Lecture #9 Examples of One‐ Dimensional FDTDemlab.utep.edu/ee5390fdtd/Lecture 9 -- Examples of 1D...
Transcript of Lecture #9 Examples of One‐ Dimensional FDTDemlab.utep.edu/ee5390fdtd/Lecture 9 -- Examples of 1D...
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Lecture 9 Slide 1
EE 5303
Electromagnetic Analysis Using Finite‐Difference Time‐Domain
Lecture #9
Examples of One‐Dimensional FDTD These notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited
Lecture Outline
Lecture 9 Slide 2
• Review of Lecture 8
– FDTD Algorithm
– Code walkthrough
• Simple Electromagnetic Structures
• Two Examples
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Lecture 9 Slide 3
Review of Lecture #8
Lecture 9 Slide 4
Typical FDTD Grid Layout
Note: A real grid would have 200 or more points.
PAB PAB
bcnbcn
max min and n n
Total‐FieldScattered‐Field
TF/SF Interface
src src and k kr r
Source Injection Point
Transmission Record Point
Reflection Record Point
Structure Being Modeled
Spacer Region
max
Spacer Region
max
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Lecture 9 Slide 5
Initializing the FDTD Simulation
Compute Source
max 0
2
0,src
src
0
2
0,src
0.5 6
exp
2 2
exp
y
r
r
x
f t
t tE t
n z tA t
c
t t tH t A
Initialize Simulation
• Initialize MATLAB• Define units• Define constants
Initialize Fourier Transforms2 0ij f t
i iK e F
Define Simulation Parameters
• Frequency range (fmax)• Device parameters• Grid parameters (NRES, etc.)
Compute Grid Resolution
min min10
min , 1dd
Nd
NN N
Initial resolution
ceil c cN d d N
Snap grid to critical dimension
Build Device on Grid
Refer to Lecture 3.
Compute Time Step
bc 02zt n c
Compute Update Coefficients
0 0 k k
Ey Hxk k
xxyy
c t c tm m
Initialize Fields to Zero
0y xE H
Initialize Boundary Terms to Zero
3 2 1 3 2 1e e e 0h h h
Lecture 9 Slide 6
The Main FDTD Loop
noDone?
, ,m i j ki i i yt t t t tF F t K E
Update Fourier Transforms
Update H (Perfectly Absorbing Boundary)
2 2
2 2
1
3e
z z
t t
zz
t t
k k k kk
Hx zt tt t
NN
Hx ztt t
x x
N Nx x
y y
y
m z k N
m
E E
E
H
zH N
H
H k
Handle H Sourcesrc
src srsrc
c
2 2
1
1 1,srct t
k
Hxk kx xt t
k
y t
mH H
zE
Record H at Boundary
2
1
2 1 1 tx t
h h h H
Update E (Perfectly Absorbing Boundary)
2 2
23
1
11 1 1
1
1
t t
t
k kk k k
Eyt t t t t
Eyt t t
y
y
x
t
y x
xy
E m H H
H h
E
E E
z k
m z k
Handle E Source
src
src sr sr
2
c c 1srct
k
k k Ey
y y
k
t xt t t t
mE E
zH
Record E at Boundary
2 1 1 z
y
N
te Ee e
Visualize Simulation
• Superimpose fields on materials• Show reflectance, transmittance and conservation
• Update only after some number of iterations
yes
z
z
f
Loop over time
t
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Lecture 9 Slide 7
The Main FDTD Loop (Pseudo Code)% MAIN FDTD LOOPfor T = 1 : STEPS
% Record H-Field at BoundaryH2Hx(1)
% Update H from Efor nz = 1 : Nz
Update Hx(nz)end
% H SourceCorrect Hx(nz_src-1)
% Record E-Field BoundaryE2Ey(Nz)
% Update E from Hfor nz = 1 : Nz
Update Ey(nz)end
% E SourceCorrect Ey(nz_src)
% Update Fourier Transformsfor nf = 1 : NFREQ
Integrate REF(nf), TRN(nf), and SRC(nf)end
% VisualizePlot fields, materials, and response
end
Lecture 9 Slide 8
Post Processing
Finished!
Compute Response
2
ref
src
2
trn
src
FFT
FFT
F fR f
E t
F fT f
E t
C f R f T f
Visualize Results
• Superimpose fields on materials• Show reflectance, transmittance and conservation
• Show response on linear and dB scale
noDone?
yes
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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
Lecture 9 Slide 10
Step 1: Define the Problem
What device are you modeling? –A dielectric slabWhat is its geometry? –1 foot thick slabWhat materials it is made from? –r=2.0, r=6.0 (outside is air)What do you want to learn? –reflectance and transmittance
from 0 to 1 GHz
2.0, 6.0r r
Air
Air
1.0 ft
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Lecture 9 Slide 11
Step 2: Compute Grid (1 of 2)
Initial Grid Resolution (Wavelength)
max
m0 s
minmax max
min
20
2.0 6.0 3.46
299792458 8.6543 cm
1.0 GHz 3.46
8.6543 cm0.4327 cm
20
r r
N
n
c
f n
N
Initial Grid Resolution (Structure)
4
30.48 cm7.6200 cm
4
d
dd
N
d
N
Initial Grid Resolution (Overall)
min , 0.4327 cmdz
2.0, 6.0r r
Air
Air
1.0 ft
Lecture 9 Slide 12
Step 2: Compute Grid (2 of 2)
Snap Grid to Critical Dimension(s)
30.48 cm70.44 cells
0.4327 cmcdNz
The number of grid cells representing the thickness of the dielectric slab is
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.
