autocorrelation and Power Spectrum

lucamirtanini · April 22, 2021, 07:22

Hi all,
usually in a lot of books of signal processing the autocorrelation of a signal

is defined as:

$\rho(\tau)=\int_{-\infty}^{+\infty}x(t)x(t+\tau)dt$

Usually the signals we work with, such as the time history of the velocity component

, are finite signals. The autocorrelation in Tennekes and Lumley book is defined as:

$\rho(\tau)=\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)dt$

where u(t) is the history of the velocity component.

If I want to find the power spectral density, doing the Fourier Transform of $\rho(\tau)$ , which should be the integral of the extremes of the Fourier Transform? the same of the integral of $\rho(\tau)$ (0 and T)?

FMDenaro · April 22, 2021, 09:19

The discrete FFT has a number of components up to the limit defined by the Nyquist frequency. Given the samplig step dt, given the period T, you have T/(2*dt) the maximum resolved wavenumber.

lucamirtanini · April 22, 2021, 09:28

If we suppose that our signal is continuous but finite which should be the extremes of the FT integral? I am trying to demonstrate the autocorrelation theorem for finite signal.

FMDenaro · April 22, 2021, 12:06

Quote:

Originally Posted by lucamirtanini

If we suppose that our signal is continuous but finite which should be the extremes of the FT integral? I am trying to demonstrate the autocorrelation theorem for finite signal.

A continuous function means you have dt->0 while a finite period T means you have a limit in the maximum wavelength.

The relation between power spectrum and correlation is described by Pope in his textbook.

LuckyTran · April 22, 2021, 12:25

When you do a Fourier transform of a finite signal, you're actually not doing it on a finite signal but a periodic extension of the signal (the same signal but repeated with a period equal to the length of the signal). Even if your actual signal in your databook is finite, the function you are transforming is infinite. There are quite a bit of nuances of Fourier stuff due to finite vs infinite signal length and discrete vs continuous.

That's why, both definitions are equivalent even though one is integrated from 0 to T and the other, minus to plus infinity. It takes some time getting used to, but they are the same...

To put it another way... If my velocity data is [1,2,3] with a length of 3. Instead of considering it as a finite signal, I instead consider the infinite data series [1,2,3,1,2,3,1,2,3,1,2,3,...] and then take the Fourier transform of that. If I do it properly, the finite Fourier transform of my original data should have the same representation as the infinite Fourier transform of my made up data series.

lucamirtanini · April 22, 2021, 12:37

I already know the relation between Power spectrum and autocorrelation.
I need to know the extremes of the fourier transform of the Power Spectrum if the signal is finite and continuous:

$PSD=\int_{?}^{?}\rho(\tau)e^{-i\omega\tau}d\tau$

In the book of Pope the definition is given for an infinite signals and the extremes of the integrals are always $[-\infty,+\infty]$

FMDenaro · April 22, 2021, 12:49

Quote:

Originally Posted by lucamirtanini

I already know the relation between Power spectrum and autocorrelation.
I need to know the extremes of the fourier transform of the Power Spectrum if the signal is finite and continuous:

$PSD=\int_{?}^{?}\rho(\tau)e^{-i\omega\tau}d\tau$

In the book of Pope the definition is given for an infinite signals and the extremes of the integrals are always $[-\infty,+\infty]$

Have a look to page 685

FMDenaro · April 22, 2021, 12:58

A "hidden" issue is about the stationary process vs. non stationary process. Clearly, the period T, repeated infinitely times is somehow a constraint

lucamirtanini · April 22, 2021, 13:07

Ok @LuckyTran your explanation is very clear! I'd never caught this nuance! Then why is the autocorrelation for a finite signal:
$\rho(\tau)=\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)dt$
while for the inifinite one:
$\rho(\tau)=\int_{-\infty}^{+\infty}u(t)u(t+\tau)dt$

Why $\frac{1}{T}$ is omitted in the infinite one?

When I apply the Fourier transform to a signal, should the signal in the period T have some requirements to be considered representative of the infinite? I had just made sure that the time average of u(t) (which are fluctuations) is 0.

