engineering maths

1. Periodic Functions and
Fourier Series

2. Periodic Functions
A function f ( θ )
is periodic
if it is defined for all real
θ
and if there is some positive number,
T
such that
f (θ + T ) = f (θ )
.

6. Fourier Series
f (θ )
be a periodic function with period
2π
The function can be represented by a
trigonometric series as:
∞
∞
n =1
n =1
f ( θ ) = a0 + ∑ an cos nθ + ∑ bn sin nθ

7. ∞
∞
n =1
n =1
f ( θ ) = a0 + ∑ an cos nθ + ∑ bn sin nθ
What kind of trigonometric (series) functions
are we talking about?
cos θ , cos 2θ , cos 3θ  and
sin θ , sin 2θ , sin 3θ 

10. We want to determine the coefficients,
an
and
bn
.
Let us first remember some useful
integrations.

11. π
∫ π cos nθ cos mθ dθ
−
1 π
1 π
= ∫ cos( n + m ) θ dθ + ∫ cos( n − m ) θ dθ
2 −π
2 −π
∫
π
cos nθ cos m θdθ = 0
−π
∫
π
cos nθ cos m θdθ = π
−π
n≠m
n=m

12. π
∫ π sin nθ cos mθ dθ
−
1 π
1 π
= ∫ sin( n + m ) θ dθ + ∫ sin( n − m ) θ dθ
2 −π
2 −π
∫
π
sin nθ cos m θdθ = 0
−π
for all values of m.

13. π
∫ π sin nθ sin mθ dθ
−
1 π
1 π
= ∫ cos( n − m )θ dθ − ∫ cos( n + m )θ dθ
2 −π
2 −π
∫
π
sin nθ sin m θdθ = 0
−π
∫
π
sin nθ sin m θdθ = π
−π
n≠m
n=m

14. Determine
a0
Integrate both sides of (1) from
−π
to
π
π
∫ π f ( θ ) dθ
−


= ∫ a0 + ∑ an cos nθ + ∑ bn sin nθ  dθ
−π
n =1
n =1


π
∞
∞

15. π
∫ π f ( θ ) dθ
−


= ∫ a0dθ + ∫  ∑ an cos nθ  dθ

−π
−π 
 n =1

π
∞
π


+ ∫  ∑ bn sin nθ  dθ

−π 
 n =1

π
π
∫π
−
∞
π
f ( θ ) dθ = ∫ a 0 dθ + 0 + 0
−π

16. ∫
π
f ( θ) dθ = 2πa0 + 0 + 0
−π
1
a0 =
2π
a0
∫
π
f ( θ ) dθ
−π
is the average (dc) value of the
function,
f (θ )
.

17. You may integrate both sides of (1) from
0
∫
to
2π
instead.
f ( θ ) dθ
0
=∫
2π
2π
0
∞
∞


a0 + ∑ an cos nθ + ∑ bn sin nθ  dθ
n =1
n =1


It is alright as long as the integration is
performed over one period.

18. ∫
2π
f ( θ ) dθ
0
2π
2π
0
0
= ∫ a0dθ + ∫
+∫
2π
0
∫
2π
0


 ∑ an cos nθ  dθ


 n =1

∞


 ∑ bn sin nθ  dθ


 n =1

∞
2π
f ( θ ) dθ = ∫ a0dθ + 0 + 0
0

19. ∫
2π
0
f ( θ ) dθ = 2πa0 + 0 + 0
1
a0 =
2π
∫
2π
0
f ( θ ) dθ

20. Determine
Multiply (1) by
an
cos mθ
and then Integrate both sides from
−π
to
π
π
∫ π f ( θ ) cos mθ dθ
−
∞
∞


= ∫  a0 + ∑ an cos nθ + ∑ bn sin nθ  cos mθ dθ
−π
n =1
n =1


π

21. Let us do the integration on the right-hand-side
one term at a time.
First term,
∫
π
a0 cos m θ dθ = 0
−π
Second term,
π
∞
∑ a cos nθ cos mθ dθ
∫
− π n =1
n

