CHAPTER 2
FOURIER SERIES
PERIODIC FUNCTIONS
A function
is said to have a period T if for all x,
positive constant. The least value of T>0 is called the period of
, where T is a
.
EXAMPLES
= sin x = sin (x + 4 ) = … Therefore the function has period 2 ,
We know that
4 , 6 , etc. However, 2 is the least value and therefore is the period of f(x).
Similarly cos x is a periodic function with the period 2
and tan x has period
.
DIRICHLET’S CONDITIONS
A function
series of the form
defined in c
x
c+2l can be expanded as an infinite trigonometric
+
provided
1.
is single- valued and finite in (c , c+2l)
2.
is continuous or piecewise continuous with finite number of finite
discontinuities in (c , c+2l).
3.
has no or finite number of maxima or minima in (c , c+2l).
EULER’S FORMULAS
If a function
series
defined in (c , c+2l) can be expanded as the infinite trigonometric
+
[ Formulas given above for
then
and
are called Euler’s formulas for Fourier coefficients]
DEFINITION OF FOURIER SERIES
The infinite trigonometric series
Fourier series of
in the interval c
+
x
is called the
c+2l, provided the coefficients are given by the
Euler’s formulas.
EVEN FUNCTION
If
=
in (-l , l) such that
=
, then
is said to be an even
function of x in (-l , l).
If
Such that
=
or
=
, then
is said to be an even function of x in
(-l , l).
EXAMPLE
y = cos x , y =
are even functions.
ODD FUNCTION
If
=
in (-l , l) such that
= -
, then
is said to be an odd
function of x in (-l , l).
If
Such that
= -
or
=-
, then
is said to be an odd function of x in
(-l , l).
EXAMPLE
y = sin x , y = x are odd functions.
FOURIER SERIES OF EVEN AND ODD FUNCTIONS
1. The Fourier series of an even function
in (-l , l) contains only cosine terms
(constant term included), i.e. the Fourier series of an even function
in (-l , l) is
given by
2
=
+
where
2. The Fourier series of an odd function
the Fourier series of an odd function
=
in (-l , l) contains only sine terms, i.e.
in (-l , l) is given by
,
where
PROBLEMS
1. Find the Fourier series of period 2l for the function
the sum of
= x(2l – x) in (0 , 2l). Deduce
=
Solution:
Let
=
+
in (0 , 2l) ..........................(1)
using Bernoulli’s formula.
=
=0
3
Using these values in (1), we have
x (2l - x) =
in (0, 2l)..................................... (2)
…
The required series
can be obtained by putting x = l in the Fourier
series in (2).
x = l lies in (0 , 2l) and is a point of continuity of the function
Sum the Fourier series in (2)
i.e.
= x(2l – x).
= f(l)
= l(2l - l)
i.e.. …
=
2. Find the Fourier series of period 2
for the function
= x cos x in 0 < x < 2 .
Solution:
Let
=
+
.……..…………(1)
if n 1
=0,
if n 1
=0
4
if n 1
=
,
if n 1
=
Using these values in (1), we get
f(x) =
3. Find the Fourier series expansion of
= sin ax in (-l , l).
Solution:
Since
is defined in a range of length 2l, we can expand
in Fourier series of
period 2l.
Also
= sin[a(-x)] = -sin ax = is an odd function of x in (-l ,
l).
Hence Fourier series of
Let
will not contain cosine terms.
= .................................................................................................... (1)
5
Using these values in (1), we get
4. Find the Fourier series expansion of
=
. Hence obtain a series for
cosec
Solution:
Though the range
is symmetric about the origin,
is neither an even function
nor an odd function.
Let
=
+
..…..…………(1)
in
6
Using these values in (1), we get
=
in
[Since x=0 is a point of continuity of f(x)]
i.e.,
i.e.,
i.e.,
7
HALF-RANGE FOURIER SERIES AND PARSEVAL’S THEOREM
(i) The half range cosine series in (0 , l) is
=
+
where
(ii) The half range sine series in (0 , l) is
=
,
where
(iii) The half range cosine series in (0 ,
=
) is given by
+
where
(iv) The half range sine series in (0 ,
=
) is given by
,
where
8
ROOT-MEAN SQUARE VALUE OF A FUNCTION
Definition
If a function y =
is defined in (c , c+2l), then
is called the root mean-
square(R.M.S.) value of y in (c , c+2l) and is denoted by
Thus
PARSEVAL’S THEOREM
If y =
can be expanded as a Fourier series of the form
+
in
in (c , c+2l), then the root-mean square value
of y =
(c , c+2l) is given by
PROOF
=
+
in (c , c+2l) ............................................ (1)
By Euler’s formulas for the Fourier coefficients,
..…………………(2)
…....……………..(3)
Now, by definition,
=
=
using (1)
=
=
,
by using (2) and (3)
9
=
EXAMPLES
1. Find the half-range (i) cosine series and (ii) sine series for
Solution:
(i) To get the half-range cosine series for
extension for
i.e. put
Now
in (
, 0).
