Correlation, energy spectrum and integral length scale

TurbJet · December 14, 2021, 23:23

In homogeneous isotropic turbulence (HIT), the correlation function

$R_{ij}(\pmb{r}) = \overline{u_i(\pmb{x})u_j(\pmb{x}')}$

and the energy spectrum tensor

$\Phi_{ij}(\pmb{k}) = \overline{\hat{u}_i(\pmb{k})\hat{u}_j^*(\pmb{k})}$

are a Fourier-transform pair, i.e.,

$\Phi_{ij}(\pmb{k}) = \frac{1}{(2\pi)^3}\int R_{ij}(\pmb{r})e^{-i\pmb{kr}}d\pmb{r}\;,\; R_{ij}(\pmb{r}) = \int \Phi_{ij}(\pmb{k})e^{i\pmb{kr}}d\pmb{r}$

where $\pmb{r} = \pmb{x}-\pmb{x}'$ , $\pmb{u}' = \pmb{U}-\bar{\pmb{U}}$ is the fluctuating part, and $\overline{(\cdot)}$ represents the averaging. Additionally, the longitudinal integral length scale is defined as

$\mathcal{L}_{11} = \int_0^{+\infty}\mathcal{R}_{11}(r_1)dr_1$

where $\mathcal{R}_{11}$ is the correlation coefficient in the longitudinal direction.

Combine all these, one can derive the integral length scale from the one-dimensional spectrum

$\Phi_{11}(k_1 = 0) = \frac{1}{2\pi}\int R_{11}(r_1)dr_1 = \frac{1}{2\pi}\cdot 2\overline{u_1^2}\mathcal{L}$

thus giving

$\mathcal{L}_{11} = \pi\frac{\Phi_{11}(k_1=0)}{\overline{u_1^2}}$

However, for HIT, $\Phi_{11}(k_1=0)$ is expected to be zero, given that the zeroth mode

for velocity should be zero mean. Hence, from above, the integral length scale will be zero as well. This doesn't seem right to me. So could anybody point out where am I missing?

Thanks.

FMDenaro · December 15, 2021, 04:58

Your case represents a full periodic box, in such a case the characteristic scales is the Taylor microscale. Clearly, while considering only fluctuations, there is no meaning in a characteristic lenght scale for the “large” structures.

TurbJet · December 15, 2021, 13:08

Quote:

Originally Posted by FMDenaro

Your case represents a full periodic box, in such a case the characteristic scales is the Taylor microscale. Clearly, while considering only fluctuations, there is no meaning in a characteristic lenght scale for the “large” structures.

Ok, let's forget about turbulence and the physical background, and consider more generic case. Let's consider a 1-D signal. Here is a zero-mean periodic brown noise on the interval $[0,2\pi]$

brown.jpeg

Clearly, it is long-range correlated, and we can extend the concept above and define an integral length scale for this signal. However, from the math I showed, the integral length scale is determined by the mean (i.e., the

mode), which will be zero in this case. This still doesn't make sense.

FMDenaro · December 15, 2021, 13:25

In isotropic homogeneous turbulence, the integral length scale

is defined as the weighted average of the inverse wavenumber, i.e.,

where

is the energy spectrum.

Have also a look to this paper
https://www.researchgate.net/publica...Numerical_Data

LuckyTran · December 15, 2021, 13:52

The velocity fluctuation signal u' has zero Fourier coefficient.

The energy spectrum is the Fourier transform of the correlation function (more technically the two-point correlation function) which is non-zero if there is any correlation. And as you say, clearly it is (long-range) correlated. Unless the signal is a trivial (zero everywhere all the time) then it will almost certainly have a finite correlation.

TurbJet · December 15, 2021, 15:25

Quote:

Originally Posted by FMDenaro

In isotropic homogeneous turbulence, the integral length scale

is defined as the weighted average of the inverse wavenumber, i.e.,

where

is the energy spectrum.

Have also a look to this paper
https://www.researchgate.net/publica...Numerical_Data

Yes, I've seen this definition. But I don't see where I made a mistake in my derivation.

Same result is shown in Eq.4.11 in here, even though it's in time.

