2.

102
ENGINEERING MATHE
Sinas as cos as
HEMATICA
F 2V? sin as-s coss
ic. () .(1)

2V? sin ss cos s

when a 1, F(6)
s
Using inverse Fourier Transform, we get

f V sin as -

as cos
as|e 15x ds

1
V27V V (sin as- as cos as |cos sx -
i sin sx] ds

r)- Sin as-as cos as

s
cos sx ds Sin as-as cos aas
sin ox ds

Sin as as cos as
cossx ds +0

[The second integral is odd and hence its value is

zero]
sin as as cos aS
cos sx ds (2)

[: the first integral is evenj

Putting a = 1, we get

(2)S) Sins SCOS s

cos sXx ds
3

- sin- t cost
Cos tr dt
FOURIER TRANSFORM 2.103

Puttingr= 0, we get
( ) 10) sint cosjt
Sint-fc 1
0

Sint-t cost dt
0

Using Parseval's identity,

F)ds SIf*|de using (1)

: V (sins-S cOs s) ds=S(1-a

-1

. ds 2S (1-jde
0

2
sin S-S cos s - 2 1+-2/]

- (0 +0-0)
=2

15 --
2
| Sin s SCOS S
i.e., J

12
ie, f Sint- t cost d-
0
 2.104

Example 2.54 Find the Fourier transform of e ax

ENQINEERING MATHEMATIC
ifa>0
Deduce thnt o (a+ a22d x -4aif a>0.
-a x
Solution: Given a) =
e

ax
if 0 sx < o
)
ax if -oo<r<0 where a>0
F Ifa)] -

-00
e a lx iSx dr
1
2t axisx d+ Sea* eisx d
0

la+is) xdt + f e-(@-is) x dr

0

| ( a + is) x
e is) x
a+is - co
- (a-is)
1
-

(a -

is),
1
4 + is a is

2V2
F - V a

By Parseval's identity
If FS) is the Fourier transform
of f(«). Then
I E RTRANSFORM 2.105
ORM

ita-V_ ds
0

ds
0

-2ax

1
2.5ds
(+a

4a3
is dummy variablel
a >O[. s
a
2

Example2.5.5: Find the Fourier transform of

1
f(x) = 1 -x if lx| <

= 0 if lx| > 1

Sin s S COS S cosds= 3z

Hence show that . 16
0 S

sin x)
Also show that X cos X dx =

Solution: See Q.No. 2.2.b(1) in problems based on Fourier

transform and its inversion formula

F) 4
Ff)) 32T Sins- coss
= s
 ENGINEERING MATMEMATKA
2.100

by
Parseval's identity
F)
Now if FV)]

Hece (1-r)d (-s CosS+ Sin

sd

eusts only in (-1, 1)

f)

- j0-ar
( - scoss +sin s)*d -1

i. -(-s coss +sins) ds

-1

1
ds = S(1-P
(-s coss + sin s 0

1
cOS X sinx
dx

T:(a-by=( -a

(-0+0)

15 10+ 3 8
15 15

(x cos x- Sin
X) d=
FOU TRANSFORM
2.107

, 56 : Find the Fourier transform of f() given by

Exanple 2 . 5

f(x) 1 for Ix| <2

= 0 for |x| >
2

and hence evaluate dx and dx.

0

[A.U. Nov/Dec.2003]

The given cquation can be written as

Colhtion
1 if -2 <x < 2
f()
=

= 0 otherwise

F ) F f) =
y27 Sf)*d

1 isx
-2 is2
ei2s 2i sin 2s
2 21 is
1 2 sin 2s
S

F () =
F If)] =
sin 2 (1)
S

Now Fourier inversion formula we have

) by

eis*ds
T) VT SF ()

Sins (cos sx - i sin sx) as

S

s sin 2 cos sx ds sin2s sin sx dr

S
2.108
ENGINEERING MATHEMATiC
cos sr dr Sin sx isan odd
function in

COsCOs sx dr coS Sx in an
cven
function in sl

S cossx dr =

putting x s
0 we gets ds
S
0

Sin25
0 S ds =
(: f) =
1 in
lal <2
Now putt =
2s
S0 t>0
dt =
2 ds
Sc t c

S Sint
o (/2)

Hence à 0 t
(ic,)sd 0 "tis
t:tis«dummy variable]
(i) Using Parseva's
identity Sf () d =S \ F () | ds
2

iIVsin 2
we get J dx
-2
ds

oo S ds

12-(-2)1 2sin 2 -

oo S ds

4
ds
 2.109
FOURIER TRANSFORM

ds 2

S 00
2s we get
putting
0
dt = 2 ds

2T

dt = 2T
2

sinEdt = 1

2 sin Edt T
[ Is an even function

-
Hence d- tis a dummy variable]

dx using transforms.
Example2.5.7: Evaluate
(+a)(+b)
M/J 2006]
[A.U. April, 2001][A.U. N/D 2005,

know that
Solution : Let f(x) =
e we

Fe IS)I =Fel e " |

Let e know that

g(x) = we

F Ig) =
Fe - VE
2.110 ENGINEERIG MATHEMATICS
By formula

Sf)s)dt - fFe lf(r)] F, ls))ds

0

V V| ds

-(a+b)x a- )| ds

- (a +b)x" 1
s
(a +b)

1 since e= 0
ds
=1
ds

(ie.)+he +) ds *

2ab (a +b)
Example 2.5.8: Find the Fourier transform of e , using

Parseval's identity show that dx

Solution: F[e"l] =

V 1+
already proved.

By Parseval's identity.

