Free Space Path Loss Diagrams

4/1/2018

Free Space Path Loss (or FSPL) represents the amount of energy that a given radio wave loses as it travels through the air away from its source. Understanding FSPL will help us understand how far a Wi-Fi signal can go. It is also widely used by Wi-Fi survey tools to predict Wi-Fi signal propagation.

It is something that we can calculate by applying this mathematical formula (d in km and f in GHz):

FSPL is not specific to Wi-Fi waves, it can be applied to any other waves using other frequencies.

In a Wi-Fi environment, FSPL refers to the amount of power a Wi-Fi signal losses as it travels away from the transmitter (it can be an AP or a station).

This loss is relative to 2 main components:

Frequency
Distance

The following diagram shows exactly how much energy a Wi-Fi wave will lose as it travels away from an AP:

Frequency

Let’s start with frequency. Here, we are talking about the frequency of the radio wave. As you can see in the formula above, the higher the frequency is, the higher the loss will be. In Wi-Fi, this means that a 2.4GHz signal is less prone to loss than a 5GHz signal. We commonly say that 2.4GHz travels further away than 5GHz.

Another factor is affecting « how far » a Wi-Fi signal is travelling. And this is related to a specific receive characteristic. Any dual band receivers will need 1 antenna to listen to 2.4GHz signals and another one to listen to 5GHz signals. Because the wavelength of a 2.4GHz signal (12 cm) is longer that the one of a 5GHz signal (6 cm), the size of the antenna needs to be bigger. This means that the 2.4GHz receiver will have a larger receive aperture and will be able to better "hear" the incoming signal.

The FSPL formula in Wi-Fi defines that, in the first meter:

A 2.4GHz signal is loosing about 40dB
- Calculation for channel 1: 20log(0.001) + 20log(2.412) + 92.45 = 40.09dB
- Calculation for channel 11: 20log(0.001) + 20log(2.462) + 92.45 = 40.28dB
- Calculation for channel 13: 20log(0.001) + 20log(2.472) + 92.45 = 40.31dB
A 5GHz signal is loosing about 47dB
- Calculation for channel 36: 20log(0.001) + 20log(5.180) + 92.45 = 46.74dB
- Calculation for channel 52: 20log(0.001) + 20log(5.260) + 92.45 = 46.87dB
- Calculation for channel 100: 20log(0.001) + 20log(5.500) + 92.45 = 47.26dB
- Calculation for channel 149: 20log(0.001) + 20log(5.745) + 92.45 = 47.64dB
- Calculation for channel 165: 20log(0.001) + 20log(5.825) + 92.45 = 47.76dB

Distance

Here, we are talking about the distance travelled by the Wi-Fi wave away from the source. Common sense tells us that it would make sense if the signal was to lose power as it travels away from its source.
The FSPL formula tells us exactly by how much following the inverse square law.

The inverse square law tells us that as the distance from the source doubles, the energy is spread out over 4 times the area. This results in the signal loosing 4 times it's original amplitude. In other words, the Wi-Fi signal is loosing 6dB every time the distance from the source is doubled (This is represented in red in the diagram above).

So, if we know the power of our signal leaving the access point, we can calculate how far this signal will go. This is assuming there is no source of attenuation between the transmitter and the receiver.

What happens when we introduce a wall between the transmitter and the receiver?

Introducing a wall

As expected, the wall will attenuate the signal. The FSPL will be applied to the signal as it travels to the wall and away from the wall as explained on this following diagram. The amount of attenuation will depend on the structure of the wall.

If we know the amount of power that the Wi-Fi signal will have leaving the AP (Equivalent Isotropically Radiation Power or EIRP), we can calculate the size of our desired cell.

Here is an example if we have an AP using an EIRP of 14dBm (or 25mW) for band frequency bands:

It is now easy for us to see that the 2.4GHz cell will be twice as big at the 5GHz cell if the same EIRP is used.

In order to obtain the same cell size for both frequency bands, we can offset the EIRP of the 2.4GHz by 7dB. The way we can do it is by decreasing the 2.4GHz transmit power by 7dB. Most Wi-Fi system will only let us increase or decrease transmit power by an increment of 3dB. Therefore, the best we can do it decreasing the 2.4GHz transmit power by 6dB:

Ressources on FSPL

Wikipedia article on Free Space Path Loss: https://en.wikipedia.org/wiki/Free-space_path_loss
Effect of Transmit power changes on AP by Nigel Bowden: http://wifinigel.blogspot.ca/2014/11/effect-of-transmit-power-changes-on-ap.html
Wi-Fi Free Space Loss Calculator by Nigel Bowden: http://wifinigel.blogspot.ca/2014/05/wifi-free-space-loss-calculator.html
FSPL diagram by Devin Akin: http://divdyn.com/so-called-ghost-frames-not-exist/

written by François Vergès

11 Comments

Glenn Cate

4/2/2018 14:37:37

Your chart is a great one to show end users who may be concerned about the strength of RF coming from an AP as being a health hazard. Let me use your calculations for the following:
Assuming a 2.4/5.0 GHz AP has an internal antenna with approximately 4 dB gain (as an example, a Cisco 3702i has this internal antenna gain for both bands), then the formula can be seen as follows:

2.4 GHz: 50 mW (17 dBm) + 4 dBm = 21 dBm (125.89 mW)
At one meter distance, FSPL loses 40 dB in signal strength
21 dBm - 40 dBm = -19 dBm (.01259 mW or 12.59 microwatts)
At two meter distance, FSPL loses 46 dB in signal strength
21 dBm - 46 dBm = -25 dBm (.00316 mW or 3.16 microwatts)

5.0 Gz: 50 mW (17 dBm) + 4 dBm = 21 dBm (125.89 mW)
At one meter distance, FSPL loses 47 dB in signal strength
21 dBm - 47 dBm = -26dBm (.00251 mW or 2.51 microwatts)
At two meter distance, FSPL loses 53 dB in signal strength
21 dBm - 53 dBm = -32 dBm (.000631 mW or .631 microwatts)

As can be readily seen, APs at normal power levels (indoors) afford negligible risk to end users. Outdoor power usage (point-to-point links) is different and that usage does not apply here. Thanks again for an excellent blog!

