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A voice-grade channel of the telephone network has a bandwidth of 3.4 KHz calculate the information capacity of the telephone channel for a signal to noise ratio of 30 dB.
- a)68 kbps
- b)34 kbps
- c)17kbps
- d)10.2kbps
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A voice-grade channel of the telephone network has a bandwidth of 3.4 ...
Capacity of a gaussian channel is
given as,
given as,
C = BW log2 (1 + SNR)
Here SNR = 30 dB (1000)
So,
So,
Most Upvoted Answer
A voice-grade channel of the telephone network has a bandwidth of 3.4 ...
Given:
- Bandwidth of the telephone channel: 3.4 KHz
- Signal to noise ratio: 30 dB
To Find:
- Information capacity of the telephone channel
Solution:
The information capacity of a communication channel can be calculated using the Shannon capacity formula:
C = B * log2(1 + SNR)
where:
C = Information capacity of the channel (in bits per second)
B = Bandwidth of the channel (in Hz)
SNR = Signal to noise ratio
Step 1: Convert the bandwidth from KHz to Hz:
Bandwidth (B) = 3.4 KHz = 3.4 * 1000 Hz = 3400 Hz
Step 2: Convert the signal to noise ratio from dB to a linear scale:
SNR (in linear scale) = 10^(SNR (in dB)/10)
SNR (in linear scale) = 10^(30/10) = 10^3 = 1000
Step 3: Calculate the information capacity using the Shannon capacity formula:
C = B * log2(1 + SNR)
C = 3400 * log2(1 + 1000)
C = 3400 * log2(1001)
Using a calculator, we can find that log2(1001) is approximately 9.9644.
C = 3400 * 9.9644
C ≈ 33861.76 bits per second
Step 4: Convert the information capacity from bits per second to kilobits per second:
Information capacity ≈ 33861.76 bits per second = 33861.76 / 1000 kilobits per second ≈ 33.86176 kbps ≈ 34 kbps (rounded to the nearest whole number)
Therefore, the information capacity of the telephone channel for a signal to noise ratio of 30 dB is approximately 34 kbps, which is option B.
- Bandwidth of the telephone channel: 3.4 KHz
- Signal to noise ratio: 30 dB
To Find:
- Information capacity of the telephone channel
Solution:
The information capacity of a communication channel can be calculated using the Shannon capacity formula:
C = B * log2(1 + SNR)
where:
C = Information capacity of the channel (in bits per second)
B = Bandwidth of the channel (in Hz)
SNR = Signal to noise ratio
Step 1: Convert the bandwidth from KHz to Hz:
Bandwidth (B) = 3.4 KHz = 3.4 * 1000 Hz = 3400 Hz
Step 2: Convert the signal to noise ratio from dB to a linear scale:
SNR (in linear scale) = 10^(SNR (in dB)/10)
SNR (in linear scale) = 10^(30/10) = 10^3 = 1000
Step 3: Calculate the information capacity using the Shannon capacity formula:
C = B * log2(1 + SNR)
C = 3400 * log2(1 + 1000)
C = 3400 * log2(1001)
Using a calculator, we can find that log2(1001) is approximately 9.9644.
C = 3400 * 9.9644
C ≈ 33861.76 bits per second
Step 4: Convert the information capacity from bits per second to kilobits per second:
Information capacity ≈ 33861.76 bits per second = 33861.76 / 1000 kilobits per second ≈ 33.86176 kbps ≈ 34 kbps (rounded to the nearest whole number)
Therefore, the information capacity of the telephone channel for a signal to noise ratio of 30 dB is approximately 34 kbps, which is option B.
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A voice-grade channel of the telephone network has a bandwidth of 3.4 KHz calculate the information capacity of the telephone channel for a signal to noise ratio of 30 dB.a)68 kbpsb)34 kbpsc)17kbpsd)10.2kbpsCorrect answer is option 'B'. Can you explain this answer?
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