Thanks a lot for this. It clarified some thoughts I had on the matter. Your entire series on Turbulence has been great so far, going to check out part 8 now. Thanks Prof. Schluter.
I think the normalization factor for the auto-correlation function should be instead of ^2 ... the latter obviously does not yield a normalized 1 at the start since rho(0) = / ^2 =/= 1 ^ it does not necessarily give 1, if not always different than 1. Similarly, for the cross-correlation with the addition of the x-component, as so: ...
Hello Dr Schluter, I would like you to kindly clarify the term in the denominator of the auto-correlation (u^2). As I understand, it should be the average of the square of the velocity, which means that the power term should be inside the average and not outside.
Hi, Prof. Schluter! I want to know how we can compute the autocorrelation numerically. For example, I have a time series of the streamwise velocity at a fixed point in a channel flow and it is represented by U(t_i). Of course, the flow has reached statistical stationary.
My intuition is it is a measure of how "correlated" the signal of current time and s time latter. If s = 0, then rho(s) = 1, it means signal is perfectly correlated to itself, it does make sense because what happens at current time fully determines itself (or 0 time latter). If s > 0, what happens at current time might "somehow" correlated to what happens at s time latter, but it might not be that "perfect". I think it might be proved to be less than 1 by math, but it is just my intuition.
in the denominator, the square should be inside. else, really helpful! thank you.
Thanks a lot for this. It clarified some thoughts I had on the matter. Your entire series on Turbulence has been great so far, going to check out part 8 now. Thanks Prof. Schluter.
Thank you so much. I really enjoyed it!
Hi professor, this is a really helpful video, thanks a lot for it
excellent video
Hi thanks for this great explanation, very very helpful.
I think the normalization factor for the auto-correlation function should be instead of ^2 ... the latter obviously does not yield a normalized 1 at the start since
rho(0) = / ^2 =/= 1
^ it does not necessarily give 1, if not always different than 1.
Similarly, for the cross-correlation with the addition of the x-component, as so: ...
Indeed, the square should be inside the average brackets. I think this is also stated in Pope's/Lumley's book.
True. I was wondering the same.
Hello Dr Schluter, I would like you to kindly clarify the term in the denominator of the auto-correlation (u^2). As I understand, it should be the average of the square of the velocity, which means that the power term should be inside the average and not outside.
Hi, Prof. Schluter! I want to know how we can compute the autocorrelation numerically. For example, I have a time series of the streamwise velocity at a fixed point in a channel flow and it is represented by U(t_i). Of course, the flow has reached statistical stationary.
QUESTION. Why rho(s) is always less than one? See diagram at 7:50. Thanks.
My intuition is it is a measure of how "correlated" the signal of current time and s time latter. If s = 0, then rho(s) = 1, it means signal is perfectly correlated to itself, it does make sense because what happens at current time fully determines itself (or 0 time latter). If s > 0, what happens at current time might "somehow" correlated to what happens at s time latter, but it might not be that "perfect". I think it might be proved to be less than 1 by math, but it is just my intuition.
becuz it is normalized, taking rho(0) the maximum value and equal to 1
Dissipation time scale?