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A telephone signal with cut-off frequency of 4 kHz is digitized into 8 bit PCM, sampled at Nyquist rate. The quantization S/N ratio is
- a)45.2 dB
- b)46.6 dB
- c)49.8 dB
- d)51.5 dB
Correct answer is option 'C'. Can you explain this answer?
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A telephone signal with cut-off frequency of 4 kHz is digitized into 8...
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Given information:
- Cut-off frequency of telephone signal = 4 kHz
- Digitized into 8 bit PCM
- Sampled at Nyquist rate
To find: Quantization S/N ratio
Solution:
1. Nyquist rate:
- The Nyquist rate is twice the maximum frequency component of the input signal, which is 4 kHz in this case.
- Therefore, the sampling frequency (Fs) = 2 x 4 kHz = 8 kHz.
2. Quantization:
- The signal is digitized into 8 bit PCM, which means there are 2^8 = 256 quantization levels.
- The quantization step size (Δ) can be calculated using the formula:
Δ = (Maximum signal amplitude) / (Number of quantization levels)
- The maximum signal amplitude of a telephone signal is typically 1 V, so Δ = 1/256 = 0.0039 V.
3. Quantization noise power:
- The quantization noise power is given by the formula:
Pq = (Δ^2) / 12
- Substituting values, Pq = (0.0039^2) / 12 = 1.276 x 10^-7 W.
4. Signal power:
- The signal power can be calculated using the formula:
Ps = (Maximum signal amplitude)^2 / 2
- Substituting values, Ps = (1/√2)^2 / 2 = 0.25 W.
5. S/N ratio:
- The S/N ratio in dB can be calculated using the formula:
S/N (dB) = 10 log10 (Signal power / Noise power)
- Substituting values, S/N (dB) = 10 log10 (0.25 / 1.276 x 10^-7) = 49.8 dB.
Therefore, the quantization S/N ratio is 49.8 dB (option C).
- Cut-off frequency of telephone signal = 4 kHz
- Digitized into 8 bit PCM
- Sampled at Nyquist rate
To find: Quantization S/N ratio
Solution:
1. Nyquist rate:
- The Nyquist rate is twice the maximum frequency component of the input signal, which is 4 kHz in this case.
- Therefore, the sampling frequency (Fs) = 2 x 4 kHz = 8 kHz.
2. Quantization:
- The signal is digitized into 8 bit PCM, which means there are 2^8 = 256 quantization levels.
- The quantization step size (Δ) can be calculated using the formula:
Δ = (Maximum signal amplitude) / (Number of quantization levels)
- The maximum signal amplitude of a telephone signal is typically 1 V, so Δ = 1/256 = 0.0039 V.
3. Quantization noise power:
- The quantization noise power is given by the formula:
Pq = (Δ^2) / 12
- Substituting values, Pq = (0.0039^2) / 12 = 1.276 x 10^-7 W.
4. Signal power:
- The signal power can be calculated using the formula:
Ps = (Maximum signal amplitude)^2 / 2
- Substituting values, Ps = (1/√2)^2 / 2 = 0.25 W.
5. S/N ratio:
- The S/N ratio in dB can be calculated using the formula:
S/N (dB) = 10 log10 (Signal power / Noise power)
- Substituting values, S/N (dB) = 10 log10 (0.25 / 1.276 x 10^-7) = 49.8 dB.
Therefore, the quantization S/N ratio is 49.8 dB (option C).
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A telephone signal with cut-off frequency of 4 kHz is digitized into 8 bit PCM, sampled at Nyquist rate. The quantization S/N ratio isa)45.2 dBb)46.6 dBc)49.8 dBd)51.5 dBCorrect answer is option 'C'. Can you explain this answer?
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