Richard Harrison wrote:
Jeff wrote:
"- 96 db seems about right for "free space" path loss."

96 seems abour the right number for a path loss at 2400 MHz at a
distance of 500 meters. To nit pick, -96 db loss is a gain.

I worked a few years with Pete Saveskie who wrote a book he called
"Propagation". It contains a formula for free space loss: 23 db is lost
in the first wavelength from the transmitter and after that 6 db loss is
added every time the distance is doubled.

With Pete`s formula, you would lose the 23 db at a distance of 0.125
meter, and at a distance of 512 meters you would lose a total of 95 db.
That`s close enough agreement for me.

Best regards, Richard Harrison, KB5WZI


Also, a lot of published descriptions of various communications schemes
(e.g. Zigbee, 802.16e) might give a maximum data rate and a maximum
range, but that doesn't mean you get both at the same time. The former
is often related to the system bandwidth, the latter to the minimum data
rate and transmitter power.

I think you can get a ballpark feel pretty quick.. use the free space
path loss (in whatever form you like). Calculate required receiver
power as
-174 dBm + 10log(datarate in bps)+ 3 + receiver NF. (use 2dB for NF if
you like)

Add path loss to required receiver power, and that's what you'll need
for EIRP from the transmitter.

Yes, assumes isotropes, and ignores coding gains, etc. But you'll be
within 10dB or so, and that's enough to know if you're even in the ballpark.

If you do the calculations and you come up with a required transmit
power of +50dBm (100W), and you're thinking small battery powered, you
know it ain't gonna work. If you come up with +10dBm, and battery
powered is the goal, you're in the ballpark, and THEN you can start
thrashing through the more detailed modeling.

On Dec 9, 12:24 pm, Jim Lux wrote:
Richard Harrison wrote:
Jeff wrote:
"- 96 db seems about right for "free space" path loss."

96 seems abour the right number for a path loss at 2400 MHz at a
distance of 500 meters. To nit pick, -96 db loss is a gain.

I worked a few years with Pete Saveskie who wrote a book he called
"Propagation". It contains a formula for free space loss: 23 db is lost
in the first wavelength from the transmitter and after that 6 db loss is
added every time the distance is doubled.

With Pete`s formula, you would lose the 23 db at a distance of 0.125
meter, and at a distance of 512 meters you would lose a total of 95 db.
That`s close enough agreement for me.

Best regards, Richard Harrison, KB5WZI

Also, a lot of published descriptions of various communications schemes
(e.g. Zigbee, 802.16e) might give a maximum data rate and a maximum
range, but that doesn't mean you get both at the same time. The former
is often related to the system bandwidth, the latter to the minimum data
rate and transmitter power.

I think you can get a ballpark feel pretty quick.. use the free space
path loss (in whatever form you like). Calculate required receiver
power as
-174 dBm + 10log(datarate in bps)+ 3 + receiver NF. (use 2dB for NF if
you like)

Add path loss to required receiver power, and that's what you'll need
for EIRP from the transmitter.

Yes, assumes isotropes, and ignores coding gains, etc. But you'll be
within 10dB or so, and that's enough to know if you're even in the ballpark.

If you do the calculations and you come up with a required transmit
power of +50dBm (100W), and you're thinking small battery powered, you
know it ain't gonna work. If you come up with +10dBm, and battery
powered is the goal, you're in the ballpark, and THEN you can start
thrashing through the more detailed modeling.

I'm within 27.5 dBm

Ginu wrote:
On Dec 9, 12:24 pm, Jim Lux wrote:
Richard Harrison wrote:
Jeff wrote:
"- 96 db seems about right for "free space" path loss."
96 seems abour the right number for a path loss at 2400 MHz at a
distance of 500 meters. To nit pick, -96 db loss is a gain.
I worked a few years with Pete Saveskie who wrote a book he called
"Propagation". It contains a formula for free space loss: 23 db is lost
in the first wavelength from the transmitter and after that 6 db loss is
added every time the distance is doubled.
With Pete`s formula, you would lose the 23 db at a distance of 0.125
meter, and at a distance of 512 meters you would lose a total of 95 db.
That`s close enough agreement for me.
Best regards, Richard Harrison, KB5WZI
Also, a lot of published descriptions of various communications schemes
(e.g. Zigbee, 802.16e) might give a maximum data rate and a maximum
range, but that doesn't mean you get both at the same time. The former
is often related to the system bandwidth, the latter to the minimum data
rate and transmitter power.

