Answer 6.7

A 64-QAM system can convey log2(64) = 6 bits per symbol. Thus for a bit rate of 256 kbps, the symbol rate is 256000 / 6 = 42667 symbols per second. For bandpass signalling, this requires a minimum bandwidth equal to the symbol rate of 42.667 kHz assuming brickwall (a = 0) filtering. For a = 0.5, the bandwidth required is increased by a factor (1 + a) or 1.5, to 64 kHz.