chapter 11 fourier series, integrals, and transforms 11.1 fourier...
TRANSCRIPT
-
Chapter 11 Fourier Series, Integrals,
and Transforms
11.1 Fourier Series
A function f (x) is called periodic if there is
some positive number p such that
Rxxfpxf ),()((1)
p is called a period of f (x)
-
constcxfxx )(,cos,sin
are periodic functions.
xexx ,, 2 are not periodic. Properties of periodic function:
1. For any integer n,
)()( xfnpxf (2)
2. If f(x) and g(x) have period p, then
)()()( xbgxafxh
also has the period p.
-
3. If a periodic function f (x) has a smallest
period p(>0), then it is often called the
fundamental period of f (x) .
The fundamental period of xx cos,sin is 2
The fundamental period of xx 2cos,2sin is
constxf )( without fundamental period.
-
2 is the common period of
,sin,cos,,2sin,2cos,sin,cos,1 nxnxxxxx
A trigonometric series has the form
1
0
22110
)sincos(
2sin2cossincos
n
nn nxbnxaa
xbxaxbxaa (4)
(4) can be used to represent any periodic
function with the period 2
We call (4) as Fourier series.
-
Assume the f(x) is a periodic function of period
that can be represented by a trigonometric
series,
2
1
0 )sincos()(n
nn nxbnxaaxf(5)
We want to determine the coefficients nn ba and
of the corresponding series (5).
-
dxxfa
dxnxbdxnxadxadxxfn
nn
)(
2
1
sincos)(
0
1
0
Similarly, we have
1
0
cossincoscos
coscos)(
n
nn mxdxnxbdxmxnxa
mxdxamxdxxf
-
0)sin(2
1)sin(
2
1
cossin
0
)cos(2
1)cos(
2
1
coscos
,0cos
dxxmndxxmn
dxmxnx
mn
mn
dxxmndxxmn
dxmxnx
dxmx
-
Thus we obtain the result
,2,1 ,cos)(1
mdxmxxfam
Finally, we got
1
0
sinsinsincos
sinsin)(
n
nn mxdxnxbdxmxnxa
mxdxamxdxxf
-
0)sin(2
1)sin(
2
1
sincos
0
)cos(2
1)cos(
2
1
sinsin
,0sin
dxxmndxxmn
dxmxnx
mn
mn
dxxmndxxmn
dxmxnx
dxmx
-
Thus we obtain the result
,2,1 ,sin)(1
mdxmxxfbm
Euler formulas:
,2,1sin)(1
)(
,2,1cos)(1
)(
)(2
1)( 0
ndxnxxfbc
ndxnxxfab
dxxfaa
n
n
(6)
-
Fourier series of f (x) with (6)
1
0 )sincos(n
nn nxbnxaa
Ex. 1 Find the Fourier coefficients of f (x)
xk
xkxf
0
0)( and
)()2( xfxf
(7)
-
Solution. f (x) is an odd function, so we have
n
nn
k
nn
knx
n
k
dxnxk
dxnxxfb
dxnxxfa
dxxfa
n
n
even 0
odd4
)cos1(2
cos2
sin2
sin)(1
0cos)(1
0)(2
1
0
0
0
-
Thus the Fourier series of f(x) is
xxx
k5sin
5
13sin
3
1sin
4
(8)
2/xFurthermore, we set
7
1
5
1
3
11
4
25sin
5
1
23sin
3
1
2sin
4
2
k
kkf
47
1
5
1
3
11
(by Leibniz, 1673).
