%C*****************************************************************
%C MoM
%C*****************************************************************
%C This is a MATLAB based program using
%C
%C I. POCKLINGTON’S INTEGRAL EQUATION (8-24)
%C II. HALLEN’S INTEGRAL EQUATION (8-27)
%C
%C That computes:
%C
%C A. CURRENT DISTRIBUTION
%C B. INPUT IMPEDANCE
%C C. NORMALIZED AMPLITUDE RADIATION PATTERN
%C
%C Of a linear symmetrically-excited dipole.
%C
%C THIS PROGRAM USES PULSE EXPANSION FOR THE ELECTRIC CURRENT MODE
%C AND POINT-MATCHING FOR THE ELECTRIC FIELD AT THE CENTER OF EACH
%C WIRE SEGMENT.
%C
%C DELTA-GAP FEED MODEL IS USED IN BOTH FORMULATIONS. IN ADDITION,
%C MAGNETIC-FRILL GENERATOR IS AVAILABLE IN THE POCKLINGTON’S
%C INTEGRAL EQUATION.
%C
%C OPTION I. POCKLINGTON’S INTEGRAL EQUATION
%C OPTION II. HALLEN’S INTEGRAL EQUATION
%C
%C
%C ** INPUT DATA:
%C
%C TL = TOTAL DIPOLE LENGTH (IN WAVELENGTHS)
%C RA = RADIUS OF THE WIRE (IN WAVELENGTHS)
%C NM = TOTAL NUMBER OF SUBSECTIONS (MUST BE AN ODD INTEGER)
%C IEX = OPTION TO USE EITHER MAGNETIC-FRILL GENERATOR OR DELTA GAP
%C
%C IEX = 1 : MAGNETIC-FRILL GENERATOR
%C IEX = 2 : DELTA-GAP FEED
%C
%C ** NOTE: IGNORE INPUT PARAMETER IEX WHEN CHOOSING OPTION II
%C (i.e., HALLEN’S FORMULATION)
%C *****************************************************************
clear;
close;
%*********************************************************
%*** OUTPUT DEVICE OPTION # : 1 —> SCREEN
% 2 —> OUTPUT FILE
%*********************************************************
disp(‘OUTPUT DEVICE OPTION FOR THE OUTPUT PARAMETERS’);
disp(’ OPTION (1):SCREEN’);
disp(’ OPTION (2):OUTPUT FILE’);
option_a=0; option_b=0; filename=0;
while((option_a=1)&(option_a=2))
option_a=input(‘OUTPUT DEVICE =’);
if(option_a==2)
filename=input(‘INPUT THE DESIRED OUTPUT FILENAME = ‘,‘s’);
end
end
%*********************************************************
%*** READING OPTION # : 1 —> POCKLINGTON’S EQUATIONS
% 2 —> HALLEN’S EQUATION
%*********************************************************
disp(‘CHOICE OF POCKLINGTON’‘S OR HALLEN’‘S EQN.’);
disp(’ OPTION (1):POCKLINGTON’‘S EQN.’);
disp(’ OPTION (2):HALLEN’‘S EQN.’);
option=input(‘OPTION NUMBER =’);
switch option
%^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
%***********************************************************************
%*** %POCKLINGTON’S INTEGRAL EQUATIONS ***
%***********************************************************************
case 1
%******************************************************
% SOME CONSTANTS AND ARRAYS
%******************************************************
nmax=200;
nmt=2nmax-1;
cgaw=zeros(nmax);
cga=cgaw(1,1:nmax);
zmnw=zeros(nmt);
zmn=zmnw(1,1:nmt);
waw=zeros(nmt);
wa=waw(1,1:nmt);
etm=zeros(361);
etmm=etm(1,1:361);
pi=3.14159265;
rad=pi/180;
beta=2.0pi;
eta=120pi;
%*****************************************************
%*** INPUT DATA
%******************************************************
nm=input(‘NUMBER OF SUBDIVISIONS =’);
tl=input(‘TOTAL DIPOLE LENGTH =’);
ra=input(‘RADIUS OF DIPOLE =’);
%*********************************************************
%*** EXITATION OPTION # : 1 —> MAGNETIC-FRILL
% 2 —> DELTA-GAP
%**********************************************************
disp(‘EXCITATION :’);
disp(’ OPTION (1):MAGNETIC-FRILL’);
