I am using an i.MX<http://i.mx/> 6SoloX development board for both the master and the slave. I start ptp4l with the following command:
ptp4l -f /etc/ptp4l_slave.conf -i eth0 -A & where ptp4l_slave.conf has the following in it: [global] twoStepFlag 1 verbose 1 time_stamping hardware slaveOnly 1 use_syslog 1 logging_level 6 Why wouldn't this create a synchronized 1PPS signal if I am not starting phc2sys? Also, if I start phc2sys with the following command: phc2sys -d /dev/pps0 -s /dev/ptp0 -w Why would that produce a 2 usec offset on 1PPS? ________________________________ From: Richard Cochran <richardcoch...@gmail.com> Sent: Monday, August 17, 2020 2:21:35 PM To: Irene Kravets <ire...@sapling-inc.com> Cc: linuxptp-users@lists.sourceforge.net <linuxptp-users@lists.sourceforge.net> Subject: Re: [Linuxptp-users] Can ptp4l be used without phc2sys? On Mon, Aug 17, 2020 at 06:07:41PM +0000, Irene Kravets wrote: > Can ptp4l be used without phc2sys on a PTP slave to produce a 1PPS > signal that's synchronized to the master's 1PPS? Yes. > If my understanding is correct, I would expect that running just > ptp4l would be sufficient to sync 1PPS signals, however, when I > tried it, the 1PPS signals were not synchronized. I thought that > phc2sys is only needed to synchronize the system clock to PHC and > that phc2sys should not be needed to synchronize 1PPS, but that's > not what I am observing. Is my understanding wrong? No, you are correct. The phc2sys program synchronizes the Linux system clock from the PHC. > When I run both ptp4l and phc2sys on the slave, the 1PPS signal is > synchronized, but there is a 2-3 usec delay between master and slave > 1PPS signal when I observe both on a scope. Why is that delay there > and is there any way to make it much smaller? I need them to be > within 20-30 nsec of each other. If the PPS signal is being produced by the PHC, then the offsets should be lower. I guess your PPS is NOT coming from the PHC? Well, I can only guess, because you did not provide any information about your hardware. Thanks, Richard
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