-
The trigonometric system
In formulas, for any integers m and n,
,sin,cos,,2sin,2cos,sin,cos,1 nxnxxxxx
is orthogonal on the interval x
) (including 0sincos
)( 0coscos
)( 0sinsin
nmdxmxnx
nmdxmxnx
nmdxmxnx
-
Given a piecewise continuous function f (x)
of period , we can get the coefficients
according to (6) and then obtain a Fourier series
of f(x), denoted as
2
1
0 )sincos(~)(n
nn nxbnxaaxf
Under what condition we get to know
?)sincos()(1
0
n
nn nxbnxaaxf
-
Theorem 1 If a periodic function f (x) is
piecewise continuous in the interval
and has a left-hand derivative and right-hand
derivative at each point of the interval, then
x
1
0 )sincos(2
)0()0(
n
nn nxbnxaaxfxf
Particularly, if f (x) is continuous at 0x
1
0000 )sincos()(n
nn nxbnxaaxf
-
Ex. 2 In example 1, f (x) has a jump at 0x
kfkf )00(,)00( . So we obtain
2
)00()00(
0
0*5sin5
10*3sin
3
10sin
4
ff
k
-
11.2 Functions of Any Period p=2L
We want to find a trigonometric series to
represent a function f (x) of period p=2L, where
. Let , then g (t) has
period . Thus, we have
0L )/()( Ltftg
2
1
0 )sincos(~)(n
nn ntbntaatg
where
-
,2,1sin)(1
)(
,2,1cos)(1
)(
)(2
1)( 0
ndtnttgbc
ndtnttgab
dttgaa
n
n
We replace t by , then Lx /
1
0 )sincos(~)()(n
nnL
xnb
L
xnaa
L
xgxf
(5)
-
with Fourier coefficients
,2,1sin)(1
)(
,2,1cos)(1
)(
)(2
1)( 0
ndxL
xnxf
Lbc
ndxL
xnxf
Lab
dxxfL
aa
L
Ln
L
Ln
L
L
(6)
-
here we only verify (b)
dxLxnxfL
LxdLxnLxg
Lxtdtnttga
L
L
L
L
n
)/cos()(1
)/()/cos()/(1
)/ setting(by cos)(1
-
Ex. 1 Find the Fourier series of
2,42
210
11
120
)(
LLp
x
xk
x
xf
Solution. It is an even function. 0nb
-
,2,1,12
)12(
2)1(20
2sin
2
2cos
2cos)(
2
124
1)(
4
1
1
1
0
2
2
1
1
2
20
mmn
m
kmn
n
n
k
dxxn
kdxxn
xfa
kdxkdxxfa
m
n
Hence the result is
,5,3,1
2
5cos
5
1
2
3cos
3
1
2cos
2
2)(
x
xxxkk
xf
-
Ex. 2 Find the Fourier series of f (x)
2,42 20
02)(
LLp
xk
xkxf
Solution. f (x) is an odd function, so we have
0/cos)(1
0)(2
10
dxLxnxfL
a
dxxfL
a
L
Ln
L
L
-
n
nn
k
nn
knx
n
k
dxnxk
dxLxnxfL
bL
Ln
even 0
odd4
)cos1(2
cos2
sin2
/sin)(1
0
0
Thus the Fourier series of f (x) is
2/5sin
5
12/3sin
3
12/sin
4xxx
k
-
Ex. 3 Find the Fourier series of
LLp
LttE
tLtu ,
22
0sin
,00)(
Solution. By formulas (6) we have
Edxx
EdttEa 0
/
00 sin
2sin
2
-
11
)1cos(
1
)1cos(
2
12sin2
])1sin()1[sin(2
)]1sin()1[sin(2
cossin
0
0
0
/
0
/
0
nn
xn
n
xnE
ndxxE
dxxnxnE
dttntnE
dttntEan
-
mnnn
Emn
mnnn
Emn
nn
n
n
nEn
2)1)(1(
2120
21
1
1
1120
11
1)1cos(
1
1)1cos(
2
10
-
10
12/
11
)1sin(
1
)1sin(
2
1]12[cos2
])1cos()1[cos(2
)]1cos()1[cos(2
sinsin
0
0
0
/
0
/
0
n
nE
nn
xn
n
xnE
ndxxE
dxxnxnE
dttntnE
dttntEbn
-
Thus
tt
Et
EEtu
4cos
53
12cos
31
12sin
2)(
-
11.3 Even and Odd Functions
Half-Range Expansions
Properties of the even or odd functions:
1.If g (x) is an even function, then
dxxgdxxgLL
L 0 )(2)(
2.If h (x) is an odd function, then
0)( dxxhL
L
-
Theorem 1 The Fourier series of an even
function of period 2L is a cosine series
1
0 cos)(n
nL
xnaaxf
with Fourier coefficients
,2,1,cos)(2
,)(1
0
00
ndxL
xnxf
La
dxxfL
a
L
n
L
(1)
(2)
-
The Fourier series of an odd function of
period 2L is a sine series
1
sin)(n
nL
xnbxf
with Fourier coefficients
,2,1,sin)(2
0 ndxL
xnxf
Lb
L
n
(3)
(4)
-
In the case of period , we give for an even
function simply 2
1
0 cos)(n
n xnaaxf
with coefficients
,2,1,cos)(2
,)(1
0
00
ndxxnxfa
dxxfa
n
(1*)
(2*)
-
In the case of period , we give for an odd
function simply 2
1
0 sin)(n
n xnbbxf
with coefficients
(3*)
,2,1,sin)(2
0 ndxxnxfbn
(4*)
-
Theorem 2 The Fourier coefficients of a sum
are the sums of the corresponding
Fourier coefficients of The Fourier
coefficients of cf are c times the corresponding
Fourier coefficients of f
21 ff
21 and ff
-
Ex. 1. Find the Fourier series of f (x)
xk
xxf
02
00)(
)()2( xfxf
Solution. )()( xgkxf where
xk
xkxg
0
0)(
is an odd function.
