intro fourier

7/26/2019 Intro Fourier

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Introduction to Fourier Series

Deian Stefan

[email protected]

January 13, 2007

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mailto:[email protected]:[email protected]


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Contents

1 Introduction 3

2 Finding the Fourier Series 32.1 Finding the Fourier Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 General Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3 Conclusion 5

4 Examples 6

4.1 Example 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64.2 Example 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

5 Acknowledgment 10

6 Copyright 10

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1 Introduction

Different methods have been used to solve linear ordinary differential equations (ODEs)of the form y +ay +by = f(t), where a and b Rand f(t) was usually an exponential,

trigonometric or polynomial function. There are, however, many functions, f(t), for whichthe solutiony(t) cannot be directly solved using the methods learned thus far. An exampleof such functions is:

f(t) =

1, < t


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Case I: Let n= sin(nt) and m= cos(mt) and = and =

sin(nt) cos(mt) dt= 0, m= n;

0, m =n.

Case II: Let n= cos(nt) and m= cos(mt) and = and =

cos(nt) cos(mt) dt=

, m= n;

0, m =n.

Case III: Letn= sin(nt) and m= sin(mt) and = and =

sin(nt) sin(mt) dt=

, m= n;

0, m =n.

All three cases show that for m =n, m(t)n(t) dt= 0, proving theorem1.2.1 Finding the Fourier Coefficients

From theorem 1 we can now determine an and bn in (2), which is rewritten below forconvenience:

f(t) =C0+n=1

(ancos(nt) +bnsin(nt))

and is expanded to

f(t) =C0+ +akcos(kt) + +ancos(nt) + +bjsin(jt) + +bnsin(nt)

which is further multiplying f(t)by cos(nt) and intergrated on the interval [, ]:

f(t)cos(nt) dt = +

0

akcos(kt)cos(nt) dt + +

0

bjsin(jt)cos(nt) dt +

+

ancos2(nt) dt

an

+ +

bnsin(nt) cos(nt) dt 0

from which:

an= 1

f(t) cos(nt) dt (5)

Similarly multiplying f(t) by sin(nt) and intergrating on the interval [, ]:

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f(t)sin(nt) dt = +0

akcos(kt) sin(nt) dt + +

0

bjsin(jt)sin(nt) dt +

+

ancos(nt)sin(nt) dt 0

+ +

bnsin2(nt) dt

bn

from which:

bn= 1

f(t)sin(nt) dt (6)

Finally, C0 is found by letting n= 0 to be:

C0= 1

2

f(t) dt= a0

2

Rewriting (2):

f(t) = a0

2 +

n=1

(ancos(nt) +bnsin(nt)) , (7)

where an and bn are defined by (5) and (6), respectively.

2.2 General Form

The Fourier series of a more general periodic functions defined on the interval (p, p) is:

f(t) =a0

2 +

n=1

ancos

n

p t

+bnsin

n

p t

, (8)

where:

an=1

p

pp

f(t)cos

n

p t

dt and bn=

1

p

pp

f(t)sin

n

p t

dt

3 Conclusion

It is important to see that as N the Fourier series of function f(t)

f(t) = a0

2 +

Nn=1

(ancos(nt) +bnsin(nt))

converges tof(t) on the defined interval. For functions with finite points of discontinuities,

such as (2), the Fourier series converges tof(t) at a point of continuity and 1

2(f(t+)+f(t))at points of discontinuity[2]. It is also interesting to note that in many cases whereN= ,but is still quite large, the Fourier series is a very close approximation and sufficient enoughfor many practical applications.

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4 Examples

4.1 Example 1

Given f(t) = 2e2

t, t (, ), find its Fourier series equivalent.

From (5):

an= 1

(2e2 t)cos(nt) dt=

0 1

2e2 cos(nt) dt1

t cos(nt) dt

Then letting u = t and dv= cos(nt) dt:

uv

v du= t sin(nt)

n

sin(nt)

n dt=

t sin(nt)

n

+cos(nt)

n2

= 0

from whichan= 0

1

0 = 0

and

a0= 1

f(t) dt= 1

2e2t

t2

2

= 4e2

From (6):

bn= 1

(2e2 t)sin(nt) dt=

0 1

2e2 sin(nt) dt1

t sin(nt) dt

Then letting u = t and dv= sin(nt) dt:

uv

v du=t cos(nt)

n

+

cos(nt)

n dt=

t cos(nt)

n

+sin(nt)

n2

0

from which

bn= 0 1

2 cos(n)

n =

2cos(n)

n =

2(1)n

n

and finally,

f(t) = a0

2 +

n=1(bnsin(nt)) = 2e

2 + 2

n=1(1)n

n sin(nt) (9)

As previously mentioned in cases whereN= , the Fourier series off(t) is a close approx-imation and can still be of practical use. As Fig. 1shows the Fourier series approximationfor (9), with increasing Nis very close to the actual function.

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Figure 1: f(t) = 2e2 t, t (, ) (black) and its Fourier series expanded with N =2(blue), 10(green), 100(red).

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4.2 Example 2

Given f(t) = |t|, t (, ) , find its Fourier series equivalent.

The function must first be rewritten as

f(t) =

t, < t


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Figure 2: f(t) = |t|, t (, ) (black) and its Fourier series expanded with N =2(blue), 10(green), 100(red).

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5 Acknowledgment

Much of this material is based on the 15th M.I.T. video lecture of Professor Arthur Mat-tucks class 18.03 [1]. Much thanks to Prof. Mattuck and the M.I.T. OpenCourseWare

group for making their lectures available.

6 Copyright

This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike2.5 License. To view a copy of this license, visithttp://creativecommons.org/licenses/by-nc-sa/2.5/ or send a letter to Creative Commons, 543 Howard Street, 5th Floor, SanFrancisco, California, 94105, USA.

References

[1] Professor Arthur Mattuck. Introduction to fourier series; basic formulas for period2(pi). Video lecture, 2003.

[2] Dennis G. Zill and Michael R. Cullen. Differential Equations with Boundary-ValueProblems, chapter 11, page 437. Thomson Brooks/Cole, 6th edition, 2005.

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