SF ds =S1sa) |de .(1)

Here f) = e"and F (6)

V
="
 FOURIER TRANSFORM
2.111

) ( ds (ekI,?dt

ds 2 (e-x dr.
0

00
1
(1+s)2 0

10-11-
1

z =
s is dummy variable

cos x, x|
Example 2.5.9: Iff(x) =
then find the Fourier
o.IH
cosa /2
transform of f(x) and hence evaluate dx using
o(1-2
Parseval's identity.

Solution We know thal

 ENGINEERING MATHEMATIC
2.112

F/ Ssme"de

cOs.re" dr | f() =
cosX, <
1/2 0, otherwise 21

cos x| cos sr + i sin sx ] dk

/2 T/2
cOs sx cos x dx + S i nsx cos x dr
V2/2
a/2

2 Jcos sx cosx dr + 0
0

Since cOs sx coSx Is an even function

sin sx sin x is an odd function.

cos (s+1)x +cos (s-1)x|

2 dx

n/2
1
| cos (s + 1)x + cos (s
-1)x]dr
1 Sin (s+1)
sin t/2 sin (s
,
-

1) z/2
S+1
S 1 +0)
- (0 +0)

Sin
2 S+1 sin-3 S-1
Sin

V27 S+1 sin-2 S-1

COS S
E
2 COs
2 7 S+
S-
 2.113
RANSFORM
UAIER TRA
FOUp

cos 2
T +1

S.
COS

S T
COS
2
F)=
*|1-A
By Parseval's identity

F P s SIfPa =

Cos t/2
2 ds = Scosxde
(1-s2 - t/2

Cos 7

( 1 -
2
2 s
= 1+cOs2 22
fltcos

/2

cos257 n/2
2
ds =
2 S 1+COS 2 [. even
2
(1-s22 0

cos t/2
2
o (1-s
ds= J(1 + cos 2x) dr
0
2.114
ENGINEERING MATHEMATICS
=
0-(0 +0)

S
cos
2
1 - 2ds

Cos2
2
[: s is dummy variable]
Example 2.5.10: Using
Fourier
00
sine transform, prove that
da
a+1) b+*) 2 (a +
b)

Sol. We know that

F[ea] =

We know by
and
F [eb
Parseval's identity
-V
SF 6) CG, (6) ds
=S f)s)dr
0

SFea" ] F, [e-bx ] ds =
f eax-bx
0

ds=S e (a+ b)x d

0

(a + b)x]
ds
(a+b)
FOURIER TRANSFORM 2.115

2
-0-
T s is dummy variable]

1
d a+b
7 +)+h
d 1
(a+1)2 +1) 2(a + b)
Example 2.5.11 Using Parseval's identity of the Fourier cosine

transform, dx, if
evaluaex(a+)
0 x(a+

f) 0, /x| >a> 0 and

g)= ea*, a>0.
1, Jxl <a
Sol. Given:f) =

0, xl>a>0
We know that Fe U)} =Fe (*) - V Sf«) cos sx dr
0

cos sx dx
0

sinsx

.V; Sin sa
S

Given:g (x) = e a |x

F.ls) V? 0
sx dx
2.116 ENGINEERING MATHEMATICS

.V 0
ax
cos sx dx
since in the interval |x| is + ve

by formula Se cos bx dr =

0 a+2
By using Parseval's identity

SFLf) 1 F. l s ) ] ds = S f ) s ) *

(V (VE ) ds Seax d
0

a sin sa

Sinax
F - e -1]
. s is a dummy variable]

Sin ax

Example 2.5.12: Using transform methods, evaluate S dx

o (x+a)

Sol. We know that F. - V? 4

By using Parseval's identity
FOURIERTRANSFORM
2.117

dr
0

ll-lo 0

a dr = eZax dr

-
dx 1

dx

transform methods, evaluate J dx

Example2.5.13 Using
+a2
where a >0. [A.U CBT Dec. 2009]

Sol. We know that F, [eax] = V

+
By using Parseval's identity

( V ) ds - /oy
0 0

ds
T +
2.118
ENGINEERING MATHEMATICS
eZax
dr
o+a2
s is a
dummy variable|

( d= 4d

Example 2.5.14 Verify convolution theorem for f(*) =

g(x)=e*.
Def: Convolution theorem for Fourier transforms. The
Fourier
transform of the convolution of f() and
g() is the product of their
Fourier transforms.

Ff s)] =
F f)} F
is))
Given f ) = 8*) = e

We know that F
)} =
Ss)e de
-isx dx

dx

V27t
 ANSFORMM
OURIER TRAN:
2.119

put=r-

dt d r

dt

put = u 0 u»0
2 dt= du | t o uo

d -d

2 Vu

- 0 ,
-in

1-s/4
FU))=Ve
Fls)] =
F|f)] =
4
:
Flst)] . Fs )] - - ()
2.120
ENGINEERING MATHEMATICS

*-u) du by convolution definition

f) &() V2 7T SSu)g
=

4 du

du

S(-du
put u-

dt=du toco

e2
V272e
0

put =y | t 0 y>0
21 dt=
dy y> *

2 vy
FOURIER TRANSFORM 2.121

2v dy
0

-xA
y
0

: S e / 2dt =
V27T V2
0

F If)'s)]
=
F 2

-FI
We know that e** is self reciprocal under Fourier transform.

. Fle2=e

: Flf) g)] - 5*2 ()

(1) (2)
Hence convolution theorem verified.

EXERCISE 2.1
I. Find the Fourier transform of

. sin x ,
0 <x<1
S) =

0,Oherwise
2. Sa) = ikx if a<x. <b
0 ifx < a &x >b

3 a) o ifOtherwise
0sx<k1
0