=
=
is even in (
,
in (
in (0 ,
………………….(1)
is given by
(ii) To get the half-range sine series of
in (0 ,
).
), we should give an odd extension
, 0).
Put
==-
Now
), we should give an even
+
in (0 ,
i.e.
)
, 0)
The Fourier half-range cosine series of
in (-
in (0 ,
).
=
for
=
is odd in (-
,
in (in (-
, 0)
, 0)
).
10
……………….(2)
=
Using this value in(2), we get the half-range sine series of
2. Find the half-range sine series of
in (0 ,
).
= sin ax in (0 , l).
Solution:
We give an odd extension for
i.e. we put
in (-l , 0).
= -sin[a(-x)] = sin ax in (-l , 0)
is odd in (-l , l)
Let
=
11
Using this values in (1), we get the half-range sine series as
3. Find the half-range cosine series of
= a in (0 , l). Deduce the sum of
.
Solution:
Giving an odd extension for
in (-l , 0),
is made an odd function in (-l , l).
Let.........................................................................................................................
(1)
f(x) =
Using this value in (1), we get
a=
Since the series whose sum is required contains constant multiples of squares of
, we apply
Parseval’s theorem.
12
4. Expand
=
-
r.m.s. value of
as a Fourier series in -1 < x < 1 and using this series find the
in the interval.
Solution:
The Fourier series of
=
in (-1 , -1) is given by
+
.………………(1)
……………….(3)
13
Substituting (2), (3), (4) in (1) we get
=
We know that r.m.s. value of f(x) in (-l , l) is
……………….(5)
From (2) we get
.………………..(6)
From (3) we get
………………..(7)
From (4) we get
..………………(8)
Substituting (6), (7) and (8) in (5) we get
5. Find the Fourier series for
=
in
Hence show that
Solution:
The Fourier series of
in (-1 , 1) is given by
=
+
14
The co-efficients
are
Parseval’s theorem is
i.e.,
=
HARMONIC ANALYSIS
The process of finding the Fourier series for a function given by numerical value is
known as harmonic analysis. In harmonic analysis the Fourier coefficients
of the
function y =
in (0 , 2 ) are given by
= 2[mean value of y in (0 , 2 )]
= 2[mean value of y cos nx in (0 , 2 )]
= 2[mean value of y sin nx in (0 , 2 )]
(i) Suppose the function
=
and now,
is defined in the interval (0 , 2l), then its Fourier series is,
+
= 2[mean value of y in (0 , 2l)]
=
15
=
(ii) If the half range Fourier sine series of
=
in (0 , l) is,
, then
=
(iii) If the half range Fourier sine series of
=
in (0 ,
) is,
, then
=
(iv) If the half range Fourier cosine series of
=
+
in (0 , l) is,
, then
= 2[mean value of y in (0 , l)]
=
(v) If the half range Fourier cosine series of
=
+
in (0 ,
) is,
, then
= 2[mean value of y in (0 ,
)]
=
.
EXAMPLES
1. The following table gives the variations of a periodic function over a period T.
x
1.98
Show that
= 0.75 + 0.37
1.3
+1.004
1.05
1.3
-0.88
-0.25
1.98
, where
Solution:
16
Here the last value is a mere repetition of the first therefore we omit that value and
consider the remaining 6 values.
n = 6.
Given .................................................................................................................... (1)
when x takes the values of 0,
,
,
,
,
,
,
takes the values 0,
,
,
. (By using (1))
Let the Fourier series be of the form
………………(2)
where
n=6
y
cos
sin
y cos
y sin
1.98
1.0
0
1.98
0
1.30
0.500
0.866
0.65
1.1258
1.05
-0,500
0.866
-0.525
0.9093
1.30
-1
0
-1.3
0
-0.88
-0.500
-0.866
0.44
0.762
-0.25
0.500
-0.866
-0.125
0.2165
1.12
3.013
4.6
Substituting these values of
in (2), we get
= 0.75 + 0.37 cos + 1.004 sin
17
2. Find the Fourier series upto the third harmonic for the function y =
(0 ,
defined in
) from the table
x
0
2.34
2.2
1.6
0.83
0.51
0.88
1.19
Solution:
We can express the given data in a half range Fourier sine series.
