15 ms·

The technical analysis you're putting out here is a bit of a canard. Yes, better filters help, but radars are really, really picky. After all, there's an inve

by mlyle 4y ago

The technical analysis you're putting out here is a bit of a canard. Yes, better filters help, but radars are really, really picky. After all, there's an inverse fourth power relationship between return strength and distance. So we have our doppler weather radar transmitting at 450kW, traveling out a big distance to a storm (inverse square), and reflecting at very low efficiency, and traveling back (inverse square). Compare to a base station putting out 40W that's closer and just subject to inverse square law. It can easily be 10 orders of magnitude stronger. You need pretty good filtering for this. > One thing is different: around 20 GHz there's lots of frequencies available (vs 5GHz) so we could have larger guard band without significant impact. You need a much larger guard band. It's easy to make a 1MHz wide filter at 10MHz, and really hard at 100000MHz. Another important thing: parasitics start to matter a whole lot, too. You can have a filter that sharply rolls off around your fundamental, but then above the resonant frequencies of your passives/filter elements become transmissive again. It's pretty hard to keep 20GHz out of a receiver that was designed for a lower frequency before 20GHz was a major concern.

beecafe 4y ago
nitpick: it's not inverse fourth power, it's just inverse square with twice the distance, which is 1/((2 * distance) ^ 2) = 1/(4 * distance ^ 2) (ignoring reflection efficiency, but that isn't determined by distance)
- mlyle 4y ago
  > nitpick: it's not inverse fourth power, it's just inverse square with twice the distance, which is 1/((2 * distance) ^ 2) = 1/(4 * distance ^ 2) No. For a diffuse reflection, the amount of light hitting your target is inverse square. And then it is scattered and inverse square back on the return journey. 1/x^2 * 1/x^2 = 1/x^4. https://en.wikipedia.org/wiki/Radar#Radar_range_equation https://en.wikipedia.org/wiki/Radar#Radar_range_equation "In the common case where the transmitter and the receiver are at the same location, Rt = Rr and the term Rt² Rr² can be replaced by R^4, where R is the range." (This all assumes that your target is smaller than the beam size, of course-- which is not as true for the two cases of a radar altimeter or a doppler radar as it is for e.g. tracking aircraft... but it's still close enough in practice).

by mlyle 4y ago

beecafe 4y ago

nitpick: it's not inverse fourth power, it's just inverse square with twice the distance, which is 1/((2 * distance) ^ 2) = 1/(4 * distance ^ 2) (ignoring reflection efficiency, but that isn't determined by distance)

mlyle 4y ago
> nitpick: it's not inverse fourth power, it's just inverse square with twice the distance, which is 1/((2 * distance) ^ 2) = 1/(4 * distance ^ 2) No. For a diffuse reflection, the amount of light hitting your target is inverse square. And then it is scattered and inverse square back on the return journey. 1/x^2 * 1/x^2 = 1/x^4. https://en.wikipedia.org/wiki/Radar#Radar_range_equation https://en.wikipedia.org/wiki/Radar#Radar_range_equation "In the common case where the transmitter and the receiver are at the same location, Rt = Rr and the term Rt² Rr² can be replaced by R^4, where R is the range." (This all assumes that your target is smaller than the beam size, of course-- which is not as true for the two cases of a radar altimeter or a doppler radar as it is for e.g. tracking aircraft... but it's still close enough in practice).