of 39
Page 1 of 39
Fourier series
Preliminaries
Periodic functions:
A function ( )f t is said to be periodic with period if there exists a positive
constant T such that ( ) ( )f t T f t for all .t
Note: If ( )f t is period with period T , then ( ) ( )f t nT f t for all integers .n
The smallest positive number T satisfying this property is called the primitive
period or simply the period of the function ( ).f t
The graph of a period function ( )f t with period T periodically repeats in an
interval of width .T Hence it is sufficient to study the properties (nature) of the function in an interval of length T , in particular in the interval [0, ],T which is
called one period of the function.
Example (1): The trigonometric functions ( ) sinf x x and ( ) cosf x x are
periodic with period 2 .T
Graph of ( ) sinf x x
Page 2 of 39
Graph of ( ) cosf x x
Example (2): The functions sin kt and coskt have period 2
,Tk
since
2 2sin ( sin( 2 ) sin .f t k t kt kt
k k
Let ( )f t be a periodic function of t with the period .T Define a new variable x
as follows.
2
x t
T or
2x t
T
i.e.,
2
Tt x
Then ( ) ( )2
Tf t f x g x
is a function of .x
Consider ( 2 ) ( 2 ) ) ( ) ( )2 2
T Tg x f x f x T f t T f t
( )2
Tf x g x
Therefore ( )g x is a periodic function of x with period 2 .
Page 3 of 39
Note: If ( )f x and ( )g x are periodic functions of x with period 1T and 2T
respectively, then 1 2( ) ( )c f x c g x is also periodic with period 1 2( , ).T lcm T T
Even an Odd functions:
A function ( )f t is said to be even if ( ) ( )f t f t and odd if ( ) ( ).f t f t
Note that the graph of an even function is symmetric about y axis, whereas the
graph of an odd function is symmetric about the origin.
Exmple (1): 2( ) ,f x x ( )f x x and ( ) cosf x x are all even functions of
.x
Graph of 2( )f x x Graph of ( )f x x Graph of ( ) cosf x x
Example (2): ( ) , sin ,f x x x 3( ) ,f x x ( ) sin ,f x x ( ) tanf x x are all
odd functions of .x
Graph of ( )f x x Graph of ( ) sinf x x
x
Page 4 of 39
Products of two even and two odd functions are even and the product of an even
function with an odd function is odd.
Also, 0
2 ( ) , if ( ) is even( )
0, if ( ) is odd
a
a
a
f t dt f tf t dt
f t
.
Generalized formula for integration by parts
If u and v are functions of ,x then
1 2 3 4' '' ''' ...uvdx uv u v u v u v
where ' ''
' , '' , ''' ,...du du du
u u udx dx dx
and
1 2 1 3 2, , ,... v vdx v v dx v v dx
Example: 2 2 cos 2 sin 2 cos 2sin 2 2 2
2 4 8
x x xx x dx x x
Orthogonality of trigonometric functions:
0, 02cos
, 0
c T
c
nnt dt
T nT
2sin 0 for all
c T
c
nt dt n
T
2 2cos sin 0 for all and
c T
c
m nt t dt m n
T T
Page 5 of 39
0, m2 2
cos cos , m
2
c T
c
nm n
t t dt TT T n
0, m2 2
sin sin , m
2
c T
c
nm n
t t dt TT T n
Also,
cos 1 ,n
n for all n , sin 0,n for all n
2
0, oddcos
2 1 , evenn
nn
n
,
1
2
0, evensin
2 1 , oddn
nn
n
Definition: A function ( )f t is said to be piecewise continuous in an interval
[ , ]a b if it is discontinuous at finite number of points in the interval and wherever
it is discontinuous, it has finite left and right hand limits.
Fourier series:
In solving many boundary valued problems involving ordinary and partial
differential equations it is required to represent some functions as a sum of
trigonometric functions cosine and sine. Such a series representation of a function
( )f t (which may be discontinuous); if exists, is called the trigonometric series
expansion of ( )f t .
Fourier introduced such an expansion of periodic functions in terms of sine and
cosine functions and hence it is called a Fourier series expansion. Many functions
including some discontinuous periodic functions can be expanded in a Fourier
series and hence are, in certain sense more universal than Taylor series expansions,
which cannot be established for discontinuous functions. Fourier series solution
method is a powerful tool in solving some ordinary and partial differential
equations given with the initial or boundary conditions.
