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Uart baudrate question

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maria_100784

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Hello forum. i have a question regarding the possible convert of the baud rate.
Example: I recorded data with the speed of 19200...but the data was coming just with 9600 speed..it is possible to convert the recorded data to get the correct data? if yes, how? in theory i belive is possible..because i`m moving from a higher to a lower baud(so all the bits must be there in the log)...but..dont know if i`m wrong..
 

Example: I recorded data with the speed of 19200...but the data was coming just with 9600 speed..it is possible to convert the recorded data to get the correct data? if yes, how? in theory i belive is possible..because i`m moving from a higher to a lower baud(so all the bits must be there in the log)...but..dont know if i`m wrong..

Your intended purpose of the above example is unclear.

What is the purpose of utilizing two different BAUD rates on a single serial connection?

Are you attempting to "recover" data previously recorded with a RX BAUD rate of 19200 and TX BAUD rate of 9600? Or vice versa?

Or are you attempting to receive data at a faster rate then it was transmitted? Or vice versa?

BigDog
 

Your intended purpose of the above example is unclear.

What is the purpose of utilizing two different BAUD rates on a single serial connection?

Are you attempting to "recover" data previously recorded with a RX BAUD rate of 19200 and TX BAUD rate of 9600? Or vice versa?

Or are you attempting to receive data at a faster rate then it was transmitted? Or vice versa?

BigDog


"Are you attempting to "recover" data previously recorded with a RX BAUD rate of 19200 and TX BAUD rate of 9600? " This is correct
 

any inputs please?...sorry..but i`m in a big big project hurry..
 

I'm sorry, i still can't get what you need as you said, "i have a question regarding the possible convert of the baud rate."
Do you .... :
a. You have recorded data with 19200 and now you want to transmit it with 9600.
b. You have saved data with wrong baud rate, and now you want to convert it to get the correct value.
c. .....(please explain it clearly)

the answers:
a. no problem. You can do that.
b. you have lost your data. Incoming data = 9600 bps but receiver is configured as 19200 bps. At a glance, we will think every bit will be recorded twice, so there is no data lost. we will think : 0101 1011 will be recorded 0011 0011 and 1100 1111, so just convert it.
But, on every data frame contains 10-bit, Start bit, 8 bit data, stop bit.
So the start bit of first frame will be supposed and saved as data and any next bits will be supposed as stop bit and start bit of second frame.
If incoming data = 19200 and receiver is configured as 9600 bps, the saved data maybe only contains odd or even bit of each real data and the last 4 bit are missing.
c. ....
 

thanks , b. was the case...so is not possible...is possible to implement a autobaud system?
 

There are many autobaud systems. They tend to be susceptable to noise on the line as well as missing the first few bytes.
 

Can you elaborate on the two devices connected by the serial connection?

In the case of microcontrollers, supported BAUD rates are dependent on several factors, device architecture, system clock frequency, etc.

BigDog
 

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