a fourier series can be defined for any function over the interval 0 x 2 l
DESCRIPTION
Similarly. A Fourier series can be defined for any function over the interval 0 x 2 L. where. Often easiest to treat n=0 cases separately. Compute the Fourier series of the SQUARE WAVE function f given by. p. 2 p. Note: f(x) is an odd function ( i.e. f(-x) = -f(x) ). - PowerPoint PPT PresentationTRANSCRIPT
A Fourier series can be defined for any function over the interval 0 x 2L
1
0 sincos2
)(n
nn L
xnb
L
xna
axf
where dxL
xnxf
La
L
n
2
0cos)(
1
dxL
xnxf
Lb
L
n
2
0sin)(
1
Ofteneasiestto treat
n=0 casesseparately
Compute the Fourier series of the SQUARE WAVE function f given by
)(xf2,1
0,1
x
x
2
Note: f(x) is an odd function ( i.e. f(-x) = -f(x) )
so f(x) cos nx will be as well, while f(x) sin nx will be even.
dxL
xnxf
La
L
n
2
0cos)(
1)(xf
2,1
0,1
x
x
dxxfa 0cos)(1 2
00
dxdx 0cos)1(0cos11 2
0
0
dxnxdxnxan
2
0cos)1(cos1
1
dxnnxdxnx ( )coscos1
00
dxnxdxnx
00coscos
1
change of variables: x x' = x-
periodicity: cos(X+n) = (-1)ncosX
for n = 1, 3, 5,…
dxL
xnxf
La
L
n
2
0cos)(
1)(xf
2,1
0,1
x
x
00 a
dxnxan
0cos
2for n = 1, 3, 5,…
0na for n = 2, 4, 6,…
change of variables: x x' = nx
dxxn
an
n
0cos
2 0
IF f(x) is odd, all an vanish!
dxL
xnxf
Lb
L
n
2
0sin)(
1)(xf
2,1
0,1
x
x
00sin)(1 2
00 dxxfb
dxnxdxnxbn
2
0sinsin
1
dxnnxdxnx ( )sinsin1
00
periodicity: sin(X±n) = (-1)nsinX
dxnxdxnx
00sinsin
1
for n = 1, 3, 5,… and vanishing for n = 2, 4, 6,…
change of variables: x xnew = xold-
)(xf2,1
0,1
x
x
00 b
dxnxbn
0sin
2for n = 1, 3, 5,…
0nb for n = 2, 4, 6,…
change of variables: x x' = nx
dxxn
n
0sin
2
dxL
xnxf
Lb
L
n
2
0sin)(
1
dxxn
0sin
2
for odd n
nxn
40cos
2 for n = 1, 3, 5,…
)5
5sin
3
3sin
1
sin(
4)( xxx
xf
1
2x
y N = 1
N= 5
http://www.jhu.edu/~signals/fourier2/
http://www.phy.ntnu.edu.tw/java/sound/sound.html
http://mathforum.org/key/nucalc/fourier.html
http://www.falstad.com/fourier/
Leads you through a qualitative argument in building a square wave
Add terms one by one (or as many as you want) to build fourier series approximation to a selection of periodic functions
Build Fourier series approximation to assorted periodic functionsand listen to an audio playing the wave forms
Customize your own sound synthesizer
Fourier transformsof one another
Two waves of slightly different wavelength and frequency produce beats.
x
x
1k
k = 2
NOTE: The spatial distribution depends on the particular frequencies involved
Many waves of slightly different wavelength can produce “wave packets.”
Adding together many frequencies that are bunched closely together
…better yet…
integrating over a range of frequencies
forms a tightly defined, concentrated “wave packet”
http://phys.educ.ksu.edu/vqm/html/wpe.html
A staccato blast from a whistle cannot be formed by a single pure frequencybut a composite of many frequencies close to the average (note) you recognize
You can try building wave packets at
The broader the spectrum offrequencies (or wave number)
…the shorterthe wave packet!
The narrower the spectrum offrequencies (or wave number)
…the longerthe wave packet!
Fourier Transforms Generalization of ordinary “Fourier expansion” or “Fourier series”
de)(g2
1)t(f ti
tdetf2
1g tiω)()(
Note how this pairs “canonically conjugate” variables and t.
