CHAPTER 2

FOURIER SERIESPERIODIC FUNCTIONS

A function is said to have a period T if for all x, , where T is a

positive constant. The least value of T>0 is called the period of .

EXAMPLES

We know that = sin x = sin (x + 4 ) = … Therefore the function has period 2 , 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 defined in c x c+2l can be expanded as an infinite trigonometric

series of the form + 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 defined in (c , c+2l) can be expanded as the infinite trigonometric

series + then

[ Formulas given above for and are called Euler’s formulas for Fourier coefficients]

DEFINITION OF FOURIER SERIES

The infinite trigonometric series + is called the

Fourier series of in the interval c x 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).

EXAMPLEy = 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).

EXAMPLEy = 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 in (-l , l) contains only sine terms, i.e.

the Fourier series of an odd function in (-l , l) is given by

= ,

where

PROBLEMS

1. Find the Fourier series of period 2l for the function = x(2l – x) in (0 , 2l). Deduce

the sum of =

Solution:

Let = + in (0 , 2l) …………(1)

using Bernoulli’s formula.

=

= 0

Using these values in (1), we have

x (2l - x) = in (0, 2l) ……………..(2)

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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 = x(2l – x).

Sum the Fourier series in (2) = f(l)

i.e. = 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 will not contain cosine terms.

Let = ………………….(1)

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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.,

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

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= +

where

(iv) The half range sine series in (0 , ) is given by

= ,

where

ROOT-MEAN SQUARE VALUE OF A FUNCTIONDefinition

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 (c , c+2l), then the root-mean square value of y =

in (c , c+2l) is given by

PROOF

= + in (c , c+2l) ....……………….(1)

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By Euler’s formulas for the Fourier coefficients,

..…………………(2)

…....……………..(3)

Now, by definition,

=

= using (1)

=

= , by using (2) and (3)

=

EXAMPLES1. Find the half-range (i) cosine series and (ii) sine series for = in (0 , )Solution:

(i) To get the half-range cosine series for in (0 , ), we should give an even

extension for in ( , 0).

i.e. put = = in ( , 0)

Now is even in ( , ).

= + ………………….(1)

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The Fourier half-range cosine series of is given by

in (0 , ).

(ii) To get the half-range sine series of in (0 , ), we should give an odd extension for

in (- , 0).

i.e. Put = - in (- , 0)

= - in (- , 0)

Now is odd in (- , ).

= ……………….(2)

Using this value in(2), we get the half-range sine series of in (0 , ).

2. Find the half-range sine series of = sin ax in (0 , l).

Solution:

We give an odd extension for in (-l , 0).

i.e. we put = -sin[a(-x)] = sin ax in (-l , 0)

is odd in (-l , l)

Let =

10

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 f(x) = ..……………(1)

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.

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4. Expand = - as a Fourier series in -1 < x < 1 and using this series find the r.m.s.

value of in the interval.

Solution:

The Fourier series of in (-1 , -1) is given by

= + .………………(1)

……………….(3)

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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

= +

The co-efficients are

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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 is defined in the interval (0 , 2l), then its Fourier series is,

= +

and now, = 2[mean value of y in (0 , 2l)]

=

=

(ii) If the half range Fourier sine series of in (0 , l) is,

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= , 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 1.3 1.05 1.3 -0.88 -0.25 1.98

Show that = 0.75 + 0.37 +1.004 , where

Solution: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)

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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

4.6 1.12 3.013

Substituting these values of in (2), we get

= 0.75 + 0.37 cos + 1.004 sin

2. 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 0.88 1.19

Solution:

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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

75.025.23

1

6

3sin23

xy

b

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

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

x 0 30 60 90 120 150 180

0 5224 8097 7850 5499 2626 0

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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

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

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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 in (c , c+2l). The

coefficient is given by

When l = , the complex form of Fourier series of in (c , c+2 ) takes the form

where

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

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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)

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

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Using this value in (1), we have

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)

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Using this value in (1), we get

in (- , ).

UNIT 2

PART – A

1. Determine the value of in the Fourier series expansion of

Ans: is an odd function.

2. Find the root mean square value of in the interval .

Ans:

RMS Vale of in is

3. Find the coefficient of in the Fourier cosine series of the function in

the interval

Ans: Here

Fourier cosine series is

= + , where

4. If and for all x, find the sum of the Fourier

series of at .

Ans: Here is a point of discontinuity.

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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 in the expansion of as a Fourier series in .

Ans: = 0

Since is an even function in .

6. If is an odd function defined in (-l , l) what are the values of

Ans: = 0

since is an odd function.

7. Find the Fourier constants for in .

Ans: = 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 .

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(ii) is periodic with period 2 .

(iii) and are piecewise continuous in .

Then the Fourier series of converges to

(a) if x is a point of continuity

(b) if x is a point of discontinuity.

11. What you mean by Harmonic Analysis?

Ans:

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 )]

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 = 0.

13. What is the constant term and the coefficient of in the Fourier expansion of

in (-7 , 7)?

Ans:

Given

The given function is an odd function. Hence are zero.

14. State Parseval’s identity for full range expansion of as Fourier series in (0 , 2l).

Ans:

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=

where

15. Find a Fourier sine series for the function = 1; 0 < x < .

Ans:

The Fourier sine series of …………………….(1)

16. If the Fourier series for the function is

Deduce that

Ans:

Putting we get

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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 is

without finding the values of find the value of

Ans:

By using Parseval’s identity,

20. Find the constant term in the Fourier series corresponding to expressed in the

interval .

Ans:

Given

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Now

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 for the function which is defined

in by means of the table of values given below. Find the series upto the third harmonic.

x 0

1.0 1.4 1.9 1.7 1.5 1.2 1.0

3.(i) Find the Fourier series of periodicity 2 for 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 in

6. Find the Fourier series for in the interval .

7. (i) Expanding as a sine series in show that

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(ii) 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

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 5 6 7 8 9 10 11

18.0 18.7 17.6 15.0 11.6 8.3 6.0 5.3 6.4 9.0 12.4 15.7

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 = sin ax in (0 , l).

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17. Expand = - as a Fourier series in -1 < x < 1 and using this series find the r.m.s.

value of in the interval.

18. The following table gives the variations of a periodic function over a period T.

x

1.98 1.3 1.05 1.3 -0.88 -0.25 1.98

Show that = 0.75 + 0.37 +1.004 , 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 0.88 1.19

20. (i) Find the half-range (i) cosine series and (ii) sine series for = in (0 , ) (ii) Find the complex form of the Fourier series of = cos ax in (- , ).

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