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Subdomain integration for effecive area calculation in PCF
Posted 15 mars 2010, 08:59 UTC−4 2 Replies
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Hi,
I am doing a project based on parametric analysis of Photonic crystal fiber. For calculating effective area, I selected the option subdomain integration. I have given Ex*conj(Ex) to calculate numerator and (Ex*conj(Ex))^2 for denominator.
My doubt is in effective area equation for numerator one square term is there.
Is it need to square the value again in numarator?
For more clarification please go through the attachment..
Thanks in advance...
Jais John
I am doing a project based on parametric analysis of Photonic crystal fiber. For calculating effective area, I selected the option subdomain integration. I have given Ex*conj(Ex) to calculate numerator and (Ex*conj(Ex))^2 for denominator.
My doubt is in effective area equation for numerator one square term is there.
Is it need to square the value again in numarator?
For more clarification please go through the attachment..
Thanks in advance...
Jais John
Attachments:
2 Replies Last Post 9 mars 2014, 09:10 UTC−4
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Posted:
1 decade ago
Hi
from the dimensions here I believe you should respect the formula, no ?
if E is in [V/m] then ¦E¦^2 is (V/m)^2, now dx and dy are both in [m] (note current version 3.5a of Comsol tends to forget this in their integration variables) so your numerator is simply of units V^4
while the denominator is (V/m)^4*m^2 = V^4/m^2, which gives, once divided one by the other a unit of [m^2] for an area, and that is what I would expect, ... no ?
But this does still not necessarily mean that the formula is correct for what you are looking for ...
Hope this helps
Ivar
from the dimensions here I believe you should respect the formula, no ?
if E is in [V/m] then ¦E¦^2 is (V/m)^2, now dx and dy are both in [m] (note current version 3.5a of Comsol tends to forget this in their integration variables) so your numerator is simply of units V^4
while the denominator is (V/m)^4*m^2 = V^4/m^2, which gives, once divided one by the other a unit of [m^2] for an area, and that is what I would expect, ... no ?
But this does still not necessarily mean that the formula is correct for what you are looking for ...
Hope this helps
Ivar
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Posted:
1 decade ago
Hello sir,
i have to do surface integration of two outputs , which are having the same geometry and mesh
plese see my attachment
i have to do surface integration of two outputs , which are having the same geometry and mesh
plese see my attachment