9/12/2016
EE 5303
Electromagnetic Analysis Using Finite‐Difference Time‐Domain
Lecture #9
Examples of One‐
Dimensional FDTD
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Lecture 9
Slide 1
Lecture Outline
• Review of Lecture 8
– FDTD Algorithm
– Code walkthrough
• Simple Electromagnetic Structures
• Two Examples
Lecture 9
Slide 2
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Review of Lecture #8
Lecture 9
Slide 3
Typical FDTD Grid Layout
Note: A real grid would have 200 or more points.
TF/SF Interface
Scattered‐Field Total‐Field
Source Injection Point
Spacer Region
max
Reflection Record Point
Structure Being Modeled
PAB
Lecture 9
Transmission Record Point
PAB
r k
nbc
Spacer Region
max
src

and  r
ksrc 
nmax and nmin
nbc
Slide 4
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Initializing the FDTD Simulation
Initialize Simulation
• Initialize MATLAB
• Define units
• Define constants
Compute Time Step
Define Simulation Parameters
• Frequency range (fmax)
• Device parameters
• Grid parameters (NRES, etc.)
Compute Source
  0.5 B
Compute Update Coefficients
k
c t
c t
k
mEy  0 k
mHx  0 k
 xx
 yy
t 
nsrc z t

2c0
2
  t  t0   t  2 
H x ,src  t   A exp   
 

 
 
Snap grid to critical dimension
N  ceil  dc 
r
r
A
N   10
Nd  1

d 
  min  min , min 
 N Nd 
t0  6
  t  t0  2 
E y ,src  t   exp   
 
    
Compute Grid Resolution
Initial resolution
 2c0 
t  nbc  z
Initialize Fields to Zero
Ey  H x  0
  dc N
Initialize Boundary Terms to Zero
Initialize Fourier Transforms
Ki  e
Build Device on Grid
 j 2  f i t
h3  h2  h1  e3  e 2  e1  0
Fi  0
Refer to Lecture 3.
Lecture 9
Slide 5
The Main FDTD Loop
yes
Done?
Update H (Perfectly Absorbing Boundary)
H x
no
k
t  2t
H xN z
z
k
 H x
t  2t
t  2t
 H xN z
  E  E  z
 m
  e  E  z

 mHx
k 1
k
k
y t
Nz
Nz
t  2t
k  Nz
y t
Hx
k  Nz
y t
3
Handle E Source
Ey
ksrc
t t
Handle H Source
H xksrc 1
t  2t
 H xksrc 1
t  2t

mHx
E y ,src
ksrc
e3  e 2
t
Record H at Boundary
t
h3  h2
h2  h1
ksrc
t t

mEy
ksrc
H xsrc
z
ksrc 1
t  2t
Record E at Boundary
ksrc 1
z
 Ey
h1  H x
e 2  e1
e1  E y
Nz
t
Update Fourier Transforms
1
Fi t t  Fi t  t   K i  E yi , j ,k
m
t  2t
t t
f
Update E (Perfectly Absorbing Boundary)
Ey
Ey
z
Lecture 9
k
t t
1
t t
k
 
  m   H
 E y  mEy
t
 Ey
1
t
k
1
Ey
H x
k
t  2t
1
x t  t
2
 H x
 h3
k 1
t  2t
 z
 z
k 1
k 1
Visualize Simulation
• Superimpose fields on materials
• Show reflectance, transmittance and conservation
• Update only after some number of iterations
Slide 6
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The Main FDTD Loop (Pseudo Code)
% MAIN FDTD LOOP
for T = 1 : STEPS
% Update H from E
for nz = 1 : Nz
Update Hx(nz)
end
% H Source
Correct Hx(nz_src-1)
% Record H-Field at Boundary
H3Hx(1)
% Update E from H
for nz = 1 : Nz
Update Ey(nz)
end
% E Source
Correct Ey(nz_src)
% Record E-Field Boundary
E3Ey(Nz)
% Update Fourier Transforms
for nf = 1 : NFREQ
Integrate REF(nf), TRN(nf), and SRC(nf)
end
% Visualize
Plot fields, materials, and response
end
Lecture 9
Slide 7
Post Processing
yes
Done?
Compute Response
no
 Fref  f  
R f   

