 other as more symbol values or different signals are used. It is not usually rea- sonable to use more than eight phases; instead, we use different amplitudes as in 16-QAM in Figure 4.9. The 16-QAM tolerates more noise than 16-PSK because with the same average signal power distances between signals can be increased. However, if we would analyze noise tolerance in more detail, we could form a general rule stating that the increase in the number of signals in use reduces noise tolerance. In low-noise channels, such as telephone voice channels, many different signals can be used but in high-interference chan- nels, such as radio channels for cellular systems, binary symbols are often the best choice. However, modulation moves the spectrum of the pulse from low fre- quencies to carrier frequencies, and the bandwidth is typically doubled when compared with baseband systems as was shown in Figure 4.2. This is why the symbol rate in radio systems is less than or equal to the transmission band- width, that is: r S BT (4.12) where r is the symbol rate in bauds and B is the transmission bandwidth in hertz. The accurate requirement of bandwidth depends on the modulation scheme in use, the study of which is beyond the scope of this book. Example 4.1 Assume that the transmission channel is an ideal lowpass channel with a bandwidth of 4 kHz. The maximum symbol rate via this channel is r 2. B = 8 kbauds; that is, we can transmit up to 8,000 independent signals, sym- bols, in a second. [To transmit the same symbol rate through a bandpass channel, we would need a bandwidth of 8 kHz according to (4.12); see also Figure 4.2.] Example 4.2 Assume that the S/N of a lowpass channel is 28 dB and its bandwidth is 4 kHz. Then S/N = 10 log S/N, S/N = 102.8 = 631. The maximum bit rate according to (4.11) is C = B log2(1 + S/N) = 4,000 log,(432) = 4,000 (log, 10 632)/log10 2 = 37.2 Kbps. In Example 4.1 we learned that the maximum symbol rate is 4 kbauds, which depends only on the bandwidth. To achieve the maximum bit rate, we transmit 4,000 