Sprint Coverage/service in SoCal?

The more antennas per sq. mi., the more calls/data sessions per
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g wrote:
> There's also a physically related issue: Although pathloss is generally
> somewhat higher at 1900 MHz than at 850 MHz (10 dB for typical paths is
> an often quoted number)

Not so.

Path loss is not frequency dependent. Antenna aperture is.
If you had an 800 MHz 8 cm antenna and a 1900 MHz 3.7 cm antenna,
it would have about one fourth the capture area, or 6 dB loss.

A two bay 1900 MHz antenna (one with just one pigtail curl in the )
middle) would have about the same area as an 800 MHz antenna and
thus same system gain.

That applies with true free space (earth to satellite) or greater
than several Fresnel zone clearances.

However you have terrestrial propagation loses where your Fresnel
clearance is often less than one zone.. Which is a whole 'nuther
animal.

DTC wrote:

> Path loss is not frequency dependent. Antenna aperture is.
> If you had an 800 MHz 8 cm antenna and a 1900 MHz 3.7 cm antenna,
> it would have about one fourth the capture area, or 6 dB loss.
>
> A two bay 1900 MHz antenna (one with just one pigtail curl in the )
> middle) would have about the same area as an 800 MHz antenna and
> thus same system gain.
>
> That applies with true free space (earth to satellite) or greater
> than several Fresnel zone clearances.
>
> However you have terrestrial propagation loses where your Fresnel
> clearance is often less than one zone.. Which is a whole 'nuther
> animal.

Path loss, as usually defined; the freespace loss between two isotropic
antennas, IS frequency dependent. However, since no energy is really
lost in free space, what is going on is that while the transmitted field
from a constant electrical size antenna doesn't change as a function of
frequency, the aperture of a similar receive antenna of constant
electrical size DOES change. The aperture of an isotropic antenna is
lambda^2/4*PI or about 8% of a square wavelength.

For one antenna of constant PHYSICAL size and the other of constant
electrical size (thus providing constant ERP), in free space there is no
change as a function of frequency. Constant aperture means that the
gain/directivity of an antenna is increasing as frequency squared, the
same rate as the distance related "loss" is decreasing. Your example
above of constant physical aperture would result in about 7 dB more gain
at PCS.

If both the receive and transmit antennas have constant physical
aperture, then ERP is also increasing as frequency squared and overall
terminal-terminal signal is increasing as frequency squared. In this
case both antennas are becoming more directional.

However, there are some additional terms related to real (indirect)
paths. For one thing foliage and obstruction loss shows a frequency
dependency. Hardwood forests have on the order of .25 dB/foot loss at
2.5 GHz and around .5 dB/foot loss at 5 GHz. Between 850 and PCS
frequencies there is also a difference. Fresnel zone issues do not
relate here, since they imply a physical LOS but with nearby obstruction.

On top of all this, the exponent associated with real terrestrial paths,
whether urban, suburban or rural, is greater than the "2" that inverse
square of free space predicts. This is a result of the inverse square
freespace "loss" (definitional result of constant electrical size
referenece antennas) and the foliage/obstruction losses (which don't
actually follow anything so nice as an exponential relationship).


g

g wrote:
> DTC wrote:
>
>> Path loss is not frequency dependent. Antenna aperture is
>>
>> That applies with true free space (earth to satellite) or greater
>> than several Fresnel zone clearances.
>>
>> However you have terrestrial propagation loses where your Fresnel
>> clearance is often less than one zone.. Which is a whole 'nuther
>> animal.
>
> Path loss, as usually defined; the freespace loss between two isotropic
> antennas, IS frequency dependent. However, since no energy is really
> lost in free space,

True, there is no energy "lost in space"; its dispersed over a larger
area such that the flux falling on a given area is smaller. It is
frequency dependent only in the sense of antenna aperture;
not space loss.

> However, there are some additional terms related to real (indirect)
> paths. For one thing foliage and obstruction loss shows a frequency
> dependency. Hardwood forests have on the order of .25 dB/foot loss at
> 2.5 GHz and around .5 dB/foot loss at 5 GHz. Between 850 and PCS
> frequencies there is also a difference.

