[srslte-users] How to guarantee the timestamp of UHD send function in srsLTE(pdsch_enodeb)

Xingguang Wei jxweixingguang at 163.com
Mon May 8 02:52:57 UTC 2017


Dear Ismael,


Thanks for your quick reply. And I have another related question, wish you can give me some suggestion. 


In the main function of pdsch_enodeb.c, there is a for loop where handle the data within one subframe each time.
          for (sf_idx = 0; sf_idx < SRSLTE_NSUBFRAMES_X_FRAME && (nf < nof_frames || nof_frames == -1); sf_idx++).


My question is how do pdsch_enodeb guarantee that the for loop handles the data of each subframe within 1ms.



From my test results, pdsch_enodeb can guarantee the processing time within about 1ms accurately. What's more, even when I add a usleep(800) in the for loop, it can still guarantee the processing time within about 1ms. I am confused about how the for loop can do that.


Sincerely.


--



xingguang Wei
Beijing University of Posts and Telecomunications
Mail: jxweixingguang at 163.com

At 2017-05-05 23:15:37, "Ismael Gomez" <ismael.gomez at softwareradiosystems.com> wrote:

Hi,


The pdsch_enodeb is just a simple example for transmitting a continuous stream of samples. It does not use timed transmissions. srsUE on the other hand does exactly what you mention for TX. 


On Fri, 5 May 2017 at 04:41 Xingguang Wei <jxweixingguang at 163.com> wrote:



Dear ismael and softwareradiosystems,


To the best of my knowledge, most of the SDR systems guarantee the UHD tx timestamp by getting the timestamp from rx function(recv). However, the srsLTE (pdsch_enodeb) only has tx procedures. And I observed that srsLTE (pdsch_enodeb)  uses the function of cuhd_send2, which has not done the time align.


So, how can the srsLTE (pdsch_enodeb) guarantee the timestamp of UHD send function ?


Any comments will be much appreciated.


Sincerely.
--



xingguang Wei
Beijing University of Posts and Telecomunications
Mail: jxweixingguang at 163.com
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