Nyquist wavenumber

lucamirtanini · June 9, 2021, 08:19

Hi all,
I am looking at this Energy spectrum in the wavenumber domain. The dashed line is the Nyquist wavenumber. My question is: how is it possible that results are shown after the Nyquist wavenumber? Shouldn't that be the limit after that there are no value?

FMDenaro · June 9, 2021, 12:09

Quote:

Originally Posted by lucamirtanini

Hi all,
I am looking at this Energy spectrum in the wavenumber domain. The dashed line is the Nyquist wavenumber. My question is: how is it possible that results are shown after the Nyquist wavenumber? Shouldn't that be the limit after that there are no value?

Right, the extension of the spectra from a numerical simulation cannot be extended for k>Kc.

You should check for more details, is that from a published paper?

lucamirtanini · June 9, 2021, 14:03

Thanks for your answer. It is the verification guide of FDS that you can find here https://pages.nist.gov/fds-smv/manuals.html. There is a chapter about turbulence from page 45 to 53. I cannot understand how can they have value after kc. Maybe they apply Taylor hypothesis?

FMDenaro · June 9, 2021, 14:24

Quote:

Originally Posted by lucamirtanini

Thanks for your answer. It is the verification guide of FDS that you can find here https://pages.nist.gov/fds-smv/manuals.html. There is a chapter about turbulence from page 45 to 53. I cannot understand how can they have value after kc. Maybe they apply Taylor hypothesis?

From what I understand, the numerical simulation has greater resolution (64^3) than the CBS (32^3) and they adopted an explicit filter. Therefore, the extension of the spectra is reported until the numerical Nyquist frequency of the finer grid.

lucamirtanini · June 9, 2021, 14:35

The filter in FDS is implicit, as attached. The CBS is the experiment of Comte Bellot and Corrsin, the two top figure are related to the 32^3 the bottom two 62^3. I really cannot understand

FMDenaro · June 9, 2021, 15:02

Quote:

Originally Posted by lucamirtanini

The filter in FDS is implicit, as attached. The CBS is the experiment of Comte Bellot and Corrsin, the two top figure are related to the 32^3 the bottom two 62^3. I really cannot understand

Actually, what I really do not understand from the figures is the number of wavenumber components reported. In the top figure, the grid is 32^3, that is the spectra should be reported for 16 wavenumbers. On the bottom (64^3), the number of components should be 32. But if you count the circle you see that they are doubled, creating a fictious doubled Nyquist frequency.My opinion is that either they wrongly performed the FFT or the number of grid in each direction has to be doubled. You should ask to the author of the verification.

lucamirtanini · June 9, 2021, 15:09

Ok. Thank you

I will do that. I wanted simply to be sure that my doubts are lecit. I have also an other question about the subgrid model, but I will open an other thread

Eifoehn4 · June 9, 2021, 16:16

I think the authors use the spherical wave number defined in 3D as

$|\mathbf{k}|=\sqrt{k_x^2+k_y^2+k_z^2}.$

Here the maximum values are:

Mesh with 32x32x32:

$|\mathbf{k}|=\sqrt{16^2+16^2+16^2}\approx 28$

Mesh with 64x64x64:

$|\mathbf{k}|=\sqrt{32^2+32^2+32^2}\approx 55$

This fits well with the presented results.

Regards

FMDenaro · June 9, 2021, 16:20

Quote:

Originally Posted by Eifoehn4

I think the authors use the spherical wave number defined in 3D as

$|\mathbf{k}|=\sqrt{k_x^2+k_y^2+k_z^2}.$

Here the maximum values are:

Mesh with 32x32x32:

$|\mathbf{k}|=\sqrt{16^2+16^2+16^2}\approx 28$

Mesh with 64x64x64:

$|\mathbf{k}|=\sqrt{32^2+32^2+32^2}\approx 55$

This fits well with the presented results.

Regards

From the figure, the Nyquist line seems at different wavenumbers. For example, in the 32^3 grid it appears exactly at the wavenumber 16

Eifoehn4 · June 9, 2021, 16:44

Yes, your right. With 32x32x32 the Nyquist wavenumber remains at the maximum value of 16. However, it is possible to make an analysis with $|\mathbf{k}|$ .

Whether that makes sense is another question. I also did this in some analyses.

FMDenaro · June 9, 2021, 16:48

Quote:

Originally Posted by Eifoehn4

Yes, your right. With 32x32x32 the Nyquist wavenumber remains at the maximum value of 16. However, it is possible to make an analysis with $|\mathbf{k}|$ .

Whether that makes sense is another question. I also did this in some analysis.

If the spectra is along r you should consider a sphere in the box, if the radius is greater than L you have spurious wavenumbers.

