Communications Test 4

Concept: 

Shannon capacity theorem states that:

\({\rm{C}} = {\rm{B\;lo}}{{\rm{g}}_2}\left( {{\rm{\;}}1 + \frac{{\rm{S}}}{{\rm{N}}}{\rm{\;}}} \right)\)

C = Channel capacity

B = Bandwidth of the channel

S = Signal power

N = Noise power

∴ It is a measure of capacity on a channel. And it is impossible to transmit information at a faster rate without error.

Calculation:

B = 3.4 kHz = 3400 Hz

Information capacity of the channel will be:

C = 4800 bits / sec

So, the signal to noise ratio for the telephone network is obtained as:

C = B log₂ (1 + SNR)

4800 = 3400 log₂ (1 + SNR)

SNR = 1.661

In decibel, we get

(SNR)_dB = 10 log₁₀ (1.661)

= 2.2 dB

C = 52 kbps

B = 4 kHz

\(\frac{{{N_0}}}{2} = 2.5 \times {10^{ - 5}}\)

N = N_o B = 4 × 10³ × 2.5 × 10^-5 × 2

\(C = B{\log _2}\left( {1 + \frac{S}{N}} \right)\)

\(\frac{C}{B} = lo{g_2}\left( {1 + \frac{S}{N}} \right)\)

\(1 + \frac{S}{N} = {2^{C/B}} = {2^{13}} = 8192\)

\(\frac{S}{N} = 8191\)

S = 8191 × 4 × 10³ × 2.5 × 10^-5 × 2 = 819.1 × 2

\({E_b} = \frac{S}{{{R_b}}}\)

\({E_b} = \frac{{819.1 \times 2}}{{{R_b}}}\)

Concept:

The entropy of a probability distribution is the average or the amount of information when drawing from a probability distribution.

It is calculated as:

\(H=\underset{i=1}{\overset{n}{\mathop \sum }}\,{{p}_{i}}{{\log }_{2}}\left( \frac{1}{{{p}_{i}}} \right)bits/symbol\)

pi is the probability of the occurrence of a symbol.

Also, the information rate of the source is given by:

R = Hfs

f_s = Sampling frequency

Calculation:

Given the bandwidth of the analog signal f_m = β

Since the signal is sampled at the Nyquist rate. So, we have the sampling frequency as:

f_m = 2f_m = 2 β samples / sec

Now, samples are quantized in 4 different levels whose probabilities are:

p₁ = p₄ = 1 / 8, p₂ = p₃ = 3 / 8

So, the entropy of the source will be:

\(H = \mathop \sum \limits_{k = 1}^4 {p_k}{\log _2}\frac{1}{{{p_k}}}\)

\(= {p_1}{\log _2}\frac{1}{{{p_1}}} + {p_2}{\log _2}\frac{1}{{{p_2}}} + {p_3}{\log _2}\frac{1}{{{p_3}}} + {p_4}{\log _2}\frac{1}{{{p_4}}}\)

\( = \frac{1}{8}{\log _2}8 + \frac{3}{8}{\log _2}\frac{8}{3} + \frac{3}{8}{\log _2}\frac{8}{3} + \frac{1}{8}{\log _2}8\)

= 1.81 bits / sample

Therefore, the information rate of the source will be:

R = Hf_s = (1.81) × (2β) = 3.6β

Each bit is transmitted thrice, majority rule is used for detection i.e. when two or three of the bits is either ‘1’ or ‘0’ then the transmitted bit respectively is ‘1’ or ‘0’.

So, when bit ‘1’ is to be sent, the error occurs when two or more times

P_e = P_3bit + P_2bit

\(\begin{array}{l} = {}_{}^3{C_2} + {}_{}^3{C_2}{P^2}\left( {1 - P} \right)\\ = 3{P^2} - 2{P^3}\end{array}\)