Re: CAN bit timing synchronization question
Tamás Fekete <[email protected]> Sun, 13 Nov 2016 17:33:31 +0100
| Newsgroups | gmane.comp.hardware.bus.can |
|---|---|
| Message-ID | <CAAz95+mKpOS_verjD=r+C9GrQSn6x22MiVo0mFKi306anZ7fRQ@mail.gmail.com> |
Thank you, I remembered that bit (no pun intended) about bit stuffing just as I was writing my question, but I wanted to ask it anyway to make it sure. I quite recently graduated and while we had to memorize these things about CAN it is only now that I had connected these two informations. thanks again, Tamás On Sun, Nov 13, 2016 at 4:08 PM, Bram Kerkhof <[email protected]> wrote: > You basically answered your own question J. Stuff bits are there to > ensure that there are sufficient edges in the frame to perform > synchronization. > > > > cheers, > > Bram > > > > *From:* [email protected] [mailto:[email protected]] *On > Behalf Of *Tamás Fekete > *Sent:* zondag 13 november 2016 15:58 > *To:* [email protected] > *Subject:* [CANLIST] CAN bit timing synchronization question > > > > **snip** > > > > So I think I understand the first point, which means that the start of > frame bit has to start with a recessive to dominant transition, and that is > a reference point for the receiver to start counting the bit times. > > > > I have trouble with the second point. Why would there be a bit edge when > for example I am transmitting two ones after each other? So how does it > know that there was supposed to be a bit edge there? Or is this why the > protocoll stuffs an extra different bit when there are too many same > consecutive bits on the line? > > > > Any help is appreciated, > > Fekete Tamás (Tamás=Thomas in english) > -- "there is no spoon"