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67 views16 pagesFourier Transform
Uploaded by
PriyanshiCopyright
© © All Rights Reserved
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Available Formats
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/ 16
——-.C-UhUlUC CO
‘The Fourler Transform
“he inves formu for Four ansorm is given by
o2 relia anon EB
THE FOURIER TRANSFORM Fey = ENS5PR= Se STFU
Part I-The Infinite Fourier Transform |
1 t0)=F (R= |, Ano
‘The inverse Formula for infinite Fourier sine transform is given by
mains (D 2 conetsion
Famke! Gora? fy 10 sinsede Se ER = 2 (Sythe
By this definition and the definition given above Tead to the same
Relationship Between Fourier Transform and Laplace|
(Kanpur B. Se. 2003; Meerut B, Sc. 2002) ater
‘We define a function F(s) as follows
‘or F, {F(x)) and is defined as,
|
= | paletMy . 20
- © sa (top tibe > FOS moto tee
Lo=F, (Ft) =|, Pecos sede -
ep sees oye rirose] roe
= eye” (Rehan
Fay=F! von =2 ff son xD © BAMA ces yt ay corner] ore
(Meera. Se. 2002; Kanpur B. Se. 2003) 3 :
Remark a [em one eo em ane
> / Some ators take SER, TE 06 tho cocticients In the two
equations fel. Del. (2.0) and (2.1 both} insload 1 and 2/n|
respectively
EEZEY Det. The infinite Fourier Transform of Fi
= a_ Gurwkut Kangri, Haridwar 2001; Kanpur 1997)
Solution. Given that
_[1. -astse
0-{0 ie
=f Rey
a
Hg diye
tems
nO eae nian)
= 33 (as c0s sa sin 30),
C/#F
(Kanpur M. Se. 2004; Meerut 91;
‘Agra 81; Roorkee M.E. 67; Purvanchal 95)
rua=[ em
litre elder
rf) de
‘The Fourier Transform 199
=f nna =
afl #7 0.p+ sf soe 2einse=tiinse Am,
gol r any
Pipe 2. (a) Find Fourier transform of
Oo ele
(Meerut B. Sc. 2000)
Solution. run=[ chpyanl, Oe de
see em
ferme
= sin as 2 cos sa)~ in.
phil? (b) Find sine and cosine transform of '«-.
(Kanpur 1981; Agra B. Se. 2001)
Solution, fx)=sten®
Fu@i=ko=fy foe e ae ao
Fptoimhor=[, e“tsinipn de
We know that f Bede 0)
a y
af eh te) dew
Taking + ip=re®, wo get
cm nLcaE
peep eee
’ WE
ante ads _ a1 sun
Salat OA ao ie Integral Transtorms
Peles tt
Ans.
aie Pe eas
(Agra 1973; Meerut 1973; ee 70; Roorkee 1966)
Solution. First Part. To determine F (fs))
For tis ee he above solved Ex.2 ie Sy 2 Sink
Second Part. Let FUfix)] =fis).
We know that if By tases Fence Fret
jo=rym=[ pea ped, (uray
—————
- fay=F-! fs)) = af Baye ds > JP fap ano
5 Peas an if
[Line weeny a wate,
= 282 by Gre par.
at Fis)
208 opr +i) dof
i
ee
s 0 it isive
5 beh) J
o unsacu st 4, ;[ sinsasins 4 {x , Ixica
gle eeeacl ise
Bie pas os eb as,
e fi sesctee gfe i tee
5 0 if Ixi>a
The Fourier Transform 195
‘Third Part, If x=0, a= 1, then the second part gives,
f. SL yan, (Forixi2. (Kanpur M, Se. 2002)
Pie reste (apse
bs x for Vexel
“7 rst
ieee eee [ote an
feceApcn ee eee
A edoded i
Leese] + nef (ease a
elie * pI
2055 5 55) 2 og ay 2 cog yy 2008 28—c055_/sin2ssins
“(CS ooze : ( . }
_2sins=2sinscoss_2(1~coss) sins
rr ? Soa
Problgft. 6. Find the Fouer transform of
ra Ge)
0. ixl>t (Allahabad 1967)
and hence evaluate
[fata
(Kanpur 1989, 82; Meerut 71; Purvanchal 96)
‘The Fourier Transform 197
Solution, Let F(x)
Also let F LF) =F)
First Part. To determine Fs)
For this see the above solved Ex. 4,
Second Part. We kaow that if
=f
Fo)=F de
then Fo)=50 Ate Foye ds
Bu Fiy=4Cscosstsing)
Using this in (2),
i A cescoss sing dds =2n Foy
sof Cemmecssn dines ty
Equating real parts, we get
tore —— $8.) Gos sx. ds=¥ Fla)
ie af fees ous
0, Ial> i.
