Example scenario:
There's a mountain between your house and the tv station and you cannot pick
up a tv signal to watch.
SO - you buy 3 high-gain direction tv antennas at radio shack.
You go up to the top of the mountain and set one antenna to point to the tv
station, and one of the other antennas to point toward your house.
The 2 antennas on the mountain feed to each other with 300-ohm ribbon cable-
like the other antenna was the tv set. (one acts like a receiver, and the
other like a transmitter.)
The third high-gain antenna is at your house - pointed to the antenna on top
of the mountain. This antenna feeds to your tv.
So yes - you need directional antennas - three of them.
The antennas need to 'see' each other.
I heard of folks using rhombic antennas on the mtn (uhf and vhf) - good
gain.


"Bryan Martin" wrote in message
...
The building location is maybe 500 feet away from where the repeater would
be. The parabolic grid is maybe 3000-3500 feet away from the repeater
location.

As for the output we will talk about the grid antenna first. It is

I keep hearing about "passive repeaters", but haven't ever seen anything
quantitative about how well they work. I had some measurements done
years ago which showed that a cell phone "passive repeater" in a sedan
doesn't do anything significant, quite obviously because the windows are
such large apertures at that frequency.

But here are some numbers to work with. Corrections are welcome.

Assuming perfect efficiency, the maximum effective aperture of an
antenna is Aem = G * lambda^2 / (4 * pi), where

lambda = the wavelength
G = the antenna's numerical power gain in the direction being analyzed

The gain in dBi = 10 * log[base10](G). For convenience, use meters for
length dimensions, so the aperture is in square meters and the
wavelength is in meters.

This means that if we have a field with power density Pdi striking an
antenna with power gain G1 (in the direction the field is coming from),
the antenna can deliver Pdi * Aem watts to a conjugately matched load.

Now let's connect this antenna to a second antenna with numerical power
gain G2, and assume that the two are perfectly matched to each other.
That is, the input impedance of the second antenna is the complex
conjugate of the impedance the first antenna has when driven. Also
assume no loss in the connecting line.

The power delivered to the second antenna is then Pdi * Aem watts.

The second antenna will radiate a field with power density equal to G2
times that of an isotropic antenna supplied with the same power, or

G2 * (Pdi * Aem) / (4 * pi * r^2)

where r is the distance from the antenna in the direction in which the
gain is measured.

Combining these to find the ratio of power density of the field radiated
from the second antenna to the power density of the field striking the
first antenna, we get:

P density ratio = G1 * G2 * (lambda / (4 * pi * r))^2

This is the numerical gain you'll see between the field at the location
of the first passive receiver antenna to the field at a distance r from
the second (radiating) passive receiver antenna. A ratio of less than
one represents attenuation rather than gain. The gain in dB is 10 * the
base 10 logarithm of this ratio, with negative results representing
attenuation. That is,

Overall gain (dB) = 10 * log[base 10](P density ratio)

Now let's look at a couple of examples. For simplicity, assume that none
of the original field arrives at the detector location; that is, there's
a perfect shield or obstruction between the original field and the
detector (technically, the detector's antenna). It doesn't matter what
kind of antenna is physically connected to the detector for this
analysis. In real life, the detector would probably be a receiver, but
I'll call it a detector so hopefully its antenna won't be confused for
either of the two passive receiver's antennas.

Let G1 = G2 = 10 -- both antennas comprising the passive repeater have
10 dBi gain (although don't forget that G1 and G2 are numerical, not dB
gains -- the two numbers just happen to be the same in the case of 10).
Wavelength = 30 meters (10 MHz frequency), and r = 10 meters -- we've
put the detector's antenna 10 meters from the second (radiating) passive
repeater antenna(*).

Working through the numbers, the power density at the detector's antenna
is 5.7 times, or 7.56 dB greater than, the power density striking the
first antenna. So the signal is stronger than it would have been if we'd
put the detector's antenna right where the first passive repeater
antenna is. Don't forget, though, that it took two 10 dBi antennas to
get that 7.56 dB gain.

