Alejandro, you want to make sure you're normalizing to the correct value of
'a'. If you want lambda to be 6e-6 and a=1e-5, the corresponding normalized
frequency should be a/lambda = 1.66.

As for resolution, the value you set is the number of grid points that 'a'
will be divided into: so a value of res=6.66 means a grid point every
a/6.66= 1.5e-6...I think your cell size is correct.

-ck


On Mon, Oct 18, 2010 at 11:00 AM, <[email protected]>wrote:

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>   1. Very lost with Meep Units (Alejandro Escu?n El?cegui)
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> ----------------------------------------------------------------------
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> Message: 1
> Date: Mon, 18 Oct 2010 17:45:22 +0200
> From: Alejandro Escu?n El?cegui
>        <[email protected]>
> Subject: [Meep-discuss] Very lost with Meep Units
> To: <[email protected]>
> Message-ID: <[email protected]>
> Content-Type: text/plain; charset="iso-8859-1"
>
>
> Hello Everybody,
>
> I'm researcher working about computacional electromagnetic and I am probing
> Meep to watch the behaviour of the software, but I don't understand very
> well the Unit in Meep.
>
> I think that I understand the meaning of 'a' and 'c', but I don't know if I
> am using it well. For example, if I want to simulate a perfect cube with the
> following characteristics:
>
> - side = 150um
> - source = continuous => Sine wave
> - frequency = 15 e13
> - lambda = 6 e-6
>
>
> Since my lambda is  6 e-6, I take 'a'= 1 e-5, so lambda_meep=0.6, and
> frequency is 1/lambda. With this value, if the size of my stucture is 150um,
> the size is 15 a's = 15*(1 e-5) = 150 e-6. Have I got any error?
>
> Moreover, I want a total of 100 Yee lattice, so I have to set the
> resolution to 6.666. Explication: 15*resolution = 100 cells, so resolution =
> 100/15, no?
>
> I'm a bit lost, please, I need your help.
>
> Thank you very much.
>
> Alejandro Escu?n
>
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