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©2000, D. L. Jaggard 1 EE 511 EE 511 EE 511: Introduction to EE 511: Introduction to Fourier Optics and Fourier Optics and Image Understanding Image Understanding Volume 2 III. Scalar Diffraction Theory (Physical Optics) IV. Fresnel and Fraunhofer Diffraction Dwight L. Jaggard University of Pennsylvania 308 Moore <[email protected]> 215.898.4411

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©2000, D. L. Jaggard 1EE 511EE 511

EE 511: Introduction to EE 511: Introduction to Fourier Optics and Fourier Optics and

Image UnderstandingImage UnderstandingVolume 2

III. Scalar Diffraction Theory (Physical Optics)IV. Fresnel and Fraunhofer Diffraction

Dwight L. JaggardUniversity of Pennsylvania

308 Moore<[email protected]>

215.898.4411

©2000, D. L. Jaggard 2EE 511EE 511

Goal

Understand foundations of scalar diffraction, perform calculations and appreciate

results

©2000, D. L. Jaggard 3EE 511EE 511

Course OutlineCourse OutlineI. History and BackgroundII. Fourier Transforms and Linear Systems

III. Scalar Diffraction Theory (Physical Optics)

IV. Fresnel and Fraunhofer Approximations

V. Vector Diffraction TheoryVI. Geometrical and Ray Optics

VII. Properties of Lens

VIII. Coherent and Incoherent Imaging

IX. Partial Coherence TheoryX. Special Topics

©2000, D. L. Jaggard 4EE 511EE 511

Two Early Contributors to Wave Propagation and Interference

Chirstiaan Huygens (1629 -1695)Contributed to wave propagation &

concept of secondary sources

Thomas Young (1773 -1829)Discovered the principle of wave

interference (double -slit experiment)

©2000, D. L. Jaggard 5EE 511EE 511

III. Scalar Diffraction TheoryA. Maxwell’s EquationsB. Wave and Helmholtz EquationsC. Green’s FunctionD. Dirichlet SolutionE. Neumann SolutionF. Solution SummaryG. Babinet’s Principle

©2000, D. L. Jaggard 6EE 511EE 511

How do waves diffract?How do waves diffract?

Diffraction of light by saw tooth

©2000, D. L. Jaggard 7EE 511EE 511

How do waves diffract?How do waves diffract?

Diffraction of light by circular aperture

©2000, D. L. Jaggard 8EE 511EE 511

A. Maxwell’s Equationsl Time-Domain Equationsl Frequency-Domain Equations

James Clerk Maxwell (1831-1879)Showed light was electromagnetic

phenomena & discovered fundamentalequations for EM fields

©2000, D. L. Jaggard 9EE 511EE 511

Time-Domain Maxwell’s Equations

l Fields and sources are functions of time t and space x = (x,y,z) [Cartesian] or x = (r,θ,φ) [spherical]

l Current density j and charge density ρ are the sourcesl Electric and magnetic fields are connectedl Boldface denotes vectors

∇ × e(t) = −∂b(t)

∂t∇ •d(t ) = ρ(t )

∇ × h(t) = j(t ) +∂d(t)

∂t∇ • b(t) = 0

where

e,d,b,h, are EM fields

j,ρ are sources

©2000, D. L. Jaggard 10EE 511EE 511

Frequency-Domain Maxwell’s Equations

l The upper case fields are the F.T. of the lower case fieldsl Harmonic waves vary as exp( –jωt) [monochromatic]l Differentiation in the time domain becomes multiplication

by [–jωt] in the frequency domain

∇ ×E(x, ω) = + jωB( x,ω ) ∇ •D(x,ω ) = ρ (x, ω)

∇ ×H(x,ω ) = J(x,ω) − jωD(x,ω) ∇•B(x,ω ) = 0where

E(x,ω) = e(x, t)−∞

∫ e+ jωtdt

e(x, t) =1

2πE(x,ω)

−∞

∫ e− jωtdω

etc.

©2000, D. L. Jaggard 11EE 511EE 511

s• Operator (Divergence)

For vector A = Ax ˆ e x + Ayˆ e y + Az ˆ e z = Ar ˆ e r + Aθ ˆ e θ + Aφ ˆ e φ :

Cartesian Coordinates x = (x, y,z):

∇ • A =∂Ax

∂x+

∂Ay

∂y+

∂Az

∂z

Sperhical Coordinates x = (r,θ ,φ ):

∇ • A =

1r 2

∂∂r

r 2 Ar[ ]+1

r sinθ∂

∂θsinθAθ[ ]+

1r sinθ

∂Aφ

∂φ

©2000, D. L. Jaggard 12EE 511EE 511

sx Operator (Curl)

For vector A = Ax ˆ e x + Ayˆ e y + Az ˆ e z = Ar ˆ e r + Aθ ˆ e θ + Aφˆ e φ :

Cartesian Coordinates x = (x, y,z):

∇ × A =∂Az

∂y−

∂Ay

∂z

ˆ e x +∂Ax

∂z−

∂Az

∂x

ˆ e y +∂Ay

∂x−

∂Ax

∂y

ˆ e z

Sperhical Coordinates x = (r,θ ,φ ):

∇ × A =

1r sinθ

∂∂θ

sinθAφ( )−∂Aθ

∂φ

e r +

1r sinθ

∂Ar

∂φ

1r

∂∂r

rAφ( )

ˆ e θ +

1r

∂∂r

rAθ( )− ∂Ar

∂θ

ˆ e φ

©2000, D. L. Jaggard 13EE 511EE 511

B. Wave and Helmholtz Equations

l Find time-domain and frequency-

domain equations for EM field

l Scalarize result for optical fields

©2000, D. L. Jaggard 14EE 511EE 511

Wave and Helmholtz Equations

l Plane waves [e.g., exp( jkz)] and spherically waves [e.g., (1/r)exp( jkr)] are solutions to the Helmholtzequation for exp(–jωt) excitation

l These are vector equations and contain polarization information

Take curl of ∇ × e(x,t ) or ∇ ×h(x,t ) equations to

find wave equation for the source- free case:

∇2 − 1c2

∂2

∂t2

e(x,t )

h(x,t )

= 0

Take curl of ∇ ×E(x,ω) or ∇ ×H(x,ω) equations to

find Helmholtz equation for the source- free case:

∇2 + k 2[ ] E(x,ω)

H(x,ω)

= 0

Here v =1 / µε and k = ω / v

©2000, D. L. Jaggard 15EE 511EE 511

Scalarizing Helmholtz Equation

l One can neglect vector nature of EM fields in certain approximations:

l Very high frequency (e.g., large aperture and observer far from aperture

l Not interested in polarization effects

l Consider scalar form of Helmholtz equation where U is a scalar component (“optical field”) of one of the fields E or H

l Consider monochromatic case and suppress time -harmonic variation exp(–jωt)

