Ordinary Differential Equations [FDM 1023]
Ordinary Differential Equations[FDM 1023]
FOURIER SERIES
Chapter 6
Learning Outcome
At the end of this section, you should be ableto:
Find the Fourier series of a periodicfunction
6. Fourier Series
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
� = 2
6. Fourier Series
Example
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
�� � = �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
�� � = �2 +� � cos 2��� + �� sin 2����
���
The first few terms of the infinite series
�� � = �2 + � cos2�� + �� sin 2�� + � cos 4��
+ �� sin 4�� +⋯
6. Fourier Series
Fourier coefficients can be obtained by
� = 2�! � � "�#
�
� = 2�! � � cos 2��� "�#
�
�� = 2�! � � sin 2��� "�#
�
�� � = �2 +� � cos 2��� + �� sin 2����
���
6. Fourier Series
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����
���
Step 1.1: Compute � � = 2�! � � "�
#
�
= 24! � � "�(
�
= 12 ! −1 "��
�+! 0"�
(
�
= −12 2 − 0 = −1
� � = $−10 < � < 202 < � < 4
= 24 ! � � "��
�+! � � "�
(
�
= 0
6. Fourier Series
= 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
= −122� sin
��2 �
�
= − 1� sin��2 �
�
= − 1� sin� 22 − sin � 02
= − 1� sin �
= 0
= 0
= 0
6. Fourier Series
�� = 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
= 1� cos
��2 �
�
= 1� cos
� 22 − cos � 02
= 1� cos � − 1
= −1 � − 1�
=1
6. Fourier Series
�� � = �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
Find the Fourier series of
� � = $0 − 5 < � < 010 < � < 5 period � =10
6. Fourier Series
Exercise
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����
���
�� � = �2 +� � cos ��+ + �� sin ��+�
���
� = 1+! � � "�,-�.
,
� = 1+! � � cos ��+ "�,-�.
,
�� = 1+! � � sin ��+ "�,-�.
,
6. Fourier Series
Fourier coefficients can be obtained by
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
6. Fourier Series
�� � = �2 +� � cos ��+ + �� sin ��+�
���
� = 1+! � � "�,-�.
,
� = 1+! � � cos ��+ "�,-�.
,
�� = 1+! � � sin ��+ "�,-�.
,
Step 1.1: Compute �
6. Fourier Series
� = 1+! � � "�,-�.
,
= 11! �"��
1�
= ��2 1�
�
= 12 −12 = 0
Step 1.2: Compute �
6. Fourier Series
� = 1+! � � cos ��+ "�,-�.
,
= 11! � cos �� "��
1�
= � sin ��� + cos ��� � 1�
�
� cos ��sin ���
−cos ��� �
1(+)
(−)0
6. Fourier Series
= � sin ��� + cos ��� � 1�
�
= 1� sin � − −sin −� + 1
� � cos � − cos −�
= 0 + 1� � cos � − cos �
= 0
Step 1.3: Compute ��6. Fourier Series
�� = 1+! � � sin ��+ "�,-�.
,
= 11! � sin �� "��
1�
= −� cos ��� + sin ��� � 1�
�
� sin ��
− cos ���−23� ��� �
1
0
(+)
(−)
6. Fourier Series
= −� cos ��� + sin ��� � 1�
�
= − 1� cos � − −cos −�
+ 1� � sin � − sin −�
= − 1� 2 cos � + 1� � 2 sin �
= −2 −1�
� = 2 −1�-�
�
= 0
� = 0, �= 0, �� = 2 −1�-�
�
= � 2 −1 �-�� sin ��
�
���
�� � = �2 +� � cos ��+ + �� sin ��+�
���
Step 2: Substitute the Fourier coefficients and +6. Fourier Series
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
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
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
For an odd function � , all amplitudes of cosine functions � are zero
�� � = �2 +� �� sin 2����
���
Fourier Sine Series
6. Fourier Series
Theorem
�� � = �2 +� �� sin ��+�
���� = 2+
�� � = �2 +� � cos 2����
���
Fourier Cosine Series
6. Fourier Series
Theorem
For an even function � , all amplitudes of sine functions �� are zero
�� � = �2 +� � cos ��+�
���� = 2+
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+ = ⇒ + =
6. Fourier Series
�� � = �2 +� � cos ��+ + �� sin ��+�
���
� = 1+! � � "�,-�.
,
� = 1+! � � cos ��+ "�,-�.
,
�� = 1+! � � sin ��+ "�,-�.
,
Step 1.1: Compute �6. Fourier Series
� = 1+! � � "�,-�.
,
= 1! ��"�7
17
= 1�53 17
7
= 13 5 − − 5
= 25
3 =23�
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
� = 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
Step 1.3: Compute ��6. Fourier Series
Since �� is an even function and based on theorem
�� = 0
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 ��+�
���