michael scalora u.s. army research, development, and engineering center redstone arsenal, alabama,...
TRANSCRIPT
![Page 1: Michael Scalora U.S. Army Research, Development, and Engineering Center Redstone Arsenal, Alabama, 35898-5000 & Universita' di Roma "La Sapienza" Dipartimento](https://reader038.vdocuments.mx/reader038/viewer/2022103111/5513e38b5503466f748b55a1/html5/thumbnails/1.jpg)
Michael Scalora
U.S. Army Research, Development, and Engineering CenterRedstone Arsenal, Alabama, 35898-5000
&Universita' di Roma "La Sapienza"
Dipartimento di Energetica
OPTICS BY THE NUMBERS
L’Ottica Attraverso i Numeri
Rome, April-May 2004
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Soluzione Numeriche di Equazioni Nonlineari Accoppiate Usando il Predictor-Corrector Algorithm:
More Examples
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2
0 0 0 0( ) ( ) ( ) ( )2
tx t t x t x t t x t
…is just a Taylor expansion for ANY function 0( )x t t
0 00 0
[ ( )] [ ( )]( ) ( )
2predictedf x t f x t t
x t t x t t
…always finds a second order accurate solution to the generic differential equation
( )[ ( )]
dx tf x t
dt
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Keep in mind that an nth order differential equation can always be reduced to n coupled equations of first order.
Example 1: can be rewritten as
the system:
2
20
1 40x x x
…with appropriate initial conditions.
2
20
( )( )
( ) 1 4( ) ( )
dx ty t
dt
dy ty t x t
dt
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24
( )( )
( ) ( )4 ( )
10
dx ty t
dtdy t y t
x tdt
0 0
40 0
1; 2 ;
10 ( )
(0) 1; (0) 0x x Con le condizioni iniziali:
Significa: …da cui…(0) (0) 0x y
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24
( )( )
( ) ( )4 ( )
10
dx ty t
dtdy t y t
x tdt
0 0 0
4 20 0 0 0
( ) ( ) ( )
( ) ( ) 10 ( ) 4 (
predicted
predicted
x t t x t y t t
y t t y t y t x t t
0 00 0
0 0 0 04 20 0
( ) ( )( ) ( )
2
( ) ( ) ( ) ( )( ) ( ) 10 4
2 2
predicted
predicted predicted
y t y t tx t t x t t
y t y t t x t x t ty t t y t t t
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Ricordiamo la Soluzione esatta:
/(2 ) 1 (0)( ) (0)cos (0) sin
2t x
x t e x t x t
1/ 2
0 2 20
11
4
0 0
40 0
1; 2 ;
10 ( )
(0) 1; (0) 0x x
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-1.0
-0.5
0
0.5
1.0
900 905 910
t
x(t)
-1.0
-0.5
0
0.5
1.0
900.5 900.7 900.9 901.1 901.3 901.5
t
x(t)
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( ) ( ) ( )
( ) ( ) ( )
x t x t y t
y t y t x t
( ) ( ) ( ) ( )x t x t x t y t
( ) ( ) ( ) ( )y t y t y t x t
Oscillatori smorzati accoppiati
Example 2:
damping restoring force
driving force
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Numerical solution
0 0 0 0( ) ( ) ( ) ( )px t t x t x t y t t
0 0 0 0( ) ( ) ( ) ( )py t t y t y t x t t
0 0
0 00 0
[ ( ) ( )]( ) ( ) / 2
[ ( ) ( )]
p
p
x t x t tx t t x t t
y t y t t
0 0
0 00 0
[ ( ) ( )]( ) ( ) / 2
[ ( ) ( )]
p
p
y t y t ty t t y t t
x t x t t
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-0.5
0
0.5
1.0
0 2 4 6 8 10
y(t)x(t)
t
==0.75; =2; =-2
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-2
-1
0
1
2
0 2 4 6 8 10
y(t)x(t)
t
=0.1; =0.3; =2; =-0.5
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Calculation of numerical derivatives
0 0 00
0
( ) ( ) ( )( ) lim
t
dx t x t t x tx t
dt t
However, in reality t cannot be zero, and so one incurs into an error:
0 0 00
( ) ( ) ( )( )
dx t x t t x tx t Error
dt t
?
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23
0 0 0 0( ) ( ) ( ) ( ) ( ) ...2
tx t t x t x t t x t t
Consider the Taylor Expansion:
Then:
20 00 0
( ) ( )( ) ( ) ( ) ...
