## Answer 2.12

#### The Shannon–Hartley theorem can be written as: C/B = log2[1 + Eb.C/N0.B]
Now for Eb/N0 = –0.6 dB, the ratio Eb/N0 = 100.6/10 = 0.871.
The bandwidth efficiency C/B that can be supported is thus:
C/B = log2[1 + 0.871C/B]
therefore:
C/B = 0.6 bits/second/Hz (approximately).
In a bandwidth of 3400 Hz, the system can thus operate at an information transfer rate of 3400 x 0.6 = 2040 bps.