Serial communication baud rate issue

Question

I am a bit confused regarding the baud rate of serial communication.

Suppose I am sending a value of 198 to the receiver with a baud rate of 9600. At the receiver, the rate is also the same. Thus isn't it possible that the data is lost? This means that after 1/9600 s a data bit is sent, and at this time the data is received as well. This means at each edge we are receiving and data can be lost since edges are not that accurate. If we have the baud rate of the receiver a bit different so it takes the data a bit ahead from the edge then it will work fine. But we usually use the same. How doesn't it create a problem?

Answer 1

This means at each edge we are receiving and data can be lost since edges are not that accurate.

A UART (as I believe you are talking about) makes use of start and stop bits to acquire synchronization: -

And, the receiver operates an internal clock (usually 16 times higher than the baud rate to ensure that data is "captured" in the middle of each symbol. As soon as the receiver detects the falling edge of the start bit it counts 8 clocks to position itself in the middle of the start bit. Then, 16 clocks later it knows it's in the middle of the 1st bit etc..

Image from here.

Answer 2

I am a bit confused regarding the baud rate of serial communication.

Your confusion appears to be about the basic clock synchronisation, not the baud rate.

so it takes the data a bit ahead from the edge

Your thinking is along the right lines: how do we make sure the receiver isn't sampling near the edge of the bit time. But your suggestion about how to do this:

If we have the baud rate of the receiver a bit different

... that won't work in the way you're imagining. We don't want different rates, we want the reciever to estimate where the sender's bit times begin. To sample reliably, we can sample in the middle of where the receiver thinks the bit will be.

The correct way uses the falling edge of the start bit to achieve this synchronisation.

The green part ensures we're in idle or a stop bit; the blue part finds the falling edge (we want to do this quickly, perhaps 16 times as fast as the bit rate), and the red part estimates the middle of the bit time, followed by getting the bits.

All the sophistication lies in either a) Finding that falling edge accurately, b) multiple sampling in order to discount spurious spikes, or c) dealing with any errors.

It's not common, but you can make systems which measure the bit rate from the signal. More common is to detect which of the common speeds is in use by trying different them and looking for recognisable patterns such as ASCII CR values. But without a doubt the most common case is simply to have both sides configured to use the same speed.

Answer 3

No, it is genrally not possible to lose data under normal conditions, as the receiver actually runs faster than at the bit rate to be able to receive bits properly at the bit rate

For 10-bit frames, i.e. 8N1 format, UARTs in practice tolerates a baud rate difference of approximately 4%. In theory, sampling the last bit can have a drift of +/- half a bit from the center of the bit so the receiver could in theory run 5% slower or faster.

Which means that one device could be about 2% slower than 9600, and the other device about 2% faster than 9600 and they would communicate successfully.

Generally accepted values for baud rate tolerances of individual devices are smaller, in the 0.5% to 1% range from the nominally reported baud rate.

This is because a clock source used may have some tolerance, and it can't be directly divided into exact bit rate. For example it is not possible to divide a 8 MHz clock soure to exactly 9600, but within 0.17% of 9600 is possible.

The specific problem you seem to be thinking is solved by oversampling, as a typical UART uses 16x oversampling for each bit slot to detect the start bit edge and sample data bits at the center of the bit time slot.

Answer 4

This answer is based on the tags in the OP's post. The Atmega microcontroller series, including the Atmega128, has built-in UART (Universal Asynchronous Receiver-Transmitter) hardware for serial communication. The term "Asynchronous" (in digital communication) pertains to a transmission technique that does not require a common clock between the communicating devices; timing signals are derived from special characters (symbols) in the data stream itself. So because start and stop are recognized at the receiver end no data can be lost as the way you described.