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[to Mr. jian] strange Radiaton pattern of antenna on GaAs ..

时间:04-01 整理:3721RD 点击:
Hello. Mr. Jian

Using IE3D, i want to see a radiation pattern of simple dipole antenna which is positioned between air(upper substrate) and [very thick] GaAs(lower substrate).

So, i've set the substrate condition like my attached figure, and made a 100mm-long dipole antenna on z=0 plane, and checked radiation pattern at 1-λ frequency (~0.9 GHz)

But the radiation pattern (in Directivity) is peculiar. there is no radiation to the below... Is that the limitation of IE3D simulation ? or my fault about setting?

thank you for reading. Mr. Jian


Hi, Insang:

1. Radiation pattern assumes that it is in air and there is no dielectric loss in the air, even in practical applications there might be slight loss. Therefore, the radiated field is:

E = F(theta, phi) * exp( -j k0 r ) / r

What a radiation pattern means is that we neglect the distance dependent factor or we define the radiation pattern as:

F(theta, phi ) = E (theta, phi, r ) * r * exp( j k0 r ) and r is approaching infinity

It becomes distance independent quantity.

2. When you have 2 different dielectrics in the forward and backward directions, if we don't assume one dielectrics has loss and the other one does not have loss, we will not be able to define the radiation pattern in the traditional way.

In reality when loss is considered for both dielectrics, we will have:

E1 = F1(theta, phi) * exp( - j k1 * r ) / r for the forward direction

E2 = F2(theta, phi) * exp( - j k2 * r ) / r for hte backward direction

Then, if we solve |E*r|, you will see the |E1*r| and |E2*r| are both dependent upon distance. We are not able to define a radiation pattern on it because it is distance dependent.

For this reason, we have to assume one dielectric is lossy and another one is not lossy. For example, the air is not lossy while the GaAs is lossy.

Then, when we take E*r*exp( - j k0 * r ) and let r approaching infinity, you will see the product becomes a non-zero constant of distance in the upper space while it becomes 0 in the lower space. That is why you get no field in the lower space because the field is absorbed by the GaAs material as ohmic loss.

Regards.

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