François Vergès

4/7/2018 10:16:47

Yes, very true. Especially when you compare it to cellular Rx powers which are usually around 1W!

Martin Ericson

4/3/2018 13:49:42

CWNA candidates may not recognise this formula
The formula express in this blog is
fspl =20log(D) + 20log(f) + 92.45 where f D is kilometer and f is in GHz. Compare this with the more common equation in CWNA -107 course books which is
20log(D) + 20log(f) + 32.45 where f is expressed in MHz,

The equations are the same but the constant differ 60 dB su to the fact that f is expressed in GHz and not in MHz in the first case. .that is a factor of 1000 which is 10 raised to three, and thus 20x3=60 which is added to the konstant to compensate for using GHz instead of MHz. same idea of change to the constant can be used to compensate for calculation in kHz for example or distance in m etc.

The origin formula is using m and Hz respectively and the the constant is then -147.55. (another 1000 for m and another 1000000 for Hz which is a total of 10 raised to nine which the equals 180 dB, 32,45-180=-147.55

Se the article in Wikipedia

4/7/2018 10:24:23

Thank you for the precisions Martin!

Lucas link

4/6/2018 09:44:06

Hi Vergès, great content.

Why on the last paragraph you want to set the same cell size on 2.4 and 5Ghz? Does it make any beneffit to have them on the same cell size?

4/7/2018 10:29:30

I like to design for 5GHz and then go back and design for 2.4GHz. When I go back to design for 2.4GHz, I don't have the flexibility to more the APs around (it would break the 5GHz design), but I have the flexibility to disable 2.4GHz radios or adjust the Tx Powers.

Some people don't like to disable 2.4GHz radios and would rather keep both radios on infor all of their APs. Then, it makes more sense and it is easier to design if both frequency bands have the same cell size. In an environment where the client density is not too high and where 2.4GHz coverage is important for the business, it can work well and make sense (I'm thinking of warehouses as an example).

I hope I have answered your questions.

Monika@10301 link

8/10/2018 04:17:19

These graphs are quite easy to understand. The only thing you need to understand first is their consecutive fields of rise and downfall.

Mike B

1/20/2019 05:37:11

You have completely misunderstood and misused FSPL. FSPL is an arbitrary (and extremely illogical) combination of spherical geometric spreading and a component of antenna gain. It only makes sense in the context of the Friis transmission equations. FSPL used by itself is not a valid measure of RF attenuation. If used at all (you should be using geometric spreading and attenuation or a full blown link budget analysis), it needs to be used in context with the old Friis equation.

Using FSPL as you have results in incorrect system design and in general a false understanding of the frequency dependence of RF attenuation.

The frequency dependence of RF loss is primarily due to attenuation (absorption and scattering) of radio waves by gasses in the atmosphere, precipitation if outdoors, dust, sand, etc. The relative importance of these attenuators is dependent on frequency. For example, at 2.4 and 5 GHz, absorption of RF by oxygen molecules in the air is most pronounced. At around 20 GHz, water vapor is the dominant absorption mechanism.

For indoor, short range WiFi applications, there is effectively NO frequency dependent loss. 2.4 GHz and 5.0 GHz will perform the same. If you have walls or other obstructions, 5.0 will generally suffer more loss due to greater attenuation as the signals propagate through those obstacles. Note: there are possibly materials with resonances at 2.4 and not 5.0 where the loss will be greater at 2.4; consider the resonances for water vapor at around 20 GHZ - loss when water is present is greater at 20 GHz than it is at 30 GHz)

Your claim that 2.4 is better than 5.0 because the antenna is twice as large is complete nonsense. The antenna must be twice as large at 2.4 in order to achieve the same gain as a 5.0 GHz antenna of a particular length. For equal size antennas, you get 3dB more gain at 5 Ghz.

Isaac Newton did not develop the "Inverse Square Law".

I lost interest in the article after that. I hate to think what other errors and misinformation might exist.

As a wireless network engineer, you should read carefully re-read the Wikipedia article on FSPL that you cite. Pay special attention to the section that begins "Despite the misleading name, the free-space path loss formula inherits two important effects from the Friis Transmission Formula". You should also look at the Wikipedia page on "link budget".

12/13/2019 20:36:34

Hi Mike,

I totally understand your point and you are totally right from the pure technical aspect. Thank you for clarifying the concept and providing a deeper scientific explanation. I really appreciate it.
I wasn’t probably clear enough about the antenna size. I wasn’t trying to convey the idea that 2.4 is better than 5 and I agree with your explanation.
The goal of my article was to explain what happens to an RF signal (especially a Wi-Fi signal) as it exits its transmitter and travels away from its source. The article is targeted to Network Engineers and is vulgarizing the process on purpose.

Thank you for your feedback.

Best Regards,

François

Zemichael Gebregziabiher

4/5/2019 10:39:07

Hello,
Signal is not a linear. if the graph is in logarithmic then you can tell how the signal behave as it travel away from source. (XX dB per decade ). As you know in free space signal is coming down at 20dB per decade.

Ze

Stef

4/21/2019 08:15:40

Thanks for this explanation, I was looking for this since weeks. Finally I understand it!