Plotting the Kolmogrov k^-5/3 slope on the Turbulent Power Spectrum
 

I was wondering how you go about plotting the k^(-5/3) slope on a power spectrum you generate from a velocity signal.

I have tested my code with data from the John Hopkins database to make sure that I have done the Fourier transform correctly and I even tested with a simple sinous wave. All is good, but I am really confused about the -5/3 slope.

I was wondering whether a simple y = mx+c would work? I have a plot of Frequency vs KE (as I have the fluctuating u, v and w components from an LES simulation). But this is really irrelevant right now as the problem lies in the -5/3 slope.

Do I pick a point on the generated frequency vs energy plot (x,y) keep the y component in memory and run a loop with this value over frequency raised to -5/3 (k^-5/3?)

For example to clarify the above would I write something like this in Matlab:

y = energy(index)
%index is the location just before where I want to begin the slope
%Energy and Frequency are thevertical and horizontal vectors on the FFT plot.
for i=1:increments_till_end_of_slope_location
.....horizontal_slope(i) = frequency(index+i);
.....vertical_slope(i) = y*(frequency(index+i)^(-5/3));
end


I have attached an image from the John Hopkins database. I personally extracted the velocity data and plotted just Hz vs E(k) I did not multiply by eta on the horizontal axis or divide by c times mu on the vertical axis. This image is just to show you what line I am referring to, the blue line.

I have tried a few things and could not reproduce the -5/3 line any suggestions would help. I did read an old thread from 2017 https://www.cfd-online.com/Forums/ma...ng-result.html that helped me get this far. My Fourier transform was fine, but the database and the comparison with the sine function from Prof. Denaro helped me confirm this.

what is exactly the problem? From your plot it appears the superimposition of the log law... you just have to plot something like C*k^(-5/3) in a loglog plot

I'm wondering what the 'k' would be. Is k the corresponding frequency in Hz if I were to be plotting Hz vs Energy?

frequency, but if the period is exactly L=2pi then it is also equivalent to the wavenumber

Perfect thank you!