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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?
Verified Answer
signal is sampled at 8 kHz and is quantized using 8 bit uniform quanti...
Bit Rate = 8k x 8 = 64 kbps
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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?
Question Description
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? for Electronics and Communication Engineering (ECE) 2025 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about 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? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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