..………………...(1)
x
y = f(0)
sin x
sin 2x
sin 3x
y sin x
y sin 2x
y sin 3x
0
2.34
0
0
0
0
0
0
30
2.2
0.5
0.87
1
1.1
1.91
2.2
60
1.6
0.87
0.87
0
1.392
1.392
0
90
0.83
1
0
-1
0.83
0
-0.83
120
0.51
0.87
-0.87
0
0.44
-0.44
0
150
0.88
0.5
-0.87
1
0.44
0.76
0.88
180
1.19
0
0
0
0
0
0
4.202
3.622
2.25
Now
Substituting these values in (1), we get
= 1.4 sin x + 1.21 sin 2x + 0.75 sin 3x
3. Compute the first two harmonics of the Fourier series for f(x) from the following data
x
0
30
60
90
120
150
180
0
5224
8097
7850
5499
2626
0
18
Solution:
Here the length of the interval is
we can express the given data in a half range
Fourier sine series
i.e.,
………………………(1)
x
y
sin x
sin 2x
0
0
0
0
30
5224
.5
0.87
60
8097
0.87
0.87
90
7850
1
0
120
5499
0.87
-0.87
150
2626
0.5
-0.87
Now
= 7867.84 sin x + 1506.84 sin 2x
4. Find the Fourier series as far as the second harmonic to represent the function given in
the following data.
x
0
1
2
3
4
5
9
18
24
28
26
20
Solution:
Here the length of the interval is 6 (not 2 )
i.e., 2l = 6 or l = 3
The Fourier series is
…………………..(1)
y
0
0
0
9
9
0
9
0
19
1
18
9
15.7
-9
15.6
2
24
-12
20.9
-24
0
3
28
-28
0
28
0
4
26
-13
-22.6
-13
22.6
5
20
10
-17.4
-10
-17.4
125
-25
-3.4
-19
20.8
Substituting these values of
in (1), we get
COMPLEX FORM OF FOURIER SERIES
The equation of the form
is called the complex form or exponential form of the Fourier series of
coefficient
When l =
in (c , c+2l). The
is given by
, the complex form of Fourier series of
in (c , c+2 ) takes the form
where
20
PROBLEMS
1. Find the complex form of the Fourier series of
=
in (0 , 2).
Solution:
Since 2l = 2 or l = 1, the complex form of the Fourier series is
Using this value in (1), we get
2. Find the complex form of the Fourier series of
= sin x in (0 ,
).
Solution:
Here 2l =
or l =
.
The complex form of Fourier series is
…………………..(1)
21
Using this value in (1), we get
in (0 ,
3. Find the complex form of the Fourier series of
=
)
in (-l , l).
Solution:
Let the complex form of the Fourier series be
Using this value in (1), we have
22
in (-l , l)
4. Find the complex form of the Fourier series of
= cos ax in (-
,
), where a is
neither zero nor an integer.
Solution:
Here 2l = 2
or l =
.
The complex form of Fourier series is
………………….(1)
Using this value in (1), we get
in (-
,
).
UNIT 2
PART – A
1. Determine the value of
Ans:
in the Fourier series expansion of
is an odd function.
2. Find the root mean square value of
in the interval
.
Ans:
RMS Vale of
in
is
23
3. Find the coefficient
of
in the Fourier cosine series of the function
in
the interval
Ans: Here
Fourier cosine series is
=
+
4. If
and
series of
Ans:
, where
at
Here
for all x, find the sum of the Fourier
.
is a point of discontinuity.
The sum of the Fourier series is equal to the average of right hand and left hand limit of the
given function at
.
i.e.,
5. Find
Ans:
Since
6. If
in the expansion of
as a Fourier series in
.
=0
is an even function in
.
is an odd function defined in (-l , l) what are the values of
24
Ans:
=0
since
is an odd function.
7. Find the Fourier constants
Ans:
for
in
.
=0
Since
is an even function in
.
8. State Parseval’s identity for the half-range cosine expansion of
in (0 , 1).
Ans:
where
9. Find the constant term in the Fourier series expansion of
in
.
Ans:
= 0 since
is an odd function in
.
10. State Dirichlet’s conditions for Fourier series.
Ans:
(i)
is defined and single valued except possibly at a finite number of points in
(ii)
is periodic with period 2 .
(iii)
and
are piecewise continuous in
Then the Fourier series of
(a)
(b)
.
