[lucamirtanini]

Hi all,

I was looking at a result in the validation guide of fds.

In particular, attached you can see the Figure 6.41 of the FDS User Guide, which is named Power spectrum.

As far as I know, this figure was produced using this script.

https://github.com/firemodels/fds/bl...wer_spectrum.m

I have 2 doubts:

1) In the Tenneke, the Power spectral density is defined as the fourier trasform of the autocorrelation. Since the autocorrelation is u(t)*u(t')/u^2, I cannot see in this script where is the t' (time difference) in the script.

2) Furthermore I cannot understand how the attached figure "Autospectral Density" can exhibit the -5/3 scaling in the inertial subrange, since I thought that the -5/3 scaling should be seen only in the nondimensionalised kinetic Energy spectrum, as you can seen in the figure "Chapman 1979" attached.

[LuckyTran]

1) The autocorrelation is not just u(t)*u'(t), there is an integration from negative to positive infinity time. The autocorrelation and power spectral density form a Fourier transform pair. People often say this as a definition, but because the autocorrelation is slow to compute, no one really computes the PSD using this route. It's faster to exploit the Fourier properties. The PSD is calculated in these two lines:

 
u = fft(W);
p = u.*conj(u)/n^2;

2). I'm not sure if u' is a velocity component or the full vector magnitude. But if it is a velocity component, u'u' (in addition to v'v' and w'w') is a part of the turbulent kinetic energy, what is the surprise? If u is the velocity magnitude (measured from a hotwire for example) then u'u' will be k. Or do you mean that the slope is -5/3 only if it is non-dimensional and not -5/3 when dimensional?

[lucamirtanini]

Originally Posted by LuckyTran

1) The autocorrelation and power spectral density form a Fourier transform pair. The autocorrelation is not just u(t)*u'(t), there is an integration from negative to positive infinity time.

As far as I know, it is not u(t)*u'(t), but u(t)*u(t'). The thing that I do not see is a different time in the second velocity term in the script.

Originally Posted by LuckyTran

2). I'm not sure if u' is a velocity component or the full vector. But if it is a velocity component, u'u' (in addition to v'v' and w'w') is a part of the turbulent kinetic energy, what is the surprise? If u is the velocity magnitude (measured from a hotwire for example) then u'u' will be k.

If you see the figure attached "spectrum" you can notice that the spectrum is normalised. If it is not normalised how can you see the -5/3 scaling, should not it been altered since it is not normalised?

Originally Posted by LuckyTran

2). Or do you mean that the slope is -5/3 only if it is non-dimensional and not -5/3 when dimensional?

Yes! I mean that as far as I understood, the slope should be -5/3 only if normalised.

[LuckyTran]

I have edited my original post. You won't find t' in the script because no one uses the autocorrelation to calculate the psd. They use fft and multiply it by its own conjugate.

[lucamirtanini]

Originally Posted by LuckyTran

I have edited my original post. You won't find t' in the script because no one uses the autocorrelation to calculate the psd. They use fft and multiply it by its own conjugate.

And is it the same thing? Is the result the same?

I have seen the editing of your post.

What about my last question. Is it ok that thee slope is -5/3 also when the spectrum is dimensional?

[LuckyTran]

Originally Posted by lucamirtanini

And is it the same thing? Is the result the same?

Mathematically/theoretically they are identical. Numerically they are sometimes not the same because the FFT is not exactly the same as a true Fourier transform.

Originally Posted by lucamirtanini

Is it ok that thee slope is -5/3 also when the spectrum is dimensional?

It is a power law and when plotted on log-log looks like lines. If you have

B*y = (A*x)^n

This is equivalent to

B*y =(A^n)*(x^n)

When you plot this on log-log, the slope is still n (but the curve is shifted up-down-left-right depending on A and B). So it doesn't matter whether it is non-dimensionalized or not if you are only looking at the slope.

[FMDenaro]

Originally Posted by lucamirtanini

Hi all,

I was looking at a result in the validation guide of fds.

In particular, attached you can see the Figure 6.41 of the FDS User Guide, which is named Power spectrum.

As far as I know, this figure was produced using this script.

https://github.com/firemodels/fds/bl...wer_spectrum.m

I have 2 doubts:

1) In the Tenneke, the Power spectral density is defined as the fourier trasform of the autocorrelation. Since the autocorrelation is u(t)*u(t')/u^2, I cannot see in this script where is the t' (time difference) in the script.

2) Furthermore I cannot understand how the attached figure "Autospectral Density" can exhibit the -5/3 scaling in the inertial subrange, since I thought that the -5/3 scaling should be seen only in the nondimensionalised kinetic Energy spectrum, as you can seen in the figure "Chapman 1979" attached.