22. Second term,
π
∞
∑a
∫π
−
n
cos nθ cos mθ dθ = amπ
n=1
Third term,
π
∞
∫ π ∑b
−
n =1
n
sin nθ cos m θ dθ = 0

23. Therefore,
π
∫ π f (θ ) cos mθ dθ = a π
m
−
1
am =
π
π
∫π
−
f ( θ ) cos mθ dθ
m = 1, 2, 

24. Determine
Multiply (1) by
bn
sin m θ
and then Integrate both sides from
−π
to
π
π
∫ π f (θ ) sin mθ dθ
−
∞
∞


= ∫ a0 + ∑ a n cos nθ + ∑ bn sin nθ  sin m θ dθ
−π
n =1
n =1


π

25. Let us do the integration on the right-hand-side
one term at a time.
First term,
π
∫πa
−
0
sin m θ dθ = 0
Second term,
π
∞
∑a
∫π
−
n =1
n
cos nθ sin m θ dθ

26. Second term,
π
∞
∑a
∫π
−
n =1
n
cos nθ sin m θ dθ = 0
Third term,
π
∞
∑b
∫π
−
n
n =1
sin nθ sin mθ dθ = bmπ

27. Therefore,
π
∫π
−
f ( θ ) sin m θdθ = bmπ
1
bm =
π
π
∫ π f (θ ) sin mθdθ
−
m = 1, 2 ,

28. The coefficients are:
1
a0 =
2π
∫
1
am =
π
π
1
bm =
π
π
f ( θ ) dθ
−π
∫π
f ( θ ) cos mθ dθ
m = 1, 2, 
π
f ( θ ) sin mθ dθ
m = 1, 2, 
−
∫π
−

29. We can write n in place of m:
1
a0 =
2π
1
an =
π
1
bn =
π
π
∫
∫π
−
π
π
f ( θ ) dθ
−π
f ( θ ) cos nθ dθ
∫ π f (θ ) sin nθ dθ
−
n = 1 , 2 ,
n = 1 , 2 ,

30. The integrations can be performed from
0
2π
to
1
a0 =
2π
1
an =
π
1
bn =
π
∫
2π
0
∫
2π
∫
2π
0
0
instead.
f ( θ ) dθ
f ( θ ) cos nθ dθ
f ( θ ) sin nθ dθ
n = 1 , 2 ,
n = 1 , 2 ,

31. Example 1. Find the Fourier series of
the following periodic function.
f ( θ) = A
= −A
when 0 < θ < π
when
f ( θ + 2π ) = f ( θ )
π < θ < 2π

