fourier series 2 marks qn&ans

7/29/2019 Fourier Series 2 Marks Qn&Ans

http://slidepdf.com/reader/full/fourier-series-2-marks-qnans 1/10

1

Two marks question & answers

Unit-I

(Fourier Series)

1. Find the constant term in the Fourier series corresponding to f(x) =cos2

x expressed in

the interval (-π,π).

Solution: Given f(x) =cos2x = ; f(-x)=cos

2(-x)=cos

2x=f(x).Therefore the

function is even. The Fourier series is given by 0

1

( ) cos2

n

n

a f x a nx

. Here the constant term is a0, ao=

ao= = =1

1

.i.e., a0=1.

2. If f(x) = x2

+ x is expanded as a Fourier series in the interval (-2,2) to which value this

series converges at x = 2 ?

Solution: Here given x = 2 is a point of discontinuity in the extremum. f(x) =

== 4: i.e., f(x)at x =2 = 4.

3. If the Fourier series of f(x) = x2

in – π ≤x ≤ π is equal to2

2

21

4( 1)cos

3

n

n

x nxn

then prove that2

21

1

6n n

Solution : Given f(x) =

2

22

1

4( 1) cos3

n

n

x nxn

.

Put x = in the above,

L.H.S. = 2 2 2 2 21 1 1[ ( ) ( )] [( ) ( ) ] [ ]

2 2 2 f f by Dirichlet’s condition

R.H.S. =2 2 2

2 2 21 1 1

4( 1) 4( 1) ( 1) 4cos

3 3 3

n n n

n n n

nn n n

Equating L.H.S. & R.H.S. we get ,

22

21

22

21

2

21

4

3

4

3

2 4

3

n

n

n

n

n

n

.

Hence2

21

1

6n n



2

4. If ( ) cos f x x is expanded in a Fourier series in the interval of ( -π < x < π ) , find a0.

Solution: Given, ( ) cos f x x

( ) cos( ) cos ( ) f x x x f x

Therefore the function is an even function. The Fourier series is

,.f(x)= + since bn = 0

To find a0. a0= , = =

a0 = =

5. State Dirichlet’s condition in Fourier series and examine whether (1/1-x) can beexpanded in a Fourier series in any interval including the point x=1.

Solution: The expansion of a function f(x) in a trigonometry is possible if it satisfies the

following condition in any interval. f(x) is well defined and single valued function.

f(x) has finite number of points of discontinuity in any one period.

f(x) has only finite number of maxima and minima.

6. Find the constant term in the Fourier series corresponding to f(x) = 2x-x2

expressed inthe interval(-2,2).Solution: The function is neither even nor odd. To find Fourier series constant,

3 3 3 3 22 2 2 2

0

1 1 1 ( ) 1 2 2(2 ) [ ] [ ( ) ] [0 ]

3 3 3 3

xa x x dx x

7. The function ( ) 00

f x for x x for x

is expanded as a Fourier series of period 2π.

What is the sum of the series at x = 0 and x = .

Solution:

Given( ) 0

0

f x for x

x for x

Sum of the series at x = 0 =(0 ) (0 )

2

f f = = .

Sum of the series at x = = ( )2 2

f

.

8. Obtain the constant term in the Fourier series corresponding to f(x)= √(1-cosx) in the

interval of(-π, π).Solution: Given f(x)= √(1-cosx) ; f(-x)= √(1-cos(-x)) = √(1-cosx) =f(x)

Therefore the function is an even function . To find the constant term

a0 = = =



3

={ ; a0=

9.If the Fourier series of f(x) = x2, 0 < x < 2 , is given by

22

21 1

4 4 4cos sin

3 n n

x nx nx

n n

Deduce the sum of the series2

1

1

n n

Solution: Given f(x) = x2, 0 < x < 2 , and the Fourier series is given by

22

21 1

4 4 4cos sin

3 n n

x nx nxn n

. put x = 0

L.H.S. =2

2 2 21 1 4[ (0) (2 )] [(0) (2 ) ] 2

2 2 2 f f

by Dirichlet’s condition

R.H.S. =2 2

2 2

1 1

4 4 4 4.1 0

3 3n nn n

Equating R.H.S. and L.H.S. we get ,

22

21

22

21

2

21

4 42

3

4 42

3

2 4

3

n

n

n

n

n

n

Hence2

21

1

6n n

10.If the Fourier series for the function f(x) = 0 , 0 < x < π= sinx , π < x < 2 π

is f(x) = ]+ .