TurbJet · December 15, 2021, 15:27

Quote:

Originally Posted by LuckyTran

The velocity fluctuation signal u' has zero Fourier coefficient.

The energy spectrum is the Fourier transform of the correlation function (more technically the two-point correlation function) which is non-zero if there is any correlation. And as you say, clearly it is (long-range) correlated. Unless the signal is a trivial (zero everywhere all the time) then it will almost certainly have a finite correlation.

Yes, I agree. But I fail to see where I made a mistake in my derivation.

LuckyTran · December 15, 2021, 15:39

If you subtract the DC part of a signal (or use any signal with zero mean), you get that the energy spectrum is 0 corresponding to a mean of 0. It says nothing about the length scale.

That is, phi is zero and u bar is zero. Any length scale satisfies that.

TurbJet · December 15, 2021, 15:57

Quote:

Originally Posted by LuckyTran

If you subtract the DC part of a signal (or use any signal with zero mean), you get that the energy spectrum is 0 corresponding to a mean of 0. It says nothing about the length scale.

That is, phi is zero and u bar is zero. Any length scale satisfies that.

Excuse me but I don't quite understand. Could you elaborate?

Take the 1-D signal I mentioned in one of the reply above for example. It's a periodic signal with zero mean, and clearly, $\Phi(k=0) = 0$ . By extending the derivation above,

$\Phi(k=0) = \frac{\overline{u^2}}{\pi}\mathcal{L} \rightarrow \mathcal{L} = 0$

FMDenaro · December 15, 2021, 16:19

Quote:

Originally Posted by TurbJet

Excuse me but I don't quite understand. Could you elaborate?

Take the 1-D signal I mentioned in one of the reply above for example. It's a periodic signal with zero mean, and clearly, $\Phi(k=0) = 0$ . By extending the derivation above,

$\Phi(k=0) = \frac{\overline{u^2}}{\pi}\mathcal{L} \rightarrow \mathcal{L} = 0$

Something in your formula makes me think about an ideal case: imagine a total random signal, that is a white spectrum. It is uncorrelated, you will see the separation point ->0. And the formula seems to become an identity like 0=0*0 ...But, on the other hand, a small correlation would produce the RHS to be non vanishing.

LuckyTran · December 15, 2021, 17:23

If your signal has zero mean then u bar is zero.

And phi is zero like you say. This does not imply that L is zero. If you re-arrange the formula for L then you make the division by 0 mistake.

I think what you are forgetting is that u bar is zero.

FMDenaro · December 15, 2021, 17:27

Quote:

Originally Posted by LuckyTran

If your signal has zero mean then u bar is zero.

$\Phi(k=0) = \frac{0}{\pi}\mathcal{L}$

And phi is zero like you say. This does not imply that L is zero. If you re-arrange the formula for L then you make the division by 0 mistake.

$\mathcal{L}_{11} = \pi\frac{\Phi_{11}(k_1=0)}{\overline{u_1^2}} = \frac{0}{0}$

I think what you are forgetting is that u bar is zero.

Well, be careful, his formula has (u^2)_bar= (u'^2)_bar.

LuckyTran · December 15, 2021, 18:16

Let me just take a step back and ask from where do we get $\Phi_{11}(k_1 = 0) = 0$

And what kinds of signals are we talking about? Their properties cannot be ignored. White noise is uncorrelated. Brown noise has a flat non-zero power spectral density. Arbitrary but otherwise statistically stationary signals have correlation.

TurbJet · December 15, 2021, 19:02

Quote:

Originally Posted by FMDenaro

Something in your formula makes me think about an ideal case: imagine a total random signal, that is a white spectrum. It is uncorrelated, you will see the separation point ->0. And the formula seems to become an identity like 0=0*0 ...But, on the other hand, a small correlation would produce the RHS to be non vanishing.

Yes, I agree. But the math I showed doesn't seem this way.

TurbJet · December 15, 2021, 19:03

Quote:

Originally Posted by LuckyTran

Let me just take a step back and ask from where do we get $\Phi_{11}(k_1 = 0) = 0$

And what kinds of signals are we talking about? Their properties cannot be ignored. White noise is uncorrelated. Brown noise has a flat non-zero power spectral density. Arbitrary but otherwise statistically stationary signals have correlation.