But now let's look at the same setup but at 150 MHz (2 meter
wavelength). The power density at the detector's antenna in this case is
just 2.5% of (or 15.96 dB less than) the power density of the field
striking the first antenna.

In all cases, if we move the detector's antenna twice as close to the
second (radiating) passive repeater antenna, that is, to 5 meters away,
we'll gain 6 dB; if we double the distance to 20 meters we lose 6 dB.
Before you say, "Aha! let's put the detector antenna 0.000001 meter from
the passive repeater antenna and get incredible gain!" you have to
realize that the antenna gain is achieved only in the far field, so
you've got to keep a good part of a wavelength away for the rules to
hold. Actually, 5 meters is almost certainly too close for the gain to
be valid at 10 MHz. The actual gain at any distance and field strength
for a given input power could easily be determined for a particular
antenna by modeling.

Why does the wavelength make such a big difference? Well, think of the
size of the field each antenna intercepts. Both the 10 MHz and 150 MHz
antennas have the same gain, so the former is dimensionally 15 times
larger than the latter, or 15^2 = 225 times the area. Consquently, the
10 MHz antenna intercepts 225 times the power that the 150 MHz antenna
does(**). 225 is exactly the ratio between the 10 and 150 MHz results of
5.7 and 0.025 (if carried out a few more places).

If you know how much signal strength margin you have at the point where
you put the first passive receiver antenna, you can use the equation
above to calculate how much antenna gain you'll need for your passive
repeaters and how close you'll have to put your receiver's antenna to
the second (radiating) passive receiver antenna, or how to trade the
two. Note that I've assumed perfect match and no loss. If, for example,
there's loss in the transmission line between the two passive repeater
antennas, that loss in dB will directly subtract from the overall
passive repeater gain (or add to the overall passive repeater attenuation).

(*) 10 meters is only 1/3 wavelengh so a bit close. The nominal antenna
gain might not actually be realized at that short distance. But I'll
assume it is.

(**) Please don't infer from this that aperture has any obvious
relationship to physical antenna area except in so-called "aperture
antennas" such as horns or parabolic antennas which have large physical
dimensions. The antennas under discussion have apertures proportional to
their physical areas only because they're assumed to be physically
identical except for a scaling factor. In general, antennas with widely
different physical areas can have the same aperture, and antennas with
the same physical area can have widely different apertures.

Roy Lewallen, W7EL

"Roy Lewallen" wrote in message
...
I keep hearing about "passive repeaters", but haven't ever seen anything
quantitative about how well they work. I had some measurements done years
ago which showed that a cell phone "passive repeater" in a sedan doesn't do
anything significant, quite obviously because the windows are such large
apertures at that frequency.

But here are some numbers to work with. Corrections are welcome.

Assuming perfect efficiency, the maximum effective aperture of an antenna
is Aem = G * lambda^2 / (4 * pi), where

lambda = the wavelength
G = the antenna's numerical power gain in the direction being analyzed

The gain in dBi = 10 * log[base10](G). For convenience, use meters for
length dimensions, so the aperture is in square meters and the wavelength
is in meters.

This means that if we have a field with power density Pdi striking an
antenna with power gain G1 (in the direction the field is coming from),
the antenna can deliver Pdi * Aem watts to a conjugately matched load.

Now let's connect this antenna to a second antenna with numerical power
gain G2, and assume that the two are perfectly matched to each other. That
is, the input impedance of the second antenna is the complex conjugate of
the impedance the first antenna has when driven. Also assume no loss in
the connecting line.

The power delivered to the second antenna is then Pdi * Aem watts.

The second antenna will radiate a field with power density equal to G2
times that of an isotropic antenna supplied with the same power, or

G2 * (Pdi * Aem) / (4 * pi * r^2)

where r is the distance from the antenna in the direction in which the
gain is measured.

