Using FDTD for Other Types of Simulation In this chapter, we will discuss the use of the FDTD method for applications other than electro ma gnetics . The firstappl ic at ion is an acousti csimula ti on problem. The math emat ic s of ac ou st ic s is similarto el ec tromag ne tics [1] ,soit is nat uralthat the develo pm en ts of FDTD are used in aco usticsimulati on[2, 3]. The secondexampleis the si mulation of the Sc hr oedinger equation ,whichis at the heartofquantum me ch anic s. Quantum me chan ic s is radica llydi ff er en t from both aco ust ics and ele ctr oma gne tic s. However, the Schroedi ng er equatio nis still a wave equ at ionandit lends its elf to si mulation withFDTD.In both cases,onl y si mpleexamplesof one- dimens iona l simula tio ns are pre sen ted . 6.1 THE ACOUSTIC FDTD FORMULATION In the fo ll owin g de velopment, we willdeal only with press urewavesand ignoreelastic waves. We start with the fi rs t- order acou st ic eq ua ti on s, a " 8 t P (x , t) = V · u (6.1a) d PoP, dtu(x,t) = Vp(x,t), (6.1b) where pix, t) is the pressurefield [F 1m2] = [kgl(m - sec 2 ) ] , u(x, t ) is the vect orvelocityfield [mlsec] , Po is the ma ss densityof water [kg/m 3 ] , P, is the relative (t o water )massdensit y of the medi um, " isthe compressibilit y of the medium [(m - sec 2 ) / kg] . (Table 6.1) Not ice that waterhas been chosenas thebackgroundmedi uminst ead of air. The compress ibilityis 1 I K=--= , P · c 2 Po · Pr . c 2 wher e c is the velocityof sound. (6.2) 133
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8/7/2019 6_Using FDTD for Other Types of Simulation
A similar procedure in Eq. (6.lb) yields three equations of the type
un+1/2(i j. k + 1/2) = un- 1/2(i j. k + 1/2)
z " z" Ilt (6.4)
+ C ' k+1/2) .. Il· [pn(i , j ,k+1)-pn(i , j ,k)] .P, r, t , Po z
Obviously, equations in the X and Y direction would be similar. Wewill limit ourselvesto a simple one-dimensional problem in the Z direction and rewrite Eqs. (6.3) and (6.4) as
follows:
pn+l/2(k) = pn-l/2(k) + ga(k). [ u ~ ( k + 1/2,) - u ~ ( k - 1/2)] (6.5a)
u ~ + l ( k + 1/2) = u ~ ( k + 1/2) + gb(k + 1/2) . [pn+l/2(k + 1) - pn+l/2(k)] , (6.5b)
where we have the parameters
ga(k) = Il t · PoPr · c2
~ ~
gb(k + 1/2) = (k 1/2) ·P + ·Po· ~
(6.6a)
(6.6b)
Notice that we have chosen to write ga in terms of the speed of sound and the pressure
rather than the compressibility since these are the parameters that are most widely used.
As before, I:1t is chosen after ~ is chosen according to
~ ~ t < -- cmax '
where emu is the fastest velocity of sound that we encounter. We will suppose that this will
be metal, in which the velocity of sound is 5900 meters per second. Just to give ourselves a
margin, we will take
~ ~ t = - .1()4
Since Po = 1000 kg/m, Eqs. (6.6) become
ga(k) = 10-1 • p, (k) c2 (k)
10-7
gb(k + 1/2) = k /'p,( + 1 2)
(6.7a)
(6.7b)
For water, which we will use as our backgroundmedium, Eqs. (6.7a) and (6.Th) tum out to be
The program us l.c at the end of the chapter is a one-dimensional acoustic simulation
program. In its present form, it is only simulating water. Figure 6.2 shows the result. A pulse
is generated at one end and subtracted out at the other. So where did the PML come from?
Here is how this program was developed: I went back and made a copy of the program
fdld_l.l.c. Using the text editor, I replaced ex with p, and hy with u. Then I went to the
program we developed in Chapter 5 to simulate microstrip antennas, and I took the PML from
the lD incident buffer and copied it over. I added the parameters ga and gb, changed someconstants, and so forth, and there it was! The entire process took about 30 minutes. The point
is, an acoustic simulation program, complete with PML, was written by using the ideas we
had already developed for EM simulation.
8/7/2019 6_Using FDTD for Other Types of Simulation
We assume we are simulating an electron and therefore we will take m to be the mass of an
electron. Wewill take ~ to be one-tenth of an Angstrom, so
m = 9.1 x 10-31 [kg]
~ = 1 X 10-11 [m].
