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.
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 !!!
@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.
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.
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
Thanks Sridhar!! Such comments are the biggest motivating factor for me to make such videos. Next one is coming on carrier aggregation . Keep waiting :)
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?
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!!
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!
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
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 ??
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
@@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
@@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.
*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
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 ?
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).
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
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
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.
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 !!!
Best RACH explanation... Thank you very much
Simple way to explain complex techniques, great video
@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.
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.
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...
Thanks Dheerendra!!
You can go to my channel and watch my other videos. Currently, I am working on VOLTE. So, keep waiting :)
Sure sir...
Beautifully explained....wonderful
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
Thank you so much for the video....Way of explanation with the diagrams was so understandable and easy too.
Thanks Sridhar!! Such comments are the biggest motivating factor for me to make such videos. Next one is coming on carrier aggregation . Keep waiting :)
sure....
where is carrier aggregation video.
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?
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!!
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!
Is it possible that enodeb decoded preamble from both successfully but it couldn't decode the Rrc request from neither?
Many thanks. Superbly explained.
awesome in depth explanation. Please post some video on csfb call flow with detailed messages!!
Thanks Vishal!! There is already one detailed video on CSFB. Please go to my channel and watch it.
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
you are Rockstar !!! thankyou Very much for making it this easy !!!
Thanks!! You made my day 😀
excellent session
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 ??
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
RA-RNTI is calculated by enb,based upon the time of reception of rach request
uE dont send RA RNTI , nerwork will find out the RA RNTI by looking on which subframe ue transmit preamble
@@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
@@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.
*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
Really Good Explanation... Can you provide all links to identify ALL LTE Related Process (Like VOLTE Call, CSFB MT call, Hand over, etc
Its very thankfull to understand rach procedure
Could you please add inter rat handover videos as well... It will very useful. Good job
Thank you for a nice video and explanation
Very good explanations thanks alot
Nice explanation........Thanks
Very informative pls post 5G videos
Beautiful explanation
ultimate information u have explained easily. thanks kindly send me link for carrier aggregation.
Please go to my channel and see the video on carrier aggregation. Enjoy :)
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 ?
Thank you so much for the excellent video.Kindly make videos related to LTE KPI's..
+Imthiyaz Ali
Sure! I will add it to my list.
good tutorial .... please explain cpri. and it's use in rach process
thanks for this useful Video
Well described👍
Excellent video
Thank you for this brief Explanations.
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
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).
thank you
How do you improve RACH STP Fails?
Hi,
what would be msg2 for RACH procedure?
very well explained
This video is helped me lot thanku but please explain indetailed about csfb mo and mt with log analysis from lte to 3g
nice explanation
Valuable points can't find in books 👍
fantastic work
thank you!
knowledge full
Hi .... very nice explanation !!!
is it possible to email me the wireshark traces ?
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
Excellent
Superb
veru good explanation
Thanks Aman!!
hi sir plez provide more tutorials further
good
excellent ,
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
hi please send this ppt Link in Comment box
nice explanation