Combining these to find the ratio of power density of the field radiated
from the second antenna to the power density of the field striking the
first antenna, we get:

P density ratio = G1 * G2 * (lambda / (4 * pi * r))^2

This is the numerical gain you'll see between the field at the location of
the first passive receiver antenna to the field at a distance r from the
second (radiating) passive receiver antenna. A ratio of less than one
represents attenuation rather than gain. The gain in dB is 10 * the base
10 logarithm of this ratio, with negative results representing
attenuation. That is,

Overall gain (dB) = 10 * log[base 10](P density ratio)

Now let's look at a couple of examples. For simplicity, assume that none
of the original field arrives at the detector location; that is, there's a
perfect shield or obstruction between the original field and the detector
(technically, the detector's antenna). It doesn't matter what kind of
antenna is physically connected to the detector for this analysis. In real
life, the detector would probably be a receiver, but I'll call it a
detector so hopefully its antenna won't be confused for either of the two
passive receiver's antennas.

Let G1 = G2 = 10 -- both antennas comprising the passive repeater have 10
dBi gain (although don't forget that G1 and G2 are numerical, not dB
gains -- the two numbers just happen to be the same in the case of 10).
Wavelength = 30 meters (10 MHz frequency), and r = 10 meters -- we've put
the detector's antenna 10 meters from the second (radiating) passive
repeater antenna(*).

Working through the numbers, the power density at the detector's antenna
is 5.7 times, or 7.56 dB greater than, the power density striking the
first antenna. So the signal is stronger than it would have been if we'd
put the detector's antenna right where the first passive repeater antenna
is. Don't forget, though, that it took two 10 dBi antennas to get that
7.56 dB gain.

But now let's look at the same setup but at 150 MHz (2 meter wavelength).
The power density at the detector's antenna in this case is just 2.5% of
(or 15.96 dB less than) the power density of the field striking the first
antenna.

In all cases, if we move the detector's antenna twice as close to the
second (radiating) passive repeater antenna, that is, to 5 meters away,
we'll gain 6 dB; if we double the distance to 20 meters we lose 6 dB.
Before you say, "Aha! let's put the detector antenna 0.000001 meter from
the passive repeater antenna and get incredible gain!" you have to realize
that the antenna gain is achieved only in the far field, so you've got to
keep a good part of a wavelength away for the rules to hold. Actually, 5
meters is almost certainly too close for the gain to be valid at 10 MHz.
The actual gain at any distance and field strength for a given input power
could easily be determined for a particular antenna by modeling.

Why does the wavelength make such a big difference? Well, think of the
size of the field each antenna intercepts. Both the 10 MHz and 150 MHz
antennas have the same gain, so the former is dimensionally 15 times
larger than the latter, or 15^2 = 225 times the area. Consquently, the 10
MHz antenna intercepts 225 times the power that the 150 MHz antenna
does(**). 225 is exactly the ratio between the 10 and 150 MHz results of
5.7 and 0.025 (if carried out a few more places).

If you know how much signal strength margin you have at the point where
you put the first passive receiver antenna, you can use the equation above
to calculate how much antenna gain you'll need for your passive repeaters
and how close you'll have to put your receiver's antenna to the second
(radiating) passive receiver antenna, or how to trade the two. Note that
I've assumed perfect match and no loss. If, for example, there's loss in
the transmission line between the two passive repeater antennas, that loss
in dB will directly subtract from the overall passive repeater gain (or
add to the overall passive repeater attenuation).

(*) 10 meters is only 1/3 wavelengh so a bit close. The nominal antenna
gain might not actually be realized at that short distance. But I'll
assume it is.

(**) Please don't infer from this that aperture has any obvious
relationship to physical antenna area except in so-called "aperture
antennas" such as horns or parabolic antennas which have large physical
dimensions. The antennas under discussion have apertures proportional to
their physical areas only because they're assumed to be physically
identical except for a scaling factor. In general, antennas with widely
different physical areas can have the same aperture, and antennas with the
same physical area can have widely different apertures.

