Extra 2 Bits During Serial Communication

[bfurr@centrak.com]

Using the attached VI, we are attempting to send two bytes, 0x11 0x11 through the serial communication lines. When using the oscilloscope to decipher what was sent, we obtained the following screenshot. First, we noticed the data was sent LSB first, is there a way to send MSB first? Second, we noticed the addition of two bits between each of the bytes, highlighted in red below. We noticed that these extra two bits exist after each byte sent. How can we get rid of these extra bits? I have attached a screenshot as well of the current VI configuration we are using. 

 

[Martin_Henz]

I cannot open your VI because it is saved in labview 2023 (or something like this). Most people here use an older versions. LabVIEW 2019 might be ok.

What can be seen on the scope depends on the transmission parameters, For me it looks like 9600 baud, 1 stop bit, 8 data bits. The extra two bits are the start bit and the stop bit. These two bit are needed to syncronize the asyncronious communication between the endpoints. That's it.

[crossrulz · Accepted solution]

Looks like a standard UART waveform.  Normal is 8 data bits, a start bit, and a stop bit, resulting in 10 bits per byte you send through the UART port.  Anything that is receiving a UART signal (microcontrollers often have this built in anymore) will interpret all of this correctly.

[Minions]

How can you tell that it is sending the LSB first when your post mentions sending 0x11 and 0x11?  Both bytes are the same.  Did you mean that you are sending 0x88 and 0x11?  The order sent is determined by the endian which is typically related to either Motorola or Intel data transfer.

Crossrulz is correct that it has the start and stop bits surrounding the data as part of the data stream.

[bfurr@centrak.com]

Thank you for the help!