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FOURIER TRANSFORMS
Introduction
Fourier Transform is a technique employed to solve ODE’s, PDE’s,IVP’s,BVP’s and Integral equations.The subject matter is divided into the following sub topics :
FOURIER TRANSFORMS
Infinite Sine Cosine ConvolutionFourier Transform Transform Theorem &Transform Parseval’s
Identity
Infinite Fourier Transform
Let f(x) be a real valued, differentiable function that satisfies the following conditions:
andinterval,finiteeveryinitiesdiscontinusimple
ofnumberfiniteaonlyhaveor,continuousarexfderivativeitsandf(x)1)
bydefinedisf(x)ofTransformFourierinfiniteTheparameter.realzero-nonbeletAlso,
exists.xfintegral the2)-
dx
dxexfxfFf xiˆ
provided the integral exists.
bydefinedisfF
bydenotedfofTransformFourierinverseTheTransform.Fourierjust theorTransformFouriercomplexcalledalsoisTransformFourierinfiniteThe
1-
defxffF xi
ˆ2
1ˆ1
Note : The function f(x) is said to be self reciprocal with respect to Fourier transform ff ˆif .
2
Basic Properties
Below we prove some basic properties of Fourier Transforms:
1. Linearity Property
For any two functions f(x) and (x) (whose Fourier Transforms exist) and any twoconstants a and b,
xbFxfaFxbxafF
Proof
By definition, we have
dxexbxafxbxafF xi
xbFxfaF
dxexbdxexfa xixi
This is the desired property.
In particular, if a = b = 1, we get xFxfFxxfF
Again if a= -b = 1, we get xFxfFxxfF
2. Change of Scale Property
have wea,constantzero-nonanyfor then,xfFfIf
aa
f
1xfF
Proof : By definition, we have
)1(dxeaxfaxfF xi
Suppose a > 0. let us set ax = u. Then expression (1) becomes
a
dueufaxfF
ua
i
)2(ˆ1
af
a
Suppose a < 0. If we set again ax = u, then (1) becomes
a
dueufaxfF a
ui
dueufa
ua
i
1
3
)3(ˆ1
af
a
Expressions (2) and (3) may be combined as
af
aaxfF
ˆ1
This is the desired property3. Shifting Properties
For any real constant ‘a’, feaxfFi ai ˆ)(
afxfeFii iax ˆ)(Proof : (i) We have
dxexffxfF xi
ˆ
Hence, dxeaxfaxfF xi
Set x-a = t. Then dx = dt.Then,
dtetfaxfF ati )(
= aie dtetf ti
= aie fii) We have
dxexfaf xai
ˆ
dxeexf xiiax
iaxiax exfxgwheredxexg )()(,
xgF
xfeF iaxThis is the desired result.
4. Modulation Property
,ˆ fxfFIf
afafaxxfFthen ˆˆ2
1cos,
where ‘a’ is a real constant.
Proof : We have
2cos
iaxiax eeax
4
Hence
2cos
iaxiax eexfFaxxfF
property.desired theisThis
.propertiesshiftandlinearityusingby,ˆˆ2
1afaf
Note : Similarly
afafaxxfF ˆˆsin21
Examples
0whereef(x)function theofTransformFourier theFind1. x-a a
For the given function, we have
dxeexfF xixa
dxeedxee xixaxixa
0
0
get we0, x-x,-xand x0 x,xfact that theUsing
dxeedxeexfF xiaxxiax
0
0
dxedxexiaxia
0
0
0
0
ia
e
ia
e xiaxia
iaia
11
22
2
a
a
2. Find the Fourier Transform of the function
ax
ax
0,
1,f(x)
where ‘a’ is a positive constant. Hence evaluate
d
xai
cossin)(
dii
0
sin)(
For the given function, we have
dxexfxfF xi)(
5
dxexfdxexfdxexf xi
a
xia
a
xia )()()(
dxe
a
a
xi
asin
2
)1(sin
2ˆ
a
fxfFThus
get weformula,inversionemployingbyfInverting
de
axf xi
sin
22
1
d
xixa sincossin1
dxa
idxa sinsincossin1
Here, the integrand in the first integral is even and the integrand in the second integralis odd. Hence using the relevant properties of integral here, we get
d
xaxf
cossin)( 1
or
)(cossin
xfdxa
ax
ax
0,
,
d
sinyields this1,a0,For x
Since the integrand is even, we have
0
sin2 d
or
20
sin
d
Transform.Fourierrespect to withreciprocalselfis22x
ef(x) thatDeduce
constant.positiveaisa'' where2x2a-ef(x)ofTransformFourier theFind3.
