eten05 electromagnetic wave propagation lecture 7: fdtd … · lecture 7: fdtd in 1d and 3d,...

ETEN05 Electromagnetic Wave PropagationLecture 7: FDTD in 1D and 3D, boundary

conditions

Daniel Sjoberg

Department of Electrical and Information Technology

September 22, 2009

Outline

1 Introduction

2 Staggered grids

3 Dispersion analysis in 3D

4 Geometry for scattering

5 Example applications

6 Conclusions

Daniel Sjoberg, Department of Electrical and Information Technology

Some questions of importance today

I How can we generalize the 1D-code to 3D?

I How should the grid be chosen?

I How can the computational region be truncated withoutcausing reflection?

I How much memory is required?


Outline

1 Introduction

2 Staggered grids




6 Conclusions


Graphical point of view


Generalization to 3D

E-field along edges, H-field along sides. Also staggered in time.Daniel Sjoberg, Department of Electrical and Information Technology

Fields on the boundary

From http://fdtd.wikispaces.com.Daniel Sjoberg, Department of Electrical and Information Technology

The staggered grid

I Invented by K. S. Yee in 1966.

I Boundary conditions involve only one field (usually E)

I Completely dominating in commercial codes

I Based on decoupling of electric and magnetic fields:∇×E = −∂B/∂t and ∇×H = ∂D/∂t: time differences ofE depend only on space differences of H and vice versa

I For bianisotropic materials, where D and B depend on bothE and H, the fields may need to be evaluated at the samepoints in space and time


Staircasing

The rectangular grid does not conform to curved surfaces:


Outline

1 Introduction

2 Staggered grids




6 Conclusions


Dispersion relation

This means derivatives can be replaced by algebraic expressions as

∂

∂t→ Dt =

2j∆t

sinω∆t

2∂

∂x→ Dx =

−2j∆x

sinkx∆x

2

and so on. Eliminating the magnetic field, Maxwell’s equationsreduce to (after quite a bit of algebra)(

D2y +D2

z −D2t /c

2 −DxDy −DxDz

−DxDy D2x +D2

z −D2t /c

2 −DyDz

−DxDz −DyDz D2x +D2

y −D2t /c

2

)(ex

ey

ez

)=(

000

)implying D2

t /c2 = D2

x +D2y +D2

z or

sin2 ω∆t/2(c∆t)2

=sin2 kx∆x/2

(∆x)2+

sin2 ky∆y/2(∆y)2

+sin2 kz∆z/2

(∆z)2


Static solutions

There is another solution for waves on the form(ex, ey, ez) = (Dx, Dy, Dz), corresponding to E = −∇ϕ, that is,static fields. This solution is

Dt = 0

corresponding to ω = 0. Thus, the staggered grid conserves staticsolutions. This means that fields that can be written as gradients(E = −∇ϕ) do not propagate through the grid.


Time step limit

In order to have a stable scheme, we need

sin2(ω∆t/2) ≤ 1

for all wave vectors k = kxx + kyy + kzz. The dispersion relationthen implies

c∆t ≤[

1(∆x)2

+1

(∆y)2+

1(∆z)2

]−1/2

For a cubic grid with ∆x = ∆y = ∆z = h,

∆t ≤ h

c√

3

Thus, the maximum time step is shorter in three dimensions thanin one.


Outline

1 Introduction

2 Staggered grids




6 Conclusions


Truncation of the computational region

In the FDTD method, we can only treat finite regions. So far wehave only used Dirichlet boundary condition, corresponding tometal walls. So how to truncate the following case?

incident field

radiating antenna

materials


Numerical scattering geometry

total fields

PEC surface

near−to−farfield transformation surface

PML volume

incident field surface


Near-to-farfield transformation

First, transform fields on the surrounding surface to frequencydomain using FFT. Then, apply the following transform:

Efar(k, r) =−jk4π

r ×∫S(M(k, r′) + η0r × J(k, r′))ejkr·r dS′

where ω = kc0, M = −n×E and J = n×H are the tangentialelectric and magnetic fields, and

E(k, r) =e−jkr

rEfar(k, r) +O(r−2), r →∞

H(k, r) =e−jkr

r

1η0

r ×Efar(k, r) +O(r−2), r →∞

The far field amplitude Efar has unit volts.


