# What are the units of RSSI, noise and SNR as defined by IEEE 802.11?

I'm a CS graduate, but to my shame have very limited knowledge of electrical engineering and especially antenna theory.

As far as I understand, RSSI determines quality of how measurer "hears" the object being measured. Noise determines environment conditions that affects measurer. And SNR is simply how much RSSI is better than Noise. This theory (assuming I got the basics right) raises only single question:

• How is it even possible for a single fixed measurer to determine both RSSI and Noise?

Now some practice. Let's say measurer is my Macbook Air running builtin Wireless Diagnostic tool. And the object being measured is my WiFi Router. Observed values are −60 dBm for RSSI and −92 dBm for Noise. Therefore SNR is 32 dB. What I completely cannot understand is:

• Why both values are negative and measured in dBm?

As far as I understand, −60 dBm means 10−9 W while −92 dBm means 10−12 W. But who radiate that power? Maybe that theory represents Noise as another "antenna"? But why is its value so small then? Or I miss some very key points here? I'll be thankful for an intuitive explanation of this stuff.

"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

• So, if my receive antenna would be another sphere circumscribed around your transmit antenna, RSSI would be very close to power your antenna radiates? Still, the value of 1 nanowatt seems very small to me… Maybe you could point me to some sheet of real world examples? Commented Nov 4, 2013 at 19:25
• No, your receive antenna would be just just a small patch on that imaginary sphere. Think of the sun radiating incredible amounts of energy in all directions. Here on Earth, every square meter facing the sun would roughly receive 1/(4*pi*r^2) fraction of the sun's power, where r is the distance from earth to the center of the sun. en.wikipedia.org/wiki/Friis_transmission_equation Commented Nov 4, 2013 at 20:34
• I've got you idea. I'm asking about another imaginary situation. Considering your example with Sun and Earth, imagine Earth and Inner Core of the Earth. In this case Earth absorbs all power radiated from the core. Am I correct? Commented Nov 4, 2013 at 21:22
• Not sure I fully understand the question... Commented Nov 5, 2013 at 2:53
• @Kentzo yes, that is correct. Actually, you can be more rigorous: it will receive 100% of the radiated signal because there is nowhere else for it to go. Commented Nov 5, 2013 at 4:22

They're negative because they are really small. The dB scale is a logarithmic scale, with 0 dBm referenced to 1 mW. Negative values are smaller and positive values are larger. Like you said -60 dBm is 1 nanowatt and -90 dBm is 1 picowatt. I'm actually not sure where the noise measurement is coming from offhand. The radio receiver does generate some noise internally that prevents it from receiving an arbitrarily small signal just due to the nature of how the receiver is built. It contains lots of electrons bouncing around and generating noise, and it's not sitting at absolute zero so things are wiggling around and generating thermal noise. Think about how small 1 picowatt is. It is 100 trillion times smaller than your standard 100 watt light bulb.

It's possible that the noise figure represents the signal level on adjacent channels in some way. Have you noticed the noise value varying at all, or is it always -92 dBm? If it is fixed at -92 dBm, then that would be considered the noise floor of the receiver, and it is not capable of receiving signals that do not have a sufficient margin above the noise floor. In this case the noise level is not being measured, it is simply a characteristic of the receiver.

If the noise value varies, then it is probably a measurement of the noise on the channel when none of the wifi radios are transmitting. In a wifi system, all nodes in a network transmit on the same frequency in a shared channel. When no nodes are transmitting, the receiver can measure the signal level on the channel for a measure of the background environmental noise. Noise in the band might be caused by other wifi networks, bluetooth devices, zigbee, microwave ovens operating at 2.4 GHz, etc.

• Noise is varying from $-92$ to $-80$. What about RSSI? Why is the value so small, or such values are common in RF and it's enough to send and receive data? Commented Nov 4, 2013 at 19:28
• It's very common. The transmitter is probably only transmitting at 10 dBm tops. And the power drops off with the inverse square of the distance, so once you're a few tens of meters away from the transmitter you are going to see a pretty low signal level. The signal is also attenuated by any obstructions - e.g. walls. Also, you have to factor in the fact that the antennas in your laptop are very small and hence rather inefficient. I will have to look in to how the receiver is measuring noise, though. I'm not sure what it's doing to come up with that number. Commented Nov 4, 2013 at 19:31
• This transmitter is transmitting at $30dBm$ + $2dBi$ antennas. It also known to work within $300m$ radius. Wouldn't it also mean that RSSI will rise significantly once I get closer to the transmitter? But it only rises up to $-10dBm$ even if I put my laptop on top of the router. Commented Nov 4, 2013 at 19:41
• These numbers make a lot of sense and it is very close to what I have observed working on these types of systems. So, -10 dBm is 50 dB (or 100 000 times power increase) as compared to -60 dBm. The other issue might be the polarization mismatch and laptop shielding. Laptop antennas are typically placed on top of the screen. The best possible reception would be to place an open laptop facing the router at the same level Commented Nov 4, 2013 at 20:46
• It's also possible that the radio's analog front end goes into saturation at -10 dBm. The receivers are designed to operate primarily on low power levels because that's what determines the maximum range. Commented Nov 4, 2013 at 21:11

The work Friis did on developing a simple formula for received power makes a basic assumption about distance - all bets are off if the transmitter and receiver are up-close. This is called the near-field and the standard equation of: -

LinkLoss (dB) = $32.45 + 20 log_{10}(F) + 20 log_{10}(D)$

..... doesn't work up-close because you are not really measuring (or receiving) a true electromagnetic wave - you'll have the E field and the H field at all sorts of odd phase angles to each other and you'll actually be loading the transmit antenna. In the far field, (several wavelengths away) you'll get something like this: -

Once you're in the far field, EM wave power quarters with distance doubling. So, plugging your numbers into the equation (where F is in MHz and D is in kilometres) we get this at 300m: -

linkloss = 32.45 + 20log(2450 for wifi) + 20log(0.3) = 32.45dB + 67.8dB -10.5dB = 89.75dB.

This is a free-space link loss and as a rough guide folk tend to add 30dB to this figure to account for fade margin giving you a link loss of 119.8dB. Your antennas steal a little back to bring it down to about 116dB and your +30dBm transmission power means that at 300m you might expect to receive: -

86dBm.

Your receiver needs more received power for a bigger bandwidth (because noise power is proportional to bandwidth) and another good rule of thumb is minimum received power required is $-154dBm + 10log_{10} (data rate) dBm$.

If data rate is 10Mbps, then your minimum receiver power is -154dBm + 70dBm = 84dBm which is pretty close I'd say. You might want to replicate the calculations at (say) 2.45m (10 wavelengths away) to see if the numbers start to tally.