07 periodic functions and fourier series

engineering maths

1. Periodic Functions and
Fourier Series

2. Periodic Functions
A function ()qf is periodic
if it is defined for all real q
and if there is some positive number,
Tsuch that ( ) ()qq fTf =+ .

6. Fourier Series
()qf be a periodic function with period p2
The function can be represented by a
trigonometric series as:
() åå
¥
=
¥
=
++=
11
0 sincos
n
n
n
n nbnaaf qqq

7. () åå
¥
=
¥
=
++=
11
0 sincos
n
n
n
n nbnaaf qqq
What kind of trigonometric (series) functions
are we talking about?


qqq
qqq
32
and32
sin,sin,sin
cos,cos,cos

10. We want to determine the coefficients,
n
aand
n
b.
Let us first remember some useful
integrations.

11. ( ) ( )òò
ò
--
-
-++=
p
p
p
p
p
p
qqqq
qqq
dmndmn
dmn
cos
2
1
cos
2
1
coscos
0=qqq
ò
p
p-
dmncoscos mn¹
p=qqq
ò
p
p-
dmncoscos mn=

12. ( ) ( )òò
ò
--
-
-++=
p
p
p
p
p
p
qqqq
qqq
dmndmn
dmn
sin
2
1
sin
2
1
cossin
0=qqq
ò
p
p-
dmncossin
for all values of m.

13. ( ) ( )òò
ò
--
-
+--=
p
p
p
p
p
p
qqqq
qqq
dmndmn
dmn
cos
2
1
cos
2
1
sinsin
0=qqq
ò
p
p-
dmnsinsin mn¹
p=qqq
ò
p
p-
dmnsinsin mn=

14. Determine
0
a
Integrate both sides of (1) from
p- top
()
qqq
qq
p
p
p
p
dnbnaa
df
n
n
n
nò åå
ò
-
¥
=
¥
=
-
ú
û
ù
ê
ë
é
++=
11
0
sincos

15. ()
qq
qqq
qq
p
p
p
p
p
p
p
p
dnb
dnada
df
n
n
n
n
òå
òåò
ò
-
¥
=
-
¥
=
-
-
÷
÷
ø
ö
ç
ç
è
æ
+
÷
÷
ø
ö
ç
ç
è
æ
+=
1
1
0
sin
cos
() 00
0 ++=òò
--
qqq
p
p
p
p
dadf

16. () 002
0++p=qq
ò
p
p-
adf
()qq
p
=
ò
p
p-
dfa
2
1
0
0
ais the average (dc) value of the
function, ()qf .

17. ()
qqq
qq
p
p
dnbnaa
df
n
n
n
nò åå
ò
ú
û
ù
ê
ë
é
++=
¥
=
¥
=
2
0
11
0
2
0
sincos
It is alright as long as the integration is
performed over one period.
You may integrate both sides of (1) from
0top2 instead.

18. ()
qq
qqq
qq
p
pp
p
dnb
dnada
df
n
n
n
n
òå
òåò
ò
÷
÷
ø
ö
ç
ç
è
æ
+
÷
÷
ø
ö
ç
ç
è
æ
+=
¥
=
¥
=
2
0
1
2
0
1
2
0
0
2
0
sin
cos
() 00
2
0
0
2
0
++=òò
qqq
pp
dadf

19. () 002
0
2
0
++=ò
adf pqq
p
()qq
p
p
dfaò
=
2
0
0
2
1

20. Determine
n
a
Multiply (1) by qmcos
and then Integrate both sides from
p- top
()
qqqq
qqq
p
p
p
p
dmnbnaa
dmf
n
n
n
nò åå
ò
-
¥
=
¥
=
-
ú
û
ù
ê
ë
é
++= cossincos
cos
11
0

21. Let us do the integration on the right-hand-side
one term at a time.
First term,
ò
p
p-
=qq 0
0 dmacos
Second term,
qqq
òå
p
p-
¥
=
dmna
n
n
coscos
1

22. pqqq
p
p
m
n
n admna =òå
-
¥
=
coscos
1
Second term,
Third term,
0
1
=òå
-
¥
=
qqq
p
p
dmcosnsinb
n
n

23. Therefore,
() pqqq
p
p
madmf =ò
-
cos
() ,2,1cos
1
==ò
-
mdmfa
m
qqq
p
p
p

24. Determine
n
b
Multiply (1) by qmsin
and then Integrate both sides from
p- top
()
qqqq
qqq
p
p
p
p
dmsinnsinbncosaa
dmsinf
n
n
n
nò åå
ò
-
¥
=
¥
=
-
ú
û
ù
ê
ë
é
++=
11
0

