y PSK (i.e., M = 8). Several important points arise here which must be made clear. First, this is how we can achieve a data-transmission system where the data rate exceeds the bandwidth. We stated that the data rate on a voice channel could not exceed 3100 bps, assigning 1 bit per hertz of bandwidth. We mentioned in passing that this held for the binary condition only. Theoretically, for QPSK, we can achieve 2 bits per hertz of bandwidth, but practically it is more like 1.2 bits per hertz. With 8-ary PSK the theoretical bit packing is 3 bits per hertz. Bandwidth is a function of transitions per second. Transitions per second are com- monly referred to as bauds. Only in the binary domain are bauds and bits per second synonymous. With QPSK operation, the baud rate is half the bit rate. For example, with a 2400-bps modem, the baud rate is 1200 bauds. The bandwidth required for a certain data transmission system is dependent on the baud rate or the number of transitions per second. For example, for BPSK a transition carries just one bit of information; for QPSK a transition carries 2 bits of information, and for 8-ary PSK a transition carries 3 bits of information. The CCITT 4800-bps modem uses 8-ary PSK and its modulation rate is then 1600 baud. Higher-rate data modems utilize a hybrid modulation scheme, combining phase modu- lation and amplitude modulation. One typical modem in this category is based on CCITT Rec. V.29 (Ref. 12) and operates at 9600 bps and its baud rate is 2400 baud. Its the- oretical bit packing capability is 4 bits per Hz of bandwidth. Each transition carries 4 Figure 10.16 A spatial representation of 8-ary PSK. 10.9 DIGITAL TRANSMISSION ON AN ANALOG CHANNEL Table 10.3 Phase Encoding for the CCITT Rec. V.29 Modem Phase Change (°) bits of information. The first three bits derive from 8-ary PSK and the fourth bit derives from using two equivalent amplitude levels. Table 10.3 gives the phase encoding for the V.29 modem and Table 10.4 shows its amplitude-phase relationships. The four bits are denomi