How can I determine the start and end of a movement?

Hello Everyone, I am trying to determine the beginning and the end of a movement , using matlab. So I have as ensr, that I used to collect data for specific movement. But before the actual movement, there is some time, where I the actual movement is not present yet, but the waiting position.

How can I get rid of that ? Any algorithm to find the start and end of a movement?

I would be greateful for any help

Answer 1

[Image Analyst]

Just crop it off starting at some index:

signal = signal(index:end)

To find out what index is, we'd need to know your signal, and your movement signal (if they're different signals). Attach it/them in a .mat file with the paper clip icon.

[Lala0099]

Hey, thank you for your answer. I am attaching the data to this comment. Thank you for helping me

[Image Analyst]

What do the columns represent? Which column is the movement number, and which is the signal you want to crop based on the movement signal?

[Lala0099]

The coloumns represent the following: coloumn1 - Time, Coloumn 2 - acceleration in X direction. Coloumn2- acceleration in Y direction. Coloumn 3 - acceleration in Z direction. Coloumn 4 - Xdirection(position), Coloumn 5 -Y dir. , coloumn 6 - z dir.

There is only one single movement, that is represented by those coloumns ( direction )

I need to crop off the values i dont need in all the coloumns.

Did I answer your question?

Thank you for helping me.

[Image Analyst]

I've done this before for an instrument. Give me some time today to see if I can adapt it to your data.

[Lala0099]

Okay, Thank you very much!

[Image Analyst]

See if the attached file works for you. I get a starting point for each of the 6 curves. You might want to get just one by taking the average of them, or taking the latest one or whatever.

[Lala0099]

Thank you so much!

there is one bit that i really dont get.

Why are those both defined by 't'?

isnt the second line oerrites the first one?

t = data(:, 1);
t = 1 : length(z);

[Image Analyst]

You can use the first one if you want. I just used the second one because when I was debugging it, it was easier to see where I was on the x axis when I used the index instead of the time.

You might also look at the findchangepts() function.

[Lala0099]

Image Analyst,

I found a problem with the code. It does not use certain lines within the function, hence does not find timePoint3.

I cannot find the reason why it cannot execute those lines.

Could you have a look?

[maxDeviation, timePoint3] = max(meanDeviationFromMean);