25. Let us do the integration on the right-hand-side
one term at a time.
First term,
ò
-
=
p
p
qq 0
0
dmsina
Second term,
qqq
p
p
dmsinncosa
n
nòå
-
¥
=1

26. 0
1
=òå
-
¥
=
qqq
p
p
dmsinncosa
n
n
Second term,
Third term,
pqqq
p
p
m
n
n
bdmnb =òå
-
¥
=
sinsin
1

27. Therefore,
() pqqq
p
p
m
bdmsinf =ò
-
() ,,mdmsinfb
m
21
1
==ò
-
qqq
p
p
p

28. ()qq
p
=
ò
p
p-
dfa
2
1
0
The coefficients are:
() ,2,1cos
1
==ò
-
mdmfa
m qqq
p
p
p
() ,2,1sin
1
==ò
-
mdmfb
m qqq
p
p
p

29. ()qq
p
=
ò
p
p-
dfa
2
1
0
We can write n in place of m:
() ,,ndncosfa
n
21
1
==ò
-
qqq
p
p
p
() ,,ndnsinfb
n 21
1
==ò
-
qqq
p
p
p

30. The integrations can be performed from
0top2 instead.
()qq
p
p
dfaò
=
2
0
0
2
1
() ,,ndncosfa
n 21
12
0
==ò
qqq
p
p
() ,,ndnsinfb
n 21
12
0
==ò
qqq
p
p

31. Example 1. Find the Fourier series of
the following periodic function.
()
p<q<p-=
p<q<=q
2
0
whenA
whenAf
( ) ()qpq ff =+2

32. ()
() ()
0
2
1
2
1
2
1
2
0
2
0
2
0
0
=
úû
ù
êë
é
-+=
úû
ù
êë
é
+=
=
òò
òò
ò
qq
p
qqqq
p
qq
p
p
p
p
p
p
p
p
dAdA
dfdf
dfa

33. ()
( )
0
11
1
1
2
0
2
0
2
0
=
ú
û
ù
ê
ë
é
-+
ú
û
ù
ê
ë
é
=
úû
ù
êë
é
-+=
=
òò
ò
p
p
p
p
p
p
p
q
p
q
p
qqqq
p
qqq
p
n
nsin
A
n
nsin
A
dncosAdncosA
dncosfa
n

34. ()
( )
[ ]ppp
p
q
p
q
p
qqqq
p
qqq
p
p
p
p
p
p
p
p
ncosncoscosncos
n
A
n
ncos
A
n
ncos
A
dnsinAdnsinA
dnsinfb
n
-++-=
ú
û
ù
ê
ë
é
+
ú
û
ù
ê
ë
é
-=
úû
ù
êë
é
-+=
=
òò
ò
20
11
1
1
2
0
2
0
2
0

35. [ ]
[ ]
oddisnwhen
4
1111
20
p
p
ppp
p
n
A
n
A
ncosncoscosncos
n
A
b
n
=
+++=
-++-=

36. [ ]
[ ]
evenisnwhen0
1111
20
=
-++-=
-++-=
p
ppp
p
n
A
ncosncoscosncos
n
A
b
n

37. Therefore, the corresponding Fourier series is
÷
ø
ö
ç
è
æ
++++ qqqq
p
7sin
7
1
5sin
5
1
3sin
3
1
sin
4A
In writing the Fourier series we may not be
able to consider infinite number of terms for
practical reasons. The question therefore, is
– how many terms to consider?

38. When we consider 4 terms as shown in the
previous slide, the function looks like the
following.
1.5
1
0.5
0
0.5
1
1.5
fq()
q

39. When we consider 6 terms, the function looks
like the following.
1.5
1
0.5
0
0.5
1
1.5
fq()
q

40. When we consider 8 terms, the function looks
like the following.
1.5
1
0.5
0
0.5
1
1.5
fq()
q

41. When we consider 12 terms, the function looks
like the following.
1.5
1
0.5
0
0.5
1
1.5
fq()
q

42. The red curve was drawn with 12 terms and
the blue curve was drawn with 4 terms.
1.5
1
0.5
0
0.5
1
1.5
q

43. The red curve was drawn with 12 terms and
the blue curve was drawn with 4 terms.
0 2 4 6 8 10
1.5
1
0.5
0
0.5
1
1.5
q