I think you can get a ballpark feel pretty quick.. use the free space
path loss (in whatever form you like). Calculate required receiver
power as
-174 dBm + 10log(datarate in bps)+ 3 + receiver NF. (use 2dB for NF if
you like)

Add path loss to required receiver power, and that's what you'll need
for EIRP from the transmitter.

Yes, assumes isotropes, and ignores coding gains, etc. But you'll be
within 10dB or so, and that's enough to know if you're even in the ballpark.

If you do the calculations and you come up with a required transmit
power of +50dBm (100W), and you're thinking small battery powered, you
know it ain't gonna work. If you come up with +10dBm, and battery
powered is the goal, you're in the ballpark, and THEN you can start
thrashing through the more detailed modeling.

I'm within 27.5 dBm

As in, the back of the envelope shows you've got 28 dB of positive
margin, or the uncertainty of your estimate is 28dB, or you're 28dB
under? It's inverse square law, after all, so 10 times the distance is
a 20dB change in power.

Zigbee is IEEE 802.15.4 at the PHY.. 250kbps at ranges of 10-100 meters
for the 2.4 GHz band.

Let's see.. -174 + 3 + 54 + 2 = -114 dBm received signal level needed,
bare minimum. Real receivers are probably more like -100 or -95.
Chipcon's CC2420 is -94dBm

The 802.15.4 only requires a sensitivity of -85dBm

Zigbee transmitters are -3dBm transmit power (minimum).. typically, 0dBm
might be more common.

So, let's look at a link budget for 10 meters
32.44 + 20log(2500)+20log(0.01) = 32.44+ 68 -40 -- about 60dB path loss
between isotropes 10m apart at 2.5GHz.

-60dBm receive power vs -100dBm sensitivity.. So, it should work ok at
250kbps and 10m (assuming no interference, multipath, etc.)

Now, bump to 100m.. That's a 20dB hit.
500m another 14dB.. now you're on the ragged edge. 6dB margin with a
-100dBm receiver and a 0dBm transmitter. And that's assuming isotropic
antennas, which may or may not be reasonable.

On Dec 9, 6:42 pm, Jim Lux wrote:
Ginu wrote:
On Dec 9, 12:24 pm, Jim Lux wrote:
Richard Harrison wrote:
Jeff wrote:
"- 96 db seems about right for "free space" path loss."
96 seems abour the right number for a path loss at 2400 MHz at a
distance of 500 meters. To nit pick, -96 db loss is a gain.
I worked a few years with Pete Saveskie who wrote a book he called
"Propagation". It contains a formula for free space loss: 23 db is lost
in the first wavelength from the transmitter and after that 6 db loss is
added every time the distance is doubled.
With Pete`s formula, you would lose the 23 db at a distance of 0.125
meter, and at a distance of 512 meters you would lose a total of 95 db.
That`s close enough agreement for me.
Best regards, Richard Harrison, KB5WZI
Also, a lot of published descriptions of various communications schemes
(e.g. Zigbee, 802.16e) might give a maximum data rate and a maximum
range, but that doesn't mean you get both at the same time. The former
is often related to the system bandwidth, the latter to the minimum data
rate and transmitter power.

I think you can get a ballpark feel pretty quick.. use the free space
path loss (in whatever form you like). Calculate required receiver
power as
-174 dBm + 10log(datarate in bps)+ 3 + receiver NF. (use 2dB for NF if
you like)

Add path loss to required receiver power, and that's what you'll need
for EIRP from the transmitter.

Yes, assumes isotropes, and ignores coding gains, etc. But you'll be
within 10dB or so, and that's enough to know if you're even in the ballpark.

