ate r in bauds (how many times the phase can change in a sec- ond), which defines a required bandwidth. For QPSK 2 bits (k = 2) are 4 Amplitude values Data sequence t/sec r=1/T 2*1/T=2*r Sampling t/sec instant Detected A/3 2A/3 value Figure 4.14 Symbol rate and bit rate. Introduction to Telecommunications Network Engineering Table 4.1 Bit Rate of a System Using Multiple Symbol Values Number of Bits, k, Encoded Number of Symbol Bit Rate Compared into Each Symbol Values, M with Symbol Rate Same as symbol rate 2 X symbol rate 3 X symbol rate 4 X symbol rate 5 X symbol rate 8 X symbol rate encoded into each symbol and the bit rate is two times the modulation rate. For 8-PSK, k =3 and rb = 3r. The 16-QAM example in Figure 4.9 used 16 combinations of carrier amplitude and phase amplitude values and the bit rate is four times the modulation rate. As we can see from Table 4.1, by increasing the number of different symbols used in the system the data rate could be increased without a limit if there were no other limitations than bandwidth. This is not possible in prac- tice because of the noise. The influence of noise is discussed next. 4.3.3 Maximum Capacity of a Transmission Channel We saw previously that the bandwidth of a channel sets the limit to the sym- bol rate in bauds but not to the information data rate. In 1948, Claude Shan- non published a study of the theoretical maximum data rate in the case of a channel subject to random (thermal) noise. We measure a noise relative to a signal in terms of the S/N. Noise degrades fidelity in analog communication and produces errors in digital communication. The S/N is usually expressed in decibels as (4.10) Taking both bandwidth and noise into account, Shannon stated that the error-free bit rate through any transmission channel cannot exceed the maximum capacity C of the channel given by: Transmission C = B log2(1+ SIN) (4.11) where C is the maximum information data rate in bits per second; B, the bandwidth in hertz; S, the signal power; N, the noise power, and SIN, the S/N