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A communication channel with AWGN operating at a signal to noise ratio of SNR >> 1 and bandwidth B has capacity C1. If the SNR is doubled keeping it constant, then the resulting capacity C2 is given by
- a)C2 ≈ 2C1
- b)C2 ≈ C1 + B
- c)C2 ≈ C1 + 2B
- d)C2 ≈ C1 + 0.3 B
Correct answer is option 'B'. Can you explain this answer?
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A communication channel with AWGN operating at a signal to noise rati...
Introduction:
The capacity of a communication channel is a measure of the maximum amount of information that can be reliably transmitted over the channel. It is determined by various factors, including the signal-to-noise ratio (SNR) and the bandwidth of the channel. In this question, we are given a communication channel with additive white Gaussian noise (AWGN) operating at an SNR >> 1 and bandwidth B, and we need to determine the resulting capacity when the SNR is doubled.
Answer:
The capacity of a communication channel is given by the Shannon capacity formula, which is given as:
C = B * log2(1 + SNR)
Where C is the capacity, B is the bandwidth, and SNR is the signal-to-noise ratio.
Given:
SNR >> 1 (i.e. SNR is very large)
Capacity C1 = B * log2(1 + SNR)
Doubling the SNR:
When the SNR is doubled while keeping it constant, the new SNR becomes 2 * SNR.
So, the new capacity C2 can be calculated as:
C2 = B * log2(1 + 2 * SNR)
Using approximations:
Since SNR >> 1, we can make the following approximations:
- log2(1 + SNR) ≈ log2(SNR) (for large SNR)
- log2(1 + 2 * SNR) ≈ log2(2 * SNR) = log2(2) + log2(SNR) = 1 + log2(SNR)
Substituting the approximations:
Using the above approximations, the new capacity C2 can be written as:
C2 ≈ B * (1 + log2(SNR))
Comparing C1 and C2:
Comparing C1 and C2, we can observe that the only difference between the two is the additional term of log2(SNR) in C2. Since SNR >> 1, log2(SNR) is a positive value but relatively small compared to 1. Therefore, the additional term does not significantly affect the overall capacity.
Conclusion:
Hence, the resulting capacity C2 when the SNR is doubled while keeping it constant is approximately equal to the initial capacity C1, i.e., C2 ≈ C1. Therefore, the correct answer is option 'B'.
The capacity of a communication channel is a measure of the maximum amount of information that can be reliably transmitted over the channel. It is determined by various factors, including the signal-to-noise ratio (SNR) and the bandwidth of the channel. In this question, we are given a communication channel with additive white Gaussian noise (AWGN) operating at an SNR >> 1 and bandwidth B, and we need to determine the resulting capacity when the SNR is doubled.
Answer:
The capacity of a communication channel is given by the Shannon capacity formula, which is given as:
C = B * log2(1 + SNR)
Where C is the capacity, B is the bandwidth, and SNR is the signal-to-noise ratio.
Given:
SNR >> 1 (i.e. SNR is very large)
Capacity C1 = B * log2(1 + SNR)
Doubling the SNR:
When the SNR is doubled while keeping it constant, the new SNR becomes 2 * SNR.
So, the new capacity C2 can be calculated as:
C2 = B * log2(1 + 2 * SNR)
Using approximations:
Since SNR >> 1, we can make the following approximations:
- log2(1 + SNR) ≈ log2(SNR) (for large SNR)
- log2(1 + 2 * SNR) ≈ log2(2 * SNR) = log2(2) + log2(SNR) = 1 + log2(SNR)
Substituting the approximations:
Using the above approximations, the new capacity C2 can be written as:
C2 ≈ B * (1 + log2(SNR))
Comparing C1 and C2:
Comparing C1 and C2, we can observe that the only difference between the two is the additional term of log2(SNR) in C2. Since SNR >> 1, log2(SNR) is a positive value but relatively small compared to 1. Therefore, the additional term does not significantly affect the overall capacity.
Conclusion:
Hence, the resulting capacity C2 when the SNR is doubled while keeping it constant is approximately equal to the initial capacity C1, i.e., C2 ≈ C1. Therefore, the correct answer is option 'B'.
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Community Answer
A communication channel with AWGN operating at a signal to noise rati...
We have C1 = Blog2(1 + S/N)
≈Blog2(S/N)
If we double the S/N ratio then
C2 ≈ Blog2(2S/N)
≈ Blog22 + Blog2(S/N)
≈ B + C1
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A communication channel with AWGN operating at a signal to noise ratio of SNR >> 1 and bandwidth B has capacity C1. If the SNR is doubled keeping it constant, then the resulting capacity C2 is given bya)C2 ≈ 2C1b)C2 ≈ C1 + Bc)C2 ≈ C1 + 2Bd)C2 ≈ C1 + 0.3 BCorrect answer is option 'B'. Can you explain this answer?
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