6_Using FDTD for Other Types of Simulation

8/7/2019 6_Using FDTD for Other Types of Simulation

http://slidepdf.com/reader/full/6using-fdtd-for-other-types-of-simulation 1/14

Using FDTD for Other Types

of Simulation

In this chapter, we will discuss the use of the FDTD method for applications other than

electromagnetics. The firstapplication is anacousticsimulationproblem.Themathematics of

acoustics is similarto electromagnetics [1], so it is naturalthat thedevelopments ofFDTDare

used in acousticsimulation[2, 3]. The secondexampleis the simulation of the Schroedinger

equation,whichisattheheartofquantummechanics. Quantummechanics isradicallydifferent

fromboth acoustics and electromagnetics. However, theSchroedingerequationis still a wave

equationand it lendsitself to simulation withFDTD.In both cases,only simpleexamplesofone-dimensional simulations are presented.

6.1 THEACOUSTIC FDTDFORMULATION

In thefollowing development, wewilldealonlywithpressurewavesandignoreelasticwaves.Westartwith the first-order acoustic equations,

a" 8t P (x , t) = V· u (6.1a)

d

PoP, dtu(x,t) =Vp(x,t), (6.1b)

where pix, t) is the pressurefield [F1m2]= [kgl(m - sec2) ] ,

u(x, t) is the vectorvelocityfield [mlsec] ,

Po is the massdensityof water [kg/m3] ,

P, is the relative(to water)massdensityof themedium,

" is the compressibility of the medium [(m - sec2) / kg] . (Table 6.1)

Notice that waterhas been chosen as thebackgroundmediuminsteadof air. The compress

ibilityis

1 IK=--= ,P · c2 Po · Pr . c2

wherec is the velocityof sound.

(6.2)

133



134 Chapter 6 • Using FDTD for OtherTypes of Simulation

TABLE6.1 Acoustic Properties of a Few Materials

Material

1. Water

2. Air

3. Metal

Speed ofSound (m/sec)

1500

343

5900

1000

1.21

7800

Compressibility (m • s2/kg)

4.4 X 10-8

7.02 X 10-6

Eq. (6.1a) can be written

dp(x, y, z. t) 1 [dUx(X, y, z, t) duy(x, y, z, t) duz(x, y, Z, t ) ]- - - -= + + .

dt K(X,y,Z) dx dy dz

We will use a differencing scheme similar to Yee's FDTD for electromagnetics [4]; however,

we will assume that the pressure locations are at nodes in a three-dimensional lattice, and that

the velocities are located between the nodes, as shown in Fig. 6.1.

z

/'

//

/'

/

III

III

(i.j.k + 1).... _----- ....

uz(i,j,k + 1/2)

uy(i,j + 1/2,k)

//'

//

/'

y

Figure 6.1 The acoustic Yeecell.

Using first-order differencing in time and space, Eq. (6.1.a) can be written

pn+I/2(i, i, k) - pn-I/2(i, i, k) = 1 [ U ~ ( i + 1/2, j , k) - u ~ ( i - 1/2, j , k)]6.t K(i, j , k) Sx

1 [U;(i, j, k + 1/2) - u;(i, j, k - 1/2)]+ «(i, j, k) !!.y

1 [ U ~ ( i , i, k + 1/2) - u ~ ( i , j, k - 1/2)] .

+ «(i, j, k) !!.Z

First, the compressibility will be replaced by Eq. (6.2). Then L).t will be moved to the right

side, giving a difference equation suitable to the FDTD formulation:

n+l/2(. . k) - n-l/2(. . k) + 6.t· POPrc2

• [ n(. + 1/2 . k) _ n(" - 1/2 "k)]p I,J, -p I ,J, 6.x Ux I .r. ux ' ,J ,

~ · POPrc2

[ n ( " . 1/2 k) n(" . 1/2 k)]+ " u y I, I + , - uy I, l - ,dy

d t . POPrc2

[n( . i. k 1/2) n ( . . k 1/2)]+ · Uz I, j , + - Uz I, j , - ·dz

(6.3)



Section 6.1 • The Acoustic FDTD Formulation 135

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

ga(k) = 10-1 . P, (k) c; (k) = 10-1. 1 . (1, 500)2 = 2.25 X 105 (6.8a)

gb(k + 1/2) = 10-7. (6.8b)

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.



