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A speech signal has a total duration of 20 sec. It is sampled at the rate of 8 kHz and then PCM encoded. The signal-to-quantization noise ratio is required to be 40dB. The minimum storage capacity needed to accommodate this signal is
- a)1.12 KBytes
- b)140 KBytes
- c)168 KBytes
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
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To calculate the minimum storage capacity needed to accommodate the given speech signal, we need to consider the sampling rate, duration of the signal, and the desired signal-to-quantization noise ratio.
Sampling Rate:
The signal is sampled at a rate of 8 kHz, which means that 8000 samples are taken per second.
Duration:
The total duration of the speech signal is given as 20 seconds.
Signal-to-Quantization Noise Ratio:
The desired signal-to-quantization noise ratio is given as 40 dB.
Now, let's calculate the number of bits required to represent each sample in the PCM encoding.
Sampling Resolution:
The signal-to-quantization noise ratio is a measure of the quantization error introduced during the PCM encoding process. It is given by the formula:
SNR = 6.02N + 1.76 dB
Where N is the number of bits per sample.
In this case, the desired SNR is 40 dB. Substituting the values into the formula:
40 = 6.02N + 1.76
Solving for N:
6.02N = 40 - 1.76
N = (40 - 1.76) / 6.02
N ≈ 6.04
Since the number of bits per sample should be an integer, we can round N up to the nearest integer:
N = 7
Now, let's calculate the number of samples in the speech signal.
Number of Samples:
The sampling rate is given as 8 kHz, which means that 8000 samples are taken per second. Since the duration of the speech signal is 20 seconds, we can calculate the number of samples as:
Number of Samples = Sampling Rate × Duration
Number of Samples = 8000 × 20
Number of Samples = 160,000
Finally, let's calculate the minimum storage capacity needed to accommodate the signal.
Minimum Storage Capacity:
The storage capacity needed is given by the formula:
Storage Capacity = Number of Samples × Number of Bits per Sample
Substituting the values:
Storage Capacity = 160,000 × 7
Storage Capacity = 1,120,000 bits
Converting bits to bytes:
Storage Capacity = 1,120,000 bits / 8
Storage Capacity = 140,000 bytes
Therefore, the minimum storage capacity needed to accommodate the given speech signal is 140 KBytes. Hence, option B is the correct answer.
Sampling Rate:
The signal is sampled at a rate of 8 kHz, which means that 8000 samples are taken per second.
Duration:
The total duration of the speech signal is given as 20 seconds.
Signal-to-Quantization Noise Ratio:
The desired signal-to-quantization noise ratio is given as 40 dB.
Now, let's calculate the number of bits required to represent each sample in the PCM encoding.
Sampling Resolution:
The signal-to-quantization noise ratio is a measure of the quantization error introduced during the PCM encoding process. It is given by the formula:
SNR = 6.02N + 1.76 dB
Where N is the number of bits per sample.
In this case, the desired SNR is 40 dB. Substituting the values into the formula:
40 = 6.02N + 1.76
Solving for N:
6.02N = 40 - 1.76
N = (40 - 1.76) / 6.02
N ≈ 6.04
Since the number of bits per sample should be an integer, we can round N up to the nearest integer:
N = 7
Now, let's calculate the number of samples in the speech signal.
Number of Samples:
The sampling rate is given as 8 kHz, which means that 8000 samples are taken per second. Since the duration of the speech signal is 20 seconds, we can calculate the number of samples as:
Number of Samples = Sampling Rate × Duration
Number of Samples = 8000 × 20
Number of Samples = 160,000
Finally, let's calculate the minimum storage capacity needed to accommodate the signal.
Minimum Storage Capacity:
The storage capacity needed is given by the formula:
Storage Capacity = Number of Samples × Number of Bits per Sample
Substituting the values:
Storage Capacity = 160,000 × 7
Storage Capacity = 1,120,000 bits
Converting bits to bytes:
Storage Capacity = 1,120,000 bits / 8
Storage Capacity = 140,000 bytes
Therefore, the minimum storage capacity needed to accommodate the given speech signal is 140 KBytes. Hence, option B is the correct answer.
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A speech signal has a total duration of 20 sec. It is sampled at the rate of 8 kHz and then PCM encoded. The signal-to-quantization noise ratio is required to be 40dB. The minimum storage capacity needed to accommodate this signal isa)1.12 KBytesb)140 KBytesc)168 KBytesd)None of the aboveCorrect answer is option 'B'. Can you explain this answer?
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