On Dec 9, 6:05 pm, Richard Clark wrote:
On Tue, 9 Dec 2008 11:26:43 -0800 (PST), Ginu
wrote:

You're trolling in my threads and, for some odd reason, appear to feel
the need to belittle my work. If you can't add anything constructive,
please don't waste my time. I would appreciate it if you refrained
from posting in my threads in the future. Thank you.

Such a tender ego, and pride of ownership is the first clue to its
easy bruising. As for trolling, your contributions have that distinct
"under the bridge" flavor. Put a notch on your keyboard and move to
the next group. Talk about wasting time.

Omar, it takes very little effort to review your trail to see you have
one foot nailed to the floor and the other on a skate. You haven't
altered your question much in 5 months (with every appearance of not
having progressed one jot), offered your heart felt thanks to all
those who "helped" you, giving you pretty much the same leads you said
you would follow up on....

And we find you at the next street corner with your hat out for more
spare change.

Hard to imagine how with so many similar responses that you haven't
gotten it yet, and more amazing you might think that the problem is
not in the question.

73's
Richard Clark, KB7QHC

Richard, again, you are adding nothing to the discussion. Please
refrain from posting in my threads from now on. Those in this thread
have given me significant information that may lead to the solution.
Something that I have not gotten previously. Thanks again and goodbye.

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.

On Tue, 9 Dec 2008 15:40:47 -0800 (PST), Ginu
wrote:

Those in this thread
have given me significant information that may lead to the solution.

You must have littered more than several dozen newsgroups with that
same proclamation - can you succinctly point out how any single
significant information advances your solution? That, too, is missing
from your identical box-car series of postings that tend to repeat
themselves for weeks into months.

What a troll.

73's
Richard Clark, KB7QHC

The overall receiver noise figure is included in the receiver
sensitivity, is it not? I'm referring to
http://www.edn.com/article/CA6442439.html
where they state that:

"the receiver sensitivity S=–174 dBm/Hz+NF+10logB+SNRMIN, where –174
dBm/Hz is the thermal noise floor, NF is the overall-receiver-noise
figure in decibels, B is the overall receiver bandwidth, and SNRMIN is
the minimum SNR. If the total path loss between the transmitter and
the intended receiver is greater than the link budget, loss of data
ensues, and communications cannot take place. Therefore, it’s
important for designers developing end systems to accurately
characterize the path loss and compare it with the link budget to
obtain initial estimations of the range."

Do you have any comments on this?

Yes, the manufacturer has told you that the receiver sensitivity is -96dBm.
(no doubt for a particular BER, which aslo equates to a particular SNR).
Without knowing more about the internals of the receiver you cannot work
back to a NF unless you acuually know what SNR equates to the BER that the
manufacturer stated the -96dBm at. Using Shannon most likely give you an
unreliable optimistic answer.


If you know the rx sensitivity is -94dBm at the data rate you require,
why
go through all of the Shannon stuff, it is not revenant. You have been
told
that the Rx sensitivity is -94dBm use that figure.

That requires a long-winded answer about my project. My study involves
optimizing a multiple technology network where each device is
optimizing their transmission. Using the Shannon theorem allows me to
perform this optimization of data rate while considering physical
layer constraints. It's something I have to work into it
unfortunately. It's functioning correctly for WiMax and ultrawideband
technology.

No it dosen't unless you know more about the receivers and the actual SNR
that they require for a specific BER. Shannon will not tell you that, it
will just give you the 'best possible case' for a particular bandwidth,
which may well be a long way from the truth.


-94 dBm = P_required + G(2.4E9, 500)

P_required is then easily calculated as -94 dBm - G(2.4E9, 500) =
0.0254 dBm. This translates to a transmit power of 1 mW. This required
power is LARGER than the maximum permissible power to transmit at 250
kbps, which is the maximum data rate for Zigbee. So how does this make
sense?

Perhaps because most quoted ranges for Zigbee are in the order of 50m not
500m????

The approximate line-of-sight range for Zigbee is 500m. That's why I
tested both 300m and 500m in my calculations.

Yes and you proved just that!! Path loss 96dB, rx sensitivity -96dBm, 0dBm
tx power, equals receiver operating at its sensitivity limit. So what's the
problem??

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.

How is -4.41dBm greater than 0dBm (1mW) ?? It is about 0.362mW.

I meant that, for my example using a distance of 300m, the transmit
power required to transmit at the maximum data rate of 250 kbps was
0.59 uW. The required power to reach the MIRS at the receiver 300m
away was 3.6211E-4. Therefore, the amount of power required to reach
the receiver (a lower bound on power) was greater than the maximum
power allowed to reach the max data rate of 250 kbps (upper bound on
power). Therefore, the transmission isn't possible. It wasn't for 0
dBm.

I a sorry I don't understand what the hell you are on about. All you need to
know is the path loss at 300m, you aready have the rx sensitivity of -96dBm
so it is a simple subtraction to find the Tx power required (which will be
less than 0dBm).

