Electronic – RSSI-Distance calculation help

Fade margin is the difference in power levels between the actual signal hitting the receiver and the bottom-line minimum signal needed by the receiver to work. It gives an indication of likely bit error rates for instance.

There is a standard formula for calculating minimum theoretical signal level needed by a receiver for a given data rate. This is -154dBm + 10\$log_{10}\$(bit rate). If data rate is 1Mbps then a receiver will need -94dBm to stand a chance of reasonably getting decent data.

If the received signal is in fact -84dBm then the fade margin is 10dB i.e. it can allow fading of the received signal up to 10dB.

To apply this to your situation means you need to understand the data rate so you can calculate minimum acceptable receiver power. Because Fm = Pr - Pm (where Pm is minimum receiver power level calculated from bit rate or maybe marked on the box) I believe you should be able to work this out based on RSSI being equivalent to Pr.

If you look in the link you provided you'll see this: -

Receive Sensitivity: 802.11b: -84dBm@11Mbps

In other words, at 11Mbps, using the formula in my answer you get a minimum receiver power required of -154 dBm + 10\$ log_{10}\$(11,000,000) dBm = -154dBm + 70.4dBm = -83.59dBm.

EDIT

I've been having a little look on this and there is a simpler formula you can use based on this document. The formula is #19 on page 3 and basically it is this: -

RSSI (dBm) = -10n \$log_{10}\$(d) + A

Where A is the received signal strength in dBm at 1 metre - you need to calibrate this on your system. Because you are calibrating at a known distance you don't need to take into account the frequency of your transmission and this simplifies the equation.

d is distance in metres and n is the propagation constant or path-loss exponent as you mentioned in your question i.e. 2.7 to 4.3 (Free space has n =2 for reference).

Your original formula - if you could supply a source for that I can check it against data I have.

"How is it even possible for a single fixed measurer to determine both RSSI and Noise?" - very good question. The noise they are talking about is receiver noise and not interfering signal. At very low powers, the noise is mostly the thermal noise of the receiver: ie, if you were to disconnect the antenna and replace it with a 50 Ohm load (most RF systems are 50 Ohm) you will measure a certain level of noise. So, even if you had all the ideal components, your noise power would be P = k*T*B*G, where k is the Boltzmann's constant, T is the temperature in K, B is the bandwidth in Hz, and G is the gain of your system. In reality, every component adds noise as specified by its noise figure (listed in the datasheet of every RF component). If you look again at the noise power equation, you will see that by reducing bandwidth, you also reduce the noise. However, high bandwidth is necessary for high data rates, which explains why you need good SNR for high data rates.

"Why both values are negative and measured in dBm" - 0 dBm means the power is 1 mW. -20 dbm means the power is .01 mW. The minus indicates the number of dB below 0 dBm. Without the minus, it would have been above 0 dBm

"But who radiate that power?" - in case of noise, it is internal, in case of signal, the transmitter. However, fundamentally it doesn't matter.

"But why is its value so small then?" - it comes from what is called Friis transmission formula. So, with several simplifications, imagine that my transmit antenna radiates power isotropically in all directions. So, your power is uniformly distributed on the surface of a sphere of radius r (and surface area 4*pi*r^2), where r is the distance from the transmit antenna. In Imagine, that your receive antenna is about 1 m^2 and it can capture all the radiation that hits its surface. Now, it can only capture 1/(4*pi*r^2) of all the radiation, making the receive power very tiny and RF engineering a complex field :). This is a very hand wavy explanation but I hope it makes sense