Deduce that × =

Solution: Given f(x) = 0 , 0 < x < π= sinx , π < x < 2 π

Put

2

x

. Since

2

x

is a point of continuity ,

0 = ] +

i.e., - = ]

i.e., ×



4

i.e., × =

11. To what value the Fourier series 2( ) f x x expressed in the interval ( 0 , 2 )

converges at x = 2 ?Solution: Given 2

( ) f x x , x = 2 is a point of discontinuity . Hence

2( ) f x x at x=2 == ; 2( ) f x x at x=2

=

81

8

12. If the Fourier series corresponding to f(x) = x in the interval (0,2π) is

0

1

( ) ( cos sin )2

n n

n

a f x a nx b nx

without finding the values of a0 ,an and bn ,

Find the value of 2

2 20

1

1[ ]

4 2n n

n

aa b

Solution: By Parseval’s theorem

2 2

2 2 2 20

10

1 1[ ( )] [ ]

2 4 2rms n n

n

a f f x dx a b

2 22 3 3 22 2 2 2 20

0

1 0 0

1 1 1 1 1 8 4[ ] [ ( )] [ ] [ ]

4 2 2 2 2 3 2 3 3n n

n

a xa b f x dx x dx

13. Find the half range sine series for f(x) = in 0 < x < 1.

Solution: Given f(x) = the half range Fourier sine series is

1

( ) sinn

n

n x f x b

l

0

2( )sin

l

n

n xb f x dx

l l

, 1herel

The Fourier series is1

( ) sinn

n

f x b n x

where1

0

2( ) sin

1n

b f x n xdx 1

0

2( ) sin

1n

b f x n xdx =

1 11

02 2 2 2

0

1

2

( 1)2 sin 2{ [sin cos ]} 2{ }

1 ( ) 1 ( ) 1 ( )1 ( 1)

21 ( )

x n x x

x

n

e e n ne n xdx n x n n x

n n ne

nn

The Fourier series is2

1

1 ( 1)( ) 2 sin

1 ( )

n x

n

f x e n n xn



5

14) The half range cosine series for f(x) = x sinx for 0 < x < is given by

2

2

1 ( 1)sin 1 cos 2 cos

2 1

n

n

x x x nxn

Deduce that1 1

1 2[ .....]1.3 3.5 2

Solution: Given2

2


2 1

n

n

x x x nxn

Put2

x

, Which is a point of continuity

22

22


2 2 2 2 1 2

( 1)1 0 2 cos

2 1 2

n

n

n

n

nn

nn

When ‘n’ is an even integer , the above reduces2

21

1

1

( 1)1 2 cos 2

2 4 1 2

cos1 2

2 (2 1)(2 1)

( 1)1 2

2 (2 1)(2 1)

m

m

m

m

m

mm

m

m m

m m

Hence expanding the sum we get,1 1

1 2[ .....]

1.3 3.5 2

15) If f(x) = sin2x , x , then find b1

2+ b2

2+ b3

2+……..

Solution : Given f(x) = sin2x; f(-x) = sin

2(-x) = (-sinx)

2= Sin

2x = f(x)

Therefore the function is even

This implies bn= 0 ; Therefore,

n 1

bn

2=

n 1

(0)

2

This implies b12+ b2

2+ b3

2+……..= 0.

16) If f(x) = x-x2

, in

1 1 x . Find the RMS value of f(x).

Solution :Given f(x) = x-x

2

;1 1 1

2 2 2 2 2 3 4

1 1 1

3 4 51

1

1 1 1[ ( )] ( ) ( 2 )

2 2 2

1 2 1 2 0 2[ ] [ ]

2 3 4 5 2 3 4 5

8

15

rms f f x dx x x dx x x x dx

x x x



6

17) State whether true or false : Fourier series of period ‘2’ for x sinx in (-1,1) contains

only sine terms. Justify your answer.

Solution :False, The Fourier series doesn’t contain sine terms

Since, f(-x) = (-x) sin (-x) = (-x)(-sinx) = x sinx = f(x). Therefore the function is evenTherefore, Fourier series 0

1

( ) cos2

n

n

a f x a n x

So, it contains only cosine terms in the series.

