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Ordinary Differential Equations [FDM 1023]
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ODE_Chapter 06 [Jan 2014](1)

May 13, 2017

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Page 1: ODE_Chapter 06 [Jan 2014](1)

Ordinary Differential Equations[FDM 1023]

Page 2: ODE_Chapter 06 [Jan 2014](1)

FOURIER SERIES

Chapter 6

Page 3: ODE_Chapter 06 [Jan 2014](1)

Learning Outcome

At the end of this section, you should be ableto:

Find the Fourier series of a periodicfunction

6. Fourier Series

Page 4: ODE_Chapter 06 [Jan 2014](1)

Periodic Function

A function � is said periodic if there exists a

smallest positive number � such that

� � + � = �(�)for all � in the domain.

Definition

6. Fourier Series

Page 5: ODE_Chapter 06 [Jan 2014](1)

� = 2

6. Fourier Series

Example

Page 6: ODE_Chapter 06 [Jan 2014](1)

Let � be a periodic function of period, �

If � can be expressed as the sum of an infinite

number of sine and cosine functions as follows

�� � = �2 +� � cos 2��� + �� sin 2����

���

then, this infinite sum is called the FOURIER SERIES of�.

6. Fourier Series

Page 7: ODE_Chapter 06 [Jan 2014](1)

�� � = �2 +� � cos 2��� + �� sin 2����

���

�, �, �� � = 1,2,… are known as Fourier coefficients

6. Fourier Series

� - the amplitudes of the cosine terms in the series

�� - the amplitudes of the sine terms in the series

Page 8: ODE_Chapter 06 [Jan 2014](1)

�� � = �2 +� � cos 2��� + �� sin 2����

���

The first few terms of the infinite series

�� � = �2 + � cos2�� + �� sin 2�� + � cos 4��

+ �� sin 4�� +⋯

6. Fourier Series

Page 9: ODE_Chapter 06 [Jan 2014](1)

Fourier coefficients can be obtained by

� = 2�! � � "�#

� = 2�! � � cos 2��� "�#

�� = 2�! � � sin 2��� "�#

�� � = �2 +� � cos 2��� + �� sin 2����

���

6. Fourier Series

Page 10: ODE_Chapter 06 [Jan 2014](1)

Find the Fourier series of

Solution

Step 1: Compute the Fourier coefficients

� � = $−10 < � < 202 < � < 4 period � = 4

6. Fourier Series

Example 1

�� � = �2 +� � cos 2��� + �� sin 2����

���

Page 11: ODE_Chapter 06 [Jan 2014](1)

Step 1.1: Compute � � = 2�! � � "�

#

= 24! � � "�(

= 12 ! −1 "��

�+! 0"�

(

= −12 2 − 0 = −1

� � = $−10 < � < 202 < � < 4

= 24 ! � � "��

�+! � � "�

(

= 0

6. Fourier Series

Page 12: ODE_Chapter 06 [Jan 2014](1)

= 0 � = 2�! � � cos 2��� "�#

= 24 ! −1 cos 2��4 "��

�+! 0 cos 2��4 "�

(

= −122� sin

��2 �

= −12 ! cos��2 "��

�+ 0

Step 1.2: Compute � � � = $−10 < � < 202 < � < 4

6. Fourier Series

Page 13: ODE_Chapter 06 [Jan 2014](1)

= −122� sin

��2 �

= − 1� sin��2 �

= − 1� sin� 22 − sin � 02

= − 1� sin �

= 0

= 0

= 0

6. Fourier Series

Page 14: ODE_Chapter 06 [Jan 2014](1)

�� = 2�! � � sin 2��� "�#

= 24 ! −1 sin 2��4 "��

�+! 0 sin 2��4 "�

(

= −12 ! sin ��2 "��

�+ 0

= 122� cos

��2 �

Step 1.3: Compute �� � � = $−10 < � < 202 < � < 4

= 0

6. Fourier Series

Page 15: ODE_Chapter 06 [Jan 2014](1)