That would be the near earth terrestrial loses or attenuation. During
a seasonal test, we noted greater than expected losses at 5.8 Gig and
the predicted losses at 2.4 Gig. Several weeks later when the prairie
grass approached 2.5 inches, the loss at 5.8 Gig was as predicted,
but the loss at 2.4 was greater than before and greater than predicted.
Generally, we see greater losses at 5.8 when shooting through trees,
but then we make sure our antennas are well over mid-point tree lines.

> Fresnel zone issues do not
> relate here, since they imply a physical LOS but with nearby obstruction.

True, they do not relate as I said "That applies with true free space
(earth to satellite) or greater than several Fresnel zone clearances,"
where an earth to satellite would not even have an Fresnel zone
excursions. But with typical cellular communications you do indeed have
Fresnel zone incursions.

Posted sometime ago in another NG:

In the past I argued the higher the frequency, the greater the
attenuation, but failed to note I was referring to terrestrial
propagation...but I'll get to that later.

Lets look at the traditional free space loss formula:
Path loss in dB = 32.4 + 20 log(f) + 20 log(d), where f is frequency
in MHz and d is distance in miles.

It implies increasing the frequency will increase the path loss
(greater attenuation). When you run a plot at 2.4 GHz and one at 5.8
GHz, you'll find there is 7.7 dB more loss at 5.8 GHz.

[added for clarity...
However its the decreased antenna aperture that accounts for this
"lose", not free-space attenuation.]

If you go backwards in the equation and see how it is derived, you'll
find the capture area (antenna aperture) defined as wavelength squared
divided by four times Pi.

[added for clarity...
The frequency portion of the formula is a simple way of accounting
for aperture loss.]

For example - Lets take two simple dipoles for 2.4 GHz and 5.8 GHz
with the ends of the dipoles at 2.45 inches and 1 inch apart
respectively. Using the above formula, we find the capture areas are
1.912 and .318 square inches. Divide the area of the first antenna by
the second antenna's area and you'll get 6, or six times smaller (so
you would express it as a negative 6)

Convert -6 to dB and you get -7.7, the same loss number you go in the
plots you ran above.

Therefore, there really isn't greater attenuation as you increase the
frequency, rather the antenna is "less sensitive".

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

Now...on to terrestrial losses.

I placed a 2.4 GHz and a 5.8 GHz transmitter with simple vertical
dipole antennas on the output connectors at 500 feet up a tower and
measured the signal level from another 200 foot tower nearby with
simple vertical dipole antennas. This is as close as you'll get to
true free space on the face of the earth. As expected, the 5.8 GHz
signal was 7.7 dB lower. I had similar signal differences from several
test points with antennas on a forty foot mast and a clear line of
site.

I ran the tests again at several locations in the county with the
transmitters on a forty foot mast and ten foot mast for the receiver
antennas. I saw some additional loss from trees as expected and a
greater loss at 5.8 GHz. The additional loss varied with the terrain
and foliage. The conclusion would be there are greater terrestrial
losses at higher frequencies.

Eventually I'll get a system set up where I can measure hourly signal
levels from several other WISP's access points from miles away over a
year period. I suspect I'll see the same results my casual testing has
shown...path fading is worst for a few weeks in the spring and a few
weeks in the fall, and about an hour after sun up after the sun warms
the earth and you have a layer of warm air under the cooler morning
air, and an hour before sun down when the air starts to cool off and
drift into low spots under the still warm evening air.

"g" <[email protected]> wrote in message
news:[email protected]...
> However, if one DID get a whole city covered with the required
> few-hundred-feet access point spacing sort of density, then there would
> indeed be far better performance available to the end users, as compared to
> 2G or even 3G mobile services.

Except for the problem that WiFi was never designed to do "handoffs," whereas
mobile services have been from day one.

Granted, if WiMax ever gets rolling, it really could drive down the cost and
availability of higher bandwidth mobile connections... although personally I'm
quite pleased with Sprint's EVDO.

