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

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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

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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

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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

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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

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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))

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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))

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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

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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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