Error while creating an EDF file

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While creating an EDF file in matlab form a matrix, matlab return an error which doesn't make sense
%getting the name from the original file
Name = string(Infos.OriginalName(1:end-4))+ ".edf";
%Extracting the data of the signal in the proper form
sigdata = [traces(1,:)', traces(2,:)', traces(3,:)'];
%getting the sampling frequency of my recordings
fs = 1000;
% t = 0:1/fs:(size(sigdata,1)-1)/fs; I used that to plot the data
%making the header for the EDF file
hdr = edfheader("EDF+");
hdr.Recording = string(Infos.RecordingDate);
hdr.StartDate = string(datetime(string(Infos.RecordingDate(1:11)),'InputFormat','dd-MMM-yyyy', 'Format','dd.MM.yy'));
hdr.StartTime = string(datetime(string(Infos.RecordingDate(13:end)),'InputFormat', 'HH:mm:ss','Format','HH.mm.ss'));
hdr.NumDataRecords = 1;
hdr.DataRecordDuration = seconds(length(sigdata(:,1))/fs);
hdr.NumSignals = 3;
hdr.SignalLabels = ["Parietal_1" "Parietal_2" "Reference"];
hdr.PhysicalDimensions = repelem("uV",3);
%% this is where my problem is
hdr.PhysicalMin = min(sigdata);
hdr.PhysicalMax = max(sigdata);
hdr.DigitalMin = [-32768 -32768 -32768];
hdr.DigitalMax = [32767 32767 32767];
EEG_signal = edfwrite(Name,hdr,sigdata,'InputSampleType',"physical");
Everything is fine until a run the last line. Matlab give me this error:
Error using signal.internal.edf.write.validateFieldBytes (line 50)
Each value of "PhysicalMin" field must be less than or equal to 8 bytes. See edfheader documentation for more information.
Error in edfwrite/createFile (line 1667)
signal.internal.edf.write.validateFieldBytes(value, fieldIndex, ...
Error in edfwrite (line 493)
[obj, fileInfo] = createFile(obj, filename, hdr,...
I don't understand because my values in the hdr.PhysicalMin and hdr.PhysicalMax are doubles. when I run the "whos" function on the individual value they are 8 bytes each. Any idea how I could solve the problem ?
  5 个评论
Hassan Farhat
Hassan Farhat 2021-11-5
Hi Kartheek Akurati
It really depends on the amplitude range of your recorded EEG.
int16() is limited between [-32768 & 32767], anynumber outside this range will be threshold to these maximum limits
ex: int32(65000) = 32767.
So if your signal is higher in amplitude, you need to use higher bits (32) (64 bits is not working)
uint32 is limited between [0 & 65535]
Threfore, if you want to stick to 16 bits you need to rescale your inputs to be limited within the above ranges.
Best Regards
Hassan Farhat
Maxime Alessandri
Hello,
I'm sorry I never had the oportunity to answer to you.
I found a solution:
My aquisition software record in volts so my hdr.PhysicalMin and hdr.PhysicalMax were like 0.0005 (whatever). So when I turned them in int32 it was just 0.
The way I managed was to scale the signal by 1000000 so that I was in microvolts. Then the int32 worked like a charm.
Thank you all for your help.
Maxime

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Federico Del Gallo
I'm having the same problem. Did you find something to fix this error?
  1 个评论
Maxime Alessandri
编辑:Maxime Alessandri 2022-4-4
Hello,
I found the solution, please see my comment above,
Cheers, Maxime

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