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Wiener Filters • Linear Filters and Orthogonality • the Time Continues Wiener Filter • the Time Discrete Wiener Filter • Example

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Wiener Filters • Linear Filters and Orthogonality • the Time Continues Wiener Filter • the Time Discrete Wiener Filter • Example Stochastic signals and processes Lec. 8 Samuel Schmidt 25-10-2011 • Introduction to Wiener filters • Linear filters and orthogonality • The time continues Wiener filter • The time discrete Wiener filter • Example How to filter a signal Y(t) so it became as similar to a desired signal S(t) as possible. Y(t) S(t) 5 Wiener 2 Y(t) 0 0 ≈ S(t) -5 filter 0 1 2 3 4 5 -2 Time (s) 0 1 2 3 4 5 L[Y(t)] Time (s) • Noise contaminated signal Y(t)=S(t)+N(t) Wiener filter • Estimate the linear filter L[·] so Ŝ(t) become as close to S(t) as possible Ŝ(t)=L[Y(t)] Se(t) 5 e(t) 5 0 Se(t) 0 e(t) -5 0 1 2 3 4 5 -5 Time (s) 0 1 2 3 4 5 Time (s) Y(t) S(t) 5 2 Y(t) 0 0 S(t) -5 0 1 2 3 4 5 -2 Time (s) 0 1 2 3 4 5 Time (s) • Introduction to Wiener filters • Linear filters and orthogonality • The time continues Wiener filter • The time discrete Wiener filter • Example • Y(t),S(t) and N(t) are zero mean stationary processes • L[·] is a linear filter • FIR filters (M-order) Y(n)=ω0 X(n)+ ω1 X(n-1)+ ω2 X(n-2)….. ωM-1 X(n-M+1) Or like 푀−1 푌 푛 = 휔푘푋(푛 − 푘) 푘=0 • In linear system the order of multiplication and addition is irrelevant. • Defined by super position Ta x [n]b x [n] aTx [n]bTx [n] 1 2 1 2 • Linear algebra: Two vectors is orthogonal if the dot product is zero • That corresponds zero correlation betweenS(t) two signals 2 0 S(t) -2 0N(t) 1 2 3 4 5 5 Time (s) 0 N(t) -5 0 1 2 3 4 5 Time (s) S(t) vs. N(t) Y(t) vs. N(t) Orthogonal (r=0.0) Non-orthogonal (r=0.69) Scatter between S(t) and N(t) Scatter between Y(t) and N(t) 4 4 3 3 2 2 1 1 N 0 0 N(t) -1 -1 -2 -2 -3 -3 -4 -4 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -5 -4 -3 -2 -1 0 1 2 3 4 5 S X(t) • If two process are orthogonal for all delays in a give range the process are also orthogonal after linear transformation. IF Then S(t) • 5 Error Se(t) 0 5 S(t) e(t) 5 ε(t)= Ŝ(t) - S(t) 0 -5 Se(t) 0 1 2 3 4 5 0 e(t) -5 Time (s) 0 1 2 3 4 5 -5 Time (s) 0 1 2 3 4 5 -Time (s) • Mean square error E[|ε(t)|2]= E[| Ŝ(t) - S(t) |2] E[|ε(t)|2]=E[| L[Y(ξ)] - S(t) |2] = E[|ε(t)|2]=E[S(t) 2 - 2 S(t)L[Y(ξ)] + L[Y(ξ)]2] 4 Y(t) e(t) 3 S(t) Se(t) 2 1 0 -1 -2 -3 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 Time (s) • The Minimum mean-square error is obtained when the error ε(t) is Orthogonal to the unfiltered signal Y(ξ), where ξ is in the range [ti, tf]. ξ ξ E[Y( ) e(t)]=E[(S(t)- Ŝ(t) ) Y( ) ]=0 1 0.8 • That means thats the 0.6 filter is optimal when the 0.4 0.2 e(t) Correlation between Y(t+k) 0 -0.2 and e(t) is low for all k -0.4 -0.6 -0.8 -5 -4 -3 -2 -1 0 1 2 3 4 5 Y(t) • The mean-square error is mean when the correlation between the e(t) and the signal is zero, that is the optimal filter Y(t) and e(t) 5 Y(t) 4 e(t) 3 2 1 0 -1 -2 -3 -4 -5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time (s) The Minimum mean-square error can be estimated. Since E[(S(t)- Ŝ(t) ) Y(ξ) ]=0 Is E[(S(t)- L[Ŝ(t)] ) L[Y(ξ)] ]=0 Thereby is the Minimum mean-square 2 2 2 em=E[|ε(t)| ]=E[S(t) - 2 S(t)L[Y(ξ)] + L[Y(ξ)] ]= E[(S(t)-L[Y(ξ)]) S(t)] em=E[(S(t)-L[Y(ξ)]) S(t)] • Introduction to Wiener filters • Linear filters and orthogonality • The time continues Wiener filter • The time discrete Wiener filter • Example 4 Y(t) e(t) 3 S(t) Se(t) 2 1 0 -1 -2 -3 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 Time (s) • Determine