مواضيع المحاضرة: FOURIER SERIES
قراءة
عرض

Lecture N o.2-9 by Hussein J. Abd ulHussein

Advanced Calculus II Al M uthanna University, College of Science

1

FOURIER SERIES

PERIODIC FUNCTIONS
If the value of each ordinate � (�) repeats itself at equal intervals in the abscissa, then � (�) is
said to be aperiodic function
If � (�)= �( then T is called the period of the function � (�)
For example : The period of sin � .cos � .sec �. ��� ����� � ��� 2��
The period of tan � ��� cot � ��� ��
sin � = sin (�+ 2��)= sin (�+ 4��)= �� sin � ��� � ���������� ��������� ���� ℎ �ℎ� ������� 2��
sin 5� = sin (5�+ 2��)= sin 5 [�+ 2��
5 ]. ������� = 2��/5
cos 3� = cos (3�+ 2��)= cos 3 [�+ 2��
3 ].������� = 2��
3
cos 2����
� = cos [2���
� + 2π ]= cos 2���
� [�+ 2���
2��� ]
= cos 2���
� [�+ �
�].������� = �

tan 2� = tan (2�+ ��)tan 2[�+ ��
2].��
2
This is also called sinusoidal periodic function
FOURIER SERIES
Here we will express a non -sinusoidal periodic function into a fundamental and its harmonics A
series and cosines of an a ngle and its multiples of the form
��0
2 + �1cos �+ �2cos 2�+ �3cos 3�+ ⋯ + ���cos �� + ⋯ + �1sin �+ �2sin 2�+
�3sin 3�+ ⋯ + ���sin �� + ⋯
= ��0
2 + ∑ ���cos �� + ∑ ���sin �� ∞��=1 ∞��=1
Is called the Fourier series, where �0.�1.�2.⋯ �1.�2.�3⋯ ���⋯ are constants
A periodic function �(�) can be expanded in a Fourier Series The series consists of the following:
(��) A constant term �0 (called � ∙�∙ co mponent in electrical work )

Lecture N o.2-9 by Hussein J. Abd ulHussein

Advanced Calculus II Al M uthanna University, College of Science

2

(����) A component at the fundamental frequency determined by the values of �1 .�1
(������ ) Component of the harmonics (multiples of the fundamental frequency determined
by �2.�3 .�2.�3 ���� �0.�1.�2⋯ . �1.�2⋯ are known as Fourier coefficients or Fourier
constants
Note (1) When the function and its derivatives are continuous then the function can be expanded
in powers of x by Maclaurin's theorem
(2) But by Fourier series we can expand continuous and discontinuous both types of functions
under certain conditions
DIRICHLET'S CONDITIONS FOR A FOURIER SERIES
If the function for the interval (−��.��)
(1)��� ������� − ������ (2)��� �������
(3)ℎ�� �� ���� ��������� ������ �� ������� ��� ��������
(4)ℎ�� ���� � �������� ������ �� ���������������
(5)��� � (�+ 2��)= �(�)�� � ������ �� � �������� [−�� .��].�ℎ��

����(�)= �0

2 + ∑ ���cos ��
��
��=1
+ ∑ ���sin ��
��
��=1

Converges to �(�) �� � → ∞ at values of x for which f (x) is continuous and the sum of
the series is equal to 1
2[ �(�+ 0)+ � (�− 0)] at points of discontinuity

ADVANTAGES OF FOURIER SERIES

1∙ Discontinuous function can be represented by Fourier series Although derivatives of
the discontinuous functions do not exist (This is true for Taylor's series)
2. The Fourier series is useful in expanding the periodic function since outside the closed
interval, there exists aperiodic exists a periodic extension of the function
3. Expansion of an discontinuous function by Fourier series gives all modes of oscillation
(fundamental and all overtones) which is extremely useful in physics
4. Fourier series of a discontinuous function is not uniformly convergent at all points
5. Term by term integration of a convergent Fourier series is always valid, and it may be
valid if the series is not convergent However, term, differentiation of a Fourier series is
not valid in most cases