44. The red curve was drawn with 20 terms and
the blue curve was drawn with 4 terms.
0 2 4 6 8 10
1.5
1
0.5
0
0.5
1
1.5
q

45. Even and Odd Functions
(We are not talking about even or
odd numbers.)

46. Even Functions The value of the
function would
be the same
when we walk
equal distances
along the X-axis
in opposite
directions.
( ) ()qqff =-
Mathematically speaking -

47. Odd Functions The value of the
function would
change its sign
but with the
same magnitude
when we walk
equal distances
along the X-axis
in opposite
directions.
( ) ()qq ff -=-
Mathematically speaking -

48. Even functions can solely be represented
by cosine waves because, cosine waves
are even functions. A sum of even
functions is another even function.
10 0 10
5
0
5
q

49. Odd functions can solely be represented by
sine waves because, sine waves are odd
functions. A sum of odd functions is another
odd function.
10 0 10
5
0
5
q

50. The Fourier series of an even function ()qf
is expressed in terms of a cosine series.
()å
¥
=
+=
1
0
cos
n
n
naaf qq
The Fourier series of an odd function ()qf
is expressed in terms of a sine series.
()å
¥
=
=
1
sin
n
nnbf qq

51. Example 2. Find the Fourier series of
the following periodic function.
( ) ()qpq ff =+2
() pp££-= xwhenxxf
2

52. ()
332
1
2
1
2
1
23
2
0
p
p
pp
p
p
p
p
p
p
=
ú
û
ù
ê
ë
é
=
==
=
-=
-- òò
x
x
x
dxxdxxfa

53. ()
úû
ù
êë
é
=
=
ò
ò
-
-
nxdxx
dxnxxfa
n
cos
1
cos
1
2
p
p
p
p
p
p
Use integration by parts. Details are shown
in your class note.

54. pn
n
a
n cos
4
2
=
odd is n when
4
2
n
a
n
-=
even is n when
4
2
n
a
n
=

55. This is an even function.
Therefore, 0=
nb
The corresponding Fourier series is
÷
ø
ö
ç
è
æ
+-+-- 
222
2
4
4cos
3
3cos
2
2cos
cos4
3
xxx
x
p

56. Functions Having Arbitrary Period
Assume that a function has
period, . We can relate angle
() with time () in the following
manner.
twq=
w is the angular velocity in radians per
second.

57. fpw2=
f is the frequency of the periodic function,
()tf
tfpq2=
T
f
1
=where
t
T
p
q
2
=
Therefore,

58. t
T
p
q
2
= dt
T
d
p
q
2
=
Now change the limits of integration.
pq-=
2
T
t-=t
T
p
p
2
=-
pq=
2
T
t=t
T
p
p
2
=

59. ()qq
p
=
ò
p
p-
dfa
2
1
0
()dttf
T
a
T
T
ò
-
=
2
2
0
1

60. () ,,ndncosfa
n 21
1
==ò
-
qqq
p
p
p
() ,2,1
2
cos
2
2
2
=÷
ø
ö
ç
è
æ
=ò
-
ndtt
T
n
tf
T
a
T
T
n
p

61. () ,,ndnsinfb
n
21
1
==ò
-
qqq
p
p
p
() ,2,1
2
sin
2
2
2
=÷
ø
ö
ç
è
æ
=ò
-
ndtt
T
n
tf
T
b
T
T
n
p

62. Example 4. Find the Fourier series of
the following periodic function.
()
4
3
42
44
T
t
T
when
T
t
T
t
T
whenttf
££+-=
££-=

63. ( ) ()tfTtf =+
This is an odd function. Therefore, 0=
na
()
() dtt
T
n
tf
T
dtt
T
n
tf
T
b
T
T
n
÷
ø
ö
ç
è
æ
=
÷
ø
ö
ç
è
æ
=
ò
ò
p
p
2
sin
4
2
sin
2
2
0
0

64. dtt
T
nT
t
T
dtt
T
n
t
T
b
T
T
T
n
÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
+-+
÷
ø
ö
ç
è
æ
=
ò
ò
p
p
2
sin
2
4
2
sin
4
2
4
4
0
Use integration by parts.

65. ÷
ø
ö
ç
è
æ
=
ú
ú
û
ù
ê
ê
ë
é
÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
=
2
sin
2
2
sin
2
.2
4
22
2
p
p
p
p
n
n
T
n
n
T
T
b
n
0=
n
b when n is even.

