Azzera filtri
Azzera filtri

Plotting with an injected numerical solution as an argument

1 visualizzazione (ultimi 30 giorni)
I have a numerical solution "a" of an equation which is obtained for a "specific value of T and k". I want to insert this solution inside a function which is then to be plotted.
In the plot function I also have same variables, and plot is for a "range of T values" (on x-axis) and fixed k. Now the numerical solution is only for a fixed T value, so it is not well synchronized with the plot function. Is there a way to tie these two (numerical solution+plot function), so I can define T and other same variables once and for all?
The code for numerical solution is
syms a
v2=-2.375; g=1; b=0.00001; e2=0.5; k=pi/2; T=1.1; w=-2*cos(k);
eqn = sin(3*k+a)/sin(2*k+a)==v2-w+(g.*T.^2)./(1+b.*T.^2)+(e2.*T.^2.*sin(k)^2)./(sin(2*k+a)^2+b*T.^2*sin(k)^2);
sol = solve(eqn,a,[0 pi]);
solutions=vpa(subs(sol),3)
The above numerical solution(s) is to be injected inside the argument of `cosine` function below and then plotted. The code for plotting function is
v0=-2.5; epsilon=0.05; v1 = v0*(1+epsilon); v2 = v0*(1-epsilon); g = 1;
T=[0:0.01:3];
k=pi/2
l=-k
w=-2*cos(k)
w1=-2*cos(l)
nu = @(T) v2-w+g.*T.^2;
delta = @(T) v1-w+g.*T.^2.*(1-2.*nu(T)*cos(k+solutions)+nu(T).^2);
nu1 = @(T) v1-w1+g.*T.^2;
delta1 = @(T) v2-w1+g.*T.^2.*(1-2.*nu(T)*cos(k+solutions)+nu(T).^2);
fplot(@(T) abs((exp(i*k)-exp(-i*k))/(1+(nu(T)-exp(i*k)).*(exp(i*k)-delta(T))))^2,[0,5],'--r')
hold on
fplot(@(T) abs((exp(i*l)-exp(-i*l))/(1+(nu1(T)-exp(i*l)).*(exp(i*l)-delta1(T))))^2,[0,5],'b')
hold off
legend('k=\pi/2','k=-\pi/2')
  4 Commenti
Torsten
Torsten il 4 Dic 2018
Use T as a second symbolic variable. Then you can substitute T in the solution by the numerical values 0:0.01:3.
AtoZ
AtoZ il 4 Dic 2018
@Torsten three errors appear:
1: Error using sym/min
Input arguments must be convertible to floating-point numbers.
2: Error in fplot (line 116)
ymin = min(y(J)); ymax = max(y(J));
3: Error in Plots (line 30)
fplot(@(T)
abs((exp(i*k)-exp(-i*k))/(1+(nu(T)................

Accedi per commentare.

Risposte (0)

Categorie

Scopri di più su Mathematics in Help Center e File Exchange

Prodotti


Release

R2014a

Community Treasure Hunt

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

Start Hunting!

Translated by