Hello. I have a toy situation in which, known the desired signal d(n), d(n) = sin(100*pi*n+pi/3); and a noisy signal, with white noise v(n) with known variance and mean x(n) = d(n) + v(n); the goal is define an optimal Wiener filter of 101 coefficients. (I call it "toy situation" because I have at the same time the value of the noisy signal and the value of the original signal) After some calculation, in this problem we have that the autocorrelation of the signal d(n) equals the correlation sequence between x(n) and d(n), r_dx(n). I calculated (with Matlab) r_d(n), the 101 pts autocorrelation sequence of d(n). I found the optimal coefficients w: w = (R_x^(-1)) * r_d(n) where R_x is the Toeplitz autocorrelation matrix of x, and the minimum error xi_min, xi_min = r_d(0) - r_d(0)'(R_x^-1)r_d(0) Then I implemented a LMS filter to compare the difference between the optimum coefficients and the coefficients estimated by LMS. According to "Statistical digital processing and modeling" by Hayes, the output error of LMS, for the same filter length, should be always greater than xi_min when the LMS filter reaches a steady state, because xi_LMS(inf) = xi_min + xi_excess(n) but in my case the LMS filter with fair values of the mu parameter (e.g. 10^-5) performs much better than the optimum filter, i.e. xi_excess is negative!! Is there an interpretation for that, or it's more likely that I did an error writing the code? I suppose that if there's a problem, it should be in the optimum filter computation... Thanks in advance. Gabriele Paganelli

# LMS vs Wiener

Started by ●June 18, 2007

Reply by ●June 18, 20072007-06-18

On Jun 19, 1:24 am, "gabinet" <gabi...@correo.ugr.es> wrote:> Hello. I have a toy situation in which, > > known the desired signal d(n), > > d(n) = sin(100*pi*n+pi/3); > > and a noisy signal, with white noise v(n) with known variance and mean > > x(n) = d(n) + v(n); > > the goal is define an optimal Wiener filter of 101 coefficients. > (I call it "toy situation" because I have at the same time the value of > the noisy signal and the value of the original signal) > > After some calculation, in this problem we have that the autocorrelation > of the signal d(n) equals the correlation sequence between x(n) and d(n), > r_dx(n). > > I calculated (with Matlab) r_d(n), the 101 pts autocorrelation sequence of > d(n). > > I found the optimal coefficients w: > > w = (R_x^(-1)) * r_d(n) > > where R_x is the Toeplitz autocorrelation matrix of x, > and the minimum error xi_min, > > xi_min = r_d(0) - r_d(0)'(R_x^-1)r_d(0) > > Then I implemented a LMS filter to compare the difference between the > optimum coefficients and the coefficients estimated by LMS. > > According to "Statistical digital processing and modeling" by Hayes, the > output error of LMS, for the same filter length, should be always greater > than xi_min when the LMS filter reaches a steady state, because > > xi_LMS(inf) = xi_min + xi_excess(n) > > but in my case the LMS filter with fair values of the mu parameter (e.g. > 10^-5) performs much better than the optimum filter, i.e. xi_excess is > negative!! > Is there an interpretation for that, or it's more likely that I did an > error writing the code? I suppose that if there's a problem, it should be > in the optimum filter computation... > > Thanks in advance. > Gabriele PaganelliIn theory a Wiener filter does not apply to periodic signals. However it wil work as will LMS. The two are the same of course as you minimise mean square error in both cases. Recent work has whosn that LMS is in fact a robust estimator which staisfies H infinity robustness ie it minimises the maximum error (minimax). H infinity criteria are more suited to periodic noise than mse techniques.