LuckyTran · April 22, 2021, 14:57

Quote:

Originally Posted by lucamirtanini

Ok @LuckyTran your explanation is very clear! I'd never caught this nuance! Then why is the autocorrelation for a finite signal:
$\rho(\tau)=\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)dt$
while for the inifinite one:
$\rho(\tau)=\int_{-\infty}^{+\infty}u(t)u(t+\tau)dt$

Why $\frac{1}{T}$ is omitted in the infinite one?

There are always two definitions for autocorrelation and these are the correct two. Which one depends on context of whether your signal is finite or infinite.

If you apply the first formula to my example...

Take the autocorrelation of a signal [1,2,3] versus [1,2,3,1,2,3] versus [1,2,3,1,2,3,1,2,3,...]. You will see that the 1/T is needed to make all three yield the same result. It might not be so readily apparent where the 1/T goes when the signal becomes infinite but keep in mind you are integrating an infinitely long signal. The integrand gets bigger in addition to 1/T driving it smaller. Or try applying this to the sine function. Integrate over 1 wavelength, 2 wavelengths, and infinite wavelengths... It looks funny but it really is just algebra.

If the function is periodic with period T then any integral of length T [t,t+T] will be the same for any t. So instead of integrating from minus infinity to plus infinity, I can divide my infinite length signal (into segments of length T) that each have an finite length integral. The number of such integrals I have will go like 1/T. It's just summation rules.

The requirements of what the signal needs to look like over the period T falls under measurement theory. You measure something and pray that the sample statistics represents the population statistics. You checked the fluctuations, but the more obvious check is that the mean velocity matches the flow characteristics.

FMDenaro · April 22, 2021, 15:20

I have doubt about the definition of autocorrelation as Int[-Inf,+Inf] () dt.
A general definition involves the averaging operator, for example see 13.7 in Kundu- Cohen.
On the other hand, without the division by a time the dimensions of the two definitions are not coincident.

LuckyTran · April 22, 2021, 16:16

Quote:

Originally Posted by FMDenaro

I have doubt about the definition of autocorrelation as Int[-Inf,+Inf] () dt.
A general definition involves the averaging operator, for example see 13.7 in Kundu- Cohen.
On the other hand, without the division by a time the dimensions of the two definitions are not coincident.

The u in Int[-Inf,+Inf] () dt must already be normalized with mean subtracted and scaled by variance. It's actually the autocorrelation coefficient in statistics. Maybe that's what is missing?

FMDenaro · April 22, 2021, 16:25

Quote:

Originally Posted by LuckyTran

The u in Int[-Inf,+Inf] () dt must already be normalized with mean subtracted and scaled by variance. It's actually the autocorrelation coefficient in statistics. Maybe that's what is missing?

I don't remember if this equation is from Lumley, it should be addressed that... However, I would use the expression as clearly stated by Kundu.

lucamirtanini · April 22, 2021, 17:31

@FMDenaro in the Tennekes and Lumley it is defined $\overline{u(t)u(t+\tau)}$ and the overbar is defined as the time average in an other section of the book. I don't have the Kindu, is there defined as a time average?

@LuckyTran So the two definitions are both correct, but the results cannot be compared between them two, isn't it?
Considering to use the second definition, should the Fourier transform of $\rho(\tau)$ be multiplied by $\frac{1}{T}$ as well? Reading other books I found also this kind of definition:
$\rho(\tau)=lim_{T->+\infty}\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)d\tau$
Is it a version of the first definition that considers an infinitely long signal?

FMDenaro · April 22, 2021, 17:44

Quote:

Originally Posted by lucamirtanini

@FMDenaro in the Tennekes and Lumley it is defined $\overline{u(t)u(t+\tau)}$ and the overbar is defined as the time average in an other section of the book. I don't have the Kindu, is there defined as a time average?

@LuckyTran So the two definitions are both correct, but the results cannot be compared between them two, isn't it?
Considering to use the second definition, should the Fourier transform of $\rho(\tau)$ be multiplied by $\frac{1}{T}$ as well?