-
For g (x), we have
n
nn
k
nn
knx
n
k
dxnxk
dxnxxgbn
even 0
odd4
)cos1(2
cos2
sin2
sin)(1
0
0
xxx
kxg 5sin
5
13sin
3
1sin
4~)(
,0,5sin
5
13sin
3
1sin
4)( xxxx
kkxf
-
Ex. 2 Find the Fourier series of f (x)
xxf )( and )()2( xfxf
Solution. xxf )( x is odd.
nn
nxn
x
dxnxxdxnxxbn
cos2
cos2
sin2
sin1
0
0
xxxxxf ,3sin
3
12sin
2
1sin2)(
-
Half-Range Expansions
Given a function Lxxf 0 ),(
we can extent to
1) the even extension; 0),()( xLxfxf
2) the odd extension; 0),()( xLxfxf
We say that both 1) and 2) are half-range
expansions.
-
Ex. 3 Find the two half-range expansions of
LxLxLL
k
LxxL
k
xf
2/),(2
2/0,2
)(
Solution. (a) Even periodic extension.
288
2
)(221
22
2
2/
2/
00
kLL
L
k
dxxLL
kdxx
L
k
La
L
L
L
-
nydyyknydyynydyy
k
nydyynydyyk
dxL
xnxL
L
kdx
L
xnx
L
k
La
n
n
L
L
L
n
cos)1(14
cos)1(cos4
cos)(cos4
cos)(2
cos22
2/
02
2/
0
2/
02
2/
2/
02
2/
2/
0
Thus 0na for the odd n
-
oddmm
kevenm
mm
k
mym
mm
k
dymym
myym
k
mydyyk
a m
22
22
2/
022
2/
0
2/
02
2/
022
40
cos12
2cos4
1sin
4
8
2sin2
12sin
2
18
2cos8
-
the even half-range expansion is
x
Lx
Lx
L
kkxf
10cos
5
16cos
3
12cos
4
2)(
222
(b) Odd periodic extension.
-
nydyyknydyynydyy
k
dxL
xnxL
L
kdx
L
xnx
L
k
Lb
n
n
L
L
L
n
sin)1(14
sin)1(sin4
sin)(2
sin22
2/
0
1
2
2/
0
12/
02
2/
2/
0
Thus 0nb for the even n
-
For the odd n
x
Lx
Lx
L
kxf
5sin
5
13sin
3
1sin
8)(
222
2sin
8sin
18
coscos18
sin8
22
2/
02
2/
0
2/
02
2/
02
n
n
kny
nn
k
dynynyyn
k
nydyyk
bn
Hence the odd half-range expansion is
-
11.4 Complex Fourier Series
The Fourier series
1
0 )sincos()(n
nn nxbnxaaxf(1)
can be written in complex form by the Euler’s
formula
nxinxe inx sincos (2)
-
inxinxinxinx eei
nxeenx 2
1sin ,
2
1cos
Thus we obtain
inx
nn
inx
nn
inxinx
n
inxinx
n
nn
eibaeiba
eebi
eea
nxbnxa
)(2
1)(
2
12
1
2
1
sincos
-
We write
nnnnnnn ccibacibaca 2/)(,2/)(,00
Then the Fourier series (1) becomes
n
inx
n
n
inx
n
inx
n
n
nn
ecececc
nxbnxaaxf
1
0
1
0
)(
)sincos()((4)
,2,1,0 ,)(2
1
ndxexfc inxn
(5)
-
For the function of period 2L, the complex
Fourier series gives as
n
Lxin
necxf/)( (6)
where
,2,1,0 ,)(2
1 / ndxexf
Lc
L
L
Lxin
n
(7)
-
Ex. 1 Find the complex Fourier series of a 2
periodic function xexfx ,)(
Solution. We note that nine )1(
sinh)1(
)1()1(
)()1)(1(2
)1()1(
)1(2
1
2
1
2n
in
eeinin
in
ein
dxeec
n
n
inxxxinx
n
-
Hence the complex Fourier series is
xen
ine inx
n
nx ,1
1)1(
sinh2
From this let us derive the real Fourier series
)sincos()sin(cos )sin)(cos1()1(
nxnxninxnnxnxinxinein inx
Omitting the imaginary part, we have
-
x
nxnnxn
nxnnxn
e
n
n
n
nx
)sin(cos1
1)1(21
sinh
)sin(cos1
1)1(
sinh
20
2
-
11.6 Approximation by Trigonometric
Polynomials
N
n
nn nxbnxaaxf1
0 )sincos()((1)
Is (1) a good approximation of f (x)?
