Calculate the Shannon capacity for a channel with a bandwidth of 1 MHz and SNR of 63.
Understand the Problem
The question asks for the calculation of the Shannon capacity using the provided bandwidth and signal-to-noise ratio (SNR). To solve this, we will apply the Shannon Capacity formula: C = B * log2(1 + SNR), where C is the channel capacity in bits per second, B is the bandwidth in Hertz, and SNR is the signal-to-noise ratio.
Answer
$6 \text{ Mbps}$
Answer for screen readers
The Shannon capacity for the channel is $6 \text{ Mbps}$.
Steps to Solve
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Identify the Given Values
The problem provides the bandwidth $B$ and the signal-to-noise ratio (SNR). Here, the bandwidth is: $$ B = 1 \text{ MHz} = 10^6 \text{ Hz} $$ And the SNR is: $$ \text{SNR} = 63 $$
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Use the Shannon Capacity Formula
The Shannon capacity formula is: $$ C = B \cdot \log_2(1 + \text{SNR}) $$ Plugging in the values for $B$ and SNR gives: $$ C = 10^6 \cdot \log_2(1 + 63) $$
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Calculate the Logarithm
First, calculate $1 + \text{SNR}$: $$ 1 + 63 = 64 $$
Now, find the logarithm: $$ \log_2(64) = 6 $$ (since $64 = 2^6$)
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Calculate the Capacity
Substitute back into the capacity formula: $$ C = 10^6 \cdot 6 = 6 \times 10^6 \text{ bps} = 6 \text{ Mbps} $$
The Shannon capacity for the channel is $6 \text{ Mbps}$.
More Information
The Shannon capacity formula quantifies the maximum data rate of a communication channel based on its bandwidth and the quality of the signal (SNR). In this case, a bandwidth of 1 MHz and an SNR of 63 yields a capacity of 6 Mbps, which is significant for many communication applications.
Tips
- Failing to convert units properly (e.g., mixing MHz and Hz).
- Incorrectly calculating the logarithm, especially not recognizing powers of 2.
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