18) The Fourier series for f(x) = x2

in – < x < will contain only cosine terms. State

whether true or false. Justify your answer.Solution: True ,the Fourier series contain only cosine terms Since, f(-x) = (-x)

2= x

2=f(x)

Therefore, the function is even. The Fourier series is given by 0

1

( ) cos2

n

n

a f x a nx

So, it contains only cosine terms in the series.

19) Find the Half range Fourier sine series of f(x) = x in (0,2) and f(x) = 1,0 < x <

(Apr ’04)

Solution: Given, f(x) = x in (0,2) .The Fourier series of f(x) is given by

2

1 0

2

02

2

1

1

1

2( ) ( ) sin sin

2 2 2

( cos ) ( sin )2 2[ 1. ]

( ) ( )2 2

( 2cos 0) 0 0 4cos 4( 1)[ 1. ]

( ) ( )2 2

4( 1)

4( 1)( ) sin

2

n n

n

n

n

n

n

n x n xi f x b where b x dx

n x n x

xn n

n n

n n n n

n

n x f x x

n

(ii) Consider f(x) =1 in 0 x The Fourier sine series of f(x) is given by



7

0

1 0

2 1

2 ( cos )( ) 1 sin 1.sin [ ]

(1 cos ) 1 ( 1)[ ].

0 ' ' int . ' ' int . 2 1, 1,2,3....1 ( 1) 1 ( 1) 2

[ ] [ ] , 1,2,3.....2 1 2 1 2 1

n n

n

n

n n

m

n

nx f x b nx where b nx dx

n

n

n n

b if n is an even eger b of n is an odd eger put n m m

b mm m m

1

.

2( ) 1 sin(2 1)

2 1m

f x m xm

20) Find the constant term in the Fourier cosine series corresponding to2

( ) int 0 32

x f x in the erval x

Solution: Given

2

( ) int 0 32

x f x in the erval x

.The Half range Fourier cosine

series of f(x) is given by

0

1

3

3 3 23

0 0

0 0

0

( ) cos2 3

(27 0)(3 0)

2 2 2 23 3( ) [ ] [ ]3 3 2 3 2 3 2

2 3 9[ ]

3 2

. ., 3

n

n

a n x f x a

x x

xwhere a f x dx dx

i e a

21) State Parseval’s theorem on Fourier coefficients.

Solution : Parseval’s (identity)Theorem : If the Fourier series of a function f(x) defined

over the interval 2c x c l converges uniformly to f(x) in the interval 2c x c l ,

then

2 2

2 2 2 20

1

1 1[ ( )] [ ]

2 4 2

c l

rms n n

nc

a f f x dx a b

l

22) State Parseval’s Identity for Half range Fourier cosine series (HRFCS) and Half

range Fourier sine series (HRFSS)Solution: If the half range Fourier cosine series of f(x) is defined over the half range

0 x l , then

2

2 2 20

1

1 1[ ( )]

4 2

l

rms n

nc

a f f x dx a

l

If the half range Fourier sine series of f(x) is defined over the half range 0 x l ,



8

2 2 2

10

1 1[ ( )]

2

l

rms n

n

f f x dx bl

23) Define root mean square value of f(x)Solution: Root Mean Square Value: The root mean square value of a function f(x)

over the interval a x b is defined and denoted as 2 21

[ ( )]

b

rms

a

f f x dxb a

24) Define complex form of Fourier series

Solution: Complex Fourier Series: The complex form of the Fourier series of the

function f(x) defined over the interval 2c x c l is given by

21( ) , ( ) 0, 1, 2....

2

in x in xc l

l ln n

cn

f x c e Where c f x e dx nl

25) What is Harmonic analysis?Solution: The process of finding Fourier series for a function f(x) given by numerical

values is known as “Harmonic Analysis”

Consider the Fourier series in an interval of f(x) as,

01 1 2 2 3 3( cos sin ) ( cos 2 sin 2 ) ( cos3 sin 3 ) ......

2

a y a x b x a x b x a x b x where,

a0 = 2[Mean value of f(x)],an = 2[Mean value of f(x) cosnx]

bn = 2[Mean value of f(x) sinnx]

26) In the interval2 2 2 2

1 4 cos cos3 cos50 1 , [ ........]

