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LTE: How much time a UE needs atleast to read PSS and SSS to get PCI?

+1 vote
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If PSS is read first then why SSS comes first in time domain?

posted Dec 6, 2014 by anonymous

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1 Answer

+2 votes

I am just doing a mathematical calculation. One LTE radio frame is 10 ms long and one radio frame has 10 sub frames, 1ms long each. One sub frame consists of two slots and each slot is 0.5ms long. In case of normal cyclic prefix, one time slot has 7 OFDM symbol, so that length of one OFDM symbol is 0.5 ms / 7 = (0.07142) ms long. PSS and SSS both are transmitted in last and second last OFDM symbols of radio sub frame '0' and '5'. In this way, we can say total of 2 OFDM symbols ( 2* 0.07142ms) time is necessary to read PSS and SSS. Once PSS and SSS are read by UE, UE known CGI.

answer Dec 7, 2014 by Neeraj Mishra
I have one query related to this. eNodeB configures meas gaps as 6 ms for inter freq meas. UE uses meas gaps to read neighbor cell measurement. In that UE reads PCI and RSPR of neighbor cell. As per above reply if UE is able to read PCI in lessthan 1ms then why eNodeB configures meas gaps as  6ms.
Correction in original comment. From PSS and SSS UE knows PCI and not CGI , for CGI ue has to read System information.

Basically there may be asynchronous system for which Ue needs to sync its time and frequency to succesfully read the Pss and Sss. Cases when Ue and eNodeB's are synchronous in that case Ue should be succesfully able to read the Pss and Sss fpr ~ 2 seconds , 1 seconds for Pss and Sss and 1 more seconds for Rf on/off. Ue has to tune to that frequency and for this Rf swithing is required. Meas gap enhancements is already proposed and new gap config is introduced with 3ms considering various results and data from different calculations.
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+1 vote

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0 votes

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+3 votes

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+1 vote

Why these two are placed side by side within "first time slot" of 0th and 5th sub frame of a radio frame ?
Is it due to any mathematical calculation or something else ?

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