Joel Koltner wrote:

> Except for the problem that WiFi was never designed to do "handoffs," whereas
> mobile services have been from day one.
>
That's generally true (original multi-accesspoint WiFi design not
withstanding). I was really only addressing the cell site/hotspot
density issue and the physical layer ramifications as a function of
frequency. For any real deployment the actual protocol and a host of
other issues are relevant as well.

> Granted, if WiMax ever gets rolling, it really could drive down the cost and
> availability of higher bandwidth mobile connections... although personally I'm
> quite pleased with Sprint's EVDO.

EVDO does indeed do a pretty good job of effectively using it's physical
resources both in terms of Shannon (information) capacity and higher
layer issues like handoff. However, both it and WiMax do have to deal
with the same physical realities. The same physics that causes shorter
radio paths (smaller cell diameters) to be fundamentally more effective
also apply to them both. WiMax at 2.5 GHz will require higher cell site
density than EVDO at 850 MHz for the same performance - though WiMax may
not be able to go slower than ~ 1 Mbps as EVDO can.

g

DTC wrote:

> If you go backwards in the equation and see how it is derived, you'll
> find the capture area (antenna aperture) defined as wavelength squared
> divided by four times Pi.
>

Which you'll see is what I posted.
> [added for clarity...
> The frequency portion of the formula is a simple way of accounting
> for aperture loss.]
....

> Therefore, there really isn't greater attenuation as you increase the
> frequency, rather the antenna is "less sensitive".

Actually "has a smaller physical aperture" might be more precise.

We are not in disagreement. There are three cases:

equal electrical antenna sizes at each end (both ends isotropic, a
dipole or similar)

terminal-terminal "loss" increases as square of frequency or -20log(F)

one antenna constant electrical size, the other constant physical size:
terminal-terminal "loss" is constant vs. frequency

both antennas of constant physical size:
terminal-terminal "loss" decreases as square of frequency or +20log(F)
>
> ---------------------------------------
>
> Now...on to terrestrial losses.
>
> I placed a 2.4 GHz and a 5.8 GHz transmitter with simple vertical
... The conclusion would be there are greater terrestrial
> losses at higher frequencies.
>

This is the situation which I was applying to radio path length,cell
site density, capacity and coverage. If one looks at the terrestrial
path loss models; such as Lee or COST231 and it's extensions, there is
indeed a frequency related term over and above the freespace situation.
The best of these models is only an approximation and shows a mean or
average loss. The greater the incremental loss (due to real world
obstructions) the greater the variation from these median values. These
models are also generally not very useful at short ranges since in these
circumstances the path may approach or become line of sight (or may
not, depending upon the particular circumstance).

> Eventually I'll get a system set up where I can measure hourly signal
> levels from several other WISP's access points from miles away over a
> year period. I suspect I'll see the same results my casual testing has
> shown...path fading is worst for a few weeks in the spring and a few
> weeks in the fall, and about an hour after sun up after the sun warms
> the earth and you have a layer of warm air under the cooler morning
> air, and an hour before sun down when the air starts to cool off and
> drift into low spots under the still warm evening air.


I did this on two different 2.4 GHz paths for over a year on one path
and a couple of years on another. For the most relevant path, which was
only about 1000' feet long but obstructed by residential clutter,
foliage etc. there was indeed a very identifiable increase in path loss
when there was a lot of water/moisture present. The worst case was
springtime right after a rain- actually after the rain stopped and I
presume the plants drew additional moisture into their stalks and
leaves. The best time was one very cold morning in winter when there
were fewest leaves and frost or ice over everything. The mean
incremental attenuation was about 35 dB worse than LOS and the variation
around that value was about +-15dB for the above circumstances.

g

I have also found US Cellular to be very good in the midwest.