Your maths is quite simple; you have the rx sensitivity, and can work out
the path loss, thats all you need to do, take one from the other and you
have the required tx power.

That's exactly what I'm doing. The NF is included in the receiver
sensitivity and the transmit power required to reach the receiver
sensitivity is greater than the power I'm allowed to transmit at
because of the 250 kbps maximum for the technology.


That is just not correct, you proved that the power required to reach
sensitivity limit at 500m was 0dBm, which is OK because that is you max tx
power.

At 300m you obviously require less tx power, so lets do the maths:

path loss at 300m, 2400MHz is 90dBm

so required tx power is 90 -96 = -6dBm which is less than your 0dBm max tx
power - no problem.
-6dBm equates to 2.5e-4W or 0.25mW.

If you are trying to say that the -6dBm figure is greater than the number
that you worked out using Shannon for a particular bandwidth and data rate
once you take off the 90dB path loss then you have to look at the accuracy
of your Shannon calculation.

You said that you Shannon Calculation produced a "reasonable transmit power
of 1.6 uW", to achieve 250kB in a 5MHz BW. Now 1.6uW = 0.0016mW = -28dBm
this does not seem reasonable when the manufacturers are quoting -96dBm!!!
You must go back and re-consider your Shannon calculations, this time taking
the actual receiver characteristics into account, such as rx NF, the real
noise bandwidth, number of levels in the modulation scheme etc., etc..

Regards
Jeff

On Wed, 10 Dec 2008 08:55:58 -0000, "Jeff" wrote:

No it dosen't unless you know more about the receivers and the actual SNR
that they require for a specific BER.

So what's the problem??

I a sorry I don't understand what the hell you are on about.

That is just not correct

you have to look at the accuracy

You must go back and re-consider

Hi Jeff,

Talking to a wall? This is the same lid who 2 years ago wanted 25 dBi
gain from a vertical and never responded to your comments.

73's
Richard Clark, KB7QHC

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.

On Dec 4, 1:15 pm, Ginu wrote:
Hello,

I've been trying to wrap my head about this problem for days and can't
seem to figure it out.

Zigbee technology allows you to transmit at 250 kbps. I'm comparing
the required power to transmit at that rate to the required minimum
power based on the path loss equations to reach the minimum receiver
sensitivity of the 802.15.4 physical layer for Zigbee of roughly -94
dBm. Simple enough right?

To analyze the required power to transmit at 250 kbps for Zigbee, I've
just re-arranged the Shannon capacity formula of Rate = w * log_2 (1 +
(Pt * G)/(w*N + I)). Here, w is the bandwidth, Pt is the transmit
power, Pt * G is the received signal power, N is the white Gaussian
noise and I is the sum of interference powers.

For Zigbee, these values are as follows:
w = 5 MHz
N (in dB) = 10*log10(k) + 10*log10(T) where k is boltzmann's constant
of 1.23E-23 and T is the temperature in Kelvin of 300 degrees. In
Watts, N is approximately 3.6900e-021.

I've assumed no interference at this stage, so I = 0. Simple enough so
far right?

So to calculate the required transmit power to reach a rate of 250
kbps, I've just re-arranged this equation to yield:
Pt = [2^(250 kbps/w) - 1]*(w * N + I)/G and plugging in the above
values gives me a reasonable transmit power of 1.6 uW.

The reason I haven't talked about G yet is that I'm addressing it
here. G is the channel quality (path loss) and is defined as follows:

G(freq,D) = GrdBi+ GtdBi- 20*log10(4*pi*freq*d0/c) -
10*pathLossExponent*log10(D/d0) (in dB)

I'm using unity gains so GrdBi = GtdBi = 0 dBi. Other parameters a
freq = 2.4 GHz
d0 = 1m as the reference distance
c = 3.0 x 10^8
pathLossExponent = 2, and
D = 500 metres

Hence, G (in Watts) * Pt is the received signal power. G(2.4E9, 500)
is the path loss transmitting at 2.4 GHz a distance of 500m away. G
(2.4E9, 500) = -94.0254 dB

Now the first thing that I mentioned was the receiver sensitivity for
Zigbee of -94 dBm, Analyzing G (in dB). The required transmit power
considering this path loss is easily calculated as:

-94 dBm = P_required + G(2.4E9, 500)

P_required is then easily calculated as -94 dBm - G(2.4E9, 500) =
0.0254 dBm. This translates to a transmit power of 1 mW. This required
power is LARGER than the maximum permissible power to transmit at 250
kbps, which is the maximum data rate for Zigbee. So how does this make
sense?

However, 1mW happens to be the maximum transmit power of Zigbee of
roughly 0 dBm, so there may be a range problem here.

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.

So again, we run into the same problem at 300m. It just doesn't make
sense to me. Can anybody shed some light on this? I hope my math is
clear. Thanks in advance.

Thanks everyone! With your feedback I was able to fix the problem.
Thanks a lot for your clarification and advice

Omar