©2000, D. L. Jaggard 16EE 511EE 511

Scalar Helmholtz Equation

l Need to find solution for optical field Ul Will use Green’s function approachl Note: λ is wavelength in the mediuml λ = λ0/nl For diffraction, usually n = 1

∇2 + k2[ ]U(x) = 0

Here the wavenumber is k = ω / v = 2π / λ

©2000, D. L. Jaggard 17EE 511EE 511

s2 Operator (Laplacian)

For scalar U (only):

Cartesian Coordinates x = (x, y,z):

∇2U (x, y, z) = ∂2

∂x2 + ∂ 2

∂y2 + ∂ 2

∂z2

U (x, y, z)

Sperhical Coordinates x = (r,θ ,φ ):

∇2U (r,θ,φ) =

1r2

∂∂r

r2 ∂∂r

+

1r2 sinθ

∂∂θ

sinθ ∂∂θ

+

1r2 sin 2 θ

∂2

∂φ 2

U(r,θ ,φ)

©2000, D. L. Jaggard 18EE 511EE 511

C. Green’s Functionl Green’s function G( x,x’) is similar

to impulse response (except Green’s function already has the shift included)

l Physically:

Green’s function [G( x,x’)] is the response of a system at x due to

a point source [ δ(x–x’)] at x’

©2000, D. L. Jaggard 19EE 511EE 511

More Green’s Functionl G(x,x’) allows fields to be written in

terms of integrals over sources f(x’) (if any) and values of fields on the surface

l For source-free fields, the optical field U(x) in a volume is uniquely found as a function of its value on the surface

l Reciprocity: G(x,x’) = G(x’,x)

©2000, D. L. Jaggard 20EE 511EE 511

Geometry – Two Views

x’ fx~x

y’

z

fy~y

U(fx,fy)U0(x’,y’)

S0

x’

y’ zU(x,y,z)

S0

S8

R

VdS’dS’

ez

λ

©2000, D. L. Jaggard 21EE 511EE 511

“Green Mathematics”

l Multiply the top equation by G( x,x’), the bottom by U(x’), and subtract

l Integrate result over volume V enclosed by surface S = S0 + S8 which encloses RHS

Let

′ ∇ 2 + k2[ ]U( ′ x ) = − f ( ′ x )

where f ( ′ x ) is a sourceThe associated Green' s function satisfies

′ ∇ 2 + k2[ ]G(x, ′ x ) = −δ(x − ′ x )

©2000, D. L. Jaggard 22EE 511EE 511

Expression for U(x)

l Volume integral vanishes in source-free regime f(x’) = 0

U(x) =G(x, ′ x ) f ( ′ x )d ′ x

V∫ +

G(x, ′ x ) ′ ∇ 2U( ′ x ) − U( ′ x ) ′ ∇ 2G(x, ′ x )[ ]d ′ x V∫

=G(x, ′ x ) f ( ′ x )d ′ x

V∫ +

G(x, ′ x ) ′ ∇ U ( ′ x )− U( ′ x ) ′ ∇ G(x, ′ x )[ ]• d ′ S S =S0 +S ∞

=G(x, ′ x ) f ( ′ x )d ′ x

V∫ +

U ( ′ x ) ′ ∇ G(x, ′ x ) − G(x, ′ x ) ′ ∇ U( ′ x )[ ]• ˆ e zdS'S0

©2000, D. L. Jaggard 23EE 511EE 511

U(x) in Source-Free Regime

lThis equation is exact for scalar Helmholtz equation

l Integration is over entire aperture plane S0 define by z = 0 [integration over S8 –> 0]

lU(x) is given anywhere in RHS z > 0 if U(x‘) and s‘U(x‘)l ez are known on the surface S 0, and if G(x,x’) is known

lUsual method is to assume U(x‘) and s‘U(x‘)l ez

U(x) = U ( ′ x ) ′ ∇ G(x, ′ x ) − G(x, ′ x ) ′ ∇ U( ′ x )[ ]• ˆ e zdS'

S0∫

©2000, D. L. Jaggard 24EE 511EE 511

Free-Space Green’s Function

l This solution is exact

l Positive (upper) sign is physically reasonable solution for causal system (it provides outgoingwaves)

lOnly valid if there are no boundaries

D.E.

∇2 + k2[ ]G(x, ′ x ) = −δ (x − ′ x )

B.C.

G(x, ′ x ) r→∞ → O1r

Solution

G(x, ′ x ) = e± jk x- ′ x

4π x- ′ x

©2000, D. L. Jaggard 25EE 511EE 511

Finding the Green’s Functionl Use the solution to the

homogenous equation (except at the singularity

l Find solution (to within a constant)

l Integrate over singularity to find the constant

l Use physical reasoning to keep outgoing wave only (i.e., discard acausal solution)

D.E. valid for x ≠ ′ x

∇ 2 + k2[ ]G(x, ′ x ) = 0

B.C.

G(x, ′ x ) r →∞ → O1r

Solution

G(x, ′ x ) = Ae+ jk x- ′ x

x - ′ x + B

e– jk x- ′ x

x - ′ x

Integrate over small volume at x = ′ x to find:

A + B = 1

4πChoose outgoing wave:

A = 1

4π and B = 0

©2000, D. L. Jaggard 26EE 511EE 511

Kirchhoff Assumptions

Reasonable physicalassumptions

l Kirchhoff B.C. approximation:

l U(x‘) and s‘U(x‘) l ez vanish on screen

l U(x‘) and s‘U(x‘)l ez are undisturbed in aperture

l Mathematical Problem:

l If U(x‘) and s‘U(x‘) l ez vanish anywhere on S, U(x) vanishes in V

©2000, D. L. Jaggard 27EE 511EE 511

Alternative Assumptionsl Dirichlet B.C.l GD(x,x’) = 0 for x’ on S0

l Estimate U(x‘) on S0

l U(x) = ?U(x‘) ?z’GD(x,x’) dS‘

l Neumann B.C.l ?z’GN(x,x’) = 0 for x’ on S0

l Estimate ?z’ U(x‘) on S0

l U(x) = −?GN(x,x’) ?z’ U(x‘) dS‘

©2000, D. L. Jaggard 28EE 511EE 511

Approximationsl This is scalar theory –– no polarizationl Usually fine for large apertures (high

frequencies)

l Usually fine if observer not too close to aperture

l Fields and derivatives are assumed undisturbed in the aperture (true only far from edges in high frequency limit)

l Solutions to scalar Helmholtz equation do not necessarily satisfy Maxwell’s equations

