CST and circular polarization

ludwig polarization

Hello. I've got a problem. I need to design a dielectric slab in a conical horn antenna for cicular polarization. I use CST, but I don't know how to see the result. How can I see in CST if the polarizator works? Axial ratio must be "1" for circular polarization, but there is only result in "negative" numbers. Which other quantities i have to see for result? The magnitude of the electric field in the probe, which is oriented in X-axis is about 50dB lower then in probe, which is oriented in Y-axis = so it's good and it's truth (basic mode TE11) - it's without dielectric slab. But with the dielectric slab, the magnitude will change only a little.
So, does anybody help me? Thanks

cst farfield coordination

Hi,

In fact, I have discussion with some guys here about the same matter two monthe ago. I don't know why I can't find it now. Anyway, you need to look for your far-field results in ludwig 3 coordination not the cartesian one. Of course, you have to have your radiation in z-direction, otherwise ludwig 3 will give no sence results.
Then, check the most potential field. If it is vertical, then it is vertical polarization, and vise versa. For circular polarization, I guess you will have the same radiation levels in both directions.

I have question for you: How I can check the Axial ratio in CST??
Let me know about your progress, it is important for me as well.

Regards

circular polarization in cst

Hello,
well, It's not me, who is doing the simulation in CST. It's my older :idea: colleague. So, I don't know exactly how to show the axial ratio :|. (I'm not sure, but I'think there is a item of menu on the left side of the window) - advanced members will help more.
We will continue on the project on monday, so We'll check your advice.
Thanks

circular polarization cst

Hi,

Sorry, I meant that how we can decide from axial ratio the polarization ? I know it is on the left window in far-field results. I didn't specify my goal clearly.

So, how we can decide if it is horizontal or vertical polarization, LHCP or RHCP ??
Could you just give me hint or link that explain this? I'm learning CST myself and without good knowledge of these terminologies.

Regards

ludwig polarizacion

Hi,
Axial ratio is the expression of elliptical polarization. AR=Emax/Emin, where Emax is the max. intensity of vector E. For linear polarization AR is infinity (Emin=0). If the Emax is in horizontal plane the polarization is called horizontal. Similarly for vertical polarization. For circular polarization AR is 1 (Emax=Emin).

something about the polarizer: h**p://www.atnf.csiro.au/observers/memos/AT39.3_106.pdf. There is a explanation of axial ratio too.

So, do I understand your question properly?

ludwig +polarization

Yes, I'm fully understand now the meaning. Wish you all the best. Let me know please when you get your problem solved.

Regards

checking circular polarization cst

hi guys,
to check the polarisation of circular antenna you should choose spherical coordinates. When this option is activated three drawings are ploted: RHCP, LHCP and Axial Ratio . If the antenna is designed for RHCP there should be a very low level of Lhcp and viceversa. In case you choose ludwig3 there should be equal field in vertical and horizontal, as abuantenna said. Be aware that the AR should be near 0 dB on the main direction of radiation but it would decrease (that is a unavoidable bad effect of circular polarised antennas) in all other directions.
I hope this will help.
best regards

axial ratio cst

Hello guys,

I think we had some discussion before for the same matter.
My question is: If we assume that we don't know the polarization and we like to let CST give us the answer.
1-Shall we start with ludwig3 and check, if we have potential horizontal part, so we have horizontal polarization, and vise versa.
2-If we have amost equal components for both of them
3-Then, we have to move to spherical coordination
If the axial ratio is 1, then we have exactly circular polarization. To check if it is left or right, we can do that from LHCP and RHCP components.

Am I right ??

Dear Abuantenna,
You are right, or at least that is what I would do if I intended to know the polarisation of an antenna. Let me remind you again that when using ludwig3 the antenna polarisation should be aligned with the x or y axes (or u, w when using an auxiliar set of axes)
best regards

Hi all
I'm back and my problem is solved.
Here is the solution:
Place at least 2 probes. Both probes are placed in the z-axis = the direction of main radiation and behind the slab (there should be circular polarization here). First probe (probe 1) is orientated in X-axis - horizontally, second probe (probe 2) in Y-axis = vertically. The key is the position of probes - must be in z-axis.

There is the dominant TE11-mode in waveguide. Without the slab, the E-field measured by the probe 1 (X-axis) is near zero.
If the slab is used and has the CORRECT LENGTH, the value of the E-field, measured by the probe 1 is the same as the value measured by the probe 2. The phase difference is 90°.

Of course, with the correct slab in the farfield result (ludwig3 coordination), has the field in the vertical and horizontal direction same value. AR is near 0dB.

Thank you for help.


-In the pictures, the probe 1 is before the slab and is Y-axis oriented, probe 2 is behind the slab (vertical). Probe 4 is behind the slab (horizontal)

Ok, great.

Just one primative question: When you put the probes. How to tell CST to calculate fields on that probes ?

Regars

Hello guys,

an anyone give me the expression linking Axial Ratio and Cross polarization level?

In my notes, I can't find it anymore..

Thanks in advance.

Lupin

Hi everyone,
do you have anyone of you any experiences with planar square corcular polarization antennas? I trying to engineer this antenna for my school project. Here is the problem, I would like simulate this problem in CST, but I don´t understood some thinks in this thread. Could anybody explame me step by step, how can i find out correctness of my results?

BTW: I'm so sorry for my english;-)