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signal is sampled at 8 kHz and is quantized using 8 bit uniform quantizer. Assuming SNRq for a sinusoidal signal, the correct statement for PCM signal with a bit rate of R is
- a)R = 32 kbps, SNRq = 25.8 dB
- b)R = 64 kbps, SNRq = 49.8 dB
- c)R = 64 kbps, SNRq = 55.8 dB
- d)R = 32 kbps, SNRq = 49.8 dB
Correct answer is option 'B'. Can you explain this answer?
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signal is sampled at 8 kHz and is quantized using 8 bit uniform quanti...
Bit Rate = 8k x 8 = 64 kbps
Most Upvoted Answer
signal is sampled at 8 kHz and is quantized using 8 bit uniform quanti...
Sampling Rate and Bit Depth:
The given signal is sampled at a rate of 8 kHz, which means that 8000 samples are taken per second. The quantization is done using an 8-bit uniform quantizer, which means that each sample can be represented by 8 bits.
Bit Rate Calculation:
The bit rate is calculated by multiplying the sampling rate with the quantization bit depth. In this case, the sampling rate is 8 kHz and the quantization bit depth is 8 bits.
Bit Rate (R) = Sampling Rate * Quantization Bit Depth
R = 8 kHz * 8 bits = 64 kbps
Therefore, the bit rate of the PCM signal is 64 kbps.
Signal-to-Quantization Noise Ratio (SNRq) Calculation:
The signal-to-quantization noise ratio (SNRq) can be calculated using the formula:
SNRq = 6.02 * Bit Depth + 1.76 dB
In this case, the bit depth is 8 bits.
SNRq = 6.02 * 8 + 1.76 dB
SNRq = 48.16 + 1.76 dB
SNRq = 49.92 dB
Therefore, the SNRq for the PCM signal is 49.92 dB.
Comparison with Given Options:
Now let's compare the calculated values with the options given.
Option A: R = 32 kbps, SNRq = 25.8 dB
The calculated values for R and SNRq are higher than the values given in option A. Therefore, option A is incorrect.
Option B: R = 64 kbps, SNRq = 49.8 dB
The calculated values for R and SNRq match exactly with the values given in option B. Therefore, option B is correct.
Option C: R = 64 kbps, SNRq = 55.8 dB
The calculated values for R and SNRq do not match with the values given in option C. Therefore, option C is incorrect.
Option D: R = 32 kbps, SNRq = 49.8 dB
The calculated values for R and SNRq do not match with the values given in option D. Therefore, option D is incorrect.
Therefore, the correct option is B: R = 64 kbps, SNRq = 49.8 dB.
The given signal is sampled at a rate of 8 kHz, which means that 8000 samples are taken per second. The quantization is done using an 8-bit uniform quantizer, which means that each sample can be represented by 8 bits.
Bit Rate Calculation:
The bit rate is calculated by multiplying the sampling rate with the quantization bit depth. In this case, the sampling rate is 8 kHz and the quantization bit depth is 8 bits.
Bit Rate (R) = Sampling Rate * Quantization Bit Depth
R = 8 kHz * 8 bits = 64 kbps
Therefore, the bit rate of the PCM signal is 64 kbps.
Signal-to-Quantization Noise Ratio (SNRq) Calculation:
The signal-to-quantization noise ratio (SNRq) can be calculated using the formula:
SNRq = 6.02 * Bit Depth + 1.76 dB
In this case, the bit depth is 8 bits.
SNRq = 6.02 * 8 + 1.76 dB
SNRq = 48.16 + 1.76 dB
SNRq = 49.92 dB
Therefore, the SNRq for the PCM signal is 49.92 dB.
Comparison with Given Options:
Now let's compare the calculated values with the options given.
Option A: R = 32 kbps, SNRq = 25.8 dB
The calculated values for R and SNRq are higher than the values given in option A. Therefore, option A is incorrect.
Option B: R = 64 kbps, SNRq = 49.8 dB
The calculated values for R and SNRq match exactly with the values given in option B. Therefore, option B is correct.
Option C: R = 64 kbps, SNRq = 55.8 dB
The calculated values for R and SNRq do not match with the values given in option C. Therefore, option C is incorrect.
Option D: R = 32 kbps, SNRq = 49.8 dB
The calculated values for R and SNRq do not match with the values given in option D. Therefore, option D is incorrect.
Therefore, the correct option is B: R = 64 kbps, SNRq = 49.8 dB.
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signal is sampled at 8 kHz and is quantized using 8 bit uniform quantizer. Assuming SNRq for a sinusoidal signal, the correct statement for PCM signal with a bit rate of R isa)R = 32 kbps, SNRq = 25.8 dBb)R = 64 kbps, SNRq = 49.8 dBc)R = 64 kbps, SNRq = 55.8 dBd)R = 32 kbps, SNRq = 49.8 dBCorrect answer is option 'B'. Can you explain this answer?
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