©2000, D. L. Jaggard 29EE 511EE 511

D. Dirichlet Solution

U (x) = U( ′ x ) ′ ∇ G(x, ′ x ) − G(x, ′ x ) ′ ∇ U ( ′ x )[ ]• ˆ e zd ′ S S0∫

UD(x) = U( ′ x ) ′ ∇ GD(x, ′ x )[ ]• ˆ e zd ′ S S0∫

= U( ′ x )∂ ′ z GD(x, ′ x )[ ]d ′ x S0 = Ap+ Screen∫ d ′ y

≈ U0( ′ x )∂ ′ z GD(x , ′ x )[ ]d ′ x Ap∫ d ′ y

l Dirichlet B.C.l GD(x,x’) = 0 for x’ on S0

l Estimate U(x‘) on S0 [replace by U 0(x‘) the undisturbed field]

l U(x) ∼ ?U ο(x‘) ?z’GD(x,x’) dS‘

©2000, D. L. Jaggard 30EE 511EE 511

Dirichlet Green’s Function

l |x-x’| and |x-x’’ | define “image points” with respect to the z = 0 plane

GD (x, ′ x ) = e+ jk x- ′ x

4π x- ′ x − e+ jk x- ′ ′ x

4π x- ′ ′ x

where

x - ′ x = (x − ′ x )2 + (y − ′ y )2 + (z − ′ z )2

x - ′ ′ x = (x − ′ x )2 + (y − ′ y )2 + (z + ′ z )2

©2000, D. L. Jaggard 31EE 511EE 511

Explicit Form for ?z’GD(x,x’)

∂′ z GD(x, ′ x )

′ z =0= e+ jk x- ′ x

2π x - ′ x z

x- ′ x

− jk + 1

x - ′ x

≈ − jke+ jk x- ′ x

2π x - ′ x z

x - ′ x

l For the observer 10 wavelengths or more away from the aperture:

|–jk| >> |x-x’|–1

©2000, D. L. Jaggard 32EE 511EE 511

E. Neumann Solution

U(x) = U ( ′ x ) ′ ∇ G(x, ′ x ) − G(x, ′ x ) ′ ∇ U( ′ x )[ ]• ˆ e zd ′ S S0

∫UN (x) = − GN (x, ′ x ) ′ ∇ U( ′ x )[ ]• ˆ e zd ′ S

S0∫

= − GN (x, ′ x )∂ ′ z U( ′ x )[ ]d ′ x S0 = Ap+ Screen∫ d ′ y

≈ − GN (x, ′ x )∂ ′ z U0 ( ′ x )[ ]d ′ x Ap∫ d ′ y

l Neumann B.C.l ?z’GN(x,x’) = 0 for x’ on S0

l Estimate ?z’ U(x‘) on S0 [replace by ?z’ U 0(x‘) the undisturbed field derivative]

l U(x) ∼ −?GN(x,x’) ?z’ U(x‘) dS‘

©2000, D. L. Jaggard 33EE 511EE 511

F. Solution Summary

U(x) ≈ kj2π

U 0( ′ x )′ z =0

e +jk x- ′ x

x- ′ x

Θ(θ, ′ θ )[ ]d ′ S

Ap∫where

Θ(θ, ′ θ ) =

12

cosθ + cos ′ θ [ ] Kirchhoff Approximation

cosθ Dirichlet Approximationcos ′ θ Neumann Approximation

l All forms are approximations l Kirchhoff form has mathematical problems but is not

restricted to planar screens/aperturesl All agree with each other and with experiment and exact

calculations in far -zone and near axis in high frequency regime

©2000, D. L. Jaggard 34EE 511EE 511

Our Usual Solution

U(x) ≈

kcosθj2π

U0 ( ′ x ) ′ z = 0

e+ jk x- ′ x

x - ′ x

d ′ S Ap∫

l This is the starting point for Fresnel and Fraunhofer diffraction

l This is a mathematical description of Huygen’s principle of secondary sources

Diffracted field is sum ofspherical waves weighted by

value in aperture [i.e, Huygen’s principle]

Constants conserve energyand provide correct units

while “j” provides phase shiftassociated with diffraction

©2000, D. L. Jaggard 35EE 511EE 511

G. Babinet’s Principlel Consider a diffracting screen S A and its

complement SB [two screens are complementaryif the opening in one corresponds to the opaque region of the other and vice versa]

l Let S0= SA+ SB [S0 could be the z = 0 plane]l Fields (from scalar diffraction theory):

l With no screen, the diffracted field is Ul With screen SA the field is U Α (integrate over opening)l With screen SB the field is U Β (integrate over opening)

l Result:

U = U Α+ U Β

Babinet’s Principle relatesdiffraction by a screen andits complement using scalar

diffraction theory sinceEach field is found by

integration over the aperture

©2000, D. L. Jaggard 36EE 511EE 511

Babinet Geometry

|UA|2

SA

SB

|UΒ|2

ΙΑ = |UΑ|2 = |UΒ|2 = ΙΒ except near z axiswhere light would be focused by lens

z

z

©2000, D. L. Jaggard 37EE 511EE 511

Two Babinet Conclusionsl If UΑ = 0 –> UΒ = Ul At points in which the intensity is zero in

the presence of one of the screens, the intensity in the presence of the other is the same as if no screen was present

l If U = 0 –> UΑ = –UΒ

l At points where U = 0, the magnitudes of UΑ and UΒ are equal and their phases are opposite –> IA = IB (e.g., case of lens in previous slide)

©2000, D. L. Jaggard 38EE 511EE 511

Babinet’s Statement

Two complementary screens produce the same illumination

(intensity) at all points in space not illuminated in their

absence

©2000, D. L. Jaggard 39EE 511EE 511

Course OutlineCourse OutlineI. History and BackgroundII. Fourier Transforms and Linear Systems

III. Scalar Diffraction Theory (Physical Optics)

IV. Fresnel and Fraunhofer Approximations

V. Vector Diffraction TheoryVI. Geometrical and Ray Optics

VII. Properties of Lens

VIII. Coherent and Incoherent Imaging

IX. Partial Coherence TheoryX. Special Topics

©2000, D. L. Jaggard 40EE 511EE 511

IV. Fresnel and FraunhoferDiffraction

A. Overview

B. Fraunhofer Approximation

C. Fresnel Approximation

D. Summary of Approximations

E. Fractal Example

F. Talbot Images

G. Scalar Diffraction Summary

©2000, D. L. Jaggard 41EE 511EE 511

Key Players in Diffraction Theory

Josef von Fraunhofer (1787 -1826)Developed diffraction gratings and

increased understanding of diffraction

Augustin Jean Fresnel (1788 -1827)Made contributions to transverse nature

of light and diffraction theory

©2000, D. L. Jaggard 42EE 511EE 511

Related Contributor

Jean-Baptiste Joseph Fourier (1768 -1830)Discovered harmonic analysis

(the basic math for Fourier Optics)