Reading other books I found also this kind of definition:
$\rho(\tau)=lim_{T->+\infty}\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)d\tau$

Is it a version of the first definition that considers an infinitely long signal?

This is exactly the definition adopted by Kundu

lucamirtanini · April 23, 2021, 07:17

@FMDenaro So based on pag 685 of the Pope, as you suggest. Also the FT is multiplied by $\frac{1}{T}$ , isnt' it?

I have checked the Tennekes and Lumley and I was wrong, the definition has before the $lim_{T->+\infty}$ . What concerns me of this definition is th fact that the signal we work with are not $T->\infty$ . Which assumption should I do in order to fit my signal to this definition? Regardless the presence of $lim_{T->+\infty}$ , we practically calculate it using the formula:
$\rho(\tau)=\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)dt$

FMDenaro · April 23, 2021, 07:54

The general assumption is that the period T is sufficiently large to describe the maximum wavelenght of the flow you are studying.

lucamirtanini · April 23, 2021, 08:08

@FMDenaro So based on pag 685 of the Pope, as you suggest. Also the FT is multiplied by $\frac{1}{T}$ , isnt' it?

FMDenaro · April 23, 2021, 08:18

Quote:

Originally Posted by lucamirtanini

@FMDenaro So based on pag 685 of the Pope, as you suggest. Also the FT is multiplied by $\frac{1}{T}$ , isnt' it?

Have a look at Appendix G, Eq.(G1-3) will show that the autocorrelation is defined by the time averaging and the spectra takes into account 1/T from the definition.

lucamirtanini · April 23, 2021, 09:04

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?

April 22, 2021, 07:22	autocorrelation and Power Spectrum	#1
lucamirtanini Senior Member luca mirtanini Join Date: Apr 2018 Posts: 165 Rep Power: 8	Hi all, usually in a lot of books of signal processing the autocorrelation of a signal is defined as: $\rho(\tau)=\int_{-\infty}^{+\infty}x(t)x(t+\tau)dt$ Usually the signals we work with, such as the time history of the velocity component , are finite signals. The autocorrelation in Tennekes and Lumley book is defined as: $\rho(\tau)=\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)dt$ where u(t) is the history of the velocity component. If I want to find the power spectral density, doing the Fourier Transform of $\rho(\tau)$ , which should be the integral of the extremes of the Fourier Transform? the same of the integral of $\rho(\tau)$ (0 and T)? Last edited by lucamirtanini; April 22, 2021 at 10:18.

April 22, 2021, 12:25		#5
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,750 Rep Power: 66	When you do a Fourier transform of a finite signal, you're actually not doing it on a finite signal but a periodic extension of the signal (the same signal but repeated with a period equal to the length of the signal). Even if your actual signal in your databook is finite, the function you are transforming is infinite. There are quite a bit of nuances of Fourier stuff due to finite vs infinite signal length and discrete vs continuous. That's why, both definitions are equivalent even though one is integrated from 0 to T and the other, minus to plus infinity. It takes some time getting used to, but they are the same... To put it another way... If my velocity data is [1,2,3] with a length of 3. Instead of considering it as a finite signal, I instead consider the infinite data series [1,2,3,1,2,3,1,2,3,1,2,3,...] and then take the Fourier transform of that. If I do it properly, the finite Fourier transform of my original data should have the same representation as the infinite Fourier transform of my made up data series. lucamirtanini likes this.

April 22, 2021, 12:58		#8
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,865 Rep Power: 73	A "hidden" issue is about the stationary process vs. non stationary process. Clearly, the period T, repeated infinitely times is somehow a constraint lucamirtanini likes this.

April 22, 2021, 15:20		#11
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,865 Rep Power: 73	I have doubt about the definition of autocorrelation as Int[-Inf,+Inf] () dt. A general definition involves the averaging operator, for example see 13.7 in Kundu- Cohen. On the other hand, without the division by a time the dimensions of the two definitions are not coincident. lucamirtanini likes this.

