Finding specific peaks and valleys
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I have a periodic signal with 6 peaks and 6 valleys. I am able to find all 6 peaks and all 6 valleys using the code below:
% plot periodic signal
plot(angle,signal)
hold on;
% fit to avoid noisy oscillations and find 6 global peaks
f = fit(angle,signal,'sin9');
yfitted = feval(f,angle);
plot(angle,yfitted,'r')
% find peaks and corresponding angles
[ypk0,idx0] = findpeaks(yfitted);
peaks = signal(idx0);
max_angles = angle(idx0);
% invert fitted signal to find 6 global valleys and corresponding angles
yfittedinv = max(yfitted) - yfitted;
[ypk,idx] = findpeaks(yfittedinv);
valleys = signal(idx);
min_angles = angle(idx);
Furthermore, I am able to find the global minimum (i.e., the lowest valley) using this line:
% find the global minimum
minangle = min_angles(valleys == min(valleys));
Both the arrays "max_angles" and "min_angles" have 6 elements. The question is how to reduce the number of elements in the array "min_angles" from 6 to 3, where I will have only the global minimum and the 2 non-neighbor valleys. The attached pic should help. I am able to find the green circle, but I also need to find the 2 red circles. Note that the global valley (green circle) is not always in the middle. It can be the first element of the array "min_angles", or it can be the last one. In the picture below, it's number 4.
2 Comments
Image Analyst
on 30 Jan 2025
Make it easy for us to help you, not hard. Attach your variables "angle" and "signal" in a .mat or text file. That way we'll have something to work with and won't have to try to create data on our own.
save('answers.mat', 'angle', 'signal');
If you have any more questions, then attach your data and code to read it in with the paperclip icon after you read this:
Aram Zeytunyan
on 31 Jan 2025
Accepted Answer
Star Strider
on 30 Jan 2025
Edited: Star Strider
on 30 Jan 2025
To find specific valleys, one option might be to use the MinPeakProminence value. The prominences are the tthird output returned by findpeaks, so you can look at those to see iif that option would work.
Another option is to get the peak values and use the mink (or maxk) function to return thee three lowest values (since that appears to be what the green circle and the red circles represent). Return both outputs from mink beecausee the second will be the index values of the minima (or maxima, depending on you you have the results returned), and you can use those indices with the returned values for the peaks and their locations to isolate the ones you want.
This is based on the figure you posted.
.
EDIT — (30 Jan 2025 at 21:33)
Example —
a = linspace(0,360).';
s = sin(2*pi*a/360*6)-sin(2*pi*a/720);
[vlys,vlocs] = findpeaks(-s)
[minpks, minpksidx] = mink(-vlys,3)
figure
plot(a, s)
hold on
plot(a(vlocs(minpksidx)), -vlys(minpksidx), 'rs')
hold off
grid
ylim([-2.5 1])
.
8 Comments
Aram Zeytunyan
on 31 Jan 2025
Edited: Aram Zeytunyan
on 31 Jan 2025
This is extemely helpful, thank you for the help. Let me try to apply it in my data analysis and get back to you. But I am suspecting that it will work for some cases and won't work for the others. For example, in the case below it won't allow to find the three green circles (which is my goal). One of the valleys found this way will probably be the red circle.
Star Strider
on 31 Jan 2025
My pleasure!
The findpeaks function has several options, among which is MinPeakDistance and.MiinPeakHeight
What are your criteria for identifying the peaks? Understanding that would help significantly.
Aram Zeytunyan
on 31 Jan 2025
If you want to find and isolate the three lowest valleys, mink should be able to do that. The peaks in my earlier eexample were vicinal, however that is not required.
Example —
a = linspace(0,360).';
angle_divisor = randi([90 360])
s = sin(2*pi*a/360*6)-sin(2*pi*a/angle_divisor);
[vlys,vlocs] = findpeaks(-s)
[minpks, minpksidx] = mink(-vlys,3)
figure
plot(a, s)
hold on
plot(a(vlocs(minpksidx)), -vlys(minpksidx), 'rs')
hold off
grid
ylim([-2.5 max(ylim)])
.
Aram Zeytunyan
on 31 Jan 2025
Considering the information in your original question, mink seems an appropriate approach.
I still don’t understand what your criteria are for identifying the peaks and valleys of interest, and your code doesn’t illuminate that.