= 1� cos

��2 �

= 1� cos

� 22 − cos � 02

= 1� cos � − 1

= −1 � − 1�

=1

6. Fourier Series

Page 16: ODE_Chapter 06 [Jan 2014](1)

�� � = �2 +� � cos 2��� + �� sin 2����

���

= −12 +� 0 cos 2��4 + −1 � − 1� sin 2��4

���

= −12 +1�

−1 � − 1� sin ��2

���

Step 2: Substitute the Fourier coefficients and �

= 0

6. Fourier Series

Page 17: ODE_Chapter 06 [Jan 2014](1)

Find the Fourier series of

� � = $0 − 5 < � < 010 < � < 5 period � =10

6. Fourier Series

Exercise

Page 18: ODE_Chapter 06 [Jan 2014](1)

Let �(�) be defined in the interval *, * + 2+

The Fourier series will now represented as

�� � = �2 +� � cos ��+ + �� sin ��+�

���

6. Fourier Series

Fourier Series when T=2L

�� � = �2 +� � cos 2��� + �� sin 2����

���

Page 19: ODE_Chapter 06 [Jan 2014](1)

�� � = �2 +� � cos ��+ + �� sin ��+�

���

� = 1+! � � "�,-�.

,

� = 1+! � � cos ��+ "�,-�.

,

�� = 1+! � � sin ��+ "�,-�.

,

6. Fourier Series

Fourier coefficients can be obtained by

Page 20: ODE_Chapter 06 [Jan 2014](1)

6. Fourier Series

Example 2

Solution

Step 1: Compute the Fourier coefficients

Find the Fourier series of the function of period, � = 2� � = �, −1 ≤ � < 1

Interval is *, * + 2+ * = −1* + 2+ = 1 ⇒ + = 1

Page 21: ODE_Chapter 06 [Jan 2014](1)

6. Fourier Series

�� � = �2 +� � cos ��+ + �� sin ��+�

���

� = 1+! � � "�,-�.

,

� = 1+! � � cos ��+ "�,-�.

,

�� = 1+! � � sin ��+ "�,-�.

,

Page 22: ODE_Chapter 06 [Jan 2014](1)

Step 1.1: Compute �

6. Fourier Series

� = 1+! � � "�,-�.

,

= 11! �"��

1�

= ��2 1�

= 12 −12 = 0

Page 23: ODE_Chapter 06 [Jan 2014](1)

Step 1.2: Compute �

6. Fourier Series

� = 1+! � � cos ��+ "�,-�.

,

= 11! � cos �� "��

1�

= � sin ��� + cos ��� � 1�

� cos ��sin ���

−cos ��� �

1(+)

(−)0

Page 24: ODE_Chapter 06 [Jan 2014](1)

6. Fourier Series

= � sin ��� + cos ��� � 1�

= 1� sin � − −sin −� + 1

� � cos � − cos −�

= 0 + 1� � cos � − cos �

= 0

Page 25: ODE_Chapter 06 [Jan 2014](1)

Step 1.3: Compute ��6. Fourier Series

�� = 1+! � � sin ��+ "�,-�.

,

= 11! � sin �� "��

1�

= −� cos ��� + sin ��� � 1�

� sin ��

− cos ���−23� ��� �

1

0

(+)

(−)

Page 26: ODE_Chapter 06 [Jan 2014](1)

6. Fourier Series

= −� cos ��� + sin ��� � 1�

= − 1� cos � − −cos −�

+ 1� � sin � − sin −�

= − 1� 2 cos � + 1� � 2 sin �

= −2 −1�

� = 2 −1�-�

= 0

Page 27: ODE_Chapter 06 [Jan 2014](1)

� = 0, �= 0, �� = 2 −1�-�

= � 2 −1 �-�� sin ��

���

�� � = �2 +� � cos ��+ + �� sin ��+�

���

Step 2: Substitute the Fourier coefficients and +6. Fourier Series

Page 28: ODE_Chapter 06 [Jan 2014](1)

