Taylor current distributions for 5x5 planar array

[kae_jolie]

i know how to calculate the Taylor current distributions for say a 5-element linear array. Let's say I want to have a 5x5 planar array, how would I calculated the Taylor current distributions for all the 25 elements? Would it simply be the same as the 5-element linear array except add 4 more rows of the same thing?

Say the current distribution for a 5-element linear array is: 1 2 3 2 1
Would the current distribution for a 5x5 planar array be:

1 2 3 2 1
1 2 3 2 1
1 2 3 2 1
1 2 3 2 1
1 2 3 2 1

Or:


1 1 1 1 1
1 2 3 2 1
1 2 3 2 1
1 2 3 2 1
1 1 1 1 1 ?

Do we need to make all the edge elements have the lowest current?

Thanks.

 

[zorro]

Hi,

It is the product of the distributions in the two axis, i.e., if the values of the distribuion in x are a1,a2,a3,... and in y are b1,b2,b3,... then for the planar array they are:

a1*b1 a2*b1 a3*b1 a2*b1 a1*b1
a1*b2 a2*b2 a3*b2 a2*b2 a1*b2
a1*b3 a2*b3 a3*b3 a2*b3 a1*b3
a1*b2 a2*b2 a3*b2 a2*b2 a1*b2
a1*b1 a2*b1 a3*b1 a2*b1 a1*b1

This is general. In your case, ai and bi are the same (provided you want similar characteristics on the two axis).
Regards

Z

 

[kae_jolie]

zorro, thanks a lot for this. Where can I read this info. i can't get my hands on it. could you send me a specific paper or book that covers this in detail? Also, is there a different algorithm for circular Taylor array distribution?

thanks.

 

[zorro]

I like these books that cover all aspects of phased arrays:

Mailloux: Phased Array Antenna Handbook, 2nd Edition (Artech House, 2005)
Hansen: Phased Array Antennas, 2nd Edition (Wiley, 2009)

There should be good information i the web as well (e.g. course notes)
For circular aperture, Taylor distribution gives a radial tapering.
Nevertheless, I don't know if for a discrete array (i.e. for example an array of elements forming an hexagonal grid inside a circle, instead of a continuous aperture), there are exact formulas like for the rectangular array.
Regards

Z

 

[kae_jolie]

zorro,

I tried the product of the distributions in the two axis and it works, I am getting the desired results. However, I need to read this in some text. I looked in both texts you referenced above and more texts I have here but I couldn't find anyone touching on this point specifically. I also looked online and couldn't find any reference. Also, I looked for the words "quadrant symmetry" and couldn't find anything, either, in reference to planar arrays. I really need a reference to put in my paper to show that I got this from a textbook or a journal paper, etc...
I would appreciate it if you could tell me which page number in any of these books I can find this information.

Also, one more related question: how would you assign the progressive phase shift to each of the 25 elements? Is it the same as linear array? For example, the 45° progressive phase shift for a 5-element linear array is:

0 45 90 135 180

For a 5x5 planar array, would it be this:

0 45 90 135 180
0 45 90 135 180
0 45 90 135 180
0 45 90 135 180
0 45 90 135 180

Thanks.

 

[zorro]

Hi, kae_jolie

I don't have the books at hand now.
Rather tan for "quadrant symmetry" look for "separable distribution" or "product distribution" for planar arrays.

With the phase distribution you show, the beam will be steered horizontally. (If we are looking the array from the front and the phases are lags, -not advances-, then the beam will point towards our right.)
Of course, if you apply the progressive phase shift in the othar axis, it will be steered vertically.
You can point the beam in other directions adding the corresponding phases needed for the horizontal and vertical steering.

Regards

Z

 

[kae_jolie]

Thanks, zorro. This was helpful. I did find some references with the correct keywords...

Now I am still stuck on the progressive phase shift part. I need to steer the 5x5 planar array at phi= 45° and theta= 30°. I came up with phase shift (phase) = -pi/(2*sqrt(2))

I constructed this phase matrix for the phase shift in Matlab:

phase_matrix = [phase*2 phase*1 phase*0 -phase*1 -phase*2;
phase*3 phase*2 phase*1 phase*0 -phase*1;
phase*4 phase*3 phase*2 phase*1 phase*0;
phase*5 phase*4 phase*3 phase*2 phase*1;
phase*6 phase*5 phase*4 phase*3 phase*2];

Is the above matrix correct? I tried to construct it so that horizontal and vertical elements have progressive phase shift, not just one axis. I am not really getting the pattern that I am looking for. The beam is slightly shifted off broadside but not centered at 30°.

Thanks.

 

[zorro]

Hi kae_jolie,

I don't see how do you get the value of "phase".
There must be two phase progressions that add at each element: one for steering the beam in the x direction (phi) and the other for the y directon (theta).
Remember that the electrical phase difference between adjacent elements is related with the geometrical steering angle (in the corresponding direction) and with d/lambda.
Regards

Z

 

[kae_jolie]

The equations used for phase shift calculations are straight from Balanis book (page 352)

beta_x = -k*dx*sin(theta0)*cos(phi0)
beta_y = -k*dx*sin(theta0)*sin(phi0)

For desired theta0 = 30 degrees and phi0 = 45 degrees and dx = dy = 0.5*lambda, beta_x and beta_y = -pi/(2*sqrt(2)) rad or 64 degrees. Balanis has the same calculation on page 357 and has the polar plot for it. My calculations match Balanis book, but my polar plot does not look the same for the scanned beam. The polar plot looks the same, though, for the broadside array on page 355. This leads me to believe that I am not distributing the current amplitudes correctly on all the 25 elements.

Thanks.

 

[zorro]

The matrix of phases should be:

for beta_x: the term beta_x multiplied by this matrix:

2 1 0 -1 -2
2 1 0 -1 -2
2 1 0 -1 -2
2 1 0 -1 -2
2 1 0 -1 -2

for beta_y: the term beta_y multiplied by this matrix:

-2 -2 -2 -2 -2
-1 -1 -1 -1 -1
0 0 0 0 0
1 1 1 1 1
2 2 2 2 2

They add, with the result (multiplied by "phase"):

0 -1 -2 -3 -4
1 0 -1 -2 -3
2 1 0 -1 -2
3 2 1 0 -1
4 3 2 1 0

That is the same as yours, with a constant displacement.
Maybe the difference with the book is the sign in an axis or a difference in the sign of the progressive phase. Its effect would be the side towards which the beam in steered.
You could try flipping this matrix along the horizontal and the vertical axes, or both. Bot you should get the same or symmetrical results.
Could you show your result?
Regards

Z