"Jerome Zelinske" <[email protected]> wrote in message
news:[email protected]...
> Like you say, it depends where you are. Here in S, SE, NE Wisc.,
> Sprint PCS covers all if verizon's coverage and then some. If you were
> out of Sprint PCS coverage and forced, or just allowed, your phone to
> roam, it would probably be on uscellular.

g <[email protected]> wrote:

>Steve Sobol wrote:
>
>> Of course not. There are a lot more people in metro areas and it's more
>> practical, from a financial standpoint, to put more antennas up.
>
>There's also a physically related issue: Although pathloss is generally
>somewhat higher at 1900 MHz than at 850 MHz (10 dB for typical paths is
>an often quoted number) once a carrier has paid the piper and put in the
>additional density required to support full coverage with the required
>shorter radio paths, that carrier is at a slight advantage. This is
>because the actual decrease in signal level as a function of distance is
>at least proportional to distance cubed and perhaps to the 4th power or
>higher, at both frequencies.
>
>Thus it may be a penalty at market entry since a higher density of basis
>are required but once that's paid, there's potential for better coverage
>at higher rates with the infrastructure. 3G performance may then be
>available over a larger percentage of the total area served.

I don't follow this. If you put up antennas at the same density -
that is, everything being equal except frequency - won't 850 MHz
perform better than 1900 MHz? I can see how 1900 MHz with more
antennas might outperform 850 MHz with fewer antennas, but I don't see
why that's a fair comparison.

Jack Hamilton wrote:

> I don't follow this. If you put up antennas at the same density -
> that is, everything being equal except frequency - won't 850 MHz
> perform better than 1900 MHz? I can see how 1900 MHz with more
> antennas might outperform 850 MHz with fewer antennas, but I don't see
> why that's a fair comparison.
>

Yes to your first question. However carriers using PCS rather than 850
MHz had to initially install at greater density just to make things
work. Having done that, they end up with more information capacity per
square meter/mile, all else equal.

Because the PCS carrier has, on average, shorter radio paths to each
user, there is less overall waste (energy lost in obstruction etc) and a
potential for higher information capacity as well as a potential for
greater spectrum reuse. Particularly with CDMA, this "extra effort" has
put him closer to next generation services.

The loss above and beyond inverse-square law (space loss) is real loss.
Transmitter power is going into heating up the environment rather than
delivering information to a user.

You could say that the initially higher investment puts the PCS carrier
closer to where he needs to be for next generation services -all else
equal, which of course it never is...

For wide area application of next generation services, radio path length
(and thus cell size) has to go down. Anyone starting with shorter paths
and higher cell density is ahead of those that don't have it.

g

Jack Hamilton wrote:
> I don't follow this. If you put up antennas at the same density -
> that is, everything being equal except frequency - won't 850 MHz
> perform better than 1900 MHz?

Yes, it will; but 1900 MHz antennas have more gain so it evens out.
Increasing the antenna gain inevitably increasing the antenna size,
and thus the antenna aperture.

DTC wrote:
> Jack Hamilton wrote:
>> I don't follow this. If you put up antennas at the same density -
>> that is, everything being equal except frequency - won't 850 MHz
>> perform better than 1900 MHz?
>
> Yes, it will; but 1900 MHz antennas have more gain so it evens out.
> Increasing the antenna gain inevitably increasing the antenna size,
> and thus the antenna aperture.


Actually, cell sites often use antennas of the same gain/directivity
but different aperture/physical size for 850 vs. PCS. Often the
directivity is such that they serve from a centrally located cell site
with three, 120 degree segments. Each segment antenna having enough
vertical dimension that it provides as narrow a beam as they can use and
still serve the target area. This means that the antennas, whether 850
or PCS, have identical azimuth (horizontal) and elevation(vertical)
beamwidth. This patten ends up prescribing the directivity/gain. The PCS
antenna is a little less than half as high/wide as the 850 MHz antenna.
It is pretty common to see the two different sizes at a single cell
site, each size in a ring of three around a central support.

Because PCS ends up with somewhat more foliage/terrain/obstruction loss
there is less field strength at a given distance. On top of that, the
user's antenna is most often omni-directional in azimuth and almost so
in elevation. Often it is effectively a dipole ( physically a monopole
against a reflecting groundplane). The net result is less signal to the
end user at a given distance and the requirement for higher base station
density to provide PCS coverage equivalent to that at 850 MHz.

g