On Sun, 7 Dec 2008 15:11:33 -0800 (PST), Ginu
wrote:

My current result doesn't
make sense because the power required to transmit at 250 kpbs for
Zigbee is less than the power required to reach the receiver at a
modest 300m away.

This is your first and most significant clue to the failure of
analysis, and it is very "path loss" oriented (the path loss
differences for your two scenarios should be almost infinitesimal).
The disparity in your computations are due to transcription error, or
math error. You should have now been able to put that to rest.

Wouldn't adding to my path loss further deteriorate
my result? I'm trying to wrap my head around this.

This is your confusion factor, and it relates to transcription error
in the abstract: you are using the wrong formulas entirely regardless
of the accuracy or correctness of arithmetic results.

The greater part of discussion has focused on Shannon-Hartley issues
which have their own application to the full mix of your original
problem.

Try unwinding the thread so that you are not trying to force a
solution out of a broken premise. None of this really sounds like
finding the missing decimal point, or the corrupted divisor is going
to solve anything.

If you think this is still path loss related, and you are showing
results in actual implementation (bread-boarded hardware, on the
bench); then you have to open up the discussion beyond the limited
math to include the conventional problems of interference and
multipath.

73's
Richard Clark, KB7QHC

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.


You seem to be missing the noise figure of the receiver in the calculation,
and making assumptions about the receiver noise bandwidth that may not be
correct.


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


-96db Seems about right for *free space* path loss.


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:

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.


-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????

So you have an rx sensitivity of -94dBm and a path loss of 94dB, so with a
tx power of 0dBm you have a receiver operating at its sensitivity limit. So
you will have a link with a Bit Error Rate of whatever the manufacturer
quoted the -94dBm sensitivity figure at (10% BER?????).


However, 1mW happens to be the maximum transmit power of Zigbee of
roughly 0 dBm,

Exactly 0dBm actually

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.

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.

Of course your formula only gives the free space path loss, reality is
likely to be considerably worse!!!

73
Jeff

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

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.

Thanks Jeff for your reply. Below are my comments:

You seem to be missing the noise figure of the receiver in the calculation,
and making assumptions about the receiver noise bandwidth that may not be
correct.

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?

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.

-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.

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.


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.

Thanks Jeff. I'd be interested in hearing your response.

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

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