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# Course 10 example application of random signals - oversampling and noise shaping

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• 1. Review of Quantization Error 8-bit quantization error In a heuristic sense, the assumptions of the statistical model appear to be valid if the signal is sufficiently complex and the quantization steps are sufficiently small, so that the amplitude of the signal is likely to traverse many quantization steps from sample to sample.
• 2. Review: Assumptions about e[n]  e[n] is a sample sequence of a stationary random process.  e[n] is uncorrelated with the sequence x[n].  The random variables of the error process e[n] are uncorrelated; i.e., the error is a white-noise process.  The probability distribution of the error process is uniform over the range of quantization error (i.e., without being clipped). The assumptions would not be justified. However, when the signal is a complicated signal (such as speech or music), the assumptions are more realistic.  Experiments have shown that, as the signal becomes more complicated, the measured correlation between the signal and the quantization error decreases, and the error also becomes uncorrelated.
• 3. Review: Quantization error analysis − ∆ / 2 < e[n] ≤ ∆ / 2 e[n] is a white noise sequence. The probability density function of e[n] is a uniform distribution
• 4. Review: quantization error vs. quantization bits The mean value of e[n] is zero, and its variance is ∆/2 1 ∆2 σ e2 = ∫ e 2 de = −∆ / 2 ∆ 12 Xm Since ∆ = B 2 For a (B+1)-bit quantizer with full-scale value Xm, the noise variance, or power, is 2 −2 B X m 2 σ e2 = 12
• 5. Review: Quantization error analysis A common measure of the amount of degradation of a signal by additive noise is the signal-to-noise ratio (SNR), defined as the ratio of signal variance (power) to noise variance. Expressed in decibels (dB), the SNR of a (B+1)-bit quantizer is σx  2  12 ⋅ 2 2 B σ x  2 SNR = 10 log10  2  = 10 log10  σ   2    e  Xm   Xm  = 6.02 B + 10.8 − 20 log10  σ    x  Hence, the SNR increases approximately 6dB for each bit added to the word length of the quantized samples.
• 6. Oversampling vs. quantization We consider the analog signal xa(t) as zero-mean, wide-sense-stationary, random process with power-spectral density denoted by φ xa xa (e jw ) and jΩ the autocorrelation function by φ xa xa (τ ) . To simplify our discussion, assume that xa(t) is already bandlimited to ΩN, i.e., φ x x ( jΩ) = 0, | Ω |≥ Ω N , a a
• 7. Oversampling  We assume that 2π/T = 2MΩN.  M is an integer, called the oversampling ratio. Decimation with ideal low-pass filterOversampled A/D conversion with simple quantization and down-sampling
• 8. Additive noise model Using the additive noise model, the system can be replaced by Decimation with ideal low-pass filter Its output xd[n] has two components, one due to the signal input xa(t) and one due to the quantization noise input e[n]. Denote these components by xda[n] and xde[n], respectively.
• 9. Signal component (assume e[n]=0) Goal: determine the ratio of signal power ε{xda2} to the quantization-noise power ε{xde2}, ε{.} denotes the expectation value. As xa(t) is converted into x[n], and then xda[n], we focus on the power of x[n] first. Let us analysis this in the time domain. Denote φxx[n] and φxx[ejw] be the autocorrelation and power spectral density of x[n], respectively. By definition, φxx[m]= ε{x[n+m]x[n]}.
• 10. Power of x[n] (assume e[n]=0) Since x[n]= xa(nT), it is easy to see that φ xx [ m] = ε {x[ n + m] x[ n]} = ε {xa (( n + m)T ) xa ( nT )} = φ x x (mT ) a a That is, the autocorrelation function of the sequence of samples is a sampled version of the autocorrelation function. The wide-sense-stationary assumption implies that ε{xa2(t)} is a constant independent of t. It then follows that ε {x 2 [ n]} = ε {xa ( nT )} = ε {xa (t )} 2 2 for all n or t.
• 11. Power of xda[n] (assume e[n]=0) Since the decimation filter is an ideal lowpass filter with cutoff frequency wc = π/M, the signal x[n] passes unaltered through the filter. Therefore, the downsampled signal component at the output, xda[n]=x[nM]=xa(nMT), also has the same power. In sum, the above analyses show that ε {xda [ n]} = ε {x 2 [ n]} = ε {xa (t )} 2 2 which shows that the power of the signal component staysthe same as it traverse the entire system from the input xa(t)to the corresponding output component xda[n].
