12.1 The Dirichlet conditions: Chapter 12 Fourier series Advantages: (1)describes functions that are...
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Transcript of 12.1 The Dirichlet conditions: Chapter 12 Fourier series Advantages: (1)describes functions that are...
12.1 The Dirichlet conditions:
Chapter 12 Fourier series
Advantages:(1) describes functions that are not everywhere continuous
and/or differentiable.(2) represent the response of a system to a period input and
depend on the frequency of the input (3) using in string vibration, light scattering, input signal
transmission in electronic circuit
(1) The function must be periodic.
(2) It must be single-valued and continuous, except possibly at a finite number of finite discontinuities.
(3) It must have only a finite number of maxima and minima within one period.
(4) The integral over one period of a function must converge.
all functions may be written as the sum of an odd and an even part
Chapter 12 Fourier series
:)(
:)(
)()()]()([2
1)]()([
2
1)(
xf
xf
xfxfxfxfxfxfxf
odd
even
oddeven
chosen as the sum of a cosine series
chosen as the sum of a sine series orthogonal properties:
the length of a period is L: )/)(2cos()/2cos(
)/)(2sin()/2sin(
00
00
LLxrLrx
LLxrLrx
pr
prL
prdxL
px
L
rx
pr
prL
prLdxL
px
L
rx
prdxL
px
L
rx
Lx
x
Lx
x
Lx
x
for 0
0for 2/
0for 0)2
sin()2
sin(
for 0
0for 2/
0for )2
cos()2
cos(
and allfor 0)2
cos()2
sin(
0
0
0
0
0
0
Fourier series expansion of the function f(x) is
Chapter 12 Fourier series
0for )2
sin()(2
0for )2
cos()(2
obtain to )2
sin()( and )2
cos()( using Similarly,
)(2
2)]}
2cos()
)(2[cos(
)]2
sin())(2
[sin({2
2
)]2
sin()2
cos([2
)(
properties orthogonal by obtained ts,coefficienFourier called are , ,
)]2
sin()2
cos([2
)(
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
x
x
0
000
00
10
1
0
0
1
0
rdxL
rxxf
Lb
rdxL
rxxf
La
dxL
rxxfdx
L
rxxf
dxxfL
a
aL
L
rx
L
Lxrb
L
rxx
L
Lxra
L
ra
L
dxL
rxb
L
rxadx
adxxf
baa
L
rxb
L
rxa
axf
Lx
xr
Lx
xr
Lx
x
L
Lx
x
r
rr
rr
Lx
x r
Lx
x
Lx
x
rr
rr
r
Ex: Express the square-wave function as a Fourier series
Chapter 12 Fourier series
2/0for 1
02/for 1)(
Tt
tTtf
f(t) is an odd function, so only the sine term survives
T
ttttf
r
rrr
rr
T
rt
r
T
Tdt
T
rt
T
dtT
rttf
Tb
r
TT
T
Tr
2for ........)
5
5sin
3
3sin(sin
4)(
even for 0
odd for 4
])1(1[2
)1(cos2
|)2
cos()2
(4
)2
sin(4
)2
sin()(2
2/0
2/
0
2/
2/
(1) At a point of finite discontinuity, , the Fourier series converges to
(2) At a discontinuity, the Fourier series representation of the function will overshoot
its value. It never disappears even in the limit of an infinite number of terms.
This behavior is known as Gibb’s phenomenon.