2 1 3 5

x x x x x

Deduce that

4

4 4

1 1

1 ........3 5 96

Solution : Given2 2

1

1 4cos(2 1)

2 (2 1)n

x n xn

Here 0 2 1 2 2

41 , 1,2,3.....

(2 1)n m

a and a a mm

By Parseval’s Identity,1 12 2

2 2 2 202 1 2 2

1 10 0

3 21

0 2 2 2 4 41 1

4

4 4 41 1

1 1 1 1 ( 4)[ ( )] . .,

1 4 2 4 2 [ (2 1) ]

1 1 ( 4) 1 1 1 16

. .,[ ] . .,3 4 2 [ (2 1) ] 3 4 2 (2 1)

8 1 1. .,

(2 1) 12 (2 1) 96

rms m

m m

m m

m m

a f f x dx a i e x dx

m

x

i e i em m

i e Hencem m

i.e.,4

4 4

1 11 ........

3 5 96



9

-

27)If 2

2

21

4( 1)( ) cos

3

n

n

f x x nx for xn

, show that

4

4 4 4

1 1 1........1 2 3 90

Solution: Here given2

2

21

4( 1)( ) cos

3

n

n

f x x nx for xn

By Parseval’s Identity,2

2 2 2 20

1

1 1[ ( )] [ ]

2 4 2rms n n

n

a f f x dx a b

.

Here2

0 2

2 4( 1),

3

n

na an

.

The function is an even function ,the Parseval’s identity reduces to2

2 2 20

10

22

4 2

210

5 4

0 41

5 4

0 41

4 4

41

4 4

41

41

1 1

[ ( )] 4 2

2( )

1 1 4( 1)3. ., [ ]4 2

1 4 1 16. ., [ ]

5 4 9 2

1 4 1 16. ., [ ]

5 4 9 2

8. .,5 9

8. .,

5 9

8 9. .,

rms n

n

n

n

n

n

n

n

a

f f x dx a

i e x dxn

xi e

n

i en

i en

i en

i en

4 45

45

Hence4

4 4 4

1 1 1........

1 2 3 90

28) Find the R.M.S. value of 2( ) , f x x in x .

Solution: 2( ) ,Given f x x in x 5 4

2 2 2 2 41 1 1 1[ ( )] ( ) [ ]

2 2 2 2 5 5rms

x f f x dx x dx x dx



10

Assignment:

1.Find the Fourier series of f (x) x; x 0 ( N / D 2008 )

x;0 x .Hence deduce that21

2 8(2n 1)n 1

2. Find the Fourier series of f(x) = x in 0 x 3 ( N / D 2008 ) .

= 6-x in 3 x 6

3.Find the Fourier series of f(x) = x sinx for 0 < x < 2 and hence deduce that

1 1 1 2

........1.3 3.5 5.7 4

( M / J 2008 )

4. Determine the Fourier series for the function f(x) =1 x; x 0

1 x;0 x

( M / J 2007)

Hence deduce that1

4(2n 1)n 1

5.Obtain the Fourier series of f(x) of period 2T givenf (x) T x; 0 x T ( N / D 2007 )

0; T x 2T

6.Obtain the Fourier series of f(x) of period 2 l given( ) , 0

0 , 2

f x l x x l

l x l

(N / D2007)

7.Find the Fourier series of the function( ) , 0 1

1 , 1 2

f x x in x

x in x

. Hence deduce that

21

2 8(2n 1)n 1

. ( N / D 2005 )

8.Find the Fourier series of periodicity 3 for 2

( ) 2 0 3 f x x x in x . (A / M 2008).

9. Find the half range cosine series of 2( ) (0, ). f x x x in Deduce that4

1

4 90nn 1

( M / J 2009)

10. Obtain the half range sine series of ( ) cos (0, ) f x x x in . ( N / D 2008).

11.Compute the first two harmonics of the Fourier series of f(x) given in the followingtables: ( A / M 2009 ,2008 )

x 0

3

2

3

4

3

5

3

2

y 1.0 1.4 1.9 1.7 1.5 1.2 1.0

12. Compute the first two harmonics of the Fourier series of f(x) given in the followingtables: ( N / D 2006)

x 0 1 2 3 4 5 6

y 9 18 24 28 26 20 9

fourier series 2 marks qn&ans

Documents