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156 views40 pagesFourier Transform Unit 2
Uploaded by
Tushar VermaCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
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/ 40
9
Fourier Transforms
1. Introduction
Fourier Transforms are use
sathematies Physics and &, ae Ree ao fad 2
or of energy si r weful in the study of the spectral
caracter of energy signals. Fourier transforms s, i
differential equations, orms can be used in the solution of integral equations
Seful Working tools in treati
‘nwineering. It is particularly
2. Fourler Transform [or infinite Fourler Transtorm|
et f(t) be a fi i efines
Let f lunction defined for ¢ « (~», 2). Then the (infinite) Fourier transform of
fit) is denoted by Fi f(t)} and is defined as J fe "dt, provided the integral exists. Thus,
j fit)e dt
ft) =
nn
Clearly this integral is a function of s and can be denoted by the symbols /(s) or simply by
fis),
Fo =FUfO)= [foe Mat a
The inverse formula for infinite Fourier transform is
[ime as of)
1
=F"1{ Fls)}=
ZA =F" f(s} ae
Note. 1. We can also use the notation f(s) for the Fourier transform of the function f(x).
2. Some authors take the coefficients of integral in (1) and (2) as - , - instead of 1 and
x’ Vr
2, taken by us.
3. Some authors define Fourier complex transforms as
Fis) = froe at 22 SPECIAL FUNCTIONS AND INTEGRAL TRANSFORMS,
and inverse transform as f= z fire “ds
2n
Infact, both the definitions lead to the same conclusion.
4. If /(0 is absolutely integrable, then the Fourier transform of f(t) defined by eq, (1) is called
Spectrum or Spectral density of ft). |F {f (#))|is called energy spectrum of f.
9.3. Fourier Sine Transform [or Infinite Fourier Sine Transform]
QD The infinite Fourier sine transform for function f(t) where, 0 < t < © is denoted by
Sf, (tO) and is defined as
CF =F, (f= Jrosin at dt (3)
3
‘The inverse formula for the infinite fourier sine transform is given as
Ft = Po) <2 J Fils)sin st a wd)
5
9.4, Fourier Cosine Transform [or Infinite Fourier Cosine Transform]
FS) The infinite Fourier cosine transform for function f(t), 0 0,
Proof. (1) We know that 7(s)= fe fix) dx A) o4 SPECIAL FUNCTIONS AND INTEGRAL TRANSFORMS
Now , Fiftax)) = j © flax) dx
Put ax =f so that adv = dt
ripen = To! po dat
(ax)) = j e :
=i fe (3) dt
ae 7(2)
a'\a
@) Weknow that 7, (6) = Fy(fe)= f fla) sin sx dx
3
Now, F (flan) = | f (ax)sin sx de
3 :
Putax=t sothat adx=dt ae
Fiftax)) = f eosin (o£) [te a>]
é aja ”
f(e)sin (2) dt
If F (6) is the Fourier transform of f(x), then show that e~* F (s) is the Fourier
_. transform of f (x - a). (K.U, 2012, 11; M.D.U. 20081
Proof. Fifte—a)) = fe flea) de
Putx—a=tso that dx =dt
Fifa) = fee Feat \URIER TRANSFORMS
Po
= fem emp ae
7 Je F(t)dt <0 " Fis).
f is i “
If f (s) is the infinite Fourier transformation of f (x), then Fourier transform of
7 15 1s
f(x) cos ax is given by ght a +3 Fiera). (KU. 2012}
Proof: FI f(x) cos ax} =
J e'* f(x) cos ax dx — (By definition of Fourier transform)
eT ese (eit eerie i
2 ve a | reas
x Jetere Fle) de + 3 feterms flx) dx
= LFe-a)+Foerar.
SOLVED EXAMPLES
Type I. Some Examples Based on Fourier Transforms.
Example 1. Find the Fourier transform of f (x) = e~*!*|, where a > 0 and x € (— ®, ~).
