electromagnetic wave propagation lecture 5: finite differences in … · 2010. 9. 13. · the nite...

Electromagnetic Wave PropagationLecture 5: Finite differences in the time

domain

Daniel Sjoberg

Department of Electrical and Information Technology

September 14, 2010

Outline

1 Finite differences

2 Generalization to 3D

3 Dispersion analysis in 3D

4 Example applications

5 Conclusions

Daniel Sjoberg, Department of Electrical and Information Technology

Numerical methods

Some problems are too complicated to do by hand, such as apacemaker in the human body. Numerical methods help.


Finite difference approximation

To make numerical simulations, we approximate the derivativeswith finite differences (where the field is evaluated in grid pointsE|nr = E(r∆z, n∆t):

∂E

∂z

∣∣∣∣nr

=E|nr+1/2 − E|

nr−1/2

∆z

∂2E

∂z2

∣∣∣∣nr

=

∂E∂z

∣∣nr+1/2

− ∂E∂z

∣∣nr−1/2

∆z=

E|nr+1−E|nr∆z − E|nr−E|nr−1

∆z

∆z

Thus, the wave equation ∂2E∂z2− 1

c2∂2E∂t2

= 0 becomes

E|nr+1 − 2E|nr + E|nr−1(∆z)2

− 1

c2E|n+1

r − 2E|nr + E|n−1r

(∆t)2= 0

We expect the error in the approximation to decrease as ∆z and∆t become small.


Time stepping

The finite difference approximation of the wave equation is

E|nr+1 − 2E|nr + E|nr−1(∆z)2

− 1

c2E|n+1

r − 2E|nr + E|n−1r

(∆t)2= 0

By solving for the field at time step n+ 1, we find

E|n+1r = 2E|nr − E|n−1r +

(c∆t

∆z

)2

(E|nr+1 − 2E|nr + E|nr−1)

Thus, the solution at time (n+ 1)∆t can be found if the solutionat two previous time steps is known.


Leap frog stencil

-

6

0

0

z

t

∆z

∆t

r∆z

n∆t

b b b b b b bb b b b b b bb b b b b b bb b b b b b bb b b b b b b

jj

j jj

Initial conditions: E|0r = fr and E|1r = gr (known functions).Boundary conditions: E|n0 = 0 and E|nR = 0 (metal walls).


Leap frog stencil

-

6

0

0

z

t

∆z

∆t

r∆z

n∆t


j

jj j

j



Leap frog stencil

-

6

0

0

z

t

∆z

∆t

r∆z

n∆t


jj

j jj



Leap frog stencil

-

6

0

0

z

t

∆z

∆t

r∆z

n∆t


jj

j j

j



Leap frog stencil

-

6

0

0

z

t

∆z

∆t

r∆z

n∆t


jj

j jj



Dispersion relation

A harmonic wave propagating through the lattice can be written

ejωn∆t−jkr∆z

For a “real” wave we should have ω = ck . Inserting theexponential function into the difference approximation we obtain

ejωn∆t−jkr∆z(e−jk∆z − 2 + ejk∆z

(∆z)2− 1

c2ejω∆t − 2 + e−jω∆t

(∆t)2

)= 0

which can be rewritten as(sin

ω∆t

2

)2

=

(c∆t

∆z

)2(sin

k∆z

2

)2

This relates the spatial frequency k to the temporal frequency ωon the numerical grid, and is called the dispersion relation.


Time step and stability: R = c∆t/∆z

R = 1: Magic time step, ω = ±ck, no dispersion.

R < 1: Stable solution, ω 6= ±ck, different frequencies travel withdifferent speed (lower than c).

R > 1: Unstable solution, growing exponentially.


Example: square wave

R = 1

R < 1

R > 1


Example: smooth wave

Upper graph: 12 points across 1/e width.Lower graph: 6 points across 1/e width.


Computation of resonance frequencies 1

Even though FDTD solves problems in the time domain, it can beused to compute resonance frequencies.In a rectangular box with metal walls (cavity), only waves withparticular resonance frequencies can exist.

Lx

Lz

Ly

fmnp =c

2

[(m

Lx

)2

+

(n

Ly

)2

+

(p

Lz

)2]1/2


Computation of resonance frequencies 2

To compute the resonance frequencies in a cavity using FDTD, dothe following:

I Discretize the cavity with finite differences.

I Set up PEC boundary conditions.

I Give random data as initialization in order to excite allfrequencies.

I Run simulation for a long time.

I Fourier transform the field, sampled at a point.

I The resulting spikes correspond to the resonances of thecavity.

Explicit matlab code in Bondeson, Rylander, Ingelstrom.


Outline





5 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

Integral interpretation

∮CE·dr =

∫S∇×E·ndS = −

∫Sµ0∂H

∂t·ndS

= −∆x∆yµ0∂

∂tHz

∮CE · dr = Ex|p+1/2,q∆x+ Ey|p+1,q+1/2∆y

− Ex|p+1/2,q+1∆x− Ey|p,q+1/2∆y

Hz can be updated from knowledge of surrounding Exy-fields.Similarly for ∇×H = ∂D/∂t.


Integral interpretation

∮CE·dr =

∫S∇×E·ndS = −

∫Sµ0∂H

∂t·ndS = −∆x∆yµ0

∂

∂tHz

∮CE · dr = Ex|p+1/2,q∆x+ Ey|p+1,q+1/2∆y

− Ex|p+1/2,q+1∆x− Ey|p,q+1/2∆y

Hz can be updated from knowledge of surrounding Exy-fields.Similarly for ∇×H = ∂D/∂t.


