[MLB-WIRELESS] Reflectors and Repeaters

Thu Jan 3 22:54:32 EST 2002

Greets again, all,

Just looking around subsequent to the math, and found the following at 
http://www.telexwireless.com/wlanfaq.htm (an excellent page):
--------------------

Q: Can I use a passive (unpowered) repeater on a hilltop in a 2.4 GHz 
NLOS PTP link?

A: If the distance to the hilltop (or obstructing building) is less than 
0.75 mile AND the total PTP distance is less than 1.5 miles, it may be 
possible to place a passive repeater on the hilltop to acheive the NLOS 
link. The Transmit EIRP should be at least +48 dBm, the passive repeater 
should consist of 2 each 36" (24 dBi) back-to-back dish antennas 
connected together with less than 1 dB feedline loss, and the Receive 
antenna should be a 36" (24 dBi) dish. The 2 antennas on each end of the 
transmit link should both be vertical polarization, and the 2 antennas 
on each end of the receive link should be horizontal polarization (or 
vice-versa).  Both links need to be aligned and clear of ANY 
obstructions for this to work.

---------------------
The reasons why the two links should be run on opposite polarisations 
kind of escape me at the moment (that is I couldn't come up with 
anything better than handwaving arguments).

Roger.

Roger Venning wrote:

>
> 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
>>
>>
>> ------------------------------------------------------------------------
>>
>>
>> ---
>> Outgoing mail is certified Virus Free.
>> Checked by AVG anti-virus system (http://www.grisoft.com).
>> Version: 6.0.312 / Virus Database: 173 - Release Date: 31/12/2001
>>
>
>

-- 
-------------------------------------------------------------
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

--
To unsubscribe, send mail to minordomo at melbwireless.dyndns.org with a subject of 'unsubscribe melbwireless'  
Archive at: http://melbwireless.dyndns.org/cgi-bin/minorweb.pl?A=LIST&L=melbwireless
IRC at: au.austnet.org #melb-wireless