Page 6 of 39
Definition:
Let ( )f t be a periodic function of t with period T , and is defined in an interval
[ , ]c c T . Then the expansion of the form
0
1
2 2( ) cos sinn n
n
n nf t a a t b t
T T
, if exists called the Fourier series or
Fourier expansion of ( )f t . Here 0, and n na a b are called Fourier coefficients.
Eulers formulae:
Given a periodic function ( )f t with period T , represented in [ , ]c c T by a Fourier
series,
0
1
2 2( ) cos sin (1)n n
n
n nf t a a t b t
T T
, to determine
the coefficients0, and n na a b , we proceed as follows,
Integrating (1) we get,
0
1
0
1
0
2 2( ) cos sin
2 2 cos sin
0
c T c T
n n
nc c
c T c T c T
n n
nc c c
n nf t dt a a t b t dt
T T
n na dt a t dt b t dt
T T
a T
0
1( ) (2)
c T
c
a f t dtT
Multiplying (1) by 2
cosm
tT
integrating, we get
2( )cos
c T
c
mf t t dt
T
Page 7 of 39
0
1
2cos
2 2 2 2 cos cos sin cos
0 02
c T
c
c T c T
n n
n c c
m
ma t dt
T
n m n ma t t dt b t t dt
T T T T
Ta
2 2( )cos (3)
c T
n
c
na f t t dt
T T
Multiplying (1) by 2
sinm
tT
integrating, we get
0
1
2 2( )sin sin
2 2 2 2 cos sin sin sin
0 0
c T c T
c c
c T c T
n n
n c c
m mf t t dt a t dt
T T
n m n ma t t dt b t t dt
T T T T
2m
Tb
2 2( )sin (4)
c T
n
c
nb f t t dt
T T
The Fourier coefficients 0, and n na a b are given by the formulae (2), (3) and (4).
These are called the Eulers formulae for Fourier coefficients.
Note:
(1) For 2
Tc
, the formulae for the Fourier coefficients
0, and n na a b becomes
2
20
20
0, if ( ) is odd
1( )
2( ) , if ( ) is even
T
T
T
f t
a f t dtT f t dt f t
T
Page 8 of 39
2
2
20
0, if ( ) is odd
2 2( )cos
4 2( )cos , if ( ) is even
T
T
n
T
f t
na f t t dt
nT T f t t dt f tT T
22
0
2
4 22 2 ( )sin , if ( ) is odd( )sin
0, if ( ) is even
TT
n
T
nn f t t dt f tb f t t dt T TT T
f t
.
Thus if ( )f t is an even function of t , then Fourier series expansion of ( )f t in
,2 2
T T
is given by
0
1
2( ) cosn
n
nf t a a t
T
where
2 2
0
0 0
2 4 2( ) and ( )cos .
T T
n
na f t dt a f t t dt
T T T
Such an expansion is called the Fourier Cosine series expansion of ( )f t . Thus an
even periodic function has Fourier Cosine series expansion.
If ( )f t is odd, then the Fourier series expansion of ( )f t in ,2 2
T T
is given by
1
2( ) sinn
n
nf t b t
T
where
2
0
4 2( )sin .
T
n
nb f t t dt
T T
This expansion is called the Fourier Sine series expansion of ( )f t . Note that an
even periodic function has Fourier Sine series expansion.
2) For 2T , the Fourier series expansion of ( )f t defined in , 2c c is given by
Page 9 of 39
01
2
0
2
2
( ) cos sin
1 where ( ) ,
2
1 ( )cos
1 and ( )sin .
n n
n
c
c
c
n
c
c
n
c
f t a a nt b nt
a f t dt
a f t ntdt
b f t ntdt
For 2
c
if f(t) is even function of t , then Fourier series expansion of f(t) is
0
1
( ) cosnn
f t a a nt
where 00 0
1 2( ) , ( )cos .na f t dt a f t nt dt
If f (t) is odd function of t , then Fourier series expansion of f(t) is
1
( ) sinnn
f t b nt
where 0
2( )sin .nb f t nt dt
3) Let f(t) has Fourier series expansion 01
2 2( ) cos sinn n
n
n t n tf t a a b
T T
Consider the terms 2 2
cos sinn nn t n t
a bT T
which is called the nth harmonic in
the Fourier series expansion of f(t) for
2 2 1cos , sin tan nn n n n n n n n n nn
ba r b r or r a b and
a
the nth harmonic
becomes 2 2 2
cos cos sin sin cosn n n n n nn t n t n t
r r rT T T
. Then
2 2
n n nr a b is called the amplitude of the nth harmonic and 1tan nn
n
b
a
is
called phase angle of the nth harmonic.
