How can I fix the error of NaN?

Hello, I want to select a range of elements in matrix p0(i,j), i started from (at-9) to (at+9) and j started from 1 to n, (n=20) and the value of at is known. then do calculation for p0, but when i run it, it gives this error: "Attempted to access p0(NaN,2); index must be a positive integer or logical. Error in static (line 27) p0(i,j)=pin;" Please, I need to fix this error. thanks in advace.

 clear all;clc;
m=360;   n=20;  d=240;  l=240;error=1e-4;RF=0.95;erin=0;
error_alt=1*pi/180;
deltheta=2*pi/m;    delz=1/n;
pin=1;
s=(d*d)/(l*l);
R=1/deltheta^2;
cor=(deltheta^2/delz^2)*s;
p0=zeros(m+1,n+1);
p=zeros(m+1,n+1);
p1=zeros(m+1,n+1);
h=zeros(m+1);theta=zeros(m+1);
ees=0.5;            iter=1000;     
for i=1:m+1
    theta(i)=(i-1)*deltheta;
    h(i)=1+ees*cos(theta(i));
end
sum(1)=0;
alt1=atan(sqrt(1.0-ees^2)/ees);
alt1=m+1-(alt1*180/pi);
at=round(alt1);
disp(at)
for k=1:iter
    sumij=0;
    for j=2:n
        for i=(at-9):(at+9);
            p0(i,j)=pin;
        end
    end
    for j=2:n
        p0(m+1,j)=p0(1,j);
    end
    for j=2:n
        for i=1:m
            if (i==1)
                h1=1+ees*cos(theta(i)+0.5*deltheta);
                h2=1+ees*cos(theta(i)-0.5*deltheta);
                conh=h1-h2;
                const=2*(1+cor)*R*h(i)^3;
                A=3*h(i)^2*R*conh/const;
                C=h(i)^3*R/const;
                E=h(i)^3*R*cor/const;
                G=-3*sqrt(R)*conh/const;
                p(i,j)=(C+A/2)*p0(i+1,j)+(C-A/2)*p0(m,j)+E*(p0(i,j+1)+p0(i,j-1))+G;           
            else
                conh=(h(i+1)-h(i-1))/2;
                const=2*(1+cor)*R*h(i)^3;
                A=3*h(i)^2*R*conh/const;
                C=h(i)^3*R/const;
                E=h(i)^3*R*cor/const;
                G=-3*sqrt(R)*conh/const;
                p(i,j)=(C+A/2)*p0(i+1,j)+(C-A/2)*p0(i-1,j)+E*(p0(i,j+1)+p0(i,j-1))+G;
            end
            if p(i,j)<0
               p(i,j)=0.0;
               sumij=sumij+p(i,j);
            else
            end
        end
    end
    for j=2:n
        p(m+1,j)=p(1,j);
    end
    for j=2:n
        for i=(at-9):(at+9)
            p(i,j)=p0(i,j);
        end
    end
    sum(k+1)=sumij;
    percentage=abs(sum(k+1)-sum(k))/abs(sum(k+1));
    if percentage < 0.0001
    for j=1:n
        for i=1:m
            p1(i,j)=p(i,j);
            if p1(i,j) < 0
                p1(i,j)=0;
            end
        end
    end
    else
        for j=2:n
            for i=1:m
                p0(i,j)=RF*p(i,j)+(1-RF)*p0(i,j);
            end
        end
    end
    wr=0;
    for j=1:n+1
        wi=0.0;
        for i=1:m
            wi=wi+p1(i,j)*cos(theta(i))*deltheta;
        end
        wr=wr+wi*delz;
    end
    wt=0.0;
    for j=1:n+1
        wi=0.0;
        for i=1:m
            wi=wi+p1(i,j)*sin(theta(i))*deltheta;
        end
        wt=wt+wi*delz;
    end
    w=sqrt(wr^2+wt^2);
    alt=atan(wt/wr);
    alt=m+1+(alt*180/pi);
    erin=abs(alt-alt1)/abs(alt);
    if erin<error_alt
        kjk=fprintf(1,'%4i\r\n',alt);
        break
    else
        alt1=0.5*alt+(1-0.5)*alt1;
        at=round(alt1);
    end
end
y=k;
figure(1)
surf(p1)
figure(2)
plot(p)

Answer 1

[Thorsten]

The code breaks when you want to compute

 alt=atan(wt/wr);

with wr = 0, resulting in alt = NaN.

Instead, use

 alt = atan2(wt, wr);

[Muhammad Abdulrasool]

Thanks Thorsten, also for the term (percentage) it has the same (i.e.NaN) can you help

[Muhammad Abdulrasool]

Thorsten , I made this correction but the iteration loop stop in the second iteration ,and the value of percentage is also NaN. where is the error?

[Thorsten]

Muhammad, you have

percentage=abs(sum(k+1)-sum(k))/abs(sum(k+1));

and when the denominator abs(sum(k+1)) == 0, percentage is NaN. I don't know what you want to compute here, so you have to fix this yourself. And note that sum is a function in Matlab, so please do not use sum as the name of a variable. Use thesum or mysum or S or whatever pleases you.

[Muhammad Abdulrasool]

OK Thorsten, I'll change that, and I want in this step to compute the error between step and the next step (in order to reach the convergence criterion).