Signal Strength help

MaximusRex · 10-01-2004

Based on this equation:

dBm = 10 log P/1mW

If the signal received by a notebook is -20 dBM, and the distance between the notebook and the access point is 10 miles, what is the signal strength expressed in mW?

I've been working on this the entire afternoon and I still don't understand. Is it right in front of me? Would I just convert -20 dBm to mW and that's the signal strength?

For instance, mW = 10 ^ (dBm/10)? - Is that even the right equation. When I work it out, its mW = 1024...But when I plug that number into the dBm = log (1024) x 10, it doesn't spit out -20.

Or should I be solving for P?

Any help would be appreciated. At the very least steer me in the right direction. I did read one of the FAQs but it didn't help me too much.

Thorn · 10-01-2004

Quote:

Originally Posted by MaximusRex

... For instance, mW = 10 ^ (dBm/10)? - Is that even the right equation. When I work it out, its mW = 1024...But when I plug that number into the dBm = log (1024) x 10, it doesn't spit out -20. ...

That's the right equation, but it looks like you forgot the negative sign.

0.01mW = -20dBm

Thusly:
mW = 10 ^ (dBm/10)
mW = 10 ^ (-20/10)
mW = 10 ^ (-2)
mW = 0.01

To get -20dBm at ten mile would require a illegal system (under FCC rules.)

By the way, check your math. Even with a positive sign, 20dBm isn't 1024mW.

mW = 10 ^ (dBm/10)
mW = 10 ^ (20/10)
mW = 10 ^ (2)
mW = 100

MaximusRex · 10-02-2004

Quote:

Originally Posted by Thorn

That's the right equation, but it looks like you forgot the negative sign.

0.01mW = -20dBm

Thusly:
mW = 10 ^ (dBm/10)
mW = 10 ^ (-20/10)
mW = 10 ^ (-2)
mW = 0.01

To get -20dBm at ten mile would require a illegal system (under FCC rules.)

By the way, check your math. Even with a positive sign, 20dBm isn't 1024mW.

mW = 10 ^ (dBm/10)
mW = 10 ^ (20/10)
mW = 10 ^ (2)
mW = 100

You're right. Haha. I think my stress made me make a stupid, stupid error. But thank you for replying.

Also, as for the 10 miles, I'm assuming the question on the worksheet was in miles. All he had was "10m" as the distance.

Thanks for your help. I'm taking a wireless communications class, thus the cry for help on this forum.

weirdjim · 10-02-2004

10m would be ten meters. Ten miles is normally abbreviated 10mi.

I'm going to my website to get the path loss equation. You might want to go to www.rstengineering.com and then to "Jim's Engineering Page" to get the path loss at 2.5 GHz. for ten meters distance.

You also have to make some assumptions...like omnidirectional antennas at both ends...to make any sense out of the equation.

I'll be right back...

Jim

weirdjim · 10-02-2004

OK, here's my hit on it...

If the received signal is at -20 dBm...
and the antennas at both ends are omnidirectional dipoles (g=2.14 dB)...
and the distance is 10 meters...

Then the path loss from transmitter to receiver is 56 dB
And the transmitter must have an output of 4 watts to make this happen.

I think a received signal of -20 dBm is rather foolish; -120 dBm would be more like it.

Jim

MaximusRex · 10-02-2004

Quote:

Originally Posted by weirdjim

OK, here's my hit on it...

If the received signal is at -20 dBm...
and the antennas at both ends are omnidirectional dipoles (g=2.14 dB)...
and the distance is 10 meters...

Then the path loss from transmitter to receiver is 56 dB
And the transmitter must have an output of 4 watts to make this happen.

I think a received signal of -20 dBm is rather foolish; -120 dBm would be more like it.

Jim

He has -20 dBm on the worksheet though as the signal received. Geez, this is a very confusing class the way he phrases things.

For instance, the next question says, "If the notebook moves away from the AP and the distance between the two becomes 20m, what will be the signal strength in dBm. If the noise level remains the same, what is the signal to noise ratio in dBm?"