Here
dxeexfF xixa 22
6
dxe xixa 22
dxeaa
iax
2
22
42
dxee a
iax
a
2
2
2
24
getwe,2a
i-ax tSetting
a
dteexfF ta
22
2
4
dteea
ta
0
422
2
21
function.gammausing,1 2
2
4
aea
2
2
4ˆ aea
f
This is the desired Fourier Transform of f(x).
2
2x-
2
2
2
2f
hence,andef(x)getwe
2x2a-ef(x)in21aFor
e
getwe,ef(x)in xputtingAlso 2x2- 2
2-e)f(
.
Hence, )f( and f are same but for constant multiplication by 2 .
f)f(Thus
reciprocalselfise)f( thatfollowsIt 22
-x
x
ASSIGNMENT
Find the Complex Fourier Transforms of the following functions :
constantpositiveaisa''where,0
,xf(1)
ax
axx
constantspositiveareb''anda''where
bx,0
bxa,1
ax,0
xf(2)
7
1,0
1,1xf(3)
x
xx
ax
axxa
,0
,xf(4)
22
constantpositiveaisa'' where)()5( xaxexf
)()6( xexf 2cos)()7( 2xxf 3sin)()8( 2xxf
2fofTransformFourierinverse theFind(9) e
FOURIER SINE TRANSFORMS
Let f(x) be defined for all positive values of x.
ThusxfFor
xdxxf
s .fby
denotedisThisf(x).ofTransformSineFourier thecalledissinintegralThe
s
0
dxxxfxfFs sinf0s
dxf s sinˆ2integralethrough th
definedisfofTransformsineFourierinverseThe
0
s
dxff
Thusf
ss
s
sinˆ2Ff(x)
.Forf(x)bydenotedisThis
0
1-s
-1s
Properties
The following are the basic properties of Sine Transforms.
(1) LINEARITY PROPERTYIf ‘a’ and ‘b’ are two constants, then for two functions f(x) and (x),
we have xgbFxfaFxbxafF sss
Proof : By definition, we have
dxxxbxafxbxafFs sin0
xbFxfaF ss This is the desired result. In particular, we have
xFxfFxxfF sss and
xFxfFxxfF sss
8
(2) CHANGE OF SCALE PROPERTY
have we0,aforthen,fxfFIf s s
aa s
f
1axfFs
Proof : We have
dxxaxf sinaxfF0s
Setting ax = t , we get
a
dtt
atf
sinaxfF
0s
aa s
f
1
(3) MODULATION PROPERTY
have we0,aforthen,ˆxfFIf s sf
afafaxxfF sss ˆˆcos21
Proof : We have
0
sincoscos dxxaxxfaxxfFs
dxxaxaxf sinsin2
10
property.Linearityusingby,ˆˆ21 afaf ss
EXAMPLES
1. Find the Fourier sine transform of
ax
ax
,0
0,1xf
For the given function, we have
a
adxxdxx sin0sinf
0s
ax
0
cos
acos1
x
ef(x)of transformsineFourier theFind2.
