Fourier series - University of Technology, Iraq

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UNIVERSITY OF TECHNOLOGYElectromechanical Department

2009‐2010

Fourier series Dr.Eng Muhammad.A.R.Yass

Sultan

M O H D _ Y A S S 9 7 @ Y A H O O . C O M

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Advance Mathematics

Fourier series

3rd Class

Electromechanical Eng.

Dr. Eng

Muhammad. A. R. Yass

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Fourier series

Definition

The series

12 cos

∞

sin

Si the Fourier series of f on ( -L , L ) when the constant are chosen to be the Fourier coefficient of “f “on (-L , L)

Where

1 cos –

0 , 1 , 2

And

1 sin –

1 , 2 , 3

And

1

Example

Let f(x) = x for –π ≤ x ≤ π . we will write the Fourier series of “f” on [ -π , π] . the coefficient are

10

1cos

1cos sin 0

4

1 sin

1sin cos 0

2cos

21

Since cos ( n ) = (-1)n if n is an integer . the fourier series of “x” on [ - , ] is

2∞

1 sin 2 sin sin 223 sin 3

12 sin 4

25 sin 5 …

Example

Find the fourier series of the periodic function

1 0 10 1 2

Solution

d = 0

2 p = 2 p = 1

2 cos …… cos …… sin …… sin

1

11 1 0

| 1 1

1 cos

11

1 cos 0 cos

1sin | 0

1sin 0 0

f(t)

1 2

1

t

5

0

1 sin

1 1 sin 0 sin

1cos |

1cos

1cos 0

11 cos

1 1 cos 2/

12

1 cos 0

13 1 2

23

14 1 4 0

The fourier series become

12

2sin 0

23 sin 3 0

35 sin 5

Example

0 0

Solution

d = -

d + 2p =

p =

1

1

1

Always sin (n ) = 0

f(t)

t

-

6

1 cos–

1 –

cos cos

Integral by partial we get 0 if n = even and if n = odd

1 sin

1 sin

0

Fourier series will be

12

∞

cos 2 2 2

Example

Let f (x) = x for - ≤ x ≤

1

12 0

1cos

1cos

1cos sin 0

1cos

1cos

period

- 3 -3

7

1sin

1sin

sin cos

2cos

21


21 sin

∞

2 sin sin 223 sin 3

12 sin 4 ……

Example

Let

0 3 0 0 3

Solution

L = 3

13

13

32

13 cos 3

13 3

3cos 3 sin 3

31 1

13 sin 3

13 sin 3

3sin 3 cos 3

31

cos 1

period

3 -3

6

-6

8

The Fourier series

34

3∞

1 1 cos 33

1 sin 3

Example

Let

2 2 find fourier series

12 dx

18

12 cos

nπx2 dx

8 164

12 sin 2

3sin 3 cos 3

3 1

The Fourier series will be

34

3 1 1

∞

cos 33

1 sin 3

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Even and odd function

Even Function

f is an even function on [-L , L ] if f(-x) = f(x) for -L ≤ x ≤ L

Odd Function

f is an odd function on [-L , L] if f(-x) = - f (x) -L≤ x ≤ L

Fig(1) Even function , cos

-3 +3

-1

Fig (2) Even function symmetrical a bout y-axis

Fig (3) Odd function , sin

Fig (4) Odd function symmetric through

the origin

+2 -2

+4

-4

+3 -3 -1 +1

-2 +2

+4

-1

10

If the function is even then

/

/

2

/

While if is odd then

/

/

0

Also even function

2

2cos 0

Then the function will be

2 cos

0 0

2 sin

sin ∞

Symmetric a bout the origin

11

Example

Find the fourier series of the function of the

2 1 01 0

2

Solution

Its an odd-function

0 0

2 sin

2 1 sin 2 sin

2 1cos

2cos

2 4cos

2cos 2

2 1 2 1

2cos 2

2

∞

1 2 1 cos 2 sin

Example

Find fourier series of the function

0 0

The function is odd then

0 ; 0

-

-2

+2

2 -

+1

-1

12

sin cos

1 10

∑ sin∞

Example

Let 1,1 find Fourier series

Solution

is an even function because f(-x) = f (x) ( i.e. example

f(-3) = f (3) on so on ) then 0

2 cos

2 cos8 6

1

the Fourier series

15 8

∞6 1 cos

Example

4 4

Solution

0

14 sin

4

12 sin 4 1 128 6/

-

h

-h

0

13


1∞

128 6 sin 4

Conclusion

Even Function


12

∞

cos

2cos 0 , 1 , 2

Odd Function


∞

sin

2sin 1 , 2

Example

Find the Fourier series if f(x) = x2 0 < x < 2

Neither even nor odd

Period = 2L = 2 L =

0 2 -2

f(t)

4 4

14

1

83

1cos

1 sin 2

– cos2

sin

4 0

1sin

1

cos 2

cos 2

sin 4 0

1 sin

1

cos2

sin2cos 4


43

4∞

cos 44

sin

#Example

“a” odd f (-x) = - f (x)

2 0 32 3 0

-2

2

3 6 -3 -6

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“b” Neither EVEN nor odd

00 2 2

“C” EVEN f (-x) = f (x)

10 0 10 10

0

1

-1

2 -

-2

0 5 10 -10

2.5

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Half Range Expansions

Half range fourier series if function f (x) is defined only in the half fourier interval (0 ) the equation of such function can be problem into other half of period (- 0) infinite way .

a) An odd b) An even c) Neither odd nor even

Example

Give f(x) = x in the interval 0 0 < x <

a) Find the Fourier series an a even function ( cos function) b) Find the Fourier series an odd function (sin function)

a- Even function bn = 0

2

2cos

U = x du = dx

dv = cos (nx) dx v = sin (nx)

2 sin

1sin

2 sin sin 0

1cos

2 1 cos cos 0

21 1

4

24 cos∞

, ,

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b – odd function 0 0

2sin

sin 1cos

2 cos

1 cos 0

2 cos 0 cos 0

2 1

2 cos

21

2

2

2cos

2 cos∞

sin

Example

Find the sine and the cosine half range series of the function series

0

a. Even function

cos cos

cos 1

23

41

∞

cos

0

f(x)

18

b. Odd function

Solved Examples

The formula for a Fourier series on an interval [c,c+T] is:

Example (1)

Find the Fourier series for , .

Following the rules from the link above,

.

19

.

So,

Example (2)

Find the Fourier series for .

FOURIER SERIES BOOKS

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Example (3)

Find the Fourier series for on

The general Fourier series on is:

The n = 0 case is not needed since the integrand in the formula for is .

In the present problem,

But since the right hand side is not defined if n = 0, the 0 index for a will have to be calculated seperately.

So the Fourier series is

for

21

Example (4) Find the Fourier series for

on



22

So the Fourier series is:

Setting x = 0 gives


on


23


on




on

Example (7) Find the Fourier series for a function

on .

Make the change of variables .

Now, look for the Fourier series of the function on

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Since ,

Example (8)

Find the Fourier series for on .

A general formula for the Fourier series of a function on an interval is:

In the current problem, and .

The function is odd, so the cosine coefficients will all equal zero. Nevertheless, should still be calculated separately.

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So the Fourier series for is


This is the general Fourier Series:

So the given function can be replaced by its Fourier expansion:

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So the solution is

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Home Work

Problem (1) Find the Fourier series of the function

Answer.

Problem (2) . Find the Fourier series of the function

Answer. We have Therefore, the Fourier series of f(x) is

Problem (3) Find the Fourier series of

Answer.

Fourier series - University of Technology, Iraq

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