Question

A voice-grade AWGN (additive white Gaussian noise) telephone channel has a bandwidth of 4.0 kHz and two-sided noise power spectral density \(\frac{{\rm{\eta }}}{2} = 2.5 \times {10^{-5}}\) Watt per Hz. If information at the rate of 52 kbps is to be transmitted over this channel with arbitrarily small it error rate, then the minimum bit-energy E_b (in mJ/bit) necessary is __________

Answer 1

Concept: For Shannon’s channel capacity theorem, the information rate is given by:

\({\rm{C}} = {\rm{Blo}}{{\rm{g}}_2}\left( {1 + {\rm{SNR}}} \right)\) Where SNR is in absolute values.

Application: Now With Bandwidth B = 4 kHz and C = 64 kbps, we can write:

\(52 × {10^3} = 4 × {10^3}{\log _2}\left( {1 + {\rm{SNR}}} \right)\)

\({\rm{SNR}} = {2^{\left( {\frac{{52 × {{10}^3}}}{{4 × {{10}^3}}}} \right)}} - 1\)

\({\rm{SNR}} = {2^{13}} - 1 = 8191 \)

So, S/N = 8191

\(\therefore S = 8191 × \frac{{\rm{η }}}{2}\times2\times Bandwidth\)

\(S=819.1\times 2 ~mJ\)

Now E_{b =} Energy per bit

\( {{\rm{E}}_{\rm{b}}} = 819.1 ×2\times \frac{{\rm{1 }}}{R_b}~ mJ/bit\)

With R_b = 52 kbps 

E_b = 31.5 mJ/bit