Timing of central RX and peripheral TX

Question

I got an easy but fundamental question: How to make sure that the central sees the advertising package of the peripheral.. 
 For example when the peripheral advertises only every 5 seconds.. The centrals scan windows would have to be >5seconds? (and also the scan interval) So ongoing RX current??? (Or does the RX current only occurs when it actually receives an advertising packet?) 
 If they would establish a connection.. could I decrease the current consumption of the central while not increasing the current consumption of the peripherals? (Do they synchronize their RX/TX or something?)

shibshab · Accepted Answer

There is no way to guarantee that a central device can see an advertising packet (due to noise etc.), and as a peripheral in essence could modify its advertising interval, this is usually not a design goal of a central. 
 The peripheral uses two scan parameters when looking for advertisement packets. As explained in the BLE Spec (v4.0 Volume 6, Part B 4.4.3): 
 
 During scanning, the Link Layer listens on an advertising channel index for the
duration of the scan window, scanWindow . The scan interval, scanInterval , is
defined as the interval between the start of two consecutive scan windows. 
 
 As long as the devices would overlap at some time you would get connected. To "guarantee" that your central would see the peripheral within a single scan window you would need a scan window of >15 seconds. 
 The RX current will occur as long as the peripheral is "listening" i.e. scanning. 
 By establishing a connection you would allow the devices to operate synchronously. This way they will both know when to enable/disable RX/TX, and you would spend less power. The maximum allowed value for connInterval is 4.0 seconds, of which their radio would only have to be active for a couple of milliseconds.

Timing of central RX and peripheral TX

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