Solid State III, Lecture 23
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Transcript Solid State III, Lecture 23
Lecture 8
Aims:
Fourier Analysis.
Fourier Theory:
Description of waveforms in terms of a
superposition of harmonic waves.
Fourier series (periodic functions);
Fourier transforms (aperiodic
functions).
Wavepackets
Convolution
convolution theorem.
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Fourier Theory
It is possible to represent (almost) any function
as a superposition of harmonic functions.
Periodic functions:
Fourier series
Non-periodic functions:
Fourier transforms
Mathematical formalism
Function f(x), which is periodic in x, can be
written:
f x
1
Ao
2
where,
2
An
l
Bn
2
l
n 1
l/2
2nx
2nx
A
cos
B
sin
n
n
l
l
2nx
f x cos
dx n 0,1,2...
l
2nx
f x sin
dx n 1,2...
l
l / 2
l/2
l / 2
Expressions for An and Bn follow from the
“orthogonality” of the “basis functions”, sin and
cos.
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Complex notation
Example: simple case of 3 terms
y cos2x l sin 2x l cos4x l
cos2x l
sin 2x l
cos4x l
y
Exponential
representation:
f x
Cn
Cn ei 2nx / l
n
1 l/2
f x e i 2nx / l d x
l l / 2
f x
with k=2n/l.
C k
3
C k eikx
n
1 l/2
f x e ikx d x
l l / 2
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Example
Periodic top-hat:
f x A l / 8 x l / 8
0 l / 2 x l / 8, l / 8 x l / 2
f(x)
Fourier transform
1 l /8
l /8
ikx
A e
l l / 8
l ik
l / 8
A
2A
eikl / 8 e ikl / 8
sin kl / 8
ikl
kl
C k
Aeikxdx
A
A
sinc kl / 8 sinc n / 4
4
4
sin x
x
N.B.
1
x 0
x
x
4
Zero when n
is a multiple of 4
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Fourier transform variables
x and k are conjugate variables.
t
Analysis applies to a periodic function in any
variable.
and w are conjugate.
F t
Cn
Cn ei 2nt / T
n
1 T /2
F t e i 2nt / T
[6.1]
w n 2n / T
T T / 2
Example: Forced oscillator
Response to an arbitrary, periodic, forcing
function F(t). We can represent F(t) using [6.1].
If the response at frequency nwf is R(nwf), then
the total response is
inw t
R nw f C n e f
n
Linear in both response and driving amplitude
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Fourier Transforms
Non-periodic functions:
limiting case of periodic function as period .
The component wavenumbers get closer and
merge to form a continuum. (Sum becomes an
integral)
1
ikx
f ( x)
g (k )
g (k )e dk
2
1
f ( x)e ikxdk
2
This is called Fourier Analysis.
f(x) and g(k) are Fourier Transforms of each
other.
Example:
f ( x) A
x / 2 x x / 2
0
x x / 2
x / 2
ikx
x
/
2
1
A e
ikx
g (k )
Ae
dk
2 x / 2
2 ik
Top hat
x / 2
2A
Ax
sin( kx / 2)
sinc( kx / 2)
2 k
2
Similar to Fourier series but now a continuous
function of k.
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Fourier transform of a Gaussian
Gaussain with r.m.s. deviation x=s.
A
x 2 / 2s 2
f ( x)
e
s 2
f ( x)dx A
Note
Fourier transform
1
A
x 2 / 2s 2 ikx
g (k )
e
e
dx
2 s 2
Integration can be performed by completing the
square of the exponent -(x2/2s2+ikx).
2
2
2 2
ik 2s k s
x
ikx
2
2
2
2s
2s
x
k 2s 2 / 2
u2
where,
x
ik 2s
u
;
du dx / 2s
2s
2
A u 2 k 2s 2 / 2
g (k )
e e
2sdu
=
2s
A
2s
2se
k 2s 2 / 2
7
e
u 2
A k 2s 2 / 2
du
e
2
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Transforms
The Fourier transform of a Gaussian is a
Gaussian.
Note: k=1/s. i.e. xk=1
Important general result:
“Width” in Fourier space is inversely related
to “width” in real space. (same for top hat)
d-function
Common functions (Physicists crib-sheet)
d-function
cosine
sine
infinite lattice
of d-functions
top-hat
Gaussian
In pictures………...
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constant
2 d-functions
2 d-functions
infinite lattice
of d-functions
sinc function
Gaussian
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Pictorial transforms
Common transforms
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Wave packets
Localised waves
A wave localised in space can be created by
superposing harmonic waves with a narrow
range of k values.
1
f ( x)
2
g (k )eikxdk
The component harmonic waves have
amplitude
1
ikx
g (k )
2
2
f ( x )e
dx
At time t later, the phase of component k will be
kx-wt, so
1
i ( kx wt )
f ( x, t )
g ( k )e
dk
Provided w/k=constant (independent of k) then
the disturbance is unchanged i.e. f(x-vt).
We have a non-dispersive wave.
When w/k=f(k) the wave packet changes shape
as it propagates.
We have a dispersive wave.
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Convolution
Convolution: a central concept in Physics.
Convolution symbol
h( x) f1 ( x) * f 2 ( x)
Convolution integral
f1(u) f2 ( x u)du
h is the convolution of f1 and f2
It is the “smearing” or “blurring” of one function
by the other.
Examples occur in all experimental situations
where the limited resolution of the apparatus
results in a measurement “broader” than the
original.
In this case, f1 (say) represents the true signal
and f2 is the effect of the measurement. f2 is the
point spread function.
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Convolution theorem
Convolution and Fourier transforms
Convolution theorem:
The Fourier transform of a PRODUCT of two
functions is the CONVOLUTION of their Fourier
transforms.
Conversely:
The Fourier transform of the CONVOLUTION of
two functions is a PRODUCT of their Fourier
transforms.
Proof:
1
h( k )
f1 ( x) f 2 ( x)e ikxdx
2
F.T.
of
f1.f2
1
2
1
2
1
2
1
iux
ikx
g
(
u
)
e
du
f
(
x
)
e
dx
1
2
2
1
g1 (u )du
f 2 ( x)e i ( k u ) x dx
2
g1 (u ) g 2 (k u )du
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Convolution
of g1 and g2
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Convolution………….
Summary:
FTf1 ( x) g1 (k ), FTf 2 ( x) g 2 (k )
1
FT
f
(
x
)
f
(
x
)
g1 (k ) * g 2 (k )
then
1
2
2
and FT f1 ( x) * f 2 ( x) 2 g1 ( k ) g 2 ( k )
If,
Examples:
Optical instruments and resolution
1-D idealised spectrum of “lines” broadened
to give measured spectrum
2-D: Response of camera, telescope. Each
point in the object is broadened in the image.
Crystallography. Far field diffraction pattern is a
Fourier transform. A perfect crystal is a
convolution of “the lattice” and “the basis”.
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Convolution Summary
Must know….
Convolution theorem
How to convolute the following functions.
d-function and any other function.
Two top-hats
Two Gaussians.
s 2 s 12 s 22
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