World Phones

SMS wrote:
>
> Paul Hovnanian P.E. wrote:
>
> > So, if I interpret this correctly:
> >
> >> * 800/850, 1800, 1900 <-- US tri-band phone (missing the Euro 900 band)
> >> * 900, 1800, 1900 <-- Euro tri-band phone (missing the US 800/850 band)
> >
> > right?
>
> Correct. These phones are pointless, since you'll need a separate phone
> anyway when you go abroad anyway. May as well get a dual band and save
> money.

You meant quad-band, right?

--
Paul Hovnanian mailto:[email protected]
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Paul Hovnanian P.E. wrote:
> SMS wrote:
>> Paul Hovnanian P.E. wrote:
>>
>>> So, if I interpret this correctly:
>>>
>>>> * 800/850, 1800, 1900 <-- US tri-band phone (missing the Euro 900 band)
>>>> * 900, 1800, 1900 <-- Euro tri-band phone (missing the US 800/850 band)
>>> right?
>> Correct. These phones are pointless, since you'll need a separate phone
>> anyway when you go abroad anyway. May as well get a dual band and save
>> money.
>
> You meant quad-band, right?

No, I mean that there is no point in a tri-band versus a dual-band, as
you'll need two tri-bands anyway. Either get two dual bands, or one quad
band.

At 07 Aug 2006 20:00:25 +0000 John Navas wrote:
> On Mon, 07 Aug 2006 12:48:36 -0700, SMS <[email protected]>
> wrote in <[email protected]>:
>
> >... you can't get by without 800 MHz in the U.S..
>
> Actually you can, particularly on T-Mobile.
>
Actually, T-Mobile signed an excrement-load of roaming agreements in the
last year or so with 850MHz carriers (including Cingular) to expand rural
coverage. All current T-Mo phones now include 850. While 1900-only
phones are ok for native T-Mo coverage (and T-Mo prepaid, which generally
disallows roaming), you get much better coverage with 850/1900 phones. I
find myself using my back-up TDMA/AMPS pre-paid handsets (using Beyond
Wireless), a lot less in the rural midwest than I had to in the past.
The rural T-Mo coverage is expanding at about the same rate the Cingular
TDMA is declining! ;-) (While I don't have true "coverage" issues with
TDMA, I get a lot more "system busy" than a year or two ago.)

As an aside, my "best" Beyond phone for rural travel is now my ancient
Nokia 232. While 800 MHz AMPS only, I can manually select the 800 carrier.
Most TDMA phones of more recent vintage automatically choose digital
over analog (a bad thing in fringe areas- static-filled calls are
preferable to audio dropouts!) and they hang on to the "preferred"
carrier even when the alternate has a far stronger signal.

--
Posted via a free Usenet account from http://www.teranews.com

Todd Allcock wrote:
> At 07 Aug 2006 20:00:25 +0000 John Navas wrote:
>> On Mon, 07 Aug 2006 12:48:36 -0700, SMS <[email protected]>
>> wrote in <[email protected]>:
>>
>>> ... you can't get by without 800 MHz in the U.S..
>> Actually you can, particularly on T-Mobile.
>>
> Actually, T-Mobile signed an excrement-load of roaming agreements in the
> last year or so with 850MHz carriers (including Cingular) to expand rural
> coverage. All current T-Mo phones now include 850. While 1900-only
> phones are ok for native T-Mo coverage (and T-Mo prepaid, which generally
> disallows roaming), you get much better coverage with 850/1900 phones.

In many areas, such as my house, there is no 1900 MHz GSM coverage.
Remember that 1900 MHz has a shorter range from the tower than 800 Mhz,
so you need a lot more towers to cover the same area. You never want to
be stuck with only 1900 MHz GSM in the U.S., unless you're in a densely
populated city (not suburbs), and never go anywhere.

On Tue, 08 Aug 2006 08:21:50 -0700, SMS <[email protected]>
wrote in <[email protected]>:

>Todd Allcock wrote:
>> At 07 Aug 2006 20:00:25 +0000 John Navas wrote:
>>> On Mon, 07 Aug 2006 12:48:36 -0700, SMS <[email protected]>
>>> wrote in <[email protected]>:
>>>
>>>> ... you can't get by without 800 MHz in the U.S..
>>> Actually you can, particularly on T-Mobile.
>>>
>> Actually, T-Mobile signed an excrement-load of roaming agreements in the
>> last year or so with 850MHz carriers (including Cingular) to expand rural
>> coverage. All current T-Mo phones now include 850. While 1900-only
>> phones are ok for native T-Mo coverage (and T-Mo prepaid, which generally
>> disallows roaming), you get much better coverage with 850/1900 phones.
>
>In many areas, such as my house, there is no 1900 MHz GSM coverage.
>Remember that 1900 MHz has a shorter range from the tower than 800 Mhz,
>so you need a lot more towers to cover the same area. ...

