Code element Rate Calculation

We can see that some people in BBS are talking about the calculation of the data transfer rate. This problem is more or less involved in the tests over the years, so I will introduce it in detail here. It is illustrated in examples.

The bitrate or modulation rate is also called the baud rate. In some books, the bit rate is measured by the number of vibrations per second in the waveform. If the data is not compressed, the baud rate is equal to the number of data transmitted per second. If the data is compressed, the number of data transmitted per second is usually greater than the modulation rate, in this way, the exchange between port and bit/s occasionally produces errors. The baud rate is the modulation rate of the carrier based on the signal. It is expressed by the number of times the carrier modulation status changes within a unit of time, and its unit is port (baud ).

The relationship between the baud rate and the bit rate is bit rate = baud rate × the binary digits corresponding to a single modulation state. In different signal modulation systems, the bit contained in each code element is different. For example, in binary digital transmission, a code dollar can carry one bit. In octal digital transmission, a code dollar can carry three bits. When a code element has eight status values, 2 ^ 3 = 8, that is, during modulation, each three bits constitute a code element, the corresponding 8 states are eight points in the constellation diagram, for example, 8 PSK, that is, the bitwise carries 3 bit information. The RB value is usually 9600, 4800, and so on.

Generally, each bitwise pulse can represent 2 M m-bit logstores. That is, the relationship between the bit rate and the baud rate is RB = RB log2m bps.

Example:
How many BITs does a code dollar contain when there are four status values? What is the information transmission rate at Port 9600? B/S?
Answer:
When there are four State values, we can see from the above reasoning (2 ^ 2 = 4) that every two bits constitute a code element. That is, the metadata element carries 2 bits of information.
Under the 9600rb condition, using the RB formula, we can directly obtain the formula = 9600 log2 4 = 9600 × log2 (4) = 9600 × 2 = 19200 bps.

The above is a basic knowledge point, no matter which kind of soft exams may be involved, please take the test into consideration.
Reference: http://bbs.51cto.com/thread-619302-1.html