However, the figures in the FDS file have no clear explanation.

lucamirtanini · June 10, 2021, 06:25

Quote:

Originally Posted by Eifoehn4

Yes, your right. With 32x32x32 the Nyquist wavenumber remains at the maximum value of 16. However, it is possible to make an analysis with $|\mathbf{k}|$ .

Whether that makes sense is another question. I also did this in some analyses.

It could be a 3D spectrum, also because if it is not, they should have used subscript such as 11 or 22. Furthermore it is a isotropic turbulence, where a 3D spectrum can be calculated, and maybe deduced from E_11.

I didn't understand a thing. Also if we make a 3D spectrum, how is it possible to see the wavenumberS after the nyquist wavenumber? shouldn't it be the limit? Does this change in a 3D contest?

FMDenaro · June 10, 2021, 07:13

In isotropic homogeneous turbulence, the statistics are the same under rotation and traslation, this way the spectrum E11,E22,E33 has the same behavior. And the spectrum computed for a sphere of radius r=L should be also the same.

If you want to assume that dr=sqrt(dx^2+dy^2+dz^2) then you have the Nyquist frequency pi/dr but this is not what I see in the figure

Eifoehn4 · June 10, 2021, 07:26

Suppose your only have two modes in each direction:

x-Direction
$k_x(:,:,1) = \left( \begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix} \right)$
$k_x(:,:,2) =\left( \begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix} \right)$

y-Direction
$k_y(:,:,1) = \left( \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right)$
$k_y(:,:,2) =\left( \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right)$

z-Direction
$k_z(:,:,1) = \left( \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right)$
$k_z(:,:,2) =\left( \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} \right)$

Spherical
$|\mathbf{k}|(:,:,1) = \left( \begin{matrix} 0 & 1~~ \\ 1 & \sqrt{2} \end{matrix} \right)$
$|\mathbf{k}|(:,:,2) =\left( \begin{matrix} 1 & \sqrt{2} \\ \sqrt{2} & \sqrt{3} \end{matrix} \right)$

Now you may accumulate your data in Fourier space according these modes.

FMDenaro · June 10, 2021, 07:31

Quote:

Originally Posted by Eifoehn4

Suppose your only have two modes in each direction:

x-Direction
$k_x(:,:,1) = \left( \begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix} \right)$
$k_x(:,:,2) =\left( \begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix} \right)$

y-Direction
$k_y(:,:,1) = \left( \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right)$
$k_y(:,:,2) =\left( \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right)$

z-Direction
$k_z(:,:,1) = \left( \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right)$
$k_z(:,:,2) =\left( \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} \right)$

Spherical
$|\mathbf{k}|(:,:,1) = \left( \begin{matrix} 0 & 1~~ \\ 1 & \sqrt{2} \end{matrix} \right)$
$|\mathbf{k}|(:,:,2) =\left( \begin{matrix} 1 & \sqrt{2} \\ \sqrt{2} & \sqrt{3} \end{matrix} \right)$

Now you may accumulate your data in Fourier space according these modes.

Not sure to understand you example, it is not for isotropic case, right?

Eifoehn4 · June 10, 2021, 07:40

I only wanted to show, what the authors may have done numerically. Perhaps the explanation is not quite well.

lucamirtanini · June 10, 2021, 07:40

Quote:

Originally Posted by FMDenaro

In isotropic homogeneous turbulence, the statistics are the same under rotation and traslation, this way the spectrum E11,E22,E33 has the same behavior. And the spectrum computed for a sphere of radius r=L should be also the same.

If you want to assume that dr=sqrt(dx^2+dy^2+dz^2) then you have the Nyquist frequency pi/dr but this is not what I see in the figure

The FDS guide says that they consider as $\Delta$ ,which they use as Filter, the geometric mean $\sqrt[3]{\Delta x\cdot\Delta y\cdot\Delta z}$ .

Actually this leads to a discrepancy in the 3d spectra the Nyquist wavenumber is not the filter limit. Could be?

acf46545 · June 10, 2021, 07:51

Quote:

Originally Posted by lucamirtanini

Hi all,
I am looking at this Energy spectrum in the wavenumber domain. The dashed line is the Nyquist wavenumber. My question is: how is it possible that results are shown after the Nyquist wavenumber? Shouldn't that be the limit after that there are no value?

Is this the theoretical Nyquist number with 2 points, or is it the numerical one for the discretization scheme? Not familiar with FDS, but if it uses an an implicit filter plus finite volume scheme, then the spectrum can contain infinite frequencies. The cut off wavenumber reported can thus be the theoretical or numerical, but depending on how the FFT is done on the finite volume solution, there can be very high frequencies present. The spectra shown here are nothing unusual for an implicit filter.

