# Array indexes must be positive integers or logical values.

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ljaseon on 12 May 2021
Commented: ljaseon on 12 May 2021
this is the code that what i input but
Array indexes must be positive integers or logical values. at the moment SUM= E1*SUM(V(3:IOUT+1,JOUT+2))....
+E1*V(2,JOUT+2)/2+E2*V(IOUT+2,2)/2;
what is the problem?
clear all;format compact;
%Output;
%
% H NT Zo
% -----------------------
% 0.25 700 69.77
% 0.1 500 65.75
% 0.05 500 70.53
% 0.05 700 67.36
% 0.05 1000 65.50
H=0.05;
NT= 100;
A=2.5; B=2.5; D=0.5; W=1.0;
ER=2.35;
EO=8.81E-12;
U=3.0E+8;
NX=A/H;
NY=B/H;
ND=D/H;
NW=W/H;
VD=100.0;
ERR=1.0;
E1=EO;
E2=EO*ERR;
%CALCULATE CHARGE WITH AND WITHOUT DIELECTRIC
L=1:2
E1=EO;
E2=EO*ERR;
%INITIALIZATION
V=zeros(NX+2,NY+2);
%SET POTENTIAL ON INNER CONDUCTOR(FIXED NODES) EQUAL TO VD
V(2:NW+1,ND+2)=VD;
%NOW,CALCULATE THE TOTAL CHARGE ENCLOSED IN A
%RECTANGULAR PATH SURROUNDING THE INNER CONDUCTOR
IOUT=round((NX+NW)/2);
JOUT=round((NY+ND)/2);
%SUM POTENTIAL ON INNER AND OUTER LOOPS
for K=1:2
SUM=0;
SUM= E1*SUM(V(3:IOUT+1,JOUT+2))....
+E1*V(2,JOUT+2)/2+E2*V(IOUT+2,2)/2;
for J=1:JOUT-1
if(J<ND)
SUM=SUM+E2*V(IOUT+2,J+2);
elseif(J==ND)
SUM=SUM+(E1+E2)*V(IOUT+2,J+2)/2;
else
SUM=SUM+E1*V(IOUT+2,J+2);
end
end
if K==1
SV(1)=SUM;
end
IOUT=IOUT-1;
JOUT=JOUT-1;
end
SUM=SUM+2.0*E1*V(IOUT+2,JOUT+2);
SV(2)=SUM;
Q(L)=abs(SV(1)-SV(2));
ERR=ER;
% FINALLY,CALCULATE Zo
Co=4.0*Q(1)/VD;
C1=4.0*Q(2)/VD;
ZO=1.0/(U*sqrt(C0*C1));
disp([H,NT,ZO])

Alan Stevens on 12 May 2021
Looks like you are using SUM when you want sum. Change to:
SUM= E1*sum(V(3:IOUT+1,JOUT+2))....
Also, in line 59, you should have C0 not Co.
ljaseon on 12 May 2021
Thank you i got it!