Non linear pendulum code

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KLETECH MOTORSPORTS
KLETECH MOTORSPORTS el 3 de Nov. de 2020
Comentada: KLETECH MOTORSPORTS el 15 de Nov. de 2020
I've been trying to run this code, for a non linear pendulum, but i get the following errors:
for the function----
Not enough input arguments.
Error in pend_l (line 4)
xdot = [x(2); -wsq*x(1)-C*x(2)];
for the script:
Error using feval
Unrecognized function or variable 'pend_n'.
Error in odearguments (line 90)
f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ode45 (line 115)
odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0,
options, varargin);
Error in pend_solve (line 10)
[t2,y] = ode45('pend_n',tspan,y0);
THE CODE
function xdot = pend_l(t1,x)
wsq = 13.56;
C = 0.1;
xdot = [x(2); -wsq*x(1)-C*x(2)];
% clf;
tspan = 0:0.01:50;
x0 = [1 ; 0];
y0 = [1 ; 0];
[t1,x] = ode45('pend_l',tspan,x0);
[t2,y] = ode45('pend_n',tspan,y0);
X1 = x(:,1);
X2 = x(:,2);
Y1 = y(:,1);
Y2 = y(:,2);
a = 2400;
% n = numel(t1);
n = 8000;
% plot(t1(a:n),0.69*X1(a:n),'linewidth',3);
% hold on;
% grid on;
%
% plot(t2(a:n)-1.81,Y1(a:n),'r','linewidth',3);
%
% legend('Linear', 'Nonlinear')
% hold off;
%
% figure(2);
% clf; % energies are scaled to 10% of its actual values
L = 1*20e-2*20e-2*Y2.*Y2 - 1*9.81*20e-2*(1-cos(Y1)); %lagrangian
E = 1*20e-2*20e-2*Y2.*Y2 + 1*9.81*20e-2*(1-cos(Y1)); %Total energy
D = 0.1*Y2.*Y2; %dissipated energy
% plot(t2,L,'linewidth',3);
% hold on;
% grid on;
% plot(t2,E,'r','linewidth',3);
% plot(t2,D,'g','linewidth',3);
% plot(t2,E-D,'k','linewidth',3); %net energy
b = 1;
f = 2;
figure(3);
clf;
plot(t1,1*20e-2*20e-2*Y2.*Y2,'linewidth',3);
hold on;
plot(t1,1*9.81*20e-2.*(1-cos(Y1)),'r','linewidth',3);
plot(t1,E,'c','linewidth',3);
% plot(t1,X1-b,'linewidth',3);
hold on;
grid on;
plot(t2,Y1-b,'r','linewidth',3);
plot(t2,L+f,'linewidth',3);
plot(t2,D+f,'g','linewidth',3);
plot(t2,(E-D)+f,'k','linewidth',3); %net energy
plot(t1,Y1*0-b-0.3,'k','linewidth',1);
plot(t1,Y1*0-b+0.3,'k','linewidth',1);
plot(t1*0+24,3.5*Y1+1,'k','linewidth',1);
text(30,-b+0.4,'\downarrow Linear Range');
% text(12,0,'\leftarrow Time instant @ which non-linear to linear');
% text(12,-0.5,'transition takes place');
% % legend('Linear', 'Nonlinear')
legend('T','V','Total Energy','Linear', 'Nonlinear','Lagrangian','Dissipated Energy','Net Energy');
figure(4);
clf;
ED = E-D;
[pks,locs] = findpeaks(L);
[pks1,locs1] = findpeaks(ED);
[pks2,locs2] = findpeaks(D);
TF = islocalmin(L);
TF1 = islocalmin(ED);
TF2 = islocalmin(D);
plot(locs*0.01,pks,'b');
hold on;
plot(locs1*0.01,pks1,'r');
plot(locs2*0.01,pks2,'k');
grid on;
plot(t1(TF),L(TF),'b')
plot(locs1*0.01,pks1,'r');
plot(t1(TF1),ED(TF1),'r');
plot(t1(TF2),D(TF2),'k');
plot(t1*0+24,1*Y1+0,'k','linewidth',1);
legend('Lagrangian','Net Energy','Dissipated Energy');
------------------------------------------------------------------------------------------------------
Any idea why it isn't working? thanks guys