If you do the calculations and you come up with a required transmit
power of +50dBm (100W), and you're thinking small battery powered, you
know it ain't gonna work. If you come up with +10dBm, and battery
powered is the goal, you're in the ballpark, and THEN you can start
thrashing through the more detailed modeling.

I'm within 27.5 dBm

As in, the back of the envelope shows you've got 28 dB of positive
margin, or the uncertainty of your estimate is 28dB, or you're 28dB
under? It's inverse square law, after all, so 10 times the distance is
a 20dB change in power.

Zigbee is IEEE 802.15.4 at the PHY.. 250kbps at ranges of 10-100 meters
for the 2.4 GHz band.

Let's see.. -174 + 3 + 54 + 2 = -114 dBm received signal level needed,
bare minimum. Real receivers are probably more like -100 or -95.
Chipcon's CC2420 is -94dBm

The 802.15.4 only requires a sensitivity of -85dBm

Zigbee transmitters are -3dBm transmit power (minimum).. typically, 0dBm
might be more common.

So, let's look at a link budget for 10 meters
32.44 + 20log(2500)+20log(0.01) = 32.44+ 68 -40 -- about 60dB path loss
between isotropes 10m apart at 2.5GHz.

-60dBm receive power vs -100dBm sensitivity.. So, it should work ok at
250kbps and 10m (assuming no interference, multipath, etc.)

Now, bump to 100m.. That's a 20dB hit.
500m another 14dB.. now you're on the ragged edge. 6dB margin with a
-100dBm receiver and a 0dBm transmitter. And that's assuming isotropic
antennas, which may or may not be reasonable.

My path loss results verify this too. The issue is this: if the range
is between 10-100 metres, how do we use a minimum of -3 dBm of
transmit power?

To transmit at 250 kbps on a 5 MHz bandwidth channel centered at 2.5
GHz to a receiver say 10 metres away, you only require 5.8009e-010
watts. That's the problem. This is the maximum power required. It is
far less than the -3 dBm quoted.

The 28 dBm that I stated was that my transmit power required to
transmit at 250 kbps is 28 dBm under the -4.4 dBm minimum transmit
power to reach 300m that I provided in my original post:

Testing other distances, say 300m, we get a maximum transmit power Pt
= 5.9181e-007 = 0.59 uW to transmit at 250 kbps, but a P_required =
-4.41 dBm = 3.6211e-004, which is our maximum transmit power Pt.

5.9181 mW corresponds to roughly -32 dBm, hence the 28 dBm under.

At the 10 metres that we are discussing here, the required transmit
power falls even further:
5.8E-10 watts = 5.8E-7 mW

This corresponds to 10*log10(5.8009e-7) = -62.365 dBm. So the margin
is actually 67 dBm.

The path loss model isn't the problem. And the Shannon capacity is
known to be able to estimate any information channel.

This is where the confusion lies.

Ginu wrote:
On Dec 9, 6:42 pm, Jim Lux wrote:


So, let's look at a link budget for 10 meters
32.44 + 20log(2500)+20log(0.01) = 32.44+ 68 -40 -- about 60dB path loss
between isotropes 10m apart at 2.5GHz.

-60dBm receive power vs -100dBm sensitivity.. So, it should work ok at
250kbps and 10m (assuming no interference, multipath, etc.)

Now, bump to 100m.. That's a 20dB hit.
500m another 14dB.. now you're on the ragged edge. 6dB margin with a
-100dBm receiver and a 0dBm transmitter. And that's assuming isotropic
antennas, which may or may not be reasonable.

My path loss results verify this too. The issue is this: if the range
is between 10-100 metres, how do we use a minimum of -3 dBm of
transmit power?

To transmit at 250 kbps on a 5 MHz bandwidth channel centered at 2.5
GHz to a receiver say 10 metres away, you only require 5.8009e-010
watts. That's the problem. This is the maximum power required. It is
far less than the -3 dBm quoted.

A question you need to ask is what's the receiver bandwidth. The
information might only be 250 kHz wide, but if the receiver is 5MHz
wide, it's seeing 13dB more noise, and it might not be able to "acquire"
the narrow band signal.

