Fourier Analysis
Thursday, 11/2/2006
Physics 158
Peter Beyersdorf
Document info
15. 1
Class Outline
Orthogonality relations
Square wave example
Fourier sine and cosine Integral
Square pulse example
Linewidth and bandwidth
15. 2
Introduction
Jean Baptiste Joseph, Baron de Fourier
Periodic functions they can be represented by a
Fourier Series
!
"
!
"
2π
2π
f (x) = c0 + c1 cos
x + φ1 + c2 cos 2 x + φ2 + . . .
λ
λ
Usually the phase terms are eliminated by
rewriting the Fourier Series in terms of even
(cosine) and odd (sine) functions
!
2π
c1 cos m x + φm
λ
"
≡ Am cos(mkx) + Bm sin(mkx)
where
Am=cm cos φm
Bm=-cm sin φm
“ a function f(x) having a spatial
period λ can be synthesized by a
sum of harmonic functions whose
period are integer sub-multiples of
l, i.e λ, λ/2, λ/3,… “
(the minus sign comes from cos(x+y)=cos(x) cos(y)-sin(x) sin(y) )
giving
f (x) = A0 /2 +
∞
!
m=1
Am cos(mkx) +
∞
!
m=1
Bm sin(mkx)
15. 3
Fourier Coefficient A0
The coefficients A0, Am and Bm can be found by
considering the following integrals:
!
λ
sin(mkx)dx =
!
λ
cos(mkx)dx = 0
0
0
so that
!
0
giving
λ
f (x)dx =
!
λ
0
2
A0 =
λ
!
A0
A0
dx =
λ
2
2
λ
f (x)dx
0
A0/2 is the mean value of f(x)
15. 4
Orthogonal Functions
The sine and cosine functions are orthogonal, meaning
!
λ
sin(akx) cos(bkx)dx =
0
0
!
!
λ
0
λ
sin(akx) sin(bkx)dx = δab λ/2
cos(akx) cos(bkx)dx = δab λ/2
0
where δab is the kronecker delta function obeying δa=b=1 and δa≠b=0
15. 5
Fourier Coefficient Am & Bm
Using
f (x) = A0 /2 +
∞
!
Am cos(mkx) +
m=1
!
λ
f (x) cos(mkx)dx =
!
Bm sin(mkx)
m=1
!
λ
λ
Am cos (mkx)dx = Am
2
2
0
0
∞
!
and all other terms in the sum evaluate to zero.
Similarly
λ
f (x) sin(mkx)dx =
0
so
Am
Bm
=
=
2
λ
2
λ
!
!
0
λ
λ
λ
Bm sin (mkx)dx = Bm
2
2
f (x) cos(mkx)dx
0
!
0
λ
f (x) sin(mkx)dx
15.
Odd and Even Functions
We can write an arbitrary periodic function as
f (x) = A0 /2 +
∞
!
m=1
Am cos(mkx) +
∞
!
Bm sin(mkx)
m=1
and find the values of A0, Am and Bm.
If f(x) is odd f(x)=-f(-x) then it must be expressed in
terms of odd functions only, so that Am=0 for all m
If f(x) is even f(x)=f(-x) then it must be expressed in
terms of even functions only, so that Bm=0 for all m
15.
Square Wave Example
Consider the square
wave shown
What are the values
for Am and Bm?
+1
+λ
−λ
-1
15. 8
Square Wave Example
Consider the square
wave shown
+1
+λ
−λ
f(x) is odd, therefor
A0=0, Am=0
Bm
=
=
=
thus
2
λ
!
-1
!
2 λ
(+1)dx +
(−1)dx
λ λ/2
0
"λ/2
"λ
"
"
1
1
cos(mkx)""
cos(mkx)""
+
mπ
mπ
0
λ/2
λ/2
1
(1 − cos(mπ))
mπ
4
f (x) =
π
!
"
1
1
sin(kx) + sin(3kx) + sin(5kx) + . . .
3
5
15. 9
Square Wave Example
4
f (x) =
π
!
"
1
1
sin(kx) + sin(3kx) + sin(5kx) + . . .
3
5
Each term added to the
series makes the sum
converge more closely to
the actual square wave.
Note that for a finite
number of terms, the
sum overshoots the
function and has ringing
+
+λ
−λ
-1
15.10
Square Wave Example
For anharmonic wave, the waveform E(x±vt) can
be expressed as a Fourier series of traveling
waves
f (x ± vt) = A0 /2 +
∞
!
m=1
Am cos(mk(x ± vt)) +
∞
!
m=1
Bm sin(mk(x ± vt)
Consider the following form of a square wave
+1
−λ
−λ/a
+λ/a
+λ
15. 11
Square Wave Example
Consider the following form of a square wave
+1
2
A0 =
λ
Am
=
=
=
=
!
λ
0
2
λ
2
λ
2
f (x)dx =
λ
!
λ
!
4
dx =
a
−λ/a
λ/a
−λ
f (x) cos(mkx)dx
0
!
λ/2
−λ/2
cos(mkx)dx
"λ/a
2 sin(mkx) ""
λ
mk "−λ/a
using
x → x − vt
−λ/a
+λ/a
+λ
v = ω/k
gives
∞
!
