Digital Signal Processing Oversampled Analog to Digital Conversion with Direct Quantization

DSP: Oversampled ADC with Direct Quantization

D. Richard Brown III

D. Richard Brown III 1 / 6 DSP: Oversampled ADC with Direct Quantization Oversampled ADC System Model

Assumptions: ◮ The input signal is band limited so that Xa(jΩ)=0 for all |Ω|≥ ΩN . ◮ The input signal is stationary with autocorrelation φxaxa (τ) and

power spectrum Φxaxa (jΩ). ◮ The C/D block oversampled by a factor of M, i.e., Ωs = 2MΩN . We have seen previously that, when M = 1, the SNR (in dB) of a B + 1 Xm bit uniform quantizer with step size ∆= 2B is 2 σx Xm SNRQ = 10 log10 2 = 6.02B + 10.8 − 20 log10 σe σx Question: What happens when M > 1? D. Richard Brown III 2 / 6 DSP: Oversampled ADC with Direct Quantization Quantization Error Noise Analysis

Under the usual white-noise statistical assumptions about the quantization jω 2 error, we have Φee(e )= σe . Hence the quantization noise propagates as

jω jω Φee(e ) Φxdexde (e ) 2 2 2 σe σe σe LPF ↓ M M ω ω ω π −2π M 2π −2π 2π and the power of the quantization noise at the output of the downsampler can be computed as π 2 2 2 1 σe σe σxde = dω = . 2π Z−π M M

D. Richard Brown III 3 / 6 DSP: Oversampled ADC with Direct Quantization Analysis of Signal Part (part 1 of 2)

We can analyze the signal component similarly:

jω Φxaxa (jΩ) Φxx(e ) c = cΩN M c T π

C/D Ω ω π π −ΩN ΩN −2π − M M 2π

jω Φxdaxda (e )

c cΩN M T = π cΩN LPF ↓ M π

ω ω π π −2π − M M 2π −2π 2π Note that the signal power is unchanged through the LPF and downsampler. D. Richard Brown III 4 / 6 DSP: Oversampled ADC with Direct Quantization Analysis of Signal Part (part 2 of 2)

Note that the power of the CT input signal is

ΩN 2 1 σxa = Φxaxa (jΩ) dΩ 2π Z−ΩN The power of the sampled DT signal is π 2 1 jω σx = Φxx(e ) dω 2π Z−π but since ∞ jw 1 ω j2πk Φxx(e )= Φxaxa j − T T T k=X−∞ we have π ∞ π/M 2 1 1 ω − j2πk 1 1 ω σx = Φxaxa j dω = Φxaxa j dω. 2π Z− T T T 2π Z− T T π k=X−∞ π/M ω We can substitute Ω= T and note that dω = TdΩ to write ΩN 2 1 2 σx = Φxaxa (jΩ) dΩ= σxa . 2π Z−ΩN Hence the power of the analog input and the discrete-time signal are identical. D. Richard Brown III 5 / 6 DSP: Oversampled ADC with Direct Quantization Oversampled ADC With Direct Quantization SNR Analysis

We can compute the SNR (in dB) of a B +1 bit uniform quantizer with step size Xm ∆= 2B and oversampling factor M as 2 σxda SNRQ = 10 log10 2 σxde 2 σx = 10 log10 2 σe /M 2B 2 12 · 2 σxM = 10 log10 2 Xm Xm =6.02B + 10.8 − 20 log10 + 10 log10(M) σx Remarks: ◮ Main idea: Oversampling by M doesn’t aﬀect the signal power but reduces the noise power by a factor of M. ◮ Doubling M and keeping B ﬁxed gives a 3 dB SNR improvement. ◮ Quadrupling M and using one less quantizer bit gives the same SNR.

D. Richard Brown III 6 / 6