32. 1 2π
a0 =
f ( θ ) dθ
∫0
2π
2π
1  π

=
f ( θ ) dθ + ∫ f ( θ ) dθ
 ∫0

π

2π 
2π
1  π

=
∫0 A dθ + ∫π − A dθ 


2π 
=0

33. 1 2π
a n = ∫ f ( θ ) cos nθ dθ
π 0
2π
1 π
=
A cos nθ dθ + ∫ ( − A) cos nθ dθ 

π


π  ∫0
1
=
π
π
2π
1
sin nθ 
 sin nθ 
 A n  + π − A n  = 0

0

π

34. 1 2π
bn = ∫ f ( θ ) sin nθ dθ
π 0
2π
1 π
=
A sin nθ dθ + ∫ ( − A) sin nθ dθ 

π


π  ∫0
π
2π
cos nθ 
1  cos nθ 

− A n  + π  A n 

0

π
A
[ − cos nπ + cos 0 + cos 2nπ − cos nπ ]
=
nπ
1
=
π

35. A
[ − cos nπ + cos 0 + cos 2nπ − cos nπ ]
bn =
nπ
A
[1 + 1 + 1 + 1]
=
nπ
4A
=
when n is odd
nπ

36. A
[ − cos nπ + cos 0 + cos 2nπ − cos nπ ]
bn =
nπ
A
[ − 1 + 1 + 1 − 1]
=
nπ
= 0 when n is even

37. Therefore, the corresponding Fourier series is
4A 
1
1
1

 sin θ + sin 3θ + sin 5θ + sin 7θ + 
π 
3
5
7

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
f(θ)
0
0.5
1
1.5
θ

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

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

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

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
θ

43. 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
0
2
4
6
θ
8
10

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

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

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

47. Odd Functions
Mathematically speaking -
f ( − θ ) = − f (θ )
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.

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.
5
0
5
10
0
θ
10

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.
5
0
5
10
0
θ
10

50. The Fourier series of an even function f ( θ )
is expressed in terms of a cosine series.
∞
f ( θ ) = a0 + ∑ an cos nθ
n =1
The Fourier series of an odd function f ( θ )
is expressed in terms of a sine series.
∞
f ( θ ) = ∑ bn sin nθ
n =1

51. Example 2. Find the Fourier series of
the following periodic function.
f ( x) = x
2
when − π ≤ x ≤ π
f ( θ + 2π ) = f ( θ )

52. 1
a0 =
2π
1
f ( x ) dx =
2π
π
∫π
−
x =π
1 x 
π
=
=
3
2π   x = −π
3
3
2
π
∫π
−
2
x dx

53. 1 π
an = ∫ f ( x ) cos nx dx
π −π
1 π 2

=
x cos nxdx
 ∫−π


π
Use integration by parts. Details are shown
in your class note.

54. 4
an = 2 cos nπ
n
4
an = − 2
n
4
an = 2
n
when n is odd
when n is even

55. This is an even function.
Therefore,
bn = 0
The corresponding Fourier series is
π
cos 2 x cos 3 x cos 4 x


− 4 cos x −
+
−
+ 
2
2
2
3
2
3
4


2

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

57. ω = 2π f
f is the frequency of the periodic function,
f (t)
θ = 2π f t
Therefore,
where
2π
θ=
t
T
1
f =
T

58. 2π
θ=
t
T
2π
dθ =
dt
T
Now change the limits of integration.
θ = −π
2π
−π =
t
T
T
t=−
2
θ =π
2π
π=
t
T
T
t=
2

59. 1
a0 =
2π
1
a0 =
T
∫
π
f ( θ ) dθ
−π
T
2
∫ f ( t ) dt
T
−
2

60. 1
an =
π
2
an =
T
π
∫ π f (θ ) cos nθ dθ
−
T
2
 2π n 
∫ Tf ( t ) cos T t dt


−
2
n = 1 , 2 ,
n = 1, 2, 

61. 1
bn =
π
2
bn =
T
π
f ( θ ) sin nθ dθ
n = 1 , 2 ,
 2π n 
f ( t ) sin
t dt
∫T
 T 
−
n = 1, 2, 
∫π
−
T
2
2

62. Example 4. Find the Fourier series of
the following periodic function.
T
T
f ( t ) = t when − ≤ t ≤
4
4
T
T
3T
= −t +
when
≤t ≤
2
4
4

63. f (t +T ) = f (t)
This is an odd function. Therefore,
T
2
 2πn 
bn = ∫ f ( t ) sin
t dt
T 0
 T 
4
=
T
∫
0
T
2
 2πn 
f ( t ) sin
t dt
 T 
an = 0

64. 4
bn =
T
4
+
T
T
4
 2πn 
t sin
t  dt
∫ T 
0
T
2
T   2πn 

t dt
 − t +  sin
∫T 
2  T 
4
Use integration by parts.

65.   T 
 nπ
 sin
 2.
 2
  2πn 

2T
 nπ 
= 2 2 sin

nπ
 2 
4
bn =
T
bn = 0
2
when n is even.





66. Therefore, the Fourier series is
2T
π2
  2π  1

 6π  1
 10π 
 sin T t  − 2 sin T t  + 2 sin T t  − 
 3

 5


 


67. The Complex Form of Fourier Series
∞
∞
n =1
n =1
f ( θ ) = a0 + ∑ an cos nθ + ∑ bn sin nθ
Let us utilize the Euler formulae.
jθ
− jθ
cos θ =
sin θ =
+e
2
e
e
jθ
−e
2i
− jθ