Whose product must be dimensionless (otherwise eit makes no sense!)
Conjugate variablestime & frequency: t,
f
2
2
What about
coordinate position & ???? r or x
inverse distance??wave number,
p2
In fact through the deBroglie relation, you can write:ph /
/iEte
/hchEc
hhp
/x
ixpe
x0
For a well-localized particle (i.e., one with a precisely known position at x = x0 )
we could write:
)(0
xx
Dirac -function
a near discontinuous spike at x=x0,(essentially zero everywhere except x=x0 )
x0
with
1)(
0dxxx
such that
dxxxxfdxxxxf )()()()(
000
f(x)≈ f(x0), ≈constant over xx, x+x
x
1x
For a well-localized particle (i.e., one with a precisely known position at x = x0 )
we could write: )(0
xx
In Quantum Mechanics we learn that the spatial wave function (x) can be complemented by the momentum spectrum of the state, found through the Fourier transform:
dxexp ixp /)(
2
1)(
Here that’s
//
00
2
1)(
2
1)(
pixixp edxexxp
Notice that the probability of measuring any single momentum value, p, is:
2
1
2
1)(
//2
2 00
pixpixeep
What’sTHAT mean?
The probability is CONSTANT – equal for ALL momenta! All momenta equally likely!The isolated, perfectly localized single packet must be comprised of an infinite range of momenta!
(k) (x)
11
2
kx
k0
2
2
2
kx
(x)(k)
k0
Remember:
…and, recall, even the most general whether confined by some potential OR free actually has some spatial spread within some range of boundaries!
Fourier transforms do allow an explicit “closed” analytic form for
the Dirac delta function
de2
1)t( )t(i
Area within1 68.26%1.28 80.00% 1.64 90.00%1.96 95.00%2 95.44%2.58 99.00%3 99.46%4 99.99%
-2 -1 +1 +2
2
2
2
)x(
e2
1x
Let’s assume a wave packet tailored to be something like aGaussian (or “Normal”) distribution
A single “damped”pulse bounded tightlywithin a few of its
mean postion, μ.
For well-behaved (continuous) functions (bounded at infiinity)
like f(x)=e-x2/22
dxexfkF ikx)(2
1)(
Starting from:
f(x) g'(x) g(x)= e+ikxik
dxxgx'fxgxf )()()()(
2
1
dxek
ix'fe
k
xif ikxikx )()(
2
1
f(x) is
boundedoscillates in thecomplex plane
over-all amplitude is damped at ±
we can integrate this “by parts”
dxex'fk
ikF ikx)(
2
1)(
)()(2
1kikFdxex'f ikx
Similarly, starting from:
dkekFxf ikx)(2
1)(
)()(2
1xixfdkek'F ikx
And so, specifically for a normal distribution: f(x)=ex2/22
differentiating: )()(2
xfx
xfdx
d
from the relation just derived: kdekF
ixf
dx
d xki ~)
~(
2
1)(
~
2 '
Let’s Fourier transform THIS statement
i.e., apply: dxeikx
21
on both sides!
dxei
kikF ikx 2
1)( 2 1
2 F'(k)e-ikxdk~ ~~
kdkFi ~
)~
(2 ' ei(k-k)xdx
~ 1 2
(k – k)~
kdkFi
kikF~
)~
()( 2 ' ei(k-k)xdx
~ 1 2
(k – k)~
)()( 2 kFi
kikF ' selecting out k=k
~
rewriting as: 2
)(
/)( kkF
dkkdF
0
k
0
k
dk''
''dk'
22
2
1)0(ln)(ln kFkF
2221
)0(
)( ke
F
kF 22
21
)0()(k
eFkF
2221
)0()(k
eFkF
22 2/)( xexf Fourier transforms
of one anotherGaussian distribution
about the origin
dxexfkF ikx)(2
1)(
Now, since:
dxxfF )(2
1)0(
we expect:
10 xie
22
1)0(
22 2/
dxeF x
2221
2)(k
ekF
22 2/)( xexf Both are of the form
of a Gaussian!
x k 1/
x k 1
orgiving physical interpretation to the new variable
x px h