©2000, D. L. Jaggard 43EE 511EE 511

Overview

l Typical Diffraction Experiment

l Regions of Validity

l Intensity

l General Approximations

©2000, D. L. Jaggard 44EE 511EE 511

Typical Diffraction Experiment

BeamExpander

DiffractingScreen

ObservationScreen

CoherentSource

©2000, D. L. Jaggard 45EE 511EE 511

Regions of Validity

l Diffraction becomes easier to calculate with distance

DiffractingScreen

z~1 mm

~10 m

~10 cm

Fraunhofer Regime(far-zone)

Easier Calculations

Fresnel RegimeDifficult Sol’n

Near ZoneVery Hard

B.V.Sol’n

VeryNearZone

~1 cm

©2000, D. L. Jaggard 46EE 511EE 511

Note On Intensity

l For EM (vector) fields:

Poynting' s Vector (= "power density" ~ "intensity")

S ≡12

Re E × H•{ } [watts/ m 2 ]

For spherical waves in the far- zone

S ≡1

2ηE

2 ˆ e r [watts/ m 2] where η =µε

[ohms]

P = S •∫ ˆ e r dS [watts]

©2000, D. L. Jaggard 47EE 511EE 511

Scalar Wave Intensity

l For scalar optical fields:

2

2

Optical field intensity

for monochromatic waves

for quasi-monochromatic waves

[total "power"]

I U

I U

P IdS

= ∫

©2000, D. L. Jaggard 48EE 511EE 511

General Approximationsl Start with exact Dirichlet formulation:

UD(x) ≈ kj2π

U( ′ x ) e+ jk x- ′ x

x - ′ x

z

x - ′ x

1+ jk x - ′ x

d ′ S Ap∫

′ z =0

l Invoke approximations:l High frequencyl Observer in far-zone

©2000, D. L. Jaggard 49EE 511EE 511

High-Frequency Approximation

l kR >> 1 where R = |x – x’|

1 + jkx - ′ x

→ 1

U( ′ x )′ z =0

→ U0 ( ′ x )′ z =0

©2000, D. L. Jaggard 50EE 511EE 511

Observer in Far-Zone

0

cos cosx - x

1 1 1 1x - x x

z

z

R r

β θ′=

= → ′

= → =′

U(x,y,z)

z

R=|x–x’|

r=|x ||x’|

θ ∼ ββ

l R >> D = |x’|max and R >> λ

Bring termoutside integral

©2000, D. L. Jaggard 51EE 511EE 511

Apply Approximationsl Approximations yield:

UD(x) ≈

kj2πr

cosθ U0 ( ′ x )e+ jk x- ′ x d ′ S Ap∫

′ z =0

l Investigate two approximations for the exponential (this is more sensitive to approximation than {|x–x’|–1} term)

lDirichlet, Neumann and Kirchhoff B.C.s all reduce to same expression for cosθ –> 1 and cosθ’ –> 1

©2000, D. L. Jaggard 52EE 511EE 511

Fraunhofer Approximationl Fraunhofer Phase Term

l Examples for Simple Aperturesl Rectangular

l Circular

l Gaussian

l Examples for Multiple Apertures

l Other Illumination

©2000, D. L. Jaggard 53EE 511EE 511

C. FraunhoferApproximation

l Fraunhofer Assumption

l Examples for Simple Apertures

l Rectangular

l Circular

l Examples for Multiple Apertures

©2000, D. L. Jaggard 54EE 511EE 511

Fraunhofer Phase Term

l Restriction is on distance (far-zone)l Few restrictions on angle θl Can use lens to cancel residual phase

R ≡ x - ′ x ′ z = 0

= x − ′ x ( )2 + y − ′ y ( )2 + z( )2

= x2 + y2 + z 2

r21 2 4 3 4 − 2 x ′ x + y ′ y ( )+ ′ x 2 + ′ y 2( )

= L

R ≈ r −x ′ x + y ′ y

r where r = x = x( )2 + y( )2 + z( )2

valid for

k( ′ x 2 + ′ y 2 )2r

<< 2π

©2000, D. L. Jaggard 55EE 511EE 511

Fraunhofer Expression

U(x) ≈k

j2πrcosθ U0( ′ x )e + jk x- ′ x d ′ S

Ap∫′ z =0

~k

j2πcosθ

e+ jkr

rU0( ′ x )e

− jkr

x ′ x + y ′ y [ ]d ′ S

Ap∫′ z = 0

U(x, y, z) ≈ kj2π

cosθ e+ jkr

rt( ′ x , ′ y )e − j2π fx ′ x + f y ′ y [ ]d ′ x d ′ y

S0∫ fx = x /λr

fy =y / λr

U(x, y, z) ≈CF t( ′ x , ′ y ){ } fx =x /λr

f y = y /λrwhere

C =k

j2πcosθ

e+ jkr

rt( ′ x , ′ y ) =U0( ′ x , ′ y ,0) in the aperture and zero elsewhere

l Fraunhofer diffraction is proportional to the Fourier transform of the undisturbed aperture field

©2000, D. L. Jaggard 56EE 511EE 511

Two-Dimensional Functions(Reminder)

circ(r / a) ≡1 r / a < 10 r / a > 1

jinc(ρ) ≡ 2J1(2πρ)

2πρ

Gaus(ax ,by) ≡ exp[−π (a 2x2 + b 2y2 )]

Gaus(ar) ≡ exp[−πa2r2 ] = exp[−πa 2(x2 + y2)]

δ (x, y) ≡ δ (x)δ (y) = δ (r)πr

comb(ax)comb(by) ≡ δ (ax − n,by − m)m = −∞

∑n= −∞

©2000, D. L. Jaggard 57EE 511EE 511

Two-Dimensional F.T. Pairs

rect (x / a)rect (y / b) ⇔ absinc(afx )sinc(bfy )

circ(r / a) ⇔ π a 2 jinc(aρ)

δ (r)πr

⇔1

δ (ax,by ) ⇔1ab

1r

⇔1ρ

1 ⇔δ ( fx, fy )

cos(πr2 ) ⇔ sin(πρ2 )

exp(± jπr2 ) ⇔ ± j exp(mjπρ 2 )

Gaus(ax,by ) ⇔1ab

Gaus( fx / a, f y / b )

Gaus(ar) ⇔ 1a 2 Gaus( fρ / a)

comb(x / a)comb(y / b) ⇔ abcomb( afx )comb(bfy)

©2000, D. L. Jaggard 58EE 511EE 511

Calculation for Rectangular Aperture

l Nulls for ax/λr = m or bλ/λr = m(m = ±1, ±2, ±3 . . )

l Maximum sidelobe level ~ –13 dB

t( ′ x , ′ y ) = rect( ′ x / a)rect( ′ y / b)

F t( ′ x , ′ y ){ } ⇔ absinc(afx )sinc(bf y)

U(x, y, z) = CF t( ′ x , ′ y ){ } fx = x / λrf y = y / λr

U(x, y, z) = Cab sinc ax / λr[ ]sinc ay / λr[ ]