Page 10 of 39
Dirichlets condition for the convergence of the Fourier series
If a periodic function f (t) of period T, is piecewise continuous in the interval
[c, c + T ] and has left and right hand derivative at each point of that interval then
The Fourier series expansion 01
2 2cos sinn n
n
n t n ta a b
T T
is convergent.
Its sum is f(t) except at a point t0 at which f(t), is discontinuous and at t0, it
converges to the average of left and right hand limit of f(t) at t0. i.e., to
0 01
f(t + 0) +f(t - 0) . 2
Parsevals identity:
If a periodic function f (t) has Fourier series expansion
0
1
2 2( ) cos sinn n
n
n t n tf t a a b
T T
which is uniformly convergent in
[c, c + T ] then 2 2 201
2( ) 2
c T
n n
nc
f t dt a a bT
Proof : Let f(t) has a Fourier series expansion
0
1
2 2( ) cos sinn n
n
n t n tf t a a b
T T
which is uniformly convergent.
Consider 2
0
1
2 2( ) ( ) cos sin
c T c T
n n
nc c
n t n tf t dt f t a a b dt
T T
f(t)
f (t0 +0)
f(t0 0)
t t0 c c + T
Page 11 of 39
0
1
2 2( ) ( ) cos ( ) sin
c T c T c T
n n
nc c c
n t n ta f t dt a f t dt b f t dt
T T
since Fourier
series is uniformly convergent
0 0
1 2 2n n n n
n
T Ta T a a a b b
2 2 201
22
n n
n
Ta a b
Therefore, 2 2 201
2( ) 2
c T
n n
nc
f t dt a a bT
which is called Parsevals identity.
Problems:
1. Expand 2( ) ,f x x x x f(x+2)=f(x) , as a Fourier series.
Solution: Let 0
1
( ) ( cos sin )n nn
f x a a nx b nx
. Then
2 2
0
0
1 2( )
2 2a x x dx x dx
since, x is odd and x
2 is even function.
3 3 2
0
1
3 3 3
x
2 20
1 2cos cosna x x nx dx x nx dx
2 2 30
2 sin cos sin2 2
nx nx nxx x
n n n
= 2 2 2 32 sin sin 0 cos cos0 2
0 2 0 sin sin 0n n n
n nn n n n n
Page 12 of 39
2 3
12 20 0 2 0 .0
n
n n
1
2 2
4 1 41
n
n
n n
.
20
1 2sin sinnb x x nxdx x nxdx
=
0
2
2
cos sin1
nx
n
nxx
n
= 22 cos 1
0 sin sinn
n non n
= 12 2 2
cos 1 1n n
nn n n
.
2
1 1
21
4 21 cos 1 sin
3
n n
n
f x nx nxn n
.
2. Obtain the Fourier series expansion of 21, 0 2
4f x x x
f(x+2)=f(x) and hence obtain 21
in
1
2
1n
iin
2
1
2 1iii
n
Solution:
Here f x is an even function.
Page 13 of 39
01
cosnn
f x a a nx
, where
2
2 22 3 3
0
0
0
3
1 1 1 1
2 4 8 24 123a x dx
x
2
2
0
1 1cos
4na x nxdx
= 2
2
2 3
0
1
4
sin cos sin2 2 1
nx nx nxx x
n n n
= 21 2
0 cos 2 cos0 04
nn
= 2 21 2 1
24 n n
2
21
1cos .
12 nf x nx
n
At x = 0, 2 2
21
10
4 12 nf
n
2 2 2
21
1................. 1
4 12 6n n
At , 0x f x
2
21
10 cos
12 nn
n
2
21
1
12
n
n n
1 2
21
1..................... 2
12
n
n n
(1) + (2) gives
Page 14 of 39
2 2 2
21
12
6 12 42 1n n
2
21
1
82 1n n
.
3. Expand
, 0 1
0, 1
2 , 1 2
x x
f x x
x x
2f x f x , as a Fourier series and
hence deduce that 1 1 1
1 ..............4 3 5 7
Solution:
f x is odd,
1
sinnn
f x b n x
Where 2
0
2sin
2nb f x n xdx
= 1 2
0 1
sin 2 sinx xdx x xdx
= 21
2 2 2 2
0 1
cos sin cos sin1 2 1
n x n x n x n xx x
n n n n
2 -2 -1
1
Page 15 of 39
= 1 cos1
cos 0 0 0 0n
nn n
= 12cos 2 2
1 1n nn
n n n
1
1
12 sin
n
n
f x n xn
1
2 2f
1
1
12 sin
2
n
n
n
n
1
2
0 is evensin
2 1 is oddn
nn
n
1
1
1sin
4 2
n
n
n
n
1
1
1
2 1
n
n n
1 1 11
3 5 7 .
