Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Taylor Distribution for Planar Arrays

Status
Not open for further replies.

mleem

Newbie level 3
Newbie level 3
Joined
Dec 9, 2008
Messages
3
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,305
If I need to achieve a certain SLL from a (square) planar array, I first was working a linear array of the same width. I achieved the result I wanted, but was unsure how to extend this to the planar array.

I've seen it done where the distribution was determined instead by using the largest dimension of the array (in this case, the diagonal) and then just rotating this around the whole array. The problem with this is that on the principal planes, you no longer get the SLL you want since it is shorter than the rotated distribution. You would have to use a much lower Taylor to achieve this, but that makes an unnecessarily large HPBW.

If I determine the distribution from the principle planes, and rotate it, what tapering would go on the corner elements? Or, would I need to modify the distribution for each angle to the length?
 

Square? Does that mean it has to be a square lattice? Can you use a triangular lattice? I think that will move some of the energy into the intercardinal planes.
 

Unfortunately, the element design forces it to be a square lattice.

So, square array (dimension-wise) and square lattice.
 

Hello,

to have circualr like side lobes with the required level you need a quadrantal symmetry in your array... In this way the corner elements is the one excited at lowest level..

Bye.
 

So, I perform a Taylor weighting in a circular form on the array and then for the corner elements that would not be included, I just weight them lower than the outer ring?

For example of weighting, see below. Blue is the circular Taylor and black would be the self-entered corner weights...

 

That seems like a heck of a lot of different power dividers?
 

No, you have to design the array directly from the optimised taylor linear array. The resulting coeffcients behaviour are similar the yours (quadrantal symmetry).

Let me know your array grid in wavelenegth and the required side lobe level, I'll give you the right set of cofficients.

However why do you need all the lobes at same level ? as pointed out the solution leads to a very difficult set of exctitaion t be implemented...

bye.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top