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From the Wikipedia, the XBee paper, the BlueBee paper, and the WeBee paper, I learned that:

  1. In the 2.4 GHz band, the bandwidth of each ZigBee channel is 2 MHz.
  2. For receiving a signal over one of these channels, a ZigBee receiver should sample the signal with a sampling interval $T_{s} = 0.5 \mu s$.

However, according to Nyquist's rule, the sampling frequency $f_{s} = 1/T_{s}$ of a ZigBee receiver should be twice ZigBee's signal bandwidth. Hence, the ZigBee's single-channel bandwidth should be $2 \times f_{s} = 2 / T_{s} = 4$ MHz, which contradicts what I learned.

Is there anything I missed?

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  • $\begingroup$ Does this answer your question? dsp.stackexchange.com/questions/672/… $\endgroup$
    – AlexTP
    Dec 28, 2021 at 20:19
  • $\begingroup$ @AlexTP Yes! Thank you very much. $\endgroup$
    – Dudada
    Dec 29, 2021 at 2:35
  • $\begingroup$ So lets close the question. :) $\endgroup$
    – AlexTP
    Dec 29, 2021 at 9:41

1 Answer 1

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Zigbee is primarily designed to be cheap, not fast or efficient. Maximum data rate is 250kb/s so 2 MHz sample rate is more than enough to capture any data signal.

The "2MHz bandwidth" number mainly refers to the RF channel layout. Zigbee channels are 2 MHz wide and spaced 5 MHz apart. This primarily determines what RF filtering needs to happen to prevent interference between adjacent channels. Here, again, there is plenty of margin, to make practical implementations as inexpensive as possible (at the expense of wasting a lot of potential bandwidth).

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