4. Expand
0, 02
( ) , ( ),
sin , 0
t
f t f t f t t
E t t
as a Fourier
series.
Solution:
( )f t is neither even nor odd.
0
Page 16 of 39
01
( ) cos sinn nn
f t a a n t b n t
where
2
0
0
2
0
1 1( ) sin
2
cos .
2
T
T
a f t dt E tdtT
E t E
2
0
2
2( )cos sin(1 ) sin(1 )
2
T
n
T
Ea f t n tdt n t n t dt
T
0
0
0
cos(1 ) cos(1 ), 1
2 (1 ) (1 )
sin 2 , 12
cos(1 ) cos(1 ) 1 1, 1
2 (1 ) (1 ) (1 ) 1
cos2, 1
2 2
E n t n tn
n n
Etdt n
E n nn
n n n n
E tn
E
22 2 2
, is even2 (1 ) 1 1 .
0, is odd
En
n n n
n
Page 17 of 39
2
0
2
0
0
2( )sin sin sin
cos( 1) cos( 1)2
sin( 1) sin( 1)1
2 ( 1) ( 1)
0, 1
W
n
W
w
Ewb f t wt dt wt nwt dt
T
Ewn wt n wt dt
Ew n wt n wtn
n w n w
n
2
1
0
0
0
2 2
sin
(1 cos2 )2
sin 2
2 2
( 0)2 2
2 cos2 cos4( ) sin .....
2 2 1 4 1
w
W
w
Ewb wt dt
Ewwt dt
Ew wtt
w
Ew E
w
E E E wt wtf t wt
5. Expand 2f(x) = x , 0 x 2, f(x+2) = f(x) as a Fourier series and hence
evaluate 2
1
1
n n
.
f (t0 +0)
f(x)
x 4 2 -2
Page 18 of 39
Solution: Note that f(x) is neither even nor odd.
01
( ) ( cos sin )n nn
f x a a nx b nx
where
22 32
0
0 0
1 1 4
2 2 3 3
xa x dx
22 2
2
2 2 3 3 2 2
0 0
2 sin cos sin 4cos 2 2
2n
x n x n x n xa x n xdx x
n n n n
22 2
2
2 2 3 3
0 0
3 3
2 cos sin ssin 2 2
2
4 2cos2 0 0 (cos2 cos0)
4.
n
x n x n x co n xb x n xdx x
n n n
n nn n
n
2 21
4 4 4( ) cos sin
3 nf x n x n x
n n
A x=0, f(x) is discontinuous
2 21
2 21
2 2
21
(0 ) (0 ) 4 4 4cos0 sin0
2 3
4 0 4 4 1
2 3
1 42
3 4 6
n
n
n
f f
n n
n
n
Page 19 of 39
Exercises : Expand the following functions as Fourier series
2
1
21
1. ( ) ,0 2 , ( 2 ) ( )
12. ( ) ,0 2 , ( 2 ) ( )andhencededuce that
2sinh 1
1 33. ( ) sin ,0 2 , ( 2 ) ( )andhencededuce that
1 4
04. ( ) ( 2 ) ( )
2 2
5.
n
x
n
n
f x x x f x f x
f x e x f x f xn
f x x x x f x f xn
x xf x f x f x
x x
f
2
( ) sin ,0 2 , ( 2 ) ( )
6. ( ) sin ,0 1, ( 1) ( )
7. ( ) 2 ,0 2, ( 2) ( )
8. ( ) 2 2 2, ( 4) ( )
19. ( ) ( ) ( )
02
2 210. ( ) ( 2 ) ( )
322
x x x f x f x
f x x x f x f x
f x x x x f x f x
f x x x f x f x
tt a
af t f t T f t
Ta t
x x
f x f x f x
x x
Page 20 of 39
2l O
y
x l -l
Half range Expansions
While solving various physical and engineering problems, there is a practical need
for expanding functions defined over a finite range. Such an expansion is possible
if functions under consideration can be extended to a periodic function which is
either even or odd.
