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Chapter 1 – Math Toolbox
Gabriel Popescu
University of Illinois at Urbana‐Champaign y
p g
Beckman Institute
Quantitative Light Imaging Laboratory
Quantitative
Light Imaging Laboratory
http://light.ece.uiuc.edu
Principles of Optical Imaging
Electrical and Computer Engineering, UIUC
ECE 460 – Optical Imaging
Objectives
 Develop a set of tools useful throughout the course
p
g
Chapter 1: Math Toolbox
2
ECE 460 – Optical Imaging
1.1 Linear Systems
1.1 Linear Systems
 Consider a simple system:
p y
 Equation of motion:
d 2x
dx
m 2   m  mo 2 x  f (t )
dt
dt
k
m
(1.1)
o 
 Define Operator: (linear differential eqs)
(
)
Lm
d
d
2
m
m




o
dt 2
dt
 L( x)  F (t )
Chapter 1: Math Toolbox
Mass
on a
spring
k
m
(1.2)
(1.3)
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ECE 460 – Optical Imaging
1.1 Linear Systems
1.1 Linear Systems
 Operator L has important properties:
p
p
p p
2
d (ax)
d (ax)
2





L
(
ax
)
m
m
m
(ax) 
a)
o
2
dt 2
dt
d ( x)
 d ( x)

 a m 2   m
 mo 2 ( x)  
dt
 dt

 aL( x)
(1.4)
d ( x  y)
d ( x  y)
m
 mo 2 ( x  y ) 
b) L( x  y )  m
dt
dt
 L( x)  L( y )
(1.5)
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.1 Linear Systems
1.1 Linear Systems
 Definition: An operator obeying properties L(ax) = aL(x) and p
y gp p
( )
( )
L(x+y)=L(x)+L(y) is called linear
 Most of the system in nature are linear; well, at least to the fi t
first approximation
i ti
 They are mathematically tractable  analytic solutions
 Consider equations:
Consider equations:
L(x1) = 0
(1.6)
L(x
( 2)) = 0
x1, x2 are solutions
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.1 Linear Systems
1.1 Linear Systems
 Continuing:
g
L(ax1+ bx2) = L(ax1) + L(bx2)
= aL(x1) + bL(x2)
(1.7)
= 0 + 0
 Any linear combination of solutions: x1, x2 is also a
solution
 The number of independent solutions = degrees of freedom
X 1 , X 2 ,,...,, X N = independent solutions if
independent solutions if
X i    j x j , for any 
(1.8)
j
j i
 Linear Differential eqs of order N allow for N independent solutions
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.2 Light‐matter
1.2 Light
matter interaction
interaction
 Classic model of atom: e‐ rotating around N ≈ planets
y
+
_
m
e, m
x(t)
x
x  x 0 cos(0t )
0
x
t
 Lorentz Model
 Analogy :
w
x
Chapter 1: Math Toolbox
x  x 0 cos(0t )
o 
k
m
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ECE 460 – Optical Imaging
1.2 Light‐matter
1.2 Light
matter interaction
interaction

 So, motion of charge follows the same eq (1.1)
,
g
q( )
d 2x
dx
m 2   m  mo 2 x  F (t )
dt
dt
(1.9)
 Incident field drives the charge: F (t )  qE (t )
 For e-, q = -e !
 Monochromatic field: E (t )  Eoeit
(1.10)
 mx   mx  mo 2  qEoeit
 This is the eq of motion for eletric charge under incident EM This is the eq of motion for eletric charge under incident EM
field. Can explain most of Optics!
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ECE 460 – Optical Imaging
1.3 Superposition principle
1.3 Superposition principle
 Suppose we have 2 fields simultaneously interacting with the pp
y
g
material (Eg. ω1, ω2):
E1
E2
E1  E1 e i t ;;qE
qE1  F1
1
E2  E2 e
 i2t
; qE2  F2
(1.11)
 Let x
Let x1, xx2 be solutions of displacements for the two forces F
be solutions of displacements for the two forces F1
and F2
L(x
( 1)) = F1((t))
(1 12)
(1.12)
L(x2) = F2(t)
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.3 Superposition principle
1.3 Superposition principle
 Consider the same solution:
L( x1  x2 )  L( x1 )  L( x2 )
 F1 (t )  F2 (t )
(1 13)
(1.13)
 So, final solution is just the sum of individual solutions. Nice!
E1
 This is the superposition principle
E
 For the 2 frequency example:
E2
E +E
1 2
E
2
x +x
1 2
x
2
E
1
x
1
 It’s as if one applies the fields one by one and sums their results
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demo available
10
ECE 460 – Optical Imaging
1.4 Green’ss function/impulse response
1.4 Green
function/impulse response
 Let the incident field, i.e driving field, have a complicated shape
h
E(t)
 arbitrary
t
i
t
 E(t) can be broken down into a succession of short pulses, i.e Dirac delta functions:
(1.14)
1, t = 0
  (t ) 
0, otherwise

