Plotting acceleration vs time on Matlab from accelerometer data that doesn't have a time column

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Adviye Irem Yuceel
评论： Duncan Po 2021-6-9
Hello, I would like to plot x y and z axes of acceleration versus time from the csv file that was outputted from my sensor, the first two rows of the data of the csv file give the timestamp and the frequency of the reading therefore they should not be read by matlab and the columns represent x y and z axes respectively. The problem I am having is that the file does not have a time column and that I need to create an abstract one so that I could plot the acceleration vs time (or do I actually need one? is there a command that I can just use?) Thank you very much in advance!!
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Adviye Irem Yuceel
Adviye Irem Yuceel 2021-6-7
There are no timestamps unfortunetly, there is only one timestamp that gives the time data collection started and then the data is sampled at 32 Hz

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回答（2 个）

Mathieu NOE
Mathieu NOE 2021-6-8
hello
as we know the sampling rate Fs and the amount of samples , there is no big difficulty to reconstruct a time vector :
samples = length(data);
dt = 1/Fs;
time_vector = (0:samples-1)*dt
Duncan Po
Duncan Po 2021-6-8
You can create a timetable, and just supply the start time and frequency. Timetable will compute the timestamps for your data. For example,
>> data = (1:10)';
>> starttime = datetime('now');
>> fs = 2; % 2Hz in this example
>> tt = timetable(data, 'StartTime', starttime, 'SampleRate', fs)
tt =
10×1 timetable
Time data
____________________ ____
08-Jun-2021 14:59:30 1
08-Jun-2021 14:59:30 2
08-Jun-2021 14:59:31 3
08-Jun-2021 14:59:31 4
08-Jun-2021 14:59:32 5
08-Jun-2021 14:59:32 6
08-Jun-2021 14:59:33 7
08-Jun-2021 14:59:33 8
08-Jun-2021 14:59:34 9
08-Jun-2021 14:59:34 10
You can then either plot directly using stackedplot, or extract the timestamps using tt.Time, and then call plot.
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Adviye Irem Yuceel
Adviye Irem Yuceel 2021-6-8
I get this error message (some of my data values are negative because the sensor gives the opposite acceleration in negative):
Error using stackedplot (line 71)
Expected vars to be positive.
Error in stress (line 32)
stackedplot(tt,y)
Is there a way to get arround this? Also can I start the starttime from the time stamp given in the csv file with the data in unix format?
Duncan Po
Duncan Po 2021-6-9
To plot the timetable using stackedplot, you can simply pass in the timetable as the only input:
stackedplot(tt)
The error you encountered is due to the second input y. Apparently y is not a valid second input.
For unix time, there is a way to convert using the datetime function:
starttime = datetime(t, 'ConvertFrom','posixTime');

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