Consider a piecewise continuous function ( ),f x defined in a finite interval
(0, ).l Then it is possible to extend ( )f x to a periodic function, which is even or
odd.
Consider the function ( )g x defined as follows:
( ), 0( ) ; ( 2 ) ( ).
( ), 0
f x x lg x g x l g x
f x l x
Then ( )g x is called an even periodic extension of ( )f x .
Graph of ( )f x
Graph of even periodic extension of f(x).
O
y
x l
Page 21 of 39
2l O
y
x l -l
The function ( )g x can be expanded as Fourier cosine series
0
1
( ) cosnn
ng x a a x
l
where 00
1( )
l
a g x dxl
and 0
2( ) cos
l
n
na g x x dx
l l
But for 0 ,x l ( ) ( ).g x f x We have
0
0
1( )
l
a f x dxl
and 0
2( ) cos
l
n
na f x x dx
l l
( ) ( )f x g x 0
1
cos ,nn
na a x
l
for 0 .x l
Such an expansion of ( )f x is called the half range Fourier Cosine series
expansion of ( ).f x
Also, if ( ), 0
( ) ; ( 2 ) ( ).( ), 0
f x x lg x g x l g x
f x l x
Then ( )g x is called an odd periodic extension of ( )f x .
Graph of ( )f x
Graph of odd periodic extension of f(x)
O
y
x l
Page 22 of 39
The function ( )g x can be expanded as Fourier Sine series
1
( ) sinnn
ng x b x
l
where
0
2( ) sin
l
n
nb g x x dx
l l
But for 0 ,x l ( ) ( ).g x f x We have
0
2( ) sin
l
n
nb f x x dx
l l
( ) ( )f x g x 1
sin ,nn
nb x
l
for 0 .x l
Which is called the half range Fourier Sine series expansion of ( ).f x
Problems:
1. Expand ( ) ,0f x x x as half range Fourier cosine and sine series. Also
draw the graph of the corresponding periodic extensions of ( ).f x
Solution:
Graph of even periodic extension of ( )f x .
0
1
( ) cosnn
f x a a nx
, where
Page 23 of 39
0
0 0
1 1( ) .
2a f x dx xdx
2
00 0
2
2
2 2 2 sin ( cos )( )cos cos (1)
0, 2 ( 1) 1
.4, odd
n
n
x nx nxa f x nxdx x nxdx
n n
n even
n nn
2 2
4 cos cos3( ) .
2 1 3
x xf x x
Graph of odd periodic extension of ( )f x .
1
( ) sinnn
f x b nx
, where
2
00 0
1
2 2 2 ( cos ) ( sin )( )sin sin (1)
2 ( 1) 0 2( 1) .
n
nn
x nx nxa f x nxdx x nxdx
n n
n n
1
1
2( ) ( 1) sin .n
n
f x x nxn
Page 24 of 39
2. Expand
, 02
( )
1 ,2
x lx
lf x
x lx l
l
as half range Fourier cosine and sine
series. Also draw the graph of the corresponding periodic extensions of
( ).f x
Solution:
Graph of even periodic extension of ( )f x .
( )f x 0
1
cosnn
na a x
l
Where
2
0
0 0
2
1 1( ) 1
l
l l
l
x xa f x dx dx dx
l l l l
22 2
02
1 11
2 2 4
l l
l
x x
l l l
2
0 0
2
2 2( )cos cos 1 cos
l
l l
n
l
n x n x na f x x dx x dx x dx
l l l l l l l
2 2
22cos cos 1
2
nn
n
1
Page 25 of 39
2 2
0, 2, 6, 10, 14, ...
8, 2, 6, 10, 14, ...
n
nn
2 2 2
1 8 1 2 1 6( ) cos cos
4 2 6f x x x
l l
.
Graph of odd periodic extension of ( )f x
( )f x 1
sinnn
nb x
l
Where
0
2( )sin
l
n
na f x x dx
l l
2
0
2
2 sin 1 sin
l
l
l
x n x nx dx x dx
l l l l l
12 2
22 2
0, even4
sin 42 1 , odd
n
nn
xn n
n
2
l
O
y
x
Page 26 of 39
2 2 2 2
4 1 1 3 1 5 ( ) sin sin sin
1 3 5f x x x x
l l l
Expand ( ) 1 ,0x
f x x ll
as a Fourier cosine and sine series. Also draw the
graph of the corresponding periodic extensions of ( ).f x
Solution:
Graph of even periodic extension of ( ).f x
Let 01
( ) cosnn
nf x a a x
l
.
