autocorrelation and Power Spectrum

FMDenaro · April 23, 2021, 09:22

Quote:

Originally Posted by lucamirtanini

In my version of the book I have not G1-3. Is The Appendix G dedicated to the "Power Low Spectra"? Which is the name of yours appendix G?

Yes, power law spectra, pag.696

LuckyTran · April 23, 2021, 14:18

The typical definition of autocorrelation is indeed:

$\rho(\tau)=lim_{T->+\infty}\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)d\tau$
For infinite duration signals, you just take the limit as the signal length goes to infinity.

I'm going to replace T with the signal length L because I like to use T for the period T. If the function is periodic over length T, then any integral of length T is the same for any t.
$\int_{t}^{t+T}u(t)u(t+\tau)dt$

If my signal is length L, its autocorrelation function is (writing the same formula as before but with L instead).
$\rho(\tau)=lim_{L->+\infty}\frac{1}{L}\int_{0}^{L}u(t)u(t+\tau)d\tau$
All I do is expand the summation using algebra.
$\rho(\tau)=lim_{L->+\infty}\frac{1}{L}( \int_{0}^{T}u(t)u(t+\tau)d\tau+\int_{T}^{2T}u(t)u(t+\tau)d\tau+...\int_{L-T}^{L}u(t)u(t+\tau)d\tau)$
All those integrals being the same... I can replace them all with integrals from 0 to T. The signal length being L, I will have L/T such integrals.
$\rho(\tau)=lim_{L->+\infty}\frac{1}{L}\frac{L}{T}( \int_{0}^{T}u(t)u(t+\tau)d\tau)$
The L's cancel and the limit becomes trivial
$\rho(\tau)=\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)d\tau$
And we are back at the original definition of autocorrelation for a finite length signal.

This is the periodic extension property I first mentioned. When you take the auto-correlation of a finite length signal, it's equivalent to taking the autocorrelation of the periodic extension of that signal.

This is why aliasing occurs if you sample a signal different from its periodicity. Also why zero padding works (and is a very good idea in certain situations), etc etc.

After writing this I realized the periodicity is not so obvious nor relevant for turbulent flows. It might be better to stick to the original definition involving 1/T for pedagogical purposes. But these properties are at play when you take Fourier transforms. I emphasize again that even if your signal is finite and not periodic, if you take the auto-correlation, you are looking at the autocorrelation of its hypothetical periodic extension.

So where did the 1/T go?

The one missing the 1/T is for finite waveforms!
$\rho(\tau)=\int_{-\infty}^{+\infty}u(t)u(t+\tau)dt$
What we mean by finite waveform is we observe a single impulse like event. For example, nothing, a single-square wave, and then nothing as opposed to a repetition of square waves.

So if your data is [1,2,3] and you use the autocorrelation formula with 1/T, you are assuming that the data would have repeated itself [1,2,3,1,2,3,1,2,3,...]. Question is whether or not your data would-have repeated itself. That's how you choose which approach.

lucamirtanini · April 23, 2021, 17:32

Thank you very much @LuckyTran ! This clarifies everything!

FMDenaro · April 23, 2021, 17:39

Of course, to make sense without the term 1/T the function must be integrable. However, the dimension of rho(tau) (if not specified differently) is different.
Moreover, consider that we are talking about a convolution product, it is easy to see the Fourier transform as the product of the transformed kernel.

April 23, 2021, 14:18		#22
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,764 Rep Power: 66	The typical definition of autocorrelation is indeed: $\rho(\tau)=lim_{T->+\infty}\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)d\tau$ For infinite duration signals, you just take the limit as the signal length goes to infinity. I'm going to replace T with the signal length L because I like to use T for the period T. If the function is periodic over length T, then any integral of length T is the same for any t. $\int_{t}^{t+T}u(t)u(t+\tau)dt$ If my signal is length L, its autocorrelation function is (writing the same formula as before but with L instead). $\rho(\tau)=lim_{L->+\infty}\frac{1}{L}\int_{0}^{L}u(t)u(t+\tau)d\tau$ All I do is expand the summation using algebra. $\rho(\tau)=lim_{L->+\infty}\frac{1}{L}( \int_{0}^{T}u(t)u(t+\tau)d\tau+\int_{T}^{2T}u(t)u(t+\tau)d\tau+...\int_{L-T}^{L}u(t)u(t+\tau)d\tau)$ All those integrals being the same... I can replace them all with integrals from 0 to T. The signal length being L, I will have L/T such integrals. $\rho(\tau)=lim_{L->+\infty}\frac{1}{L}\frac{L}{T}( \int_{0}^{T}u(t)u(t+\tau)d\tau)$ The L's cancel and the limit becomes trivial $\rho(\tau)=\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)d\tau$ And we are back at the original definition of autocorrelation for a finite length signal. This is the periodic extension property I first mentioned. When you take the auto-correlation of a finite length signal, it's equivalent to taking the autocorrelation of the periodic extension of that signal. This is why aliasing occurs if you sample a signal different from its periodicity. Also why zero padding works (and is a very good idea in certain situations), etc etc. After writing this I realized the periodicity is not so obvious nor relevant for turbulent flows. It might be better to stick to the original definition involving 1/T for pedagogical purposes. But these properties are at play when you take Fourier transforms. I emphasize again that even if your signal is finite and not periodic, if you take the auto-correlation, you are looking at the autocorrelation of its hypothetical periodic extension. So where did the 1/T go? The one missing the 1/T is for finite waveforms! $\rho(\tau)=\int_{-\infty}^{+\infty}u(t)u(t+\tau)dt$ What we mean by finite waveform is we observe a single impulse like event. For example, nothing, a single-square wave, and then nothing as opposed to a repetition of square waves. So if your data is [1,2,3] and you use the autocorrelation formula with 1/T, you are assuming that the data would have repeated itself [1,2,3,1,2,3,1,2,3,...]. Question is whether or not your data would-have repeated itself. That's how you choose which approach. FMDenaro and lucamirtanini like this.

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April 23, 2021, 17:32		#23
lucamirtanini Senior Member luca mirtanini Join Date: Apr 2018 Posts: 165 Rep Power: 8	Thank you very much @LuckyTran ! This clarifies everything!

April 23, 2021, 17:39		#24
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,897 Rep Power: 73	Of course, to make sense without the term 1/T the function must be integrable. However, the dimension of rho(tau) (if not specified differently) is different. Moreover, consider that we are talking about a convolution product, it is easy to see the Fourier transform as the product of the transformed kernel.