Overestimation of mean with trapz

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Emily
Emily le 19 Juin 2017
Commenté : Star Strider le 19 Juin 2017
Hi,
I'm using trapz(x,y) to find the mean value of a portion of a signal.
x = [0.3 0.3 0.3 0.4 0.5 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.2 1.3 1.4 1.4 1.5 1.5];
y = [14.0 15.8 15.8 16.3 19.7 23.5 25.2 27.1 28.4 29.1 30.3 30.6 30.6 30.5 30.5 30.4 30.4 30.4 30.3];
The maximum value in y is 30.6. However, trapz(x,y) = 32.3.
Why is the mean value estimated by trapz greater than the maximum value of the dataset? I thought maybe the spacing was too coarse and tried interpolating on a much finer grid, but the trapz mean still exceeded the maximum value of y.
Thanks for any help!

Réponse acceptée

Star Strider
Star Strider le 19 Juin 2017
The mean is defined as the integral of ‘y(x)’ from ‘x=a’ to ‘x=b’ divided by ’(b-a)’:
trapz_mean = trapz(x,y)/(max(x)-min(x));
trapz_mean =
26.9292
  2 commentaires
Emily
Emily le 19 Juin 2017
Thanks! I thought the division by the range was implicit.
Star Strider
Star Strider le 19 Juin 2017
My pleasure!
In the mean function, it is. Calculating it using numerical (or symbolic) integration, it is not.

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