Answer 6.9

  1. A 256-QAM system can convey n = log2(256) = 8  bits per symbol resulting in a symbol rate of 4 million symbols per second for a 32 Mbps data rate. For a bandpass modulation system, this requires a minimum bandwidth of 4 MHz for brick-wall filtering.

    As the actual bandwidth used is 7 MHz, this implies the use of pulse shaping with a filter having an a given by:

    (1 + a) = 7 / 4 = 1.75

    Therefore, a = 0.75.
  2. The Shannon-Hartley equation gives us the required relationship between channel capacity in bits/second, the bandwidth and the signal to noise ratio as follows:

    Channel capacity CB log2(S / N + 1) bits/second

    Therefore, C = 7 · 106 log2(10000 + 1) bits/second = 93 Mbps