LEG (ay @. «
Solution. Here f(s) = [fe de = fesiles de
5
= -alz| p-ise a |x| g-isx
Jesters de + fe el de
= fee ie dey ferterans a
" d 4 26 SPECIAL FUNCTIONS AND INTEGRAL TRANSFORMS
© asst”
& le 2. Find the Fourier transform of the function
or fey {hl Oar ce ~
0, otherwise
Solution. 7 (6) =Fif@= [fade ™ de
= [reve de + [roe ax + free ax
-= o a
= foe ae + fret acs fore ae [Given}
= ao a
04k [eax +0
é
-.(22] .
—# Jye0
+ ,
=F foro) = (e-isa_y),
is 3
Example 8. Find the Fourier transform of the step function
van
| Binet! pa
21
0, |x|>0 po
. _p -A 4 &
Solution, FI/@) = [fe fz)de *e
fe) =
+
1
= few Fladde+ fe fa)dx+ fe rade
: i
[irene ays foe aes oem de
4 i oust ‘TRANSFORMS
97
dx +0
. Jo. ays fo oa
a ample 4. Find the Fourier transform of the function ky de
(-1, -a
dle { 0, x<0-
Solution. f(s) = F [f(x)] = : Flee dx
= Trove aes] pee a
a a.
°
Joe ars fre
-s °
=0+ frcetems PS Untegrating by parts]
. 7 00
J, cue [e - 0}
Laid |
— :
etsins
je ari?
Lim
1 x | 4 [
= tis sso ler oi | 14 ig)? [x
1
ae
; x
~ hy he [un Fon raa|
+ 1
em | e+ is?
[+ e® = cos 0 +i sin 6]
~— 4, Lim
asi? 38
Lim oumims|
Ese @F x cosax+isin ox
t
ae | i ‘so
(+is)? |=>* ¢F (cos sx tisinss) | * ayin)
1 1 ae
a -0-~—> 941 _
l+is (1+ is? (1+ is)?
Lim © = Lim © = ping 1 o|
roe So (ey mS
1 upléR TRANSFORMS
K
a Ee
99
exercise 9.1
' 1 zen
L rear eentens reine He eo 2
Oy (xfer
uu : = ¢
ne
Ka! pS a 0G eet
| fe? dada, | “ i
{ 4 trove hat he Pour tanaform af pp ( F\i0
2
=F). 7 FO)?
t
1-=, for O uo : :
Hp 7 UPL oor U7 Uy
ANSWERS
eilo-ab _ gilw-sia
T= s)
Gp Bsinuos iy Ziups? Dvn ge + 2Hgs cs 081
= ‘ ee SPECIAL FUNCTIONS AND INTEGRAL TRANSFORMS
be usit ese results
Note. Students are advised to remember the following, results. We shall be using thes 8 in
the following solved examples
pola sin bx +0)-b eos x0)
ue fet sinrsede =
fa) [c™ costbx +6) dx =
J a +8
From these it follows that
{acos (bx +c)+ sin (bx +0))
i eo meaas)
fewsineeds she fe het Tenn)? 2
y P48 2 eptcius com Lim bt) aa
4 © cos bx dx a 4 »
i“ J Fae 9 B74 42
Type Il. Some Examples Based on Fourier Sine and Cosine Transforms.
Example 1. Find the sine and cosine Fourier transform of 2e-% + 5e~*,
~ Solution. (i) Here f(x) = 2¢-8 4 5¢-2
Now, Fils) = [fe rsin sede = Ff 00)
0 . -
Fs) = [Qe + 5e-%) sin sx dx
é
so. «
=2 fe sin Sede +5 fe* sin sx dx
o “0 °
- s 8
© 2 a (Refer note above]
- [ 2 5
zs
W+s? S44
« {
@) Fis) = | (2e"* + 50") cos sx dx
3 YS DY NAA
unfit TRANSFORMS
a
Find the Fourier ne transfor
tu
T sform of
x
Solution. Here fy. 2"
®
£9) =F, faxy
Differentiating w.r+, s, using Le fer
8, using Leibni
ibnitz rule of di
— on
wrt
a, YOu dbte, byt
ay,
of differentiation under integral sign, we have
d (>, Te
ath) = [ :
a’
Integrating w.r-t ‘s’, we have
ee
() = a--tan* 5 4¢
aa
= tant S4¢
a
Putting s = 0 in (1), we get
7, @=0
Putting s = 0 in (2), we got
he (0) =¢
e=0
Hence from (2), we have
6 Fi) = tan =.
‘ .ple 3, Find the Fourier cosine transform of e-**.