136 Chapter 6 • Using FDTD for Other Types of Simulation

OL--_--L__ __ ___J .__ ___J""___L...___ L____ .L _ .___ . .L._.__ ..L__ ___J

10 20 30 40 50 60 70 80 90 100

O t - - - ~ - - ~ - - ~ - - ~ - - ~ - - - -10 20 30 40 50 60 70 80 90 100

10000000

Cells

T= 660

40000

o . - . - ~ - - ~ - - ~ - - ~ - - ~ ~ - - ~ - - ~ - - . . . . . . - - " "

Figure6.2 Simulation ofan acoustic pulse. The pulse is generated atcell IS and propagates

outward in both directions. After 200 time steps, the part that propagated to

the left is being absorbed by the PML. Similarly, after 660 time steps, the part

that propagated to the right is being absorbed.

PROBLEM SET 6.1

1. Develop your own program usl. c using the procedure described and get it running.

2. Put a metal sheet of about 10 cells in the center of the problem space and see what happens.

Then put in an air pocket of 10 cells and see what happens.

3. Repeat the above problem, but for a sinusoid waveform of 2.5 MHz in a Gaussian envelope.

In determining the cell size remember: 10points per wavelength of the shortest wavelength in

the problem. Since metal has the highest speed, it will have the shortest wavelength.

6.2 SIMULATION OF THE SCHROEDINGER EQUATION

In this section a simple formulation of the Schroedinger equation in FDTD is described. The

Schroedinger equation is the basis of quantum mechanics. If you do not know any quantum

mechanics, this section may prove to be a little puzzling. (An excellent text for beginning

quantum mechanics is the one by Griffiths [5].) Problems involving quantum mechanics,

unlike the previous section where it wasn't difficult to pick up the basics of acoustics, are

quite different. For instance, the parameter used in our wave equations is 1/I(r, t) , which, like

the electric field or the pressure, is a function of space and time. But unlike the electric field

or pressure, it is a complex number, even though it is a time domain parameter. It is also

difficult to attribute any distinct physical meaning to 1/1. It is the product of 1/1 with its complex

conjugate, 1/J(r, t)*1/J(r, t), that indicates the probability of a particle being at the point r attime t.



Section 6.2 • Simulation of the Schroedinger Equation

6.2.1 Formulating the Schroedinger Equation into FDTD

The time-dependent Schroedinger equation is given by [5-8]

01/l(r, t) n2in = --V 2

1/1 (r , t) + VCr, t) ·1/I(r, t)at 2m

ora1/l(r, t) n i

=i -\721/1(r, t) - t;"V (r, t) ·1/I(r, t) ,at 2m n

137

(6.9)

(6.10)

where m is the mass of the particle [kg],

Ii = 1.054 X 10-34 [J - sec] is Plank's constant,

V is the potential [J]. (It will often bemore convenient to express

the potential in electron volts (eV), where 1 eV = 1.602 x 10-19J.)

Toavoid using complex numbers, we will split the variable 1/I(r, t) into its real and imaginary

components

1/1(r, t) = 1/1real(r, t) + i1/Iimag (r, t) . (6.11)

Inserting Eq. (6.11) into Eq. (6.10) and separating real and imaginary parts results in the

following coupled set of equations:

al/Jreal (r, t) n 2 1at = - 2m . V 1/Iimag(r, t) + hV 1frimag(r, t) (6.12a)

a1/limag(r, t) Ii 2 1at = 2m · V 1frreal(r, t) - hV 1/Ireal(r, t) . (6.12b)

We begin by assuming a one-dimensional space, arbitrarily taking the z direction. Starting

with Eq. (6.12a), the finite difference approximations in space and time result in

1/I:zeal (k) - 1 / I ~ ~ 1 (k)~ -a,,,n-I/2(k 1) 2-a,,,n-I/2(k) -a,,,n-l/2(k 1

Ii o/imag + - o/imag + o/imag -) 1 n-I/2= - 2m (.6.z)2 + h, V(k) '1/!imag (k),

from which we get

1/J:'eal (k) = 1 / J ~ - ; / (k)

_~

!!:-. [-a'I'!,-1/2(k+ 1) _ 2-a'I'.n-l/2(k) + -a'I'!,-1/2(k - 1)]~ Z 2m o/tmag o/tmag o/tmag (6.13)

~ n-I/2+r;V(k)·1/I imag (k).

