I'm really not sure this question at all.  UART4 Tx a single wire which is
high by default. Put a multimeter on the uart4tx pin to confirm this. If
you send data to it then it goes out in blocks of N bits (N normally being
10 - start bit, 8 data bits, stop bit), then returns to the default state
of high until the next transmission.
http://en.wikipedia.org/wiki/Asynchronous_start-stop

Do you want to send binary data bit-by-bit, as in: write 1 to it, the pin
goes high, write 0 and the pin goes low? Then that's not what a UART does -
sounds like GPIO to me.

Do you want to send the byte 0, as in: start bit, 8 x low bits, stop bit?
Then "echo -n \x00 > /dev/ttyO4" should do it. But much easier to do this
in C (or any other language) as the ttyO4 is just a file: open it, set it
up as a TTY (in C with functions tcsetattr, cfmakeraw etc), then write the
zero byte to the stream.


On 26 November 2013 14:42, Andrei <andrej.kondrat...@gmail.com> wrote:

> Hello Rod,
>
> Thanks  you for the help on this, was really helpful.
> But do you know how can I send the binary instead of ascii?  (Do I need to
> put maybe prefix of SB before the value to send binary?)
>
> After many unsuccessful attempts of sending binary over UART,  I'm looking
> at possibility to change the UART MODE described in *TI AM335x ARM A8
> Microprocessors technical reference manual  ,  *maybe this will allow me
> to send the binary then.
>
>
> For my project, the transceiver that I wan to connect to UART needs
> CMOS/TTL logic.
>
>
> On Monday, 25 November 2013 21:05:16 UTC+1, rod calabio wrote:
>>
>> Andrei
>>
>>    ascii for "1"  is 00110001
>>    ascii for Line feed is 00001010
>>
>>   so you will get
>>    st=0
>>    d0 = 1
>>    d1 = 0
>>    d2 = 0
>>    d3 = 0
>>    d4 = 1
>>    d5 =1
>>    d6 = 0
>>    d7 =0
>>    sp = 1
>>    st = 0
>>    d0 = 0
>>    d1 = 1
>>    d2 = 0
>>    d3 = 1
>>    d4 - d7  =0
>>     sp = 1
>>     idle = 1
>>
>>  --
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