1
Putting x=5 + we get
aff Eesosssing
1
ale
a
ol
creme sing 4,
=i
c (=x cos.x+sins)
os lo ?
oe i (Bete)
fl
sig vit
g
ii
i
&
i
alg ale 2A
at
mre
:
siz ae Integral Transforms ‘The Fourier Transform veo
‘constants, then
‘Theorem 1. Linear Property. If ¢, and ¢2 are arbi aI ‘aF() 00s (sat) dt = of, (as).
F (cy FG) +0 Ga} =, FARO) +62 F (GO) =
Proof, F (ey Fla) +e) Gl) ‘Tacorem 4 To prove that
: wenn
= fey Fa) + ¢9 Gay} dx
pp reraree | “The proof is left as as exerise for students
{ IE a tee | ‘Theorem 5, Shifting property. If fls) is the Fourier transform of Fl),
=] tee Ft eg CE) de | ‘hen
| = j 7! fs) is the Fourier transform of F(x—a).
foae* rar] ae omar Pee Figeara (mene
we] ct rmare |, eM cae
ee = el pe a, x-a=t
=e; FUR) +09 F (G0) Proved. -
‘Change of Scale Property : fem mane | Mand
Theorem 2. If) the Fourier wanform of Fs hen (3) ste earn Pree
| 6 Modulation Theorem. If Fs) has the Fourier transform
Fourier transform of Fa) (Kanpur B. Se. 2002; Meerut B. Se. 2003; eee
i Gara Kangr Haridwar 200, | BG) con Oe Dat Sa.
meen = 24) ee (Kanpur M. Se. 2003, 2004)
e Proof. Recall that cos 0= —
> Pir =[, € Fe dempo) oye 2
Sar F (Fla) cos ax) =|__ F(x) cos adr
2 riran=[.e rand=[_ Emp text ie
ise tee
=f etn aa
A jeeceee nites)
; 1h p010) by detinition Proved. Jen ieee es dima
Theorem 3. If f(s) is the Fourier cosine transform of F(x), then oot -
d ier cesne ransorm of Fl), then show eS Ji cte-oe roar f elU+Ox pa de
‘hat Fourier cotine transform a) eee
| e ’
s =!Y-a+fs+a). Ans.
Proof. ee = 2
fe fr Gi} -f *(i}os axdx | Problem. State Modulation theorem, (Kanpur M. Se, 2003, 2008)
is ‘Theorem 7. Det ye Theorem. The Fourier transform of F’(x), the
= Jy Fi cosas ade, Ep
derivative of F(x), is ifs), where fs) is the Fourier transform of Fix).
| (Kanpur B. Sc. 2004)
a ae ee ————————————— ee = Integral Transforms
Proof. By defiition
F (Ax) -[ MF) dx integrating by parts)
[roe [=f re eines
Boke (a norm aengateere
if).
FF") =i fo) if Fa) 9 0 as xt,
‘Theorem 8 (Extension of Theorem 7). To show that
+ {et ar A), where F (F(a) =19),
I the first (a1) derivative of Fs) vanish identically as x0,
(Vikram 1996; MeeratB. Se, Hons. 1973, 72)
root. Suppose the frst (n~ 1) dervaives of Fx) vanish as x40
Some (integrating by pars)
Similarly we can show that
a*r|
FLE cape
| | (oy F
Note. If we adopt the definition,
d*F\ [7 ar
fi ride jer
er fF ee me,
Proved.
then we shall get the eau,
FEE iat fe
“a
Convolution,
rf
Foe =| rin cic apde,
ee-= PWG -wya
The Fourier Transform eat
Theorem 9, Convolution Theorem. if F ({:)) and F (g(2)) are the
Fourier transforms. of functions f(x) and g(x) respectively ‘then Fourier
transform of the convolution of fx) and gx) isthe product of the their Fourier
transforms, ie.
F (i+ ela)) =F Uf) F (xt)
(Raj, 1983; Meerut 82; B.H.U. 72)
Proof. F (ffx) * #60)
= [Wo ain oa
(On changing the onder of integration)
& oe (ee
ChE encom eaecalene
2 [toa ema
-[
am fe
=F (Uf). F {9G}.
Genter tr tas of Be Rilowing dete tees
du Fig)
+e nat
eran Ce
a
ye LZ} ee
ie ofa le
(Kanpur 1990; Kashmir 64) [Pf ospt dt =e PO f
-b
yee
tyme ER) = (rr ris cosptett Compe =) et
- Integral Transforms ‘The Fourier Transtorm
202 by innate Laplus bash Fler” cospes wah =
° oh
eels E Kor [7 cosde
Sotution. fy fa) on sxde= a es “ee or gel mae Proved.