clear
close all
% [fn,pn] = uigetfile('*.csv;*.mat;*.xlsx*','Select saved data file','MultiSelect', 'on');
% [fn,pn] = uigetfile('*.csv;*.mat;*.xlsx*','Select saved data file');
% cd(pn);
files = dir('*.csv')
k = 1;
% for k = 1:numel(cellstr(fn))
% if numel(cellstr(fn))>=2
% loaddata = importdata([pn,char(fn(k))]);
% signal(:,k) = loaddata(:,2);
% else
% loaddata = importdata([pn,char(fn)]);
% signal = loaddata(:,2);
% end
% end
% if numel(cellstr(fn))>=2
% for k = 1:numel(cellstr(fn))
% loaddata = importdata([pn,char(fn(k))]);
loaddata = readmatrix(files.name)
signal(:,k) = loaddata(:,2);
angle(:,k) = loaddata(:,1);
plot(angle(:,k),signal(:,k),'LineWidth',1);
hold on;
% end
% else
% loaddata = importdata([pn,char(fn)]);
% angle = loaddata(:,1);
% signal = loaddata(:,2);
% plot(angle,signal)
% hold on;
% end
% angle = loaddata(:,1);
% app.angle = angle;
% app.signal = signal;
%
% signal_test = app.signal;
% size(signal_test,2);
% numel(cellstr(fn));
% plot(angle,signal)
% hold on
% [globalmin,globalminidx] = min(signal)
% globalminangle = angle(signal == globalmin)
% globalminangle1 = angle(globalminidx)
% plot(angle,signal)
f = fit(angle,signal,'sin9');
yfitted = feval(f,angle);
plot(angle,yfitted,'r')
xlabel('angle (deg)')
ylabel('signal (mV)')
xlim([0 360])
hold on;
[ypk0,idx0] = findpeaks(yfitted);
peaks = signal(idx0);
max_angles = angle(idx0);
yfittedinv = max(yfitted) - yfitted;
% plot(app.UIAxes_2,angle,yfitted,'r','LineWidth',1);
[ypk,idx] = findpeaks(yfittedinv);
% [ypk1,idx1] = mink(ypk,3);
% valleys = yfitted(idx);
valleys = signal(idx);
min_angles = angle(idx);
minangle = min_angles(valleys == min(valleys));
minvalley = valleys(valleys == min(valleys));
idx2 = idx(valleys == min(valleys));
idx3 = find(idx == idx2);
if rem(idx3, 2) == 0 % global min index is even
idxv = [2 4 6];
if yfitted(2)-yfitted(1)>0 % curve is ascending
idxp = [2 3 4 5 6 1];
else % curve is descending
idxp = [1 2 3 4 5 6];
end
else % global min index is odd
idxv = [1 3 5];
if yfitted(2)-yfitted(1)>0 % curve is ascending
idxp = [1 2 3 4 5 6];
else % curve is descending
idxp = [6 1 2 3 4 5];
end
end
% plot(max_angles(idxp),peaks(idxp),'o','MarkerSize',12)
% hold on;
min3_angles = min_angles(idxv);
valleys3 = valleys(idxv);
plot(min3_angles,valleys3,'x','MarkerSize',12)
hold on
% plot(minangle,minvalley,'.r','MarkerSize',30)
% hold on
newpeaks = peaks(idxp);
newmax_angles = max_angles(idxp);
plot(newmax_angles(1:2),newpeaks(1:2),'.r','MarkerSize',15)
hold on;
plot(newmax_angles(3:4),newpeaks(3:4),'.g','MarkerSize',15)
hold on;
plot(newmax_angles(5:6),newpeaks(5:6),'.b','MarkerSize',15)
hold on;
newpeaks1 = (newpeaks(1:2:end) + newpeaks(2:2:end))/2;
delta = newpeaks1-valleys(idxv);
idx4 = find(delta == min(delta));
plot(minangle,minvalley,'.r','MarkerSize',30)
hold on
plot(min3_angles(idx4),valleys3(idx4),'.g','MarkerSize',30)
hold on
xline(min3_angles(idx4),'-g', LineWidth = 1.5);
hold on
yline(valleys3(idx4),'-g', LineWidth = 1.5);
.
If you describe the problem you want to solve, it would likely be straightforward to devise a more efficient solution.
Lackng that description, I’ll delete my Answer in a few days.
.
Aram Zeytunyan
on 1 Feb 2025
Edited: Aram Zeytunyan
on 1 Feb 2025
I need to find 3x valleys out of 6. The starting/reference point is the lowest valley (the large red dot in the plot, which is easy to find). The other two cannot be the neighboring valleys as that's forbidden by physics. mink will find valleys labeled 2, 4, and 5. I need to find valleys 2,4, and 6.
I am still not certain what you want, since I am still uncertain of the criteria.
See if this works —
loaddata = readmatrix('testdata.csv');
angle = loaddata(:,1);
signal = loaddata(:,2);
[pks,plocs,pprm] = findpeaks(signal, MinPeakProminence=10);
[vys,vlocs,vprm] = findpeaks(-signal, MinPeakProminence=10);
[minvly,minidx] = min(-vys); % Minimum Valley & Associated Valley Index
idxrng = [max(1,(minidx-2)) minidx min((minidx+2),numel(vlocs))] % Valley Indices ±2 Positions From Minimum
Results = table(angle(vlocs(idxrng)), signal(vlocs(idxrng)), VariableNames=["Angle","Value"])
figure
plot(angle, signal, DisplayName='Signal')
hold on
plot(angle(plocs), pks, 'vg', DisplayName='Peaks')
plot(angle(vlocs), -vys, '^r', DisplayName='Valleys')
plot(angle(vlocs(idxrng)), signal(vlocs(idxrng)), 'ms', MarkerSize=12, DisplayName='Selected')
hold off
grid
xlabel('Angle (°)')
ylabel('Signal (mV)')
legend(Location='best')
The ‘idxrng’ calculation contains a ‘fail safe’ to prevent finding indices outside the allotted range.
.
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