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.

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.

Just a bit of clarification. The node needs to transmit above
P_required for successful reception at the receiver above its minimum
receiver sensitivity, but this P_required is greater than the power
that correlates to a data rate of 250 kbps. To me this doesn't make
sense. I thought I should add in that conclusion. Thanks.

On Dec 4, 1:23 pm, Ginu wrote:
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.

Just a bit of clarification. The node needs to transmit above
P_required for successful reception at the receiver above its minimum
receiver sensitivity, but this P_required is greater than the power
that correlates to a data rate of 250 kbps. To me this doesn't make
sense. I thought I should add in that conclusion. Thanks.

anyone?

On Thu, 4 Dec 2008 21:59:34 -0800 (PST), Ginu
wrote:

anyone?

You appear to be contradicting yourself:
P_required... translates to a transmit power of 1 mW.
....
However, 1mW happens to be the maximum transmit power
However?

As for:
I hope my math is clear.
If your conclusions for these case tests depart from rational
expectations; then you have one of two possibilites:

1. Your math may be clear, but wholly invalid (transcription error of
complex formulas into ascii text is a bummer);

b. Your math may be clear or murky, and otherwise entirely correct,
but you did it wrong.

iii. You suffer from transcription AND "plug-n-chug" errors;

four. Zigbee technology doesn't work.

If this is a classroom problem, you need to resolve it yourself. If
this is an engineering problem, what did the Zigbee help-line offer?
There's probably a simple nomograph posted at their website or in
their application notes.

73's
Richard Clark, KB7QHC

On Dec 5, 2:10 am, Richard Clark wrote:
On Thu, 4 Dec 2008 21:59:34 -0800 (PST), Ginu
wrote:

anyone?

You appear to be contradicting yourself:P_required... translates to a transmit power of 1 mW.
...
However, 1mW happens to be the maximum transmit power

However?

As for: I hope my math is clear.

If your conclusions for these case tests depart from rational
expectations; then you have one of two possibilites:

1. Your math may be clear, but wholly invalid (transcription error of
complex formulas into ascii text is a bummer);

b. Your math may be clear or murky, and otherwise entirely correct,
but you did it wrong.

iii. You suffer from transcription AND "plug-n-chug" errors;

four. Zigbee technology doesn't work.

If this is a classroom problem, you need to resolve it yourself. If
this is an engineering problem, what did the Zigbee help-line offer?
There's probably a simple nomograph posted at their website or in
their application notes.

73's
Richard Clark, KB7QHC

This isn't a classroom problem. It is a research study that I am
conducting.

For the first case, I used a distance of 500 metres, which
corresponded to what happens to be the maximum allowed power of 1 mW.
Hence, I stated that I may be having a range issue because I was
testing the maximum distance case. That is why I provided a case for
300 metres as well.

If I'm doing something incorrectly, this is why I posted this question
to the board. Stating that my math may just be wrong, doesn't really
help the situation.

I will check into the Zigbee forums thank you.

So at 5AM I may not be too sharp...

The formula you used doesnt seem to allow for the required demod s/n.
QAM64 for example (from memory) is around 30dB so this has to be "added"
to the path loss. Differing mod/demod/FEC have differing margin
requirmeents. Somewhere in there too you have to stipulate the bit loss
rate "allowed".

In the past I use to calculate the noise floor at b/w, apply RX noise
figure and then add the margin.

When I last did this I upset my employer by saying that their marketing
hype was flawed. They had specified a lowest usable sensitivity that
after applying the required demod s/n was below the calculated noise
floor for that b/w. We were in fact consistently getting RX sensitivty
issues during the manufacturing phase for this reason... Marketing had
created a spec that wasnt possible!

Also, I dont know about 802.15 but I do know that 802.11 has variable
data rates (ie bandwidth) that can be advertised something like "54Mbps
and -87dBm sensitivity". The numbers dont actually go together. At
54Mbps you might get -77dBm and you have to use 11Mbps to get -87dBm!

I'll admit I havent studied your maths in detail....

The above probably not a lot of use to you... Sorry!

Cheers Bob VK2YQA

Ginu wrote:
Hello,

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

On Dec 5, 6:10 am, Bob Bob wrote:
So at 5AM I may not be too sharp...

The formula you used doesnt seem to allow for the required demod s/n.
QAM64 for example (from memory) is around 30dB so this has to be "added"
to the path loss. Differing mod/demod/FEC have differing margin
requirmeents. Somewhere in there too you have to stipulate the bit loss
rate "allowed".

Can you explain this further please?

In the past I use to calculate the noise floor at b/w, apply RX noise
figure and then add the margin.

When I last did this I upset my employer by saying that their marketing
hype was flawed. They had specified a lowest usable sensitivity that
after applying the required demod s/n was below the calculated noise
floor for that b/w. We were in fact consistently getting RX sensitivty
issues during the manufacturing phase for this reason... Marketing had
created a spec that wasnt possible!

Interesting. So these margin requirements are specified in the
standard?

