Calculating scan duty cycle vs advertisement detection probability:
A BLE gateway scans for beacons that advertise every 1000 ms. What scan duty cycle is needed to catch >95% of advertisements?
Beacon advertising:
- Advertising interval: 1000 ms (1 second)
- Advertisement duration: ~2 ms (includes 3 advertising channels)
Scanner with 10% duty cycle:
- Scan interval: 1000 ms
- Scan window: 100 ms
- Probability of catching advertisement: $ P_{} = = = 0.102 = 10.2% $
Wait, that’s not right. Let me recalculate properly:
For a beacon advertising every 1000 ms with 2 ms advertisement duration, a scanner with scan window of 100 ms will catch it if the advertisement falls within any of the scan windows.
Since beacon advertisements happen at random offsets within the 1000 ms interval (due to ± 10ms randomization), the probability is: $ P_{} = = = 10% $
But we scan every second, so over 10 scan intervals (10 seconds): $ P_{} = 1 - (1 - 0.10)^{10} = 1 - 0.35 = 65% $
For >95% detection within 3 seconds, we need:
Let scan window be \(w\), with scan interval = 1000 ms. For 3 scans to have >95% cumulative detection: $ 1 - (1 - )^3 $
Solving: $ (1 - )^3 - $
Therefore \(w \geq 632\) ms. This means 63% duty cycle is needed for 95% detection within 3 seconds.
Power comparison (12 mA scan current, 2 mA idle):
100% duty cycle: 12 mA continuous 63% duty cycle: \((12 \times 0.63) + (2 \times 0.37) = 7.56 + 0.74 = 8.3\) mA 10% duty cycle: \((12 \times 0.10) + (2 \times 0.90) = 1.2 + 1.8 = 3.0\) mA
The 10% duty cycle reduces power by 64% \((1 - 3.0/8.3)\) compared to 63% duty cycle, but detection latency increases from 3 seconds to ~10 seconds.