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Companding; delta modulator numerical (5 kHz, 200 kHz)

Short answer

Companding compresses weak signals before uniform quantization to maintain constant SQNR. For DM: minimum step size 0.157 V, SQNR 2431.7 (33.86 dB), minimum bandwidth 100 kHz.

What is companding? A Delta modulator is used to encode speech signal band-limited to 5 kHz with a sampling frequency of 200 kHz. For maximum signal amplitude of Amax = 1, find a) Minimum step size to avoid slope overloading b) Assuming the speech signal to be sinusoidal, find signal to quantization noise ratio c) Determine the minimum transmission bandwidth

Companding

Companding = compressing + expanding. At the transmitter the signal is passed through a compressor that amplifies weak signals more than strong ones; it is then uniformly quantized; at the receiver an expander applies the inverse characteristic. The overall effect is non-uniform quantization: small steps for weak signals and large steps for strong signals, so SQNR stays nearly constant over a wide dynamic range. Standard laws: µ-law (μ=255\mu = 255, North America/Japan) and A-law (A=87.6A = 87.6, Europe/India/Nepal).

Given

fm=5f_m = 5 kHz, fs=200f_s = 200 kHz, Amax=1A_{max} = 1, Ts=1/fsT_s = 1/f_s.

a) Minimum step size to avoid slope overload

The staircase can rise at most Δ/Ts=Δfs\Delta/T_s = \Delta f_s. For m(t)=Asin⁡2πfmtm(t) = A\sin 2\pi f_m t the maximum slope is 2πfmA2\pi f_m A. No slope overload requires

Δfs≥2πfmAmaxΔmin=2πfmAmaxfs=2π×5000×1200 000=0.157 V\begin{aligned} \Delta f_s &\ge 2\pi f_m A_{max} \\ \Delta_{min} &= \frac{2\pi f_m A_{max}}{f_s} = \frac{2\pi \times 5000 \times 1}{200\,000} = 0.157\ \text{V} \end{aligned}

b) SQNR for a sinusoidal signal

  • Signal power at the limit of slope overload: S=A2/2S = A^2/2 with A=Δfs/(2πfm)A = \Delta f_s/(2\pi f_m), so S=Δ2fs28π2fm2S = \dfrac{\Delta^2 f_s^2}{8\pi^2 f_m^2}.
  • Granular noise is uniform in (−Δ,Δ)(-\Delta, \Delta): power Δ2/3\Delta^2/3, spread evenly over 00 to fsf_s; the output LPF (bandwidth fM=fmf_M = f_m) passes the fraction fM/fsf_M/f_s: Nq=Δ23fMfsN_q = \dfrac{\Delta^2}{3}\dfrac{f_M}{f_s}.
SQNR=38π2 fs3fm2fM=38π2(fsfm)3=0.0380×(2005)3=0.0380×64 000=2431.7SQNRdB=10log⁡102431.7=33.86 dB\begin{aligned} \text{SQNR} &= \frac{3}{8\pi^2}\,\frac{f_s^3}{f_m^2 f_M} = \frac{3}{8\pi^2}\left(\frac{f_s}{f_m}\right)^3 \\ &= 0.0380 \times \left(\frac{200}{5}\right)^3 = 0.0380 \times 64\,000 = 2431.7 \\ \text{SQNR}_{dB} &= 10\log_{10}2431.7 = 33.86\ \text{dB} \end{aligned}

c) Minimum transmission bandwidth

DM sends one bit per sample, so Rb=fs=200R_b = f_s = 200 kbps. The minimum (Nyquist) bandwidth for binary data is Rb/2R_b/2:

BT,min=Rb2=2002=100 kHzB_{T,min} = \frac{R_b}{2} = \frac{200}{2} = 100\ \text{kHz}

Answer: Δmin=0.157\Delta_{min} = 0.157 V; SQNR = 2431.7 (33.86 dB); BT=100B_T = 100 kHz (if rectangular pulses with BT=RbB_T = R_b are assumed, 200 kHz).

✓ Answer checked by Claude Opus 5.5 (high effort)

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