of Fourier Series



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



Periodic Physical Phenomena:

Forces on the needle

Motions of ponies



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:



Mathematical expressions for periodical signals from an oscilloscope

by Fourier series:



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



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)



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)



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.



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)



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.



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



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



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)]



(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:



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



( ) 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



( ) 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!



( ) 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



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.



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 =

chapter 6 fourier series

Documents