[MLB-WIRELESS] Reflectors and Repeaters
Roger Venning
r.venning at telstra.com
Thu Jan 3 22:38:32 EST 2002
No, adding the two antenna gain & subtracting connecting cable loss is
certainly not correct. This set up is certainly feasible though. I have
heard that a number of the big 2m+ dishes out on Canterbury Rd are doing
just this, where they are linked by waveguide and essentially just put
on a high point to bend the signal around the crest of a hill /
curvature of the earth. In summary they present a very large amount of
'loss' though, with typical 'gain' of a well engineered reflector going
to be down around -40dB when bouncing stuff between two 1km equi-distant
endpoints.
If we imagine that we had a transmitter/antenna combination with an EIRP
of 1W/m^2. Taking this over some intervening amount of free space and
obstructions imposes attenuation x dB and then resulting in incident
power of 10^(-x/10) W/m^2. If the receiving antennae has an effective
receiver area of y m^2 (closely related to the gain.... hmm... dig
through some old notes: yes, gain to effective receiver area is the same
for all antenna, and the ratio (G:A) is 4*pi/lamba^2), then it will pick
up power equal to y*10^(-x/10) W or ..... from Friis' idealised
Transmission Equation:
Pr = Pt*lambda^2*Gr*Gt / (4*pi*R)^2
Pr = Power recieved
Pt = Power transmitted
lambda = wavelength
Gr = Gain of receiving antenna (receiving half of the reflector)
Gt = Gain of transmitting antenna (our original source)
R = distance between antenna
Combining Friis' eqn for the other half, assuming 1/L of the power
couples into the second antenna gives us
Pr = Pt*lambda^4*Gt*Gr1*Gr2*Gr / (L * (4*pi)^4 * R1^2 * R2^2)
recognising the difference between this and the original Friis formula
without intervening gives us the 'gain' of the intervening reflector
(yes, a long winded way to go about it, but this is the first time I've
thought about it) and assuming that R1 = R2 => R1 = R/2 = R2 =>
(R1^2*R2^2) = (R^4/16)
P' / P (ie. the 'gain' of the gunk in the middle) = lambda^2*Gr1*Gr2 /
(L * pi^2 * R^2)
in the normal dB terms and taking Gtot as the sum of two antenna gain in
dB, also Cl (coupling loss) as the value of (1/L) in dB and working at
2.4 GHz => lambda = 0.125m
gain of reflector = Gtot + 2*10*log(0.125/pi) - Cl - 2*10*log(R)
= Gtot - 28dB - Cl - 20*log(R)
The 20*log(R) term is all about the difference in suffering the 6dB loss
every time you double the distance over two stretches instead of just
one when you place something in the way. As an example, if we had two
24dB gain monsters, a 2km distance, the reflector halfway at 1km, and no
connector loss to simplify things, we end up with a 'gain' of the system
at (24+24) - 28 - 20*log(1000) = 48 - 28 - 60 = -40dB. You could do you
normal link budget stuff with 2km of separation etc. and then 'just
whack' this huge attenuation of -40dB in the middle. Note if there was
only 100m of separation but a huge obstacle intervened, this would this
time be just -20dB.
Tony: does this seem correct? It finds the ~32dB difference you spoke
about if you combined the -28dB with a typical Cl of 4dB (I love fudge
factors!), but the 20*log(R) term is new here...
Roger.
Tony Langdon, VK3JED wrote:
> At 07:37 PM 3/01/2002 +1100, you wrote:
>
>> Tony, can you find out the details of this? I can't see why there
>> should be any 32 dB constant involved. I think it should be
>> simple as receive antenna gain minus cable loss plus transmit
>> antenna gain.
>
>
> I'll see if I can, but that's what I recall from seeing the formula.
>
>> This arrangement can even appear to have some gain -- which
>> is possible if the transmit antenna has narrower beamwidth
>> than the receive antenna.
>
>
> Well, if both antennas have a lot of gain, it will definitely appear
> to have more gain...
>
> 73 de Tony, VK3JED
> http://www.qsl.net/vk3jed
>
>
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-------------------------------------------------------------
Roger Venning \ Do not go gentle into that good night
Melbourne \ Rage, rage against the dying of the light.
Australia <r.venning at bipond.com> Dylan Thomas
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