What is the minimum $\frac{{{E_b}}}{{{N_0}}}$ required to achie

$What is the minimum $\frac{{{E_b}}}{{{N_0}}}$ required to achie$

| What is the minimum $\frac{{{E_b}}}{{{N_0}}}$ required to achieve a spectral efficiency of 6 bps/Hz?

A. 5.2

B. 5.3

C. 10.5

D. 15.8

Please scroll down to see the correct answer and solution guide.

Concept:

Channel capacity of communication channel is:

C = W log₂ (1 + SNR) bits/sec

Where,

W = bandwidth of channel (Hz)

SNR = Signal to Noise ratio

$SNR = \frac{{Signal\;power}}{{Noise\;Power}}$

$ = \frac{{{P_s}}}{{{N_0}W}} = \frac{{Energy\;per\;second}}{{Noise\;power}}$

Spectral efficiency (P) is defined as:

$P = \frac{{Data\;rate\;supported\;by\;channel}}{{Bandwidth\;of\;channel}}\left( {\frac{{bit\;per\;sec}}{{Hz}}} \right)$

$\frac{C}{W}\;bps/Hz$

Calculation:

Given spectral efficiency is:

$6\frac{{bps}}{{Hz}} = \frac{C}{W}$

∵ C = W log₂ (1 + SNR)

$\therefore SNR = {2^{\frac{C}{W}}} - 1$

$\frac{{{E_b}}}{{{T_b}{N_0}W}} = {2^{\frac{C}{W}}} - 1$

$\frac{{{E_b}{R_b}}}{{{N_0}W}} = {2^{\frac{C}{W}}} - 1$

Where,

E_b = bit energy

R_b = bit rate or data rate (in bits per sec)

Since units of C and R_b both are same i.e. bits per sec.

$\therefore \frac{{{E_b}C}}{{{N_0}W}} = {2^{\frac{C}{W}}} - 1$

$\therefore \frac{{{E_b}}}{{{N_0}}} = \frac{W}{C}\;\left[ {{2^{\frac{C}{W}}} - 1} \right]$

$= \frac{1}{6}\left[ {{2^6} - 1} \right]$

$= \frac{1}{6} \times 63$

$\frac{{{E_b}}}{{{N_0}}} = 10.5$

So, the minimum value of $\frac{{{E_b}}}{{{N_0}}}$ will be 10.5 to achieve a spectral efficiency of 6 bps/Hz.

What is the minimum \(\frac{{{E_b}}}{{{N_0}}}\) required to achie