round 71 cells
30.48 cm0.4293 cm
71c
N N
dz
N
2.0, 6.0r r
Air
Air
1.0 ft
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Lecture 9 Slide 13
Step 2: Build Device on the Grid (1 of 2)
Determine Size of Grid
71 2 10 cells 3 94 cellszN
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.
slab will go herespace space
record transmission
record reflection
TF/SF Source
TRN/REF/SRC cells
spacer regions
cells for device
Lecture 9 Slide 14
Step 2: Build Device on the Grid (2 of 2)
Compute Position of Materials on Grid
,1
,2 ,1
2 10 1 13
round 1 13 71 1 83
z
z z
n
n n d z
,1zn ,2zn
Add Materials to Grid
,1zn ,2zn
UR(nz1:nz2) = ur;ER(nz1:nz2) = er;
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Lecture 9 Slide 15
Step 2: Initialize FDTD (1 of 2)
Compute the Time Step
12bc
m0 s
1.0 0.4293 cm7.1599 10 sec
2 2 299792458
n zt
c
Compute Source Parameters
10
max
90
1 15.00 10 sec
2 2 1 GHz
6 3.00 10 sec
f
t
0 6 Rule of thumbt
Compute Number of Time Steps
9maxprop m
0 s
10 9 8prop
3.46 94 0.4293 cm4.6629 10 sec
299792458
12 5 12 5 10 s 5 4.6597 10 s 2.9314 10 sec
STEPS round 4095
zn N zt
c
T t
T
t
prop12 5 Rule of thumbT t
STEPS must be an integer
Time it takes a wave to propagate across the grid.
Lecture 9 Slide 16
Step 2: Initialize FDTD (2 of 2)
Compute the Source Functions for Ey/Hx Mode
src
src
11src
0
2 2
0 0
3 1.01.0740 10 sec 1
2 2 2 1.0
exp exp
krkr
y x
n z t tt A
c
t t t t tE t H t A
Initialize the Fourier Transforms
% INITIALIZE FOURIER TRANSFORMSNFREQ = 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);
% COMPUTE GAUSSIAN SOURCE FUNCTIONSt = [0:STEPS-1]*dt; %time axisdelt = nsrc*dz/(2*c0) + dt/2; %total delay between E and HA = - sqrt(ersrc/ursrc); %amplitude of H fieldEsrc = exp(-((t-t0)/tau).^2); %E field sourceHsrc = A*exp(-((t-t0+delt)/tau).^2); %H field source
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Lecture 9 Slide 17
Step 3: Run FDTD (3 of 3)
Lecture 9 Slide 18
Step 4: Analyze the Data
Normalize the Data to the Source Spectrum
t
src
src
r
rn
2
2
ef
FFT
FFT
F fR
E t
E t
f
R f
F fT
f
f
fC T
% COMPUTE REFLECTANCE % AND TRANSMITTANCEREF = abs(REF./SRC).^2;TRN = abs(TRN./SRC).^2;CON = REF + TRN;
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Lecture 9 Slide 19
Simple Electromagnetic
Structures
Lecture 9 Slide 20
Reflection and Transmission at an Interface
2 1 2
2 1 2 1
2 r t
Reflection and Transmission CoefficientsAt 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.
Reflectance and TransmittanceThe reflectance and transmittance quantify the fraction of power that is reflected from, or transmitted through, an interface.
2 2 R r T t
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%.
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Lecture 9 Slide 21
Anti‐Reflection Layer
1 1,n 2 2,n
1 1,n 2 2,n ar ar,n
L
ar 1 2
0
ar4L
n
General Case
ar 1 2
0
ar4
n n n
Ln
No magnetic response
Lecture 9 Slide 22
Bragg Gratings
Ln Ln Ln LnHnHnHnHn
LLHL LLHL LLHL LLHL
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.
0
0
4
4
LL
HH
Ln
Ln
stop band
0
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Lecture 9 Slide 23
Example #1: The Invisible Slab
Lecture 9 Slide 24
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.
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Lecture 9 Slide 25
A Solution
Add anti‐reflection layers to both sides of the radome.
1
2
2
Lecture 9 Slide 26
The Design
To match the slab material to air on both sides, the dielectric constant and thickness of the anti‐reflection layers should be
1
2
2
2d
2d
2 1 air
2 2
m0 s
0 90
ms
9
02
2
12 1 3.46
3.46 1.86
299792458
2.4 10 Hz
299792458 12.49 cm
2.4 10 Hz12.49 cm
1.6779 cm4 4 1.86
n
c
f
dn
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Lecture 9 Slide 27
FDTD Simulation
Lecture 9 Slide 28
Final Transmittance/Reflectance/Conservation
Frequency (GHz)
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Lecture 9 Slide 29
Example #2: The Blinded Missile
Lecture 9 Slide 30
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).
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Lecture 9 Slide 31
A Solution
Use a Bragg grating with alternating layers of SiO2 and SiN.
2n
1n
1n
1n
1n
2n
2n
2n
2SiO
SiN
1.5
2.0
n
n
SiN
SiO2
SiN
SiO2
SiN
SiO2
SiN
SiO2
2d
1d
Lecture 9 Slide 32
The Design
2n
1n
1n
1n
1n
2n
2n
2n
2d
1d
01
1
02
2
980 nm163 nm
4 4 1.5
980 nm122 nm
4 4 2.0
dn
dn
SiN
SiO2
SiN
SiO2
SiN
SiO2
SiN
SiO2
But how many layers?
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Lecture 9 Slide 33
Number of Layers for 30 dB Suppression
10 periods, barely 20 dB 20 periods, 45 dB! Yikes!!
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 34
FDTD Simulation Results