Chapter 12 Fourier series
dx
)]()([lim2
10
dd xfxf
1 term 2 terms
3 terms
20 terms
overshooting
12.4 Discontinuous functions
Chapter 12 Fourier series
12.5 Non-periodic functions:
period=L, no particular symmetry
period=2L, antisymmetry; odd fun
period=2L, symmetry; even fun
Ex. : Find the Fourier series of 20 ,)( 2 x xxf
46
16
3
4116
3
42
20for )2
cos()1(
163
4
3
8
4
20
1for )1(16
cos16
)2
cos(8
|)2
cos(8
)2
sin(4
|)2
sin(2
)2
cos()4
2cos(
4
2
0 ,22 ,4L period function even an is )(
),()4( and )()(
2
21
222
122
2
2
0
22
2
20
2222
2
0222022
2
0
20
2
2
0
22
2
2
r
r
r
r
r
r
rxx
xrx
rx
dxxdxxar
rr
rr
dxrx
r
rxx
r
dxrx
xr
rxxr
dxrx
xdxrx
xa
bxxf
Ikxfkxfxfxf
Chapter 12 Fourier series
(1) make the function periodic and symmetric
(2) make the function periodic and antisymmetric
Chapter 12 Fourier series
4not zero to converges seriesFourier the 2at
)2
sin()]1)1(()(
2)1(
1[8
)1)1(()(
16)1(
8|)
2cos()
2(
8)1(
8
])2
sin(2
|)2
sin(2
[4
)1(8
)2
cos(22
|)2
cos(2
)2
sin(
))(2
sin())((2
1)
2sin(
2
1
)4
2sin()(
4
2)
4
2sin(
4
2
0 ,4 ;)()4( and )()(
31
2
32022
2
0
20
2
0
20
22
0
2
0
2
22
0
2
0
2
22
0
2
x
rx
rrx
rr
rx
rrr
dxrx
r
rxxrrr
dxrx
xr
rxx
rdx
rxx
dxrx
xdxrx
x
dxrx
xdxrx
xb
aLxfxfxfxf
rr
r
rrr
r
r
r
Integration and differentiation (1) The Fourier series of f(x) is integrated term by term then the resulting Fourier series converges to the integral of f(x). (2) f(x) is a continuous function of x and is periodic then the Fourier series that results from differentiating term by term converges to f(x).
)2()2
sin()1(
323
4)
2sin()
1(
)1(32
3
4
3 133
122
3
Crx
rxC
rx
rrx
x
r
r
r
r
Chapter 12 Fourier series
Ex: Find the Fourier series of 20 ,)( 3 x xxf
Sol: from the previous example
integrate (1) term by term
)1()2
cos()1(
163
4
122
2
rx
rx
r
r
integrate (1) term by term
put (3) into (2)
)3()2
sin()1(
821
rx
rx
r
r
00)0(for
)2
sin()1(
96)2
sin()1(
161
331
3
Cf
Crx
r
rx
rx
r
r
r
r
12.7 Complex Fourier series
rxirxirx sincos)exp(
* ,)(2
1 ,)(
2
1
is and to relations the has
for 0
for )2
exp()2
exp(
:ityorthogonal the using
)2
exp()(1
)2
exp()(
0
0
0
0
rrrrrrrr
rrr
Lx
x
Lx
xrr
r
ccibacibac
bac
pr
prLdxL
irx
L
irx
dxL
irxxf
Lc
L
irxcxf
Complex Fourier series expansion is:
Chapter 12 Fourier series
Chapter 12 Fourier series
Ex: Find a complex Fourier series for in the range xxf )( 22 x
)2
exp()1(2
)1(2
sin2
cos2
|)2
exp(1
)]exp()[exp(1
)2
exp(2
1|)
2exp(
2)
2exp(
4
1
0 (2)
04
1
0 (1)
,0
22
2222
2
2
22
2
2
2
20
irx
r
ix
r
ir
r
ir
r
i
irx
ririr
irx
dxirx
ir
irx
ir
xdx
irxxc
r
xdxc
r
rr
r
r
r
general proof:
Chapter 12 Fourier series
12.8 Parseval’s theorem:
)(2
1)
2
1(|||)(|
1 2
1
220
220
0r
rr
rr
Lx
xbaacdxxf
L
22
**
**
**
|||)(|1
)()(set we if
])2
exp()(1
[
)2
exp()()()(1
)2
exp()()()(
)2
exp()( ),2
exp()(
0
0
0
0
0
0
0
0
rr
Lx
x
rr
r
Lx
xr
r
Lx
xr
rLx
x
rr
rr
rr
cdxxfL
xfxg
cdxL
irxxg
Lc
dxL
irxxg
L
cdxxgxf
L
L
irxxgcxgxf
L
irxxg
L
irxcxf
Ex: Using Parseval’s theorem and the Fourier series for
evaluate the sum
Chapter 12 Fourier series
22 ,)( 2 x-xxf
1
4
r
r
90256)
9
16
5
16(
1
5
1616
2
1)
3
4()(
2
1)
2
1(
)2
cos()1(
163
4 using and
5
16
4
1
44
14
144
222
1
220
122
22
2
4
r
rr
rr
r
r
r
rbaa
rx
rxdxx