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




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

Twoquantities ofimportance inquantummechanics aretheexpectedvaluesofthekinetic

energyand thepotentialenergy. Theyare calculated from"" (x, t) as follows:

Kinetic energy. Theexpected valueof thekineticenergyis givenby

( Ili2

(2

1) li2

10

( 82

1/1 )T) = 1/1 --- 1/1 =-- 1/1*- dz.2m 8z2 2m - 0 0 8z2



Problems

The Laplacian operator ::2 is approximated by

a2

1/Jreal2(k) [1/Freal (k - 1) - 21/Freal (k) + 1/Freal(k + 1)] / l:1z2

az

a21/Jimag(k) [k 'III' k 'III' k 1)] / 2--a-z-

2- = 1/Jimag( - 1) - 2tpimag( ) + tpimag( + L\z .

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?




4005000

PE=OeV

250

r - · _ · _ · _ · _ · _ _ · _ · _ · _ ·_ ·· V= lOOeV

I

20050

KE= 146eV

100

T= l

50

o. . . - - - - - - . .1-+,

0.2

-0.2

400500050

r - · _ · _ · _ · _ · - _ · _ · _ · _ · _ ·· V= lOOeV,

200SO00

T=350

50

0.. . ---------"

0.2

-0.2

r-·_·-·_·_·--·_·_·_·-·· V= lOOeV

I

T= 1300

ot------v:\. .1iJ-\

0.2

-0.2 KE=67 eV PE=7geV

50 100 150 200 250 300 350 400

Figure 6.3 Simulationof an electronmoving in free space and then hitting a potential.

While the electron is in free space, all the energy is kinetic. After it strikes

the potential, part of the energy has been convertedto potential energy. The

waveformat T =1300indicatesthatthere is someprobabilitythat theelectronwasreflectedand someprobabilitythatit penetratedthepotentialbarrier.

REFERENCES

[1] L. Beranek, Acoustics, New York: McGraw-Hill, 1954.

[2] J. J. Bowman, T. B. A. Senior, and P. L. E. Uslenghi, Electromagnetic andAcoustical Scattering by

Simple Shapes. New York: Hemisphere, 1987.

[3] J. G. Maloney and K. E. Cummings, Adaptation of FDTD techniques to acoustic modeling, 11th

AnnualReview ofProgress inAppliedComputational Electromagnetics,Monterey, CA, vol. 2,March

1995, pp. 724-731.[4] K. S. Vee, Numerical solution of initial boundary value problems involving Maxwell's equations in

isotropic media, IEEE Trans. Antennas and Propagat., vol. 17, 1966, pp. 585-589.

[5] D. J. Griffiths, Introduction to Quantum Mechanics. Englewood Cliffs, NJ: Prentice-Hall, 1995.

[6] C. Cohen-Tannoudji, B. Diu, and F. Laloe, QuantumMechanics. New York: Wiley, 1977.

[7] S. Brandt and H. D. Dahmen, The Picture Book ofQuantum Mechanics. Berlin: Springer Verlag,

1995.

[8] S. Borowitz, Fundamentals ofQuantum Mechanics. New York: W. A. Benjamin, 1967.



C Programs 141

/* us l . c . 1D FDTD Acoustic s imu la ti on * /

# include <math.h>

# include <stdlib.h>

# i nc lude <stdio. h>

#define KE 100

main ()

{

f loa t p[KE] ,u[KE];

in t n,k,kc,ke,NSTEPS,n-pml;

f loa t ddx,dt,T;

f loat to,spread,pulse,xn,XXD,npml;

FILE *fp, *fopen(};

f loa t g2[KE] ,g3[KE] ,f2[KE] ,f3[KE];

/* In i t ia l ize */

for ( k=l; k < KE; k++ )

{ P [k ] = 0 . ;

u[k] = 0 .;

g2 [k ] 1 . i

g3 [k] 1 . ;

f2 [k ] 1 . i

f3 [k ] 1 . ;

}

/* PML */

npml = 10;

nJ>ml = npml;

fore k=O; k < n-pml; k++ ) {

xxn = (npml-k)/npml;

xn = .33*pow{xxn,3.) i

g2[k] = l . / ( l .+xn) ;

g2[KE-k-l] = 1./{1.+xn);

g3[k] = ( l . -xn)/{l .+xn);

g3[KE-k-l] = (1.-xn)/(1.+xn) ixxn = (npml-k-.S)/npmli

xn = .33*pow(xxn,3.);

f2[k] = 1. / ( l .+xn) i

f2[KE-k-2l = 1./(1.+xn);

f3[k] = ( l . -xn) /( l .+xn) ;

f3[KE-k-2] = ( l . -xn) /( l .+xn) ;