66. Therefore, the Fourier series is
ú
û
ù
ê
ë
é
-÷
ø
ö
ç
è
æp
+÷
ø
ö
ç
è
æp
-÷
ø
ö
ç
è
æp
p
t
T
t
T
t
T
T 10
5
16
3
122
222
sinsinsin

67. The Complex Form of Fourier Series
() åå
¥
=
¥
=
++=
11
0
sincos
n
n
n
n
nbnaaf qqq
Let us utilize the Euler formulae.
2
qq
q
jj
ee
cos
-
+
=
i
ee
sin
jj
2
qq
q
-
-
=

68. The nth harmonic component of (1) can be
expressed as:
i
ee
b
ee
a
nbna
jnjn
n
jnjn
n
nn
22
sincos
qqqq
qq
--
-
+
+
=
+
22
qqqq jnjn
n
jnjn
n
ee
ib
ee
a
--
-
-
+
=

69. qq
qq
jnnnjnnn
nn
e
jba
e
jba
nbna
-
÷
ø
ö
ç
è
æ+
+÷
ø
ö
ç
è
æ-
=
+
22
sincos
Denoting
÷
÷
ø
ö
ç
ç
è
æ-
=
2
nn
n
jba
c ÷
ø
ö
ç
è
æ+
=
-
2
nn
n
jba
c,
and
00ac=

70. qq
qq
jn
n
jn
n
nn
ecec
nsinbncosa
-
-
+=
+

71. The Fourier series for ()qf
can be expressed as:
() ( )
å
å
¥
-¥=
¥
=
-
-
=
++=
n
jn
n
n
jn
n
jn
n
ec
ececcf
q
qq
q
1
0

72. The coefficients can be evaluated in
the following manner.
() () qqq
p
qqq
p
p
p
p
p
dnf
j
dnf
jba
c
nn
n
sin
2
cos
2
1
2
òò
--
-=
÷
ø
ö
ç
è
æ-
=
()( )ò
-
-=
p
p
qqqq
p
dnjnf sincos
2
1
()ò
-
-
=
p
p
q
qq
p
def
jn
2
1

73. () () qqq
p
qqq
p
p
p
p
p
dnf
j
dnf
jba
c
nn
n
sin
2
cos
2
1
2
òò
--
-
+=
÷
ø
ö
ç
è
æ+
=
()( )ò
-
+=
p
p
qqqq
p
dnjnf sincos
2
1
()ò
-
=
p
p
q
qq
p
def
jn
2
1

74. .
÷
ø
ö
ç
è
æ-
=
2
nn
n
jba
c ÷
ø
ö
ç
è
æ+
=
-
2
nn
n
jba
c
n
c
-
nc
Note that is the complex conjugate of
.Hence we may write that
()
ò
p
p-
q-
qq
p
= defc
jn
n
2
1
,,,n 210 ±±=

75. The complex form of the Fourier series of
()qf p2 with period is:
()å
¥
-¥=
=
n
jn
necf
q
q

76. Example 1. Find the Fourier series of
the following periodic function.
()
p<q<p-=
p<q<=q
2
0
whenA
whenAf
( ) ()qpq ff =+2

77. A5:=
fx() A 0x£ p<if
A- px£ 2p×£if
0otherwise
:=
A0
1
2p
0
2p
xfx()
ó
ô
õ
d×:=
A00=

78. n18..:=
An
1
p
0
2p
xfx()cosnx×()×
ó
ô
õ
d×:=
A10= A20= A30= A40=
A50= A60= A70= A80=

79. Bn
1
p
0
2p
xfx()sinnx×()×
ó
ô
õ
d×:=
B16.366= B20= B32.122= B40=
B51.273= B60= B70.909= B80=

80. Complex Form
()å
¥
-¥=
q
=q
n
jn
necf ()
ò
p
p-
q-
qq
p
= defc
jn
n
2
1
,,, 210±±=n
Cn()
1
2p
0
2p
xfx()e
1i-n×x×
×
ó
ô
õ
d×:=

81. C0()0= C1()3.183i-= C2()0= C3()1.061i-=
C4()0= C5()0.637i-= C6()0= C7()0.455i-=
C1-()3.183i= C2-()0= C3-()1.061i=
C4-()0= C5-()0.637i= C6-()0= C7-()0.455i=
Cn()
1
2p
0
2p
xfx()e
1i-n×x×
×
ó
ô
õ
d×:=