Solution, Here f(x) =e"
Fis) = FA foo =I (say)
[Refer note above]
(2)
(3)
Ad)
[From (3)and (4)] ae SPECIAL FUNCTIONS AND. INTEGRAL TRANSFORMS
t= fe
é
Differentiating w.r.t ‘s’, using Leibnitz rule of differentiation under integral sign, we have
vos sx dx: (1)
ae = ri
ao)” sin ax-x dx ‘ -
Pa olay ep LLAT
a2 Teorey se
=a! 2x e*) sins de ’
ache a sy
* ds 2 (Using (D)
fret a
a ad Funuion | ds (8
BR Fe Ot) I 2
Integrating Wrt's’, we get
or
or
(2)
o
Putting s = 0 in (2), we have
Isc ip TRANSFORMS
nt
ot!
From (2), Ie
plence the result.
example 4. (0 If f.(s) is the Fourier sine transform of f(x), then prove that the Fourier
cinetransform of f (2) is af, (as), where a > 0,
(ii) If f(s) és the Fourier cosine transform of f (x), then prove that the Fourier cosine
sransform of f (2) is a f,(as), where a >0.
Solution. (i) We know that f(s) =
= resin sx ae CQ)
i -{r(B}
Put X=1 sothat dx =adt
a
jin
‘Thus, we have F, {r (2) =a | fw sin (ast) at
3 ion
=af,(as)
(ii) Proceed similarly as in part (i).
6 3 Ble : sins, O<
\ Bxampleg. Eisdthe sine transform of the function f(x) = { 3 2 a
3 ig
) Find the Foorer gine and mine ans 78 =x. Inm= oe
b, ocece F2= ( A
x asxsb.
On =>b
3,_-Find the Fourier sine transform off) #£ f@= |
SQ
cos x, when 0er0
Hence from (2), we have
fe) = 2 tant =
ar
Deduction:
Puttinga =0in f(x) = 2 tan =, we have
ne Oe |
FOURIER TRANSFORMS
() 3 tant on)
|ava 0 * a}
=o
x
x
Fay
loo BES nay
he Fourier transform of Fla) = {3 for |xjc1 -1444)
. 0 for \x\>1 >
Hence evaluate jm ide IM.D.U, 2012, 11; K.U. 2011)
38
Solution, Fi fq James dx
“1
fre oF dee Ire dx+ { Fla) e™ dx
> AA i
1
0+ [fre devo
Ue f@) =I for |x| <1]
= 28iS for's 20
s so
1
for s=0 5-0, | fea
; a
FUG) =
2sins) i 540
Thus,
if s=0
inversion formula,
Now using 9.22
or [PsB8 oe ds
.
ie, [P22 0% as
Putting x =0, we have
or
or
oY 2 Ghoti the Fourier transform of
fe = nae [eeilter x3, |x|<1
and hence evaluate J ( Zoos ®
é
ae Iat
SPECIAL FUNCTIONS AND INTEGRAL TRANSFORMS
sin s
=f js e ds,s#0
-a{h for |x|<1
©, for |x|>1
Newest
0, for|x|>1
ae
Gs integrand is an even function of s
we
MDU
“\A%41
% Ikl>t -|>%>I
5 °
SS
ee lo
eS ow tae 60 ' “p.0. 2007
2
Solution. Here fore {ie pee
0,
Now, |
F(s) =Ff@)}=
*<-lorx>1
fe* reas ESSEX Grade by NAAC
FOURIER TRANSFORMS, |
= Je ronae Jeu
-5 Je" rndes few pods
~ - ®
elayet F /
[a He 2s ae
‘
(-is}
(By general rule of integration by parts]
(o ation (3)
rae
t peck
Ascoss +4sin s] = 4,[sins— 5 cos s} 4)
Deduction :
Weknow that fG)= 2 f Fie as
on *
{By inverse transformation}
~it4 jes
fa) = 5 J isin scons ds (By)
Reversing the sides, we have
+] Aisin o cose ds =f)
an)
1 4 Gi tex*, |al0
xe Lex? 2
ons Find the Fourier transform of f(x) defined by
FQ) = {
Vv
QB jee . (M.D.U. 20071
Now, Fif@) = fe (F(x) dx
= fee pedes je“ fonds fo nd
ett we
is
_ oo Sinas rot)
ie, fi) = anes
Deductions : i .
