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The output signal-to-quantization-noise ratio of a 10-bit PCM was found to be 30 dB. The desired SNR is 42 dB. It can be increased by increasing the number of quantization level.In this way the fractional increase in the transmission bandwidth would be (assume log210 = 0.3)
- a)20%
- b)30%
- c)40%
- d)50%
Correct answer is option 'A'. Can you explain this answer?
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The output signal-to-quantization-noise ratio of a 10-bit PCM was foun...
= log C + 20 nlog2 = α + 6 ndB. This equation shows that increasing n by one bits increase the by 6 dB.
Hence an increase in the SNR by 12 dB can be accomplished by increasing 9is form 10 to 12, the transmission bandwidth would be increased by 20%
Hence an increase in the SNR by 12 dB can be accomplished by increasing 9is form 10 to 12, the transmission bandwidth would be increased by 20%
Most Upvoted Answer
The output signal-to-quantization-noise ratio of a 10-bit PCM was foun...
Solution:
To increase the signal-to-quantization-noise ratio (SQNR), we need to increase the number of quantization levels. Let's analyze the given information step by step.
Given:
- Output SQNR = 30 dB
- Desired SNR = 42 dB
- log2(10) = 0.3
Step 1: Calculating the SQNR for a 10-bit PCM
The SQNR for a PCM system is given by the formula:
SQNR = 6.02N + 1.76 dB
Where N is the number of bits used for quantization. For a 10-bit PCM, the SQNR can be calculated as follows:
SQNR_10bit = 6.02 * 10 + 1.76 dB
= 60.2 + 1.76 dB
= 62.96 dB
Step 2: Calculating the fractional increase in the number of quantization levels
To increase the SQNR from 30 dB to 42 dB, we need to find the fractional increase in the number of quantization levels.
Let's assume the initial number of quantization levels in the 10-bit PCM system is L1, and the increased number of quantization levels is L2.
Using the formula for SQNR, we can write:
SQNR_10bit = 6.02 * log2(L1) + 1.76 dB
SQNR_desired = 6.02 * log2(L2) + 1.76 dB
Since the desired SNR is 42 dB, we can convert it to SQNR as follows:
SQNR_desired = SNR_desired - 1.76 dB
= 42 dB - 1.76 dB
= 40.24 dB
Now, equating the two equations for SQNR, we get:
6.02 * log2(L1) + 1.76 dB = 6.02 * log2(L2) + 1.76 dB
6.02 * log2(L1) = 6.02 * log2(L2)
log2(L1) = log2(L2)
L1 = L2
This implies that the number of quantization levels remains the same (L1 = L2).
Step 3: Calculating the fractional increase in the transmission bandwidth
To calculate the fractional increase in the transmission bandwidth, we need to find the ratio of the increase in the number of quantization levels to the initial number of quantization levels.
Let's assume the fractional increase in the number of quantization levels is x.
Then, the increased number of quantization levels can be written as:
L2 = (1 + x) * L1
Since L1 = L2, we have:
L2 = (1 + x) * L2
1 = 1 + x
x = 0
Therefore, the fractional increase in the transmission bandwidth is 0.
Conclusion:
The fractional increase in the transmission bandwidth is 0, which means there is no increase in the transmission bandwidth when the number of quantization levels is increased to increase the SQNR. Hence, the correct answer is option 'A' (20%).
To increase the signal-to-quantization-noise ratio (SQNR), we need to increase the number of quantization levels. Let's analyze the given information step by step.
Given:
- Output SQNR = 30 dB
- Desired SNR = 42 dB
- log2(10) = 0.3
Step 1: Calculating the SQNR for a 10-bit PCM
The SQNR for a PCM system is given by the formula:
SQNR = 6.02N + 1.76 dB
Where N is the number of bits used for quantization. For a 10-bit PCM, the SQNR can be calculated as follows:
SQNR_10bit = 6.02 * 10 + 1.76 dB
= 60.2 + 1.76 dB
= 62.96 dB
Step 2: Calculating the fractional increase in the number of quantization levels
To increase the SQNR from 30 dB to 42 dB, we need to find the fractional increase in the number of quantization levels.
Let's assume the initial number of quantization levels in the 10-bit PCM system is L1, and the increased number of quantization levels is L2.
Using the formula for SQNR, we can write:
SQNR_10bit = 6.02 * log2(L1) + 1.76 dB
SQNR_desired = 6.02 * log2(L2) + 1.76 dB
Since the desired SNR is 42 dB, we can convert it to SQNR as follows:
SQNR_desired = SNR_desired - 1.76 dB
= 42 dB - 1.76 dB
= 40.24 dB
Now, equating the two equations for SQNR, we get:
6.02 * log2(L1) + 1.76 dB = 6.02 * log2(L2) + 1.76 dB
6.02 * log2(L1) = 6.02 * log2(L2)
log2(L1) = log2(L2)
L1 = L2
This implies that the number of quantization levels remains the same (L1 = L2).
Step 3: Calculating the fractional increase in the transmission bandwidth
To calculate the fractional increase in the transmission bandwidth, we need to find the ratio of the increase in the number of quantization levels to the initial number of quantization levels.
Let's assume the fractional increase in the number of quantization levels is x.
Then, the increased number of quantization levels can be written as:
L2 = (1 + x) * L1
Since L1 = L2, we have:
L2 = (1 + x) * L2
1 = 1 + x
x = 0
Therefore, the fractional increase in the transmission bandwidth is 0.
Conclusion:
The fractional increase in the transmission bandwidth is 0, which means there is no increase in the transmission bandwidth when the number of quantization levels is increased to increase the SQNR. Hence, the correct answer is option 'A' (20%).
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The output signal-to-quantization-noise ratio of a 10-bit PCM was found to be 30 dB. The desired SNR is 42 dB. It can be increased by increasing the number of quantization level.In this way the fractional increase in the transmission bandwidth would be (assume log210 = 0.3)a)20%b)30%c)40%d)50%Correct answer is option 'A'. Can you explain this answer?
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