error calling a function

5 ビュー (過去 30 日間)
Susan Santiago
Susan Santiago 2018 年 5 月 13 日
コメント済み: Susan Santiago 2018 年 5 月 13 日
function f = Mu(t,y,ti,tf,n)
m1 = 55;
m2 = 400;
m3 = 100;
k1 = 230000;
k2 = 30000;
k3 = 50000;
k4 = 3000;
n2 = 1500;
n3 = 4000;
n4 = 700;
tm = (ti+tf)/2;
t1 = tf/4;
L0 = 5;
v = 15;
A = 0.03;
M =[0 1 0 0 0 0;
-(k1+k2)/m1 -(n2+n4)/m1 k2/m1 n4/m1 0 n4/m1;
0 0 0 1 0 0;
k2/m2 n2/m2 -(k3+k2)/m2 -(n3+n2)/m2 k3/m2 n3/m2;
0 0 0 0 0 1;
k4/m3 n4/m3 k3/m3 n3/m3 -(k3+k4)/m3 -(n3+n4)/m3];
switch n
case 1
u = (A/2)*(1-cos(2*pi*v*t/L0));
case 2
u = 1*(t<tm)+(-1)*(t>=tm);
case 3
u = 0*(t<tm)+1*(t>=tm);
case 4
u = 1*(t<tm)+0*(t>=tm);
case 5
u = (A*t)*(t<=tm)+0*(t>tm);
case 6
u1 = A*t;
u2 = A*(t-2*tm);
u = u1.*(t<=tm)+u2.*(t>tm);
case 7
u1 = A*t;
u2 = A*(-t+4*t1);
u3 = A*(-t+3*t1);
u = u1.*(t<=t1)+u2.*((t>=t1)&(t<=2*t1)) +u3.*(t>=2*t1);
case 8
u = A.*((t>=t1)&(t<=2*t1));
vect=[0;-k1*u/m1;0;0;0;0];
f = M*y+vect;
end
here is the main file where i'm trying to call the above function
% initial conditions
IC=[0 0 0 0 0 0];
%initial time
ti=0;
%final time
tf=20;
%interval where the problems can be solved
tspan=[ti,tf];
%function Matrix*Y+u(t)
fMu=@(t,y) Mu(t,y,ti,tf,1);
[t,y]=ode45(fMu,tspan,IC)
The error I'm receiving says "Output argument "f" (and maybe others) not assigned during call to "Mu"." I'm not sure what that means or how to fix it. Please help
  2 件のコメント
Jan
Jan 2018 年 5 月 13 日
From the view point of numerical maths it is a professional blunder to use ODE45 to integrate a non-smooth function. See http://www.mathworks.com/matlabcentral/answers/59582#answer_72047. It is like using a drilling machine to beat a screw into the wall. Of course, it is in afterwards and you get a final number. But you have driven the tool against the specifications and have no control if the result is dominated by rounding errors.
The stable way is to run the integration in steps over the smooth intervals and to use the final value of one step as initial value of the next one.
Susan Santiago
Susan Santiago 2018 年 5 月 13 日
Thank you for the advice although what I'm working on is a project in which I was specifically asked to solve this function with this method. I will remember this for the future, though

サインインしてコメントする。

採用された回答

Stephan
Stephan 2018 年 5 月 13 日
Hi Susan,
you should finish your switch case command with an end command.
function f = Mu(t,y,ti,tf,n)
m1 = 55;
m2 = 400;
m3 = 100;
k1 = 230000;
k2 = 30000;
k3 = 50000;
k4 = 3000;
n2 = 1500;
n3 = 4000;
n4 = 700;
tm = (ti+tf)/2;
t1 = tf/4;
L0 = 5;
v = 15;
A = 0.03;
M =[0 1 0 0 0 0;
-(k1+k2)/m1 -(n2+n4)/m1 k2/m1 n4/m1 0 n4/m1;
0 0 0 1 0 0;
k2/m2 n2/m2 -(k3+k2)/m2 -(n3+n2)/m2 k3/m2 n3/m2;
0 0 0 0 0 1;
k4/m3 n4/m3 k3/m3 n3/m3 -(k3+k4)/m3 -(n3+n4)/m3];
switch n
case 1
u = (A/2)*(1-cos(2*pi*v*t/L0));
case 2
u = 1*(t<tm)+(-1)*(t>=tm);
case 3
u = 0*(t<tm)+1*(t>=tm);
case 4
u = 1*(t<tm)+0*(t>=tm);
case 5
u = (A*t)*(t<=tm)+0*(t>tm);
case 6
u1 = A*t;
u2 = A*(t-2*tm);
u = u1.*(t<=tm)+u2.*(t>tm);
case 7
u1 = A*t;
u2 = A*(-t+4*t1);
u3 = A*(-t+3*t1);
u = u1.*(t<=t1)+u2.*((t>=t1)&(t<=2*t1)) +u3.*(t>=2*t1);
case 8
u = A.*((t>=t1)&(t<=2*t1));
end
vect=[0;-k1*u/m1;0;0;0;0];
f = M*y+vect
end
I guess what you wanted is shown above. Like you did it, f only gets a value if n=8.
In your example n=1 - that is the reason for f getting no value.
Best regards
Stephan
  2 件のコメント
Susan Santiago
Susan Santiago 2018 年 5 月 13 日
That was my issue, thank you! It's my first time using a switch function
Stephan
Stephan 2018 年 5 月 13 日
My pleasure

サインインしてコメントする。

その他の回答 (1 件)

Ngoc Thanh Hung Bui
Ngoc Thanh Hung Bui 2018 年 5 月 13 日
編集済み: Ngoc Thanh Hung Bui 2018 年 5 月 13 日
ode45(fMu,tspan,IC)
try ode45(@fMU,...) or ode45('fMU',...) instead
  1 件のコメント
Susan Santiago
Susan Santiago 2018 年 5 月 13 日
I tried the first option you suggested and got this message ""fMu" was previously used as a variable, conflicting with its use here as the name of a function or command."

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeData Type Conversion についてさらに検索

Community Treasure Hunt

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

Start Hunting!

Translated by