passive repeater
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passive repeater
Does anyone have the length or the formula for the jumpers for a passive repeater in 800 mhz?
Can anyone please explain why both the question and the answer don’t make sense and if it does make sense, why?
If I understand the question correctly, than the cable between the two antennas should be equal to the amount of waves between the two antennas plus any number of full waves as needed. Taking into account the velocity factor of the cable used.
As an example, if the distance is 10.3 waves, the cable needs to be this long as well times the velocity factor (electrical wave length). And since this is now to short, any amount of additional full waves (electrical wavelength) should be added.
Nand.
If I understand the question correctly, than the cable between the two antennas should be equal to the amount of waves between the two antennas plus any number of full waves as needed. Taking into account the velocity factor of the cable used.
As an example, if the distance is 10.3 waves, the cable needs to be this long as well times the velocity factor (electrical wave length). And since this is now to short, any amount of additional full waves (electrical wavelength) should be added.
Nand.
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- Joined: Wed Jun 11, 2003 1:52 pm
passive repeater
I apologize
, if i was unclear in the post.
if i understood , the distance for a 860 mhz will be
(3X10e8/8.610e8)(0.84)= 0.2930 mts (where 0.84 is the
factor for a Heliax 1/4", if i need more distance it will be
(0.2930)(2) or (0.2930)(3)...etc.
what others factors i must consider.
Thanks for your time,
Lenny

if i understood , the distance for a 860 mhz will be
(3X10e8/8.610e8)(0.84)= 0.2930 mts (where 0.84 is the
factor for a Heliax 1/4", if i need more distance it will be
(0.2930)(2) or (0.2930)(3)...etc.
what others factors i must consider.
Thanks for your time,
Lenny
Re: passive repeater
Well, the math sector of my brain just exploded. Is that a factor?lenny kravitz wrote:
(3X10e8/8.610e8)(0.84)= 0.2930 mts (where 0.84 is the
factor for a Heliax 1/4", if i need more distance it will be
(0.2930)(2) or (0.2930)(3)...etc.
what others factors i must consider.




"I'll eat you like a plate of bacon and eggs in the morning. "
- Some loser on rr.com
eBay at it's finest:
Me: "What exactly is a 900Mhz UHF CB?"
Them: "A very nice CB at 900Mhz speed!"

- Some loser on rr.com
eBay at it's finest:
Me: "What exactly is a 900Mhz UHF CB?"
Them: "A very nice CB at 900Mhz speed!"

