This video explains LTE Random Access Process in detail. Contention based random access process is explained with signalling flow.
Пікірлер: 79
@nehashrivastav45054 жыл бұрын
I never put comments in any video but your 14:16min video made me to do so. A big thank you for this video. every point is explained neatly. you cleared my doubts and even now I can talk about all those logs parameters in interview. Again thanks allot. Please keep uploading more videos.
@tejasvishe29742 жыл бұрын
The Best Explanation for RACH procedure till date. Loved the way you explained difference in Contention based and Non contention based procedures. Cleared all my doubts regarding timing advance and multiple uses of RACH procedure. More power to you my friend !!!
@Aishugannerla11 ай бұрын
Best RACH explanation... Thank you very much
@abhisheknrai7 жыл бұрын
Simple way to explain complex techniques, great video
@raghavsamba4 жыл бұрын
Beautifully explained....wonderful
@dhiruiitk8 жыл бұрын
Thank you very much... I got it fully. You made it easier for me to understand RACH otherwise i have to hunt books. Eager to watch your next video...
@LTEVideoTutorials8 жыл бұрын
Thanks Dheerendra!! You can go to my channel and watch my other videos. Currently, I am working on VOLTE. So, keep waiting :)
@dhiruiitk8 жыл бұрын
Sure sir...
@bhavinjoshi67802 жыл бұрын
Many thanks. Superbly explained.
@sanaullahkhan23548 жыл бұрын
you are Rockstar !!! thankyou Very much for making it this easy !!!
@LTEVideoTutorials8 жыл бұрын
Thanks!! You made my day 😀
@sridharvennala73098 жыл бұрын
Thank you so much for the video....Way of explanation with the diagrams was so understandable and easy too.
@LTEVideoTutorials8 жыл бұрын
Thanks Sridhar!! Such comments are the biggest motivating factor for me to make such videos. Next one is coming on carrier aggregation . Keep waiting :)
@sridharvennala73098 жыл бұрын
sure....
@keshavsharma62756 жыл бұрын
where is carrier aggregation video.
@sckmr6 жыл бұрын
Its very thankfull to understand rach procedure
@khanparvez400 Жыл бұрын
Very good explanations thanks alot
@vishalece7 жыл бұрын
awesome in depth explanation. Please post some video on csfb call flow with detailed messages!!
@LTEVideoTutorials7 жыл бұрын
Thanks Vishal!! There is already one detailed video on CSFB. Please go to my channel and watch it.
@simranrajgupta84625 жыл бұрын
Thank you for a nice video and explanation
@syamvadapalli91728 жыл бұрын
Thanks for the wonderful video.. I have a question regarding collision.. @10:24 - you said that there are two possibilities : 1. None of the RACH preamble heard by network 2. Only one of the RACH preamble received by network (i.e. UE A, per your example) Why can't both the RACH preambles (from UE A & UE B) be decoded by network? could you pls. clarify?
@LTEVideoTutorials8 жыл бұрын
First of all thanks for such a positive feedback! See, when two UEs choose different RACH preambles then contention never arises. In this case, eNodeB can easily decode RACH preambles from both UE as all the RACH Preambles sequences are orthogonal to each other. But when two UEs choose same RACH preamble then problem arises at eNodeB end. In this case, depending on the timing alignment of RACH preambles from both UE's, eNodeB could decode either any one UE or none of them. Decoding will depend on their timing alignment and strength of the received signal at eNodeB end. It is like, if two people scream same sentence at the same time, you can either get complete sentence from one of them or both will distort each others voice. Hope, I was able to clarify your doubt!!
@syamvadapalli91728 жыл бұрын
Thanks a lot for the clarification! Initially I was thinking in the same direction, that two messages (sent by UE A & UE B) with same code would interfere at node B and would sound like noise. However, I wanted a confirmation from an expert like you!
@muhammadanwarhussain41947 жыл бұрын
Is it possible that enodeb decoded preamble from both successfully but it couldn't decode the Rrc request from neither?
@deeptanudatta723511 ай бұрын
Hi @LTE Video Tutorials, could you throw some more light on random access use cases during data download or upload when UL synchronization status is 'non-synchronized' @13:30
@kalpeshksanghani7 жыл бұрын
Really Good Explanation... Can you provide all links to identify ALL LTE Related Process (Like VOLTE Call, CSFB MT call, Hand over, etc
@varun2842 жыл бұрын
excellent session
@amoljadhao474 жыл бұрын
Nice explanation........Thanks
@shovanchoudhury7 жыл бұрын
@11:15 >> While 2 UE send same preamble at the same time to same enodeB, what happens if by chance both UE A and UE B again pick up same random number while sending RRC connection request? Then, enodeB will not able to decode the RRC request from UE B because of mismatch of timing advance. But the RRC setup message sent from eNB for UE A will be received by UE B also and the random number will match.
@conradlichota73104 жыл бұрын
I think its because the eNB will not be able to decode the RRC connection request because the timing advance number will be wrong. That was returned with the original RAR with RA RNTI for UE-A. So there will be timing incoherency.
@chamankapoor76104 жыл бұрын
Please make more videos like this.....I have seen each of your videos atleast 5-8 times......thanks Please make video on LTE layers protocol with logs and IEs that needed to be see
@mustafasheik57268 жыл бұрын
Could you please add inter rat handover videos as well... It will very useful. Good job
@abebebelayadege2997 жыл бұрын
Thank you for this brief Explanations.