J Flore ds Wefinition ofinverse transform)
(i) Weknow that fa) = FZ)
— anpey= J fre ds
(Using)
1 oe eit ds
o myers 1
—— | % EGRAL TRANSFORMS
228 SPECIAL FUNCTIONS AND INTEGI
1 if [xfea [By reversing the sides)
or ds = 2x .
0, if [xfoa
7 2sinas [2m iflela
Equating real paris on both sides, we get
[sina cos 2 {[m if lela
(‘Taking x = 0 and a = 1 in result of part (i), we have
ee ds =n |0}<1
3
or 2 J So [Property of definite integral]
s
4
or “ sin s <2
[meas 2 Ay
Example 7. Solve the integral equiation
J f(x) 08 ox de
4
Solution. Given { f(s) cosexdx= 7 (9) = { 1-8, for Ol
By inversion formula,
fx)
2 | Fle) 8 sx ds
i Oo FOURIER TRANSFORMS 9.27
2|' «
<3 J F.(s) cos sx ds + J 7.18) cos sx ds
i
1
=2
= 2 Joeman +0
| Untegrating by parts}
= 2(1-c0s x
= (|S).
Example 8. Show that the Laplace integral
o -fu)
Sin ee gs, x>0, @>0 is equal to © ea,
dass? 2
Solution. Let f(z) Se", x>0,a>0
0
Fy Uf) f(x) sin sx dx
4
e a
Ir
J
[: fe sin bx dx
é
By sine inversion formula,
)
=F 3
fo = E355)
-2 ft se ain sx ds
aia
3
a 27 ;
s e a on Grin sx ds
“Tssin st gy. Lm nu yy
on Sat = Fe" .2>0,a>0
ao 9.28
HONS AND INTEGRAL TRANSFORMS
WAL EU
EXERCISE 9.3
if's< 2a
1. Find f(x) ifits cosine transform is /.(s) =
if'se 2a
a, What
2. Find the cosine transform of a function of x which is unity for 0 8 (n +1) 0, where tano = * & fix) ——
nite a noma x?) FOURIER TRANSFORMS,
29
9.6. Convolution (Definition)
“The convolution of 4
wo functions
by Fw and is defined ag "HOM £0) and K(x) over the interval (.#
a
1
Also, fg naap
follows
v so that du =— dy
>) is denoted
which can be easily proved as
In (1), put x-us
Fr@e = | pe - gua)
J a fe-vdv
=@*f)(x)
& lence, fegagep
Convolution Theorem for Fourier Transform Fite
If Fi f(@)} and Fig()} are the Fourier transforms of the functions f(x) and ©
42) respectively, then the Fourier transform of convolution of f(x) and g (x) is the
product of their Fourier transforms. $
ov Qols
Proof : By definition of Fourier transform, :
FI (x) * g()]
Je™ F@)+ ao) ae
1
(By def. of convolution)
845 ey) wd
= jr] fe exw) al
(By changing the order of integration)
= [rel Jon g(x —u) ew tx
= fem rl fe FOS Bx) ax
Lu = SPECIAL FUNCTIONS AND_ IN’
fe ‘tn git) dt
du
fe Flu)
| Putting x —u = t 80 that dx = dt)
fe ‘ f(u) du-Flgit))
4
= fet paydx- Fiat)
=FIf@). Fle)
a8. Fourier Transform of the Derivative
‘ Theorem 1. If the Fourier transform of f(x) is f (s), then the Fourier transform of
f'(z) is given bylis F (s). ALEC) - fls
* Sf
Proof. FC) = fe fade a ‘
=
po] - fois) pyar
=O+is J e f(x) dx, provided f(x) > 0 asx > +0
Sin 2. (Extension of the above result)
e an pone
f(x) and first (n - 1) derivatives of f(x) vanish identically as x — + ©, then
“| =U" Fi).
