MATLAB Answers

cutting up the signal into repeating parts

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I was wondering how can I cut up the signal from the data into repeating parts(each containing a single action potential) and then plot all these action potentials on a single graph, at the end shift all these action potentials in time so they are aligned along the x-axis (time ). data is attached

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per isakson
per isakson 18 May 2018
"a single action potential" where does it start and end? Start: The signal is increasing and passes -40 or something of that kind?

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Star Strider
Star Strider 18 May 2018
편집: Star Strider 18 May 2018
I seem to have seen this waveform somewhere else recently!
Try this:
D = xlsread('Star11.xls');
tv = D(:,1);
ap = D(:,2);
[pks, locs] = findpeaks(ap, 'MinPeakHeight',100, 'MinPeakDist',20); % Determine Peaks & Indices
figure(1)
plot(tv, ap)
hold on
plot(tv(locs), pks, '+r')
hold off
grid
for k1 = 1:numel(locs)-1
apc{k1} = ap(locs(k1)-10:locs(k1+1)-10); % Define MUAP Frames
tvc{k1} = tv(locs(k1)-10:locs(k1+1)-10);
end
figure(2)
hold all
for k1 = 1:numel(apc)
plot(tvc{k1}-tvc{k1}(1), apc{k1}) % Plot MUAP Frames
end
hold off
grid
That should get you started.
Experiment to get the result you want.
EDIT Added plot:

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sara bitarafan
sara bitarafan 21 May 2018
Thank you :)
Star Strider
Star Strider 21 May 2018
As always, my pleasure!

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추가 답변(1개)

John BG
John BG 18 May 2018
편집: John BG 18 May 2018
Hi Sara Bitarafan
please have a look at the attached script. .
1.-
Acquiring signal:
clear all;clc;close all
A=xlsread('Star11.xls')
t=A(:,1)'
s=A(:,2)'
figure(1)
plot(t,s);grid on
Removing DC
min_s=min(s)
s=s+abs(min_s)
center signal, closest to dominant cycle to a sin cos signal
max_s=max(s)
s=s-max_s/2
hold on
plot(t,s);grid on;axis tight
For this sample there's not much DC, but it may the case that without removing DC the sought eye diagram is difficult to get.
2.
Calculating the dominant cycle with fft:
.
S=fft(s)
absS=abs(S);
figure(2);
plot(absS);grid on;
max_absS=max(absS);
n_maxS=find(absS==max_absS)
.
2nd value is just FFT mirror
n_maxS=n_maxS(1)
.
If n_maxS = max_amount_cycles this would be the highest discernible requency, with the FFT: it would be just 2 time samples per cycle.
max_amount_cycles=floor(numel(t)/2)
.
n_maxS: amount of cycles
.
hold on;plot(n_maxS,max_absS,'ro')
.
nT: amount of samples per dominant cycle:
nT=floor(numel(t)/n_maxS)
.
dominant cycle T
t1=t([1:nT])
T=t1(end)
amount of lost samples ignoring .2
mod(nT,n_maxS) % samples lost, not a crisis
.
instead of a for loop:
% sc=zeros(n_maxS,T)
% for ,,
s2=s([1:1:end-(numel(t)-n_maxS*nT)])
.
But the dominant cycle is not that constant.
figure(3);
for k=1:1:3 % n_maxS
plot(s2(k*[1:nT]))
hold on;
end
% 003
.
For the same basic cycle t(nT) we see 1,2 and 3 peaks
This is caused by a 2nd tone almost half way the dominant tone in amplitude and really close to the dominant.
fft approach is more reliable when there's a clear frequency peak but no high tones anywhere, particular near the dominant.
Yet the variable nT is going to be useful in next point.
Also, assuming the single action potential refers to a single peak
3.-
When the dominant cycle is not that constant we can use squelch:
Set a threshold that includes all peaks above a given amplitude.
close all
s=A(:,2)'
th1=100
figure(1);plot(t,s);grid on
[pks,locs]=findpeaks(s,'MinPeakHeight',th1)
hold on
plot(t(locs),pks,'bd')
avoid false peaks imposing min distance between found peaks, that may be for instance: nT
.
[pks,locs]=findpeaks(s,'MinPeakHeight',th1,'MinPeakDistance',floor(nT/2))
hold on
plot(t(locs),pks,'rd')
% 004
.
So far what can we assert about this signal and the 'potential' events you have mentioned in the question?
How often do the events take place
mean(diff(locs))
.
How much jitter suffer such events understanding jitter ~ standard deviation of locs
var(diff(locs))^.5
4.-
Plotting the centered diagram with all events:
Let be dt the + - span left and right of each peak event
dt=floor(.5*mean(diff(locs)))
And the eye diagram:
sc2=zeros(numel(locs),2*dt+1) % + - window span around each peak
for k=1:1:numel(locs)
if locs(k)<dt % 1st cycle, with 1st peak closer than dt to beginning of signal.
s0=s([1:locs(k)+dt])
sc2(k,:)=[zeros(1,2*dt+1-numel(s0)) s0]
end
if locs(k)>numel(t)-dt % last cycle with peak closer than dt to end of signal.
sc2(k,:)=s([locs(k)-dt:end])
end
if locs(k)<numel(t)-dt && locs(k)>dt
sc2(k,:)=s([locs(k)-dt:locs(k)+dt])
end
end
% 4. and the eye diagram is:
figure(2);
for k=1:1:numel(locs)
plot(sc2(k,:))
hold on
end
grid on
% 005
.
.
Sara
if you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?
To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link
thanks in advance for time and attention
John BG

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sara bitarafan
sara bitarafan 21 May 2018
thank you for your answer. it it very well explained and neat.
John BG
John BG 21 May 2018
Thanks Sara

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