.
converges to
if x is a point of continuity
if x is a point of discontinuity.
11. What you mean by Harmonic Analysis?
Ans:
25
The process of finding the Fourier series for a function given by numerical value is
known as harmonic analysis. In harmonic analysis the Fourier coefficients
function y =
of the
in (0 , 2 ) are given by
= 2[mean value of y in (0 , 2 )]
= 2[mean value of y cos nx in (0 , 2 )]
= 2[mean value of y sin nx in (0 , 2 )]
12. In the Fourier expansion of
in
. Find the value of
,
the coefficient of sin nx.
Ans:
Since
is an even function the value of
13. What is the constant term and the coefficient of
= 0.
in the Fourier expansion of
in (-7 , 7)?
Ans:
Given
The given function is an odd function. Hence
14. State Parseval’s identity for full range expansion of
are zero.
as Fourier series in (0 , 2l).
Ans:
=
where
26
15. Find a Fourier sine series for the function
= 1; 0 < x <
.
Ans:
…………………….(1)
The Fourier sine series of
16. If the Fourier series for the function
is
Deduce that
Ans:
Putting
we get
27
17. Define Root mean square value of a function?
Ans:
If a function y =
is defined in (c , c+2l), then
is called the root mean-
square(R.M.S.) value of y in (c , c+2l) and is denoted by
Thus
18. If
is expressed as a Fourier series in the interval (-2 , 2), to which value this
series converges at x = 2.
Ans:
Since x = 2 is a point of continuity, the Fourier series converges to the arithmetic mean of
at x = -2 and x = 2
i.e.,
19. If the Fourier series corresponding to
in the interval
without finding the values of
is
find the value of
Ans:
By using Parseval’s identity,
20. Find the constant term in the Fourier series corresponding to
interval
expressed in the
.
Ans:
Given
Now
28
PART B
1. (i) Express
as a Fourier series in
(ii) Show that for 0 < x <l,
. Using root mean square
value of x, deduce the value of
2. (i) Find the Fourier series of periodicity 3 for
in 0 < x < 3.
(ii) Find the Fourier series expansion of period 2
in
for the function
which is defined
by means of the table of values given below. Find the series upto the third harmonic.
x
0
1.0
1.4
1.9
3.(i) Find the Fourier series of periodicity 2
1.7
for
1.5
1.2
1.0
for 0 < x < 2 .
(ii) Show that for 0 < x <l,
. Deduce that
4. (i) Find the Fourier series for
. Hence deduce the sum to infinity of
the series
(ii) Find the complex form of Fourier series of
in the form
and hence prove that
5. Obtain the half range cosine series for
6. Find the Fourier series for
7. (i) Expanding
as a sine series in
in
in the interval
.
show that
(ii) Find the Fourier series as far as the second harmonic to represent the function given in the
following data.
29
x
0
1
2
3
4
5
9
18
24
28
26
20
8. Obtain the Fourier series for
of period 2l and defined as follows
Hence deduce that
9. Obtain the half range cosine series for
in
10. (i) Find the Fourier series of
(ii) Obtain the sine series for the function
11. (i) Find the Fourier series for the function
and
for all x.
(ii) Determine the Fourier series for the function
12. Obtain the Fourier series for
in
. Deduce that
13. Obtain the constant term and the first harmonic in the Fourier series expansion for
where
is given in the following table.
x
0
1
2
3
4
18.0 18.7 17.6 15.0 11.6
5
8.3
6
7
8
9
10
11
6.0 5.3 6.4 9.0 12.4 15.7
30
14. (i) Express
as a Fourier series in
(ii) Obtain the half range cosine series for
in the interval 0 < x < 2.
15. Find the half range sine series of
in
16. (i) Find the Fourier series expansion of
(ii) Find the half-range sine series of
17. Expand
value of
=
-
=
= sin ax in (0 , l).
as a Fourier series in -1 < x < 1 and using this series find the r.m.s.
in the interval.
18. The following table gives the variations of a periodic function over a period T.
x
1.98
Show that
1.3
= 0.75 + 0.37
1.05
+1.004
1.3
-0.88
-0.25
1.98
, where
19. Find the Fourier series upto the third harmonic for the function y =
defined in (0 ,
)
from the table
x
0
2.34
2.2
1.6
0.83
0.51
20. (i) Find the half-range (i) cosine series and (ii) sine series for
=
(ii) Find the complex form of the Fourier series of
= cos ax in (-
0.88
in (0 ,
, ).
1.19
)
31