Roy Lewallen, W7EL
This is way beyond me. Think I'll take up knitting. (if I can get the VSWR
right).

Regards Mike.

Thanks for the tutorial. The only time I was (remotely) involved in a
similar set up was in some very early cell phone demos inside buildings
where the cell site was not very close. Two antennas were used, but two
sets of filters and amplifiers were used to get considerable power gain in
both directions. in/out of the demo room.
73, Steve, K9DCI


"Roy Lewallen" wrote in message
...
I keep hearing about "passive repeaters", but haven't ever seen anything
quantitative about how well they work. I had some measurements done
years ago which showed that a cell phone "passive repeater" in a sedan
doesn't do anything significant, quite obviously because the windows are
such large apertures at that frequency.

But here are some numbers to work with. Corrections are welcome.

Assuming perfect efficiency, the maximum effective aperture of an
antenna is Aem = G * lambda^2 / (4 * pi), where

lambda = the wavelength
G = the antenna's numerical power gain in the direction being analyzed

The gain in dBi = 10 * log[base10](G). For convenience, use meters for
length dimensions, so the aperture is in square meters and the
wavelength is in meters.

This means that if we have a field with power density Pdi striking an
antenna with power gain G1 (in the direction the field is coming from),
the antenna can deliver Pdi * Aem watts to a conjugately matched load.

Now let's connect this antenna to a second antenna with numerical power
gain G2, and assume that the two are perfectly matched to each other.
That is, the input impedance of the second antenna is the complex
conjugate of the impedance the first antenna has when driven. Also
assume no loss in the connecting line.

The power delivered to the second antenna is then Pdi * Aem watts.

The second antenna will radiate a field with power density equal to G2
times that of an isotropic antenna supplied with the same power, or

G2 * (Pdi * Aem) / (4 * pi * r^2)

where r is the distance from the antenna in the direction in which the
gain is measured.

Combining these to find the ratio of power density of the field radiated
from the second antenna to the power density of the field striking the
first antenna, we get:

P density ratio = G1 * G2 * (lambda / (4 * pi * r))^2

This is the numerical gain you'll see between the field at the location
of the first passive receiver antenna to the field at a distance r from
the second (radiating) passive receiver antenna. A ratio of less than
one represents attenuation rather than gain. The gain in dB is 10 * the
base 10 logarithm of this ratio, with negative results representing
attenuation. That is,

Overall gain (dB) = 10 * log[base 10](P density ratio)

Now let's look at a couple of examples. For simplicity, assume that none
of the original field arrives at the detector location; that is, there's
a perfect shield or obstruction between the original field and the
detector (technically, the detector's antenna). It doesn't matter what
kind of antenna is physically connected to the detector for this
analysis. In real life, the detector would probably be a receiver, but
I'll call it a detector so hopefully its antenna won't be confused for
either of the two passive receiver's antennas.

Let G1 = G2 = 10 -- both antennas comprising the passive repeater have
10 dBi gain (although don't forget that G1 and G2 are numerical, not dB
gains -- the two numbers just happen to be the same in the case of 10).
Wavelength = 30 meters (10 MHz frequency), and r = 10 meters -- we've
put the detector's antenna 10 meters from the second (radiating) passive
repeater antenna(*).

Working through the numbers, the power density at the detector's antenna
is 5.7 times, or 7.56 dB greater than, the power density striking the
first antenna. So the signal is stronger than it would have been if we'd
put the detector's antenna right where the first passive repeater
antenna is. Don't forget, though, that it took two 10 dBi antennas to
get that 7.56 dB gain.

But now let's look at the same setup but at 150 MHz (2 meter
wavelength). The power density at the detector's antenna in this case is
just 2.5% of (or 15.96 dB less than) the power density of the field
striking the first antenna.