U(x, y, z) =k

j2πcosθ

e+ jkr

rab

sin(πax / λr )(πax / λr )

sin(πby / λr)(πby / λr)

©2000, D. L. Jaggard 59EE 511EE 511

Calculation for Circular Aperture

l Nulls at zeros of first -order Bessel function (non-periodic)

l Maximum sidelobe level ~ –17 dB

t( ′ x , ′ y ) = circ ′ x 2 + ′ y 2 / a[ ]= circ[ ′ r / a]

F t( ′ x , ′ y ){ } ⇔ a 2 jinc a fx2 + fy

2[ ]= a 2 j inc aρ[ ]

U(x, y, z) = CF t( ′ x , ′ y ){ } fx = x/ λrf y = y/ λr

U(x, y, z) = Cπ a 2 jinc aρ[ ]ρ = x

2+y

2/ λr

U(x, y, z) = kj2π

cosθ e+ jkr

rπ a 2

J1 2πa x 2 + y2 / λr[ ]aπ x 2 + y2 / λr[ ]

jinc a x 2 + y2 / λr[ ]1 2 4 4 4 3 4 4 4

©2000, D. L. Jaggard 60EE 511EE 511

Intensity Results

I( x, y, z) = U(x, y, z) 2

For rectangular aperture:

I( x, y, z) =k cosθ

2πr

2

ab 2sin2(πax / λr)(πax / λr) 2

sin2(πby / λr)(πby / λr)2

sinc 2 ax/ λr[ ]sinc 2 by / λr[ ]1 2 4 4 4 4 4 3 4 4 4 4 4

For circular aperture:

I( x, y, z) = k cosθ2πr

2

π 2 a 4J1

2 2πa x 2 + y2 / λr[ ]π 2a2( x2 + y 2) / ( λr)2

jinc2 a x2 +y 2 /λr[ ]1 2 4 4 4 3 4 4 4

©2000, D. L. Jaggard 61EE 511EE 511

sinc and jinc Fuctionsjinc(r/2)

jinc2(r/2)sinc2(x)

sinc(x)

Note similarities and differences between sinc and jinc functions

©2000, D. L. Jaggard 62EE 511EE 511

Square Aperture Diffraction

©2000, D. L. Jaggard 63EE 511EE 511

Circular Aperture Diffraction

©2000, D. L. Jaggard 64EE 511EE 511

Experimental Results for Rectangular and Circular

Aperture Diffraction

©2000, D. L. Jaggard 65EE 511EE 511

Gaussian “Aperture”and Apodization

l Aperture edges create high spatial frequencies [remember F.T. properties]

l To reduce sidelobes , apertures are apodized or tapered

l Consider Gaussian transmission function or “aperture” with unit area and its Fraunhofer diffraction

©2000, D. L. Jaggard 66EE 511EE 511

Calculation for Gaussian “Aperture”

t( ′ x , ′ y ) =1

2πσexp − ′ r 2 / 2σ 2[ ]= Gaus(a ′ r ) with a =

12πσ

F t( ′ x , ′ y ){ } ⇔ exp −2π 2σ 2( fx2 + f y

2 )[ ]U(x,y, z) = CF t( ′ x , ′ y ){ } fx = x /λr

f y = y /λr

U(x,y, z) = C exp −2π 2σ 2 (x 2 + y2 ) / λ2r2[ ]I (x,y, z) = kcosθ

2πr

2

exp −4π 2σ 2 (x 2 + y2 ) / λ2r2[ ]

l No nulls – need to use half -power points to determine width

l Low (non-existent) sidelobes obtained at expense of increased size of main beam

©2000, D. L. Jaggard 67EE 511EE 511

Equivalent Widths

l If two functions have equal height and the same area, this defines their “equivalent width”

l There are other similar “widths”

∆xy ≡g(x , y)dxdy

−∞

∫−∞

∫g(0,0 )

and ∆ f x f y≡

G( fx , fy)dfxdfy

−∞

∫−∞

∫G(0,0)

I.C.B.S.T. ∆ xy∆ f x f y=1 (equivalent width in x - y × equivalent

bandwidth in fx − fx = 1)Also:

∆rmsxy ≡

x 2y2 g(x, y) 2 dxdy−∞

∫−∞

g(x, y) 2 dxdy−∞

∫−∞

1/2

and ∆rmsfx f y ≡

fx2 fy

2 G( fx , fx ) 2 dfxdfx−∞

∫−∞

G( fx , fx ) 2 dfxdfx−∞

∫−∞

1/2

I.C.B.S.T. ∆ rmsxy∆rms

f x f y ≥14π

(uncertainty relation)Assumes functionscentered on their

centroids

©2000, D. L. Jaggard 68EE 511EE 511

Equivalent WidthDiffraction Example

l For same equivalent widths, Guassian aperture is broader

l For same equivalent widths, Gaussian aperture diffraction is narrower

l Tapering aperture can reduce sidelobes at the expense of increased aperture diameter or main beam size

ˆ Ψ ( ′ r )Circular

= circ( π ′ r )

ˆ Ψ ( ′ r )Gaussian

= exp(−π ′ r 2 )

these have same equiv. widths

Ψ(x , y,z)Circular

=k cosθ

2πr

J1 2π x2 + y2( ) / π λr[ ]x2 + y2( )/ π λr[ ]

Ψ(x, y,z )Gaussian

=k cosθ

2πr

exp −π x2 + y2( )/ λ2 r2[ ]

diffracted fields

©2000, D. L. Jaggard 69EE 511EE 511

Two Aperture Diffraction

l Consider diffraction from two circular apertures

x’

y’

2d

b

©2000, D. L. Jaggard 70EE 511EE 511

Two Aperture Intensity

I( x, y, z)I(0,0, z)

=J1

2 2πb fx2 + fy

2( )π 2b2 ( f x

2 + fy2 )

cos2(2π fy d)12[1 +cos( 4 πf y d )

1 2 4 3 4 fx = x /λr

fy = y/ λr

l Result yields fringes along y axis modulated by jinc function

l First zero of J1 when argument = 3.832l First zero of cosine when argument = π/2

©2000, D. L. Jaggard 71EE 511EE 511

Diffraction Gratings and Multiple Apertures

l For N multiple identical apertures, use geometric progression and sum

l Result is similar to antenna array factorl Can be used for diffraction grating calculations

Let g(x) ⇔ G( f x )

F g(x − na)-

N -12

N -12

= G( f x)

Usually slowlyvarying due tosinge aperture

1 2 3 sin(Nπ fx a)sin(πfx a)

