plot the graph of function

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shiv gaur
shiv gaur on 29 Jan 2022
Commented: Walter Roberson on 31 Jan 2022
l=633;
k0=2*pi/l;
ea=1.39;
da=1:300;
theta=1:70;
y=zeros(length(theta),length(da));
np=1.5;
x=(2*pi/l)*np*sind(theta);
for i=1:length(theta)
for j=1:length(da)
ka=k0*sqrt(ea-x.^2);
f=(1-exp(2*1i.*da(j).*ka(i)))/(1+exp(2*1i.*da(j).*ka(i)));
%pfms=2*atan(1i*a.*b);
y(i,j)=2*atan(1i*c.*f);
disp(y)
end
end
plot(da,y )
error message pl plot between da vs y and dy/da vs da
Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 1-by-100.
  3 Comments
shiv gaur
shiv gaur on 29 Jan 2022
how to plot dy/dda vs da

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Accepted Answer

Kevin Hanekom
Kevin Hanekom on 29 Jan 2022
clc;clear;close all;
l=633;
c = 1;
k0=2*pi/l;
ea=1.39;
da=1:300;
theta=1:70;
y=zeros(length(theta),length(da));
np=1.5;
x=(2*pi/l)*np*sind(theta);
for i=1:length(theta)
iter = 0;
for j=1:length(da)
iter
ka=k0*sqrt(ea-x.^2);
f=(1-exp(2*1i.*da(j).*ka(i)))/(1+exp(2*1i.*da(j).*ka(i)));
%pfms=2*atan(1i*a.*b);
y(i,j)=2*atan(1i*c.*f);
% disp(y);
iter = iter + 1;
end
end
plot(da,y )
What exactly is wrong?
  2 Comments
Walter Roberson
Walter Roberson on 31 Jan 2022
That code never calls the function f
Reminder: your y is a 2D array that is changing in size as you go (since you do not pre-allocate). And each time you assign to a new y(i,j) location, you are plotting using the constant-sized vector da as the independent variable, and using all of the changing-size y as the dependent variable. That is going to run into size mismatches, and is also going to be attempting to plot locations in y that have not been explicitly given a value yet.
What is the point of doing the plot inside the double nested loop?
It might make sense to plot at the end of the double nested loop.
... but it would probably make more sense to return y to the calling function and let the calling function plot whatever it wants.

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