E (t ) 


E (t ' ) (t  t ' )dt '
(1.15)

Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.4 Green’ss function/impulse response
1.4 Green
function/impulse response
 If we know the response of the system to a short pulse, , p
y
p
,  (t ) ,
the problem is solved
 Let h(t) be the solution to  (t )
 The final solution for an arbitrary force F (t )  qE (t ) is:
x(t ) 


E (t ' )h(t  t ' )dt '
(1.16)

 This is the Green’s method of solving linear problems
h(t) = Green
Green’ss function or impulse response of the system
function or impulse response of the system
 h(t)  Complicated problems become easily tractable!
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ECE 460 – Optical Imaging
1.5 Fourier Transforms
1.5 Fourier Transforms
 Very efficient tool for analyzing linear (and non‐linear) processes

 Definition:  i 2  xf x
[ f ( x)] 


f ( x )e
dx
 F ( f x )  f ( )
(1.17)
 F is the Fourier transform of f
 ( , ,  )
(
x
,
y
,
z
)

 f :   ;    , f must satisfy:
a)  f   - modulus integrable
b)) f has finite number of discontinuities in the finite
domain 
c) f has no infinite discontinuities
 In
I practice,
i
some off these
h
conditions
di i
are sometimes
i
relaxed
Chapter 1: Math Toolbox
demo available
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ECE 460 – Optical Imaging
1.5 Fourier Transforms
1.5 Fourier Transforms
 Inverse Fourier Transforms:
1    f ( x)   


f ( )e  i 2 xf df
x

 f ( x)
  [ ( f )]  f
1
x
(1.18)
(1.19)
 Meaning of F.T: reconstruct a complicated signal by summing Meaning of FT: reconstruct a complicated signal by summing
sinusoidals with proper weighting
Chapter 1: Math Toolbox
demo available
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ECE 460 – Optical Imaging
1.5 Fourier Transforms
1.5 Fourier Transforms
 Fourier transform is a linear operator:
p
[af ( x)  bg ( x)] 


 i 2 x
[
af
(
x
)

bg
(
x
)]
e
dx 






 a  f ( x)e i 2 x dx  b  g ( x)e i 2 x dx
 a[ f ( x)]  b[ g ( x)]
Chapter 1: Math Toolbox
(1.20)
demo available
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ECE 460 – Optical Imaging
1.6 Basic Theorems with Fourier Transforms
1.6 Basic Theorems with Fourier Transforms
a)) Shift Theorem: if f ( )  [ f ( x)]
{ f ( x  a )}  f ( )e i 2 a
(1.21)
 Easy to prove using definition
 Eq 1.21 suggest that a shift in one domain corresponds to a linear phase ramp in the other (Fourier) domain
linear phase ramp in the other (Fourier) domain
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.6 Basic Theorems with Fourier Transforms
1.6 Basic Theorems with Fourier Transforms
b)) Parseval’s theorem: if [ f ( x)]  f ( )




f ( x) dx  
2
f ( ) 2 d
(1.22)

 Conservation of total energy
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.6 Basic Theorems with Fourier Transforms
1.6 Basic Theorems with Fourier Transforms
c)) Similarity theorem: if y
f
[ f ( x)]  f ( f x ) , i.e. is the F.T of f
1  
[ f (ax)] 
f 
|a| a
(1.23)
 Theorem 1.23 provides intuitive feeling for F.T
 Let’s consider
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.6 Basic Theorems with Fourier Transforms
1.6 Basic Theorems with Fourier Transforms
c)) Similarity theorem:
y
 Let’s consider:
f ( f )
x
f ( x)

Δx
Δfx
(
(Figures pageII‐8)
)
x
fx
f ( f )
x
f ( x)

2Δx
Δfx
2
x
Chapter 1: Math Toolbox
fx
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ECE 460 – Optical Imaging
1.6 Basic Theorems with Fourier Transforms
1.6 Basic Theorems with Fourier Transforms
! Broader functions in one domain imply narrower functions in py
the other and vice‐versa
 Eg. To obtain short temporal pulses of light, one needs a broad spectrum (Ti: Saph
t
(Ti S h laser)
l
)