Usually sharply peakedfrom multiple apertures

1 2 4 3 4

©2000, D. L. Jaggard 72EE 511EE 511

Scaling and Shiftingfor Circular Apertures

©2000, D. L. Jaggard 73EE 511EE 511

Multiple Circular Apertures

©2000, D. L. Jaggard 74EE 511EE 511

Amplitude Diffraction Grating

l Periodic grating gives characteristic 0 th and 1 st

order diffractionl Phase grating is more efficient

t( ′ x , ′ y ) = 12

+ m2

cos 2πf0 ′ x ( )

rect ′ x

a

rect ′ y

b

U( x, y, z) = CF t( ′ x , ′ y ){ } f x =x / λrf y =y / λr

= C ab sinc bf y( ) sinc afx( ) + m2

sinc a f x − f0( )[ ]+ m2

sinc a fx + f0( )[ ]

f x =x / λr

f y =y / λr

I( x,y, z) ~ C2 ab 2 sinc 2 bf y( ) sinc 2 af x( ) + m 2

4sinc 2 a f x − f0( )[ ]+ m 2

4sinc 2 a fx + f0( )[ ]

f x = x / λr

f y= y / λr

Diffraction Efficiencies:

η0 =1 / 4

η+1 = η−1 = m2 / 16

What happens forsquare-wave grating?

Maximum efficiencyof ~6%

©2000, D. L. Jaggard 75EE 511EE 511

Other Illuminationsl Oblique Incidencel Center of diffracted beam is moved

(think “Modulation Theorem”)l Oblique incidence produces the

expected result

l Spherical Wave Sourcel Quadratic phase approximation

©2000, D. L. Jaggard 76EE 511EE 511

Oblique Incidence

l Both Fraunhofer and Fresnel diffraction patterns are shifted as expected

Diffraction patternis centered here

r r sin α

If t( ′ x , ′ y ) = e jk xsinα + z cosα[ ] and

U(fx , fy) = CF t( ′ x , ′ y ){ }= T( fx , fy ) for α = 0

Then U(x ,y ,z) = CΨ fx −ksinα

2π, fy

for α ≠ 0

©2000, D. L. Jaggard 77EE 511EE 511

Spherical Wave Incidence

For sperhical source e jkRs

Rs

at (xs , ys ,zs) with

Rs = x − xs( )2+ y − ys( )2

+ z − zs( )2

The diffraction integral is evaluated at

x = ′ x , y = ′ y ,z = ′ z = 0

Rs ≈ zs 1+′ x − xs( )2

2 zs2 +

′ y − ys( )2

2zs2

z

Source at (xs, ys, zs)

Rs

©2000, D. L. Jaggard 78EE 511EE 511

D. Fresnel Approximations

l Fresnel Assumptions

l Example for Slit Diffraction

l Example for Edge Diffraction

l Examples for Circular Aperture

©2000, D. L. Jaggard 79EE 511EE 511

Fresnel Assumptionsl Fraunhofer result breaks down for

knife edge diffraction

l Need better approximation for the phase term exp{j2πk|x–x’|}

©2000, D. L. Jaggard 80EE 511EE 511

Knife Edge Example

U0 ( ′ x , ′ y ) =1 + sgn( ′ x )

2

F U0( ′ x , ′ y ){ }=12

δ ( fx )+1

jπf x

fy

U(x,y,z) =C2

δ( fx ) +1

jπf x

f y

f x =x / λr

fy =y / λr

Ψ(x,z)

z

x’ x

Does this Fraunhofer

resultmake sense?

©2000, D. L. Jaggard 81EE 511EE 511

Fresnel Phase Term

l Restriction on z (not R)l Observer must be close to axisl Valid in “near zone” as well as far zone

R ≡ x - ′ x ′ z = 0

= x − ′ x ( )2 + y − ′ y ( )2 + z( )2

= z 1+x − ′ x ( )2 + y − ′ y ( )2

z 2

= L

R ≈ z +(x - ′ x )2 + (y − ′ y 2 )[ ]

2z

valid for

k (x - ′ x )2 + (y − ′ y 2 )[ ]8z 2 << 2π

©2000, D. L. Jaggard 82EE 511EE 511

Fresnel Expression

U(x) ≈ kj2πr

cosθ U0( ′ x )e+ jk x- ′ x d ′ S Ap∫

′ z =0

≈k

j2πe+ jkz

zU0( ′ x )e

+jk

2 z( x− ′ x )2 +( y − ′ y ) 2[ ]

d ′ S Ap∫

′ z = 0

U(x, y, z) ≈ kj2π

e+ jkz

zt( ′ x )e

+jk2z

(x − ′ x )2 + (y − ′ y )2[ ]d ′ x d ′ y S0

l Fresnel diffraction has quadratic phase in integrandl The phase expansion is valid near the z axis

©2000, D. L. Jaggard 83EE 511EE 511

Fresnel Diffraction by Rectangular Aperture -I

U(x, y, z) ≈k

j2πe+ jkz

zC

1 2 4 3 4 t ( ′ x )e

+jk

2 z(x − ′ x ) 2 + ( y− ′ y )2[ ]

d ′ x d ′ y S0

= C e+

jk2z

( x− ′ x )2 + (y − ′ y ) 2[ ]d ′ x d ′ y where C = ke jkz

j2πz− Lx / 2

Lx /2

∫− Ly /2

L y / 2

Let ξ ≡ kπz

( ′ x − x) and dξ = kπz

d ′ x

Let η ≡ kπz

( ′ y − y) and dη = kπz

d ′ y

Then ξ1

2

= m kπz

( Lx

2± x) and η1

2

= m kπz

(Ly

2± y)

U(x, y, z) = Cπzk

e+

jπ2

ξ2

dξ πzk

e+

jπ2

η2

dηη1

η2

∫ξ 1

ξ 2

©2000, D. L. Jaggard 84EE 511EE 511

Fresnel Diffraction by Rectangular Aperture -II

U(x, y, z) = C πzk

e+

jπ2

ξ2

dξ πzk

e+

jπ2

η2

dηη1

η2

∫ξ1

ξ2

where ξ ≡ kπz

( ′ x − x) and η ≡ kπz

( ′ y − y)

Note: close to aperture ξ and η are large

far from aperture ξ and η are small

l Calculations given in terms of tabulated Fresnel Integrals

©2000, D. L. Jaggard 85EE 511EE 511

Fresnel Integral Definitions

I(ξ ) ≡ e+ jπ

2t 2

dt 0

ξ

∫ [Complex Fresnel Integral I]

I(ξ ) = C(ξ )+ jS(ξ ) [Fresnel Integrals C and S]

C(ξ ) = cosπ2

t2

dt

0

ξ

S(ξ ) = sin π2

t2

dt

0

ξ

©2000, D. L. Jaggard 86EE 511EE 511

Fresnel Integrals & Cornu Spiral

©2000, D. L. Jaggard 87EE 511EE 511

Plot of Fresnel Integrals

©2000, D. L. Jaggard 88EE 511EE 511

Cornu Spiral

©2000, D. L. Jaggard 89EE 511EE 511

Fresnel Approximations

For ξ → 0

e+

jπ2

t2

→1 + jπ2

ξ2 −12!