2
x t t x t tx t x t t
t
Calculating the numerical derivative as:
0 00
( ) ( )( )
x t t x tx t
t
Implies an error of order t …
0( )2
tError x t
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Usando la matematica del buon senso
La derivata al punto puo essere definita in almeno due modi:
0t t0t 0t t
0 00
0 00
( ) ( )( )
( ) ( )( )
x t t x tx t
tx t x t t
x tt
0t
0 0 0 00
( ) ( ) ( ) ( )( )
2 2
x t x t x t t x t tx t
t
mediando le due soluzioni…
.. .
…si presume con un errore piu piccolo.
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5
2 3( ) ( ) ( ) ( ) ( )0 0 0 0 02 6
4( ) ( ) ..0 24
.
t tx t t x t x t t x t x t
tx t t
Consider the Taylor Expansions:
2 3
0 0 0 0 0
54
( ) ( )0 24
( ) ( ) ( ) ( ) ( )2 6
tx t t
t tx t t x t x t t x t x t
35
0 0 0 0( ) ( ) 2 ( ) ( ) ( ) ...3
tx t t x t t x t t x t t
20 0
0 0
( ) ( )( ) ( )
2 6
x t t x t t tx t x t
t
Subtract…
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20 0 0 0
4( )0 12
( ) ( ) 2 ( ) ( )t
x tx t t x t t x t x t t
20 0 0
0 2( )0 12
( ) 2 ( ) ( )( )
tx t
x t t x t x t tx t
t
0 0 0 0 0
50
2 3( ) ( ) ( ) ( ) ( )
2 64
( ) ( ) ..24
.
t tx t t x t x t t x t x t
tx t t
2 3
0 0 0 0 0
54
( ) ( )0 24
( ) ( ) ( ) ( ) ( )2 6
tx t t
t tx t t x t x t t x t x t
Add…
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1. Increasing precision requires more information, storage
0 00
( ) ( )( ) ( )
x t t x tx t t
t
20 0
0
( ) ( )( ) ( )
2
x t t x t tx t t
t
2. An nth order differential equation can always be reduced to n coupled equations of first order.
2
20
1 40x x x
2
20
( )( )
( ) 1 4( ) ( )
dx ty t
dt
dy ty t x t
dt
3. Calculating derivatives and integrating differential equations is more of an art than a science, and one uses whatever works. Care should be exercised when considering functions that vary rapidly in space or time.
Ex.:
In general,
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2 2 2
2 2 2
( , ) ( , )E z t n E z t
z c t
We have all the elements for a simple one dimensional electromagnetic pulse propagation algorithm
0
0.2
0.4
0.6
0.8
1.0
0 0.2 0.4 0.6 0.8 1.0
z (distance)
t (t
ime)
0 0( , )z t
0 0( , )z t t
0 0( , )z t t
0 0( , )z z t 0 0( , )z z t
0t
0z
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22 20 0 0 0 0 0
2 2 2 2
20 0 0 0 0 0 0 02
0 0
2
( , ) 2 ( , ) ( , )( ,
( , ) ( , ) 2 ( , ( )
)
) ,
E z t E z t E z t tn n
c t c t
E z t E z z t E z t E z z t
z z
E z t t
Combine and solve for 0 0( , )E z t t
0 0 0 0
2 2
0 0
0 0 0 0 0 02 2
2 ( , ) ( , )
( , ) 2 ( , ) , )
,
(
( ) E z t E z t t
c tE z z t E z t E z z
z
E z t
tn
t
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2 20( ) /
0 0 0( , 0) cos( )z z wE z t E e k z
Initial Condition
0.2
0.6
1.0
10 20
Position in m
Fie
ld Inte
nsity
Input Intensity 00
2k
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0
0.2
0.4
0.6
0.8
1.0
0 25 50 75
Position in m
Fie
ld I
nten
sity
Propagation in free Space
1 20t t t t t
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0
0.2
0.4
0.6
0.8
1.0
0 10 20 30 40
position in m
Fie
ld I
nte
nsit
y
For clarity only envelopes are shown
n=1 n=2
Incident
transmitted
reflected
|E|2
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2 2 2
2 2 2
( , ) ( , )E z t n E z t
z c t
Another point of view:
L’equazione di secondo grado:
Diventa due equazioni accoppiate di primo grado…
2
2
( , ) ( , )
( , ) ( , )
E z t c B z t
t n zB z t E z t
t z
…che sono le equazioni di Maxwell in una dimensione e il tempocioe il punto di partenza.
