1
$\begingroup$

I'm making my way through the Linesman-related section's of Gough's Watching the Skies. Linesman was developed to counter the carcinotron jammer.

The main solution was the Type 85, which had 12 frequencies switching each pulse at random.

However, they also deployed a second radar, the Type 84. This was a normal single-frequency pulse radar, but in the L-band rather than the S-band. This was to be used as an early-warning system.

I'm trying to understand why... for one, as a single-frequency system it would be highly vulnerable to jamming, and for another, the longer wavelength would mean lower resolution or much larger antennas.

Gough mentions a single possible reason for this, in passing, that L-band was less susceptible to clutter. I'm not sure why this would be.

Does anyone have any ideas on why the L-band would be better for early warning?

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ nothing to do with clutter; search radars are long range radars, using L-band has to do with "spreading" loss that is lower for lower frequencies, see Friis's formula en.wikipedia.org/wiki/Friis_transmission_equation $\endgroup$ – hyportnex Jan 23 '18 at 17:23
  • $\begingroup$ But isn't this the inverse of the directivity? IE, given that the receiver and transmitter is the same antenna, you get a D^2 term that decreases at the same rate as the improvement in this formula? $\endgroup$ – Maury Markowitz Jan 23 '18 at 18:36
  • $\begingroup$ This question is not really about "Physics" and the answer is in radar system design and for that there many engineering considerations. For example, making the directivity too high and having a very narrow beam will slow down the target acquisition below acceptable level for an airdefense system. My comment was really aimed only to minimize the importance of clutter that you mentioned explicitly in the context of a ground-to-air search radar such as Lineman of which I know next to nothing. $\endgroup$ – hyportnex Jan 23 '18 at 18:45
1
$\begingroup$

Ahh, found the answer in a totally unrelated document.

The L-band has less interaction with rain, hail, snow and clouds, and therefore reduces clutter in the long-range role.

So it was a physics reason.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Feel free to accept your own answer if it did solve the question. $\endgroup$ – Fingolfin Jan 30 '18 at 17:35
  • $\begingroup$ Check out Mie scattering to see why S-Band sees rain and L-band doesn't--it's all about raindrop (snowflake) size relative to the wavelength. $\endgroup$ – JEB May 7 '18 at 13:18
  • $\begingroup$ At ~2.5 cm wavelength, wouldn't Mie be small and Rayleigh larger? $\endgroup$ – Maury Markowitz Dec 10 '18 at 14:47
1
$\begingroup$

If the radar’s mission is to search a given solid angle in a given period of time, then there is no real advantage (other than accuracy) to using a higher frequency, because a narrower beam will not be able to dwell as long on any given position. If the design features a phased array antenna, it is less costly to build it from lower-frequency modules, since the number of modules required per unit area scales as the square of frequency. L-band technology offers a nice compromise between cost and accuracy.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

S-Band is used for weather radar (signal for NWS, clutter for DOD)--that's in long range application. It's used for final approach in air traffic track surveillance. L-band is used for air-route surveillance (ARSR) and defense application (e.g. AN/FPS-117) in the long range (200 nautical miles).

Range resolution is determined by bandwidth, which is typical around 1 MHz, and that can easily be achieved in L or S band.

Regarding frequency hopping, not having it is a weakness (in military applications).

Aside: L-Band is currently allocated for Earth observations, surveillance, and geo-navigation (GPS), so if you want to do Earth science from space, you have to worry about RFI from the other applications.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.