 FFT  Esrc  t   



2
 Ftrn  f  
T f 

 FFT  Esrc  t   



C  f   R f  T  f 
2
Visualize Results
• Superimpose fields on materials
• Show reflectance, transmittance and conservation
• Show response on linear and dB scale
Finished!
Lecture 9
Slide 8
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Outline of Steps for FDTD Analysis
• Step 1: Define problem
–
–
–
–
What device are you modeling?
What is its geometry?
What materials is it made of?
What do you want to learn about the device?
• Step 2: Initialize FDTD
–
–
–
–
Compute grid resolution
Assign materials values to points on the grid
Compute time step
Initialize Fourier transforms
• Step 3: Run FDTD
• Step 4: Analyze the data
Lecture 9
Slide 9
Step 1: Define the Problem
What device are you modeling? What is its geometry? What materials it is made from? What do you want to learn? –A dielectric slab
–1 foot thick slab
–r=2.0, r=6.0 (outside is air)
–reflectance and transmittance
from 0 to 1 GHz
1.0 ft
Air
r  2.0,  r  6.0
Air
Lecture 9
Slide 10
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Step 2: Compute Grid (1 of 2)
Initial Grid Resolution (Wavelength)
N   20
nmax  r  r 
299792458 ms
c0

 8.6543 cm
f max nmax 1.0 GHz  3.46 
min
N

8.6543 cm
 0.4327 cm
20
Air
1.0 ft
min 
 
 2.0  6.0   3.46
Initial Grid Resolution (Structure)
r  2.0,  r  6.0
Air
Nd  4
d 
d
30.48 cm

 7.6200 cm
Nd
4
Initial Grid Resolution (Overall)
z   min    ,  d   0.4327 cm
Lecture 9
Slide 11
Step 2: Compute Grid (2 of 2)
Snap Grid to Critical Dimension(s)
The number of grid cells representing the thickness of the dielectric slab is
N 
dc
30.48 cm

 70.44 cells

z 0.4327 cm
It is impossible to represent the thickness of the slab exactly with this grid resolution.
To represent the thickness of the slab exactly, we round N’ up to the nearest integer and then calculate the grid resolution based on this quantity.
N  round   N   71 cells
d c 30.48 cm

 0.4293 cm
N
71
Air
1.0 ft
z 
r  2.0,  r  6.0
Air
Lecture 9
Slide 12
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Step 2: Build Device on the Grid (1 of 2)
Determine Size of Grid
We need to have enough grid cells to fit the device being modeled, some space on either side of the device (10 cells for now), and cells for injecting the source and recording transmitted and reflected fields.
N z  71  2 10 cells   3  94 cells
TRN/REF/SRC cells
spacer regions
cells for device
space
space
slab will go here
TF/SF Source
record transmission
record reflection
Lecture 9
Slide 13
Step 2: Build Device on the Grid (2 of 2)
Compute Position of Materials on Grid
nz ,1  2  10  1  13
nz ,2  nz ,1  round  d z   1  13  71  1  83
nz ,1
nz ,2
Add Materials to Grid
nz ,1
nz ,2
UR(nz1:nz2) = ur;
ER(nz1:nz2) = er;
Lecture 9
Slide 14
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Step 2: Initialize FDTD (1 of 2)
Compute the Time Step
t 
nbc z 1.0  0.4293 cm 

 7.1599 1012 sec
2c0
2  299792458 ms 
Compute Source Parameters

1
1

 5.00  1010 sec
2 B 2 1 GHz 
t0  6  3.00  109 sec
t0  6 Rule of thumb
Compute Number of Time Steps
tprop 
nmax N z z  3.46  94  0.4293 cm 

 4.6629  10 9 sec
c0
 299792458 ms 
Time it takes a wave to propagate across the grid.
T  12  5tprop  12  5 1010 s   5  4.6597  109 s   2.9314  108 sec
T 
STEPS  round     4095
 t 
T  12  5tprop Rule of thumb
STEPS must be an integer
Lecture 9
Slide 15
Step 2: Initialize FDTD (2 of 2)
Compute the Source Functions for Ey/Hx Mode
t 
nsrc z t 3t


 1.0740  1011 sec
2c0
2
2
  t  t0  2 
E y  t   exp   
 
    
% COMPUTE GAUSSIAN SOURCE FUNCTIONS
t
= [0:STEPS-1]*dt;
delt = nsrc*dz/(2*c0) + dt/2;
A
= - sqrt(ersrc/ursrc);
Esrc = exp(-((t-t0)/tau).^2);
Hsrc = A*exp(-((t-t0+delt)/tau).^2);
A
 r k
r k
src

src


1.0
 1
1.0
  t  t0   t  2 
H x  t   A exp   
 

 
 
%time axis
%total delay between E and H
%amplitude of H field
%E field source
%H field source
Initialize the Fourier Transforms
% INITIALIZE FOURIER TRANSFORMS
NFREQ = 100;
FREQ = linspace(0,1*gigahertz,NFREQ);
K
= exp(-i*2*pi*dt*FREQ);
REF
= zeros(1,NFREQ);
TRN
= zeros(1,NFREQ);
SRC
= zeros(1,NFREQ);
Lecture 9
Slide 16
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Step 3: Run FDTD (3 of 3)
Lecture 9
Slide 17
Step 4: Analyze the Data
Normalize the Data to the Source Spectrum
 Fref  f  
R f   

