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Understanding LMS Algorithm: Coefficient Error Vector and Covariance Matrix Analysis

This lecture focuses on the Least Mean Squares (LMS) algorithm, delving deeper into the coefficient error vector and covariance matrix. We analyze the transition from k to k+1 in error estimation, discussing the covariance matrix's steady-state behavior and expressing the mean square error (MSE). The relationship between misadjustment and convergence time is explored, highlighting how increasing the step size improves convergence but may lead to higher misadjustment. Theoretical predictions are compared with simulated results, underscoring the nuances in the LMS algorithm's performance.

austin-rosales
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Understanding LMS Algorithm: Coefficient Error Vector and Covariance Matrix Analysis

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  1. ELG5377 Adaptive Signal Processing Lecture 6: LMS Algorithm Continued

  2. Coefficient Error Vector Covariance Matrix • c(k) = w(n)-wo. • cov[c(k)] = E[c(k)cH(k)] = K(k). • Recall that • c(k+1) = [I - mx(k)xH(k)]c(k) +mx(k)eo*(k). • K(k+1) = E{[I - mx(k)xH(k)]c(k)cH(k) [I - mx(k)xH(k)]H} + E{[I - mx(k)xH(k)]mxH(k)eo(k)} + E{mx(k)eo*(k) {[I - mx(k)xH(k)]} + m2E[|eo(k)|2x(k)xH(k)]. • K(k+1)= [I - mR]K(k)[I - mR]H + m2JminR. • K(k+1)= [I - mR]K(k)[I - mR] + m2JminR.

  3. Coefficient Error Vector Covariance Matrix 2 • At steady state (or for large k), K(k+1)≈K(k). • Therefore • K(k)= [I - mR]K(k)[I - mR] + m2JminR. • 0 = -mK(k)R-mRK(k)+m2RK(k)R+m2JminR. • K(k)R+RK(k) = mJminR.

  4. Mean Square Error • e(k) = d(k)-y(k) = d(k)-wH(k)x(k). • e(k) = d(k)-y(k) = d(k)-(w(k)-wo)Hx(k)-woHx(k). • e(k) = eo(k)-cH(k)x(k). • E[|e(k)|2]=E[|eo(k)|2] + E[cH(k)x(k)xH(k)c(k)]. • E[|eo(k)|2]= Jmin. • E[cH(k)x(k)xH(k)c(k)] = E[tr{cH(k)x(k)xH(k)c(k)}] = E[tr{c(k)cH(k)x(k)xH(k)}] = tr{E[c(k)cH(k)x(k)xH(k)]} ≈ tr{K(k)R]. • tr{K(k)R} = tr{RK(k)}. • K(k)R+RK(k) = mJminR. • tr{K(k)R+RK(k)}=mJmintr{R}. • Therefore tr{K(k)R} = mJmintr{R}/2

  5. Mean Square Error 2 • Therefore the MSE at the output of the LMS filter is • J = Jmin + mJmintr{R}/2. • J = Jmin[1+(m/2)Sli] • Suppose R has a dominant eigenvalue (lmax >> li) • J ≈ Jmin(1+ (mlmax/2)).

  6. Excess Mean Square Error • Jex = J – Jmin. • Jex = mJmintr{R}/2 = Jmin(m/2)Sli. • If R has a dominant eigenvalue, then • Jex ≈Jmin(mlmax/2).

  7. Misadjustment • M = Jex/Jmin. • For LMS Filters, • M = (m/2)tr{R} = (m/2)Mr(0) = (m/2)Sli. • M ≈ (mlmax/2) • In our example in the previous lecture, Jmin = 0.0985. • For the LMS filter with m = 0.1, the misadjustment should be • 0.05* 3.57 = 0.1785 • Simulated misadjustment = (0.1255-0.0985)/0.0985 = 0.274. • For LMS filter with m = 0.3, • Theoretical = 0.536 • Simulated = 2.57

  8. Conclusion • Performance of LMS algorithm as a function of m. • Increasing m improves convergence time at a cost of increasing the misadjustment. • Misadjustment and convergence time are inversely proportional.

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