-ax
Here
9
0s
sinf
x
dxxe ax
Differentiating with respect to , we get
0s
sinf
x
dxxe
d
d
d
d ax
0
sin dxxx
axe
performing differentiation under the integral sign
0
cos dxxxx
e ax
022
sincos xxaa
e ax
22
a
a
Integrating with respect to , we get
ca
1s tanf
00fBut s when
c=0
a
1
s tanf
3. Find f(x) from the integral equation
2,0
21,2
10,1
sinxf0
xdx
Let () be defined by
2,0
21,2
10,1
Given
0
ˆsin s
fxdxxf
Using this in the inversion formula, we get
dxx
sin2
xf0
dxdxdx sinsinsin
2
2
1
1
02
10
2
1
1
0
2 0sin2sin
dxdx
xxx
2cos2cos12
ASSIGNMENT
Find the sine transforms of the following functions
ax
axxa
xx
xf
,0
1,
10,
)1(
0,)2( axexf ax
ax
axxxf
,0
0,sin)3(
givenf(x)forSolve(4)
1,0
10,1sinxf
0
dxx
Find the inverse sine transforms of the following functions :
0,ˆ)5(
ae
fa
s
2
ˆ)6( sf
FOURIER COSINE TRANSFORMS dxxxf cosintegralThe x.of valuespositivefordefinedbef(x)Let
0
Thus.Forfbydenotedisandf(x)ofTransformCosineFourier thecalledis cc xf
xdxxfxf
cos2
Ff0cc
Thus.orxfbydenotedisThis.2
integralthe
throughdefinedisofTransformCosineFourierinverseThe
cf
1-cFcosˆ
cf
0
dxfc
0
1-c cosˆ2ˆFxf
dxff c
Basic Properties
The following are the basic properties of cosine transforms :
(1) Linearity property
xxfxafFhave we(x),andf(x)functionsfor two thenconstants, twoareb''anda''If
c
cc bFaFxb
11
(2) Change of scale property
aa
cc
cc
f1
axfF
have we0,aforthen,fxfFIf
(3) Modulation property
aaax
ccc
cc
ff2
1cosxfF
have we0,aforthen,fxfFIf
The proofs of these properties are similar to the proofs of the correspondingproperties of Fourier Sine Transforms.
Examples(1) Find the cosine transform of the function
2,0
21,2
10,
x
xx
xx
xf
We have
xdxxdxxxdxx
xdxxf
cos0cos2cos
cosf
2
2
1
1
0
0c
Integrating by parts, we get
2
12
1
0
2c
cos1
sin2
cossinf
xx
xxx
x
2
12coscos2
0 22
-ax
ax
kxcosevaluateHence0.a,ef(x)of transformcosine theFind(2)
dx
Here
22c
022
0c
f
sincosa
cosf
a
a
Thus
xxae
xdxe
ax
ax
Using the definition of inverse cosine transform, we get
xda
x cosa
2f
0 22
12
or
0 22
cos
2
da
xe
aax
Changing x to k, and to x, we get
a
edx
a
kx ax
2x
cos0 22
(4) Solve the integral equation
aedxxxf cos
0
Let () be defined by() = e-a
c0fcosGiven
dxxxf
Using this in the inversion formula, we get
22
00 22
0
0
2
sincos2
cos2
cos2
f
xa
a
xxaxa
e
xde
xdx
a
ASSIGNMENT
Find the Fourier Cosine Transforms of the following functions :
4,0
41,4
10,4
f(x)(1)
x
xx
xx
0,)2(2 aexf ax
ax
axxxf
,0
0,cos)()3(
0,)()4( axexf ax
21
1)()5(
xxf
21
2cos)()6(
x
xxf
13
givenf(x)forSolve)7(
1,0
10,1cos)(
dxxxf0
thatShow(8)
afaf
afaf
cc
ss
ˆˆaxf(x)sinF(ii)
ˆˆaxf(x)sinF(i)
21
s
21
c
CONVOLUTIONexist.)(and)(such thatfunctions twobeg(x)andf(x)Let
dxxgdxxf
Then the integral
dttgtxf
is called the convolution of f(x) and g(x), and is denoted by f * g. Thus
dttgtxfgf
*
Note that f * g is a function of x
Properties
hfgf **hg*f2.f*gg*f1.
Convolution Theorem
)(g)(fg*fF
Thenly.respectiveg(x)andf(x)ofTransformsFourier thebe)(g)and(fLet
The convolution theorem may also be rewritten as
gfFg*f 1-Parseval’s Identity
A direct consequence of convolution theorem is Parseval’s identity. The Parseval’sidentities in respect of Fourier transforms, sine transforms and cosine transforms areas indicated below :
Fourier Transforms:
dxxfdfb
dxxgxfdgf
22ˆ)(
ˆˆ(a)
14
Fourier Sine Transforms:
dxxfdfb
dxxgxfdgf
s
s
22ˆ)(
ˆˆ(a)
Fourier CosineTransforms:
dxxfdfb
dxxgxfdgf
c
c
22ˆ)(
ˆˆ(a)
Examples
(1) Employ convolution theorem to find the inverse Fourier Transform of
9
1ˆ,4
1ˆ94
1
22
22
gfLet
We recall the result
a
e
aF
ora
aeF
xa
xa
221
22
1
For a=2, 3, we get
)(3
ˆ9
1
)(2
ˆ4
1
3
21
2
21
xge
gF
xfe
fF
x
x
Convolution theorem is
dte
dtee
dttgtxfgfgfF
ttx
ttx
32
32
1
12
13
1
2
1
*ˆˆ
1,0
1,1f(x)given that
sinevaluateoidentity tsParseval'Employ2.
