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.
- How many symbol states are required in order to
achieve this level of performance?
- 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?
- What is the theoretical capacity for the ASK
system if the S/N ratio on the telephone link is 33 dB?
Solution
-
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.
-
For the baseband equivalent,
28800 = 2 x 3100 log2M
Thus
M = 25.02 or 32 states to the nearest power of 2.
-
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.