AWGN Channel Parameters

The problem specifies a voice-grade AWGN channel with the following characteristics:

• Bandwidth ($B$): $4.0 \text{ kHz} = 4.0 \times 10^3 \text{ Hz}$.
• Two-sided noise power spectral density ($\frac{\eta}{2}$): $2.5 \times 10^{-5} \text{ Watt/Hz}$. This implies the one-sided power spectral density is $\eta = 2 \times (2.5 \times 10^{-5}) = 5.0 \times 10^{-5} \text{ Watt/Hz}$.
• Information rate ($R$): $52 \text{ kbps} = 52 \times 10^3 \text{ bps}$.
• Goal: Determine the minimum bit-energy ($E_b$) for transmitting data with an arbitrarily small bit error rate (BER).

Shannon Capacity and Minimum Eb

Achieving an arbitrarily small BER in an AWGN channel requires the transmission rate $R$ to be less than the channel capacity $C$. The minimum $E_b$ needed for this is found by considering the scenario where $R$ approaches $C$.

The channel capacity $C$ is given by the Shannon-Hartley theorem:

$ C = B \log_2 \left( 1 + \frac{S}{N} \right) $

Here, $S$ is the average signal power and $N$ is the average noise power. These can be expressed in terms of $E_b$, $R$, and the noise spectral density $\eta$:

• Signal power: $S = R \times E_b$.
• Total noise power in bandwidth $B$: $N = \eta B$.

For the condition $R \le C$ to hold with minimum $E_b$, we set $R = C$:

$ R = B \log_2 \left( 1 + \frac{R E_b}{\eta B} \right) $

Rearranging the equation to solve for $E_b$:

$ \frac{R}{B} = \log_2 \left( 1 + \frac{R E_b}{\eta B} \right) $

$ 2^{R/B} = 1 + \frac{R E_b}{\eta B} $

$ 2^{R/B} - 1 = \frac{R E_b}{\eta B} $

$ E_b = \frac{\eta B}{R} \left( 2^{R/B} - 1 \right) $

Calculation of Minimum Bit-Energy

Substitute the given parameter values into the derived formula:

• $B = 4.0 \times 10^3 \text{ Hz}$
• $R = 52 \times 10^3 \text{ bps}$
• $\eta = 5.0 \times 10^{-5} \text{ Watt/Hz}$

Calculate the ratio $\frac{R}{B}$:

$ \frac{R}{B} = \frac{52 \times 10^3 \text{ bps}}{4.0 \times 10^3 \text{ Hz}} = 13 $

Calculate the term $\frac{\eta B}{R}$:

$ \frac{\eta B}{R} = \frac{(5.0 \times 10^{-5} \text{ Watt/Hz}) \times (4.0 \times 10^3 \text{ Hz})}{52 \times 10^3 \text{ bps}} = \frac{0.2 \text{ Watt}}{52 \times 10^3 \text{ bps}} = \frac{1}{260000} \text{ J/bit} $

Calculate $2^{R/B} - 1$:

$ 2^{R/B} - 1 = 2^{13} - 1 = 8192 - 1 = 8191 $

Compute $E_b$:

$ E_b = \left( \frac{1}{260000} \text{ J/bit} \right) \times 8191 $

$ E_b \approx 0.031503846 \text{ J/bit} $

Convert $E_b$ to millijoules per bit (mJ/bit):

$ E_b \approx 0.031503846 \times 1000 \text{ mJ/bit} $

$ E_b \approx 31.50 \text{ mJ/bit} $

Final Result

The minimum bit-energy $E_b$ required is approximately 31.50 mJ/bit.