0 2
2
x
x
dxx
x
For the given function, we have
15
sin2
2
1
2
1)1(
2
1ˆ1
1
1
1 i
edxef
xixi
Parseval’s identity for Fourier Transforms is
0 2
2
0 2
2
2
2
2
2
21
1
2
22
2
sin
get weby x,Replacing
even.isL.H.S.on theintegrand theas,2
sin
sin
sin22
sin2
2
11
ˆ
dxx
x
d
or
d
or
d
or
ddx
or
dfdxxf
ASSIGNMENT
22
1-
2
x-
1
1Ffind to
n theoremconvolutioemploy,1
1eFGiven that1.
2. Use Parseval’s identity to prove the following :
0,
4ax
x(iii)
41x
dx(ii)
1241x
dx(i)
0222
20
22
022
adx
x
0
4
2
6
xcos-1 thatProve,
1,0
1,1)()(
dx
xx
xxxfIfiv
16
Z – TRANSFORMS
Introduction
The Z-transform plays an important role plays in the study of communications,sample data control systems, discrete signal processing , solutions of differenceequations etc.
Definition
Let un = f(n) be a real-valued function defined for n=0,1,2,3,….. and un = 0 for n<0.Then the Z-transform of un denoted by Z(un) is defined by
0
)()(n
nnn zuuZzu (1)
The transform also is referred to as the one sided Z-transform or unilateral Z-transform. Next, we define un = f(n) for n=0, 1, 2, …… ∞.
The two-sided Z-transform is defined by
nnn zuuZ )( (2)
The region of the Z-plane in which the series (1) or (2) converges is called the regionof convergence of the transform.
Properties of Z-transform
1. Linearity property
Consider the sequences {un} and {vn} and constants a and b. Then
Z[aun + bvn ] = aZ(un) + bZ(vn)
Proof : By definition, we have
Definition Properties Examples
Z - TRANSFORMS
17
)()(
][][
0 0
0
nn
n n
nn
nn
n
nnnnn
vbZuaZ
zvbzua
zbvaubvauZ
In particular, for a=b=1, we get
Z[un+vn] = Z(un) + Z(vn)and for a=-b=1, we get
Z[un - vn] = Z(un) - Z(vn)
2. Damping property
Let Z(un) = )(zu . Then (i) azuuaZ n
n )( (ii) )()( azuuaZ nn
Proof : By definition, we have
a
zu
a
zuzuauaZ
n
n
nn
nn
nn
n
0 0
)()(
Thus
a
zuuaZ n
n )(
This is the result as desired. Here, we note that that if Z(un) = )(zu , then
][)( zuuaZ nn
aZZ
=
a
zu
Next,
)(
)()()(00
azu
azuzuauaZn
nn
n
nn
nn
n
Thus
)()( azuuaZ nn
This is the result as desired.
3. Shifting property
(a) Right shifting rule :
If Z(un) = )(zu , then Z(un-k) = z-k )(zu where k>0
Proof : By definition, we have
18
0n
nknkn zuuZ
Since un = 0 for n<0, we have un-k = 0 for n=0,1,……(k-1)
Hence
)(
].......[
.......
0
110
)1(10
zuz
zuz
zuuz
zuzu
zuuZ
k
n
nn
k
k
kk
kn
nknkn
Thus
Z(un-k) = z-k )(zu
(b)Left shifting rule :
]......)([)( )1(1
22
110
k
kk
kn zuzuzuuzuzuZ
Proof :
0
1
0
0
)(
0
)(
0
,
n
k
n
nn
nn
k
n
nkkn
k
n
nkkn
k
n
nknkn
zuzuz
knmwherezuzzuz
zuuZ
= ]......)([ )1(1
22
110
kk
k zuzuzuuzuz
Particular cases :
In particular, we have the following standard results :
1. ])([)( 01 uzuzuZ n
2. ])([)( 110
22
zuuzuzuZ n
3. ])([)( 22
110
33
zuzuuzuzuZ n etc.