At this point, we don' t know what the time step, ~ t should be. Unfortunately, there is
no specific Courant condition to guide us, as was the case in EM simulation. We will startby assigning what seems to be a "reasonable" number, i.e., something that will likely keep
8/7/2019 6_Using FDTD for Other Types of Simulation
138 Chapter 6 • Using FDTD for Other Types of Simulation
Eq. (6.13)fromblowing up. Let's just take 1/8for now:
~ Ii I- -=- ,~ Z 2m 8
(6.14)
whichmeans
(6.16a)
(6.15a)
(6.16b)
(6.15b)
1 2m 2 1 9.1 x 10-
31
kg ( - 11 )2 -1 9~ = 8 T ~ Z = 4· 1.05 X 10-34 J . sec· 10 m =2.165 x 10 sec.
Nowthe constantin frontof thepotentialtermcanbecalculated:
/1t = 2.165 x 1O-19sec = 2 053 X 1015 -1 . (1.602 X 10-19J) = 3 85 -4 -1
Ii 1.055 X 10-34J .sec· J leV .2 x 10 eV ·
The twocoupledequations can then bewritten
1 / I ~ ( k ) = 1/I:e;/ (k)
- [Y,:;;:2(k + 1) - 2y,:;;:2(k) + y,:;;:2(k - 1)]
+ ~ V(k) . y , : : a ~ / 2 ( k )1 / J r : a ~ / 2 ( k ) = 1 / J ~ ~ / 2 ( k )
l[ n n n k ]+"8 1/Ireal(k + 1) - 21/1real(k) + 1/!real( - 1)
~ k n ·- h V( ) ·1/Ireal(l).
Tosimulate a particlemovingin free space,it is necessary to specifyboth the real and
imaginarypartsin thespatialdomain. Forexample, wewill initiatea particleat a wavelengthof A in a Gaussian envelope ofwidtha with the following twoequations:
y, (k) - - . 5 · ( ~ r (21r(k - ko»)real - e cos A
' " (k) - - . 5 · ( ~ r · (21r(k - ko»)real - e sin A '
whereko is thecenterof thepulse. Once theseequations arecalculated, the amplitudes mustbe normalized [5], to satisfy
i: y, * (x)·y,(x) dx = 1. (6.17)
6.2.2 CalCUlating the Expectation Values of the Observables
The kinetic energy in the simulation is calculated by
(T ) = - 2 .hmet {[/Freal(k) - i 1/Fimag (k)] • [ a 2 1 / F ~ ~ (k) + i a 2 1 / F ~ ~ (k) ] }.
Potential energy. The expected value of the potential energy is
(V ) = (1/FIVI1/F) =i: V (z) 11/F (z, t)12dz,
which is calculated by
N
(V) =L V (k) · [1/F?em(k) + 1 / F i ~ g ( k ) ] .i=1
where V (k) is the potential at that cell.
6.2.3 Simulation of an Electron Striking a Potential Barrier
139
(6.18)
(6.19)
Figure 6.3 shows a simulation of an electron moving in free space next to a region with
a potential of 100 eV. It is initiated at 146 eV, which is pure kinetic energy because there is
no potential. At 350 time steps, is has propagated to the right. The waveform has begun to
spread, but the total kinetic energy remains the same. After 1300 time steps, it has struck thepotential. Part of the waveformhas penetrated the potential barrier, and parthas been reflected.
Notice that the total energy has remained the same, but now part of it is in the form of potential
energy, because part of the waveform is at a potential of 100 eV. Does this mean the electron
has split into two parts? No. It means there is a certain probability that it has penetrated the
barrier and a certain probability that it has been reflected. These probabilities are calculated
by the following:
Probability of reflection =i: 1/F*(x)·1/F(x) dx (6.20a)
Probability of transmission =
1:1/F*(X)·1/F(X) dx, (6.20b)
where XPot. is the location where the potential begins. For this particular problem, Eq. (6.20a)
calculated 0.206 and Eq. (6.19b) calculated 0.794. Of course, the two must sum to one.
PROBLEM SET 6.2
1. The program se.c at the end of this section was used to do the simulation in Fig. 6.3. Get this
program running and duplicate the results.
2. Repeat the simulation of problem 6.2.1, but use a potential of -100 eV.
3. Repeat the simulation of problem 6.2.1 using a potential of 1000eVe
What happens? Does thismake sense?
8/7/2019 6_Using FDTD for Other Types of Simulation