See the discussion here Energy Spectrum !!! and search for similar answer.
You will find also a matlab script.

[lucamirtanini]

Originally Posted by LuckyTran

It is a power law and when plotted on log-log looks like lines. If you have

B*y = (A*x)^n

This is equivalent to

B*y =(A^n)*(x^n)

When you plot this on log-log, the slope is still n (but the curve is shifted up-down-left-right depending on A and B). So it doesn't matter whether it is non-dimensionalized or not if you are only looking at the slope.

Thank you... I have understood!!!

Originally Posted by LuckyTran

Mathematically/theoretically they are identical. Numerically they are sometimes not the same because the FFT is not exactly the same as a true Fourier transform.

Can you tell me a reference that I can study that can explain why they are identical?

[LuckyTran]

See Spectral density

and Convolution Theorem


The autocorrelation is a self-convolution (convolution of a signal with itself). The Fourier transform of a convolution is equivalent to the product of Fourier transforms of the individual signals. Hence, to get the power spectral density, you take the Fourier transform (or fft in this script) and multiply it by its own conjugate (because autocorrelation is self-convolution).


And when you do this using a fast fourier transform, it is faster than actually calculating the autocorrelation. Just try it if you don't believe. It is night and day the difference in speed. Just take a signal of random noise and try to calculate both. Or you can use the signal as in the script.

[lucamirtanini]

Sorry but I have other questions
In the tenneke it is often cited that E is proportional to k^(-5/3). Can it be assumed that the same scaling law can be found if we have not wavenumber but frequency?

If I have as output of my analysis a velocity component in 2 points of the space (instead of time as done in the power spectral density) can I produce a one-dimensional spectrum, by doing an autocorrelation?
If I do the the fft of the turbulent kinetic energy do I obtain the three-dimensional spectrum? If not I cannot understand which practical calculation should I do

[FMDenaro]

Originally Posted by lucamirtanini

Sorry but I have other questions
In the tenneke it is often cited that E is proportional to k^(-5/3). Can it be assumed that the same scaling law can be found if we have not wavenumber but frequency?

If I have as output of my analysis a velocity component in 2 points of the space (instead of time as done in the power spectral density) can I produce a one-dimensional spectrum, by doing an autocorrelation?
If I do the the fft of the turbulent kinetic energy do I obtain the three-dimensional spectrum? If not I cannot understand which practical calculation should I do

The frequency and the wavenumber (time or space no matter) are simply linked by k=n*2*pi/L where L is the period.


If you want to perform a spectral analysis in space, you need of a complete spatial distribution, at least in one periodic direction. Only two distinct value cannot define the minimum wavelenght.


No, the fft of the kinetic energy does not produce the energy spectra when you multiply the Fourier coefficient for its conjugate, it is no longer dimensionally congruent to the energy

[lucamirtanini]

1. Is F11 (as cited in the Tenneke and Lumley book, attached pages) equal to E11 (cited in the Pope book and in this paper http://storm.colorado.edu/~jweiss/50...avalli1994.pdf ) ?

2. Is F33 not cited in the Tenneke because it should be considered longitudinal as well?

3. So, is F11 related to the velocity component u (or v ) and F22 to the velocity component w?

4. I still cannot understand how, starting from the output of some velocity component device, i can obtain the 3d spectrum? Can you help telling me the step-by step procedure?

[FMDenaro]

Originally Posted by lucamirtanini

1. Is F11 (as cited in the Tenneke and Lumley book, attached pages) equal to E11 (cited in the Pope book and in this paper http://storm.colorado.edu/~jweiss/50...avalli1994.pdf ) ?

2. Is F33 not cited in the Tenneke because it should be considered longitudinal as well?

3. So, is F11 related to the velocity component u (or v ) and F22 to the velocity component w?

4. I still cannot understand how, starting from the output of some velocity component device, i can obtain the 3d spectrum? Can you help telling me the step-by step procedure?

F11 is the longitudinal spectrum equivalent to E11. You can also find in literature the explicit nomenclature such as Euu, Evv, Eww.


The 3D spectrum is more for theoretical study in isotropic turbulence than for practical applications. In the literature you will often find only one-dimensional spectra.