Also, I dont know about 802.15 but I do know that 802.11 has variable
data rates (ie bandwidth) that can be advertised something like "54Mbps
and -87dBm sensitivity". The numbers dont actually go together. At
54Mbps you might get -77dBm and you have to use 11Mbps to get -87dBm!

Thank you. I'll look into this!

I'll admit I havent studied your maths in detail....

The above probably not a lot of use to you... Sorry!

Cheers Bob VK2YQA

Thanks a bunch. A reply to my above queries would be very helpful.

"Ginu" wrote in message
...
On Dec 5, 6:10 am, Bob Bob wrote:
So at 5AM I may not be too sharp...

The formula you used doesnt seem to allow for the required demod s/n.
QAM64 for example (from memory) is around 30dB so this has to be "added"
to the path loss. Differing mod/demod/FEC have differing margin
requirmeents. Somewhere in there too you have to stipulate the bit loss
rate "allowed".

Can you explain this further please?

http://www.satsig.net/lnb/ebno-calculator.htm is one of many sites dealing
with considerations of Energy Per Bit to Noise Density Ratio, which may help
you.

The Eb/No -- or to some, Eb/N0 -- influences the bit error rate (BER). Let
the Eb/No go too low and the link is not error-free. Generally, a BER of
better than one error in ten-to-the-eighth bits is considered to be an
error-free link.

Navy satellite data links of my acquaintance usually ran at 256 kbps and, if
memory serves, needed an Eb/No around 8. The modem reported the Eb/No and
that value was the first thing we looked at if the mux started taking hits.

I think Sal has answered this well enough. Suffice to say that if you
start with the noise floor or amount of noise energy in the bandwidth
you are going to use, you have to have a "margin" above that for the
radio system to be able to transfer information. This even applies to
morse code and voice transmission and thus the human brain's ability to
do the filtering and demod! Morse code for example can actually be heard
below the noise floor (or if you like a negative margin) because you can
concentrate on the 500Hz odd tone rather than the wide band noise.
Repeating the message that is sent also lowers the margin as it is a
kind of forward error correction that might give you a few extra dB.
Even a voice you know vs one you dont know lowers the margin. A trained
brain is remarkably good and has a huge dynamic range as well.

Even something like a human shout of warning has "greater range" because
it is a very narrow bandwith data stream. ie the message sent is binary
(yes vs no) or very short (like "help" "fire" "911" or "000") rather
than something like "There is a fire down here in the trees"

The margin required is more or less linked to the specification of the
overall modulation method. I quoted QAM64. When you add FEC to QAM64 the
margin becomes less at the expense of less data bandwidth (as more bits
are sent) I saw an FEC calc for an amateur radio satellite telemetry of
6-7dB. I browsed sround and found 802.11a is around 12dB for a BER of 1
in 10E6. Data stuff tends to be layered. ie they have a basic radio
modulation method (eg QAM, QPSK etc) and the FEC on top of that (eg Reed
Solomon & Verterbi encoding) Above that you may also have resends at a
higher layer. If you are running TCP/IP for example, TCP ensures that a
packet is received and reassembled within a certain timeout period. If
not it request the data again.

Hope this helps. apologies for the excessive analogies..

Cheers Bob W5/VK2YQA

On Dec 7, 7:30 am, Bob Bob wrote:
I think Sal has answered this well enough. Suffice to say that if you
start with the noise floor or amount of noise energy in the bandwidth
you are going to use, you have to have a "margin" above that for the
radio system to be able to transfer information. This even applies to
morse code and voice transmission and thus the human brain's ability to
do the filtering and demod! Morse code for example can actually be heard
below the noise floor (or if you like a negative margin) because you can
concentrate on the 500Hz odd tone rather than the wide band noise.
Repeating the message that is sent also lowers the margin as it is a
kind of forward error correction that might give you a few extra dB.
Even a voice you know vs one you dont know lowers the margin. A trained
brain is remarkably good and has a huge dynamic range as well.

Even something like a human shout of warning has "greater range" because
it is a very narrow bandwith data stream. ie the message sent is binary
(yes vs no) or very short (like "help" "fire" "911" or "000") rather
than something like "There is a fire down here in the trees"

The margin required is more or less linked to the specification of the
overall modulation method. I quoted QAM64. When you add FEC to QAM64 the
margin becomes less at the expense of less data bandwidth (as more bits
are sent) I saw an FEC calc for an amateur radio satellite telemetry of
6-7dB. I browsed sround and found 802.11a is around 12dB for a BER of 1
in 10E6. Data stuff tends to be layered. ie they have a basic radio
modulation method (eg QAM, QPSK etc) and the FEC on top of that (eg Reed
Solomon & Verterbi encoding) Above that you may also have resends at a
higher layer. If you are running TCP/IP for example, TCP ensures that a
packet is received and reassembled within a certain timeout period. If
not it request the data again.

Hope this helps. apologies for the excessive analogies..

Cheers Bob W5/VK2YQA

Thanks for the explanation. In your previous post you suggested that I
have to "add" the margin to the path loss. 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. Wouldn't adding to my path loss further deteriorate
my result? I'm trying to wrap my head around this.