! Only an infinite spectrum allows for ‐function pulses

s(ω)
ω
I(t)
t
 Physically Impossible
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.6 Basic Theorems with Fourier Transforms
1.6 Basic Theorems with Fourier Transforms
 Before we present the last theorems, we introduce the p
,
definitions of convolution and correlation
 Let

g(x) 
G( )
h(x) 
H( )
 Convolution of g and h:
Convolution of g and h:
gh


g ( x ' )h( x  x ' )dx '
(1.24)

 Correlation of g and h
g h


g ( x ' )h( x '  x)dx '
(1 25)
(1.25)

Chapter 1: Math Toolbox
demo available
21
ECE 460 – Optical Imaging
1.6 Basic Theorems with Fourier Transforms
1.6 Basic Theorems with Fourier Transforms
v

 Difference between and is h(x‐x’) vs h(x’‐x), i.e. flip vs ( )
( ),
p

non‐flip of h
 Particular case:
 Autocorrelation: g=h
gg


g ( x ' ) g ( x '  x)dx '
(1.26)

 Exercise: Use PC to show:
x
=
(Figures page II‐9)
x
Gauss
Chapter 1: Math Toolbox
=
x Gauss = Gauss
demo available
22
ECE 460 – Optical Imaging
1.6 Basic Theorems with Fourier Transforms
1.6 Basic Theorems with Fourier Transforms
d)) Convolution theorem:
v h ]  GH
[ g 
(1.27)

'
'
'

[
g
(
x
)
h
(
x

x
)
dx
]  G ( ) H ( )
i.e


 Convolution in one domain corresponds to a product in the other. Nice!
Multiplication is always easy to do
 Multiplication is always easy to do
 Recall Green’s function: h(t) = the response to a ‐function 
light pulse
Chapter 1: Math Toolbox
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23
ECE 460 – Optical Imaging
1.6 Basic Theorems with Fourier Transforms
1.6 Basic Theorems with Fourier Transforms
 We found (Eq 1.16):
( q
)
x(t ) 


E (t ' ) h(t  t ' ) dt '

i.e the response to an arbitrary field E(t) is the convolution E h!
 Let’s take the F.T:
x( )  E ( )h( )
(1.28)
 It doesn
It doesn’tt get any simpler than this
get any simpler than this
i.e if we know the impulse response h(t), (or the Green’s 
 response to any function) take F.T  h(ω) transfer function
field E is:
x(t )  [ E ( )h( )]
Chapter 1: Math Toolbox
(1.29)
demo available
24
ECE 460 – Optical Imaging
1.6 Basic Theorems with Fourier Transforms
1.6 Basic Theorems with Fourier Transforms
e)) Correlation theorem:
  differs from only by minus sign  similar theorem:

i.e [ 
[ g  h]  GH *
g ( x ' )h( x '  x)dx ' ]  G ( ) H ( )*
(1.30)

 Particular case: g = h (auto correlation):
[ g  g ]  GG  G
*
Chapter 1: Math Toolbox
2
(1.31)
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ECE 460 – Optical Imaging
1.6 Basic Theorems with Fourier Transforms
1.6 Basic Theorems with Fourier Transforms
e)) Correlation theorem:
 Eg: F.T of an auto correlation is the power spectrum
 Very important for both time and space fluctuating fields:
(t ) 

 E (t ') E (t ' t )dt = auto correlation

t
[(t )]  E ( ) E * ( )  S ( ) = power spectrum
(1.32)
(Wiener–Khinchin theorem)
 We
We’llll meet them again later!
meet them again later!
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.7 Differential equations and Fourier Transforms
f
 Let f be a function of time:

f (t ) 

F ( )e  it d  1 ( F )

