How to resolve interp1 error?

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);

[Walter Roberson]

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;

Answer 1

[Jan]

% 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.

[SARNENDU JATI]

The result should give me a decaying ramp. I tired the first solution and it gave me the same error as the size of t is not same as x1_opt. I'm sorry I didn't mention t=0:0.001:4.452;. I tired changing the size but that didn't work.

[Jan]

Use the debugger to find out, which variables have which size. See:

t = 0:0.001:4.452;
size(t)
t_opt = linspace(0,4452*0.001,4452)'
size(t_opt)

I do not understand why you want to interpolate at all here. But the actual error seems to be simple. The message

 X and V must be of the same length.

is clear. So care for the sizes of these two inputs.

[SARNENDU JATI]

Yeah, I did. And it is still giving me the same error for some reason.

[Walter Roberson]

length(t) is 4453, not 4452.... just as I suggested might be the problem.

[Jan]

@SARNENDU JATI: You mentioned, that you cared for X and V having the same size. In the code you have posted this is not the fact and the error message tells you clearly that they have different sizes. So please explain, why you think, that these vectors have matching sizes?

Post the output of these commands:

t = 0:0.001:4.452;
size(t)
t_opt = linspace(0,4452*0.001,4452)'
size(t_opt)

Use the debugger also. Type e.g. this in the command window:

dbstop if error

Now run the code again until Matlab stops at the error. Now check the sizes of the inputs of interp1.

[SARNENDU JATI]

@Jan: This is the adjusted code

 function xinit = xinit_fcn(t)
 t_opt=linspace(0,4452*0.001,4453);
 Tf=4.453;
 t=0:0.001:4.452;
 x1_opt= -(pi/Tf)*t + pi;
 figure,plot(x1_opt)
x1_opt= -(pi/Tf)*t + pi;
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*

[Walter Roberson]

(I am away from my desk for a few hours and will not be able to test until later)

[Jan]

@SARNENDU JATI: Does it work now?

[Walter Roberson]

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*

Answer 2

[Walter Roberson]

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 .