Let a trigonometric polynomial of degree N
N
n
nnN nxBnxAAxF1
0 )sincos()((2)
is a best approximation of f (x), What is ? )(xFN
-
Or what is in (2), such that nn BAA ,,0
dxFfE N
2)((3)
achieves the minimum for all trigonometric
polynomials of degree N.
(3) is called the total square error of relative
to the function f (x) on the interval
)(xFN
x
dxFdxfFdxf
dxFfE
NN
N
22
2
2
)(
-
where
N
n
nn
N
n
nn
N
n
nnN
BAA
dxnxBnxAA
dxnxBnxAAdxF
1
222
0
1
22222
0
2
1
0
2
)(2
sincos
sincos
N
n
nnnn
N
n
nnN
bBaAaA
dxnxfBnxfAfAdxfF
1
00
1
0
)(2
sincos
-
Thus we have
N
n
nnnn
N
n
nn
N
n
nn
N
n
nnnn
bBaAaA
baadxf
BAA
bBaAaAdxfE
1
222
00
1
222
0
2
1
222
0
1
00
2
])()[()(2
)(2
)(2
)(22
-
It is clear that E reaches the minimum
0)(2min*1
222
0
2
N
n
nn baadxfEE
(6)
if and only if NN bBbBaAaA ,,,, 111100
Hence we prove the theorem 1 below.
-
Theorem 1 The total square error of in
(2) relative to the function f (x) on the interval
is minimum if and only if the
coefficients of in (2) are the Fourier
coefficients of f (x). The minimum value is
given by (6).
x
)(xFN
)(xFN
-
The Bessel inequality
.any for ,1
)(2 2
1
222
0 NdxfbaaN
n
nn
(7)
The Parseval’s identity
dxfbaan
nn
2
1
222
0
1)(2(8)
-
Ex. 1 Compute the total square error of
with N=3 relative to
)( )( xxxf
)(xFN
Solution. xxf )( We have
nn
nxn
x
dxnxxdxnxxbn
cos2
cos2
sin2
sin1
0
0
-
xxxxF 3sin
3
12sin
2
1sin2)(3
567.3]2[
])3/2(122[)(*
94923
38
22222
dxxE
-
Ex. 2 Compute the value of
222
1
3
1
2
11
n
Solution. xxxf ,)( We have
nn
nxn
x
dxnxxdxnxxbn
cos2
cos2
sin2
sin1
0
0
xxxxf 3sin
3
12sin
2
1sin2~)(
-
According to the Parseval’s identity
3
2
3
2114
2
0
32
12
xdxxnn
So we have
6
1
3
1
2
11
2
222
n
-
11.7 Fourier Integrals
Consider a periodic function of period
2L that can be represented by Fourier series
)(xfL
L
nwxwbxwaaxf n
n
nnnn
,)sincos()(1
0
According to the Euler’s formula, we have
L
Lnn
n
L
LnLn
L
LLL
vdvwvfxw
vdvwvfxwL
dxxfL
xf
sin)(sin
cos)(cos1
)(2
1)(
1
-
We now set LL
n
L
nwww nn
)1(1
dwwvdvvfwx
vdvwvfwxxf
vdvwvfwxw
vdvwvfwxw
dxxfL
xf
L
Lnn
n
L
LnLn
L
LLL
sin)(sin
cos)(cos1
)(
sin)()sin(
cos)()(cos1
)(2
1)(
0
1
(1)
(3)
-
We denote
vdvwvfwB
vdvwvfwA
sin)(1
)(
,cos)(1
)(
(4)
Then we can rewrite this in the form
dwwxwBwxwAxf sin)( cos)()(0
(5)
(5) is called a representation of f (x) by a
Fourier integral.