 What is What is f ?
(1.33)
f
 
 [  F ( )eit d ] 
t t 

 it
  F ( ) [e ]d 
t

t

  [i F ( )]eit d 

 1 i F 
 So, f  F & Chapter 1: Math Toolbox
f
t
 i F
(1.34)
 Very Useful!
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ECE 460 – Optical Imaging
1.7 Differential equations and Fourier Transforms
f
 Great:
[ f (t )]  F ( )  Then
f (t )
[
]  i F ( )
t
2 f
 f
 Now [ 2 ]  [ ( )]  ii F ( )
t
t t
  2 F ( )
n f
In others words: [ n ]  i n n F ( )
t
(1.34)
(1.35)
 Differentiation theorem
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.7 Differential equations and Fourier Transforms
f
 Why 1.35 result is important? Because linear differential y
p
equations are resolved in the frequency domain more easily
 Eg: Recall our e‐ revolving around nucleus under field ill i ti E(t)
illumination E(t)
d 2 x(t )
dx(t )
2
m
m
m
x(t )  qE (t )




o
2
dt
dt
(1.36)
 The solution is x(t). But we can solve for x(ω)= [ x (t )] and take in the end
1
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.7 Differential equations and Fourier Transforms
f
 So,, let’s take F.T of 1.36,, using
g the differentiation
theorem:
m[ 2 x( )]  i mx( )  mo 2 x( )  q
qE ( )
x( )[ m 2  i m  mo 2 ]  qE ( )
Since q=
q=-e:
e:
e
E ( )
x( )  2 m
  i  o 2
Chapter 1: Math Toolbox
(1.37)
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ECE 460 – Optical Imaging
1.7 Differential equations and Fourier Transforms
f
 Exercise: use PC to take  of Eq.
q 1.37
1
 “damped”
damped oscilation,
oscilation  = damping factor
! Problem solved
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.8 Refraction and absorption. Dispersion
1.8 Refraction and absorption. Dispersion
 Given the electron displacement
p
as a function of
frequency, x(ω), we can define the dipole moment:
pq
qx
p  ex
(1.38)
 The dipole moment is a microscopic quantity; we need a
macroscopic counterpart:
Ne 2

E
P  N p  2 m2
  o  i
Chapter 1: Math Toolbox
(1.39)
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ECE 460 – Optical Imaging
1.8 Refraction and absorption. Dispersion
1.8 Refraction and absorption. Dispersion
 P  induced p
polarization
 N  concentration [m-3]
 But P relates to the macroscopic response of the
material , i.e. eletric susceptibility:





P  o E
 o = permeability of vacuum
Finally,    r  1  n 2  1
 r = relative permeability
 11
n = refractive index






If
, as opposed
pp
to
 21
Chapter 1: Math Toolbox

 31
(1.40)
(1.41)
12
 22
 32
13 
 233  , material is
 33  isotropic
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ECE 460 – Optical Imaging
1.8 Refraction and absorption. Dispersion
1.8 Refraction and absorption. Dispersion
 So,, combining
g 1.39 and 1.40:
Ne 2 / m o
2

 1 
 2
n
2
(  o )  i
(1.42)
 For low-n materials, such as rarefied gases,
n 2  1  (n  1)(n  1)  2(n  1)
Ne 2
1
 n 1
2m o ( 2  o 2 )  i
 n'  in"
Chapter 1: Math Toolbox
(1.43)
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ECE 460 – Optical Imaging
1.8 Refraction and absorption. Dispersion
1.8 Refraction and absorption. Dispersion

2
2
2
Ne



o
n'  1 
2m o ( 2  o 2 )   2 2
2
Ne

"
n 
2m o ( 2  o 2 )   2 2
(1.44a)
(1.44b)
 n’’ = Re(n)
( ) = refractive
f
index
 n” = Im(n) = absorption index
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.8 Refraction and absorption. Dispersion
1.8 Refraction and absorption. Dispersion
 Eg:
g Plane wave:
E  E o e i k  r ; k  nk o
E  Eoe
ik
 Eoe
ink o r
E  Eoe
o

r ( n '  in " )
 n " k o r in ' k o r
absorption
e
refraction
  n k o  absorption coefficient
"
Chapter 1: Math Toolbox
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ECE 460 – Optical Imaging
1.8 Refraction and absorption. Dispersion
1.8 Refraction and absorption. Dispersion
 Definition:
n’((ω)) = variation of refractive index with
frequency
= dispersion
n”(ω) = absorption line shape
n’’(ω)
n’,n’’
1
(Figure page II-16)
dn’
d >0
dω
Chapter 1: Math Toolbox
ω
0
dn’’
d
dω <0
n’(ω)
dn
dn’
dω > 0
ω
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ECE 460 – Optical Imaging
1.8 Refraction and absorption. Dispersion
1.8 Refraction and absorption. Dispersion
 Note the line shape:
p

1

2
2
2
2
(  o )   