disp(’ OPTION (2):DELTA-GAP’);
iex=input(‘OPTION NUMBER :’);
hl=tl/2;
nmh=0.5*(nm+1);
dz=2hl/nm;
zm=hl-0.5dz;
b=0.5dz;
a=-0.5dz;
n=79;
hd=(b-a)/(n+1);
%*********************************************************************
% THE IMPEDANCE MATRIX HAS A TOEPLITZ PROPERTY, THEREFORE ONLY
% NM ELEMENTS NEED TO BE COMPUTED, AND THE MATRIX IS FILLED IN A
% FORM THAT CAN BE SOLVED BY A TOEPLITZ MATRIX SOLVING
%**********************************************************************
for I = 1: nm
zn=hl-(I-0.5)dz;
za1=zn-zm+a;
%*****************************************************************
% FAST ALGORITHM FORM OF THE SIMPSON’S INTEGRAL ROUTINE
%******************************************************************
recgp=sqrt(rara+za1za1);
cgp1=exp(-jbetarecgp)((1.0+jbetarecgp)(2.0recgprecgp-3.0rara)+(betararecgp)2)/(2.0*beta*recgp5);
zb1=zn-zm+b;
roc=sqrt(rara+zb1zb1);
cgp2=exp(-jbetaroc)((1.0+jbetaroc)(2.0rocroc-3.0rara)+(betararoc)2)/(2.0*beta*roc5);
crt=cgp1+cgp2;
for k = 1: n
xk=a+khd;
zx1=zn-zm+xk;
r=sqrt(rara+zx1zx1);
cgp3=exp(-jbetar)((1.0+jbetar)(2.0rr-3.0rara)+(betarar)2)/(2.0*beta*r5);
if mod(k,2)~=0
crt=crt+4.0cgp3;
else
crt=crt+2.0cgp3;
end
end
crt=crthd0.33333;
zmn(I)=crt;
if I~=1
zmn(nm+I-1)=crt;
end
end
rb=2.3ra;
tlab=2.0log(2.3);
for i = 1: nm
zi=hl-(i-0.5)dz;
r1=betasqrt(zizi+rara);
r2=betasqrt(zizi+rbrb);
j=0.0+j1.0;
if iex==1
cga(i)=-jbeta^2/(etatlab)(expm(-jr1)/r1-expm(-jr2)/r2);
else
if i~=nmh
cga(i)=0;
else
cga(i)=-jbeta/(etadz);
end
end
end
%******************************************************************
% INPUT:
% ©A(2M - 1) THE FIRST ROW OF THE T-MATRIX FOLLOWED BY
% ITS FIRST COLUMN BEGINNING WITH THE SECOND
% ELEMENT. ON RETURN A IS UNALTERED.
% ©B(M) THE RIGHT HAND SIDE VECTOR B.
% ©WA(2M-2) A WORK AREA VECTOR
% (I)M ORDER OF MATRIX A.
% OUTPUT:
% ©B(M) THE SOLUTION VECTOR.
% PURPOSE:
% SOLVE A SYSTEM OF EQUATIONS DESCRIBED BY A TOEPLITZ MATRIX.
% A * X = B
% ******************************************************************
t2=length(zmn);
t1=length(wa);
m=nm;
a=zmn;
b=cga;
wa=wa;
m=nm;
a1=a;
a2=a(m+1:t2);
t=b;
c1=wa;
c2=wa(m-1:t1);
r1=a1(1);
t(1)=b(1)/r1;
if m ~=1
for n=2:m
n1=n-1;
n2=n-2;
r5=a2(n1);
r6=a1(n);
if n ~= 2
c1(n1)=r2;
for i1=1:n2
i2=n-i1;
r5=r5+a2(i1)c1(i2);
r6=r6+a1(i1+1)c2(i1);
end
end
r2=-r5/r1;
r3=-r6/r1;
r1=r1+r5r3;
if n~=2
r6=c2(1);
c2(n1)=0.0+j0.0;
for i1=2:n1
r5=c2(i1);
c2(i1)=c1(i1)r3+r6;
c1(i1)=c1(i1)+r6r2;
r6=r5;
end
end
c2(1)=r3;
r5=0.0+j0.0;
for i1=1:n1
i2=n-i1;
r5=r5+a2(i1)t(i2);
end
r6=(b(n)-r5)/r1;
for i1=1:n1
t(i1)=t(i1)+c2(i1)r6;
end
t(n)=r6;
end
end
cga=t;
%***************************************************************
% Output files for curr-MoM.dat
%******************************************************************
fid=fopen(‘Curr-MoM_m.dat’,‘wt’);
fprintf(fid,‘CURRENT DISTRIBUTION ALONG ONE HALF OF THE DIPOLE\n’);
fprintf(fid,‘POSITION Z: MAGNITUDE: REAL PART: IMAGINARY PART: SECTION #:\n\n’);
%*****************************************************************************
% OUTPUT THE CURRENT DISTRIBUTION ALONG OF THE DIPOLE
%*****************************************************************************
for i=1:nmh
xi=hl-(i-0.5)dz;
yi=abs(cga(i));
table2=[xi,yi,real(cga(i)),imag(cga(i)),i];
fprintf(fid,’%1.4f %1.6f %1.6f %1.6f %2.1f \n\n’,table2);