In the example above, I used a brown noise. And I believe brown noise has energy spectrum $E(k)\propto k^{-2}$ , not a flat one; flat one is for white noise.

TurbJet · December 15, 2021, 19:04

Quote:

Originally Posted by LuckyTran

If your signal has zero mean then u bar is zero.

And phi is zero like you say. This does not imply that L is zero. If you re-arrange the formula for L then you make the division by 0 mistake.

I think what you are forgetting is that u bar is zero.

Well, the denominator is $\overline{u^2}$ , the

here is the fluctuating part, i.e., $u = U-\bar{U}$ , as I already indicated in the post; indeed, $\bar{U} = 0$ . Anyway, I don't think the denominator will be zero.

And for why $\Phi(k=0) = 0$ . Recall that I consider a 1-D zero-mean brown noise above, and in 1-D the energy spectrum is

$\Phi(k) = \hat{u}(k)\hat{u}^*(k)$

in which $\hat{u}(k)$ is the Fourier transform of the fluctuating part $u(x) = U(x) - \bar{U}(x)$ . Since the noise is zero mean, $\hat{u}(k=0) = 0$ , and so $\Phi(k=0) = \hat{u}(k=0)\hat{u}^*(k=0) = 0$ .

LuckyTran · December 15, 2021, 21:14

You're right. Brown noise is the integral of white noise. The power spectral density for white noise is a constant. The power spectrum for brown noise goes like 1/f^2 or 1/k^2 since we're talking about space. Either way, neither are identically zero.

The Fourier transform of a constant is not 0.

TurbJet · December 15, 2021, 21:40

Quote:

Originally Posted by LuckyTran

You're right. Brown noise is the integral of white noise. The power spectral density for white noise is a constant. The power spectrum for brown noise goes like 1/f^2 or 1/k^2 since we're talking about space. Either way, neither are identically zero.

The Fourier transform of a constant is not 0.

What constant are you referring to?

For a zero-mean signal, isn't the zeroth-mode (or the DC component) $\hat{u}(k=0)=0$ ? Then the spectrum $\Phi(k=0) = \hat{u}(k=0)\hat{u}^*(k=0) = 0$ ?

LuckyTran · December 15, 2021, 22:08

You're probably thinking that $\hat{u}$ is the thingy you see when you plot an FFT or the amplitude when you look at a Fourier plot. That's only the real part of the complex Fourier representation. There is also an imaginary part.

If consider a constant signal with non-zero mean, a simple DC signal, the Fourier transform of a constant DC signal is a Dirac delta function. The special case of the constant being 0 has very peculiar properties. Basically, $\hat{u}$ ain't 0.

TurbJet · December 16, 2021, 19:01

Quote:

Originally Posted by LuckyTran

You're probably thinking that $\hat{u}$ is the thingy you see when you plot an FFT or the amplitude when you look at a Fourier plot. That's only the real part of the complex Fourier representation. There is also an imaginary part.

If consider a constant signal with non-zero mean, a simple DC signal, the Fourier transform of a constant DC signal is a Dirac delta function. The special case of the constant being 0 has very peculiar properties. Basically, $\hat{u}$ ain't 0.

This doesn't make sense to me. For a real-value signal, the zero mode has to be real as well, and its value is equal to the mean of the signal. I am not talking about a constant signal here, I am talking about a signal with zero mean.