"There you go again."

In metro areas there is no significant technology difference between 850
and 1900 bands (due to less than maximum range spacing), and even in
non-metro areas the difference, due to shorter range from lower
permitted maximum power for 1900 MHz, tends to be relatively small:

* Maximum power in the 800 band is 3 watts.
* Maximum power in the 1900 band is 2 watts.

It's not intuitively obvious, but that's only about 18% less range for
1900, and then only when range is limited only by power (not by
terrain).

--
Best regards, FAQ FOR CINGULAR WIRELESS:
John Navas <http://en.wikibooks.org/wiki/Cingular_Wireless_FAQ>

John Navas wrote:
>
> On Tue, 08 Aug 2006 08:21:50 -0700, SMS <[email protected]>
> wrote in <[email protected]>:
>
> >Todd Allcock wrote:
> >> At 07 Aug 2006 20:00:25 +0000 John Navas wrote:
> >>> On Mon, 07 Aug 2006 12:48:36 -0700, SMS <[email protected]>
> >>> wrote in <[email protected]>:
> >>>
> >>>> ... you can't get by without 800 MHz in the U.S..
> >>> Actually you can, particularly on T-Mobile.
> >>>
> >> Actually, T-Mobile signed an excrement-load of roaming agreements in the
> >> last year or so with 850MHz carriers (including Cingular) to expand rural
> >> coverage. All current T-Mo phones now include 850. While 1900-only
> >> phones are ok for native T-Mo coverage (and T-Mo prepaid, which generally
> >> disallows roaming), you get much better coverage with 850/1900 phones.
> >
> >In many areas, such as my house, there is no 1900 MHz GSM coverage.
> >Remember that 1900 MHz has a shorter range from the tower than 800 Mhz,
> >so you need a lot more towers to cover the same area. ...
>
> "There you go again."
>
> In metro areas there is no significant technology difference between 850
> and 1900 bands (due to less than maximum range spacing), and even in
> non-metro areas the difference, due to shorter range from lower
> permitted maximum power for 1900 MHz, tends to be relatively small:
>
> * Maximum power in the 800 band is 3 watts.
> * Maximum power in the 1900 band is 2 watts.
>
> It's not intuitively obvious, but that's only about 18% less range for
> 1900, and then only when range is limited only by power (not by
> terrain).

What about the increased attenuation of the higher frequencies (shorter
wavelengths)?

> --
> Best regards, FAQ FOR CINGULAR WIRELESS:
> John Navas <http://en.wikibooks.org/wiki/Cingular_Wireless_FAQ>

--
Paul Hovnanian mailto:[email protected]
------------------------------------------------------------------
definition: recursion; see recursion.

On Wed, 09 Aug 2006 15:25:31 -0700, "Paul Hovnanian P.E."
<[email protected]> wrote in <[email protected]>:

>John Navas wrote:

>> In metro areas there is no significant technology difference between 850
>> and 1900 bands (due to less than maximum range spacing), and even in
>> non-metro areas the difference, due to shorter range from lower
>> permitted maximum power for 1900 MHz, tends to be relatively small:
>>
>> * Maximum power in the 800 band is 3 watts.
>> * Maximum power in the 1900 band is 2 watts.
>>
>> It's not intuitively obvious, but that's only about 18% less range for
>> 1900, and then only when range is limited only by power (not by
>> terrain).
>
>What about the increased attenuation of the higher frequencies (shorter
>wavelengths)?

Attenuation is a function of distance, not frequency.
See <http://www.sss-mag.com/pdf/1propagation.pdf>.
Methinks you have in mind reflection, diffraction, and scattering?

--
Best regards, FAQ FOR CINGULAR WIRELESS:
John Navas <http://en.wikibooks.org/wiki/Cingular_Wireless_FAQ>

Paul Hovnanian P.E. wrote:

> What about the increased attenuation of the higher frequencies (shorter
> wavelengths)?

John is wrong of course, but he knows that.

"Frequencies at the lower end of the spectrum tend to have a long range
with higher levels of interference, while those at the higher end tend
to have shorter ranges with a low level of interference and the ability
to penetrate structures. Further, losses in the signal strength over a
distance (propagation) are greater at higher frequencies, and the
signals are absorbed by the outer atmosphere instead of being reflected
(Kuruppillai 1997). These characteristics of higher frequency
transmission restrict the PCS network design to smaller sized cells that
are primarily direct line-of-sight. This "line-of-sight" is not
necessarily a visible line-of-sight, but the obstructions between the
user and the antenna need to be relatively few to minimize degradation
of the signal caused by passing through structures (attenuation). There
are several other issues significant to the cel l size. As with
cellular, these include terrain characteristics, physical obstructions
such as buildings and customer penetration levels. In general, though,
PCS cell sites are smaller than cellular."