π2

2

ξ 4 − j13!

π2

3

ξ6 L

∴ I(ξ ) = e+

jπ2

t2

dt 0

ξ

∫ → ξ + j11!

π2

ξ3

3−

12!

π2

2 ξ5

5− j

13!

π2

3 ξ7

7L

C(ξ) = cosπ2

t2

dt

0

ξ

∫ → ξ 1 −1

2!5π2

ξ2

2

+1

4!9π2

ξ2

4

L

S(ξ ) = sin π2

t2

dt

0

ξ

∫ → ξ 11!3

π2

ξ2

− 1

3!7π2

ξ 2

3

+ 15!11

π2

ξ 2

5

L

For ξ → ∞

C(ξ) → ±12

+sin πξ2 / 2( )

πξL for z

>

<0

S(ξ )→ ±12

−cos πξ2 / 2( )

πξL for z

><

0

©2000, D. L. Jaggard 90EE 511EE 511

Fresnel Intensity byRectangular Aperture

U(x, y, z) = Cπzk

[C (ξ2 ) − C(ξ1 )] + j[ S(ξ2 ) − S(ξ1 )]{ }Ae jψ

1 2 4 4 4 4 4 4 3 4 4 4 4 4 4 o

[C(η2 ) − C(η1)] + j[S(η2 ) − S(η1)]{ }Be jχ

1 2 4 4 4 4 4 4 3 4 4 4 4 4 4

I (x, y, z) =14

A2 B2 (for I0 = 1)

l Can use Cornu spiral to find characteristics– A is the length on the Cornu spiral from ξ1 to ξ2

– B is the length on the Cornu spiral from η1 to η2

©2000, D. L. Jaggard 91EE 511EE 511

Fresnel Cornu Calculations

l Let ∆v = ξ2 – ξ1 = Lx(k/ πz)1/2 = constant for given slitlAnd ξ0 = (ξ2 – ξ1)/2 = x(k/ πz)1/2 – moves as x (the center

of the aperture) moves

Ae j ψ(ξ0= 0)

with

∆v = 2.0

ψ

A

©2000, D. L. Jaggard 92EE 511EE 511

Case: ∆v = 1.5

lHere ∆v =15 = ξ2 – ξ1 = Lx(k/πz)1/2

lAnd ξ0 = (ξ2 – ξ1)/2 = x(k/ πz)1/2

∆v = 1.5

Α(ξ0=0) ~ 1.45Α(ξ0=0.25) ~ 1.37Α(ξ0=1.00) ~ 0.50Α(ξ0= 2.00) ~ 0.36

A(ξ 0 = 2.00)

A(ξ 0 = 1.00)

A(ξ 0 = 0)

A(ξ0 = 0.25)

lFor ξ0 =0 (aperture centered), draw line ∆v =15 from u = –0.75 to u = +0.75lFind A (optic

field) and A 2

(intensity)lNext move the

aperture ξ0, repeat and plot

©2000, D. L. Jaggard 93EE 511EE 511

Cases: ∆v = 0.6 to ∆v = 10.1A2 vs ξ 0

ξ0 = (ξ2 – ξ1)/2 = x(k/πz)1/2 –>

∆v = 0.6 = ξ2 – ξ1= L(k/πz)1/2

∆v = 1.5

∆v = 2.4

∆v = 3.7

∆v = 5.5

∆v = 6.2

∆v = 10.1

∆v = 4.7

©2000, D. L. Jaggard 94EE 511EE 511

Notes on Fresnel Diffractionl For ∆v > 10, pattern approaches geometrical

optics patternlRipples always appearlRipples become finer with increasing ∆v

l For ∆v < 1, pattern approaches Fraunhofer diffraction pattern

l For ∆v << 1, pattern spreads greatly in accordance with far-zone beam divergence

l For 1 < ∆v < 10, pattern undergoes pronounced change

l Here ∆v = [8NF]1/2, where NF (= Lx2/4λz) is

the Fresnel number used by Goodman

©2000, D. L. Jaggard 95EE 511EE 511

Fresnel Slit

Diffraction Evolution

∆v = 0.283

∆v = 2.83

Fourier transform-likeresult near far-zone

∆v = 5.65

∆v =8.94

Sloughing off of diffraction“shoulders” with distance

Geometric “image” ofslit near aperture

©2000, D. L. Jaggard 96EE 511EE 511

Infinite ApertureLet x dimensions → ±∞

Ae jψ = [C(ξ1) − C(ξ2 )]+ j[S(η1 ) − S(η2 )]x→ ±∞

= [C(∞) − C (−∞)] + j[S(∞) − S(−∞ )]

= 1+ j

= 2e jπ / 4

If y dimensions also → ±∞

Be jχ = 2e jπ / 4

Then

U(x, y, z ) =e jkz

2 j2ejπ /4 2e jπ /4 Ψ0

U(x, y, z ) = e jkzΨ0

This validates the constants infront of the diffraction integral

©2000, D. L. Jaggard 97EE 511EE 511

Fresnel Knife Edge Diffraction

DiffractingKnife Edge

z

ξ0 = 0

ObservationScreen

k=2π/λ

x’

l Here:l ξ2 –> –8l η2 –> –8 and η1 –> +8

©2000, D. L. Jaggard 98EE 511EE 511

Fresnel Knife Edge Diffraction

DiffractingKnife Edge

z

ξ0 < 0

ObservationScreen

k=2π/λ

x’

x

Shift coordinateupward by x

l Now, move coordinate system upward and find A

©2000, D. L. Jaggard 99EE 511EE 511

Knife Edge CalculationsLet y dimensions → ±∞

Be jχ = 2e jπ / 4

Then

U (x, y,z ) =e jkz

2 jAe jψ 2e jπ / 4U0

I (x, y,z ) =12

A 2 (for I0 = U0

2=1)

Here Ae jψ = [C (ξ1) − C(∞)]+ j[S(ξ1) − S(∞)]

= [C (∞) − C(ξ1 )]+ j[S(∞) − S(ξ1 )]

if ξ1 taken negative Use Cornu spiral to see

the physics of diffraction

©2000, D. L. Jaggard 100EE 511EE 511

Case: Knife EdgelLet ξ2 –> –8

lVary ξ1 (< 0)

lFind A (optical field) and A 2

(intensity)

lCalculate I = A2/2

A(ξ 0 = –0.5)

Max. V

alue

A(ξ 1 = 0)

A(ξ0= –1.22)

©2000, D. L. Jaggard 101EE 511EE 511

Fresnel Knife Edge Diffraction

©2000, D. L. Jaggard 102EE 511EE 511

Fresnel Strip Diffraction

l Think Babinet!

