how can i plot this code ?

2 visualizaciones (últimos 30 días)
Tomer Segev
Tomer Segev el 30 de Sept. de 2020
Comentada: Rik el 1 de Oct. de 2020
hey, I want to plot this code the ri veriable is a scalar and not a vector, also how can I define Z1 and Z2 without the operator @ ?
s= [0,5*10^-2,1*10^-1,1.5*10^-1,2*10^-1,2.5*10^-1,3*10^-1,3.5*10^-1,4*10^-1,4.5*10^-1,5*10^-1,5.5*10^-1,6*10^-1,6.5*10^-1,7*10^-1,7.5*10^-1,8*10^-1,8.5*10^-1,9*10^-1,9.5*10^-1,9.9*10^-1,9.99*10^-1,1,1,1,1,1];
omega= [3.74*10^-1, 3.8*10^-1,3.87*10^-1,3.94*10^-1,4.02*10^-1,4.11*10^-1,4.2*10^-1,4.29*10^-1,4.4*10^-1,4.51*10^-1,4.64*10^-1,4.78*10^-1,4.94*10^-1,5.12*10^-1,5.33*10^-1,5.57*10^-1,5.86*10^-1,6.23*10^-1,6.72*10^-1,7.46*10^-1,8.71*10^-1,9.56*10^-1,9.68*10^-1,9.72*10^-1,9.75*10^-1,9.8*10^-1,9.86*10^-1];
plot(s,omega,'.');
hold on
Z1=@(s) (1+ ((1-s.^2)^(1/3))*((1+s).^1/3+(1-s).^1/3));
Z2=@(s,Z1) ((3*s.^2+Z1.^2).^1/2);
p=2;
w=3;
c= 3*10^8;
G=6.67384*10^-11;
M=1;
ri= (G*M./c^2).*(3.+Z2+sqrt(9+6.*Z2-(Z1).^2-2*(Z2.*Z1)));
plot(ri,omega,'r');

Respuestas (1)

Alan Stevens
Alan Stevens el 30 de Sept. de 2020
You need to plot them separately as they have completely different 'x' values (indeed the ri values seem unbelievably small!):
s= [0,5*10^-2,1*10^-1,1.5*10^-1,2*10^-1,2.5*10^-1,3*10^-1,3.5*10^-1,4*10^-1,4.5*10^-1,5*10^-1,5.5*10^-1,6*10^-1,6.5*10^-1,7*10^-1,7.5*10^-1,8*10^-1,8.5*10^-1,9*10^-1,9.5*10^-1,9.9*10^-1,9.99*10^-1,1,1,1,1,1];
omega= [3.74*10^-1, 3.8*10^-1,3.87*10^-1,3.94*10^-1,4.02*10^-1,4.11*10^-1,4.2*10^-1,4.29*10^-1,4.4*10^-1,4.51*10^-1,4.64*10^-1,4.78*10^-1,4.94*10^-1,5.12*10^-1,5.33*10^-1,5.57*10^-1,5.86*10^-1,6.23*10^-1,6.72*10^-1,7.46*10^-1,8.71*10^-1,9.56*10^-1,9.68*10^-1,9.72*10^-1,9.75*10^-1,9.8*10^-1,9.86*10^-1];
Z1= (1+ ((1-s.^2).^(1/3)).*((1+s).^1/3+(1-s).^1/3));
Z2= ((3*s.^2+Z1.^2).^1/2);
p=2;
w=3;
c= 3*10^8;
G=6.67384*10^-11;
M=1;
ri= (G*M./c^2).*(3.+Z2+sqrt(9+6.*Z2-(Z1).^2-2*(Z2.*Z1)));
subplot(2,1,1);
plot(s,omega,'.');
xlabel('s'),ylabel('omega')
subplot(2,1,2)
plot(ri,omega,'r');
xlabel('ri'),ylabel('omega')
  1 comentario
Rik
Rik el 1 de Oct. de 2020
Comment posted as answer by Tomer Segev:
Hey, thank you for your help!

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