No, it isn’t that simple. The problem is that as a phased array radar scans off its boresite, if the element spacing is too large, you will start to get grating lobes, which are essentially additional main beams, which causes all sorts of problems. There are techniques that can be used to suppress these grating lobes such as those used in the Sea-Based X-Band radar (this is why the different sections of the antenna in Figure 3 do not line up exactly, even so its electronic scan is limited to about plus or minus 12 degrees), but it is desirable to avoid the problem altogether. For a full scan angle of plus or minus 60 degrees, with elements on a square grid, the maximum allowable element spacing is about 0.536λ which gives an area per element of 0.287λ^2. (I see in my previous response I mistakenly said 0.278 instead of 0.287.)

This is not a calculation I did myself, I simply looked up the answer, specifically in Theodore E. Cheston and Joe Frank, “Chapter 7: Phased Array Radar Antennas,” in Merrill I. Skolnik, Radar Handbook, 2nd ed. (1990), pp. 7-17 to 7-21. ]]>

If you don’t mind, one more question. How did you derive the element area as function of lambda from module spacing? It doesn’t appear to be as simple as squaring the distance.

]]>I wrote out a derivation, but unfortunately when I paste it into the reply box, it loses all formatting. So I’ll just try to explain it here. If you still have a question, let me know I can try to write it out here or email it to you as a Word file.

The key point here is that the area of each antenna is given by A = N*k*λ^2 ,

Where k is a constant that depends on the element arrangement and the required electronic field of view. As discussed earlier in the post, for a square array with a plus or minus 60 degree electronic scan, k = 0.278. Since k will be the same for both S and X band, so we can omit it. Then we have:

A = N*λ^2 and then A^2 = N^2*λ^4

Substituting for Ax^2 and As^2 in the equation just above the one you cite:

(Nx^3*λx^4)/λx^2 = (Ns^3*λs^4)/λs^2

So then:

Nx^3*λx^2 = Ns^3*λs^2

Which gives: Nx = Ns*(λs/λx)^2/3. ]]>

I think we have to assume the defense will make use of all of the sensor information and computer processing capabilities available to it. It is known that missile defense radars such as the X-band TPY-2 and the Aegis SPY-1 do use inverse synthetic aperture techniques to obtain small cross-range resolutions (I refer to this as Doppler processing in the post). So it is a certainty that radars such as the LRDR do also. Other techniques, such the time variation of the return from individual scatterers, will also be used. The main point here is that most, if not all, of these techniques work better as the bandwidth increases (and hence range resolution improves), and bandwidth will be significantly better at X-band than at S-band.

However, signal strength is also very important for these techniques, and for a fixed amount of money to build a radar, you will get significantly better signal strength at S-band than at X-band.

The most important thing in the post, and the reason I wrote it, is that we now have the people that did the classified trade studies for the LRDR admitting that a X-band would have given superior discrimination performance and that the reason S-band was chosen was to save money.

Given that discrimination is generally recognized as the most difficult problem for above-the- atmosphere missile defenses, maybe spending the extra money would have been a good idea. Hard to say anything definitive, because we don’t know how big the cost difference would have been to build a X-band LRDR capable of obtaining a signal strength equal to an S-band version. It would almost certainly be at least twice as much and probably much more.

]]>One possibility would be using inverse synthetic aperture, using the motion of the warheads, to increase the effective spatial resolution, at least in one dimension, well beyond that possible by pure optics. Another could be the use of spectral signatures of the radar reflections. Who knows what else, but the point is that the improved resolution from reducing the wavelength from S to X bands may not be as important as it would have been with older techniques, before the era of massive parallel computing power.

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