Error using interp1>re​shapeAndSo​rtXandV (line 424) X and V must be of the same length.

Hello,

I geht this error message when I try to derive my acceleration of the car. I think there is a size problem, but I don't know how to solve it.

The data is measured from a car.

Answer 1

[Stephen Cobeldick]

Assuming that meas.acc.data and meas.acc.time are vectors with the same number of elements, then diff on the data vector will return a numeric vector with one element fewer than the time vector, so meas.jerk.data and meas.jerk.time will likely have mismatching sizes.

One easy solution is to take the time as the mean of adjacent time points, e.g.:

meas.jerk.time = (meas.acc.time(1:end-1) + meas.acc.time(2:end)) ./ 2;

[Megan]

I tried it like this:

disp(meas.acc.data)
disp(gradient(meas.vel.data))

It should be the same values but it's not. So gradient is not correct in my case.

-Why did you change your answer should I try it again with diff?

[Megan]

I dont get your answer :/ Can you please help me out

[Stephen Cobeldick]

"Why did you change your answer"

I deleted my other answer because it was unlikely to give you the result that you expect.

"I dont get your answer"

Because you used diff your data vector is one element shorter than your time vector, so essentially you need to think of a way to match your time point to the new (shorter) data vector. There is no single solution to this, but something like this is easy to do:

1. average adjacent time points (what I showed in my answer).
2. ignore the first time point.
3. etc.

Note that I cannot tell you which to select because which method is appropriate depends on the features and properties of your data and your data analyses.

[Megan]

meas.jerk.time = (meas.acc.time(1:end-1) + meas.acc.time(2:end)) ./ 2;

Could you maybe explain this part of the code? I didn't get it.

What makes 1:end-1? What is end and what do you mean with it?

Thanks!

I got, that my vektor is smaler then my time vektor. I got this part. How do you get the average adjacent time points? I mean could you explain this little code :) I'm really new with matlab I'm sorry

[Stephen Cobeldick]

"What is end and what do you mean with it?"

end is the index of the last element in that vector.

https://www.mathworks.com/help/matlab/ref/end.html

Because end is the index of the last element, end-1 is the index of the second-to-last element. Thus

meas.acc.time(1:end-1)

uses basic MATLAB indexing to return a subvector of meas.acc.time, from its first element to its second-to-last element. Thus the subvector has one fewer element than meas.acc.time, and therefore exactly the same number of elements as the output of diff.

"How do you get the average adjacent time points?"

Simply by using the definition of arithmetic mean of two values:

https://en.wikipedia.org/wiki/Arithmetic_mean

which is commonly written as (val1 + val2) / 2

Compare with my answer, which has the form (vec1 + vec2) ./ 2 , where vec1 is the subvector from the first element to the second-to-last element, and vec2 is the subvector from the second element to the last element. Then these two vectors are added together (i.e. corresponding elements are summed), and then divided by two to give the arithmetic mean.

V          % [a,b,c,d,e]
V(1:end-1) % [a,b,c,d]
V(2:end)   % [b,c,d,e]
 V(1:end-1)+V(2:end)    % [a,b,c,d]+[b,c,d,e] = [a+b,b+c,c+d,d+e]
(V(1:end-1)+V(2:end))/2 % [a+b,b+c,c+d,d+e]/2 = [(a+b)/2,(b+c)/2,(c+d)/2,(d+e)/2]

Tip: to use MATLAB you need to learn how to use indexing, and work with vectors/arrays.

Note that you do no have to use the arithmetic mean. As I wrote in my answer, only you can decide if such an averaging is appropriate, or if another type of averaging would be better.

[Megan]

Thank you very much!