end
%*****************************************************************
% Output figure for curr-MoM.dat
%******************************************************************
i=[1:nmh];
xi=hl-(i-0.5)*dz;
yi=abs(cga(i));
stem(xi,yi)
grid on;
xlabel(‘Position (z)’)
ylabel(‘Magnitude’)
title(‘CURRENT DISTRIBUTION ALONG ONE HALF OF THE DIPOLE’)
figure;
fclose(fid);
%******************************************************
% COMPUTATION OF THE INPUT IMPEDANCE
%******************************************************
zin=1.0/cga(nmh);
%******************************************************************
% COMPUTATION OF AMPLITUDE RADIATION PATTERN OF THE ANTENNA
%******************************************************************
for i=1:181
theta=(i-1.0)rad;
cth=cos(theta);
sth=sin(theta);
if abs(cth)<0.001
ft=1;
else
ft=sin(betadzcth0.5)/(betadzcth0.5);
end
crt=0;
for m=1:nm
zm=hl-(m-0.5)dz;
crt=crt+exp(jbetazmcth)ftcga(m)dz;
end
ptt=abs(crt)sthstheta0.5;
etmm(i)=ptt;
end
amax=etmm(1);
for i=2:181
if etmm(i)>=amax
amax=etmm(i);
end
end
for i=1:181
ptt=etmm(i)/amax;
if ptt<=0.00001
ptt=0.00001;
etmm(i)=20log10(ptt);
end
etmm(i)=20log10(ptt);
end
%******************************************************************
% Output files for Patt-MoM_m.dat
%******************************************************************
pok=fopen(‘Patt-MoM_m.dat’,‘wt’);
fprintf(pok,‘RADIATION PATTERN vs OBSERVATION ANGLE THETA\n’);
fprintf(pok,‘THETA (in degrees) MAGNITUDE(in dB)\n’);
for i=1:181
xi=i-1;
table3=[xi,etmm(i)];
fprintf(pok,’%3.1f %3.3f \n’,table3);
end
%******************************************************************
% Output figure for Patt-MoM_m.dat
%******************************************************************
i=[1:181];
xi=i-1;
% Polar Plot
etmm1=[etmm(1:181),fliplr(etmm(1:180))];
q=polar_dB([0:360],etmm1,-40,0,4,’-’);
set(q,‘linewidth’,1.5);
% plot(xi,etmm(i),‘linewidth’,2);
% axis ([0 180 -60 0]);
% grid on;
% xlabel(‘Theta(degrees)’);
% ylabel(‘Magnitude(dB)’);
title(‘RADIATION PATTERN vs OBSERVATION ANGLE’)
for i=182:361
xi=i-1;
table4=[xi,etmm(362-i)];
fprintf(pok,’%3.1f %3.3f \n’,table4);
end
%^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
%***********************************************************************
%*** HALLEN’S INTEGRAL EQUATION ***
%***********************************************************************
case 2
%******************************************************
% SOME CONSTANTS AND INPUT DATA
%******************************************************
nm=input(‘NUMBER OF SUBDIVISIONS =’);
tl=input(‘TOTAL DIPOLE LENGTH =’);
ra=input(‘RADIUS OF DIPOLE =’);
n=nm;
l=tl;
rho=ra;
rtod=180/pi;
eta=120pi;
bk=2pi;
cj=0.0+j1.0;
dz=l/(2(n-1));
dz1=l/(2n);
%C***********************************************************************
%C PERFORM COMPLEX INTEGRATION USING SIXTEEN POINT
%C GAUSSIAN QUADRATURE WITH INCREASING ACCURACY SET
%C BY INTEGER NO
%C***********************************************************************
absica=[-0.095012509837637,-0.281603550779259;
-0.458016777657227,-0.617876244402644;
-0.755404408355003,-0.865631202387832;
-0.944575023073233,-0.989400934991650;
0.095012509837637,0.281603550779259;
0.458016777657227,0.617876244402644;
