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N_eff for 2D mode analysis

Posted 1 apr 2012, 02:39 GMT-4 RF & Microwave Engineering, Studies & Solvers Version 4.1, Version 4.2, Version 4.2a 7 Replies

Tee Din Chai
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Anyone know why the Neff obtained from below two method is not the same??

i) mphglobal(model, 'emw.neff'')
ii) [sz,ndofs,list,name,type]=mphgetp(model,'solname','sol1') and then use "list" to calculate neff with neff=real(1j*list/(2*pi/Lambda)),

for example,
the neff i obtained by method (i) is 1.436800701048935
by method (ii) is 1.437795377426004

I am using comsol 4.2

Thanks
7 Replies Last Post 3 apr 2012, 04:54 GMT-4
Alexander Kuznetsov
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Posted: 1 decade ago 2 apr 2012, 12:50 GMT-4
what's your exact definition of Lambda?
what's your exact definition of Lambda?
Deepak Jain
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Posted: 1 decade ago 2 apr 2012, 14:59 GMT-4
Hi

i do not understand what are these two methods. Can you please a bit elaborate it or share your .mph files.

regards
deepak
Hi i do not understand what are these two methods. Can you please a bit elaborate it or share your .mph files. regards deepak
Alexander Kuznetsov
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Posted: 1 decade ago 2 apr 2012, 16:36 GMT-4
I don't know what the methods you posted are.
I asked about Lambda because it's mentioned in the second method. It's important how you calculate it. My suspicion is that lambda was defined as Speed of light over frequency. Now if you use 3*10^8 instead of 299,792,458 m/s you get exactly the difference between 1.436800701048935 and 1.437795377426004
I don't know what the methods you posted are. I asked about Lambda because it's mentioned in the second method. It's important how you calculate it. My suspicion is that lambda was defined as Speed of light over frequency. Now if you use 3*10^8 instead of 299,792,458 m/s you get exactly the difference between 1.436800701048935 and 1.437795377426004
Tee Din Chai
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Posted: 1 decade ago 2 apr 2012, 23:58 GMT-4

Hi

i do not understand what are these two methods. Can you please a bit elaborate it or share your .mph files.

regards
deepak
[/QUOT



that two methods are for Matlab comsol.
one is to extract the propagation constant (beta) from matlab comsol..
another one is to extract effective refractive index from matlab comsol...

thx
[QUOTE] Hi i do not understand what are these two methods. Can you please a bit elaborate it or share your .mph files. regards deepak [/QUOT that two methods are for Matlab comsol. one is to extract the propagation constant (beta) from matlab comsol.. another one is to extract effective refractive index from matlab comsol... thx
Tee Din Chai
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Posted: 1 decade ago 3 apr 2012, 00:01 GMT-4

I don't know what the methods you posted are.
I asked about Lambda because it's mentioned in the second method. It's important how you calculate it. My suspicion is that lambda was defined as Speed of light over frequency. Now if you use 3*10^8 instead of 299,792,458 m/s you get exactly the difference between 1.436800701048935 and 1.437795377426004


Hi,

i use matlab comsol, i change the Lambda value each time for different simulation.
so, i defined the lambda value by myself, example, from 1000nm to 2000nm with increment of 10nm.

i think because of the way comsol calculate lambda as C/F, and the C i used is 3e8, while, comsol used 2.998e8.
you exactly help me to solve this.

By the way, do you know how many decimal point comsol use?
if i put 1.23456789 , Comsol will treat it as 1.23456789 or 1.235??
thx
[QUOTE] I don't know what the methods you posted are. I asked about Lambda because it's mentioned in the second method. It's important how you calculate it. My suspicion is that lambda was defined as Speed of light over frequency. Now if you use 3*10^8 instead of 299,792,458 m/s you get exactly the difference between 1.436800701048935 and 1.437795377426004 [/QUOTE] Hi, i use matlab comsol, i change the Lambda value each time for different simulation. so, i defined the lambda value by myself, example, from 1000nm to 2000nm with increment of 10nm. i think because of the way comsol calculate lambda as C/F, and the C i used is 3e8, while, comsol used 2.998e8. you exactly help me to solve this. By the way, do you know how many decimal point comsol use? if i put 1.23456789 , Comsol will treat it as 1.23456789 or 1.235?? thx
Deepak Jain
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Posted: 1 decade ago 3 apr 2012, 04:42 GMT-4
Hi,

decimal digits depends on tolerance value in solver configuration.


Deepak
Hi, decimal digits depends on tolerance value in solver configuration. Deepak
Tee Din Chai
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Posted: 1 decade ago 3 apr 2012, 04:54 GMT-4

Hi,

decimal digits depends on tolerance value in solver configuration.


Deepak


i tried for the tolerance value of 1e-12,
but, it show me WARNING
""
The relative error is greater than the relative tolerance, for the direct solver.
Relative residual: 6.9e-013.
Relative error: 1.6e-010.
""

Do you knw what is this error?
[QUOTE] Hi, decimal digits depends on tolerance value in solver configuration. Deepak [/QUOTE] i tried for the tolerance value of 1e-12, but, it show me WARNING "" The relative error is greater than the relative tolerance, for the direct solver. Relative residual: 6.9e-013. Relative error: 1.6e-010. "" Do you knw what is this error?

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