Peak area through Findpeak

Question

iqra kiran 2023-7-11

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1994448-peak-area-through-findpeak

评论： Star Strider 2023-7-11

Hello Everyone,

I have the following data: as attched m file. I can fine the peak location, width through peakfind command. Now i want to find the area under the peak. would you please guide me how i can find using findpeaks.

[pks20_4,loc20_4,w20_4,p20_4] =findpeaks(Y20_4,X4);

Thank you,

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Answer 1

Star Strider 2023-7-11

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1994448-peak-area-through-findpeak#answer_1271158

在 MATLAB Online 中打开

It depends on how you define ‘peak’.

LD1 = load('X4.mat');

X4 = LD1.X4;

LD2 = load('Y20_4.mat');

Y20_4 = LD2.Y20_4;

[pks20_4,loc20_4,w20_4,p20_4] =findpeaks(Y20_4,X4)

pks20_4 = 274.4520

loc20_4 = 3.5596

w20_4 = 0.1954

p20_4 = 273.9922

[pks20_4,loc20_4ix,w20_4ix,p20_4] =findpeaks(Y20_4)

pks20_4 = 274.4520

loc20_4ix = 491

w20_4ix = 111.0153

p20_4 = 273.9922

ixv = find(diff(sign(Y20_4-105)))

ixv = 1×2

421 572

PkArea = trapz(X4(ixv(1):ixv(2)), Y20_4(ixv(1):ixv(2)))

PkArea = 50.5773

figure

plot(X4, Y20_4)

grid

hold on

xp = X4(ixv(1):ixv(2));

yp = Y20_4(ixv(1):ixv(2));

patch([xp flip(xp)], [yp zeros(size(yp))], [ 1 1 1]*0.75, 'FaceAlpha',0.5)

hold off

xline(X4(ixv), '--r')

text(mean(X4(ixv(1):ixv(2))), 105, sprintf('Area = %.3f', PkArea), 'Horiz','center')

.

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Star Strider 2023-7-11

在 MATLAB Online 中打开

The default value for the width id ‘half prominence’, so if you want full-width-half-maximum, you would need to specify 'WidthReference','halfheight'.

My analysis —

LD1 = load('X4.mat');

X4 = LD1.X4;

LD2 = load('Y20_4.mat');

Y20_4 = LD2.Y20_4;

[pks20_4,loc20_4,w20_4,p20_4] =findpeaks(Y20_4,X4)

pks20_4 = 274.4520

loc20_4 = 3.5596

w20_4 = 0.1954

p20_4 = 273.9922

[pks20_4,loc20_4ix,w20_4ix,p20_4] =findpeaks(Y20_4)

pks20_4 = 274.4520

loc20_4ix = 491

w20_4ix = 111.0153

p20_4 = 273.9922

ixv = find(diff(sign(Y20_4-105)))

ixv = 1×2

421 572

PkArea = trapz(X4(ixv(1):ixv(2)), Y20_4(ixv(1):ixv(2)))

PkArea = 50.5773

figure

plot(X4, Y20_4)

grid

hold on

xp = X4(ixv(1):ixv(2));

yp = Y20_4(ixv(1):ixv(2));

patch([xp flip(xp)], [yp zeros(size(yp))], [ 1 1 1]*0.75, 'FaceAlpha',0.5)

hold off

xline(X4(ixv), '--r')

text(mean(X4(ixv(1):ixv(2))), 105, sprintf('Area = %.3f', PkArea), 'Horiz','center')

[pks20_4,loc20_4,w20_4,p20_4] =findpeaks(Y20_4,X4);

x4=loc20_4(1)-w20_4(1)/2

x4 = 3.4619

x_4=loc20_4(1)+w20_4(1)/2

x_4 = 3.6572

area_4 = trapz([x4,x_4], [pks20_4(1), pks20_4(1)])

area_4 = 53.6179

ixv = find(diff(sign(Y20_4-p20_4/2))) % Find 'Half Prominence' Indices

ixv = 1×2

434 546

ixr = [1 loc20_4ix numel(X4)];

for k = 1:numel(ixr)-1

idxrng = ixr(k) : ixr(k+1);

hp(k) = interp1(Y20_4(idxrng), X4(idxrng), p20_4/2); % Calculate Half-Prominence 'X4' Values

end

hp = 3.4607

hp = 1×2

3.4607 3.6568

HalfPromWidthIdx = X4(ixv(2)) - X4(ixv(1)) % Half-Prominence Width: Indices

HalfPromWidthIdx = 0.1971

HalfPromWidthItp = hp(2) - hp(1) % Half-Prominence Width: Interpolated Values

HalfPromWidthItp = 0.1961

PkArea = trapz(X4(ixv(1):ixv(2)), Y20_4(ixv(1):ixv(2))) % Half-Prominence Area: IUndices

PkArea = 42.4737

figure

plot(X4, Y20_4)

grid

hold on

xp = X4(ixv(1):ixv(2));

yp = Y20_4(ixv(1):ixv(2));

patch([xp flip(xp)], [yp zeros(size(yp))], [ 1 1 1]*0.75, 'FaceAlpha',0.5)

hold off

xline(X4(ixv), '--r')

text(mean(X4(ixv(1):ixv(2))), 105, sprintf('Area = %.3f', PkArea), 'Horiz','center')

Using my code with the half-prominence indices results in a different (about 20% smaller) area (42.4737) than ‘area_4’ (53.6179).

Use whatever calculation you wish.

.

iqra kiran 2023-7-11

Thanks alot.

Star Strider 2023-7-11

As always, my pleasure!

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Peak area through Findpeak

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