Question

The capacity of band-limited additive white Gaussian Noise (AWGN) channel is given by \(C = W{\log _2}\left[ {1 + \frac{P}{{{\sigma ^2}w}}} \right]\) bits per second (bps), where W is the channel Bandwidth, P is the average power received and σ² is the one-sided power spectral density of the AWGN.

For a fixed \(\frac{P}{{{\sigma ^2}}} = 1000\), the channel capacity (in kbps) with infinite Bandwidth (W → ∞) is approximately

1. 1.44
2. 1.08
3. 0.72
4. 0.36

Answer 1

Correct Answer - Option 1 : 1.44

Concept: we can remember the relation  \(y = \mathop {\lim }\limits_{x \to \infty } x\;{\log _2}\left[ {1 + \frac{1}{x}} \right]\) = log2 e = 1.442

Application: It is given that the capacity of Band-limited AWGN is given by:

\(C = W{\log _2}\left[ {1 + \frac{P}{{{\sigma ^2}W}}} \right]\)

It is given W →∞

\(C = \mathop {\lim }\limits_{W \to \infty } W{\log _2}\left[ {1 + \frac{P}{{{\sigma ^2}W}}} \right]\)

\(C = \mathop {\lim }\limits_{W \to \infty } \frac{{W{\sigma ^2}}}{P}\frac{P}{{{\sigma ^2}}}{\log _2}\left[ {1 + \frac{P}{{{\sigma ^2}W}}} \right]\)

\(C = \mathop {\lim }\limits_{W \to \infty } \frac{P}{{{\sigma ^2}}}\left( {\frac{{W{\sigma ^2}}}{P}} \right){\log _2}\left[ {1 + \frac{P}{{{\sigma ^2}W}}} \right]\)

Let \(\frac{{W{\sigma ^2}}}{P} = x\)

Since W →∞ then x →∞

\(C = \mathop {\lim }\limits_{x \to \infty } \frac{P}{{{\sigma ^2}}}\;x\;lo{g_2}\left[ {1 + \frac{1}{x}} \right]\) 

\(C = \frac{P}{{{\sigma ^2}}}\mathop {\lim }\limits_{x \to \infty } {\log _2}{\left[ {1 + \frac{1}{x}} \right]^x}\)

Considering:

\(y = \mathop {\lim }\limits_{x \to \infty } {\log _2}{\left[ {1 + \frac{1}{x}} \right]^x}\)

\(y = \mathop {\lim }\limits_{x \to \infty } x\;{\log _2}\left[ {1 + \frac{1}{x}} \right]\)

\(y = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\log }_2}\left[ {1 + \frac{1}{x}} \right]}}{{1/x}}\)

Putting x →∞, we are getting \(\frac{0}{0}\) form, hence using the L-hospitals rule, we can write:

\(y = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{1 + \frac{1}{x}}}\frac{{{{\log }_2}e}}{{\left( {\frac{{ - 1}}{{{x^2}}}} \right)}}\;\left( {\frac{{ - 1}}{{{x^2}}}} \right)\)

y = log₂ e

\(C = \frac{P}{{{\sigma ^2}}}{\log _2}e\) 

\(C = 1.44\frac{P}{{{\sigma ^2}}}\)

C = 1.44 × 1000 bps

C = 1.44 kbps