Then 00
2 1( ) .
2
l
a f x dxl
0 0
2 2
2 2
2 2( )cos 1 cos
0, 2
1 ( 1) .4,
l l
n
n
n x na f x xdx xdx
l l l l l
n even
n n oddn
2 21
1 4 (2 1)( ) cos
2 (2 1)n
n xf x
n l
.
-l l
Page 27 of 39
Graph of odd periodic extension of ( )f x
Let 1
( ) sin .nn
nf x b x
l
Then 0 0
2 2 2( )sin (1 )sin
l l
n
n x nb f x xdx xdx
l l l l l n
Thus 1
2( ) sin .
n
nf x x
n l
Problems:
Obtain the half range Fourier Cosine and sine series expansions of the
functions. Also draw the graph of corresponding periodic extensions.
1) ( ) sin ,0f x x x x .
2) ( ) ( ),0f x x x x .
3)
, 02
( )
,2
x x
f x
x x
.
4) ( ) 2 ,0 2f x x x .
5) ( ) cos ,0f x x x ll
.
6) ( ) sin ,0f x x x ll
.
7) If ( )f x is piecewise continuous for 0 ,x l having half range expansions
l -l
0
2l
Page 28 of 39
01 1
( ) cos & sinn nn n
n nf x a a x f x b x
l l
then show that
2 22 2 2
0
1 10 0
2 2( ) 2 & ( )
l l
n n
n n
f x dx a a f x dx bl l
.
Fourier integral representation:
Let f(x) be a piecewise continuous and absolutely integrable function of x in
( , ). , ., ( ) .i e f x dx exists
Then f(x) can be represented by an integral as
0
1( ) ( ) cos ( )sin (1)
( ) ( )cos , ( ) ( )sin .
f x A S sx B s nx ds
where A s f t st dt B s f t st dt
Such an integral representation is called the Fourier integral representation of f(x).
The integral on RHS of (1) converges to f(x0) if f(x) is continuous at x0 and to
average of left and right hand limits if f(x) is discontinuous at x0.
Proof: Consider a periodic function fl(x) defined in (-l, l) such that fl(x) =f(x) for
l x l
Then, fl(x) can be represented by a Fourier series as
0
1
( ) cos sin (2)l n nn
n x n xf x a a b
l l
Where
Page 29 of 39
0
1 1 1( ) , ( )cos , ( )sin
2
l l l
l n l n l
l l l
n t n ta f t dt a f t dt b f t dt
l l l l l
Substituting in (2) we get,
1
1 1 1( ) ( ) ( )cos cos ( )sin sin
2
l l l
l l l l
nl l l
n t n x n t n xf x f t dt f t dt f t dt
l l l l l l l
1
1 1( ) ( ) ( )(cos cos sin sin )
2
l l
l l l
nl l
n t n x n t n xf x f t dt f t dt
l l l l l l
n
nLet s
l
Then 11 n
n n n
ss s s or
l l
On substitution we get
1
1 1( ) ( ) ( )(cos cos sin sin )
2
l l
nl l l n n n n n
nl l
sf x f t dt f t s t s x s t s x dt s
1
1 1( ) ( ) cos ( )cos sin ( )sin )
2
l l l
l l n n l n n l n n
nl l l
f x f t dt s s x f t s t dt s x f t s t s
Let , then ( ) ( )ll f x f x .
( ) 0 0.
l
l n n
l
f t dt s s
Now taking the limit as , we getl
Page 30 of 39
0
0
1( ) cos ( )cos sin ( )sin
1( )cos ( )sin
f x sx f t stdt sx f t stdt ds
A s sx B s sx ds
Where ( ) ( )cos A s f t st dt
( ) ( )sin B s f t st dt
Note (1): The Fourier integral representation of ( )f x can also be expressed as
0
1( ) ( ) cos cos sin sin f x f t st sx st sx dt ds
1( ) cos ( ) (3)
2
cos ( ) is an even function of .
