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Angle of incidence, RF modul, Comsol 4.2a

Dear All,

I have issue. I have one metal block on substrate, and I want to illuminate it but at some angle (picture).

One half sphere is substrate, one half air.

Now I want to apply x polarization, scattered field.


I have two cases, first when substrate is also air, and second when substrate is glass (with index refraction ns).

In the case when substrate is air I applied equation: E0*exp(-j*(emw.k0*x*sin(phi)+emw.k0*z*cos(phi)).

In the case when substrate is glass I applied equation: E0*exp(-j*(emw.k*x*sin(phi)+emw.k*z*cos(phi)).


Or maybe in second case I have to write:

(1-sign(z))*E0*(-j*2*pi*x*sin(phi)/Lambda-j*2*pi*z*cos(phi)/Lambda)/2 +

+ (1+sign(z))*E0*(-j*2*pi*x*sin(phi)*ns/Lambda-j*2*pi*z*cos(phi)*ns/Lambda)/2


Tell me please for any of these equations are they correct.


Thank very much in advanced.


5 Replies Last Post 2012年11月18日 GMT-0500下午11:58
Posted: 5 years ago 2012年11月17日 GMT-0500上午11:17
I want to know that you use the scattering boundary condition or the Port to input the light, and how do you set the wave direction or the propagation constant
I want to know that you use the scattering boundary condition or the Port to input the light, and how do you set the wave direction or the propagation constant

Posted: 5 years ago 2012年11月17日 GMT-0500上午11:53

I want to know that you use the scattering boundary condition or the Port to input the light, and how do you set the wave direction or the propagation constant


I used scattering boundary condition, but I am nor sure that I understoood second question about equation.
[QUOTE] I want to know that you use the scattering boundary condition or the Port to input the light, and how do you set the wave direction or the propagation constant [/QUOTE] I used scattering boundary condition, but I am nor sure that I understoood second question about equation.

Posted: 5 years ago 2012年11月18日 GMT-0500上午12:44
I think you can type E0*exp(-j*(emw.k0*x*sin(phi)+emw.k0*z*cos(phi)) in the "incident field", or you can just set kx=k0*sin(phi), ky=k0*cos(phi) in the "wave direction"
I think you can type E0*exp(-j*(emw.k0*x*sin(phi)+emw.k0*z*cos(phi)) in the "incident field", or you can just set kx=k0*sin(phi), ky=k0*cos(phi) in the "wave direction"

Nikola Bralovic
Posted: 5 years ago 2012年11月18日 GMT-0500上午9:42

I think you can type E0*exp(-j*(emw.k0*x*sin(phi)+emw.k0*z*cos(phi)) in the "incident field", or you can just set kx=k0*sin(phi), ky=k0*cos(phi) in the "wave direction"


Hello,

I am not sure about second way. I used first way: E0*exp(-j*(emw.k0*x*sin(phi)+emw.k0*z*cos(phi)). I always use incidetn field, not wave direction.

Cheers
[QUOTE] I think you can type E0*exp(-j*(emw.k0*x*sin(phi)+emw.k0*z*cos(phi)) in the "incident field", or you can just set kx=k0*sin(phi), ky=k0*cos(phi) in the "wave direction" [/QUOTE] Hello, I am not sure about second way. I used first way: E0*exp(-j*(emw.k0*x*sin(phi)+emw.k0*z*cos(phi)). I always use incidetn field, not wave direction. Cheers

Posted: 5 years ago 2012年11月18日 GMT-0500下午11:58
Hello
Do you find that if you use the same incident field for example E0*exp(-j*(emw.k0*x*sin(phi)+emw.k0*z*cos(phi)), but select the different boundary as the input, the wave direction is different, do you know why?
Cheers!
Hello Do you find that if you use the same incident field for example E0*exp(-j*(emw.k0*x*sin(phi)+emw.k0*z*cos(phi)), but select the different boundary as the input, the wave direction is different, do you know why? Cheers!

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