 FFT  Esrc  t   



2
 Ftrn  f  
T f 

 FFT  Esrc  t   



C  f   R f T  f 
2
% COMPUTE REFLECTANCE
% AND TRANSMITTANCE
REF = abs(REF./SRC).^2;
TRN = abs(TRN./SRC).^2;
CON = REF + TRN;
Lecture 9
Slide 18
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Simple Electromagnetic Structures
Lecture 9
Slide 19
Reflection and Transmission at an Interface
Reflection and Transmission Coefficients
At normal incidence, the field amplitude of waves reflected from, or transmitted through, an interface are related to the incident wave through the reflection and transmission coefficients.
r
 2  1
2  1
t
2 2
 2  1
Reflectance and Transmittance
The reflectance and transmittance quantify the fraction of power that is reflected from, or transmitted through, an interface.
R r
2
Tt
2
1
2
Useful Special Cases
For r=9.0 and r=1.0, R=25% and T=75%.
For r=1.0 and r=9.0, R=25% and T=75%.
For r=9.0 and r=9.0, R=0% and T=100%.
Lecture 9
Slide 20
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Anti‐Reflection Layer
n1 ,1
n2 ,2
nar ,ar
n1 ,1
L
ar  1 2
General Case
n2 , 2
L
nar  n1n2
0
L
4nar
0
No magnetic response
4nar
Lecture 9
Slide 21
Bragg Gratings
LH
LL
LH
LL
LH
LL
LH
LL
LL 
nH
nL
nH
nL
nH
nL
nH
nL
LH 
0
4nL
0
4nH

A Bragg grating is typically composed of alternating layers of high and low refractive index. Each layer is /4 thick. Higher index contrast provides wider stop band. More layers improves suppression in the stop band.
stop band
0
Lecture 9
Slide 22
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Example #1: The Invisible Slab
Lecture 9
Slide 23
Design Problem
A radome is being designed to protect an antenna operating at 2.4 GHz. For mechanical reasons, it must be constructed from 1 ft thick plastic with dielectric constant 12. How could you modify the design to maximize transmission through the radome? Simulate the design using 1D FDTD.
Lecture 9
Slide 24
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A Solution
Add anti‐reflection layers to both sides of the radome.
2
1
2
Lecture 9
Slide 25
The Design
To match the slab material to air on both sides, the dielectric constant and thickness of the anti‐reflection layers should be  2  1 air 
12 1 
3.46
n2   2  3.46  1.86
0 
c0 299792458 ms

f0
2.4 109 Hz
299792458 ms

 12.49 cm
2.4 109 Hz

12.49 cm
d2  0 
 1.6779 cm
4n2
4 1.86 
Lecture 9
2
d2
1
2
d2
Slide 26
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FDTD Simulation Results
Frequency (GHz)
Lecture 9
Slide 27
Example #2: The Blinded Missile
Lecture 9
Slide 28
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Design Problem
A heat‐seeking missile is vulnerable to jamming from high power lasers operating at 0=980 nm. Design a multilayer cover that would prevent this energy from reaching the infrared camera. The design should provide at least 30 dB of suppression at 980 nm. Simulate the design using 1D FDTD. The only materials available to you are SiO2 (nSiO2 = 1.5) and SiN (nSiN = 2.0).
Lecture 9
Slide 29
A Solution
Use a Bragg grating with alternating layers of SiO2 and SiN.
d2
d1
nSiO2  1.5
nSiN  2.0
SiN
SiO2
SiN
SiO2
SiN
SiO2
SiN
SiO2
Lecture 9
n2
n1
n2
n1
n2
n1
n2
n1
Slide 30
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The Design
0
980 nm
d1 

 163 nm
4n1 4 1.5 
d2 
0
4n2

d2
d1
980 nm
 122 nm
4  2.0 
SiN
SiO2
SiN
SiO2
SiN
SiO2
But how many layers?
SiN
SiO2
n2
n1
n2
n1
n2
n1
n2
n1
Lecture 9
Slide 31
Number of Layers for 30 dB Suppression
20 periods, 45 dB! Yikes!!
10 periods, barely 20 dB 
15 periods, 30 dB. 
In practice, you may want to include a few extra layers as a safety margin. Manufacturing inaccuracies often degrade performance.
Lecture 9
Slide 32
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FDTD Simulation Results
Lecture 9
Slide 33
17