}

pr in t f ( Iff & 9 \n lf) i

for ( k=Oi k < KE; k++ ) {

pr in t f ( "%2d %6.4f %6.4f \n",k,g2[kl ,g3[k]) i

pr in t f ( If %6.4f %6.4f \n lf,f2[k],f3[k])i}



142

kc = KE/2;

to = 150.0;

spread 50;

T = 0;

NSTEPS 1;

while ( NSTEPS > 0 ) {

pr in t f ( "NSTEPS - -> I I ) ;

scanf("%d", &NSTEPS);

printf("%d \n", NSTEPS);

n= 0;

for ( n=l; n <=NSTEPS n++)

{

T = T + 1;

/* Main FDTD Loop */

Chapter 6 • Using FDTD for Other Types of Simulation

/* Calcula te t he Pressure */

for ( k=l; k < KE; k++ )

{ P [k] 93 [k] *p [k] + 92 [k] *. (2 . 2SeS) * ( u [k-1] - u [k] )

/* Put a Gaussian pulse in the middle */

pulse = exp(-.S*(pow( (to-T)/spread,2.0) » ;

p[n-pml+S] = pulse;

pr in t f ( "%5.1f %6.2f\n",tO-T,p[nyml+S]);

/* Calcula te th e veloci ty */

for ( k=O; k < KE-l; k++ )

{ u [k] = f3 [k] *u [k] + f2 [k] * (6 . 67e- 8) * ( p [k] - P [k+l] )

/* End of the Main FDTD Loop */

/* At the end of the calculat ion, pr in t outthe pressure and veloci ty f ie ld s */

for ( k=1; k <= KE; k++ )

{ pr in t f ( "%3d %6.2f %6.2f\n",k,p[k],u[k]);

/* Write the P f ield out to a f i le "Pres" */

fp = fopen ( "Pres II , "w") ;

for ( k=l; k <= KE; k++ )

{ fpr in t f ( fp," %6.2f \n" ,p[k]) i

fclose(fp) ;

/* Write the U f ield out to a f i le "Vel" */

fp = fopen( "Vel II , "w") ;

for ( k=l; k <= KE; k++ )



C Programs 143

{ fpr intf ( fp," %6.2f \n" ,u[k]) ; }

fclose (fp) ;

pr in t f ( liT = %5. Of\n" ,T) ;




/* se .c . 10 FOTO Schroedinger simulation */

# include <math.h>

# include <stdlib.h>

# i nc lude <stdio.h>

#define KE 400

main ()

{

f loa t psi_rl[KE] ,psi_im[KE] i

f loa t Vpot,vp[KE] i

in t n,k,kc,ke,kstart ,kcenter,NSTEPS,n-pml;

f loa t pi,melec,hbar;

f loa t lambda,sigma,ptot;

f loat ddx,dt , ra ,Ti

f loa t pulse,xn,xxn,npml;

FILE *fp, *fopen{);

f loa t lap_rl,lap_im,ke_rl,ke_im,kine,PE;

/* In i t ia l ize */

for ( k=Oi k < KE; k++ )

{ psi_rl[k] = 0 .;

psi_im[kl = 0 .;

vp [k] = 0 . ;

}

pi = 3.14159;

melec = 9.2e-31; /* Mass of an electron */

hbar = 1.0S5e-34; /* P lank's constant * /

ddx = .1e-lO; /* Cell s ize */

ra = 1 . /8 . ;

pr in t f ( "ra %e \n" , ra) ;

dt = .25* (melec/hbar)*pow{ddx,2.) ;

pr in t f ( " dt = %e \n" ,dt) i

printf{ " ks ta rt,Vpot ( in eV) --> I I ) ;

scanf{lI%d ",&kstart ) ;

scanf{"%f", &Vpot)i

for { k=kstart i k < KE; k++

{ vp[k] = Vpot*1.602e-19 ; /* Convert eV to Joules */

/* Write the Potent ia l to a f i le "Vpot" */

fp = fopen ( "Vpot", "w") ;

for ( k=O; k < KE; k++ )