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Power Spectral Density - Energy Spectrum	lucamirtanini	Main CFD Forum	34	April 8, 2022 20:40
Power spectrum.	Jake.	Main CFD Forum	1	October 9, 2004 16:18

April 22, 2021, 09:19		#2
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,865 Rep Power: 73	The discrete FFT has a number of components up to the limit defined by the Nyquist frequency. Given the samplig step dt, given the period T, you have T/(2*dt) the maximum resolved wavenumber.

April 22, 2021, 09:28		#3
lucamirtanini Senior Member luca mirtanini Join Date: Apr 2018 Posts: 165 Rep Power: 8	If we suppose that our signal is continuous but finite which should be the extremes of the FT integral? I am trying to demonstrate the autocorrelation theorem for finite signal.

April 22, 2021, 12:37		#6
lucamirtanini Senior Member luca mirtanini Join Date: Apr 2018 Posts: 165 Rep Power: 8	I already know the relation between Power spectrum and autocorrelation. I need to know the extremes of the fourier transform of the Power Spectrum if the signal is finite and continuous: $PSD=\int_{?}^{?}\rho(\tau)e^{-i\omega\tau}d\tau$ In the book of Pope the definition is given for an infinite signals and the extremes of the integrals are always $[-\infty,+\infty]$

April 22, 2021, 13:07		#9
lucamirtanini Senior Member luca mirtanini Join Date: Apr 2018 Posts: 165 Rep Power: 8	Ok @LuckyTran your explanation is very clear! I'd never caught this nuance! Then why is the autocorrelation for a finite signal: $\rho(\tau)=\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)dt$ while for the inifinite one: $\rho(\tau)=\int_{-\infty}^{+\infty}u(t)u(t+\tau)dt$ Why $\frac{1}{T}$ is omitted in the infinite one? When I apply the Fourier transform to a signal, should the signal in the period T have some requirements to be considered representative of the infinite? I had just made sure that the time average of u(t) (which are fluctuations) is 0.

April 22, 2021, 17:31		#14
lucamirtanini Senior Member luca mirtanini Join Date: Apr 2018 Posts: 165 Rep Power: 8	@FMDenaro in the Tennekes and Lumley it is defined $\overline{u(t)u(t+\tau)}$ and the overbar is defined as the time average in an other section of the book. I don't have the Kindu, is there defined as a time average? @LuckyTran So the two definitions are both correct, but the results cannot be compared between them two, isn't it? Considering to use the second definition, should the Fourier transform of $\rho(\tau)$ be multiplied by $\frac{1}{T}$ as well? Reading other books I found also this kind of definition: $\rho(\tau)=lim_{T->+\infty}\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)d\tau$ Is it a version of the first definition that considers an infinitely long signal?

April 23, 2021, 07:17		#16
lucamirtanini Senior Member luca mirtanini Join Date: Apr 2018 Posts: 165 Rep Power: 8	@FMDenaro So based on pag 685 of the Pope, as you suggest. Also the FT is multiplied by $\frac{1}{T}$ , isnt' it? I have checked the Tennekes and Lumley and I was wrong, the definition has before the $lim_{T->+\infty}$ . What concerns me of this definition is th fact that the signal we work with are not $T->\infty$ . Which assumption should I do in order to fit my signal to this definition? Regardless the presence of $lim_{T->+\infty}$ , we practically calculate it using the formula: $\rho(\tau)=\frac{1}{T}\int_{0}^{T}u(t)u(t+\tau)dt$

April 23, 2021, 07:54		#17
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,865 Rep Power: 73	The general assumption is that the period T is sufficiently large to describe the maximum wavelenght of the flow you are studying.

April 23, 2021, 08:08		#18
lucamirtanini Senior Member luca mirtanini Join Date: Apr 2018 Posts: 165 Rep Power: 8	@FMDenaro So based on pag 685 of the Pope, as you suggest. Also the FT is multiplied by $\frac{1}{T}$ , isnt' it?

April 23, 2021, 09:04		#20
lucamirtanini Senior Member luca mirtanini Join Date: Apr 2018 Posts: 165 Rep Power: 8	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?