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How to resolve interp1 error?

Asked by SARNENDU JATI on 12 Jun 2018
Latest activity Commented on by Walter Roberson
on 14 Jun 2018
function xinit = xinit_fcn(t)
t_opt=linspace(0,4452*0.001,4452)';
Tf=4.4520;
x1_opt= -(pi/Tf)*t + pi;
x2_opt=[0,0];
x3_opt=[0,0];
x4_opt=[0,0];
xinit = [interp1(t_opt,x1_opt,t);
interp1(t_opt,x2_opt,t);
interp1(t_opt,x3_opt,t);
interp1(t_opt,x4_opt,t);] ;
I'm getting the error:
Error using interp1>reshapeAndSortXandV (line 423)
X and V must be of the same length.
Error in interp1 (line 92)
[X,V,orig_size_v] = reshapeAndSortXandV(varargin{1},varargin{2});
Error in void>xinit_fcn (line 51)
xinit=[interp1(t_opt,x1_opt,t);

  1 Comment

One thing (but probably not the reason for the error) is I suspect you want
t_opt=linspace(0,4452*0.001,4453)';
instead of 4452 there. Better yet would be
t_opt = (0:4452)/1000;

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2 Answers

Answer by Jan
on 12 Jun 2018

% Failing:
t_opt = linspace(0,4452*0.001,4452)';
x1_opt = -(pi/Tf)*t + pi;
xinit = interp1(t_opt, x1_opt, t);
Now x1_opt has the same number of elements as t. As the error message tells, interp1 needs the 1st and 2nd input to have the same number of elements, not the 2nd and the 3rd. So maybe you want:
xinit = interp1(t, x1_opt, t_opt);
Or
x1_opt = -(pi/Tf) * t_opt + pi;
Perhaps the hard coded limit of 4452 is a bad idea and you need to let the limits depend on the length of the input t.
The solution is hard to guess, because you did not mention the intention of the code.

  9 Comments

(I am away from my desk for a few hours and will not be able to test until later)
Jan
on 14 Jun 2018
@SARNENDU JATI: Does it work now?
You have
x2_opt=0;
x3_opt=0;
x4_opt=0;
xinit=[interp1(t_opt,x1_opt,t); %x1*
interp1(t_opt,x2_opt,t); %x2*
interp1(t_opt,x3_opt,t); %x3*
interp1(t_opt,x4_opt,t);] ; %x4*
Your x2_opt, x3_opt, and x4_opt are each scalar 0, not zeros the same size as t_opt. And with them being constant 0, there is little point doing interpolation: just output 0 without interpolation.
x2_opt=0;
x3_opt=0;
x4_opt=0;
xinit=[interp1(t_opt,x1_opt,t); %x1*
x2_opt; %x2*
x3_opt; %x3*
x4_opt;] ; %x4*

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Answer by Walter Roberson
on 14 Jun 2018

Your xinit_fcn is invoking interp1 with an x1_opt value derived from the t value that is passed into the function. However, bvpinit invokes the function with individual t values, not with the entire list of t values, so x1_opt will be a scalar. You are then trying to interp1(1 x 4452, 1 x 1, 1 x 1) . That fails because the second parameter must be the same length as the first parameter.
If you did have the entire original list of time values passed in, and were calculating x1_opt= -(pi/Tf)*All_t + pi and wanting to interpolate at the particular time, t, that was passed in to the xinit_fcn, then the result would always be the same as -(pi/Tf)*t + pi where t is the individual t. So it is not clear why you do not just have your xinit_fcn return -(pi/Tf)*t + pi .

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