4 −� = −4(�)A function is said to be an odd function if

Examples: 4 � = sin �, 4 � = � , 4 � = �5

If � is an odd function, then

! 4 � "� = 06

16

6. Fourier Series

Definition

Theorem

Odd Function

Page 29: ODE_Chapter 06 [Jan 2014](1)

4 −� = 4(�)Examples: 4 � = cos � , 4 � = 1 , 4 � = ��

! 4 � "� = 2! 4 � "�6

6

16

6. Fourier Series

A function is said to be an even function if

Theorem

If � is an even function, then

Definition

Even Function

Page 30: ODE_Chapter 06 [Jan 2014](1)

If both � and 4 are odd functions, then �4 is an even

function

If both � and 4 are even functions, then �4 is an even

function

If � is an odd function and 4 is an even function, then �4is an odd function

6. Fourier Series

Page 31: ODE_Chapter 06 [Jan 2014](1)

For an odd function � , all amplitudes of cosine functions � are zero

�� � = �2 +� �� sin 2����

���

Fourier Sine Series

6. Fourier Series

Theorem

�� � = �2 +� �� sin ��+�

���� = 2+

Page 32: ODE_Chapter 06 [Jan 2014](1)

�� � = �2 +� � cos 2����

���

Fourier Cosine Series

6. Fourier Series

Theorem

For an even function � , all amplitudes of sine functions �� are zero

�� � = �2 +� � cos ��+�

���� = 2+

Page 33: ODE_Chapter 06 [Jan 2014](1)

6. Fourier Series

Example 3

Solution

Step 1: Compute the Fourier coefficients

Find the Fourier series of the following function

Interval is *, * + 2+

� � = ��, − ≤ � < , � = 2

* = −* + 2+ = ⇒ + =

Page 34: ODE_Chapter 06 [Jan 2014](1)

6. Fourier Series

�� � = �2 +� � cos ��+ + �� sin ��+�

���

� = 1+! � � "�,-�.

,

� = 1+! � � cos ��+ "�,-�.

,

�� = 1+! � � sin ��+ "�,-�.

,

Page 35: ODE_Chapter 06 [Jan 2014](1)

Step 1.1: Compute �6. Fourier Series

� = 1+! � � "�,-�.

,

= 1! ��"�7

17

= 1�53 17

7

= 13 5 − − 5

= 25

3 =23�

Page 36: ODE_Chapter 06 [Jan 2014](1)

Step 1.2: Compute �

6. Fourier Series

� = 1+! � � cos ��+ "�,-�.

,

= 1! �� cos �� "�7

17

�� cos ��sin ���

−9:2 ����

2�

2

(+)(−)

−sin ���50

(+) � = 2! �� cos �� "�

7

= 2�� sin ��

� + 2� cos ���� − 2 sin ���5 �

7

Page 37: ODE_Chapter 06 [Jan 2014](1)

� = 2�� sin ��

� + 2� cos ���� − 2 sin ���5 �

7

= 2� � sin � − 0 +

4�� cos � − 0

− 4�5 sin � − 0

= 4�� cos �

= 4�� −1 � = −1 � 4��

6. Fourier Series

= 0

= 0

Page 38: ODE_Chapter 06 [Jan 2014](1)

Step 1.3: Compute ��6. Fourier Series

Since �� is an even function and based on theorem

�� = 0

Page 39: ODE_Chapter 06 [Jan 2014](1)

The Fourier series is

Fourier Cosine Series

= 122�3 +� −1 � 4�� cos

��

���

= �3 + 4� −1 �

���

cos ����

= �3 + 4 −cos � + cos 2�4 − cos 3�9 +⋯

6. Fourier Series

Step 2: Substitute the Fourier coefficients and +

�� � = �2 +� � cos ��+�

���