If a channel has a bandwidth of 2 MHz and an SNR of 1,000, what is its theoretical capacity using the Shannon formula?

Steps to Solve

1. Identify given values

The given values are:

• Bandwidth, $B = 2 \text{ MHz} = 2 \times 10^6 \text{ Hz}$
• Signal-to-Noise Ratio, $SNR = 1000$

2. Apply the Shannon Capacity Formula

The Shannon capacity formula is given by:

$$ C = B \cdot \log_2(1 + SNR) $$

Substituting the known values:

$$ C = 2 \times 10^6 \cdot \log_2(1 + 1000) $$

3. Calculate logarithm

First, compute $1 + 1000 = 1001$.

Now, find $\log_2(1001)$ using the change of base formula:

$$ \log_2(1001) = \frac{\log_{10}(1001)}{\log_{10}(2)} $$

Using approximate values:

• $\log_{10}(1001) \approx 3.000434 \approx 3$
• $\log_{10}(2) \approx 0.30103$

Thus,

$$ \log_2(1001) \approx \frac{3}{0.30103} \approx 9.96578 $$

4. Calculate final capacity

Now substitute back into the capacity formula:

$$ C \approx 2 \times 10^6 \cdot 9.96578 $$

Calculating this gives:

$$ C \approx 19.93156 \times 10^6 \text{ bits per second} $$

5. Convert to Mbps

To convert bits per second to megabits per second (Mbps):

$$ C \approx 19.93156 \text{ Mbps} \approx 20 \text{ Mbps} $$