Assuming that an FSK signal over voice-grade phone lines can send 1200 bits per second, it has a bit rate of l200. Each frequency shift represents a single bit so it requires 1200 signal elements to send 1200 bits. Its baud rate, therefore, is also 1200. Each signal variation in an 8-QAM system, however, represents three bits. So a bit rate of 1200, using 8-QAM, has a baud rate of only 400. As Figure 4.3-17 shows, a dibit system has a baud rate of one-half the bit rate. A tribit system has a baud rate of one-third the bit rate. And a quadbit system has a baud rate of one-fourth the bit rate.

Table 4.1 shows the comparative bit and baud rates for the various methods of digital-to-analog encoding.

Encoding |
Units |
Bits/Baud |
Baud rate |
Bit rate |

ASK, FSK, 2PSK |
Bit |
1 |
N |
N |

4PSK, 4-QAM |
Dibit |
2 |
N |
2N |

8-PSK, 8-QAM |
Tribit |
3 |
N |
3N |

Quadbit |
4 |
N |
4N | |

Pentabit |
5 |
N |
5N | |

64- QAM |
Hexabit |
6 |
N |
6N |

128- QAM |
Septabit |
7 |
N |
7N |

Octabit |
8 |
N |
A constellation diagram consists of eight equally spaced points on a circle. If the bit rate is 4800 bit/s, what is the baud rate? ## SolutionThe constellation indicates 8-PSK encoding with the points 45 degrees apart. Since 23=8, three bits are transmitted with each signal element. Therefore, the baud rate is 4800/3=1600 baud Example 4.14 Compute the bit rate for a 1000 baud 16-QAM signal. Solution A 16-QAM signal means that there are four bits per signal element since T=16. Thus, Example 4.15 Compute the baud rate for a 72,000 bit/s 64-QAM signal. Solution A 64-QAM signal means that there are six bits per signal element since 26=64. Thus, 72,000/6=12,000 baud |

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