Non linear pendulum code

조회 수: 9 (최근 30일)
KLETECH MOTORSPORTS
KLETECH MOTORSPORTS 2020년 11월 3일
댓글: KLETECH MOTORSPORTS 2020년 11월 15일
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

채택된 답변

Stephan
Stephan 2020년 11월 3일
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!
  댓글 수: 5
KLETECH MOTORSPORTS
KLETECH MOTORSPORTS 2020년 11월 4일
KLETECH MOTORSPORTS
KLETECH MOTORSPORTS 2020년 11월 4일
i tried the code you edited, and it worked. This is pend_solve.

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추가 답변 (1개)

KLETECH MOTORSPORTS
KLETECH MOTORSPORTS 2020년 11월 4일
  댓글 수: 10
KLETECH MOTORSPORTS
KLETECH MOTORSPORTS 2020년 11월 15일
Thanks Rik, will check out.

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