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0 0 0 0
2 20 0 0 0 0 0
2
0
2
0( , ) ( ,( , )
2
( , ) ( , ) ( , )
2
)E z t E z t t
t t
B z t B z
E z t
z t B z z tc c
n z n z
t
0 0
2
0 0 0 0 0 02( , ) ( , )( ) ( ,, )
c tE z t t B z z t B z z tt t
n zE z
Algebraically Solve for 0 0( , )E z t t
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0 00 0 0 0
0 0 0 0 0 0
( , ) ( , )
2( , ) ( , ) ( , )
2
( , )B z t B z t t
t tE z t E z z t E
B z
z
t t
z t
z z
0 0 0 0 0 00 0( , ) ( , ) ( , ) ( , )t
B z t t E z z t EB zz z tz
t t
Algebraically Solve for 0 0( , )B z t t
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1. The SVEA, the Wave Equation, and Diffraction
2. Spectral Methods and Free-Space Diffraction
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Diffraction
The Bending of Light Around Corners
a
0 a
a
0 ~ aL
L
ray opticswave optics
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0
0.2
0.4
0.6
0.8-40-30
-20-10
010
2030
40
transverse coordinate
intensity
a
0 a
a
0 ~ aL
ray opticswave optics
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2 22
2 2
n
c t
E
E
( )ˆ ( , , ) .i kz tE x y z e c c E x
x
z
y
E
2 2 22 2
2 22 t
E E nik k E E E
z z c
2 22
2 2t x y
k n
c
This term decribes diffraction
Assumption: the beam envelpe does not vary in timeso-called CW (continuous wave) beam
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2
22
E Eik
z z
( )( , ) .i kz tE E x z e c c
For simplicity, let’s assume only one transverse dimension:
Assume…
2
22
E Eik
z x
22
2t x
E(x,z) e’ un inviluppo che varia lentamente nello spazio rispetto a e nel tempo rispetto a
/c.
2
22
E i E
z k x
k nc
2 2 2 22
2 2 22
E E E nik k E E
z z x c
Apply the SVEA…
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/ /z L x x L
We may also define the Fresnel number as…0
4 nLF
0E
F
F small
Il fascio non diffrange: Ray optics..
Regime di Forte diffrazzione.
2
22
E i E
z k x
2
2
E i E
F x
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2 2 2
2 2 2
( , ) ( , )E z t n E z t
z c t
How to calculate diffraction using the wave equation
Confrontiamo…
Con…2
2
E i E
F x
20 0 0 0 0 0 0 02 2
( , ) ( , ) 2 ( , ) ( , )E x z E x x z E x z E x x z
x x
20 0 0 0 0 0 0 02 2
( , ) ( , ) 2 ( , ) ( , )E z t E z z t E z t E z z t
z z
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0 0 0 0 0 0( , ) ( , ) ( , )
2
E x E x E x
20 0 0 0 0 0 0 02 2
( , ) ( , ) 2 ( , ) ( , )E x E x x E x E x x
x x
Combine and solve for 0 0( , )E x
0 0 0 0
0 0 0 0 0 02
( , ) ( , )
( , ) 2 ( , ) ( , )2
E x E x
E x x E x E x x
x
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0
0 0
0 0
0 0
0 02
0
( , )
( , ) 2 (( ,
( ,
) )
)
,2
E x
E x
E x x E x x
E
x
x
We can proceed as follows:
0 0 00 0
0( , )
2
( ))
,( ,
E x xE x
E
Substitute and solve for 0 0( , )E x
0 0
0 0
0 0
0 0 0 0 0 02
( , )
( , ) ( , )
( , )
( , ( )2
) ,
E x
E x
E x
E x x E x E x x
x
Let…
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0 0
0 0 0 0 0 02 2
0 0 ( , )
( , ) ( , ) ( , ) 2
(1 2
( , )
/ )
E x
E x x E x E
x
x
x x
E
x
0 0
0 0
0 0
0 0 0 0 0 02
( , )
( , ) ( , )
( , )
( , ( )2
) ,
E x
E x
E x
E x x E x E x x
x
Ci sono altri modi di procedere che non richiedono un’impostazione cosi onerosa dal punto di vista del numero di varibili e vettori di cui
tener conto. Ci occuperemo di metodi cosidetti spettrali.