[lucamirtanini]

Hi, sorry for the new post, but I am still having a lot of doubts:

1. Which is the unit of the three-dimensional spectrum? Which is the unit of E11?

2. In order to obtain the autocorrelation of a given velocity component, I cannot simply do the FFT of v^2, isnt’ it?

3. I am studying the Pope, in order to understand how to calculate the 3d energy spectrum starting form the result a simulation done with fds. I cannot understand why, when Pope explain how to calculate it, he needs to assume the isotropic turbulence. Is it possible to calculate it without this assumption?

4. Can I calculate the spectrum just doing
E(k,t) = 0.5*(e11+e22+e33)*4*PI*k^2 ?
where eij is the Fourier transform of the two point velocity correlation.

5.

Originally Posted by LuckyTran

I have edited my original post. You won't find t' in the script because no one uses the autocorrelation to calculate the psd. They use fft and multiply it by its own conjugate.

I cannot find the theorem that state tat the autocorrelation is equal to use fft and multiply it by its own conjugate. Can you tell me the name of the theorem

[FMDenaro]

Originally Posted by lucamirtanini

Hi, sorry for the new post, but I am still having a lot of doubts:

1. Which is the unit of the three-dimensional spectrum? Which is the unit of E11?

2. In order to obtain the autocorrelation of a given velocity component, I cannot simply do the FFT of v^2, isnt’ it?

3. I am studying the Pope, in order to understand how to calculate the 3d energy spectrum starting form the result a simulation done with fds. I cannot understand why, when Pope explain how to calculate it, he needs to assume the isotropic turbulence. Is it possible to calculate it without this assumption?

4. Can I calculate the spectrum just doing
E(k,t) = 0.5*(e11+e22+e33)*4*PI*k^2 ?
where eij is the Fourier transform of the two point velocity correlation.

5.
I cannot find the theorem that state tat the autocorrelation is equal to use fft and multiply it by its own conjugate. Can you tell me the name of the theorem

Have a look at this discussion https://www.researchgate.net/post/Ho...ent_flow_field


some useful brief notes are here https://www.io-warnemuende.de/tl_fil...Chap4_WS08.pdf where you clearly see the dimension of the Eii(k).



Specifically, you can follow exactly the use of the Fourier transform of the autocorrelation and compare to the use of the multiplication of the Fourier coefficient by its conjugate (for example this way of doing the spectral analysis is used in Matlab, see https://www.mathworks.com/help/matla...-analysis.html).

[lucamirtanini]

Thank you for the answer.
I am searching for the theorem (or explained theory) that state that the autocorrelation is equal to use fft and multiply it by its own conjugate.

[FMDenaro]

Originally Posted by lucamirtanini

Thank you for the answer.
I am searching for the theorem (or explained theory) that state tat the autocorrelation is equal to use fft and multiply it by its own conjugate.

That is explained here https://pubweb.eng.utah.edu/~rstoll/.../Lecture04.pdf

[lucamirtanini]

Originally Posted by FMDenaro

That is explained here https://pubweb.eng.utah.edu/~rstoll/.../Lecture04.pdf

Hi, this slide can be very useful. But unfortunately they confuse me more

1. At which pag can I find what I've asked?

2. In the last slide states

• Typically (when written as) E(k) we mean the contribution to the turbulent kinetic energy (tke)=½(u2+v2+w2) and we would say that E(k) is the contribution to tke for motions of the scale (orsize) k .

So I cannot understand why, starting from the (tke)=½(u2+v2+w2) calculated in every point of the space, doing a fft, cannot I find E(k)?

[FMDenaro]

Originally Posted by lucamirtanini

Hi, this slide can be very useful. But unfortunately they confuse me more

1. At which pag can I find what I've asked?

2. In the last slide states

• Typically (when written as) E(k) we mean the contribution to the turbulent kinetic energy (tke)=½(u2+v2+w2) and we would say that E(k) is the contribution to tke for motions of the scale (orsize) k .

So I cannot understand why, starting from the (tke)=½(u2+v2+w2) calculated in every point of the space, doing a fft, cannot I find E(k)?

When you look at the autocorrelations, you see each component of the main diagonal of a tensor. The total kinetic energy is the trace of such tensor.

Depending on the homogeneity of the flow problem, we can analyse in separate way each contribution to the tke along wavenumber ki.

That is what is explained in slide #10 and #11.

[lucamirtanini]

Originally Posted by FMDenaro

When you look at the autocorrelations, you see each component of the main diagonal of a tensor. The total kinetic energy is the trace of such tensor.

Depending on the homogeneity of the flow problem, we can analyse in separate way each contribution to the tke along wavenumber ki.

That is what is explained in slide #10 and #11.

But, why, starting from the (tke)=½(u2+v2+w2) calculated in every point of the space, doing a fft, cannot I find E(k)?