"Subscript indices must either be real positive integers or logicals" error while using trapz function

2 ビュー (過去 30 日間)
Hello, I am using following code to calculate area under the peak.
D = amplitude(:);
l = 59;
chg(1) = 1;
chg(2) = l;
for k1 = 1:numel(chg)-1
segment_area_1k1(a,j) = trapz(chg(k1:k1+1), D(chg(k1:k1+1)));
end
end
The code upon run shows error "subscript indices must either be real positive integers or logicals". I have attached amplitude data for reference. Can anyone solve this problem?

回答 (2 件)

Ameer Hamza
Ameer Hamza 2018 年 5 月 19 日
編集済み: Ameer Hamza 2018 年 5 月 19 日
What are the values of a and j? They must be a positive integer. Also if you just want to calculate the area under the curve, then pass an entire vector to trapz()
area = trapz(D);
If you want to get segmented area, you can avoid for loop altogether. A better approach is to do it like this
segmentedArea = diff(cumtrapz(D))
  4 件のコメント
ishita agrawal
ishita agrawal 2018 年 5 月 19 日
Actually, this code worked perfectly for other datasets. This time, it is working for some values in the dataset and then this error occurred.
There is something I want to ask. When I tried your code (trapz(D)), the value of calculated area was different than what I calculated using my code. I couldn't understand the reason.
Image Analyst
Image Analyst 2018 年 5 月 19 日
Give your complete code, including the code where you read in the data file, and your complete error message, including line numbers and ALL the red text.

サインインしてコメントする。


Image Analyst
Image Analyst 2018 年 5 月 19 日

カテゴリ

Help Center および File ExchangeNumerical Integration and Differentiation についてさらに検索

タグ

Community Treasure Hunt

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

Start Hunting!

Translated by