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PCM 7-bit at 50 Mbps: maximum message bandwidth

Short answer

Sampling rate fs = Rb/n = 50×10⁶/7 ≈ 7.143 MHz. Maximum message bandwidth for Nyquist sampling is W_max = fs/2 ≈ 3.57 MHz.

A PCM system uses a uniform quantizer followed by a 7 bit binary encoder. The bit rate of the system is equal to 50×10⁶ bits/sec. What is the maximum message signal bandwidth for which the system operates satisfactorily?

Given: n=7n = 7 bits/sample, Rb=50×106R_b = 50 \times 10^6 bps, uniform quantizer.

For satisfactory operation the sampling rate must be at least the Nyquist rate, fs≥2Wf_s \ge 2W.

  1. Bit rate relation:
Rb=nfs  ⇒  fs=Rbn=50×1067=7.143 MHzR_b = nf_s \;\Rightarrow\; f_s = \frac{R_b}{n} = \frac{50 \times 10^6}{7} = 7.143\ \text{MHz}
  1. Maximum message bandwidth (sampling exactly at Nyquist rate):
Wmax=fs2=Rb2n=50×10614=3.571 MHzW_{max} = \frac{f_s}{2} = \frac{R_b}{2n} = \frac{50 \times 10^6}{14} = 3.571\ \text{MHz}

Answer: Maximum message bandwidth ≈ 3.57 MHz.

✓ Answer checked by Claude Opus 5.5 (high effort)

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