0.755404408355003,0.865631202387832;
0.944575023073233,0.989400934991650];
wght=[0.189450610455068,0.182603415044924;
0.169156519395002,0.149595988816577;
0.124628971255534,0.095158511682493;
0.062253523938648,0.027152459411754;
0.189450610455068,0.182603415044924;
0.169156519395002,0.149595988816577;
0.124628971255534,0.095158511682493;
0.062253523938648,0.027152459411754];
%***********************************************************
% FILL THE MATRIX AND EXCITATION VECTOR OF THE SYSTEM:
% [ZMATRX] and [ELECUR]
%************************************************************
for i=1:n
z=(2i-1)dz1/2.0;
zmatrx(i,n)=-cos(bkz);
elecur(i)=-cjsin(bkz)/(2.0eta);
for j=1:n-1
lower=(j-1)dz1;
upper=(j)dz1;
%***************************************************************
% PERFORM NUMERICAL INTEGRATION OF THE KERNEL FOR HALLEN’S
% INTEGRAL EQUATION
%*****************************************************************
no=10;
del=(upper-lower)/(2.0no);
sum=0.0+j0.0;
for t=1:no
s=lower+(2t-1)del;
for q=1:16
x=s+absica(q)del;
%*********************************************************************
%C KERNEL PROVIDES THE KERNEL OF HALLEN’S EQN FOR INTEGRATION
%C SYMMETRY IS USED TO REDUCE THE SYSTEM OF EQUATIONS AND
%C HENCE ALTERS THE KERNEL
%***********************************************************************
r1=sqrt(rhorho+(z-x)(z-x));
r2=sqrt(rhorho+(z+x)(z+x));
kernel=exp(-cjbkr1)/(4.0pir1)+exp(-cjbkr2)/(4.0pi*r2);
sum=sum+wght(q)kernel;
end
end
res=sumdel;
zmatrx(i,j)=res;
end
end
%*********************************************************************************
% DECOMPOSE AND SOLVE THE SYSTEM FOR THE CURRENT DISTRIBUTION
%*********************************************************************************
%****************************
% GET SCALING INFO.
%****************************
for i=1:n
zmax=0.0;
for j=1:n
caz=abs(zmatrx(i,j));
if caz >= zmax
zmax=caz;
end
end
scal(i)=1.0/zmax;
end
%****************************
% CROUT’s algorithm.
%****************************
for j=1:n
for i=1:j-1
for k=1:i-1
zmatrx(i,j)=zmatrx(i,j)-zmatrx(i,k)zmatrx(k,j);
end
end
zmax=0.0;
%****************************************
% SEARCH FOR LARGEST PIVOT ELEMENT.
%*****************************************
for i=j:n
for k=1:j-1
zmatrx(i,j)=zmatrx(i,j)-zmatrx(i,k)zmatrx(k,j);
end
problem=scal(i)abs(zmatrx(i,j));
if problem >=zmax
imax=i;
zmax=problem;
end
end
%*********************************
% INTERCHANGE THE ROWS.
%***********************************
if j~=imax
for k=1:n
temp=zmatrx(imax,k);
zmatrx(imax,k)=zmatrx(j,k);
zmatrx(j,k)=temp;
end
scal(imax)=scal(j);
end
iperm(j)=imax;
%***********************************
% DIVIDE BY PIVOT ELEMENT.
%***********************************
if j~=n
for i=j+1:n
zmatrx(i,j)=zmatrx(i,j)/zmatrx(j,j);
end
end
end
%******************************************************************
% SOLVES LINEAR SYSTEM GIVEN THE LU DECOMPOSITION FROM LUDEC
% FORCING VECTOR IS REPLACED WITH SOLUTION VECTOR UPON EXIT
%*****************************
% FORWARD SUBSTITUTION.
%******************************************************************
for i=1:n
temp=elecur(iperm(i));
elecur(iperm(i))=elecur(i);
for j=1:i-1
temp=temp-zmatrx(i,j)elecur(j);
end
elecur(i)=temp;
end
%*******************************
% BACKWARD SUBSTITUTION.