Respuesta aceptada

Stephan
Stephan el 3 de Nov. de 2020
You mixed up the functions and the calling code. Also there is no function for pend_n - to build a working code i just copied the same code as used for pend_l and used it as pend_n.
tspan = 0:0.01:50;
x0 = [1 ; 0];
y0 = [1 ; 0];
[t1,x] = ode45(@pend_l,tspan,x0);
[t2,y] = ode45(@pend_n,tspan,y0);
X1 = x(:,1);
X2 = x(:,2);
Y1 = y(:,1);
Y2 = y(:,2);
a = 2400;
% n = numel(t1);
n = 8000;
% plot(t1(a:n),0.69*X1(a:n),'linewidth',3);
% hold on;
% grid on;
%
% plot(t2(a:n)-1.81,Y1(a:n),'r','linewidth',3);
%
% legend('Linear', 'Nonlinear')
% hold off;
%
% figure(2);
% clf; % energies are scaled to 10% of its actual values
L = 1*20e-2*20e-2*Y2.*Y2 - 1*9.81*20e-2*(1-cos(Y1)); %lagrangian
E = 1*20e-2*20e-2*Y2.*Y2 + 1*9.81*20e-2*(1-cos(Y1)); %Total energy
D = 0.1*Y2.*Y2; %dissipated energy
% plot(t2,L,'linewidth',3);
% hold on;
% grid on;
% plot(t2,E,'r','linewidth',3);
% plot(t2,D,'g','linewidth',3);
% plot(t2,E-D,'k','linewidth',3); %net energy
b = 1;
f = 2;
figure(3);
clf;
plot(t1,1*20e-2*20e-2*Y2.*Y2,'linewidth',3);
hold on;
plot(t1,1*9.81*20e-2.*(1-cos(Y1)),'r','linewidth',3);
plot(t1,E,'c','linewidth',3);
% plot(t1,X1-b,'linewidth',3);
hold on;
grid on;
plot(t2,Y1-b,'r','linewidth',3);
plot(t2,L+f,'linewidth',3);
plot(t2,D+f,'g','linewidth',3);
plot(t2,(E-D)+f,'k','linewidth',3); %net energy
plot(t1,Y1*0-b-0.3,'k','linewidth',1);
plot(t1,Y1*0-b+0.3,'k','linewidth',1);
plot(t1*0+24,3.5*Y1+1,'k','linewidth',1);
text(30,-b+0.4,'\downarrow Linear Range');
% text(12,0,'\leftarrow Time instant @ which non-linear to linear');
% text(12,-0.5,'transition takes place');
% % legend('Linear', 'Nonlinear')
legend('T','V','Total Energy','Linear', 'Nonlinear','Lagrangian','Dissipated Energy','Net Energy');
figure(4);
clf;
ED = E-D;
[pks,locs] = findpeaks(L);
[pks1,locs1] = findpeaks(ED);
[pks2,locs2] = findpeaks(D);
TF = islocalmin(L);
TF1 = islocalmin(ED);
TF2 = islocalmin(D);
plot(locs*0.01,pks,'b');
hold on;
plot(locs1*0.01,pks1,'r');
plot(locs2*0.01,pks2,'k');
grid on;
plot(t1(TF),L(TF),'b')
plot(locs1*0.01,pks1,'r');
plot(t1(TF1),ED(TF1),'r');
plot(t1(TF2),D(TF2),'k');
plot(t1*0+24,1*Y1+0,'k','linewidth',1);
legend('Lagrangian','Net Energy','Dissipated Energy');
function xdot = pend_l(~,x)
wsq = 13.56;
C = 0.1;
xdot = [x(2); -wsq*x(1)-C*x(2)];
end
function xdot = pend_n(~,x)
wsq = 13.56;
C = 0.1;
xdot = [x(2); -wsq*x(1)-C*x(2)];
end
So this appears to work, bit you have to care for the correct pend_n code to get correct results!
  5 comentarios
KLETECH MOTORSPORTS
KLETECH MOTORSPORTS el 4 de Nov. de 2020
KLETECH MOTORSPORTS
KLETECH MOTORSPORTS el 4 de Nov. de 2020
i tried the code you edited, and it worked. This is pend_solve.

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KLETECH MOTORSPORTS
KLETECH MOTORSPORTS el 4 de Nov. de 2020
  10 comentarios
KLETECH MOTORSPORTS
KLETECH MOTORSPORTS el 15 de Nov. de 2020
Thanks Rik, will check out.

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