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  1. In case of using equalizers with ZigBee systems, what type of fading does the channel show, flat or frequency selective, fast or slow?

  2. Is the inter-symbol interference (ISI) problem encountered in ZigBee systems?

I am asking this question because I konw that ZigBee Systems transmit data at low rate (250 kb/sec), and ISI problem and hence frequency selective fading appear with high data rate applications. I found some papers employing equalizers with ZigBee under Rayleigh fading conditions, but the type of fading was not indicated.

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In the $2.4 \textrm{ GHz}$ band, the data rate is indeed just $250\textrm{ kbit/s}$ but the direct-sequence spread spectrum (DSSS) coding results in a bandwidth of $2\textrm{ MHz}$. This means that there will be flat fading only if the channel's coherence bandwidth is significantly greater than $2\textrm{ MHz}$, say around $10\textrm{ MHz}$ (this is just a fuzzy estimate). This corresponds to a maximum delay spread of approximately $1/(10\textrm{ MHz})=100\textrm{ ns}$.

Consequently, any environment with a delay spread greater than approximately $100\textrm{ ns}$ will result in frequency-selective fading. So in relatively small indoor areas the fading will be mostly flat, but in large office spaces or outdoors you can expect frequency-selective fading.

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  • $\begingroup$ Thank you for your answer. If four multipath components with path delays [0, 1, 5, 10] ns and average path gains of [0, -2, -3, -5] dB are used for simulation, does this mean that the delay spread is 10ns-0ns=10ns? (difference between the latest component and the earliest component? $\endgroup$ – Noha Aug 18 '18 at 9:05
  • $\begingroup$ @Noha: Yes, that would be a reasonable assumption. $\endgroup$ – Matt L. Aug 18 '18 at 12:33
  • $\begingroup$ But when I simulated the Zigbee System under these conditions, the ISI problem appeared, although the symbol period is greater than the delay spread, why? $\endgroup$ – Noha Aug 18 '18 at 16:01
  • $\begingroup$ Continuing the previous comment. For 2 Mchips per second, the chip period is the reciprocal of the chip rate, which leads .5 microseconds. Since OQPSK is employed, then the symbol duration is 4x .5= 2 microseconds which is greater than 10 nanoseconds. Are my calculations correct? if so, then what is the reason for the ISI problem? $\endgroup$ – Noha Aug 18 '18 at 16:15
  • $\begingroup$ @Noha: It's hard to tell what's going on in the simulation. If you decrease the delay spread of the channel, does the ISI become smaller? Can you detect perfectly if you remove all paths but one from the channel? $\endgroup$ – Matt L. Aug 18 '18 at 21:41

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