Absorbing boundary conditions

There have been many suggestions on how to absorb waves,particularly in one spatial dimension. The most successful is thePerfectly Matched Layers, PML (Berenger 1994). It is based onthe idea of introducing an artificial magnetic conductivity

∇×E = −(µ0∂H

∂t+ σ∗H

)∇×H =

(ε0∂E

∂t+ σE

)The time harmonic impedance is then

ηc =√µc

εc=

√µ0 + σ∗/jωε0 + σ/jω

=√µ0

ε0if

σ∗

µ0=σ

ε0


Perfectly matched layers

The reflection coefficient for a wave normally incident on such asemi-infinite medium is

Γ =η0 − ηc

η0 + ηc

Thus, since we can choose ηc = η0 by choosing σ and σ∗ properly,the front reflection can be made small, and the losses in thematerial make the wave attenuate so that the reflection againstthe final PEC surface is negligible. Typical design issues:

I How thick does the layer have to be?

I What should the conductivity profile σ(z) be?

For oblique incidence, more care needs to be taken since we cannotmatch the reflection perfectly for every angle, see the literature.


Handin 2

In the second handin, you are requested to investigate properties ofa PML in one dimension.

I Starting code is provided (matlab)

I You need to modify the code and make numerical experiments


Outline

1 Introduction

2 Staggered grids




6 Conclusions


Typical flow chart for an FDTD code

I PreprocessingI Load grid and geometryI Memory allocationI Declare constantsI Compute parameters

I Main loop (iterate the following steps)I Evaluate sourcesI Update internal H fieldsI Update internal E fieldsI Update boundary conditionsI Save some data, typically on the boundary

I PostprocessingI Fourier transform to frequency domainI Near-to-farfield transformationI Scattering coefficientsI . . .


Typical complexity

I Size of computational region in wavelengths

I At least 10–15 points per wavelength

I Small geometries may require high discretization

I Number of time steps (becomes large at resonances)


Wireless communication with a pacemaker

I Frequency?

f = 402–405 MHz

I Permittivity?

εr ≈ 64

I Conductivity?

σ ≈ 0.94 S/m

I Wavelength?

λ = c/f ≈ c0/(f√εr) =

3·108

400·106·8 m ≈ 0.09 m

I Geometry?

1 · 1 · 2 m3 = 2m3

I Sampling?

λ/15 ≈ 0.006 m

I Number of unknowns?

6 20.0063 = 55 · 106 Refined mesh for antenna

and organsThe number 6 stems from 3 components in E and 3 componentsin H.



I Frequency? f = 402–405 MHzI Permittivity?

εr ≈ 64

I Conductivity?

σ ≈ 0.94 S/m

I Wavelength?

λ = c/f ≈ c0/(f√εr) =

3·108

400·106·8 m ≈ 0.09 m

I Geometry?

1 · 1 · 2 m3 = 2m3

I Sampling?

λ/15 ≈ 0.006 m






I Frequency? f = 402–405 MHzI Permittivity? εr ≈ 64I Conductivity? σ ≈ 0.94 S/mI Wavelength?

λ = c/f ≈ c0/(f√εr) =

3·108

400·106·8 m ≈ 0.09 m

I Geometry?

1 · 1 · 2 m3 = 2m3

I Sampling?

λ/15 ≈ 0.006 m






I Frequency? f = 402–405 MHzI Permittivity? εr ≈ 64I Conductivity? σ ≈ 0.94 S/mI Wavelength?λ = c/f ≈ c0/(f

√εr) =

3·108

400·106·8 m ≈ 0.09 mI Geometry?

1 · 1 · 2 m3 = 2m3

I Sampling?

λ/15 ≈ 0.006 m







√εr) =

3·108

400·106·8 m ≈ 0.09 mI Geometry? 1 · 1 · 2 m3 = 2m3

I Sampling?

λ/15 ≈ 0.006 m







√εr) =

3·108

400·106·8 m ≈ 0.09 mI Geometry? 1 · 1 · 2 m3 = 2m3

I Sampling? λ/15 ≈ 0.006 mI Number of unknowns?

6 20.0063 = 55 · 106

Refined mesh for antennaand organs

The number 6 stems from 3 components in E and 3 componentsin H.




√εr) =

3·108

400·106·8 m ≈ 0.09 mI Geometry? 1 · 1 · 2 m3 = 2m3



and organs

The number 6 stems from 3 components in E and 3 componentsin H.