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How to calculate diffraction using the wave equation…
Useful properties of the Fourier Transform: Derivatives…
2 22
2 2
( , ) ( , ) ( , )
( , ) ( , ) ( , )
( , ) ( , ) ( , )
( , ) ( , ) ( ) ( , )n
iqx
iqx
iqx
niqx n
n n
FT E x E q E x e dx
FT E x E x e dx iqE qx x
FT E x E x e dx q E qx x
FT E x E x e dx iq E qx x
2
2
E i E
F x
21,2,3,4...
x
q j jL
xL N x
N is the number of points used to discretize the space x in units x.
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Some Advantages of using Spectral (FT) methods
(i) calculation of all kinds of derivatives is simple(ii) derivatives are extremely accurate(iii) simple algorithm
Some Disadvantages
(i) It is slower(ii) Functions should be very smooth (which usually are)
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2( , )( , )
E q iqE q
F
2
( , ) ( ,0)iq
FE q e E q
…is the solution in q-space. To find the solution in x-space, where things are observable, we need to take the inverse transform:
2
2
( , ) ( , )E x i E x
F x
Taking the FT of both sides with respect to x…
And…
2
1( , ) ( , ) ( ,0)iq
iqxFE x FT E q e E q e dq
( ,0) ( ,0)E q FT E x
2
1( , ) ( ,0)iq L
FE x L FT e FT E x
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2
1( , ) ( ,0)iq L
FE x FT e FT E x
Algorithm:
(i) Trasformata di Fourier (FT) del campo iniziale
(ii) Moltiplicazione per il propagatore
(iii) Transformata inversa di tutto
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Examples: single slit
0
0.2
0.4
0.6
0.8
-40 -30 -20 -10 0 10 20 30 40
transverse coordinate
inte
nsity
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Direction of Propagation
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-15 -5 5 15
TRANSVERSE COORDINATE
0
10
20
30
40
50
LO
NG
ITU
DIN
AL
C
OO
RD
INA
TE
0.0
1.0 1.1
1.1
Direction of Propagation
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0
0.2
0.4
0.6
0.8
1.0
-60 -20 20 60
Example2: double slit
Poisson Spot
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-25 -15 -5 5 15 25
TRANSVERSE COORDINATE
0
10
20
30
40
50
LO
NG
ITU
DIN
AL
C
OO
RD
INA
TE
0.0 0
.0
0.1 0.1
1.0 1.0
1.0 1.0 1.1 1.1 1
.1
1.1
1.1
1.1 1.2
1.2
1.2
1.2
1.2 1.2
1.3 1
.3
1.3 1.3
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2 22
2 2t x y
2 2
2 2
( , , )( , , )
E x y iE x y
F x y
Per aperture con geometrie piu complicate, e.g.Apertura Circolare o quadrata, e’ necessario
Ritornare alle due dimensioni trasverse:
2 2( , , ) ( )( , , )x y x yx y
E q q i q qE q q
F
2 2( )
( , , ) ( , ,0)x yi q q
Fx y x yE q q e E q q
2
1 2 2 2( , , ) ( , ,0)iq L
Fx yE x y L FT e FT E x y q q q
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Circular Aperture
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Square Aperture
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10 60 110 160 210
Y
10
60
110
160
210
X
0.0
0.0
0.0
0.0
0.0 0.0
0.1 0.1 0.1 0.1 0.1
0.1
0.1
0.1 0.1
Square Aperture
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Algorithm:
(i) Trasformata di Fourier (FT) del campo iniziale
(ii) Moltiplicazione per il propagatore
(iii) Transformata inversa di tutto
2
1 2 2 2( , , ) ( , ,0)iq L
Fx yE x y L FT e FT E x y q q q
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2
22
E Eik
z x
2
22
E i E
z k x
2 2 2 22
2 2 22
E E E nik k E E
z z x c
Removing the second order spatial derivative also meansmaking the Paraxial Wave Approximation:
the beam radius cannot be too smallor inconsistencies with experimental observations
may arisem since the wave tends to diffract very fast,contrary to expectations.
This problem may be partially cured as follows:
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2 2 2 22
2 2 22
E E E nik k E E
z z x c
2 2 2 22
2 2 22
E E n Eik k E E
z x c z
2 2
2 22
E E Eik
z x z
2 2
2 2
E i E i E
F x F
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22
2 2
i E iE
F
E
x F
2 3
2 3
2
2
E i E i E
F x F
2 2 3
2 2 3
E i E i i i E
F x F F
E
x F
2 2 32 2
2 2 2 2 3
E i E i i i E
F x F F
i E
x
i
F x F F
E
4
3
2 2 2 3
2 3 2 2 34
E i E i E i E
F x F
E
x FF x
i