Te (ig Oe
Td is Tat
> Oasx>tx
af . er
i yay oe dx" 7 FOURIER TRANSFORNS
“.
= Gay? pa 2e)
a
In general, applying general rule of
_ ee . integration by parts, we have
=e P'3) de
ot “e] = [emer escialymt fomigt) yo
a De® (is) (Fen) | = +n" Jercisr f(x) de
As f(x) and first (n ~ 1) derivatives +0 asx + +0
{ee =(isy Fs).
Theorem 3. If the Fourier sine and cosine transforms of f(x) and f’ (x) exist and
fx) Ove then
(f'@)] =-9 Ff)
Se [f@1-FO.
Proof: (i) F,{f" Wl = J F'@)sin sx de (By definition]
a
= {i sin sx f(x) dx Untegrating by parts]
é
= [sin sx. f@)]!") - J scos ex. f3) de
3
=0-0-s J F008 sx ae [- f@) > 0as z+]
é
Hence, F,{f"ol =F F00 SPECIAL FUNCTIONS AND INTEGRAL TRANSFORMS
922
7 [By definition}
wa) FL GI] = | f'C2) cos sx dx
é
_ j cos sx f'(x) de Untegrating by parts]
= [eossx. f(a)]" 5 - J -ssin se. f(2) de
3
=0-(c0s0).fO)+s ffG@sinseds — [v fla) >0asx>a]
é
=-/(0)+sF, [f@l
Hence, FL f"(@)] = F, #@-F).
Note. In th¢ similar way, we can also prove the following results
N ye Ff" @)) =-s F, [f'@)) =-s Is F, (fF) -F0)
SF, [fG0]+8 (0)
Ff" @)) =sF, (f'@1-F'0)
=s[-sF, {fiwl-f' 0) =-s? F, f)) -f' (0).
elation Between Fourier and Laptace Transform (M.D.U. 2011]
eo, t>0
7 tK<0
Define a function f(t) such thay’ f(t) = {
Fife) = fem fat
° 5
= fem rma fem piorar
Es d /
° s
= fem odes fee onrae
a .
= fered
j
= fe“ omar, where x + iy =5
4
=Li ow)
Hence, Ff@) = Ll 9 (). b
y
FOURIER TRANSFORMS
If
@
Gi
Parsevar’
'S Ideny
Y fo
(s) and GO) de, nek ourler Transtorm
eu
‘ou
a Ir _ rier transform Of f(x) and g (x), then
(8) Gis) dy |
[re ate
re fire den fire an
where bar indicate,
the com,
Plex conju,
gate,
Proof : (i) [roe ae
(ii
= 1
= J Fx) {x i Gis) ot alas
By inverse transform for #(x)
= tae! 7 =
- = 32 J G@) [rere axbas
[Changing the order of integration]
"
175
On J Gls) - Fis) ds [By definition of transform of f (x)|
) Putting g (x) = f(x) in part (i), we have
= JFPOFeas = fre@f@dc= fir@p ax
= -o 0
or & fireras = JiscoP ax
Qn s a
9.11. Parseval’s Identities for Fourier Sine and Cosine Transform {K.U. 2011)
eerie denote the Fourier cosines transforms of
QOL. If F,(s) and G,(s) resp
f(x)'and g (x) respectively, th
] ) de
2 7 F,(s):Ge(s) ds = [reece
na 3
= {[pol? a
Fear]
3 4 SPECIAL FUNCTIONS AND INTEGRAL TRANSFORMS,
Proof. [ fix) giydy = [rvo| fre cow are} et
x
j J
By use of inverse cosine transform|
: 7 ;
2 how | J Ft 008 5x ala
® 0 0
(By changing the order of integration)
"
2 aw Reds
™ 0
(H) Putting g(x) = f() in the result obtained in part (i), we have
2 Rie ds = Tray?
= JiR ds = [treo ax,
™ 0 0
9.11.2. IfF, (s) and G,(s) respectively denote the Fourier sine transforms of f(x) and
& (x) respectively, then
Nw 2 Trwe,eds = [fe ge)de
j 3
067% [or on? ds = [UpGo ax
é 3
Proof. Left as an exercise for the students,
SOLVED EXAMPLES
/ Example 1. Find the Fourier sine and cosine transforms of the function em", m>Oby
fing its second derivative.