John Navas wrote:
> Attenuation is a function of distance, not frequency.
> See <http://www.sss-mag.com/pdf/1propagation.pdf>.
> Methinks you have in mind reflection, diffraction, and scattering?

Wrong...fundamentally WRONG...SOOO WROOOOONNNNGGGGG!

Path Loss = 20* log(12.57*r/w) dB, where
r = distance between transmitter and receiver
w = wavelength (the inverse of the frequency)

On Wed, 09 Aug 2006 17:43:49 -0700, SMS <[email protected]>
wrote in <[email protected]>:

>Paul Hovnanian P.E. wrote:
>
>> What about the increased attenuation of the higher frequencies (shorter
>> wavelengths)?
>
>John is wrong of course, but he knows that.

I am actually correct.

>"Frequencies at the lower end of the spectrum tend to have a long range
>with higher levels of interference, while those at the higher end tend
>to have shorter ranges with a low level of interference and the ability
>to penetrate structures. Further, losses in the signal strength over a
>distance (propagation) are greater at higher frequencies, and the
>signals are absorbed by the outer atmosphere instead of being reflected
>(Kuruppillai 1997). These characteristics of higher frequency
>transmission restrict the PCS network design to smaller sized cells that
>are primarily direct line-of-sight. This "line-of-sight" is not
>necessarily a visible line-of-sight, but the obstructions between the
>user and the antenna need to be relatively few to minimize degradation
>of the signal caused by passing through structures (attenuation). There
>are several other issues significant to the cel l size. As with
>cellular, these include terrain characteristics, physical obstructions
>such as buildings and customer penetration levels. In general, though,
>PCS cell sites are smaller than cellular."

That old chestnut. Unattributed (as is often the case with you),
probably because you know the source isn't authoritative. Here's the
link: <http://www.freenet.msp.mn.us/people/brose/papers/pcs.html>
And I quote:

In partial fulfillment of the requirements for
the degree of Master of Planning in Public Affairs

In other words, not an authoritative source. Some parts patently wrong
(e.g., the part about propagation losses being greater at higher
frequencies). Dredged up by Philip Koenig about six months ago.
<http://groups.google.com/group/ba.internet/msg/b81d3e40125aebdc>
Haven't you been able to come up with anything better in the meantime?

For a solid technical assessment of the issues, see CS 294-7: Radio
Propagation by Prof. Randy H. Katz, CS Division, University of
California, Berkeley <http://www.sss-mag.com/pdf/1propagation.pdf>.
Those interested will find that frequency isn't an issue in outdoor
range, and is a relatively minor issue in indoor penetration.

The actual dominant outdoor difference between 800/850 MHz and 1900 MHz
is that the latter isn't allowed as much power. I've dealt with this
issue before:

Maximum power in the 800 band is 3 watts.
Maximum power in the 1900 band is 2 watts.

It's not intuitively obvious, but that's only about 18% less range
for 1900, or a maximum of about 20% more towers along a flat rural
highway strip, or a maximum of 50% more towers in area coverage, and
then only when range is limited only by maximum power, which is
rarely the case in metro areas. Tower spacing is only near maximum
in flat rural areas (and current small handsets don't come close to
maximum power levels)...

--
Best regards, FAQ FOR CINGULAR WIRELESS:
John Navas <http://en.wikibooks.org/wiki/Cingular_Wireless_FAQ>

John Navas wrote:
> On Wed, 09 Aug 2006 17:43:49 -0700, SMS <[email protected]>
> wrote in <[email protected]>:
>
>> Paul Hovnanian P.E. wrote:
>>
>>> What about the increased attenuation of the higher frequencies (shorter
>>> wavelengths)?
>> John is wrong of course, but he knows that.
>
> I am actually correct.

No...you are not. And I noticed that you had no response to the general
math formula as it applies to path loss.

Obviously higher math and radio propagation theory is not you
forte...and don't say it actually is, you just proved it isn't.

Have a nice day!

On Fri, 11 Aug 2006 01:12:54 GMT, DecaturTxCowboy <[email protected]> wrote in
<[email protected]>:

>John Navas wrote:
>> On Wed, 09 Aug 2006 17:43:49 -0700, SMS <[email protected]>
>> wrote in <[email protected]>:
>>
>>> Paul Hovnanian P.E. wrote:
>>>
>>>> What about the increased attenuation of the higher frequencies (shorter
>>>> wavelengths)?
>>> John is wrong of course, but he knows that.
>>
>> I am actually correct.
>
>No...you are not. And I noticed that you had no response to the general
>math formula as it applies to path loss.
>
>Obviously higher math and radio propagation theory is not you
>forte...and don't say it actually is, you just proved it isn't.

