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Angle of incidence, RF modul, Comsol 4.2a
Posted 3 nov. 2012, 07:31 UTC−4 RF & Microwave Engineering Version 4.2a 5 Replies
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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.
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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.
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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
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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！