FMDenaro · June 10, 2021, 07:58

The Nyquist frequency is identified by the grid size, no matter about which numerical method is used. The filter width identifies the Nyquist frequency only for spectral methods, otherwise pi/Delta is not equal to pi/h.

acf46545 · June 10, 2021, 08:06

Quote:

Originally Posted by FMDenaro

The Nyquist frequency is identified by the grid size, no matter about which numerical method is used. The filter width identifies the Nyquist frequency only for spectral methods, otherwise pi/Delta is not equal to pi/h.

I agree. But the question is different: is the line in the plot: Nyquist number (theoretical) or cut off number of the FDS scheme? Also, second question, how is the finite volume solution interpolated to do FFT?

June 9, 2021, 15:09		#7
lucamirtanini Senior Member luca mirtanini Join Date: Apr 2018 Posts: 165 Rep Power: 8	Ok. Thank you I will do that. I wanted simply to be sure that my doubts are lecit. I have also an other question about the subgrid model, but I will open an other thread FMDenaro likes this.

June 9, 2021, 16:16		#8
Eifoehn4 Senior Member - Join Date: Jul 2012 Location: Germany Posts: 184 Rep Power: 13	I think the authors use the spherical wave number defined in 3D as $\|\mathbf{k}\|=\sqrt{k_x^2+k_y^2+k_z^2}.$ Here the maximum values are: Mesh with 32x32x32: $\|\mathbf{k}\|=\sqrt{16^2+16^2+16^2}\approx 28$ Mesh with 64x64x64: $\|\mathbf{k}\|=\sqrt{32^2+32^2+32^2}\approx 55$ This fits well with the presented results. Regards __________________ Check out my side project: A multiphysics discontinuous Galerkin framework: Youtube, Gitlab.

June 9, 2021, 16:44		#10
Eifoehn4 Senior Member - Join Date: Jul 2012 Location: Germany Posts: 184 Rep Power: 13	Yes, your right. With 32x32x32 the Nyquist wavenumber remains at the maximum value of 16. However, it is possible to make an analysis with $\|\mathbf{k}\|$ . Whether that makes sense is another question. I also did this in some analyses. __________________ Check out my side project: A multiphysics discontinuous Galerkin framework: Youtube, Gitlab.

June 10, 2021, 07:13		#13
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,781 Rep Power: 71	In isotropic homogeneous turbulence, the statistics are the same under rotation and traslation, this way the spectrum E11,E22,E33 has the same behavior. And the spectrum computed for a sphere of radius r=L should be also the same. If you want to assume that dr=sqrt(dx^2+dy^2+dz^2) then you have the Nyquist frequency pi/dr but this is not what I see in the figure lucamirtanini likes this.

June 10, 2021, 07:26		#14
Eifoehn4 Senior Member - Join Date: Jul 2012 Location: Germany Posts: 184 Rep Power: 13	Suppose your only have two modes in each direction: x-Direction $k_x(:,:,1) = \left( \begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix} \right)$ $k_x(:,:,2) =\left( \begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix} \right)$ y-Direction $k_y(:,:,1) = \left( \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right)$ $k_y(:,:,2) =\left( \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right)$ z-Direction $k_z(:,:,1) = \left( \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right)$ $k_z(:,:,2) =\left( \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} \right)$ Spherical $\|\mathbf{k}\|(:,:,1) = \left( \begin{matrix} 0 & 1~~ \\ 1 & \sqrt{2} \end{matrix} \right)$ $\|\mathbf{k}\|(:,:,2) =\left( \begin{matrix} 1 & \sqrt{2} \\ \sqrt{2} & \sqrt{3} \end{matrix} \right)$ Now you may accumulate your data in Fourier space according these modes. __________________ Check out my side project: A multiphysics discontinuous Galerkin framework: Youtube, Gitlab.

June 9, 2021, 14:03		#3
lucamirtanini Senior Member luca mirtanini Join Date: Apr 2018 Posts: 165 Rep Power: 8	Thanks for your answer. It is the verification guide of FDS that you can find here https://pages.nist.gov/fds-smv/manuals.html. There is a chapter about turbulence from page 45 to 53. I cannot understand how can they have value after kc. Maybe they apply Taylor hypothesis?

June 10, 2021, 07:40		#16
Eifoehn4 Senior Member - Join Date: Jul 2012 Location: Germany Posts: 184 Rep Power: 13	I only wanted to show, what the authors may have done numerically. Perhaps the explanation is not quite well. __________________ Check out my side project: A multiphysics discontinuous Galerkin framework: Youtube, Gitlab.

June 10, 2021, 07:58		#19
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,781 Rep Power: 71	The Nyquist frequency is identified by the grid size, no matter about which numerical method is used. The filter width identifies the Nyquist frequency only for spectral methods, otherwise pi/Delta is not equal to pi/h.

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