Lecture N o.2-9 by Hussein J. Abd ulHussein

Advanced Calculus II Al M uthanna University, College of Science

3

USEFUL INTEGRALS

The following integrals are useful in Fourier Series

(��)∫ sin �� �� = 0

2��
0
(����)∫ cos �� �� = 0
2��
0

(������ )∫ sin 2�� �� = ��
2��
0
(��� )∫ cos 2�� �� = ��
2��
0

(�)∫ sin �� ∙sin �� �� = 0
2��
0
(��� )∫ sin �� cos �� �� = 0
2��
0

(����� )∫ sin �� ∙cos �� �� = 0
2��
0
(������� )∫ sin �� cos �� �� = 0
2��
0

(��� )[�� ]1= ��1− �′�2+ �′′ ��3+ �′′′��4+ ⋯

Where

�1= ∫ � �� . �2= ∫ �1 �� ��� �� �� �′= ��
�� .�′′= �2�
�� 2 ��� �� �� ���
(�)sin ��� = 0.cos ��� = (−1)�� �ℎ��� � �� �
DETERMINATION OF FOURIER COEFFICIENTS (EULER'S FORMULE)
� (�)= �0
2 + �1���� + �2cos �+ ⋯ + ���cos �� + ⋯
+ �1sin �+ �2sin 2�+ ���sin �� + ⋯ ⋯ (1)
(��) To find : Integrate both sides of (1) from � = 0 �� � = 2��

∫ �(�)�� = �0

2 ∫ �� + �1
2��
0
2��
0
∫ �����
2��
0
+ �2 ∫ ��� 2��� + ⋯
2��
0
+ ���∫ cos �� �� + ⋯ + �1
2��
0
∫ sin ��� + �2
2��
0
∫ sin 2��� + ⋯
2��
0
+ ���∫ sin �� �� + ⋯ + �0
2 ∫ ��
2��
0

2��

0

(other integrals = 0 by formulae (��) and (����))
∫ �(�)�� = �0
2 ∫ 2��
2��
0
2��
0
⇒ �0= 1
�� ∫ �(�)��
2��
0
⋯ (2)
(����) To find: Multiply each side of (1) by cos �� and integrate from � = 0 �� � = 2��

Lecture N o.2-9 by Hussein J. Abd ulHussein

Advanced Calculus II Al M uthanna University, College of Science

4

∫ �(�)cos �� �� = ��0
2∫ cos �� �� + �1 2��
0
2��
0 ∫ ���� cos �� �� 2��
0 + ⋯ +
�1∫ sin �cos �� �� + �2 2��
0 ∫ sin 2�cos �� �� + ⋯ 2��
0
= ���∫ cos 2 �� �� = �����
2��
0
(�� ℎ�� ���������� = 0.)

���= 1

�� ∫ �(�)cos �� ��
2��
0
⋯ (3)
By taking � = 1.2 we can find the values of �1 .�2 ⋯
(������ )To find ��� : Multiply each side of (1) by sin �� and integrate from � = 0 �� � = 2��

∫ �(�)sin�� �� = ��0
2∫ sin �� �� + �1∫ ���� sin �� �� + ⋯ + 2��
0
2��
0
2��
0
���∫ cos �� sin �� �� + ⋯ + �1∫ sin �sin �� �� + ⋯ + ⋯ + ���∫ ���� 2 �� �� + 2��
0
2��
0
2��
0

= ���∫ ���� 2 �� ��
2��
0

= ���π

���= 1
�� ∫ �(�)cos �� ��
2��
0
⋯ (4)
Note: To get similar formula of �0 .1
2 has been written with �0 in Fourier series
Example 1. Find the Fourier series representing
�(�)= � 0 < � < 2��
And sketch its graph from � = − 4�� �� � = 4��
Solution Let �(�)= ��0
2 + �1cos �+ �2cos 2�+ ⋯ + �1����� + �2sin 2�+ ⋯ (1)
Hence �0= 1
��∫ �(�)�� = 1
��
2��
0 ∫ ��� = 1
��
2��
0 [��2
2]2��
0 = 2��

���= 1

��∫ �(�)cos �� �� = 1
��
2��
0
∫ ���� ����
2��
0

= 1
��[�sin ��
� − 1(− cos ��
�2 )]2��
0 = 1
π[cos 2 ���
�2 − 1
�2]= 1
�2��(1− 1)= 0

���= 1

��∫ �(�)sin �� �� 2��
0 = 1
��∫ �sin �� �� 2��
0
= 1
��[�(− cos ��
� )− 1(− sin ��
�2 ) ]2��
0 = 1
��[−2��cos 2���
� ]= − 2

