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Hi,
Your feed look like strange. But to begin, you can only, with your software (CST ?) feed with ports at 90 phase shift you antenna array. Then , size you antenna to obtain the right resonant frequency. Better, size only one patch and after build your array. I think that at 9 Ghz the...
For PAE :
In fact, it depend on the class of you amplifier.
Class B , class C or class AB.
For the Noise as we have previously said : the noise is not taken into account for a power amplifier.
In fact for an amplifier, if you choose to minimize the noise, the gain will not be maximum and...
Hi,
For maximum power : the criterion is the matching to the load and source charge.
For bandwidth : you must check that the amplifier is stable on whole bandwidth and beyond, that is matched too and that the gain is almost constant.
Bye
Hi,
For all i know, the quality factor of a cavity is a function of the loss. And an ideal cavity without loss gives modal solutions different from a real cavity (see Maxwell equations which give : Laplacien(E) + k² E = 0 where k is a function of material values and gives rise to the eigen value...
Hi,
In the "compute antenna parameter" module, the gain is in linear value. so you have to take 10 log(gain).
In a 3D polar plot, it's you that define the representation of the gain, dB value, linear value ... If you have choose dB and if the result is 1.5e+001 for the red area, the gain is 15...
Hi,
There are a lot's of references.
However, to begin, I suggest you to read this book writen by Matthew N.O. Sadiku : "numerical Techniques in Electromagnetics".
It's a very good reference to understand the basic elctromagnetics methods.
have a good reading.
Bye
Hi mirou,
Good news.
I had a look to your array-patches.
In fact your geometry had several mistakes.
I have send you on your email adresse an explanation of the correction.
The result is :
A good gain : about 12 dBi that's the maximum for this four patch array.
A frequency resonnance of : 52...
Hi Mirou,
Well, are you sure of the patch impedance (50 Ohms). Did you determine it by simulation ? Or calculus ? You can, in a first time simulate the feeder alone by placing a 50 Ohms impedance instead of the patch element.
This will permit you to know if the feeder is well done ...
If you...
Hi,
No because, 50 ohms in paralell with 50 ohms gives rise to 25 ohms so you have to transform 25 to 100 ohms and not 50 to 100. Consequently the transformer used is a 50 ohms caracteristic impedance ...do the calculus...
Bye
Hi,
In fact, You have to take into account the loss of your dielectric (tan(delta)) and the relative permittivity. The higher the permittivity the smaller the antenna. The lower the tan(delta) the higher efficiency of your antenna. The thickness of the substrate is of great importance too. The...
Hi,
In order to obtain circular polarization you must have two modes spacially orthogonal and with a 90° phase shift. These two modes are usually distributed in the same manner in the space (for the electric or the magnetic field). The magnetic field can give us the surface current ( Js = H^n)...
Hi,
FDTD is for time varying field and FEM is for harmonic field (time is frozen).
FDTD is rather for CEM for example.
FEM work well for antennas.
About computer memory, the two simulations are volumic meshes so need huge memory (depend on the structure simulated).
Regards
Hi,
Two slots on circular patch give rise to a circular polarized antenna...
I suppose you've only one feed ? Perhaps your slots are not optimized and your two degenerate mode are not properly excited.
Regards
Hi mirou,
first : Have you made any optimization with HFSS ?
second : why did you choose a 70 Ohms lambda/4.It's possible but a little bit complicated, I think.
If your two patches are 50 Ohms at the "T" junction you've got 25 Ohms so after the "T" you must change this 25 Ohms impedance with a...
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