1
 (  o )2o 
1 




2

1

1
   o 
1 



/
2
o 

2
 Lorentz function:
f

1
1
( )  
; a = width
2 
a  1  ( / a ) 
Chapter 1: Math Toolbox
38
ECE 460 – Optical Imaging
1.8 Refraction and absorption. Dispersion
1.8 Refraction and absorption. Dispersion
 Thus the asorption
p
line is a Lorentzian:
 ( ;o ) 
1

[ ( )]  e
1
2 ;   
   o 
1 






 t
e
-(t)

t
 The Fourier transform of a Lorentzian is an exponential!
Chapter 1: Math Toolbox
39
ECE 460 – Optical Imaging
1.8 Refraction and absorption. Dispersion
1.8 Refraction and absorption. Dispersion
Connect to quantum mechanics:
q
 2 level system:
1  E1
E  E1  Eo  (1  o )
0  Eo
 Probability of spontaneous emission/absorption:
/ lif i
 ‐t/tlifetime
 p(t) e
 exponential decay
 Linewidth is Lorentz = natural linewidth
! The model of e‐ on springs was introduced by... Lorentz
Chapter 1: Math Toolbox
40
ECE 460 – Optical Imaging
1.9 Maxwell’ss Equations
1.9 Maxwell
Equations
 Fully describe the propagation of EM fields
y
p p g
 Quantify how and generate each other
Ê
Ĥ
B
I
 E  
t
D
II
 H 
j
(1.45a)
t
D  
III
B  0
Chapter 1: Math Toolbox
IV
41
ECE 460 – Optical Imaging
1.9 Maxwell’ss Equations
1.9 Maxwell
Equations
 Plus material equations
q
D  oE  P
V
E
B  o H  M
VI
 Definitions:
  charge density
E  Eletric field vectors
Magnetic field vectors j   E = current density
= current density
H  Magnetic field vectors
D  Eletric displacement P  polarization
Magnetic inductance M  magnetization
B  Magnetic inductance Chapter 1: Math Toolbox
42
ECE 460 – Optical Imaging
1.9 Maxwell’ss Equations
1.9 Maxwell
Equations




Let’s combine I and II (assume no free charge: = 0, =0)
(
g  , j )
Use property:   (  E )  (E )   2 E
2


0


E

0



(


E
)


E
Since Take   ( EqI ) :
 B 
  (  E )    
(1.46)

 t 

  2 E  (  B ) 
t

  ( o  H ) 
t
(see next slide)


  ( o ) 
t
t
Chapter 1: Math Toolbox
43
ECE 460 – Optical Imaging
1.9 Maxwell’ss Equations
1.9 Maxwell
Equations

  E  (  B ) 
t

  ( o  H ) 
t


  ( o ) 
t
t
2E
  2
t
2
2
E

2
Thus:  E   2  0
t
(1.47)
 Wave Equation
Chapter 1: Math Toolbox
44
ECE 460 – Optical Imaging
1.9 Maxwell’ss Equations
1.9 Maxwell
Equations
Note: 1
  2 ;
v
c
v  ; c = speed of light in vacuum
n

n
 o o
 The wave equation describes the propagation of a atime‐
dependent field (eg. pulse)
Solution: plane wave: E  Eo e i (t  k r )
 Solution: plane wave:  k  2    n  ; k = wave equation

v
c
Chapter 1: Math Toolbox
45
ECE 460 – Optical Imaging
1.9 Maxwell’ss Equations
1.9 Maxwell
Equations
 Phase of the field:
(1.48)
  t  k r

 Note: = constant describes a surface
that moves with a certain velocity
t  k  r  constant eq of planes k
  dt  kdr  0
k
dr 

  vp
(1.49)
dt k
wave front
The surface of constant phase is traveling with velocity:

v p   phase velocity
k
Chapter 1: Math Toolbox
46
ECE 460 – Optical Imaging
1.9 Maxwell’ss Equations
1.9 Maxwell
Equations
 What is the counterpart of the wave equation in the frequency p
q
q
y
domain?
 
 i
 Well, remember
t
 Upon Fourier transforming, Eq. 1.47 becomes:
1
 E  2 (i.i ) E ( )  0
v
1
  2 E  2 ( 2 ) E ( )  0
v
2
 Note: k 

v
  2 E ( )  k 2 E ( )  0
Chapter 1: Math Toolbox
(1.50)
47
ECE 460 – Optical Imaging
1.9 Maxwell’ss Equations
1.9 Maxwell
Equations
  2 E ( )  k 2 E ( )  0
 The equation above is the “Helmholtz equation”
 Describes how each frequency ω propagates
Chapter 1: Math Toolbox
48