Document info 15.
Fourier AnalysisThursday, 11/2/2006
Physics 158Peter Beyersdorf
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15.
Class Outline
Orthogonality relations
Square wave example
Fourier sine and cosine Integral
Square pulse example
Linewidth and bandwidth
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15.
Introduction
Jean Baptiste Joseph, Baron de Fourier
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“ 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,… “
Periodic functions they can be represented by a Fourier Series
Usually the phase terms are eliminated by rewriting the Fourier Series in terms of even
(cosine) and odd (sine) functions
(the minus sign comes from cos(x+y)=cos(x) cos(y)-sin(x) sin(y) ) giving
whereAm=cm cos φm Bm=-cm sin φm
f(x) = c0 + c1 cos!
2!
"x + #1
"+ c2 cos
!22!
"x + #2
"+ . . .
c1 cos!
m2!
"x + #m
"! Am cos(mkx) + Bm sin(mkx)
f(x) = A0/2 +!!
m=1
Am cos(mkx) +!!
m=1
Bm sin(mkx)
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Fourier Coefficient A0
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The coefficients A0, Am and Bm can be found by considering the following integrals:
so that
A0/2 is the mean value of f(x)
giving
! !
0sin(mkx)dx =
! !
0cos(mkx)dx = 0
! !
0f(x)dx =
! !
0
A0
2dx =
A0
2!
A0 =2!
! !
0f(x)dx
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Orthogonal Functions
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The sine and cosine functions are orthogonal, meaning
where δab is the kronecker delta function obeying δa=b=1 and δa≠b=0
! !
0sin(akx) cos(bkx)dx = 0
! !
0sin(akx) sin(bkx)dx = !ab"/2
! !
0cos(akx) cos(bkx)dx = !ab"/2
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Fourier Coefficient Am & Bm
and all other terms in the sum evaluate to zero.Similarly
so
Using f(x) = A0/2 +!!
m=1
Am cos(mkx) +!!
m=1
Bm sin(mkx)
! !
0f(x) cos(mkx)dx =
! !
0Am cos2(mkx)dx =
!
2Am
! !
0f(x) sin(mkx)dx =
! !
0Bm sin2(mkx)dx =
!
2Bm
Am =2!
! !
0f(x) cos(mkx)dx
Bm =2!
! !
0f(x) sin(mkx)dx
15.
Odd and Even Functions
We can write an arbitrary periodic function as
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
f(x) = A0/2 +!!
m=1
Am cos(mkx) +!!
m=1
Bm sin(mkx)
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Square Wave Example
Consider the square wave shown
What are the values for Am and Bm?
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-1
+1
−λ
+λ
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Square Wave Example
Consider the square wave shown
f(x) is odd, therefor A0=0, Am=0
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-1
+1
−λ
+λ
Bm =2!
! !/2
0(+1)dx +
2!
! !
!/2(!1)dx
=1
m"cos(mkx)
""""!/2
0
+1
m"cos(mkx)
""""!
!/2
=1
m"(1! cos(m"))
f(x) =4!
!sin(kx) +
13
sin(3kx) +15
sin(5kx) + . . .
"thus
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Square Wave Example
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f(x) =4!
!sin(kx) +
13
sin(3kx) +15
sin(5kx) + . . .
"
-1
+
−λ
+λ
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
15.
Square Wave Example
For anharmonic wave, the waveform E(x±vt) can be expressed as a Fourier series of traveling waves
Consider the following form of a square wave
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f(x± vt) = A0/2 +!!
m=1
Am cos(mk(x± vt)) +!!
m=1
Bm sin(mk(x± vt)
+1
−λ +λ+λ/a−λ/a
15.
Square Wave Example
Consider the following form of a square wave
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+1
−λ +λ+λ/a−λ/a
E(x, t) =4a
!!
m=1
4asinc(2!m/a) cos(mkx!m"t)
x! x" vt v = !/kusing
gives
Am =2!
! !
0f(x) cos(mkx)dx
=2!
! !/2
!!/2cos(mkx)dx
=2!
sin(mkx)mk
""""!/a
!!/a
=4a
sin(2"m/a
2 pim/a=
4asinc(2"m/a)
A0 =2!
! !
0f(x)dx =
2!
! !/a
!!/adx =
4a
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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:
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.
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Am
A1=
sin(2!m/a)sin(2!/a)
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Examples
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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
becomes
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f(x) = A0/2 +!!
m=1
Am cos(mkx) +!!
m=1
Bm sin(mkx)
Note: the definition of the Fourier integrals is inconsistent between textbooks, it often includes a normalization factor of π, or √π
f(x) =1!
! !
0A(k) cos(kx)dk +
1!
! !
0B(k) sin(kx)dk
15.
Square Wave Example
Consider a single (non repeating square pulse)
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A(k) =! !
"!f(x) cos(kx)dx =
! L/2
"L/2E0 cos(kx)dx
B(k) = 0
A(k) =E0
ksin(kx)
!!!!L/2
!L/2
=2E0
ksin(kL/2)
= E0Lsin(kL/2)
kL/2= E0Lsinc(kL/2)
giving
15.
Square Wave Example
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A(k) = E0Lsinc(kL/2)
therefor the Fourier integral representation of f(x) is
π 2π 3π 4π
f(x) =E0L
!
! !
0sinc(kL/2) cos(kx)dk
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Diffraction from a slit of width L
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
The Square Aperture
Note: ky=(2π/λ)sinθ≈2πθ/λ, and a similar expression holds in x
y
f(|x| < L) = E0 cos(k0x)f(|x| > L) = 0
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Finite length harmonic wave
-L LNote that f(x) is an even function so B(k)=0
The Square Envelope
What are A(k) and B(k) for this wave?
A(k) =! L
LE0 cos(k0x) cos(kx)dx
= E0/2! L
LE0 (cos((k0 + k)x) cos((k0 ! k)x)) dx
= E0L
"sin((k0 + k)L)
(k0 + k)L)+
sin((k0 ! k)L)(k0 ! k)L)
#
= E0L (sinc((k0 + k)L) + sinc((k0 ! k)L))
15.
The Square Envelope
k0-π/L k0+π/L
k0-k0This is for a
“forward” going wave
This is for a “backward” going
wave
f(|x| < T ) = E0 cos(!0t)f(|x| > T ) = 0
A(!) = E0T (sinc((!0 + !)T) + sinc((!0 ! !)T))
15.
Look at the time function
By analogy
What is the width of the transform?
Linewidth and Bandwidth
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Given the transforms we’ve found, we have the “uncertainty relations”
ΔkΔx=4πΔωΔt=4π
more generally we have
ΔkΔx≥πΔωΔt≥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
Uncertainty Relations
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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.
Coherence Length
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
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