f t s x t dt ds
s x t s
Also, 1
( ) sin ( ) 0 (4) 2
f t s x t dt ds
sin ( ) is an odd function of .s x t s
(3) (4)i gives
1
( ) ( ) cos ( ) sin ( 2
f x f t s x t i s x t dt ds
( )1 ( )
2
is x tf t e dt ds
Page 31 of 39
Which is called the complex form of Fourier integral representation of ( ).f x
Note (2): If ( )f x is an even function of ,x then
0
( ) ( )cos 2 ( )cos A s f t st dt f t st dt
( ) ( )sin 0B s f t st dt
Therefore Fourier integral becomes
0
2( ) ( )cos f x A s sx ds
0 0
2( )cos cos f t st sx dt ds
Which is called the Fourier Cosine integral representation of ( ).f x
If ( )f x is an odd function of ,x then
( ) ( )cos 0A s f t st dt
0
( ) ( )sin 2 ( )sin A s f t st dt f t st dt
0
2( ) ( )sin f x B s sx ds
0 0
2( )sin sin f t st sx dt ds
Which is called the Fourier Sine integral representation of ( ).f x
Page 32 of 39
Fourier transforms
Consider the Fourier integral representation of the function ( )f x given by
( )1( ) ( ) 2
is x tf x f t e dt ds
)1 1 ( )
2 2
isx iste f t e dt ds
Let 1
( ) ( ) (1)2
istF s f t e dt
Then 1
( ) ( ) (2)2
isxf x F s e ds
The integral defined by (1) is called the Fourier transform of the function ( )f x
and is denoted by ( ) .F f x Given ( ) ( ) ,F s F f x the formula (2) defined ( ),f x which is called the inverse Fourier transform of ( )F s and is denoted by
1 ( ( )F F s .
Note (1): A function ( )f x is said to be self-reciprocal under Fourier transforms if
( ) ( ).F f x F s
Note (2): If ( ) ( ),F f x F s then 1( ) ( )f x F F s is called spectral representation of ( )F s and ( )F s is called spectral density of ( ).f x Here s is
called the frequency of the transform. The graph of ( )F s is called amplitude
spectrum of ( )f x and 2
( )F s is called energy of the spectrum.
Properties of the Fourier transforms:
(1) Fourier transform is linear
i.e., if 1 2 and c c are constants then
1 2 1 2( ) ( ) ( ) ( )F c f x c g x c F f x c F g x
Page 33 of 39
Proof: Follows from definition and linearity of the integral.
(2) If ( ) ( )F f x F s then ( ) ( )iaxF e f x F s a
Proof: Consider ( )1( ) ( )
2
i s a tF s a f t e dt
1
( )2
ist iatf t e e dt
1
( )2
iat istf t e e dt
( )iaxF e f x (3) If ( ) ( )F f x F s then ( ) ( )isaF f x a e F s
Proof: Exercise
(4) If ( ) ( )F f x F s then 1
( )s
F f ax Fa a
Proof: Exercise
(5) If ( ) ( )F f x F s then
(i) 1
( )cos ( ) ( )2
F f x ax F s a F s a
(ii) ( )sin ( ) ( )2
iF f x ax F s a F s a
(iii) ( ) ( )F f x F s (iv) ( ) ( )F f x F s
(v) ( ) ( )F f x F s
(vi) ( ) ( )n
n n
n
dF x f x i F s
ds
(vii) ( ) ( ) ( ) ( )n nF f x is F f x
Page 34 of 39
Convolution:
For functions ( ) & ( ),f x g x we define the convolution of ( ) & ( ),f x g x denoted by
( )f g x as 1
( ) ( ) ( ) ;2
f g x f t g x t dt
provided the integral exists.
Note that .f g g f
Convolution Theorem:
( ) ( ) ( ) .F f g x F f x F g x
Proof:
Consider
( )
1( ) ( )( )
2
1 1( ) ( )
2 2
1 1( ) ( )
2 2
isx
isx
ist is x t
F f g x f g x e dx
f t g x t dt e dx
f t e g x t e dx dt
Let x t u . Then
( )
( )
1 1( ) ( ) ( )
2 2
1 1( ) ( )
2 2
( ) ( ) .
ist is u
ist is u
F f g x f t e g u e du dt
f t e dt g u e du
F f x F g x
Parsevals Identity:
If ( ) ( )F f x F s then 2 2| ( ) | | ( ) |f x dx F s ds
.
Proof:
Page 35 of 39
1
( ) ( ), { ( )} ( ),
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
1 1. ., ( ) ( ) ( ) ( )
2 2
isx
If F f x F s F g x G s then
F s G s F f x g x or
f x g x F F s G s
i e f t g x t dt F s G s e ds
For 0,x we get
( ) ( ) ( ) ( )f t g t dt F s G s ds
( ) ( ) ( ) ( )Let g x f x or g x f x .
2 2
( ) ( ) ( ) ( ) ( ).
