3
\$\begingroup\$

I’m doing a crowdsensing experiment and I'm capturing a handful of probe requests packets in my Raspberry Pi 3B in my vehicle. To test them, I capture a handful of packets of my own phone. Whenever I did I log the timestamp and the received strength signal in dBm, as shown below. My question is: why does the signal strength varies so much? Is because of interference w.r.t. the vehicle movement?

+-----------+-------------------+-----+
| Timestamp |        MAC        | dBm |
+-----------+-------------------+-----+ 
| 09:29:49  | 18:89:5b:36:ef:4f | -48 |
| 09:29:49  | 18:89:5b:36:ef:4f | -50 | 
| 09:29:49  | 18:89:5b:36:ef:4f | -54 | 
| 09:29:49  | 18:89:5b:36:ef:4f | -55 |
| 09:29:49  | 18:89:5b:36:ef:4f | -72 | 
| 09:29:49  | 18:89:5b:36:ef:4f | -82 | 
| 09:29:49  | 18:89:5b:36:ef:4f | -49 |
| 09:29:49  | 18:89:5b:36:ef:4f | -49 |
| 09:29:49  | 18:89:5b:36:ef:4f | -55 |
+-----------+-------------------+-----+
\$\endgroup\$

1 Answer 1

2
+100
\$\begingroup\$

There are a few factors that can affect wifi signal strength measurements:

  • Signal Path
  • Other devices
  • Distance

If other objects are moving, then this will affect the signal path and the signal strength will vary.

Wifi bands are inherently noisy, a big factor is other devices, especially those that are transmitting in the same band as your device. They are all sending packets, but at different intervals. If you were to look at the wireless spectrum in your area, there could be a lot of noise from other devices.

Another factor is the antenna pattern of each device, which varies according to the spheical angle of the antenna (ie at one point you might get -40dBm, move 5deg and you might get -20dBm). Here is a picture of a 3d antenna pattern from a rasp pi:

enter image description here
Source: https://hackaday.com/2018/04/19/raspberry-pi-w-antenna-analysis-reveals-clever-design/

You should expect the signal strength to vary considerably with any wireless connection, especially those in noisy environments.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.