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Bits per sample and information rate (4.5 kHz, 8 levels)

Short answer

Nyquist rate is 9 kHz, so sampling rate is 18 kHz. For 8 levels, minimum fixed-length bits per sample is 3. Information rate using entropy (2.862 bits/sample) yields 51.52 kbps.

A signal of bandwidth 4.5 kHz is sampled at the double rate given by Nyquist, the signal is quantized in 8 levels, the probability of occurrence of the level are 0.1, 0.15, 0.15, 0.05, 0.2, 0.05, 0.18, 0.12. Find the minimum no of bits per sample and information rate.

Given: fm=4.5f_m = 4.5 kHz; sampled at twice the Nyquist rate; 8 levels with P={0.1,0.15,0.15,0.05,0.2,0.05,0.18,0.12}P = \{0.1, 0.15, 0.15, 0.05, 0.2, 0.05, 0.18, 0.12\} (sum = 1).

Sampling rate: Nyquist rate =2fm=9= 2f_m = 9 kHz, so

fs=2×9 kHz=18 000 samples/sf_s = 2 \times 9\ \text{kHz} = 18\ 000\ \text{samples/s}

Minimum number of bits per sample: with a fixed-length binary code, 8 levels need

n=log⁡28=3 bits/samplen = \log_2 8 = 3\ \text{bits/sample}

Entropy (the theoretical minimum average number of bits per sample with ideal source coding):

Symbolpip_ilog⁡2(1/pi)\log_2(1/p_i)pilog⁡2(1/pi)p_i\log_2(1/p_i)
Q10.13.32190.3322
Q20.152.73700.4105
Q30.152.73700.4105
Q40.054.32190.2161
Q50.22.32190.4644
Q60.054.32190.2161
Q70.182.47390.4453
Q80.123.05890.3671
Total12.8622
H=2.8622 bits/sampleH = 2.8622\ \text{bits/sample}

Information rate:

R=fsH=18 000×2.86224=51 520.24 bits/s≈51.52 kbpsR = f_s H = 18\ 000 \times 2.86224 = 51\ 520.24\ \text{bits/s} \approx 51.52\ \text{kbps}

(With 3-bit PCM words the line bit rate is 18 000×3=5418\ 000 \times 3 = 54 kbps. If "the double rate given by Nyquist" is read as the Nyquist rate itself, fs=9f_s = 9 kHz and R≈25.76R \approx 25.76 kbps.)

Answer: 3 bits/sample (entropy limit 2.862 bits/sample); information rate ≈ 51.52 kbps.

✓ Answer checked by Claude Opus 5.5 (high effort)

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