DiffractingSlit

z

Case: ∆v = L(k/πz)1/2 = 0.6

L

ObservationScreen

k=2π/λ

©2000, D. L. Jaggard 103EE 511EE 511

Fresnel Strip Calculations

©2000, D. L. Jaggard 104EE 511EE 511

Fresnel Strip Diffraction

l Note: this is different than diffraction by slit with ∆v = 0.6l Babinet’s Principle is relation for amplitudes

©2000, D. L. Jaggard 105EE 511EE 511

Fresnel Diffraction by Circular Apertures

©2000, D. L. Jaggard 106EE 511EE 511

Fresnel Zone

l Radius of first Fresnel zone is a = (λr)1/2

l Keeping aperture such that a << (λr)1/2 is equivalent to Fresnel approximation

Aperture

zr0

k=2π/λ

x’

r0 + λ/2a

©2000, D. L. Jaggard 107EE 511EE 511

Fresnel Zone Platel Consider radial transmission plate t(r’)

with absorption corresponding to the Fresnel zones

l Such plates can focus light due to constructive interference

t( ′ r ) =12

1 + cos(α ′ r 2 )[ ]circ( ′ r / a) [sinusoidal variation]

or

t( ′ r ) = 12

1 + sgn{cos(α ′ r 2)}[ ]circ( ′ r / a) [square- wave variation]

©2000, D. L. Jaggard 108EE 511EE 511

Fresnel Zone Plate & Lens

z

r0

x’

k=2π/λ

r0 +λ

r0+2λ

Focused lightfrom constructive

interference

l Fresnel zone plate has openings so that constructive interference occurs at distance r 0

l Similar idea can be applied to a lens

r0 +nλ

an

Radii of openings:an

2 =[r0+nλ]2 – r02

©2000, D. L. Jaggard 109EE 511EE 511

Fresnel On-Axis Diffraction

l Intensity along z axis can be found for circular aperturelCalculations can be done for

complementary disk

Aperture

z

k=2π/λ

x’

a

©2000, D. L. Jaggard 110EE 511EE 511

E. Fractal Resultsl Fractals are self-similar objects:l Are often “grown” through iterationl Have roughness on all scalesl Look similar at various

magnifications

l Fractal dimension D is measure of roughness or “area filling” nature

l For curves:l D –> 1 means smooth curvel D –>2 means very rough curve

©2000, D. L. Jaggard 111EE 511EE 511

FractallySerrated

Aperturesl Analytic and

experimental results available

l Fractal dimension Dis measure of perimeter “roughness”

l For lines: 1 < D < 2

D = 1.01

D = 1.50

D = 1.99

©2000, D. L. Jaggard 112EE 511EE 511E

xam

ple

s o

f Fra

ctal

A

per

ture

s

©2000, D. L. Jaggard 113EE 511EE 511

Stage 0

Stage 1 Stage 2

CantorSquare

Diffraction

l Stage 0 result is for square aperture

l Diffraction for higher stages can be found by iteration

l Scaling and complexity of aperture is reflected in diffraction pattern

©2000, D. L. Jaggard 114EE 511EE 511

Selected Other Fractal Results Using Scalar &

Vector Theoriesl Antenna array analysis/synthesis

l Aperture array analysis/synthesis

l Antenna element design

l Rough surface scattering

l Beam propagation & diffraction in turbulence

©2000, D. L. Jaggard 115EE 511EE 511

F. Talbot Images

x’ fx~x

y’

z

fy~y

U(fx,fy)t(x’,y’)

L

l Consider diffraction by infinite periodic grating

l Use Fresnel appoximation

t( ′ x , ′ y ) =12

+m2

cos 2π ′ x / L( )

©2000, D. L. Jaggard 116EE 511EE 511

Alternative Fresnel Method

U(x, y, z) =k

j2πe+ jkz

zC

1 2 4 3 4 t( ′ x , ′ y )e

+ jk2 z

( x− ′ x )2 +( y− ′ y )2[ ]d ′ x d ′ y

S0

Or in the frequency domain,

we write, the transfer function

H( fx , fy ) = Fke jkz

j2πzexp j

k2 z

x2 + y2( )

= e jkz exp −jπλz fx2 + fy

2( )[ ]→

ignore phase term{ exp − jπλz fx

2 + fy2( )[ ]

Now write transmission functionin frequency domain and multiply

by transfer function H( fx,fy ) –then take inverse transform

Easiest methodfor periodic

input

©2000, D. L. Jaggard 117EE 511EE 511

Talbot Calculation - I

F t( ′ x , ′ y ){ }= F 12

1+ m cos 2π ′ x / L( )[ ]

= 12

δ ( fx , fy ) + m4

δ ( fx − 1L

, fy ) + m4

δ ( fx + 1L

, fy )

Note:

H ± 1L

, 0

= exp − j πλz

L2

After propagation over distance z

F t( ′ x , ′ y ){ }=12

δ ( fx , fy ) +m4

e− j

π λz

L 2 δ ( fx −1L

, fy ) +m4

e− j

π λz

L2 δ ( fx +1L

, fy )

Now take the inverse transformto find the field

©2000, D. L. Jaggard 118EE 511EE 511

Talbot Calculation - II

U(x, y, z) =12

+m4

e− j

πλz

L2e

+ j 2πxL +

m4

e− j

πλz

L2e

− j 2πxL

=12

1+ me− j

πλz

L2 cos2πx

L

Or:

I(x, y, z) =14

1+ 2mcosπλzL2

cos2πxL

+ m 2 cos2 2πxL

This result produces perfect“Talbot mages” or “self -images”

of the original grating

©2000, D. L. Jaggard 119EE 511EE 511

Talbot Calculation - IIICase I:

πλzL2 = 2nπ (n is an integer)

I(x ,y,z) =14

1 + m cos2πxL

2

Case II: πλzL2 = (2n + 1)π (n is an integer)

I(x ,y,z) =14

1 − m cos2πxL

2

Case III: πλzL2 = (2n − 1)π / 2 (n is an integer)

I(x ,y,z) =14

1 + m2 cos2 2πxL

=14

1 +m2

2

+

m2

2cos

4πxL

Produces perfect“Talbot images” or “self -images”

Produces contrast reversed“Talbot images” or “self -images”

Produces lower contrast“Talbot sub-images” with

twice the spatial frequency

©2000, D. L. Jaggard 120EE 511EE 511

G. Scalar Diffraction Summary

©2000, D. L. Jaggard 121EE 511EE 511

Course OutlineCourse OutlineI. History and BackgroundII. Fourier Transforms and Linear Systems

III. Scalar Diffraction Theory (Physical Optics)

IV. Fresnel and Fraunhofer Approximations

V. Vector Diffraction TheoryVI. Geometrical and Ray Optics

VII. Properties of Lens

VIII. Coherent and Incoherent Imaging

IX. Partial Coherence TheoryX. Special Topics