Lecture N o.2-9 by Hussein J. Abd ulHussein

Advanced Calculus II Al M uthanna University, College of Science

5

Substi tuting the values of �0.�1. �2⋯ .�1.�2⋯ ��� (1). we get

� = �� − 2[sin �+ 1

2sin 2�+ 1
3sin 3�+ ⋯ ]
Example 2. Given that �(�)= �+ �2 ��� − �� < � < �� . find the Fourier expression
of �(�) Deduce that ��2
6 = 1+ 1
22+ 1
32+ 1
42+ ⋯

Solution Let �+ �2= ��0

2 + �1cos �+ �2cos 2�+ ⋯ + �1 ����� + �2sin 2� ⋯ (1)

�0= 1

��∫ �(�)�� = 1
��
��
−��
∫ (�+ �2)��
��
−��

= 1
��[�2
2 + �3
3] ��
−�� = 1
��[��2
2 + ��3
3 + ��2
2 + ��3
3]= 2��2
3

���= 1

��∫ �(�)cos �� �� =
��
−��
∫ (�+ �2)cos �� ��
��
−��

Fourier Series

= 1
��[(�+ �2)sin ��
� − (2�+ 1)(− cos �� )
�2 + (2)(− sin ��
�3 )] ��
−��

= 1
��[(2�� + 1)cos ���
�2 − (−2π + 1)cos (− ��� )
�2 ]= 1
��[4��cos ���
�2 ]= 4(−1)��
�2

���= 1

��∫ �(�)sin �� �� = 1
��
��
−��
∫ (�+ �2)sin �� ��
��
−��

= 1
��[(�+ �2) (− cos ���
� )− (2x + 1)(−sin ���
�2 )+ 2cos ��
�3 ] ��
−��

= 1
��[−(��+ ��2)cos ���
� + 2cos ���
�3 + (−��+ ��2)cos ���
� + 2cos ���
�3 ]

= 1
��[2��
� cos ��� ] = − 2
�(−1)��
Substituting the values of �0.�1. �2⋯ .�1.�2⋯ ��� (1). we get

�+ �2= ��2

3 + 4 [− cos �+ 1
22cos 2�− 1
22cos 3�+ ⋯ ]

Lecture N o.2-9 by Hussein J. Abd ulHussein

Advanced Calculus II Al M uthanna University, College of Science

6

−2[− sin �+ 1

2sin 2�− 1
3sin 3�+ ⋯ ]⋯ (2)
Putting � = ��.(2) Becomes �� + ��2= ��2
3 + 4[1+ 1
22+ 1
32+ 1
42+ ⋯ ]⋯ (3)

Putting � = −��.(2) Becomes ��+ ��2= ��2
3 + 4[1+ 1
22+ 1
32+ 1
42+ ⋯ ]⋯ (4)
Adding (3) and (4), 2��2= 2��2
3 + 8[1+ 1
22+ 1
32+ 1
42+ ⋯ ]

4��2

3 + 8[1+ 1
22+ 1
32+ 1
42+ ⋯ ]

��2

6 = 1+ 1
22+ 1
32+ 1
42+ ⋯ = ∑ 1
�2

��=1

Example 3. Find the Fourier series expansion for �(�)= �+ ��2
4 .−�� ≤ � ≤ ��

Solution. Let �+ ��2

4 = ��0
2 + �1cos �+ �2cos 2�+ ⋯ + �1sin ��2sin 2�+ ⋯
Where
�0= 1
��∫ �(�)�� = 1
��
��
−�� ∫ (�+ ��2
4)�� ��
−��

= 1
��[��2
2 + ��3
12] ��
−�� = 1
��[��2
2 + ��3
12 − ��2
2 + ��3
12 ]= 1
�� [2��3
12 ]= ��2
6

���= 1

��∫ �(�)cos �� �� = 1
��
��
−��
∫ (�+ �2
4)cos �� ��
��
−��

= 1
��[(�+ �2
4)(sin ��
� )− (1+ 2�
4 )(−cos ��
�2 )+ 1
2(− sin ��
�3 )] ��
−��

= 1
��[(�� + ��2
4)(sin ���
� )− (1+ 2��
4 )(−cos n��
�2 )− 1
2(− sin ���
�3 )− (−�� + ��2
4)(sin (−��� )
� )
− (1− 2��
4 )(cos (−��� )
�2 + 1
2
sin (−��� )
�3 ]
= 1
��[ (1+ ��
2) (−1)��
�2 − (1− ��
2)(−1)��
�2 ]= (−1)��
�2�� [1+ ��
2− 1+ ��
2]= (−1)��
�2�� (��)= (−1)��
�2
�1= −1. �2= 1
4 . �3= − 1
9 . �4= 1
16 ⋯ ⋯

