matlab area calculation under fitted curve

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Ancalagon8
Ancalagon8 am 16 Sep. 2020
Kommentiert: Rik am 15 Okt. 2020
I have a signal and want to calculate the area under the produced fft with step of 5Hz. So far i created an fft, normalized and smoothed it. The problem is that i want to make a loop with decimal step, because the signals' length is 1070 so 1 Hz is 1070/45= 23,77 and step (5Hz) = 118.58
So far my code is:
[x1Data, y1Data] = prepareCurveData(f1, p1norm);
% Set up fittype and options.
ft1 = fittype( 'smoothingspline');
opts1 = fitoptions( 'Method', 'SmoothingSpline');
opts1.SmoothingParam = 0.95;
% Fit model to data.
[fitresult1, gof1] = fit(x1Data, y1Data, ft1, opts1);
% Plot fit with data.
figure
h1 = plot(fitresult1, x1Data, y1Data,'b' );
h = get(gca, 'Children');
Any ideas how to proceed with the loop?
  1 Kommentar
Rik
Rik am 25 Sep. 2020
Comment posted as answer (and erroneously flagged by the spam filter) by Sirius8:
Noone? Come on guys a little help!

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Rik
Rik am 25 Sep. 2020
Bearbeitet: Rik am 28 Sep. 2020
You can use trapz for a numerical approximation of an integration. By selecting your bounds you can use this to calculate the area under a curve.
Edit:
Using the code you posted and adding the loop I'm getting something similar to what you describe. If you don't want the legend entries, you should use explicit handles in your call to legend.
S=load('test1.mat');
[x1,y1,y11]=deal(S.x1,S.y1,S.y11);
figure(1),clf(1)
subplot(2,1,1)
plot(x1,y1)
grid on
hold all
plot(x1,y11,'m')
grid on
hold all
legend ('normalized fft','smoothed')
subplot(2,1,2)
plot(x1,y11,'m')
grid on
hold all
legend ('smoothed')
cmap=colormap('prism');n_col=0;
freq_range=5;
for x_start=0:freq_range:max(x1)
x_end=x_start+freq_range;
L=x1>=x_start & x1<=x_end;
if sum(L)<=1,continue,end%skip if there is only 1 point
X=[x_start x1(L) x_end];
Y=[0 y11(L) 0];
A=trapz(x1(L),y11(L));
n_col=n_col+1;C=cmap(n_col,:);
patch(X,Y,C,'DisplayName',sprintf('A(%d-%dHz)=%.2f',x_start,x_end,A))
end
  38 Kommentare
Ancalagon8
Ancalagon8 am 15 Okt. 2020
Thank you again!
Why you think that the legend is not correct?
Rik
Rik am 15 Okt. 2020
Because A is a vector once you put it in sprintf to create the legend entry. You should also change the frequency part from '%d' to something like '%.1f'.

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