In SX Microcontrollers, SX/B Compiler and SX-Key Tool, Peter Van der Zee wrote:

Hello Michael;

I did something like the timing diagram you suggested, and found indeed you are correct!

In my example, I show two half-clocks per bit; SS is "start", 11 through 88 are the "data" bits 1 through 8, and PP is the "stop" bit.




[code]
Foreign data stream SS1122334455667788PPSS1122334455667788PPSS1122334455667788PPSS1122334455667788PPSS1122334455667788PP 

Receiving UART                          SS112233445566778...SS112233445566778...SS112233445566778...SS112233445566778... 

Retransmit UART                                          SS1122334455667788PPSS1122334455667788PPSS1122334455667788PP 
[/code]
So, I can now see that regardless of when the re-transmitter starts sending, it always needs to transmit 10 bits per character continuously, and if the foreign stream is also continuous at a slightly faster clock rate, the re-transmitter eventually can't get rid of it in time, and data corruption results.
 

So what this implies is that one can never reliably "repeat" data at the same rate as it comes in without shortening up the repeated stop bit......

 

Sorry to have been such a pain about this, I was totally convinced of my position. Now I know better.....I learned several things today!

 

Cheers,

 

Peter (pjv)

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