{ fpr in t f ( fp," %6.2f \n",vp[kl*6.2415e18); /* Write in eV*/pr in t f (" %6.2f \n" , (d t /hbar)*vp[k]) ; }

fclose(fp);



C Programs

/* In i t ia l ize the pulse */

pr in t f ( "lambda,sigma - -> ");

scanf("%f ",&lambda ) ;

scanf ( "%f" , &sigma);

kcenter = ksta r t /2 ;

pr in t f ( "kcenter = %3d \n lt, kcenter) ;

ptot = 0 .;

for ( k=l; k < ks tar t ; k++ )

{ psi_r l[kl = cos (2*pi* (k-kcenter)/lambda)

* exp( -.S*pow( (k-kcenter)/sigrna,2.) );

psi_im[kl = sin(2*pi*(k-kcenter) / lambda)

* exp( -.S*pow( (k-kcenter)/sigma,2.) );

ptot = ptot + pow(psi_rl[kl ,2.) + pow (psi_im[k] ,2 . ) ;

pr in t f ( " ptot= %f \n II , ptot) ;

/* Normalize the waveform */

for ( k=l; k < ks tar t ; k++ )

{ ps i_ rl [k l p si_ rl[k l/sq rt (p to t) ;

psi_im[kl = psi_im[kl /sqrt (ptot ) ;

kc = KE/2;

T = 0;

NSTEPS = 1;

while ( NSTEPS > 0 ) {

pr in t f ( "NSTEPS - -> I I ) ;

scanf ( II %d", &NSTEPS);

pr in t f ("%d \n", NSTEPS);

n= 0;

for ( n=l; n <=NSTEPS n++)

{

T = T + 1;

/* Main FDTD Loop */

/* Calc ula te th e Real part */

for { k=l; k < KE-l; k++

{ psi_rl[kJ = psi_rl[kJ

- ra*( psi_im[k+l] - 2.*psi_im[k] + psi_im[k-l] )

+ (dt/hbar)*vp[k]*psi_im[k] ; }

/* Calc ula te t he imaginary par t */

for ( k=l; k < KE-l; k++ )

{ psi_im[kJ = psi_im[kJ

+ ra*{ psi_rl[k+l] - 2.*psi_rl[k] + psi_r l [k- l ] )

145




- (dt/hbar)*vp[k] *psi_r l [k] ; }

/* End of the Main FDTD Loop */

/* Print out the rea land

imaginary par ts */for ( k=O; k < KE; k++ )

{ printf{ "%3d %6.2£ %6.2f\n",k,psi_rl[k] ,psi_im[k]);

/* Write the rea l pa rt to a f i le "prl" */

fp = fopen{ "pr l" , "w");

for ( k=O; k < KE; k++ )

{ fpr int f{ fp," %6.2f \n" ,ps i_r l [k] ) ;

fclose(fp) ;

/* Write the imaginary par t to a f i le "pim" */

fp = fopen ( "pim", "w") ;

for ( k=O; k < KE; k++ )

{ fprint£{ fp," %6.2f \n",psi_im[k]);

fclose{fp);

/* Calcula te th e KE and PE */

ke_rl = 0 .;

ke_im = O .i

PE = 0 .;

for { k=O; k < KE; k++ {

lap_rl = psi_r l[k+ll - 2.*psi_rl[k] + psi_r l[k- l ] ; /* Laplacian */

lap_im = psi_im[k+ll - 2.*psi_im[kl + psi_im[k-ll ;

ke_rl = ke_rl + psi_rl [k]*lap_rl + psi_im[kl*lap_im;

ke_im = ke_im + psi_rl[kl*lap_im - psi_im[kl*lap_rl;

PE = PE + vp[kl * {pow(psi_r l [kl ,2.) + pow{psi_im[k] , 2 . » ;

kine = .S*(hbar/melec)*(hbar/pow(ddx,2.»

*sqrt(pow(ke_rl ,2.) + pow(ke_im,2.»;

print f{ "k e %6.3f eV \n",6.23elS*kine); /* convert J to eV */print f{ " pe = %6.3f eV \n",6.23elS*PE);

pr in t f ( "T = %5.Of\n", T) ;

Date post:	08-Apr-2018
Category:	Documents
Upload:	bennett09
View:	230 times
Download:	0 times

6_Using FDTD for Other Types of Simulation

Documents