Lecture N o.2-9 by Hussein J. Abd ulHussein

Advanced Calculus II Al M uthanna University, College of Science

7

���= 1

��∫ �(�)sin �� �� = 1
��
��
−��
∫ (�+ �2
4)sin �� ��
��
−��

= 1
��[(�+ �2
4)(cos ��
� )− (1+ 2�
4 )(sin ��
�2 )+ 1
2(cos ��
�3 )] ��
−��
= 1
��[(�� + ��2
4)(cos ���
� )− (1+ 2��
4 )(sin ���
�2 )+ 1
2(− cos (− ��� )
�3 )+ 1
4+ 1
��[cos �
1
− cos3 �
3 + cos5 �
5 + ⋯ + sin �
1 + sin 2�
2 + sin 3�
3 + ⋯ ]

1
��[− (�� + ��2
4)(−1)��
� + 1
2((−1)��
�3 )+ (�� + ��2
4)+ (−1)��
� − 1
2
(−1)��
�3 ]

= (−1)��

�� [− (�� + ��2
4)1
�+ (��2
4 − ��)1
�]= (−1)��[−1
� − ��
4�+ ��
4�− 1
�]= −2(−1)��
��
�1= 2
1. �2= 1. �3= 2
3. �4= 1
2 ⋯

Hence, Fourier series of the given function

�(�)= �0
2 + ∑ ���cos �� + ∑ ���sin �� = ��2
12

��=1

��=1
+ ∑ (−1)��
�2 cos �� − ∑ 2
� (−1)�� sin ��

��=1

�(�)= ��2

12 − cos �+ 1
4cos 2�− 1
9cos 3�+ 1
16 cos 4�+ ⋯ + 2sin �− sin 3�+ 2
3sin 4�− 1
2sin 4�+ ⋯
Problems
��∙ Find a Fourier series to represent, �(�)= �� − � ��� 0 < � < 2��
���� ∙2[ ����� + 1
2sin 2�+ 1
3sin 3�+ ⋯ + 1
�sin �� + ⋯
2. Find a Fourier series to represent �− �2 ���� � = −�� �� �� ��� �ℎ�� �ℎ��
��2
12 = 1
12− 1
22+ 1
32− 1
42+ ⋯
Ans − ��2
3 + 4[cos �
12 − cos 2�
22 + cos 3�
32 − cos 4�
42 ]+ 2[sin �
1 − sin2 �
2 + sin3 �
3 − sin4 �
4 + ⋯ ]

Lecture N o.2-9 by Hussein J. Abd ulHussein

Advanced Calculus II Al M uthanna University, College of Science

8

3. Find a Fourier series to repre sent the function �(�)= ���.��� –�� < � < �� and hence derive a
series for ��
sin ℎ �� ���� 2sin ℎ ��
�� [1
2− 1
12+1cos �+ 1
22+1cos 2�− 1
32+1cos 3�+ ⋯ +
1
12+1sin �− 2
22+1sin 2�− 3
32+1sin 3�⋯ .]. ��
sin ℎ ��= 1+ 2[− 1
2+ 1
5− 1
10 + ⋯ ]
4. Obtain the Fourier series for �(�)= ���. in the interval 0≤ � < 2��
���� 1−��−2��
�� [1
2+ 1
2cos �+ 1
5cos 2�+ 1
10 cos 3�+ 1
2sin �+ 2
5sin 2�+ 3
10 sin 3�+ ⋯ ]
5∙�� � (�)= (��− �
2 )
2
.0< � < 2��.�ℎ�� �ℎ�� �(�)= ��2
12 + ∑ cos ��
�2

��=1

6∙����� �ℎ�� �2= ��2
3 + 4∑ (−1)��cos ��
�2

��=1
.−�� < � < �� ����� �ℎ�� �ℎ��
(��)∑ 1
(2�− 1)2= ��2
8 (����)∑ 1
�4= ��4
90
7. If � (�) is aperiodic function defined over a period (0.2�� )�� �(�)= (3��2−6���� +2��2)
12
Prove that �(�)= ∑ cos ����
��2 ∞��=1 and h ence show that ��2
6 = 1+ 1
22+ 1
32+ ⋯


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