Vectors must be the same length

Ploting this equation numerically
and this is what I've done (below) but it says vector must be the same length. Thank you for your help
v = (0.01*2.0e-12) ;u = 5.0;
K = 8.617e-5;
T = 300; hBar = 1;
no = 10e16; dg = 1;
trig = (sin(4)*cos(4));
t = 2.0e-12; e = -1;
deltag = 0.0156;
n=[0.00 0.036 0.072];
Betag = linspace(0,10, 30); % However many you want.
p = (0.9*pi);
delta1 = deltag./(K.*T);
I0 = besseli(0,delta1); I1= besseli(1, delta1);
I2 = (I0./I1);
jog = ((no*e*deltag*dg)/hBar).*I2;
[mu] = meshgrid(-10000:100:10000);
legendStrings = cell(length(n), 1);
for k1 = 1:length(n)
thisN = n(k1);
for i = 1:length(Betag)
B1 = Betag(i);
J1 = besselj(mu,B1);
J = (J1.^2);
R = (v + (mu.*u));
S = ((v + (mu.*u)).^2);
Z = (1+thisN.*((I2.*exp(1i.*mu.*p))+1)).*trig;
tmp = ((J.*R)./(1+(2.*thisN)+(thisN.^2)+S)).*Z;
J(i) = jog.*sum(tmp(:));
end
legendStrings{k1} = sprintf('n = %.2f', thisN);
plot(Betag, real(J), '.-', 'LineWidth', 2, 'MarkerSize', 15);
hold on;
drawnow;
end
grid on;
fontSize = 20;
xlabel('\beta_\gamma', 'FontSize', fontSize)
ylabel('\it j_\gamma', 'FontSize', fontSize)
title('\it j_\gamma vs. \beta_\gamma', 'FontSize', fontSize)
legend(legendStrings, 'Location', 'best');

 Risposta accettata

[mu] = meshgrid(-10000:100:10000);
mu is now a 201 x 201 array.
J1 = besselj(mu,B1);
mu is 201 x 201 so J1 is 201 x 201.
J = (J1.^2);
J is the same size as J1, 201 x 201.
J(i) = jog.*sum(tmp(:));
one of the first 30 linear-indexed elements of J is replaced, leaving it 201 x 201.
plot(Betag, real(J), '.-', 'LineWidth', 2, 'MarkerSize', 15);
You plot the vector of length 30 against the 201 x 201 array.
I suggest
for i = 1:length(Betag)
B1 = Betag(i);
J1 = besselj(mu,B1) .^ 2;
R = (v + (mu.*u));
S = ((v + (mu.*u)).^2);
Z = (1+thisN.*((I2.*exp(1i.*mu.*p))+1)).*trig;
tmp = ((J1.*R)./(1+(2.*thisN)+(thisN.^2)+S)).*Z;
J(i) = jog.*sum(tmp(:));
end

2 Commenti

I have tried it but still having same problem
after the correction, this is what I had but the graphs are not distinct (see below). How do we make it distinct as in the legend
(see below)

Accedi per commentare.

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