how can i plot this code ?

1 Ansicht (letzte 30 Tage)
Tomer Segev
Tomer Segev am 30 Sep. 2020
Kommentiert: Rik am 1 Okt. 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');

Antworten (1)

Alan Stevens
Alan Stevens am 30 Sep. 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 Kommentar
Rik
Rik am 1 Okt. 2020
Comment posted as answer by Tomer Segev:
Hey, thank you for your help!

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Graphics finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by