4
4
E(x, t) =
sinc(2πm/a) cos(mkx − mωt)
a m=1 a
4 sin(2πm/a
4
= sinc(2πm/a)
a 2 pim/a
a
15. 12
Approximations
How many terms in the Fourier series are
necessary to accurately represent the original
function?
Consider the relative magnitude of the mth term,
relative to the 1st:
Am
sin(2πm/a)
=
A1
sin(2π/a)
The first term to have a value of zero occurs for
a=m/2. Thus the narrower the pulse, the more
frequency components (of significant amplitude)
it contains.
15.13
Examples
15.
Fourier Integral
As λ→∞, the Fourier series terms go from a
discrete set of k-values to a smooth function in
k-space called the spectral density.
The Fourier series
f (x) = A0 /2 +
∞
!
∞
!
Am cos(mkx) +
m=1
becomes
1
f (x) =
π
!
0
∞
1
A(k) cos(kx)dk +
π
Bm sin(mkx)
m=1
!
∞
B(k) sin(kx)dk
0
Note: the definition of the Fourier integrals is inconsistent between
textbooks, it often includes a normalization factor of π, or √π
15.15
Square Wave Example
Consider a single (non repeating
square pulse)
A(k) =
B(k)
= 0
giving
A(k)
!
=
∞
−∞
f (x) cos(kx)dx =
!
L/2
E0 cos(kx)dx
−L/2
!L/2
!
E0
2E
0
!
sin(kx)!
sin(kL/2)
=
k
k
−L/2
sin(kL/2)
= E0 Lsinc(kL/2)
= E0 L
kL/2
15. 16
Square Wave Example
A(k) = E0 Lsinc(kL/2)
therefor the Fourier integral
representation of f(x) is
E0 L
f (x) =
π
!
∞
sinc(kL/2) cos(kx)dk
0
π
2π
3π
4π
15.17
The Square Aperture
Diffraction from a slit of width L
y
The amplitude of the
diffraction pattern is
sinc(kyL/2), the intensity
is sinc2(kyL/2).
sinc(kyL/2)=0 at kyL/2=π±mπ
The first zero is at ky=±2π/L
Note: ky=(2π/λ)sinθ≈2πθ/λ, and a similar expression holds in x
15.
The Square Envelope
Finite length harmonic wave
f (|x| < L) = E0 cos(k0 x)
f (|x| > L) = 0
Note that f(x) is an
even function so B(k)=0
-L
L
What are A(k) and B(k) for this wave?
15.
The Square Envelope
A(k) =
!
L
L
E0 cos(k0 x) cos(kx)dx
!
L
= E0 /2
E0 (cos((k0 + k)x) cos((k0 − k)x)) dx
L
#
"
sin((k0 + k)L) sin((k0 − k)L)
+
= E0 L
(k0 + k)L)
(k0 − k)L)
= E0 L (sinc((k0 + k)L) + sinc((k0 − k)L))
-k0
This is for a
“backward” going
wave
k0
This is for a
“forward” going
wave
k0-π/L k0+π/L
15.
Linewidth and Bandwidth
Look at the time function
f (|x| < T ) = E0 cos(ω0 t)
f (|x| > T ) = 0
By analogy
A(ω) = E0 T (sinc((ω0 + ω)T) + sinc((ω0 − ω)T))
What is the width of the transform?
15.
Uncertainty Relations
Given the transforms we’ve found, we have the
“uncertainty relations”
ΔkΔx=4π
ΔωΔt=4π
more generally we have
ΔkΔx≥π
ΔωΔt≥π
if the equality in the above relations hold the
pulse is said to be transform limited.
For example a Q-switched laser pulse with a
pulse dureation of Δt=50 ns has a transform
limited bandwidth of Δf=Δω/2π=2/Δt≈40 MHz
15.
Coherence Length
Define the coherence time Δtc as Δtc=2/Δf, where Δf is the frequency
bandwidth of the radiation. Hence the coherence length is Lc=cΔtc
This is the distance in space for a phase slip of π due to the frequency
bandwidth. Ambient white light has a bandwidth of about
Δf=(0.4-0.7)x1015 Hz, corresponding to a coherence time of about 3 fs.
The corresponding coherence length is about 1 μm or about 2λ of green
light.
For a discharge lamp lines in the visible the typical bandwidth (due to
Doppler broadening) is Δf≈1 GHz Δx≈1 m.
Michelson used the Cd red line at λ=643.847 nm ± 0.0065 nm to
measure the meter stick. The corresponding bandwidth for this line
width is 725 MHz and the coherence length is 0.83 m.
15.
Summary
An arbitrary periodic function f(x) can be expressed by
an infinite sum of sinusoidal waves having frequencies
that are integer multiples of the frequency of f(x)
When only a finite number of components are included
in the approximation of f(x) the resulting waveform has
overshoot and ringing
The narrower the pulse (compared to the period) the
more frequency components are necessary to
accurately describe the pulse
As the period approaches ∞ the Fourier sum becomes
an integral
15.24