Solution. Let fi) =e-™, m>0 KY 201 13
f'@)=-me-m
and f(z) =me-m
To find F, [f()] ie, F,(e-™*):
We have Lf" Gl =—s? FFG +s (0) [Refer note after Theorem 3, Art. 9.8]
; F, (m® em) =— 52 B (em) 4.5 60
i mF, (e-") + s?F, (e-™) = 5
- | F(e-m
can FOURIER TRANSFORMS
Fo" Gos
PRY £0)
Feb em) wh Bom
or
Come)
MIR (0-4 Kg ™=m
|R(e-my. om] mea
oa | my Dram
A Brampe ne Using Parseval’s identity, prove the following
dx x xidx x
ie + 4 :
a i
(@ Parseval's identity for cosine transform is
——Seval's identity for cosine transform is
2 Tor ds = fIroor dx
Taking f(x) =e-*, x50
FAG) = fe* cos sx de
. é
= pb eos ex osinsel?
_
9.35
[Refer note after Theorem 3, Art. 9.81
Roe cue
IK.U. 2012)
4
[ Ay
ft)
wor
wo
vw
=
ars "4
f dx
Gy gee? 4
(Changing the variable] IAL FUNCTIONS AND INTEGRAL TRANSFORMS
ia = Jurors
Taking fw’
2 ‘we have
fo‘ sin sede |
é
Fs)
|
a
“Tes CO Te
From (2), we have
£ = fer)? de
ot
or 3 j =
é
=o
4
Replace sox
Hence, dx a [Changing the variable]
EXERCISE 9.4 |
i, Evaluate . ,
t(1-cos x)? sin? x
@ Jl 4) ae oy ards
& { = “Gasp sin ax
[was 2ab (a+b) a) 1855
é
[ne Use Parseval's identity for sine and cosine transform of (f(s { : Ose<1 |
eaet
NDU 201%
Using Parseval's identity, prove that
x(a? +x?)
[Hint. (@) Take f(2)=e-" and g(x)=e-b :
1, Oa ¢
fin Take f(x)=e™ and sore { :
FOURIER TRANSFORMS
identity show that fea
\MD.U. 2008)
. 0K x. 0 MDU oe é
4. rae {% L letes a ju, od
Axbs a ‘®) the Fourier transform of f(x) be 28if ee 19-0, then prove that
AY
grax. convolution theorem for f(x) =g (x) = ¢ *?
dy = 74
2
L Og o£
WW
gp Ths rhnte s fe and Cosine Transform of f(x) _ Simm = O wn
Definition. The finite Fourier sine transform of f(x) where 0 < 2. a bonded by
F, (f@)} or 7,(s) and is defined as
;
Fire) = 700 = J reersin 22 ae
a
where s is a positive integer.
Function f (x) is called the inverse finite Fourier sine transform of f,(s) and is given as
7 =F EF} =
Definition : The finite Fourier cosine transform of f(x) is defined as
Fu) = Fe (FO) = [rene Ba
where s is a positive integer or zero and inverse finite Fourier cosine transform of /,(s) is
given as
=F (Fw}=2 Ao+2
Fe =F (Fe)}=7 0+ ee Fits) cos
1A
Above formulae are obtained from Fourier sine and cosine series. Students are advised to
Note.
remember these formulae. 9.38
SPECIAL FUNCTIONS AND INTEGRAL TRANSFOR45
SOLVED EXAMPLES
O/cman 1. Find the finite Fourier sine and cosine transform of f (x) = 1
Solution. If range of x is not given, then we take the usual range [0,x]_ KU 2016
F,(s) = J feo sin
é
ieee
4
“cos (nx +8)=(-1)" cos6
sin(nx+6)=(-L)" sind
f(s)
6 Fas) = ff) S28 ay
3 *
= 1.cos xs dx Liam
3
- [= ‘|
°
sinns) (sino ae
= (7 —}-| 57] <0-0=0.
Example 2. Find the finite sine transform of f(x) = 2x, where U