-
Theorem 1 If f (x) is piecewise continuous in
very finite interval and has a right-hand
derivative and a left-hand derivative at every
point and if the (4) is well defined, then f (x)
can be represented by a Fourier integral (5).
At a point where f (x) is discontinuous the value
of the Fourier integral equals the average of the
left- and right-hand limits of f (x) at the point.
-
Ex. 1 Find the Fourier integral representation
of the function
1||0
111)(
x
xxf
Solution. From (4) 0)( wB
w
wvdvwvdvwvfwA
sin2cos
2cos)(
1)(
1
0
and (5) gives the answer
dww
wwxxf
0
sincos2)(
-
10
14/
102/sincos
0
x
x
x
dww
wwx
We set 0x then
2/sin
0
dww
w
-
If f (x) is even, then B(w)=0, and then
vdvwvfwA
0
cos)(2
)(
(5) then reduces to the Fourier cosine integral.
dwwxwAxf
0
cos)()((11)
(10)
If f (x) is odd, then A(w)=0, and then
vdvwvfwB
0
sin)(2
)(
dwwxwBxf
0
sin)()((13)
(12)
(5) then reduces to the Fourier sine integral.
-
If f (x) is defined on a half range, we can find
the even or odd representation, respectively.
Evaluation of integrals
Ex. 2 Find the Fourier cosine and sine integrals
of )0,0(,)( kxexf kx
Solution. (a) from (10) we have
-
)(
2
cossin)(
2
cos2
)(
22
0
22
0
wk
k
wvwvk
we
wk
k
vdvwewA
kv
kv
The Fourier cosine integral representation
)0,0( ,cos2
)(0 22
kxdwwk
wxkexf kx
)0,0( ,2
cos
0 22
kxekdw
wk
wx kx
-
(b) from (12) we have
)(
2
cossin)(
2
sin2
)(
22
0
22
0
wk
w
wvwvk
we
wk
w
vdvwewB
kv
kv
The Fourier sine integral representation
)0,0( ,sin2
)(0 22
kxdwwk
wxwexf kx
)0,0( ,2
sin
0 22
kxedwwkwxw kx
-
0
02/
00
1
sincos
0 2
xe
x
x
dww
wxwwx
x
Ex. 3 Evaluation of Integrals for the follows
Solution. From (4) we have for the rhs function
20 1
1coscos)(
1)(
wvdvwevdvwvfwA x
20 1sinsin)(
1)(
w
wvdvwevdvwvfwB x
-
So we have
0
02/
00
1
sincos
0 2
xe
x
x
dww
wxwwx
x
-
11.8 Fourier Cosine and Sine Transforms
For an even function f (x), we have
dwwxwAxf
0
cos)()((1)
with vdvwvfwA
0
cos)(2
)(
Then
xdxwxfwfc
0
cos)(2
)(ˆ
and
(3)
(2)
dwwxwfxf c
0
cos)(ˆ2
)(
-
)(ˆ wfc is called the Fourier cosine transform of
f (x), and f (x) is called the inverse Fourier
cosine transform of )(ˆ wfc
For an odd function f (x), we have
dwwxwBxf
0
sin)()((4)
with Then vdvwvfwB
0
sin)(2
)(
-
xdxwxfwfs
0
sin)(2
)(ˆ
and
(6)
(5)
dwwxwfxf s
0
sin)(ˆ2
)(
)(ˆ wfs is called the Fourier sine transform of
f (x), and f (x) is called the inverse Fourier
sine transform of )(ˆ wfs
We also denote
ffFffFffFffF scsscc sc }ˆ{ ,}ˆ{ ,ˆ}{ ,ˆ}{ 11
-
Ex. 1 Find the Fourier cosine and Fourier sine
transforms of the function
ax
axkxf
0
0)(
Solution. From (2) and (5) we have
w
awkxdxwkwf
a
c
sin2cos
2)(ˆ
0
w
awkxdxwkwf
a
s
cos12sin
2)(ˆ
0
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Ex. 2 Find }{x
c eF
Solution. By (2) we get
2
02
0
1
/2
)sincos(1
2
cos2
}{
w
wxwwxw
e
xdxweeF
x
xx
c
-
It is easy to show that the Fourier cosine and
sine transforms are linear operations,
}{}{}{}{}{}{
gbFfaFbgafFgbFfaFbgafF
sss
ccc
(7)
Theorem 1 Let f (x) be continuous and
absolutely integrable on the x-axis, let f ’(x) be
piecewise continuous on each finite interval,
and let , then xasxf ,0)(
)}({)}({
)0(2
)}({)}({
xfwFxfF
fxfwFxfF
cs
sc
(8)
