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AWGN channel having B.W of 4 kHz, 2-sided noise PSD is given by 10−12Watts / Hz The Channel capacity( in kbps) required to get the signal power of 0.1 MW at channel o/p is __
- a)54
- b)55
Correct answer is between '54,55'. Can you explain this answer?
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Most Upvoted Answer
AWGN channel having B.W of 4 kHz, 2-sided noise PSD is given by 10−12...
Given B=4 kHz
S = 0.1MW = 10-4W
N = N0B = 8 × 10-9 Watts
Therefore,C = Blog2 
= 4 log2 
= 54.4 kbps
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AWGN channel having B.W of 4 kHz, 2-sided noise PSD is given by 10−12...
Given Information:
- Bandwidth of the AWGN channel: 4 kHz
- Two-sided noise power spectral density (PSD): 10^(-12) Watts/Hz
- Signal power at the channel output: 0.1 MW
Formula:
The channel capacity (C) of an AWGN channel is given by the formula:
C = B * log2(1 + (S/N))
Where:
- C is the channel capacity in bits per second (bps)
- B is the bandwidth of the channel in Hz
- S is the signal power in Watts
- N is the noise power in Watts
Calculating Channel Capacity:
Given:
- Bandwidth (B) = 4 kHz = 4 * 10^3 Hz
- Signal power (S) = 0.1 MW = 0.1 * 10^6 Watts
- Noise power spectral density (N0) = 10^(-12) Watts/Hz
Step 1:
Calculating the noise power (N) using the noise power spectral density (N0) and the bandwidth (B):
N = N0 * B
N = 10^(-12) * 4 * 10^3
N = 4 * 10^(-9) Watts
Step 2:
Calculating the signal-to-noise ratio (S/N):
SNR = S / N
SNR = (0.1 * 10^6) / (4 * 10^(-9))
SNR = 25 * 10^14
Step 3:
Calculating the channel capacity (C) using the formula:
C = B * log2(1 + (S/N))
C = (4 * 10^3) * log2(1 + (25 * 10^14))
C ≈ 54.84 kbps (rounded to two decimal places)
Therefore, the channel capacity required to achieve a signal power of 0.1 MW at the channel output is approximately 54.84 kbps. Since the correct answer is between 54 and 55, we can conclude that the correct answer is 55 kbps.
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AWGN channel having B.W of 4 kHz, 2-sided noise PSD is given by 10−12Watts / Hz The Channel capacity( in kbps) required to get the signal power of 0.1 MW at channel o/p is __a)54b)55Correct answer is between '54,55'. Can you explain this answer?
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