the impulse response h(ξ) of the filter. • Use the Minimum mean-square error principle which requires the error S(t)-Ŝ(t) to be Orthogonal (correlation=0) to Y (ξ) • If we assume stationarity and time invariance • Thereby we “just” need to solve for h(ξ) Fourier transform cross correlation between S(t) and Y(t) power spectrum of Y(t) Therefor H(f) is the cross correlation between S(t) and Y(t) dived with the power spectrum of Y(t) • If S(t) and N(t) are statistically independent and zero mean Therefore H(f) can be determined from the power spectrums of noise and the desired signal PSD S(f) 20 0 -20 -40 S(f) dB/Hz -60 -80 0 2 10 10 Frequency (Hz) PSD S (f)+S (f) ss nn 20 0 -20 (f) (f) dB/Hz nn -40 (f)+S ss S -60 -80 0 2 10 10 Frequency (Hz) PSD H(f) 0 -20 -40 H(f) dB/Hz H(f) -60 -80 0 2 10 10 Frequency (Hz) • The minimum mean-square error can be calculated: • The h(t) is not causal and thereby is the filter not realizable in real-time • There solutions based on spectral factorizations ∞ −훼|휏| −푗2휋푓휏 2훼 푆푠푠 푓 = 푒 푒 푑휏 = 2 2 2 Wolfram −∞ 훼 + 4휋 푓 ∞ 푁0 −푗2휋푓휏 푁0 푆푛푛 푓 = 훿 휏 푒 푑휏 = −∞ 2 2 2훼 4훼 훼2+4휋2푓2 푁0 퐻 푓 = 2훼 푁0=4훼 2 2 2 + 푁 +훼 +4휋 푓 훼2+4휋2푓2 2 0 2훼 2 2 2 4훼 푎 퐻 푓 = 훼 +4휋 푓 = 푁0 = 2훼 푁0 4훼 +훼2+4휋2푓2 푏2+4휋2푓2 2 2 2+ 푁0 훼 +4휋 푓 2 • Inver Fourier transform: • Introduction to Wiener filters • Linear filters and orthogonality • The time continues Wiener filter • The time discrete Wiener filter • Example Ŝ(n)=ω0 Y(n)+ ω1 Y(n-1)+ ω2 Y(n-2)….. ωM-1 Y(n-M+1) • Different methods for estimation of filter coefficients: – Spectrum based • (Estimate impulse response h(n)) Y(n)=h(0) X(n)+ h(1) ω1 X(n-1)+ h(2) X(n-2)….. h(M-1)X(n-M+1) – Mean square error • (Estimate filter coefficients ω(n) ) ω(n) Y(n)=ω0 X(n)+ ω1 X(n-1)+ ω2 X(n-2)….. ωM-1 X(n-M+1) • As in the continues domain the transfer function of H(z) is the cross spectrum between S(t) and Y(t) dived by the PSD of Y(t) (Real cases) • The discrete filter • Determine the coefficients 휔 which minimize the Mean square error Error: 푒 푛 = 푆 푛 − 푆(푛) Mean square error: 퐶 휔 = 퐸 푒 푛 푒 푛 = 퐸 푒 푛 2 = 퐸[(푆 푛 − 푆 푛 )2] • Denote the filter coefficients 푇 휔 = 휔0 휔1 휔2 … 휔푀−1 • Denote the signal 푌(푛) = 푌 푛 푌 푛 − 1 푌 푛 − 2 … . 푌(푛 − 푀) 푇 • Thereby filter equations can be described using the dot product of the filter coefficients and the signal. 푌 푛 푌 푛 − 1 푆 푛 = 휔푇푌 푛 = 휔 휔 휔 … 휔 0 1 2 푀−1 푌 푛 − 2 푌(푛 − 푀) Ŝ(n)=ω0 Y(n)+ ω1 Y(n-1)+ ω2 Y(n-2)….. ωM-1 Y(n-M+1) 푆 푛 = 휔푇푌 푛 =푌 푛 푇휔 (continued) • Since: 푆 푛 = 휔푇푌 푛 =푌 푛 푇휔 • Mean square error is 퐶 휔 = 퐸 푒 푛 푒 푛 = 퐸[ 푆 푛 − 휔푇푌 푛 (푆 푛 − 푌 푛 푇휔)] • Which can be expanded to 퐶 휔 = 퐸 푆 푛 푆 푛 − 휔푇퐸[푌 푛 푆(푛)] − 퐸[푆 푛 푌(푛)푇]휔+ 휔푇E[푌 푛 푌(푛)푇]휔 (continued) • Expanded Mean square error explained 퐶 휔 = 퐸 푆 푛 푆 푛 − 휔푇퐸[푌 푛 푆(푛)] − 퐸[푆 푛 푌(푛)푇]휔+ 휔푇E[푌 푛 푌(푛)푇]휔 Transformed σ 2 Signal variance s Cross correlation vector version of R Autocorrelation matrix of ys Y(n) between Y(n) and S(n) Rys =[rys(0) rys(1) ……..rys(M-1) ] ryy(0) ryy(1) ryy(2) rys(τ)=E[Y(n+τ)S(n)] , τ=0,1,2,…M-1 Ryy= ryy(−1) ryy(0) ryy(1) r (−2) r (−1) r (0) yy yy yy • Compact form ryy(τ)=E[Y(n+τ)Y(n)] , It an Toeplitz matrix 퐶 휔 = σ 2 − 휔푇R − R 푇 휔+ 휔푇R 휔 s ys ys yy Page 432 (continued) • The Mean square error is a second order equation. 2 푇 푇 푇 퐶 휔 = σs − 휔 Rys − Rys 휔+ 휔 Ryy휔 • Thereby does the equation has a global minimum which can be identified by differentiation according to 휔 and solving for zero Differentiating equation components: 푑 푇 풅푪(흎) 풅 (휔 R )= R = (σ 2 − 휔푇R − R 푇 휔+ 휔푇R 휔) = −2R +ퟐR 휔 = 0 푑휔 ys ys 풅흎 풅흎 s ys ys yy ys yy 푑 (R 푇 휔)= R 푑휔 ys ys 푑 • The filter coefficients is then easily solved.
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