Recognise specific pattern in timetable

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Mads Petersen
Mads Petersen on 9 May 2022
Answered: Mads Petersen on 9 Jul 2022
I have a timetable logfile from an electrical motor.
This motor runs a variable rounds per minute, according to manual adjustment.
if i plot the entire timeline vs rpm i get what like in attached picture "1".
The intresting part of that data is where you see the rpm drops from ~2250 rpm to ~1300rpm. please see picture "2".
Is it possible to make a code that is recognising this specific pattern automatically, and isolate the time where this is happening? in this case it must be isolated from ~60s to ~100s to isolate this specifik range.
Mads Petersen
Mads Petersen on 10 May 2022
Thank you that is a very cool function i did not know. Do you have experience with it? got some pattern recognition, but a lot of failures also. I dont think the matlab documentation i explaining clear enough what the different steps do.
i made a vector that look a lot like the pattern i am intrested in. see image3
got pattern script setup to find pattern in data. see here:
looks promising, but is finding more pattern than it should. following matlab documentation more. looks better
i still have a lot of errors. Specially when finding pattern of different log files, where typical the time and lograte (hz) is changed. Rpm pattern starting from ~2200 and down to ~1400 is always the same. see code:
dt = data;
dt(t>24&t<25) = 1900;
dt(t>25&t<26) = 800;
findsignal(dt,signal,'TimeAlignment','dtw','Metric','absolute') %Let the x-axes stretch if the stretching results in a smaller absolute distance between the closest data segment and the signal.
dt(t>0&t<0.1) = 2400;
dt(t>0.1&t<0.12) = 1900;
Can yo explain more on how this is working?

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Answers (3)

Mitch Lautigar
Mitch Lautigar on 10 May 2022
After you grab the data, run a simple check to see if the rpm's drastically decrease. This can look something like the following
%rpms_val is the name i'm use for your "y" data points.
%rpms_time is the name i'm going to use for your "x" data points.
rpms_slope = [];
for i = 1:length(rpms_val)-1
rpms_slope(i) = (rpms_val(i+1) - rpms_val(i)) / (rpms_time(i+1)-rpms_time(i));
rpms_check = find(rpms_slope < -5) %-5 can be whatever tolerance you want.
from here, you can code in any flags you wish. You can even change the color of the data that is negative.
Mitch Lautigar
Mitch Lautigar on 16 May 2022
rpms_slope(i) = (rpms_val(i+1) - rpms_val(i)) ./ (rpms_time(i+1)-rpms_time(i))
forgot it was vector math. Should fix your issue.

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Les Beckham
Les Beckham on 10 May 2022
Edited: Les Beckham on 10 May 2022
One approach without requiring the Signal Processing Toolbox:
load('answers.mat') % an approximation of your data
x = data(:,1);
y = data(:,2);
startidx = find(y>2000, 1, 'first'); % you may need to adjust this
% The following line may also need to be adjusted if the desired region is
% not strictly decreasing
stopidx = find(diff(y(startidx:end))>1, 1, 'first') + startidx - 1
stopidx = 26
hold on;
plot(x(startidx:stopidx), y(startidx:stopidx), 'r.')

Mads Petersen
Mads Petersen on 9 Jul 2022
Thank you very much for all the nice answers. All of them are very intresting. i went with the diff. solution which i simple and works for now.
I thnik the findsignal is more intresting but required more fine tuning.
Thanks everyone


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