-
Proof. This follows from the definitions
)};({)0(2
sin)(cos)(2
cos)(2
)}({
00
0
xfwFf
wxdxxfwxwxf
wxdxxfxfF
s
c
)}({0
cos)(sin)(2
sin)(2
)}({
00
0
xfwF
wxdxxfwxwxf
wxdxxfxfF
c
s
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It is easy to show by theorem 1 that
)0(2
)}({
)}('{)}({
)0(2
)}({
)0(2
)}({)}({
2
2
wfxfFw
xfwFxfF
fxfFw
fxfwFxfF
s
cs
c
sc
(9)
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Ex. 3. Find the Fourier cosine transform of axexf )( where a>0
Solution. By differentiation, )()(2 xfaxf
22
22
2)}({
)},({2
)}({ )}({
wa
axfF
xfFaxfFwxfF
c
ccc
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11.9 Fourier Transform
We consider the Fourier integral
dwwxwBwxwAxf sin)( cos)()(0
vdvwvfwBvdvwvfwA
sin)(
1)( ,cos)(
1)(
Combining together, we have
dvdwxvwvf
dvdwwxwvwxwvvfxf
)](cos[)(1
sinsin coscos)(1
)(
0
0
-
And then
dvdwxvwvfxf )](cos[)(2
1)(
(1)
We also get that
0 )](sin[)(2
1
dvdwxvwvf
(2)
(1)-i(2) and resulting
dvdwevf
dvdwxvwixvwvfxf
xviw
)()(2
1
)](sin[)](cos[)(2
1)(
(4)
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Here (4) can be rewritten as
dwedvevfxf iwxiwv
)(
2
1
2
1)(
(5)
Furthermore
dxexfwf iwx)(2
1)(ˆ
(6)
is called the Fourier transform of f (x), and
dwewfxf iwx)(ˆ
2
1)(
(7)
is called the inverse Fourier transform of )(ˆ wf
)(ˆ)}({ wfxfF )()}(ˆ{1 xfwfF
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Sufficient for the existence of the Fourier
transform (6) are the following two conditions:
1. f (x) is piecewise continuous.
2. f (x) is absolutely integrable on the x-axis.
Ex. 1 Find the Fourier transform of f (x)=k if
0
-
Ex. 2 Find the Fourier transform of 2
)( axexf , where a>0. Solution.
dxa
iw
a
iwxa
dxiwxaxeF ax
22
2
22exp
2
1
)exp(2
1}{
2
-
a
w
aaa
w
dvvaa
w
dxa
iwxa
a
iw
4exp
2
11
4exp
2
1
)exp(1
4exp
2
1
2exp
2exp
2
1
22
22
22
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Theorem 1 The Fourier transform is a linear
operation. That means
)}({)}({)}()({ xgbFxfaFxbgxafF
Proof. It is easy to get according to the
definition of Fourier transform.
Theorem 2 Let f (x) be continuous on the x-asis
and Furthermore, let
f ’(x) be absolutely integrable on the x-asis. Then
|| 0)( xasxf
)}({)}({ xfiwFxfF (9)
-
Proof. From the definition, we have
)}({)(2
1
)()(2
1
)(2
1)}({
xfiwFdxexfiw
dexfexf
dxexfxfF
iwx
iwxiwx
iwx
)}({)}({)}({ 2 xfFwxfiwFxfF
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Ex. 3. Find the Fourier transform of 2
)( xxexf
Solution. Let 2
)( xexg by Ex. 2, we have
4/22
2
1}{)}({ wx eeFxgF
Notice that )(22)(2
xfxexg x
4/2
22)}({
2)}({
2
1)}({ we
iwxgF
iwxgFxfF
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The convolution f*g of f (x) and g (x) is defined
by
dvvgvxfdvvxgvfxgf )()()()())(*(
(11)
Theorem 4. Suppose that f (x) and g (x) are
piecewise continuous, bounded, and absolutely
integrable on the x-axis. Then
}{}{2}*{ gFfFgfF (12)
-
Proof. By the definition
}{){2
)()(2
1
)()(2
1
)()(2
1
)()(2
1}*{
)(
gFfF
dpepgdvevf
dvdpepgvf
dvdxevxgvf
dxdvevxgvfgfF
iwpiwv
pviw
iwx
iwx
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By taking the inverse Fourier transform,
)(ˆ)(ˆ))(*(
dwewgwfxgf iwx(13)