How to calculate area under each peak in plot from sampled data

1 回表示 (過去 30 日間)
Lukas Poviser
Lukas Poviser 2021 年 4 月 3 日
コメント済み: Star Strider 2021 年 4 月 4 日
Hi, I would like to know area of each peak from my sampled data [time,flow] plot as shown in picture. Area of each peak should be written to matrice [time_of_peak, area], so I would be able to plot graph.

採用された回答

Star Strider
Star Strider 2021 年 4 月 3 日
Try this:
D = load('flow_inspi3.txt');
t = D(:,1);
s = D(:,2);
brk = find(ischange(s,'linear','Threshold',1E-8),1); % First Peak Start
brk1 = [brk; find(islocalmin(s, 'FlatSelection','first'))]; % Peak Start
brk2 = [find(islocalmin(s, 'FlatSelection','last')); numel(s)]; % Peak End
for k = 1:size(brk,1)
idxrng = brk1(k):brk2(k);
AUC(k) = trapz(t(idxrng), s(idxrng));
end
locs = find(islocalmax(s));
pks = s(locs);
figure
plot(t,s)
grid
text(t(locs), pks/2, compose('Area = %9.3f',AUC), 'Horiz','center', 'Vert','middle', 'Rotation',90)
xlim([1.6E+5 1.8E+5]) % Display First Four Peaks (Delete Later)
The file only has information for positive deflections, not negative as in the original plot image. Data with positive and negative deflections would require a slightly different approach.
A mnore robust version:
D = load('flow_inspi3.txt');
t = D(:,1);
s = D(:,2);
zrx = find(diff(sign(s)));
for k = 1:numel(zrx)
idxrng = max(1,zrx(k)-2):min(zrx(k)+2,numel(s));
B = [t(idxrng) ones(size(idxrng(:)))] \ s(idxrng);
intcpt(k) = -B(2)/B(1);
end
mindifft = min(diff(t));
for k = 1:numel(intcpt)-1
if (intcpt(k+1)-intcpt(k)) > mindifft
idxrng = t>=intcpt(k) & t<=intcpt(k+1);
AUC(k) = trapz(t(idxrng), s(idxrng));
[pks(k),locs1] = max(s(idxrng));
locs(k) = locs1*mindifft+round(intcpt(k));
end
end
posidx = AUC > 0;
figure
plot(t,s)
grid
text(locs(posidx), pks(posidx)/2, compose('Area = %9.3f',AUC(posidx)), 'Horiz','center', 'Vert','middle', 'Rotation',90)
xlim([1.6E+5 1.8E+5])
.
  6 件のコメント
Lukas Poviser
Lukas Poviser 2021 年 4 月 4 日
It works perfectly! Thank you!
Star Strider
Star Strider 2021 年 4 月 4 日
As always, my pleasure!

サインインしてコメントする。

その他の回答 (1 件)

darova
darova 2021 年 4 月 3 日
編集済み: darova 2021 年 4 月 3 日
Here is ax example
x = 0:20;
y = sin(x);
[xc,yc] = polyxpoly(x,y,[0 20],[0 0]); % find '0' points intersections
x1 = [xc' x]; % merge together
y1 = [yc' y];
[xx,ix] = sort(x1); % sort x coordinate
yy = y1(ix); % order y coordinate
ix1 = find(abs(yy)<0.01); % zero point indices
plot(xx,yy,'marker','.')
for i = 1:length(ix1)-1
ii = ix1(i):ix1(i+1); % indices of area
s = trapz(xx(ii),yy(ii)); % calculate area
text(xx(ix1(i)),yy(ix1(i)),num2str(s))
end
  4 件のコメント
Lukas Poviser
Lukas Poviser 2021 年 4 月 3 日
Matlab respond is:
Error using horzcat
Dimensions of arrays being concatenated are not consistent.
Error in Volume_mathworks (line 33)
x1 = [xc' x]; % merge together
Maybe I am not sure how to edit your code to fit to my data. Variable x is for time of sample and y is for value of sample? Like that?
x = sample_time;
y = sample_value;
Do I need to change something else in your code?
darova
darova 2021 年 4 月 4 日
xc and x are row and column. Try this
x1 = [xc(:); x(:)]; % merge together
y1 = [yc(:); y(:)];

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeScopes and Data Logging についてさらに検索

製品


リリース

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by