problem in plot for two function

Hi

i want to plot two function in the same axis

i rreceive to message Escattheory(rho,phi))

Error using plot

Data must be numeric, datetime, duration, categorical, or an array convertible

to double.

function y = Escattheory(rho,phi)
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
syms  k
global k0
global ra
Z1=k0.*rho
Z2=k0.*ra;
scale=1;
factor1=besselh(k,2,Z1).*besselj(k,Z2)./besselh(k,2,Z2);
factor2=(-1i).^k.*cos(k.*phi);
factor=factor1.*factor2.*en(k);
y=vpa(symsum(factor,k,0,inf),3)
end

and for the second i receive the message

Unable to perform assignment because the indices on the left side are not

compatible with the size of the right side.

Error in Escat (line 15)

RR(jj)=sqrt((x-xx(jj)).^2+(y-yy(jj)).^2);

Error in mas_closett_linsolve_2_2 (line 80)

plot(rho,abs(Escat(x,y)))

function y = Escat(x,y)
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
global Is
global  Nc
global k0
global Z0
global a 
global Is
sum=0;
for  jj=1:Nc
    xx(jj)=a.*cos(2.*pi.*(jj-1)./Nc);
    yy(jj)=a.*sin(2*pi.*(jj-1)./Nc);
    RR(jj)=sqrt((x-xx(jj)).^2+(y-yy(jj)).^2);
    sum=sum-Is(jj).*k0.*Z0.*besselh(0,2,k0.*RR(jj))./4;
end
y=sum;
end

what happen ??

thank you

George

the main code is

clear all
clc
global  ra
global  N
global  a
global  f
global Is
global  N
global Nc
global k0
global Z0
global  ra
global  a 
Z0=120.*pi;
c0=3e8;
f= input('Give the freuency  = ');
%mumber  of  auxiliary  source 
N=input('Give the total number of the AS : ');
Nc=input('Give the the test  point: ');
ra=input('Give the  radius  of cylinder  : ');
a=input('Give the inside auxilary source radious : ');
lambda=c0./f;
x=zeros(1,N);
y=zeros(1,N);
xc=zeros(1,Nc);
yc=zeros(1,Nc);
k0=2*pi./lambda;
for j=1:N
    for  i=1:Nc
        x(j)=a.*cos(2*pi.*(j-1)./N);
        y(j)=a.*sin(2*pi.*(j-1)./N);
        xc(i)=ra.*cos(2*pi.*(i-1)./Nc);
        yc(i)=ra.*sin(2*pi.*(i-1)./Nc);
        R(i,j)=sqrt((x(j)-xc(i)).^2+(y(j)-yc(i)).^2);
        
    end
end
WI=zeros(N,N);
Source=zeros(1,N);
Is=zeros(1,N);
%N for the matching points
for j=1:N
    % nomizo einai i sinartisi Electricfieldz ousiastika edo pairnei to kuma diegersis
    for i=1:Nc
        WI(i,j)=(1./4).*k0.*Z0.*besselh(0,2,k0.*R(i,j));
        Source(i) =Efieldin(xc(i),yc(i));
    end
    W = linsolve(WI,Source');
end
for i=1:N 
    Is(i)=W(i); 
end 
for  k=1:10.*N
    xc(k)=ra.*cos(2*pi.*(k-1)./(10.*Nc));
    yc(k)=ra.*sin(2.*pi.*(k-1)./(10.*Nc));
    %z(k)=abs(Escat(xc(k),yc(k))+Efieldin(xc(k),yc(k)));
    z(k)=abs(Escat(xc(k),yc(k)));
    final=z';
end
phi=pi;
rho=0:0.1:ra
%plot(rho, Escattheory(rho,phi));
hold on
x=rho.*cos(phi);
y=rho.*sin(phi);
plot(rho,abs(Escat(x,y)))
xlabel('$\rho$','Interpreter','latex')
ylabel('$|E_{s}|$','Interpreter','latex' )
hold  off

Answer 1

[Walter Roberson]

function y = Escattheory(rho,phi)
%...
y=vpa(symsum(factor,k,0,inf),3)

Regardless of anything else, because of the vpa(), Escattheory is going to be returning something symbolic.

%plot(rho, Escattheory(rho,phi));

plot() cannot plot anything symbolic. plot() is restricted to numeric, datetime, duration, categorical, or an array convertible to double. plot() does not automatically convert symbolic to double.

phi=pi;
rho=0:0.1:ra

rho will be a vector.

x=rho.*cos(phi);
y=rho.*sin(phi);

Because rho is a vector and you have rho.* in the code, x and y will end up as vectors.

plot(rho,abs(Escat(x,y)))

vector x and vector y will be passed into Escat

function y = Escat(x,y)
%...
    RR(jj)=sqrt((x-xx(jj)).^2+(y-yy(jj)).^2);

x and y are vectors. vector x minus scalar xx(jj) value gives a vector result; likewise vector y minus scalar yy(jj) gives a vector result. As a result, the right hand side is calculating a vector. The result of the vector calculation is being assigned to the scalar location RR(jj) which is not going to fit.

    sum=0;

We firmly recommend against naming a variable sum. It happens a lot that people use a variable named sum and then also want to use the function sum in the same section of code. Also, sum is such a common function, that people who glance at the code are likely to misread the code as if it is referring to function sum instead of variable sum . It is our job as programmers to write code that is not open to easy misinterpretation.

[george veropoulos]

thank you

how can solve this problem ?

[Walter Roberson]

Well, there are two possibilities:

1. The results of the symsum() are something that can be computed numerically (without convergence problems.) In this case, instead of y=vpa(symsum(factor,k,0,inf),3) you would use y=double(symsum(factor,k,0,inf),3)
2. OR, the results of the symsum() are something that cannot be computed numerically without convergence problems. In this case, there is nothing you can do that will make plotting work.

As for the other problem:

If you examine your code

for  jj=1:Nc
    xx(jj)=a.*cos(2.*pi.*(jj-1)./Nc);
    yy(jj)=a.*sin(2*pi.*(jj-1)./Nc);
    RR(jj)=sqrt((x-xx(jj)).^2+(y-yy(jj)).^2);
    sum=sum-Is(jj).*k0.*Z0.*besselh(0,2,k0.*RR(jj))./4;
end

you will note that you never use xx(jj) or yy(jj) or RR(jj) outside of the loop. There is no point in remembering their values. You might as well use

for  jj=1:Nc
    xx=a.*cos(2.*pi.*(jj-1)./Nc);
    yy=a.*sin(2*pi.*(jj-1)./Nc);
    RR=sqrt((x-xx).^2+(y-yy).^2);
    sum=sum-Is(jj).*k0.*Z0.*besselh(0,2,k0.*RR)./4;
end

and this way you avoid all issues about assigning the wrong number of elements.