• 12. Power of the noise component According to previous studies, let us assume that e[n] is a wide-sense-stationary white-noise process with zero mean and variance ∆2 2 σe = 12 Consequently, the autocorrelation function and power density spectrum for e[n] are, white noise φee [ m] = σ e2δ [ m] The power spectral density is the DTFT of the autocorrelation function. So, φee (e jw ) = σ e2 , −π < w < π
• 13. Power of the noise component (assume xa(t)=0) Although we have shown that the power in either x[n] or e[n] does not depend on M, we will show that the noise component xde[n] does not keep the same noise power.  It is because that, as the oversampling ratio M increases, less of the quantization noise spectrum overlaps with the signal spectrum, as shown below.
• 14. Illustration of frequency and amplitude scaling Since oversampled by M, the power spectrum of xa(t) and x[n] in the frequency domain are illustrated as follows.
• 15. Illustration of frequency for noise By considering both the signal and the quantization noise, the power spectra of x[n] and e[n] in the frequency domain are illustrated as
• 16. Noise component power Then, by ideal low pass with cutoff wc=π/M in the decimation, the noise power at the output becomes 1 π /M σ 2 ε {e [ n]} = ∫π σ dw = 2 2 e 2π e − /M M
• 17. Powers after downsampling Next, the lowpass filtered signal is downsampled, and as we have seen, the signal power remains the same. Hence, the power spectrum of xda[n] and xde[n] in the frequency domain are illustrated as follows:
• 18. Noise power reduction Conclusion: Thus, the quantization-noise power ε{xde2[n]} has been reduced by a factor of M through the filtering and downsampling, while the signal poweer has remained the same. 1 π σ e2 σ e2 ∆2 ε {xde } = ∫π dw = = 2 2π − M M 12 M For a given quantization noise power, there is a clear tradeoff between the oversampling factor M and the quantization step ∆.
• 19. Oversampling for noise power reduction Xm Remember that ∆ = B 2 Therefore 1 Xm 2 ε {x } = 2 de ( B) 12 M 2 The above equation shows that for a fixed quantizer, the noise power can be decreased by increasing the oversampling ratio M. Since the signal power is independent of M, increasing M will increase the signal-to-quantization-noise ratio.
• 20. Tradeoff between oversampling and quantization bits Alternatively, for a fixed quantization noise power, 1 Xm 2 Pde = ε {x } = 2 de ( B) 12 M 2 the required value for B is 1 1 1 B = − log 2 M − log 2 12 − log 2 Pde + log 2 X m 2 2 2 From the equation, every doubling of the oversampling ratio M, we need ½ bit less to achieve a given signal-to- quantization-noise ratio. In other words, if we oversample by a factor M=4, we need one less bit to achieve a desired accuracy in representing the signal.
• 21. Further reduction of quantization error Oversampling and Noise Shaping Previously, we have shown that oversampling and decimation can improve the signal-to-quantization-noise ratio. The result is remarkable, but if we want to make a significant reduction, we need very large sampling ratios.  Eg., to reduce the number of bits from 16 to 12 would require M=44=256. The basic concept in noise shaping is to modify the A/D conversion procedure so that the power density spectrum of the quantization noise is no longer uniform.
• 22. Delta-Sigma Modulator The noise-shaping quantizer, generally referred to as a sampled-data Delta-Sigma modulator, is roughly shown as the following figures.  Analog form
• 23. Oversampled Quantizer with Noise Shaping  Can be represented by the discrete-time equivalent system as follows:  Discrete-time formMinus the delayed feedback:delta Accumulator (like an integrator): Sigma
• 24. Modeling the quantization error As before, we model the quantization error as an additive noise source. Hence, the above figure can be replaced by the following linear model: additive noise
• 25. Output of a linear system This linear system has two inputs, x[n] and y[n]. According to the linearity, we can get the output y[n] by 1. set x[n]=0, find the output y[n] w.r.t. e[n] 2. set e[n]=0, find the output y[n] w.r.t. x[n] 3. add the above two outputs.
• 26. Transfer functions Consider the output in the z-domain. We denote the transfer function from x[n] to y[n] as Hx(z) and from e[n] to y[n] as He(z). When e[n]=0: 0 (X(z)-Z-1Y(z))/(1-z-1) X(z)-Z-1Y(z) X(z) Z-1Y(z)
• 27. Output when e[n]=0 We have −1 X [ z] − z Y ( z) Y [ z] = −1 1− z −1 −1 So Y [ z ] − z Y [ z ] = X [ z ] − z Y ( z ) That is Y [ z] = X [ z] when E[z] is zero.
• 28. Transfer functions when x[n]=0 When x[n]=0: -Z-1Y(z)/(1-Z-1) E(Z) -Z-1Y(z) 0 Z-1Y(z)
• 29. Output when x[n]=0 We have z −1Y ( z ) Y [ z ] = e( z ) − −1 1− z So −1 −1 −1 Y [ z ] − z Y [ z ] = E[ z ] − z E[ z ] − z Y ( z ) That is −1 Y [ z ] = (1 − z ) E[ z ] when X[z] is zero.