68. n
th harmonic component of (1) can be
The
expressed as:
an cos nθ + bn sin nθ
= an
= an
e
jnθ
e
jnθ
+e
2
− jnθ
+e
2
− jnθ
+ bn
e
− ibn
jnθ
e
−e
2i
jnθ
− jnθ
−e
2
− jnθ

69. an cos nθ + bn sin nθ
 an − jbn  jnθ  an + jbn  − jnθ
=
e + 
e
2 
2 


Denoting
 a n − jb n
cn = 

2

and
c0 = a0




,
c−n
 a n + jbn 
=

2



70. a n cos nθ + bn sin nθ
= cne
jnθ
+ c− n e
− jnθ

71. The Fourier series for
can be expressed as:
∞
(
f (θ )
f ( θ ) = c0 + ∑ c n e
n =1
=
∞
∑c e
n
n= −∞
jnθ
jnθ
+ c− ne
− jnθ
)

72. The coefficients can be evaluated in
the following manner.
 an − jbn 
cn = 

2 

1 π
j
=
∫−π f ( θ ) cos nθdθ − 2π
2π
1
=
2π
1
=
2π
π
π
∫ π f ( θ ) sin nθ dθ
−
∫ π f (θ )( cos nθ − j sin nθ ) dθ
−
π
∫π
−
f (θ ) e
− jnθ
dθ

73.  an + jbn 
c− n = 

2 

1 π
j
=
∫−π f ( θ ) cos nθdθ + 2π
2π
1
=
2π
1
=
2π
π
∫π
−
π
∫π
−
π
∫ π f (θ ) sin nθ dθ
−
f ( θ )( cos nθ + j sin nθ ) dθ
f (θ ) e
jnθ
dθ

74.  an − jbn 
cn = 

2 

Note that
cn .
c−n
is the complex conjugate of
Hence we may write that
1
cn =
2π
.
c− n
 an + jbn 
=

2 

∫
π
f ( θ) e
−π
− jnθ
dθ
n = 0 , ± 1, ± 2 , 

75. The complex form of the Fourier series of
f ( θ ) with period
f (θ ) =
2π
∞
∑c e
n = −∞
n
is:
jnθ

76. Example 1. Find the Fourier series of
the following periodic function.
f ( θ) = A
= −A
when 0 < θ < π
when
f ( θ + 2π ) = f ( θ )
π < θ < 2π

77. A := 5
f ( x) :=
A if 0 ≤ x < π
−A if π ≤ x ≤ 2 ⋅π
0 otherwise
2π
1 ⌠
A0 :=
⋅ f ( x) dx
2π ⌡0
A0 = 0

78. n := 1 .. 8
2π
1 ⌠
An := ⋅ f ( x) ⋅cos ( n⋅x) dx
π ⌡0
A1 = 0
A2 = 0
A3 = 0
A4 = 0
A5 = 0
A6 = 0
A7 = 0
A8 = 0

79. 2π
1 ⌠
Bn : ⋅
=  f ( x) ⋅ ( n⋅) d
s
i
n
x x
⌡
π0
B1 = 6.366
B2 = 0
B3 = 2.122
B4 = 0
B5 = 1.273
B6 = 0
B7 = 0.909
B8 = 0

80. f ( θ) =
∞
∑
cn e
jnθ
n = −∞
1
cn =
2π
∫
π
f ( θ) e
−π
n = 0, ± 1, ± 2, 
2π
1 ⌠
−n⋅
1⋅x
i
C(n) :
= ⋅ f (x) ⋅
 e
d
x
⌡
2π
0
− jn θ
dθ

81. 2π
1 ⌠
−n⋅
1⋅x
i
C(n) :
= ⋅ f (x) ⋅
 e
d
x
⌡
2π
0
C ( 0) = 0
C ( 1) = − 3.183i
C ( 2) = 0
C ( 3) = − 1.061i
C ( 4) = 0
C ( 5) = − 0.637i
C ( 6) = 0
C ( 7) = − 0.455i
C ( − 1) = 3.183i
C ( − 2) = 0
C ( − 3) = 1.061i
C ( − 5) = 0.637i
C ( − 6) = 0
C ( − 7) = 0.455i
C ( − 4) = 0