The question asks us to find the channel capacity of a band-limited Additive White Gaussian Noise (AWGN) channel when the bandwidth (W) approaches infinity. We are given the formula for channel capacity and a specific ratio of power to noise spectral density.

Channel Capacity Formula and Given Values

The capacity of a band-limited AWGN channel is given by the Shannon-Hartley theorem:

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

Where:

• \(C\) is the channel capacity.
• \(W\) is the channel bandwidth in Hertz (Hz).
• \(P\) is the average received power in Watts (W).
• \({\sigma ^2}\) is the one-sided power spectral density of the AWGN in Watts per Hertz (W/Hz).

We are provided with the following information:

• The ratio \(\frac{P}{{{\sigma ^2}}} = 1000\).
• We need to find the channel capacity as \(W \to \infty\).

Capacity Calculation with Infinite Bandwidth

First, substitute the given ratio \(\frac{P}{{{\sigma ^2}}} = 1000\) into the capacity formula:

\[C = W{\log _2}\left[ {1 + \frac{1000}{W}} \right]\]

Now, we need to evaluate the limit of this expression as \(W \to \infty\):

\[C_{\text{infinite bandwidth}} = \lim_{W \to \infty} W{\log _2}\left[ {1 + \frac{1000}{W}} \right]\]

To solve this limit, we can use a standard limit property related to the mathematical constant 'e'. Recall that \(\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1\).

Let's rewrite the \(\log_2\) term in terms of natural logarithm (\(\ln\)):

We know that \({\log _b}a = \frac{{{\log _c}a}}{{{\log _c}b}}\), so \({\log _2}x = \frac{{\ln x}}{{\ln 2}}\).

Applying this to our capacity formula:

\[C = W \frac{\ln\left(1 + \frac{1000}{W}\right)}{\ln 2}\]

Rearrange the terms to fit the limit form:

\[C = \frac{1}{\ln 2} \times \frac{\ln\left(1 + \frac{1000}{W}\right)}{\frac{1}{W}}\]

Multiply the numerator and denominator inside the fraction by 1000 to match the form \(\frac{\ln(1+x)}{x}\), where \(x = \frac{1000}{W}\):

\[C = \frac{1000}{\ln 2} \times \frac{\ln\left(1 + \frac{1000}{W}\right)}{\frac{1000}{W}}\]

As \(W \to \infty\), let \(x = \frac{1000}{W}\). Then \(x \to 0\). The limit becomes:

\[C_{\text{infinite bandwidth}} = \lim_{x \to 0} \frac{1000}{\ln 2} \times \frac{\ln(1+x)}{x}\]

Since \(\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1\), the expression simplifies to:

\[C_{\text{infinite bandwidth}} = \frac{1000}{\ln 2} \times 1\]

Now, we need to calculate the numerical value. We know that \(\ln 2 \approx 0.6931\).

\[C_{\text{infinite bandwidth}} \approx \frac{1000}{0.6931}\]

\[C_{\text{infinite bandwidth}} \approx 1442.66\text{ bps}\]

Converting Capacity to kbps

The question asks for the channel capacity in kilobits per second (kbps). To convert bps to kbps, we divide by 1000:

\[C_{\text{kbps}} = \frac{1442.66}{1000}\text{ kbps}\]

\[C_{\text{kbps}} \approx 1.44266\text{ kbps}\]

Rounding to two decimal places, the channel capacity is approximately 1.44 kbps.

Conclusion

Based on our calculation, the channel capacity with infinite bandwidth for the given conditions is approximately 1.44 kbps.