• 30. Remark: feedback system In fact, feedback systems have been widely used (serves as a fundamental architecture) in control engineering. E Generally: X Y G H Formula: Y ( z ) G( z) E ( z) 1 = ; = X ( z) 1 + G( z) H ( z) X ( z) 1 + G( z) H ( z)
• 31. Another way of derivation From the feedback system formula, we can also obtain that 1 H x ( z) = Y ( z) = G( z) = 1 − Z −1 = 1 X ( z) 1 + G( z) H ( z) Z −1 1+ 1 − Z −1 Y ( z) 1 1 H e ( z) = = = = 1 − Z −1 E ( z) 1 + G( z) H ( z) Z −1 1+ 1 − Z −1
• 32. Time domain relation Hence, in the time domain, we have y x [ n] = x[ n] ye [ n] = e[ n] = e[ n] − e[ n − 1] ˆ Therefore, the output y[n] can be represented equivalently as y[ n] = y x [ n] + ye [ n] = x[ n] + e[ n] ˆ The quantization noise e[n] has been modified as ˆ e[ n]
• 33. Power spectral density of the modified noise To show the reduction of the quantization noise, let’s consider the power spectral density of e[ n] ˆ Since we have the input-output relationship between e[n] and e[ n] as ˆ Ye ( z ) = (1 − Z −1 ) E ( z ) In the frequency domain, we have E (e jw ) = Ye (e jw ) = (1 − e − jw ) E (e jw ) ˆ
• 34. Equivalent system
• 35. Power spectral density of the modified noise  The power-spectral-density relation between the modified and original quantization noises is thusφeˆeˆ (e ) =|| 1 − e jw || φee (e ) − jw 2 jw Model the original quantization error as white =|| 1 − e || σ e − jw 2 2 noise with this variance = (1 − e − jw )(1 − e jw )σ e2 = (1 − e − jw − e − jw + 1)σ e2 = ( 2 − (e − jw +e − jw ))σ = (2 − 2 cos(w))σ 2 e 2 e = (2 sin( w / 2)) σ 2 2 e p.s.d. of the original noise p.s.d. of the modified noise
• 36. Quantization-noise power Remember that the downsampler does not remove any of the signal power, the signal power in xda[n] is Pda = ε {xda [ n]} = ε {x 2 [ n]} = ε {xa (t )} 2 2 The quantization-noise power in the final output is 1 π 2π ∫−π Pde = φ xde xde (e jw )  See the following illustration for its computation
• 37. Before decimation Power spectral density of modified noiseThe modified noise density is non-uniform and lower in the effective band region
• 38. After decimation Downscaled by M Down-scaled by M and also stretched by M
• 39. Quantization-noise power Hence, the quantization-noise power in the final output is 1 π 2π ∫−π Pde = φ xde xde (e jw ) 1 ∆2 π 2 2 = 2π 12 M ∫−π (2 sin( 2 M )) dw Assume that M is sufficiently large, we can approximate that w w = sin( )≈ 2M 2M
• 40. Bits and quantization tradeoff in noise shaping With this approximation, 1 ∆2π 2 Pde = 36 M 3 For a (B+1)-bit quantizer and maximum input signal level between plus and minus Xm, ∆= Xm/2B. To achieve a given quantization-noise power Pde, we have 3 1 1 B = − log 2 M + log 2 (π / 6) − log 2 Pde + log 2 X m 2 2 2 We see that, whereas with direct quantization a doubling of the oversampling ratio M gained ½ bit in quantization, the use of noise shaping results in a gain of 1.5 bits.
• 41. Second-order noise shaping The noise-shaping strategy can be extended by incorporating a second stage of accumulation, as shown in the following:
• 42. Second-order (i.e., 2-stage) noise shaping In the two-stage case, it can be derived that H e ( z ) = (1 − z −1 ) 2 φeˆeˆ (e jw ) = σ e2 (2 sin( w / 2)) 4 In general, if we extend the case to p-stages, the corresponding noise shaping is given by φeˆeˆ (e jw ) = σ e2 (2 sin(w / 2)) 2 p
• 43.  By evaluation, with p=2 and M=64, we obtain almost 13 bits of increase in accuracy, suggesting that a 1-bit quantizer could achieve about 14-bit accuracy at the output of the decimator. Although multiple feedback loops promise greatly increased quantization-noise reduction, they are not without problems. Specifically, for large values of p, there is an increased potential for instability and oscillations to occur.