Rules of Reynolds Averaging
5 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
William Kett
am 2 Okt. 2020
Beantwortet: Alan Stevens
am 2 Okt. 2020
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/369547/image.png)
I tried to create an array, u:
u = [1,3,6,8,24];
uprime = diff(u)
which output a vector of differences (not sure if this is what I wanted).
Then I tried to take an average of uprime using the following code below:
mean(uprime)
According to the rule, I should have gotten 0, but I got an average of the numbers in the vector which was nonzero. Am I misinterpreting the rule or is the diff function not actually taking a derivative?
0 Kommentare
Akzeptierte Antwort
Alan Stevens
am 2 Okt. 2020
uprime doesn't mean du/dt here. It means u - umean.
u = [1,3,6,8,24];
umean = mean(u);
uprime = u-umean;
mean(uprime)
ans =
-3.5527e-16
0 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Spectral Measurements finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!