E

8 8 e8

8
 FOURIER TRANSFORM
2.83

cos (s + a) x + cos (s a) x] dr
0

S)cos (s +a)xdr +V} Sfx) cos (s -a)xde

0 0

=
Fe 6 +a) + Fe (-a)

(qi) F S C)
sin a r ] =
VË Jfx) sin ax cos sx dr
0

[sin (a + s) x + sin (a s)x]dr

-

V fa) sin (a +5) xdr +V? Sf) sin (a - s)x dr|

0

-Fa+) +F,(a -s) 1

()F.LfG) cos ar] = V£ Sf) cosaxcossx dr

V S) cos sr cos ar dr

V cos (s + a) x + cos
(s-a)x ]dr

V Jfx) cos (s + a)xdr+ \ Sf) cos (s a)x dr| -

IF(6 +a) Fe (6 a) 1
+ -
2.84 ENGINEERING MATHEMATIcs

3.FI(ax)1 = :

Proof: F IS(ar)] Js (ax) sin sr dr

put ar = x 0 0D
a dr =dt

- VIro sin

a V ro sia 0

Similarly Fe Lf (ax)] =

Fe
4. F, [f' (x)] =
-s Fe (s), if f(x) > 0 as
x co.

Proof: F.S')I =

Vron dx

-V sin sx d |f «)|

V (sin sxf )-sSfa) cos sx de1

Sfx) cos sx dr |Assuning
(¢)-0 as x>
»|
=-s Fe (s)
 ANSFORM
2.85
o U R I E R T R A N

V?..
=
- f()+s F, (s) if f(x) --0 as x

FIf

F. 1f
}
=

VË Js'(x) cos sx dr
Proof:

d [f () ]|
V cos sx

cos sxf() b +sJf«) sin sx dr ]

0

VE [0-f(0)]+s F,() Assuming ft)-0 as x -

f0) + s F, (G)

&F, xt(1
=
- F . ()1
OS

Proof We know that

F. If) = VSf) cos sxdr

Diferentiating both sides w.r.to s

FAS)=V Sfe) cos sx dr

a (cos sx) dr

- f«) (-xsin sx) dr

2.86 ENGINEERING MATHEMATICs

- VF SS)x sin sx de
0

-F, Ixf)]

(i.c..) F, [xf)1 = F.1)

7. F xt) =
E,)
Proof We know that

F)] -V f) sin sx dx
Differentiating both sides w.r.to s we get

ds
fe) sin sx de
0

V Is (sinsa) d
=
So) cos sx (x) dx
0

Sx) x cos sx dr

= Fekf )

i.c., Fe kS)) =
F
as UN
 FOURIER TRANSFORM
2.87

IX PROBLEMS BASED ON PROPERTIES OF F.T, F.C.T AND

F.S.T.
Example 2.4.1: Find the Fourier sine and cosine transformations
of xe

Solution: (G) We know that F, Ixf¢)] = F. 1S)]

ds

Vl FIe* ] =V.
- V

-V
-V (+a
2as

() We know that Fe lxf)] =

F, f)1
ds

Sol. F.[xe" = - F , I"]

ds

- -V2+d0-s(2
(G+a2
= - V2+-22
 2.88 ENGINEERING MATHEMATICs

Example 2.4.2: Find Fourier cosine transform of e * and hence

find F, [xea*]. [A.U. N/D 2006]

Solution : See Example 2.3 a(14)

Fea = /43

Now F, Ixe*)- Fds i

es/4a
2
4a
e-s/4a
s43

22a3

transform of e" and hence deduce

Example 24.3: Find Fourier
2as
that () F [xea |xl =
VT (s+a22
cos xt d
=e-a /x|
2007]
oa24?
[A.U. Oct.2001, M/J

Solution: We know that

FIf)

1
2.
FOURIER TRANSFORM
2.89

V27T ea (cos sx + isin sx) dr

1
V27T 2eax
0
cos sx dr

Since ea 1x cos sx is an even function

eax cos sx dx =
2Je cos sx d
0

ea x sin sx is an odd function

Ceaxsinsx dx = 0

V 0
ax cos sx dx

by formula Se a* cos bx dr =

a+b
Using inversion formula we get

1
V F If«)] e ds

1 SX ds

V27 s2+2

V
2.90 ENGINEERING MATHEMATICSs

-
cOs sr- i sin sx ds

-P
cos Sx
Since 1 5 an even function.

COS SX
f ds =
2 f
COS SX

+a2

Sin sr
i s an odd function.

sin S ds =0

f) f COs sx
ds
os+2

COs x
dt=ea x|[: s is
dummy variable]
F [xea k] =
-i F )

ds
FOURIER TRANSFORM
2.91

-iaV 0-2s
+a

-N +
2as

Example 2.4.4: Find the Fourier sine transform of e a and hence

ax
find the Fourier cosine transform of

Solution: We know that

F,f6)]-V? Is9sinsd 0

-ax
sin ax dx

-V : Ssinbx d
0
=

To find Fe [xe]

-[ V
2.92 ENGINEERING MATHEMATICS

VE )-1

V -

V
Example 2.4.5:Find the Fourier transform of e* and hence find
the Fourier transform of e cos 2x

Solation: We know that F

[f)] =

S)E dr

F =

S de

2Se (cos sx + i sin sx) dr

1 00
VTSe cossx dx +-
Sesin sx dr
2
ecos sx dr +0
Since e cos sx is an even function.
e sin sx is an odd function.
FOURIER TRANSFORM
2.93

F () V S"cos
0
.sx dx

V :e"cos bx dk +
a

To find F [elcos 2

By Modulation theorem,

Ff)cosa) ; =
[F (6-a)+ F
(s+a)]

