## Example 5.1

#### An Amplitude Shift Keying format is used for transmitting data at a rate of 28.8 kbps over a telephone channel with bandwidth extending from 300 Hz to 3400 Hz.

1. How many symbol states are required in order to achieve this level of performance?
2. What would be the equivalent number of symbol states needed if the channel passband extended from 0 Hz to 3100 Hz and baseband M-ary signalling was used?
3. What is the theoretical capacity for the ASK system if the S/N ratio on the telephone link is 33 dB?

### Solution

####

1. The capacity of a bandpass ASK channel is given by:
   CASK = B log2M
   compared with
   Cbaseband = 2B log2M
   Hence,
   28800 = (3400 – 300) log2M
   and
   M = 626.1 or 1024 states to nearest power of 2.
2. For the baseband equivalent,
   28800 = 2 x 3100 log2M
   Thus
   M = 25.02 or 32 states to the nearest power of 2.
3. Applying the Shannon-Hartley equation,
   C = B log2(S/N + 1)
   we obtain:
   C = (3400 – 300) log2(103.3 + 1) = 33.996 kbps.

Note: The Shannon-Hartley expression is valid for both baseband and bandpass channels – it is the number of symbol states that must be increased on a bandpass channel, but as we will see for QPSK, this does not necessarily imply that the Eb/N0 performance will be degraded compared to a baseband link.