December 14, 2021, 23:23	Correlation, energy spectrum and integral length scale	#1
TurbJet Senior Member Join Date: Oct 2017 Location: United States Posts: 233 Blog Entries: 1 Rep Power: 10	In homogeneous isotropic turbulence (HIT), the correlation function $R_{ij}(\pmb{r}) = \overline{u_i(\pmb{x})u_j(\pmb{x}')}$ and the energy spectrum tensor $\Phi_{ij}(\pmb{k}) = \overline{\hat{u}_i(\pmb{k})\hat{u}_j^(\pmb{k})}$ are a Fourier-transform pair, i.e., $\Phi_{ij}(\pmb{k}) = \frac{1}{(2\pi)^3}\int R_{ij}(\pmb{r})e^{-i\pmb{kr}}d\pmb{r}\;,\; R_{ij}(\pmb{r}) = \int \Phi_{ij}(\pmb{k})e^{i\pmb{kr}}d\pmb{r}$ where $\pmb{r} = \pmb{x}-\pmb{x}'$ , $\pmb{u}' = \pmb{U}-\bar{\pmb{U}}$ is the fluctuating part, and $\overline{(\cdot)}$ represents the averaging. Additionally, the longitudinal integral length scale* is defined as $\mathcal{L}_{11} = \int_0^{+\infty}\mathcal{R}_{11}(r_1)dr_1$ where $\mathcal{R}_{11}$ is the correlation coefficient in the longitudinal direction. Combine all these, one can derive the integral length scale from the one-dimensional spectrum $\Phi_{11}(k_1 = 0) = \frac{1}{2\pi}\int R_{11}(r_1)dr_1 = \frac{1}{2\pi}\cdot 2\overline{u_1^2}\mathcal{L}$ thus giving $\mathcal{L}_{11} = \pi\frac{\Phi_{11}(k_1=0)}{\overline{u_1^2}}$ However, for HIT, $\Phi_{11}(k_1=0)$ is expected to be zero, given that the zeroth mode for velocity should be zero mean. Hence, from above, the integral length scale will be zero as well. This doesn't seem right to me. So could anybody point out where am I missing? Thanks.

December 15, 2021, 04:58		#2
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,896 Rep Power: 73	Your case represents a full periodic box, in such a case the characteristic scales is the Taylor microscale. Clearly, while considering only fluctuations, there is no meaning in a characteristic lenght scale for the “large” structures. PENGGEGE777 likes this.

December 15, 2021, 22:08		#19
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	You're probably thinking that $\hat{u}$ is the thingy you see when you plot an FFT or the amplitude when you look at a Fourier plot. That's only the real part of the complex Fourier representation. There is also an imaginary part. If consider a constant signal with non-zero mean, a simple DC signal, the Fourier transform of a constant DC signal is a Dirac delta function. The special case of the constant being 0 has very peculiar properties. Basically, $\hat{u}$ ain't 0. Eifoehn4 likes this.

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December 15, 2021, 13:25		#4
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,896 Rep Power: 73	In isotropic homogeneous turbulence, the integral length scale is defined as the weighted average of the inverse wavenumber, i.e., where is the energy spectrum. Have also a look to this paper https://www.researchgate.net/publica...Numerical_Data

December 15, 2021, 13:52		#5
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	The velocity fluctuation signal u' has zero Fourier coefficient. The energy spectrum is the Fourier transform of the correlation function (more technically the two-point correlation function) which is non-zero if there is any correlation. And as you say, clearly it is (long-range) correlated. Unless the signal is a trivial (zero everywhere all the time) then it will almost certainly have a finite correlation.

December 15, 2021, 15:39		#8
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	If you subtract the DC part of a signal (or use any signal with zero mean), you get that the energy spectrum is 0 corresponding to a mean of 0. It says nothing about the length scale. That is, phi is zero and u bar is zero. Any length scale satisfies that.

December 15, 2021, 17:23		#11
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	If your signal has zero mean then u bar is zero. And phi is zero like you say. This does not imply that L is zero. If you re-arrange the formula for L then you make the division by 0 mistake. I think what you are forgetting is that u bar is zero.

December 15, 2021, 18:16		#13
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	Let me just take a step back and ask from where do we get $\Phi_{11}(k_1 = 0) = 0$ And what kinds of signals are we talking about? Their properties cannot be ignored. White noise is uncorrelated. Brown noise has a flat non-zero power spectral density. Arbitrary but otherwise statistically stationary signals have correlation.

December 15, 2021, 21:14		#17
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	You're right. Brown noise is the integral of white noise. The power spectral density for white noise is a constant. The power spectrum for brown noise goes like 1/f^2 or 1/k^2 since we're talking about space. Either way, neither are identically zero. The Fourier transform of a constant is not 0.