2 rae Problem 9/ Solve the integreal equation,
By def, FUO)= I fis) cos sx d= aaah. =) me 1-2 for OSAS1
and. =no=2[, Fuoreos sx ds. (3) (Kanpur 2003; Purvanchal 1996)
‘ Solution. By def,
Equa ma -
2
m=5. ‘Then
a = F The inverse formula relative to
Ef etonnaed ts [eel commen zsy]
Z
Mas | From which,
yblem 8. Sheee - ’ ~
are {esrb ap=-net frw=f FO) cos ix
Pe ne |
(Roorkee ME. 1967; Purvanchal 95) ¥v conncanef ee ce as
Solution. We know that esata
mat a J a= Dowie. cos Ae dh
J, «cos be=a et combat =n
a8 ae]
eae! il . Peoshrdea ts A (ettatat- 1 (ate 2) cos Ax th 0
+
Ser ates Pe faaee mann -ysts, fsa]
= Spee, 50-6 3 a
eli fo) 608 Pde Q (1-19 ind _ cose]
eee
Veet, Sie =
we get pe fae Fi < Sb: he aa ery
Horo by lnuarte Ce: aes _ 160s x
tha tris tarts renting fis, , ood 0 (ose 1) =F
a 2
Fe! GON) =. =n -2[) FeO) 008 da dh ea | or Fu) = S(t en he
eee a mu | Problem 10/ Solve for F(x) the integral equation
ee - 1, ostet
pete costap [P revanepacele | 151<2
z PE [0 , 122.
—
PA a | ge telig Syne dg =A |e tarp
| ar Trans al
are Integral Transforms | ‘The Fourier Transform 4° “ged a
| dete Ae
Solution. By det, | aff ePcotnte-t1 pene th -@
Fytr)= fp Rosinaxde=f0 0) a ac. Vee
tien) a1. gy
1, ostet | 2 §
lo | za |
; on rede. 62/4 A
‘The sin inversion formula relative to (1) is C- i (3 a © vt
~ when s=0, then T=}, &* (cosO)dr=)o e* d=.
| p= (ire. \ , pee
e Fors () ges =
| From which "ee)=[, 69 sin cas alad or AE
1 2 5 128k iM, Ans.
Bacio areata
1 2 es HPP ig ge? :
$2) = 6° 7 is Vie", (Agra 1982; Meer B. Se. 2001)
Seed anncelPacne, stn.
-~24] sa(ae) Pr | = [nota fo eth et gy
Fhe CF be | ‘
+cos x~2e0s ¥ =) eursiagtn
x + cos x~2 cos 2x 7 2, ; xtis
ory Leet teo [leet wa, wee Say
2
o y= 2 (1+c08- 20082). Ai ~ is |
PO ay ; = atl ty antete ll ory
Problems Related to Fourier Transform : 2 él
Poy 1. Find she Fourier cosine transform of e-” RT ane? Rachie, Ans.
(Meerut 1977; Kanpur 96, 94; Agra 82,75) | On date are:
Solution. We are to determine F, Samick Geept de Pun i eas ae |
Me res, ra
gfret’ Z
By def, Fe? aie oF agate det (say) Fghe rae + Problem 13. Prove the she Fourier monform of the step function
if er to" PY ef ets gi (an
a inv (F@)) =) 21
0
Pen ia
BE in
= fmt ence | ~ [eae fet®
Fag ce soe Integral Transforms
Solution, F ran=[,
-[lave was,
-[-rnerws[,
f oe oF |
ao NE Ba AE
us
$50, (Kanpur 1998)
4 Se) DF fe) + SF (oY
eee
ar
Be oe,
i) F, (24 50°}
enicrverienl eee stl
eres (ize 3
FP Fae [P4058 Pea
Ans, sine transform is s|—2— + —5.
P45 Ped
1
‘cosine transform is 10
L P44
Problem 18. Find F, (Kanpur 1976)
Solution. F,~' (
O™ sin sede
ae SS—~ 0 and x belongs to (-%.2). (Meerut 1970)
Daf coy tmayeen fp cht ta sein iie[4 255
os Taf, fi
‘ 0 » Tae[sortan[ eee
f = f em dee a @ ie f
[Equating real and imaginary pasts on both sides, we get
eee eons
j "tence de= ec) t . =
Tin) pi afar cerns
[for tsncanar= PO (7) |
- nt
Fae =() |
Fer sis «() Ans. ~
ii ae Problem 39 Find the inverse Fourier transform of fp) =<°'?"? where
Pte 2. Fd fon an er orem nea Sahoo
heed) boi aie
nonfses)eose mh
Solution. Let F, Uf)} =f.) FG) =% a. Fo) e™ dp
0-2) oreo anes |
Pe lieeimes efi b ann ar
ie ;
-2f, a(t} Pe
Pa sin saul,
alt x ie ales ie L 4 °
aA]! sin 20-2 sin aac +1 (908s |
“alt sei staat a(- is ia |
| 1
= (4 cos 20x) = Ste el =aglyor tte)
Sip I-00 209 = Am | anlye