@punniyamoorthy51277 жыл бұрын
thanks for this useful Video
@keshavsharma62756 жыл бұрын
ultimate information u have explained easily. thanks kindly send me link for carrier aggregation.
@LTEVideoTutorials6 жыл бұрын
Please go to my channel and see the video on carrier aggregation. Enjoy :)
@pawansingh-ub2ej5 жыл бұрын
Beautiful explanation
@imthiyazali8 жыл бұрын
Thank you so much for the excellent video.Kindly make videos related to LTE KPI's..
@LTEVideoTutorials8 жыл бұрын
+Imthiyaz Ali Sure! I will add it to my list.
@haianeeshm6 жыл бұрын
As per my understanding the only because of TA difference the eNb is unable to decode the UE-B random number.. isnt it?? what will be the outcome if both the UEs having the same TA ??
@sachdevaharkirat6 жыл бұрын
Hi At 11:02 you mentioned ENB will send RA-RNTI but As per my understanding RA-RNTI is sent by UE in Request message and ENB will revert back with TC-RNTI(Temporary Cell) in RAR
@veerojumanasa79486 жыл бұрын
RA-RNTI is calculated by enb,based upon the time of reception of rach request
@yuvrajpatil17106 жыл бұрын
uE dont send RA RNTI , nerwork will find out the RA RNTI by looking on which subframe ue transmit preamble
@veerojumanasa79486 жыл бұрын
@@yuvrajpatil1710 "UE shall monitor the PDCCH for Random Access Response identified by the RA-RNTI" ,WHICH MEANS THAT UE MUST HAVE SENT THE RA-RNTI..........is it not the correct justification
@yuvrajpatil17106 жыл бұрын
@@veerojumanasa7948 after transmitting Preamble UE serch for responce with in RAR window . this RAR window will always start from 3rd subframe after the premble send. the responce window (RAR window) size is configured upto 10subframe and it informed in SIB2 so for responce UE will serch only the RAR window.
@veerojumanasa79486 жыл бұрын
*yeah in the 3rd subframe it reads pdcch first... *in dat pdcch it reads ra-rnti,which it has sent while transmitiing request........... *once if it finds its own ra-rnti,den it starts reading pdsch where info abt uplink grant is present *so now in the whole scenario,,,ra-ranti is transmited by ue bassed upon the time at which it is sending
@vijaym25127 жыл бұрын
good tutorial .... please explain cpri. and it's use in rach process
@satishkumarsingh731010 ай бұрын
Very informative pls post 5G videos
@vaibhavhatwar4 жыл бұрын
Well described👍
@kurapatisrikanth84555 жыл бұрын
This video is helped me lot thanku but please explain indetailed about csfb mo and mt with log analysis from lte to 3g
@mustafasheik57268 жыл бұрын
Excellent video
@pushpendradixit5 жыл бұрын
very well explained
@ofektoladuno95596 жыл бұрын
fantastic work thank you!
@bitcooin6 жыл бұрын
nice explanation
@mohammadfayaz91065 жыл бұрын
In Contention bases random access . Group A and B are there. Group B is of good coverage Preambles and Group A are poor coverage Preambles. Am i right ?
@aishajaved20888 жыл бұрын
knowledge full
@FARAZKHAN-ch1qx3 жыл бұрын
Valuable points can't find in books 👍
@veerojumanasa79486 жыл бұрын
hi all,can any1 explain this....... at 11:39 how cum UE B send rrc connection request,when it was not assigned any uplink resource
@Mahimammu6 жыл бұрын
Hi its a Control Plane Message No Need Pre established UL Connection Because RRC Connection Procedure itself for Creating UL Connection(Radio Bearer Setup). for Attach with N/W UE sends RRC Connection request with EnB for a Radio Bearer Setup. RRC Connection is an Uplink Message/request spends through PUL-SCH-->UL-SCH->CCCH (UL).
@veerojumanasa79486 жыл бұрын
thank you
@nikhilgoyal48404 жыл бұрын
Hi, what would be msg2 for RACH procedure?
@Luis-dm3qg5 жыл бұрын
How do you improve RACH STP Fails?
@bibhubehera6497 жыл бұрын
Superb
@joyoussikh8 жыл бұрын
veru good explanation
@LTEVideoTutorials8 жыл бұрын
Thanks Aman!!
@ravikiranreddy65297 жыл бұрын
Hi .... very nice explanation !!! is it possible to email me the wireshark traces ?
@hsar255 жыл бұрын
Nice one...sir please solve this doubt...there is no problem with UE as well as network side..but still UE unable to camp on the network..what might be the reason.. please solve
@syedmohtada8 жыл бұрын
Excellent
@bindureddy68817 жыл бұрын
hi sir plez provide more tutorials further
@northsamsung73688 жыл бұрын
excellent ,
@ArunK-ix2ol Жыл бұрын
good
@cationii6 жыл бұрын
11:27 both A B will choose random number as initial identity and send RRC connection request and start the timer t300 but you know we will not be able to decode message from B is using the timing advance value that was intended for A 兩個都會收到msg 4 both you will decode this message as it is addressed by TC RNTI " eNodeb will include random number in this message" that was sent by A 這是A的隨機號碼 although both UE s will decode this message but random number sent by embassy by B will be will mismatch only at this stage you will be will come to know that it has lost out to some other ue in contention resolution then B will be will start random access process again and from very beginning contention free random access procedure did instances went because of timing restrictions contentions are not acceptable B因為msg4帶的隨機號碼錯誤而知道自己沒有得到競爭