( ) ( ) ( ) ( ) ( ) ( )
| ( ) | | ( ) |
ThenG s F g x F f x F f x F s
f t g t dt f t f t dt F s F s ds
f t dt F s ds
Problems:
1) Find the Fourier transform of 1, | |
( )0, | |
x af x
x a
. Hence deduce that
2sin sin
2 2
t tdt and dt
t t
.
Solution:
1
( ) ( )2
1 1 2 sin1 ( )
2 2
isx
aa isxisx
a a
F f x f x e dx
e ase dx F s
is s
.
Page 36 of 39
0
0
1 1 2 sin( ) ( ) cos sin
2 2
2 sin sincos sin is an odd function of
0, | |
sincos ( ) , | |
2 2
1
2 2
isx asf x F s e sx i sx dss
as assxds sx s
s s
x a
assxds f x x a
s
0 , | |4
x a
.
0
For 0, ( ) 1.
sin.
2
x f x
asds
s
0
0
Let or . .
On substitution, we get
sin
/ 2
sintor .
t 2
t dtas t s ds
a a
t dt
t a a
dt
2 2
2 2
0
2
0
2
0
By Parseval's identity,
| ( ) | | ( ) |
2 sin 4 sini.e., 1
sin
2
sinFor 1, we get .
2
a
a
f x dx F s ds
as asdx ds ds
s s
as ads
s
ta ds
t
Page 37 of 39
2) Find the Fourier transform of , 0a x
e a
and hence evaluate 2 2
0
cos xtdt
a t
and
{ }a x
F xe
0
2 2 0
2 2 2 2
2 2
2 2
0
2 2 2
0
1 2{ } cos
2
2cos sin
2 2(0 ( )) ( )
1( ) ( )
2
1(cos sin )
2
2 cos
cos cos
a x a x isx a x
ax
o
isx
F e e e dx e sxdx
ea sx s sx
a s
e aa F s
a s a s
f x F s e ds
asx i sx ds
a s
a sxds
a s
sx xtds
a s a
20
( )
2 2
a xf xdt e
t a a
Since
1
2 2 2 2 2
{ ( )} ( )
2 2 21, { } { }
( )
nnn
n
a x a x
dF x f x i F s
ds
d d a asfor n F xe i F e i i
ds ds a s a s
3) Find the Fourier transform of 2 2
, 0a xe a and hence show that
2
2
x
e
is
self-reciprocal under Fourier transform.
Page 38 of 39
2 2 2 2 2 2
2 2
2
2
22
( )
2 4
42
1 1{ }
2 2
1
2
2
a x a x isx a x isx
is sax
a a
sis
a axa
F e e e dx e dx
e dx
ee dx
Let 2
isax t
a then
dtdx
a
=
2 2
2 2
2 24 42
2 2
s s
a at te dt ee e dt
a a
Put
2
1
21
2
t y or t y
dt y dy
2 2 2 2
2 2 2 214 4 4 42
0
12
2 22 2 2 2
s s s s
ya a a ae e e e ey dy
a a a
For 1
2a or 2 1a
2 22 2
2 2{ } { }x s
a xF e F e e
2
2s
e
is self-reciprocal under Fourier transform.
4) Find the Fourier transform of 1 1
( )0 1
x xf x
x
and hence evaluate
2
0
sin xdx
x
Page 39 of 39
1
1
1
0
1
2 2
0
2
1 1{ ( )} ( ) (1 )(cos sin )
2 2
2(1 )cos (1 )sin .
2 sin cos 2 (1 cos )(1 ) ( )
1( ) ( )
2
1 2 (1 cos )(cos
2
isx
isx
F f x f x e dx x sx i sx dx
x sxdx x sx is an odd function of x
sx sx sx F s
s s s
f x F s e ds
ssx
s
2
0
sin )
(1 ) 12 (1 cos )cos ( ) 2
20 1
i sx ds
x xssxds f x
sx
20
2
20
22
20 0
(1 cos )0, (1 0)
2 2
2sin ( / 2)
2
/ 2 2 2
2 sin sin2
4 2 2
sxFor x ds
s
sds
s
Let s t or s t ds dt
t tdt or dt
t t
Exercises:
1. Obtain the Fourier transform of 2
3
0
1 1 cos sin( ) and hence evaluate cos( / 2)
0 1
x x x x xf x x dx
x x
2. Find the Fourier transform of ( )0 0
x x af x
x a
3. Find the Fourier transform of
2 2
( )0
a x x af x
x a
