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Serial-data speed is usually stated in terms of bit rate. However, another oft-quoted measure of speed is the baud rate. Though the two aren’t the same, similarities exist under some circumstances. This tutorial will make the difference clear.
Most data communications over networks occur via serial-data transmission. Data bits transmit one at a time over some communications channel, such as an RS-232 cable or a wireless path. Figure 1 typifies the digital-bit pattern from a computer or some other digital circuit. This data signal is often called the baseband signal. The data switches between two voltage levels, such as +3 V for a binary 1 and +0.2 V for a binary 0. Other binary levels are also used. In the non-return-to-zero (NRZ) format (Fig. 1, again), the signal never goes to zero as that of return-to-zero (RZ) formatted signals.
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1. Non-return to zero (NRZ) is the most common binary data format. Data rate is indicated in bits per second (bits/s).
Bit Rate
The speed of the data is expressed in bits per second (bits/s or bps). The data rate R is a function of the duration of the bit or bit time (TB) (Fig. 1, again):
R = 1/TB
Rate is also called channel capacity C. If the bit time is 10 ns, the data rate equals:
R = 1/10 x 10–9 = 100 million bits/s
This is usually expressed as 100 Mbits/s.
Overhead
Bit rate is typically seen in terms of the actual data rate. Yet for most serial transmissions, the data represents part of a more complex protocol frame or packet format, which includes bits representing source address, destination address, error detection, and correction codes, and other information or control bits. In the protocol frame, the data is called the “payload.” Non-data bits are known as the “overhead.” At times, the overhead may be substantial—up to 20% to 50% depending on the total payload bits sent over the channel.
For example, an Ethernet frame can have as many as 1542 bytes or octets, depending on the data payload. The payload can range from 42 to 1500 octets. With a maximum payload, the overhead is only 42/1542 = 0.027, or about 2.7%. It would be even greater if the payload was anything smaller. This relationship is usually expressed as a percentage of the payload size to the maximum frame size, otherwise known as the protocol efficiency:
Protocol efficiency = payload/frame size = 1500/1542 = 0.9727 or 97.3%
Typically, the actual line rate is stepped up by a factor influenced by the overhead to achieve an actual target net data rate. In One Gigabit Ethernet, the actual line rate is 1.25 Gbits/s to achieve a net payload throughput of 1 Gbit/s. In a 10-Gbit/s Ethernet system, the gross data rate equals 10.3125 Gbits/s to achieve a true data rate of 10 Gbits/s. The net data rate also is referred to as the throughput, or payload rate, of an effective data rate.
The term “baud” originates from the French engineer Emile Baudot, who invented the 5-bit teletype code. Baud rate refers to the number of signal or symbol changes that occur per second. A symbol is one of several voltage, frequency, or phase changes.
NRZ binary has two symbols, one for each bit 0 or 1, which represents voltage levels. In this case, the baud or symbol rate is the same as the bit rate. However, it’s possible to have more than two symbols per transmission interval, whereby each symbol represents multiple bits. With more than two symbols, data is transmitted using modulation techniques.
When the transmission medium can’t handle the baseband data, modulation enters the picture. Of course, this is true of wireless. Baseband binary signals can’t be transmitted directly; rather, the data is modulated onto a radio carrier for transmission. Some cable connections even use modulation to increase the data rate, which is referred to as “broadband transmission.”
By using multiple symbols, multiple bits can be transmitted per symbol. For example, if the symbol rate is 4800 baud and each symbol represents two bits, that translates into an overall bit rate of 9600 bits/s. Normally the number of symbols is some power of two. If N is the number of bits per symbol, then the number of required symbols is S = 2N. Thus, the gross bit rate is:
R = baud rate x log2S = baud rate x 3.32 log10S
If the baud rate is 4800 and there are two bits per symbol, the number of symbols is 22 = 4. The bit rate is:
R = 4800 x 3.32 log(4) = 4800 x 2 = 9600 bits/s
If there’s only one bit per symbol, as is the case with binary NRZ, the bit and baud rates remain the same.
Hope to help you!
