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PCM, DPCM and DM comparison; SQNR of PCM

Short answer

PCM quantizes each sample with n bits; DPCM codes the prediction error with fewer bits; DM sends 1-bit sign of difference with oversampling. For uniform PCM, SQNR = (3P/mp²) 2^{2n}. For a full-scale sinusoid, SQNR_dB = 1.76 + 6.02n dB; each additional bit increases SQNR by about 6 dB.

Compare Pulse Code Modulation (PCM), Differential Pulse Code Modulation (DPCM) and Delta Modulation. Find the Signal to Quantization Noise ratio (SQNR) of Pulse Code Modulation (PCM).

Comparison of PCM, DPCM and DM

PointPCMDPCMDM
What is codedEach sample valueDifference between sample and its predictionSign of difference only
Bits per samplenn (e.g. 8)Fewer than PCM (e.g. 4)1
Sampling rate≥2W\ge 2W (Nyquist)≥2W\ge 2WMuch higher than 2W2W (oversampling)
PredictorNoneLinear predictorOne-sample delay (accumulator)
Quantization noiseGranular, Δ2/12\Delta^2/12Granular, smaller (prediction gain)Granular + slope overload
SQNR1.76+6.02n1.76 + 6.02n dBPCM value + 10log⁡Gp10\log G_p3fs3/(8π2fm2W)3f_s^3/(8\pi^2 f_m^2 W)
BandwidthHighest (nWnW)Less than PCMCan be less, but needs high fsf_s
ComplexityModerateMost complexSimplest
UseTelephony (64 kbps), CDSpeech, video codingSimple voice links

SQNR of PCM

Let m(t)m(t) have peak value mpm_p and be quantized uniformly into L=2nL = 2^n levels.

Step size: Δ=2mpL\Delta = \dfrac{2m_p}{L}

Noise power: the error qq is uniform in (−Δ/2,Δ/2)(-\Delta/2, \Delta/2):

Nq=∫−Δ/2Δ/2q2Δ dq=Δ212=mp23L2N_q = \int_{-\Delta/2}^{\Delta/2}\frac{q^2}{\Delta}\,dq = \frac{\Delta^2}{12} = \frac{m_p^2}{3L^2}

SQNR: with signal power PP,

SQNR=PNq=3L2Pmp2=3Pmp2 22n\text{SQNR} = \frac{P}{N_q} = \frac{3L^2P}{m_p^2} = \frac{3P}{m_p^2}\,2^{2n}

For a full-scale sinusoid, P=mp2/2P = m_p^2/2:

SQNR=1.5×22n⇒SQNRdB=1.76+6.02 n\text{SQNR} = 1.5\times 2^{2n} \quad\Rightarrow\quad \text{SQNR}_{dB} = 1.76 + 6.02\,n

Each extra bit adds about 6 dB. Example: 8-bit PCM gives 1.76+48.16=49.921.76 + 48.16 = 49.92 dB.

✓ Answer checked by Claude Opus 5.5 (high effort)

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