√εr) =

3·108

400·106·8 m ≈ 0.09 mI Geometry? 1 · 1 · 2 m3 = 2m3





Scattering of light against a blood cell

I Wavelength in vacuum?

λ0 ≈ 0.632µm

I Refractive index plasma?

np ≈ 1.34

I Refractive index in cell?

nc ≈ 1.41

I Minimum wavelength?

λ = λ0/nc ≈ 0.632µm1.41 ≈ 0.45µm

I Diameter?

7.76µm ≈ 17λ

I Height?

2.50µm ≈ 6λ

I Geometry?

82 · 3 (µm)3 = 192 (µm)3

I Sampling?

λ/15 ≈ 0.4515 µm ≈ 0.03µm


6 192(0.03)2

≈ 42 · 106 Refined mesh formembrane



I Wavelength in vacuum?λ0 ≈ 0.632µm

I Refractive index plasma?

np ≈ 1.34

I Refractive index in cell?

nc ≈ 1.41


λ = λ0/nc ≈ 0.632µm1.41 ≈ 0.45µm

I Diameter?

7.76µm ≈ 17λ

I Height?

2.50µm ≈ 6λ

I Geometry?

82 · 3 (µm)3 = 192 (µm)3

I Sampling?

λ/15 ≈ 0.4515 µm ≈ 0.03µm


6 192(0.03)2





I Refractive index plasma? np ≈ 1.34I Refractive index in cell?

nc ≈ 1.41


λ = λ0/nc ≈ 0.632µm1.41 ≈ 0.45µm

I Diameter?

7.76µm ≈ 17λ

I Height?

2.50µm ≈ 6λ

I Geometry?

82 · 3 (µm)3 = 192 (µm)3

I Sampling?

λ/15 ≈ 0.4515 µm ≈ 0.03µm


6 192(0.03)2





I Refractive index plasma? np ≈ 1.34I Refractive index in cell? nc ≈ 1.41I Minimum wavelength?

λ = λ0/nc ≈ 0.632µm1.41 ≈ 0.45µm

I Diameter?

7.76µm ≈ 17λ

I Height?

2.50µm ≈ 6λ

I Geometry?

82 · 3 (µm)3 = 192 (µm)3

I Sampling?

λ/15 ≈ 0.4515 µm ≈ 0.03µm


6 192(0.03)2





I Refractive index plasma? np ≈ 1.34I Refractive index in cell? nc ≈ 1.41I Minimum wavelength?λ = λ0/nc ≈ 0.632µm

1.41 ≈ 0.45µmI Diameter?

7.76µm ≈ 17λ

I Height?

2.50µm ≈ 6λ

I Geometry?

82 · 3 (µm)3 = 192 (µm)3

I Sampling?

λ/15 ≈ 0.4515 µm ≈ 0.03µm


6 192(0.03)2






1.41 ≈ 0.45µmI Diameter? 7.76µm ≈ 17λI Height?

2.50µm ≈ 6λ

I Geometry?

82 · 3 (µm)3 = 192 (µm)3

I Sampling?

λ/15 ≈ 0.4515 µm ≈ 0.03µm


6 192(0.03)2






1.41 ≈ 0.45µmI Diameter? 7.76µm ≈ 17λI Height? 2.50µm ≈ 6λI Geometry?

82 · 3 (µm)3 = 192 (µm)3

I Sampling?

λ/15 ≈ 0.4515 µm ≈ 0.03µm


6 192(0.03)2






1.41 ≈ 0.45µmI Diameter? 7.76µm ≈ 17λI Height? 2.50µm ≈ 6λI Geometry? 82 · 3 (µm)3 = 192 (µm)3

I Sampling?

λ/15 ≈ 0.4515 µm ≈ 0.03µm


6 192(0.03)2







I Sampling? λ/15 ≈ 0.4515 µm ≈ 0.03µm


6 192(0.03)2

≈ 42 · 106

Refined mesh formembrane






I Sampling? λ/15 ≈ 0.4515 µm ≈ 0.03µm

I Number of unknowns?6 192

(0.03)2≈ 42 · 106 Refined mesh for

membrane


Outline

1 Introduction

2 Staggered grids




6 Conclusions


Pros and cons of the FDTD method

+ Easy to implement

+ Can handle advanced material models

+ Easy analysis

+ Broad band

− Stair case approximation of geometry

− Numerical dispersion


eten05 electromagnetic wave propagation lecture 7: fdtd … · lecture 7: fdtd in 1d and 3d,...

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