Q: Compare the maximum data rate of a noiseless 4-kHz
channel using
(a) Analog encoding (e.g.,QPSK)with 2 bits per sample.
(b) The T1 PCM system.
Solution:
a) Maximum sampling rate = 2 x 30 = 60 Kbps.
4 level PCM will need 2 bits/sample.
Maximum datarate = 60 x 2 =120
Kbps
b)Maximum sampling rate = 2 x 30 = 60 Kbps.
6 level PCM will need 3 bits/sample.
Maximum datarate = 60 x 3 =180
Kbps
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