FIel cos2x]=5 1
(6+2)+1
1 1
|2-4s +5 s+45+ 5
+ 4+5+s-45 +5
V27
+5-(4
2+5)
V27T4 10s+25- 16s
=
+5

A-6s+25
2.5 CONVOLUTION THEOREM- PARSEVAL'S IDENTITY
2.5.1 Definition: Convolution
The convolution of two functions f ) and g (1) is defined as

8 ) SO8-1) dt.
2T
 2.94 ENGINEERING MATHEMATICs
2.5.2 Convolution Theorem:
The Fourier ansform of the convolution of f(x) and g ) is the

product of their Fourier transforms. [AU Trichy ND 2009

(i.e.) F Lf) *s)] =
F () G 6) =
F [f] F ls()]
Proof: We know that F [f)] = S f )e dr

Ff)*8) ] =
Ss)*g )] d

27

- ros-9aa
by changing the order of integration we get

SfOs -) ad
c

1
V2Tf() Fls-)] dt
1
VSf)" G ()dt

by shifting property Ff*-a)] = e s F ( )

 rOURIER TRANSFORM
2.95

- G () F (6) =
F (6) G (6)

Note FF () G ) =f)*8)
-

FF )] F [G (6)]
25.3 PARSEVAL'S IDENTITY
IfF () is the Fourier transform of f(x). Then

S1fdr = S | F(¢) |ds

- 00

[A.U. CBT Dec. 2008]

Proof: By convolution theorem

F f ) * s ) ] = F (). G ()

fo)*8) = F[F (6). G ()]

S08-)dt = F) G 6)»
ds (1)
putI=0 and g(-1) =
f) there it follows that
G ) =
F()

(0)Sf)() d = f F() F() ds

Note: In the same

way we can prove Parseval's identity for Fourier
Since and
cosine
transforms.
Ff)] F, I and =

F. l8 )] =
Fe («) then
 2.96 ENGINEERING MATHEMATICS
00

SIs) |*dr =
f
b
IF ()|ds and
0

(i) S I s ) |*dt = f IFe()ds.

0 0

Problems based on Convolution theorem and Parseval's

X.
identity
Example 2.5.1 Find the Fourier transform of

- t
if xl 2where
a
a
is a positive real number.

Hence deduce that (i) in at= and (i) dt=

[A.U. March, 1996] [A.U N/D 2007, A/M 2008]

Solution : The given function can be written as

if <x < a
10 otherwise -a

fo)

F(s)= F f«)]: Sf)e d

-00

1
- de
V27t -a

1as e ias
1
V2 Tt IS

ea elas
1 2i sin as . s i n ax =
2
V2 is

F() V2 sinas .(1)

Fourier inversion formula we have

(i) Now, by

S)= V22SFo)e-ixds
OURIER TRANSFORM
2.07

sin as
(cos sx- i sin x) ds using (1)1
Sinas Sin as
cos sx ds-T sin sx ds
S

c o scos sx d s 0 ) sin as sin sx is an odd function in s]

cos srds :i cos sx is an cven function in s

sin &s cos sx ds S ) . (2)

putting , we get J ds
S
0

Now, put = as. Then we have s =and ds

ds =

Further 0 as s 0 and t o as s > 0o

Gi) Using Parseval's identity S Isc)|*d« =

S |F()12s
2

We get 1d Sinas ds
S

2
ds

2
Sin as
ds= a n

Putting t= as we ge
2.98 ENGINEERING MATHEMATIcs

sindt
a

2SSdt =: is an even function]

dt

Example 25.2 Find the Fourier Transform of

Hence deduce that

Oif ixi >1
1

-
[AU. April, 2001, May, 2001]
[AU. A/M 2005, N/D 2005, M/J 2006, N/D 2007, CBT Dec.
2008]
[A.U N/D 2009]
Solution: Given f() = * l , -1<x<1
,X<-1 and x >1
Tne Fourier transform of f) is

F)

S(1 -

|x|) (cos sx +i sin sx) dx

V2T -1 (1 l) cos sx de + S(1 -

xl) sin sr de
-1

S (1- |xl) cos sx dr +0

-1 [ (1- Jx|) sin sx is odd
FOURIER TRANSFORM
2.99

( x ) cosstde . (1- x|) cossx is even]

[As x>0, Ixl =x ]

Va--- cos sx

va- Sin COS Sx

0

.V[o5 -
V cosS

-V -COS S|

(1-coss
Hence IFC)2 -

Using Parseval's identity S |F()|* ds =

S |(x)1
-00
dx
00

we have
-cos s ds - S(1-xl)
-1

=
2S(1 -

x) d
asx>0, rl =x
]|
[ 1 -rl is even]
2.100
ENGINEERING MATHEM
3 0
-[0-
2
-10-1
00 Sin
COS s ds
2

0
as
- 00

ds

co Sin
ds put 2=s

sin'! (2dt) 2424 dt d s

o (2)
2 d t = ds

Example 2.53: Show that the Fourier transform of

)= a-1, |x| <a sin as a s cos as

|0, V
Hence deduce that ( Sint-tcost
dt==Using Parseval's
identity show that sint- tcos t
0 dts
A.U. March 196, Nov. 2001, A/M 2004]|[AU. N/D 2008, A/M 2009

Solution: Proof: We know that

F O U R I E R T R A N S F O R M

2.101

0 + S (a*-x)e dr +0 -a

5x

-(-2x)
27

27 S is -a

0+2 2-a
1
0- is is
2 a es a

+
2e1sa 2ae 2 isa
s i
2 a ei s a e i s a

1 2ae'sa
2iesa
V27 s
1 -3-")
1 12 cosas]+sin as
V27
V27T
t e I X
-ix
- sin r =

. cosx =

2
,
2

1 -4as cos as + 4 sinas

V2T