%********************************
for i=1:n
ii=n-i+1;
temp=elecur(ii);
for j=ii+1:n
temp=temp-zmatrx(ii,j)elecur(j);
end
elecur(ii)=temp/zmatrx(ii,ii);
end
%**************************************************************************
% COMPUTATION OF THE INPUT IMPEDANCE AND CURRENT DISTRIBUTION
%***************************************************************************
zin=1.0/elecur(1);
%******************************************************************
% Output files for curr-MoM_m.dat
%******************************************************************
fid=fopen(‘Curr-MoM_m.dat’,‘wt’);
fprintf(fid,‘CURRENT DISTRIBUTION ALONG ONE HALF OF THE DIPOLE\n’);
fprintf(fid,‘POSITION Z: CURRENT MAGNITUDE: CURRENT PHASE: SECTION:\n’);
for i=1:n-1
%******************************************************************
% THIS FUNCTION IS COMPUTES THE ARCTANGENT GIVEN X,Y. IT IS
% SIMILAR TO ATAN2 EXCEPT IT AVOIDS THE RUN TIME ERRORS ON
% SOME MACHINES FOR SMALL ARGUMENTS.
%******************************************************************
smlt=0.0000001;
if (abs(x)<=smlt) & (abs(y)<=smlt)
btan2=0.0;
else
btan2=atan2(real(elecur(i)),imag(elecur(i)));
end
cur=abs(elecur(i));
pha=rtodbtan2;
table=[idz-dz/2.0,cur,pha,i];
fprintf(fid,’%2.6f %2.6f %3.6f %3.1f\n\n\n\n’,table);
end
%******************************************************************
% Output figure for curr-MoM_m.dat
%******************************************************************
i=[1:n-1];
smlt=0.0000001;
if (abs(x)<=smlt) & (abs(y)<=smlt)
btan2=0.0;
else
btan2=atan2(real(elecur(i)),imag(elecur(i)));
end
cur=abs(elecur(i));
pha=rtodbtan2;
stem(idz-dz/2.0,cur)
grid on
xlabel(‘Position (z)’)
ylabel(‘Current magnitude’)
title(‘CURRENT DISTRIBUTION ALONG ONE HALF OF THE DIPOLE’)
fclose(fid);
figure;
%*********************************************************************************
% CALCULATE THE RADIATION PATTERN OF THE ANTENNA
%*********************************************************************************
maxang=181;
pmax=-1.0;
for j=1:maxang
theta=(j-1)/(maxang-1)pi;
%*******************************************************************
% CALCULATES THE RADIATED POWER LEVEL AT ANGLE THETA RADIANS
% TO THE DIPOLE AXIS. SINCE THE PATTERN IS NORMALIZED TO THE
% MAXIMUM RADIATED POWER, COMMON CONSTANTS ARE REMOVED
%********************************************************************
sth=sin(theta);
cth=cos(theta);
arg=pidzcth;
if abs(arg) <= 0.001
ft=1.0;
else
ft=sin(arg)/arg;
end
n2=n-1;;
crt=0.0+j0.0;
for k=1:n2
argp=pi(-l+(k-1)2.0dz+dz)cth;
crt=crt+exp(cjargp)ftelecur(k);
argp=pi*(-l+(n2+k-1)2.0dz+dz)cth;
crt=crt+exp(cjargp)ftelecur(k);
end
power=abs(crt)sthsth;
pwr(j)=power;
if pwr(j)>=pmax
pmax=pwr(j);
end
end
%******************************************************************
% Output files for Patt-MoM_m.dat
%******************************************************************
hel=fopen(‘Patt-MoM_m.dat’,‘wt’);
fprintf(hel,‘RADIATION PATTERN vs OBSERVATION ANGLE THETA\n’);
fprintf(hel,‘THETA (in degrees) MAGNITUDE (in dB)\n’);
%*******************************************************************
% WRITE THE RADIATION PATTERN IN dB