In all cases, if we move the detector's antenna twice as close to the
second (radiating) passive repeater antenna, that is, to 5 meters away,
we'll gain 6 dB; if we double the distance to 20 meters we lose 6 dB.
Before you say, "Aha! let's put the detector antenna 0.000001 meter from
the passive repeater antenna and get incredible gain!" you have to
realize that the antenna gain is achieved only in the far field, so
you've got to keep a good part of a wavelength away for the rules to
hold. Actually, 5 meters is almost certainly too close for the gain to
be valid at 10 MHz. The actual gain at any distance and field strength
for a given input power could easily be determined for a particular
antenna by modeling.

Why does the wavelength make such a big difference? Well, think of the
size of the field each antenna intercepts. Both the 10 MHz and 150 MHz
antennas have the same gain, so the former is dimensionally 15 times
larger than the latter, or 15^2 = 225 times the area. Consquently, the
10 MHz antenna intercepts 225 times the power that the 150 MHz antenna
does(**). 225 is exactly the ratio between the 10 and 150 MHz results of
5.7 and 0.025 (if carried out a few more places).

If you know how much signal strength margin you have at the point where
you put the first passive receiver antenna, you can use the equation
above to calculate how much antenna gain you'll need for your passive
repeaters and how close you'll have to put your receiver's antenna to
the second (radiating) passive receiver antenna, or how to trade the
two. Note that I've assumed perfect match and no loss. If, for example,
there's loss in the transmission line between the two passive repeater
antennas, that loss in dB will directly subtract from the overall
passive repeater gain (or add to the overall passive repeater
attenuation).

(*) 10 meters is only 1/3 wavelengh so a bit close. The nominal antenna
gain might not actually be realized at that short distance. But I'll
assume it is.

(**) Please don't infer from this that aperture has any obvious
relationship to physical antenna area except in so-called "aperture
antennas" such as horns or parabolic antennas which have large physical
dimensions. The antennas under discussion have apertures proportional to
their physical areas only because they're assumed to be physically
identical except for a scaling factor. In general, antennas with widely
different physical areas can have the same aperture, and antennas with
the same physical area can have widely different apertures.

Roy Lewallen, W7EL

Hal Rosser wrote:
"(one acts like a receiver, and the other acts like a transmitter.)"

Back-to-back antennas both act like transmitters. Neither cares if its
energy came from radiation sweeping it or fed to it from its feedline.

The most energy any antenna can extract from a passing wave is 50%. That
is only when the receiving antenna is perfectly matched to its load. In
this case the load is an identical antenna, so the chances for a match
are good. But, at least 50% of the energy captured by the receiving
antenna is going to be reradiated by that antenna right back in the
direction it came from.

Another problem is the rapid decline with distance in the first
wavelength from the antenna. It losses 22 dB in field strength in the
first wavelength from the transmitting antenna. After traveling through
a second wavelength from the antenna (distance doubled from the antenna)
the field strength declines by another 6 dB. It loses 6 dB every time
the distance doubles. So, after 4 wavelengths, total loss will be 34 dB.

At great distances form the transmitting antenna, double the distance is
such a great distance that the field strength hardly varies at all even
when traveling directly toward or away from the transmitter.

It is obvious that path loss is a function of frequency and distance if
only from continuous expansion of the sphere of the electromagnetic
wavefront. There are less watts per square meter when the watts are the
same but the number of square meters is growing. Our initial loss
distance was determined by wavelength which is inversely related to the
frequency.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
. . .
The most energy any antenna can extract from a passing wave is 50%. That
is only when the receiving antenna is perfectly matched to its load. In
this case the load is an identical antenna, so the chances for a match
are good. But, at least 50% of the energy captured by the receiving
antenna is going to be reradiated by that antenna right back in the
direction it came from.
. . .

This is a bit misleading. Typically, an antenna extracts only a tiny
fraction of the energy from a passing wave. The actual amount of energy
it does extract is expressed as its "capture area" or "effective
aperture". A recent posting I made includes a way to calculate this if
you know the antenna's gain. Because the "capture area" is usually a
tiny fraction of the wave's total area, only a tiny fraction of the
wave's energy is extracted.

Roy Lewallen, W7EL