Passive Repeater== A device that passes signals to and from a poor signal location, redirects signals into another aera.
Usally the Passive Repeater is just two directional antennas hooked to back to back by the best cable one can get as it is only PASSIVE.. There needs to be isolation between the antennas, usally physical seperation.
One antenna is placed to pickup signals from the real repeater or cell site and the other is placed in a building or tunnel to help signals to get in and out of a structure/ tunnel.
Usally the Passive Repeater is just two directional antennas hooked to back to back by the best cable one can get as it is only PASSIVE.. There needs to be isolation between the antennas, usally physical seperation.
One antenna is placed to pickup signals from the real repeater or cell site and the other is placed in a building or tunnel to help signals to get in and out of a structure/ tunnel.
I likely wasn’t to clear either. If you want to feed the inside antenna with the exact phase of the wave at a point so as not to have any cancellation because of the original signal, you need to use a cable that has the electrical length of this distance. And since the cable will be physically shorter in that case, you can add additional full waves to make it longer. Full waves do not change the phase.
Your calculation for figuring out the electrical wavelength in meters of the cable is correct (300,000,000) / (frequency in Hz) X (cable velocity factor).
First you need the distance in meters between the two antennas and figure out the true wavelength for that path.
Assume for example that the measured distance is 10.3 meters.
The real wavelength at 860 MHz is 0.3488 meters.
This gives you the distance between the two antennas as 28.67 waves (10.3 / 0.3488).
Now since the cable with this electrical wavelength is shorter than the real distance between antennas because of the velocity factor, you can add any number of additional electrical full wavelengths to this cable without changing the phase relationship. Like 40.67 waves or 56.67 waves. This also allows you to route the cable around corners etc.
If there isn’t any or very little coverage inside, then there is no point in doing it this way. You would only go to this extend if the direct signal competes with the passive radiated signal. It may be easier in that case to move the antenna in small increments until a good spot is found.
As far as cable loss goes, you don’t need to use Heliax for short distances. It all depends on the how strong a signal you have to work with outside and how far you need to go into the building. And again, if you have to go a long distance where there is no penetration from the outside, then the cable length is not important.
Also, if you want to use ¼ inch Heleiax, you will save some money using LMR400 instead. There is no good reason to use Heliax in this case. All you need is a low loss cable if you go the distance. Try to keep the loss below 2 dB with a gain antenna outside and a dipole inside or a small beam if you need more directivity.
Nand.
Your calculation for figuring out the electrical wavelength in meters of the cable is correct (300,000,000) / (frequency in Hz) X (cable velocity factor).
First you need the distance in meters between the two antennas and figure out the true wavelength for that path.
Assume for example that the measured distance is 10.3 meters.
The real wavelength at 860 MHz is 0.3488 meters.
This gives you the distance between the two antennas as 28.67 waves (10.3 / 0.3488).
Now since the cable with this electrical wavelength is shorter than the real distance between antennas because of the velocity factor, you can add any number of additional electrical full wavelengths to this cable without changing the phase relationship. Like 40.67 waves or 56.67 waves. This also allows you to route the cable around corners etc.
If there isn’t any or very little coverage inside, then there is no point in doing it this way. You would only go to this extend if the direct signal competes with the passive radiated signal. It may be easier in that case to move the antenna in small increments until a good spot is found.
As far as cable loss goes, you don’t need to use Heliax for short distances. It all depends on the how strong a signal you have to work with outside and how far you need to go into the building. And again, if you have to go a long distance where there is no penetration from the outside, then the cable length is not important.
Also, if you want to use ¼ inch Heleiax, you will save some money using LMR400 instead. There is no good reason to use Heliax in this case. All you need is a low loss cable if you go the distance. Try to keep the loss below 2 dB with a gain antenna outside and a dipole inside or a small beam if you need more directivity.
Nand.
- Big Blue TO/\/\
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OK, much lower-tech here...
Am working with an ambulance manager, trying to solve a cell-phone dillemma he has. He has gone out and spent >300.00 on a passive cell antenna for the ambulance. The phone inside is a flip-phone, .6 watt
I used one of these passive dealies, way back when, with very little if any result.
Anyone out there have any info that they actually work decent now, especially with Digital Cell Signal ?
Next, the price? did he get scr%$#D ? they were much cheaper back when...
Tom
Am working with an ambulance manager, trying to solve a cell-phone dillemma he has. He has gone out and spent >300.00 on a passive cell antenna for the ambulance. The phone inside is a flip-phone, .6 watt
I used one of these passive dealies, way back when, with very little if any result.
Anyone out there have any info that they actually work decent now, especially with Digital Cell Signal ?
Next, the price? did he get scr%$#D ? they were much cheaper back when...
Tom
KE4RXM
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- Big Blue TO/\/\
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OK, what's a powered BDA ?
I know that there are now amplifiers for digital cell... but what he described to me was a passive glassmount antenna...
And the amplifier really won't fix the problem. (topic for another thread)
I know that there are now amplifiers for digital cell... but what he described to me was a passive glassmount antenna...
And the amplifier really won't fix the problem. (topic for another thread)
KE4RXM
You don't drink beer? Use Motorola RSS, you WILL soon...
PLEASE use EMAIL over PM. Much Easier for Me. Thanx!
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PLEASE use EMAIL over PM. Much Easier for Me. Thanx!
[quote="Big Blue TO/\/\"]OK, what's a powered BDA ?[/quote]
Bi-Directional-Amplifier - basically, a pair of band-limited amplifiers back-to-back in a "69" configuration. One amp will provide gain on the outbound frequency, one on the inbound frequency.
If you have a shielded site - one where RF on the inside doesn't leak out to the outside very much - you can set up a BDA with an antenna outside and an antenna inside. However, you have to have more attenuation from the inside to the outside than you have gain, or you have a dandy oscillator.
Bi-Directional-Amplifier - basically, a pair of band-limited amplifiers back-to-back in a "69" configuration. One amp will provide gain on the outbound frequency, one on the inbound frequency.
If you have a shielded site - one where RF on the inside doesn't leak out to the outside very much - you can set up a BDA with an antenna outside and an antenna inside. However, you have to have more attenuation from the inside to the outside than you have gain, or you have a dandy oscillator.
This is my opinion, not Aeroflex's.
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I WILL NOT give you proprietary information. I make too much money to jeopardize my job.
I AM NOT the Service department: You want official info, manuals, service info, parts, calibration, etc., contact Aeroflex directly, please.