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It was found that a sampling rate 20% above the rate would be adequate. So, the maximum SNR (in dB) that can be realised without increasing the transmission bandwidth would be _________. (Answer up to one decimal place)
Correct answer is '50.1'. Can you explain this answer?
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It was found that a sampling rate 20% above the rate would be adequat...
Nyquist Rate = 2 MHz
50% higher rate = 3 MHz, L = 256 = 28
Thus, transmission bandwidth is 3 MHz × 8 = 24 Mbits/s.
New sampling rate is at 20% above the Nyquist rate.
Sampling rate = 1.2 × 2 = 2.4 MHz.
Level = 210 = 1024
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It was found that a sampling rate 20% above the rate would be adequat...
Sampling Rate and SNR
The sampling rate refers to the number of samples taken per second in an analog-to-digital conversion process. It is an important factor in determining the quality of the digital signal representation. On the other hand, the Signal-to-Noise Ratio (SNR) is a measure of the quality of a signal, representing the ratio of the power of the signal to the power of the noise.
Sampling Rate and Adequacy
It has been found that a sampling rate 20% above the required rate is adequate for most applications. This means that increasing the sampling rate beyond this margin does not significantly improve the quality of the digital signal representation. Therefore, the given question assumes that the required sampling rate has already been achieved.
Maximum SNR without Increasing Bandwidth
To determine the maximum SNR that can be realized without increasing the transmission bandwidth, we need to consider the Nyquist-Shannon sampling theorem. According to this theorem, the minimum sampling rate required to accurately represent a signal is twice the highest frequency component of the signal.
If we assume that the highest frequency component of the signal is f, then the minimum required sampling rate is 2f. However, since the sampling rate is already 20% above the required rate, the actual sampling rate is 1.2 times the minimum required rate, which is 2.4f.
The maximum SNR without increasing the transmission bandwidth can be calculated using the formula:
SNR = 6.02N + 1.76 dB
Where N is the number of bits per sample. In this case, we assume that the signal is represented using 16-bit samples (N = 16).
Using the formula, we can calculate the maximum SNR:
SNR = 6.02 * 16 + 1.76 dB
= 96.32 + 1.76 dB
= 98.08 dB
Therefore, the maximum SNR that can be realized without increasing the transmission bandwidth is approximately 98.1 dB.
However, the correct answer provided is '50.1'. It is unclear how this answer was obtained based on the given information. The calculation and explanation provided above indicate a different result.
The sampling rate refers to the number of samples taken per second in an analog-to-digital conversion process. It is an important factor in determining the quality of the digital signal representation. On the other hand, the Signal-to-Noise Ratio (SNR) is a measure of the quality of a signal, representing the ratio of the power of the signal to the power of the noise.
Sampling Rate and Adequacy
It has been found that a sampling rate 20% above the required rate is adequate for most applications. This means that increasing the sampling rate beyond this margin does not significantly improve the quality of the digital signal representation. Therefore, the given question assumes that the required sampling rate has already been achieved.
Maximum SNR without Increasing Bandwidth
To determine the maximum SNR that can be realized without increasing the transmission bandwidth, we need to consider the Nyquist-Shannon sampling theorem. According to this theorem, the minimum sampling rate required to accurately represent a signal is twice the highest frequency component of the signal.
If we assume that the highest frequency component of the signal is f, then the minimum required sampling rate is 2f. However, since the sampling rate is already 20% above the required rate, the actual sampling rate is 1.2 times the minimum required rate, which is 2.4f.
The maximum SNR without increasing the transmission bandwidth can be calculated using the formula:
SNR = 6.02N + 1.76 dB
Where N is the number of bits per sample. In this case, we assume that the signal is represented using 16-bit samples (N = 16).
Using the formula, we can calculate the maximum SNR:
SNR = 6.02 * 16 + 1.76 dB
= 96.32 + 1.76 dB
= 98.08 dB
Therefore, the maximum SNR that can be realized without increasing the transmission bandwidth is approximately 98.1 dB.
However, the correct answer provided is '50.1'. It is unclear how this answer was obtained based on the given information. The calculation and explanation provided above indicate a different result.
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It was found that a sampling rate 20% above the rate would be adequate. So, the maximum SNR (in dB) that can be realised without increasing the transmission bandwidth would be _________. (Answer up to one decimal place)Correct answer is '50.1'. Can you explain this answer?
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