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.