19
Some Standard Z-Transforms :
1.Transform of an
By definition, we have
.....1
)(
0
2
0
n
n
n
nnn
z
a
z
a
z
a
zaaZ
The series on the RHS is a Geometric series. Sum to infinity of the series is
az
zor
za 1
1Thus,
az
zaZ n
)(
In particular, when a=1, we get Z(1) =1z
z
2. Transform of ean
Here)()( nan kZeZ where k = ea
aez
z
kz
z
Thus
aan
ez
zeZ
)(
3. Transform of np , p being a positive integer
We have,
0
)1(1
0
)(
n
np
n
npp
nznz
znnZ
Also, we have by defintion
0
11)(n
npp znnZ
Differentiating with respect to z, we get
0
)1(1
0
11
)(
)(
n
np
n
npp
znn
zndz
dnZ
dz
d
Using this in (1), we get
)]([)( 1 pp nZdz
dznZ
(1)
20
Particular cases of )( pnZ :-
1. For p = 1, we get
2)1(1
)1()(
z
z
z
z
dz
dzZ
dz
dznZ
Thus,
2)1()(
z
znZ
2. For p = 2, we get
3
2
22
)1()1()()(
z
zz
z
z
dz
dznZ
dz
dznZ
Thus,
3
22
)1()(
z
zznZ
3. For p = 3, we get
4
233
)1(
4)(
z
zzznZ
4. Transform of nan
By damping property, we have
22
2
)(1
)1()()(
az
az
a
za
z
z
znZnaZ
aZZ
aZZ
n
, in view of damping
propertyThus,
2)()(
az
aznaZ n
5. Transform of n2an
We have,
a
ZZ
a
ZZ
n
z
zznZanZ
3
222
)1()()(
Thus,
3
222
)()(
az
zaazanZ n
21
6. Transforms of coshn and sinhnWe have
)(2
1)(cosh
2cosh
nn
nn
eeZnZ
een
)()(2
1 nn eZeZ , by using the linearity property
1cosh2
cosh
1)(2
)(
1)(2
2
1
2
22
zz
zz
eezz
eez
zeezz
ezezz
ez
z
ez
z
Next,
1cosh2
sinh
1cosh22
11
2)(sinh
2sinh
2
2
zz
z
zz
eez
ezez
znZ
een
nn
7. Transforms of cosn and sinn
We have
)(2
1)(cos
2cos
inin
inin
eeZnZ
een
1cos2
]cos[
1)(
)(2
2
2
1
2
2
zz
zz
eezz
eezz
ez
z
ez
z
inin
inin
ii
22
Next,
1cos2
sin
1cos22
2
1)(sin
2sin
2
2
zz
z
zz
ee
i
z
ez
z
ez
z
inZ
i
een
ii
ii
inin
Examples :
Find the Z-transforms of the following :
1.nn
nu
4
1
2
1
We have,
nn
n ZuZ4
1
2
1)(
)14)(12(
)38(2
14
4
12
2
4
1
2
1
4
1
2
1
zz
zz
z
z
z
z
z
z
z
z
ZZnn
2.!
1
nun
Here
z
n
n
n
nn
ezz
nz
znn
ZuZ
1
2
00
........!2
1
!1
1
1
!
1
!
1
!
1)(
,
, by exponential theorem
23
3. nau nn cos
We have
1cos2
)cos()(cos
2
zz
zznZ
By using the damping rule, we get
22
2
2
2
)cos(
1cos2
cos
1cos2
)cos()cos(
aazz
azz
a
z
a
z
a
z
a
z
zz
zznaZ
aZZ
n
4. un =)!2(
1
n
Let us denote vn =!
1
n, so that vn+2 =
)!2(
1
n= un
Here
zn evZ
1
)( Hence
z
vvvZzvZuZ nnn
10
22 )()()( , by left shifting rule
zez
zez
z
z
11
!1
1.
1
!0
1
12
12
ASSIGNMENT
Find the Z-transforms of the following :1. un = cos(2n+3)2. un = cosh2n3. un = n4
4. un = an coshn5. un = an sinhn6. un = e-an cosn7. un = e-an sinn8. un = a-n n2
9. un = (n-2)3
10. un = (n+1)4