%*******************************************************************
for i=1:maxang
theta=(i-1)/(maxang-1)180;
pattrn=pwr(i)/pmax;
if pattrn <=0.00001
pattrn=-100;
else
pattrn=20log10(pattrn);
end
table1=[theta,pattrn];
fprintf(hel,’%3.1f %3.3f \n’,table1);
end
%******************************************************************
% Output figure for Patt-MoM_m.dat
%******************************************************************
i=[1:181];
theta=(i-1);
pattrn=pwr(i)/pmax;
pattrn=20log10(pattrn);
% Polar Plot
pattrn=[pattrn,fliplr(pattrn(1:180))];
q=polar_dB([0:360],pattrn,-40,0,4,’-’);
set(q,‘linewidth’,1.5);
% plot(theta,pattrn,‘linewidth’,2);
% axis ([0 180 -60 0]);
% grid on;
% xlabel(‘Theta(degrees)’);
% ylabel(‘Magnitude(dB)’);
title(‘RADIATION PATTERN vs OBSERVATION ANGLE’)
for i=182:361
theta=(i-1)/(maxang-1)180;
pattrn=pwr(362-i)/pmax;
if pattrn <=0.00001
pattrn=-100;
else
pattrn=20log10(pattrn);
end
table2=[theta,pattrn];
fprintf(hel,’%3.1f %3.3f \n’,table2);
end
fclose(hel);
end
%******************************************
% OUTPUT FILE
%*******************************************
if(option_a2)
diary(filename);
end
%*******************************************
% FORMAT STATEMENTS
%*******************************************
disp(sprintf(’\n\n\n\n\n\n\n\n’));
disp(strvcat(‘WIRE ANTENNA PROBLEM’,’====’));
if (option1)&(iex1)
disp(sprintf(‘POCKLINGTON’‘S EQUATION AND MAGNETIC FRILL MODEL’));
elseif (option1)&(iex2)
disp(sprintf(‘POCKLINGTON’‘S EQUATION AND DELTA GAP MODEL’));
elseif (option2)
disp(sprintf(‘HALLEN’‘S EQUATION’));
end
disp(sprintf(’\nLENGTH = %4.4f (WLS)’,tl));
disp(sprintf(‘RADIUS OF THE WIRE = %4.4f (WLS)’,ra));
disp(sprintf(‘NUMBER OF SUBSECTIONS = %2.2f\n’,nm));
ZR=real(zin);
ZI=imag(zin);
if ZI>0
output=sprintf(‘INPUT IMPEDANCE: Z= %5.1f +j %5.1f (OHMS)\n’,ZR,ZI);
else
output=sprintf(‘INPUT IMPEDANCE: Z= %5.1f -j %5.1f (OHMS)\n’,ZR,abs(ZI));
end
disp(output);
disp(strvcat(’*** NOTE:’,…
’ THE DIPOLE CURRENT DISTRIBUTION IS STORED IN Curr-MoM_m.dat’,…
’ THE AMPLITUDE RADIATION PATTERN IS STORED IN Patt-MoM_m.dat’,…
’ =========================================================’));
diary off;
warning off;
MOM
OUTPUT DEVICE OPTION FOR THE OUTPUT PARAMETERS
OPTION (1):SCREEN
OPTION (2):OUTPUT FILE
OUTPUT DEVICE =1
CHOICE OF POCKLINGTON’S OR HALLEN’S EQN.
OPTION (1):POCKLINGTON’S EQN.
OPTION (2):HALLEN’S EQN.
OPTION NUMBER =2
NUMBER OF SUBDIVISIONS =12
TOTAL DIPOLE LENGTH =24
RADIUS OF DIPOLE =10
MOM
OUTPUT DEVICE OPTION FOR THE OUTPUT PARAMETERS
OPTION (1):SCREEN
OPTION (2):OUTPUT FILE
OUTPUT DEVICE =1
CHOICE OF POCKLINGTON’S OR HALLEN’S EQN.
OPTION (1):POCKLINGTON’S EQN.
OPTION (2):HALLEN’S EQN.
OPTION NUMBER =2
NUMBER OF SUBDIVISIONS =12
TOTAL DIPOLE LENGTH =24
RADIUS OF DIPOLE =36
WIRE ANTENNA PROBLEM
HALLEN’S EQUATION
LENGTH = 24.0000 (WLS)
RADIUS OF THE WIRE = 36.0000 (WLS)
NUMBER OF SUBSECTIONS = 12.00
INPUT IMPEDANCE: Z= 6872533449.1 +j 19260575406.8 (OHMS)
*** NOTE:
THE DIPOLE CURRENT DISTRIBUTION IS STORED IN Curr-MoM_m.dat
THE AMPLITUDE RADIATION PATTERN IS STORED IN Patt-MoM_m.dat
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