Page 5 of my citation:

...in free space, power degrades by 1/d2

Have a nice day.

--
Best regards, FAQ FOR CINGULAR WIRELESS:
John Navas <http://en.wikibooks.org/wiki/Cingular_Wireless_FAQ>

On Thu, 10 Aug 2006 03:04:45 GMT, DecaturTxCowboy <[email protected]> wrote in
<[email protected]>:

>John Navas wrote:
>> Attenuation is a function of distance, not frequency.
>> See <http://www.sss-mag.com/pdf/1propagation.pdf>.
>> Methinks you have in mind reflection, diffraction, and scattering?
>
>Wrong...fundamentally WRONG...SOOO WROOOOONNNNGGGGG!
>
>Path Loss = 20* log(12.57*r/w) dB, where
>r = distance between transmitter and receiver
>w = wavelength (the inverse of the frequency)

See <http://en.wikipedia.org/wiki/Friis_transmission_equation> for the
actual Friis transmission equation. "The Friis transmission equation is
used in telecommunications engineering, and gives the power transmitted
from one antenna to another under idealized conditions." In other
words, it's a model of available antenna power under ideal conditions,
not just free space signal attentuation. Furthermore, "The ideal
conditions are almost never achieved in ordinary terrestrial
communications, due to obstructions, reflections from buildings, and
most importantly reflections from the ground."

--
Best regards, FAQ FOR CINGULAR WIRELESS:
John Navas <http://en.wikibooks.org/wiki/Cingular_Wireless_FAQ>

John Navas wrote:
>>>>> What about the increased attenuation of the higher frequencies (shorter
>>>>> wavelengths)?
>>>> John is wrong of course, but he knows that.
>>> I am actually correct.
>> No...you are not. And I noticed that you had no response to the general
>> math formula as it applies to path loss.
>>
>> Obviously higher math and radio propagation theory is not you
>> forte...and don't say it actually is, you just proved it isn't.
>
> Page 5 of my citation:
>
> ...in free space, power degrades by 1/d2
>
> Have a nice day.

Obviously you place more faith in your Googled citations than common
sense. But then, since you aren't a professional in the radio
communications landscape, you would not be able to discern the
difference between an unduplicated study (your citation) and what the
rest of communication professionals see duplicated everyday in path loss
issue.

You still haven't shown anything that contradicts the general math of
path loss.

Path Loss = 20* log(12.57*r/w) dB, where
r = distance between transmitter and receiver
w = wavelength (the inverse of the frequency)

This formula is used all the time and actual daily practice bears out
the predictions.

For your convenience, you may pick one of the numbered replies:

1) Rubbish
2) Not from my experience
3) We'll have to agree to disagree
4) Not according to my sources

On Fri, 11 Aug 2006 02:09:30 GMT, DecaturTxCowboy <[email protected]> wrote in
<uFRCg.4554$%[email protected]>:

>John Navas wrote:
>>>>>> What about the increased attenuation of the higher frequencies (shorter
>>>>>> wavelengths)?
>>>>> John is wrong of course, but he knows that.
>>>> I am actually correct.
>>> No...you are not. And I noticed that you had no response to the general
>>> math formula as it applies to path loss.
>>>
>>> Obviously higher math and radio propagation theory is not you
>>> forte...and don't say it actually is, you just proved it isn't.
>>
>> Page 5 of my citation:
>>
>> ...in free space, power degrades by 1/d2
>>
>> Have a nice day.
>
>Obviously you place more faith in your Googled citations than common
>sense. But then, since you aren't a professional in the radio
>communications landscape, you would not be able to discern the
>difference between an unduplicated study (your citation) and what the
>rest of communication professionals see duplicated everyday in path loss
> issue.
>
>You still haven't shown anything that contradicts the general math of
>path loss.
>
>Path Loss = 20* log(12.57*r/w) dB, where
>r = distance between transmitter and receiver
>w = wavelength (the inverse of the frequency)
>
>This formula is used all the time and actual daily practice bears out
>the predictions.
>
>For your convenience, you may pick one of the numbered replies:
>
>1) Rubbish
>2) Not from my experience
>3) We'll have to agree to disagree
>4) Not according to my sources

5) See prior response.

--
Best regards, FAQ FOR CINGULAR WIRELESS:
John Navas <http://en.wikibooks.org/wiki/Cingular_Wireless_FAQ>