Before, he stated the SNR was 50 dB and based on the -20 dBm, I said the noise strength was -70 dBm. So I wonder how changing the distance by 10m effects anything. :-(

weirdjim · 10-02-2004

A. (S)He very well MAY have - 20 dBm as the signal received, but the required transmitter signal at 10 meters still has to be 4 watts.

B. Doubling the distance between any transmitter and receiver implies a 6 dB path loss; the signal strength will be -26 dBm.

C. If the instructor really gave a SNR in dBm, then (s)he has absolutely no idea what (s)he is talking about.

Hey, I teach this stuff in college for a living, and I would run (not walk) as fast as I could away from this sort of crap from a professor.
********************************

He has -20 dBm on the worksheet though as the signal received. Geez, this is a very confusing class the way he phrases things.

For instance, the next question says, "If the notebook moves away from the AP and the distance between the two becomes 20m, what will be the signal strength in dBm. If the noise level remains the same, what is the signal to noise ratio in dBm?"

Before, he stated the SNR was 50 dB and based on the -20 dBm, I said the noise strength was -70 dBm. So I wonder how changing the distance by 10m effects anything. :-(

10-01-2004	#1 (permalink)
MaximusRex Registered Member Join Date: Oct 2004 Posts: 3	Signal Strength help Based on this equation: dBm = 10 log P/1mW If the signal received by a notebook is -20 dBM, and the distance between the notebook and the access point is 10 miles, what is the signal strength expressed in mW? I've been working on this the entire afternoon and I still don't understand. Is it right in front of me? Would I just convert -20 dBm to mW and that's the signal strength? For instance, mW = 10 ^ (dBm/10)? - Is that even the right equation. When I work it out, its mW = 1024...But when I plug that number into the dBm = log (1024) x 10, it doesn't spit out -20. Or should I be solving for P? Any help would be appreciated. At the very least steer me in the right direction. I did read one of the FAQs but it didn't help me too much.

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10-02-2004	#4 (permalink)
weirdjim Registered Member Join Date: Oct 2004 Posts: 5	10m would be ten meters. Ten miles is normally abbreviated 10mi. I'm going to my website to get the path loss equation. You might want to go to www.rstengineering.com and then to "Jim's Engineering Page" to get the path loss at 2.5 GHz. for ten meters distance. You also have to make some assumptions...like omnidirectional antennas at both ends...to make any sense out of the equation. I'll be right back... Jim

10-02-2004	#5 (permalink)
weirdjim Registered Member Join Date: Oct 2004 Posts: 5	OK, here's my hit on it... If the received signal is at -20 dBm... and the antennas at both ends are omnidirectional dipoles (g=2.14 dB)... and the distance is 10 meters... Then the path loss from transmitter to receiver is 56 dB And the transmitter must have an output of 4 watts to make this happen. I think a received signal of -20 dBm is rather foolish; -120 dBm would be more like it. Jim

10-02-2004	#7 (permalink)
weirdjim Registered Member Join Date: Oct 2004 Posts: 5	A. (S)He very well MAY have - 20 dBm as the signal received, but the required transmitter signal at 10 meters still has to be 4 watts. B. Doubling the distance between any transmitter and receiver implies a 6 dB path loss; the signal strength will be -26 dBm. C. If the instructor really gave a SNR in dBm, then (s)he has absolutely no idea what (s)he is talking about. Hey, I teach this stuff in college for a living, and I would run (not walk) as fast as I could away from this sort of crap from a professor. ******************************** He has -20 dBm on the worksheet though as the signal received. Geez, this is a very confusing class the way he phrases things. For instance, the next question says, "If the notebook moves away from the AP and the distance between the two becomes 20m, what will be the signal strength in dBm. If the noise level remains the same, what is the signal to noise ratio in dBm?" Before, he stated the SNR was 50 dB and based on the -20 dBm, I said the noise strength was -70 dBm. So I wonder how changing the distance by 10m effects anything. :-(