Fields on the boundary

From http://fdtd.wikispaces.com. For PEC boundary, n×E = 0and n ·H = 0.


The staggered grid

I Invented by K. S. Yee in 1966.

I Boundary conditions involve only half of the field components(usually n×E and n ·H)

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:


Grid refinement

Simple method: refinealong planes where thegeometry has small scale

Advanced method: implicit time step-ping in regions with unstructured grid(Rylander & Bondeson 2000)


Outline





5 Conclusions


Dispersion relation

This means derivatives can be replaced by algebraic expressions as

∂

∂t→ Dt =

2j

∆tsin

ω∆t

2∂

∂x→ Dx =

−2j∆x

sinkx∆x

2

and so on. Eliminating the magnetic field, Maxwell’s equationsreduce to the wave equation ∇2E − 1

c2∂2

∂t2E, or(

D2x +D2

y +D2z −

1

c2D2t

)E = 0

Using the expressions for Dx, Dy, Dz, and Dt, we find

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


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

]−1/2For a cubic grid with ∆x = ∆y = ∆z = h,

∆t ≤ h

c√3≈ 0.577

h

c

Thus, the maximum time step is almost half as short in threedimensions than in one.


Discretization and error

From the dispersion relation

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

we can compute ω(k) and hence the phase and group velocities byTaylor series (allowing 1% error):

vp =ω

k= c

(1− (kh)2

36+ O

((kh)4

))10.5 cells per λ

vg =∂ω

∂k= c

(1− (kh)2

12+ O

((kh)4

))18 cells per λ

The total phase error in domain of diameter L is

ephase = (knum−k)L =

(ω

c(1− (kh)2/36 + · · · )− ω

c

)L ≈ k3h2L

36

meaning the cell size must scale as ω−3/2 to control the totalphase error.


Outline





5 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 = 402–405MHz

I Permittivity?

εr ≈ 64

I Conductivity?

σ ≈ 0.94 S/m

I Wavelength?

λ = c/f ≈ c0/(f√εr) =

3·108400·106·8 m ≈ 0.09m

I Geometry?

1 · 1 · 2m3 = 2m3

I Sampling?

λ/15 ≈ 0.006m

I Number of unknowns?

6 20.0063

= 55 · 106 Refined mesh for antennaand organs

The number 6 stems from 3 components in E and 3 components inH. With double precision floats, this means 55 · 106 · 8 = 0.4GB.



I Frequency? f = 402–405MHz

I Permittivity?

εr ≈ 64

I Conductivity?

σ ≈ 0.94 S/m

I Wavelength?

λ = c/f ≈ c0/(f√εr) =

3·108400·106·8 m ≈ 0.09m

I Geometry?

1 · 1 · 2m3 = 2m3

I Sampling?

λ/15 ≈ 0.006m


6 20.0063






I Permittivity? εr ≈ 64

I Conductivity? σ ≈ 0.94 S/m

I Wavelength?

λ = c/f ≈ c0/(f√εr) =

3·108400·106·8 m ≈ 0.09m

I Geometry?

1 · 1 · 2m3 = 2m3

I Sampling?

λ/15 ≈ 0.006m


6 20.0063








I Wavelength?λ = c/f ≈ c0/(f

√εr) =

3·108400·106·8 m ≈ 0.09m

I Geometry?

1 · 1 · 2m3 = 2m3

I Sampling?

λ/15 ≈ 0.006m


6 20.0063









√εr) =

3·108400·106·8 m ≈ 0.09m

I Geometry? 1 · 1 · 2m3 = 2m3

I Sampling?

λ/15 ≈ 0.006m


6 20.0063









√εr) =

3·108400·106·8 m ≈ 0.09m

I Geometry? 1 · 1 · 2m3 = 2m3

I Sampling? λ/15 ≈ 0.006m


6 20.0063

= 55 · 106

Refined mesh for antennaand organs








√εr) =

3·108400·106·8 m ≈ 0.09m

I Geometry? 1 · 1 · 2m3 = 2m3

I Sampling? λ/15 ≈ 0.006m

I Number of unknowns?6 20.0063




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


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 ≈ 1.34


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 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 Minimum wavelength?λ = λ0/nc ≈ 0.632µm

1.41 ≈ 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








1.41 ≈ 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








1.41 ≈ 0.45µm


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 ≈ 0.45µm



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 ≈ 0.45µm



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







1.41 ≈ 0.45µm



I Geometry? 82 · 3 (µm)3 = 192 (µm)3

I Sampling? λ/15 ≈ 0.4515 µm ≈ 0.03µm

I Number of unknowns?6 192(0.03)2



What about larger problems?

Sometimes the problem is too large to fit in one computer. Howdoes FDTD perform on clusters?

Platon cluster in Lund,216 × 2 × 4 = 1728 pro-cessors, 64 bit 2.26GHz3GB per core, startednov 2009.

I Each cell “communicates” only with its nearest neighbors; ablock of N3 cells only exchanges N2 data with others.

I Well suited for parallelization!I Current trend: computation on GPU:s (hardware adapted to

parallel operations).


Outline





5 Conclusions


Conclusions

I FDTD is a simple numerical scheme for time domain.

I It is vital to choose a smaller discretization in time than infrequency. The relation between time and space determinesthe numerical dispersion.

Some pros and cons:

+ Easy to implement

+ Can handle advanced material models (next lecture!)

+ Easy analysis

+ Broad band

+ Parallelizable

− Stair case approximation of geometry

− Global time step (but local implicit time stepping is possible)

− Numerical dispersion


electromagnetic wave propagation lecture 5: finite differences in … · 2010. 9. 13. · the nite...

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