60dB path loss, 250kHz BW is 54dBHz, so kTB noise floor is -120dBm, for
a "real" receiver and cabling, probably 5dB worse, call it -115dBm. Add
the 60dB, and you need an EIRP of -55dBm.. (You calculated 6E-10W,
-62dBm.. that's reasonably close)

But, if the receiver is seeing the full 5MHz (or more) BW, then you'll
need more.. 13dB at least (-100dBm at the receiver.. Obviously, most
receivers don't do this well, or they're counting some S/N margin, to
get a spec of -94dBm)

Then, you'd need -34dBm (with the same 60dB path loss)

But I'll bet you actually need more...

Non-ideal antennas
Losses in cabling
Mismatch (the Tx may put out 0dBm, but if the antenna presents a 2:1
mismatch, it's not radiating 0dBm, etc.

And, the "acquisition" threshold might be different than the
"communicate" threshold.

On Thu, 11 Dec 2008 09:43:05 -0800, Jim Lux
wrote:

To transmit at 250 kbps on a 5 MHz bandwidth channel centered at 2.5
GHz to a receiver say 10 metres away, you only require 5.8009e-010
watts. That's the problem. This is the maximum power required. It is
far less than the -3 dBm quoted.

A question you need to ask is what's the receiver bandwidth. The
information might only be 250 kHz wide, but if the receiver is 5MHz
wide, it's seeing 13dB more noise, and it might not be able to "acquire"
the narrow band signal.


Really Jim,

Do you think the vendor would specify a bit rate capacity and then
fail to supply the needed bandwith?

Omar has a peculiar habit (much like our own home-grown trolls) of
focusing on issues that have been solved, and complaining (through the
veil of supposing "what-if") about the math behind them creating
confusion.

Part of this veil is Omar holds all the cards while revealing nothing
(another trolling technique). Simply consult:
http://focus.ti.com/docs/prod/folders/print/cc2431.html
and take any of the several links to specifications, applications,
block diagrams, schematics; and it becomes painfully obvious that any
confusion that Omar suffers, is that advice already posted in
abundance devolves to rather simpler issues than Nyquist, Shannon,
Hartley, or more exotic sources.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
On Thu, 11 Dec 2008 09:43:05 -0800, Jim Lux
wrote:

To transmit at 250 kbps on a 5 MHz bandwidth channel centered at 2.5
GHz to a receiver say 10 metres away, you only require 5.8009e-010
watts. That's the problem. This is the maximum power required. It is
far less than the -3 dBm quoted.
A question you need to ask is what's the receiver bandwidth. The
information might only be 250 kHz wide, but if the receiver is 5MHz
wide, it's seeing 13dB more noise, and it might not be able to "acquire"
the narrow band signal.


Really Jim,

Do you think the vendor would specify a bit rate capacity and then
fail to supply the needed bandwith?

Sure.. it's the other way around though.. they might have a wideopen
front end or IF (e.g. 5 MHz wide) and be looking for a narrow band
signal in that big band.

Particularly if you have a system that supports multiple rates, and you
want a single hardware design without adjustable bandwidth, your
hardware has to support the widest band.

You're left with two design choices:
1) have a "minimum detectable signal" threshold that corresponds to the
higher noise floor
or
2) Have a narrow(er) band detector (implemented in software or hardware).

the first is cheaper, and can be overcome by marketing


Part of this veil is Omar holds all the cards while revealing nothing
(another trolling technique). Simply consult:
http://focus.ti.com/docs/prod/folders/print/cc2431.html
and take any of the several links to specifications, applications,
block diagrams, schematics; and it becomes painfully obvious that any
confusion that Omar suffers, is that advice already posted in
abundance devolves to rather simpler issues than Nyquist, Shannon,
Hartley, or more exotic sources.


Well, this IS true.

The whole thing is quite well documented, although if one just reads the
ad copy, it is confusing. Worse, if your manager asks why you can't do
the high rate and max distance at the same time, and you have to resort
to "laws of physics" arguments.