How can I extracting a signal pulse from a data

I am basic user of in such coding, I Will be grateful for the help. I have a data file (measured signal) that contains few pulses. I want to extract the first pulse (which is also the highest pulse) from this measured signal and save it as another file for further use. The example is shown in the attachment.I have 50 data files ((measured signal).

I want to extract all the first pulses from these 50 measured signal. There is not very hard requirement for the number of samples extracted related to this pulse, may be start to end (rising zero to falling zero etc) or any thing like this.

[dpb]

Do you have Signal Processing TB? If so, findpeaks will be quite useful here...let us know before we invest time elsewhere.

Answer 1

[dpb]

"extract the first pulse (which is also the highest pulse)"

Presuming you can be assured of the assumption, the first is the max, then it's pretty simple...

[pk,ipk]=max(s);                            % largest peak, location
ilo=find(s(1:ipk)<pk/2,1,'last');           % lower half-max location
ihi=find(s(ipk:end)<pk/2,1,'first')+ipk;    % upper half-max location
pkS=s(ilo:ihi);

ADDENDUM

With the second pdf that looks like baseline-crossing is the desired result,

load('Signal.mat');
[pk,ipk]=max(s);                            % largest peak, location
BL=0;                                       % set baseline level
ilo=find(s(1:ipk)<=BL,1,'last');            % lower boundary location
ihi=find(s(ipk:end)<=BL,1,'first')+ipk-1;   % upper boundary location
pkS=s(ilo:ihi);

See what are results--

>> [ilo,ihi]
ans =
   613    682
>> s([ilo ihi]).'
ans =
  -8.9407e-12        -0.0004
>>

This is within one element of your location, you can modify the test somewhat by looking for within some eps of BL to get nearest to BL or even interpolate for fractional location or whatever it is thought to be most suitable result.

[Image Analyst]

I think you mean /4 or so, since 2 is about a quarter of 8, not half of it.

[dpb]

I just did FWHM (approx) IA; OP can use whatever thresholding desired.

The idea is to locate crossings from the peak with the assumption(*) as stated that the first peak that is desired is known a priori to be the largest.

(*) It's also presumed there will be sufficient undershoot to isolate the peak uniquely.

As for the thresholding, for the real signal I'd probably compute the pretrigger mean and select some rise over it in actual use.

It's these things that make findpeaks handy, it's got lots of such selective criteria already coded-up so can just set some input parameters judiciously and not worry too much about "how"...

[Image Analyst]

Yes, findpeaks() is indeed powerful, but it's not so easy to figure out what all those parameters mean. It's so complicated, it confuses even me. What I like about your approach is that it's very simple so anyone can understand it in seconds. It's slightly different from my two approaches in general, but for this data set, they should all give the same results, though for others they won't.

[dpb]

Indeed, it is simplicity itself; of course it gets that simplicity almost 100% by relying on the assumption regarding the shape of the waveform. :)

I'll agree with the complexity of findpeaks; over the years I've used it enough I've gradually come to grips with the interface so I may underestimate the level of effort to find the "judicious" input parameters for those who come to it for the first time or only on occasion.

[Khan Engr]

Thanks dpb for your attention and help. I have used the code you mentioned and first I get below;

[pk,ipk]=max(s);                        % largest peak, location
ilo=find(s(1:ipk)<mx/2,1,'last');       % lower half-max location
ihi=find(s(ipk+1:end)<mx/2,1,'first');  % upper half-max location
pkS=s(ilo:ihi);
Undefined function or variable 'mx'.

I assume by 'mx' you mean 'max'? doing this i got as below;

[pk,ipk]=max(s);                        % largest peak, location
ilo=find(s(1:ipk)<max/2,1,'last');       % lower half-max location
ihi=find(s(ipk+1:end)<max/2,1,'first');  % upper half-max location
pkS=s(ilo:ihi);

Error using max

Not enough input arguments.

[Image Analyst]

And where did you post s? I can't find it attached to any of your postings here. Where did you put it?

[dpb]

Oh, another case where I changed variable names to try to be more readable and missed a location where the old name was used.

'mx' formerly was what 'pk' is now; the value returned by max().

I'll edit Answer to correct.

[Khan Engr]

Thanks for updating, I have used the updated code,

[pk,ipk]=max(s);                        % largest peak, location
ilo=find(s(1:ipk)<pk/2,1,'last');       % lower half-max location
ihi=find(s(ipk+1:end)<pk/2,1,'first');  % upper half-max location
pkS=s(ilo:ihi);

and the outcome is;

ipk     638
ihi     16
ilo     624
pk      0.0112
pks     []

Still not working as required

What I am looking for is attached in PDF and I have also attached the data file I am using.

[dpb]

There's a "+ipk-1" missing on the location for "ihi"; find is counting from the peak location plus one so the absolute location is from that base point.

You'll have to decide about a thresholding level; as noted in the comments with IA, I just illustrated the search mechanism using FWHM as an example.

If you want more of the peak than that (the pdf is nebulous as to a real definition), you'll have to use a baseline estimate or someother level.

Again, being able to set other parameters in the selection besides a fixed threshold is why I asked about SP Toolbox and findpeaks initially; this is a very simple-minded searching algorithm.

[Khan Engr]

Thank you very much dpb. Your modified code (below) is working well now :)

load('Signal.mat');
[pk,ipk]=max(s);                            % largest peak, location
BL=0;                                       % set baseline level
ilo=find(s(1:ipk)<=BL,1,'last');            % lower boundary location
ihi=find(s(ipk:end)<=BL,1,'first')+ipk-1;   % upper boundary location
pkS=s(ilo:ihi);

Best regards

Answer 2

[Image Analyst]

This should do it

mask = signal > 2e-3;  % Threshold the signal to find regions higher than 2e-3.
% If you get more than one region, use the code below to get only the largest region, otherwise skip it.
mask = bwareafilt(mask, 1);
maskedSignal = signal(mask);

[Image Analyst]

If you want the first peak, regardless of how wide it is, use this

mask = signal > 2e-3;  % Threshold the signal to find regions higher than 2e-3.
% If you get more than one region, use the code below to get only the largest region, otherwise skip it.
labeledPeaks = bwlabel(mask);
mask = ismember(labeledPeaks, 1); % Extract peak #1 only.
maskedSignal = signal(mask)

[Khan Engr]

Image Analyst, may be Image Processing Toolbox is missing in my matlab, the outcome of the code applied is;

 mask = S > 2e-3; 
>> mask = bwareafilt(mask, 1);
maskedSignal = signal(mask);
To use 'bwareafilt', you might need:
  bwareafilt - Image Processing Toolbox

[Image Analyst]

Yep, you're missing that toolbox.

[Khan Engr]

Thanks for your help, I will try to update my Matlab for IPT.