Hi, Suppose I have the autocorrelation function which results from Gaussian white noise passing through a 1st order LTI system, analytically: Ryy(tau) = 1/RC sigma^2/2 exp(-|tau|/RC) How does Rzz(tau) look when z=y^3 ? Peter PS: In the absense of the LTI system, trivially Rzz=delta(0) * 15*sigma^6/2 ... from E((x^3)^2)=E(x^6)=15*sigma^6

# What happens to the autocorrelation when the underlying signal is squared/cubed?

Started by ●May 30, 2018

Posted by ●May 30, 2018

On 5/30/18 12:43 AM, Peter Mairhofer wrote:> Hi, > > Suppose I have the autocorrelation function which results from Gaussian > white noise passing through a 1st order LTI system,i think they call this an Ornstien-Ulenback process.> analytically: > > Ryy(tau) = 1/RC sigma^2/2 exp(-|tau|/RC) > > How does Rzz(tau) look when z=y^3 ?that's painful. but you should be able to derive a new p.d.f. for z. and you will have to make some assumptions about joint probability with past (or future) samples.> > > PS: In the absense of the LTI system, trivially Rzz=delta(0) * > 15*sigma^6/2 ... from E((x^3)^2)=E(x^6)=15*sigma^6but what you need to derive is some expression for E{ z(t) z(t+tau) } from E{ y(t) y(t+tau) } = Ryy(tau) i think that you can come up with some bayesian p.d.f. p(y | y(t-tau)) = some guassian with mean that approaches y(t-tau) and variance that goes to zero as tau -> 0. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."

Posted by ●May 30, 2018

On 2018-05-30 10:12, robert bristow-johnson wrote:> On 5/30/18 12:43 AM, Peter Mairhofer wrote: >> Hi, >> >> Suppose I have the autocorrelation function which results from Gaussian >> white noise passing through a 1st order LTI system, > > i think they call this an Ornstien-Ulenback process >> analytically: >> >> Ryy(tau) = 1/RC sigma^2/2 exp(-|tau|/RC) >> >> How does Rzz(tau) look when z=y^3 ? > > that's painful. but you should be able to derive a new p.d.f. for z.Are you sure? I do not think so. First, I don't think z^3 exists (e.g. [1]) Second, I don't think this is necessary because the autocorrelation function only depends on second order moments. The one PDF that's in the system is the Gaussian white input.> and you will have to make some assumptions about joint probability with > past (or future) samples.That's exactly the task of the autocorrelation function.>> PS: In the absense of the LTI system, trivially Rzz=delta(0) * >> 15*sigma^6/2 ... from E((x^3)^2)=E(x^6)=15*sigma^6 > > but what you need to derive is some expression for > > E{ z(t) z(t+tau) } > > from > > E{ y(t) y(t+tau) } = Ryy(tau)Exactly. But as I hope (and think) I do not need the PSD - second order moments of the cubed Gaussian, i.e., E((x^3)^2) should be enough. Actually, to make things worse - I am not dealing with a stationary process. Hence my autocorrelations are actually Rxx(t1,t2), Ryy(t1,t2) and Rzz(t1,t2). If I just try Rzz = (15*sigma^4*Ryy)^(3/2) the result is suspiciously close - but not too close: N = 2^22; sigma = 0.1; x = sigma*randn(N,1); [b,a] = butter(1,0.01); y = filter(b,a,x); [rxx,lags] = xcorr(x);rxx=rxx/length(rxx); [ryy,lags] = xcorr(y);ryy=ryy/length(ryy); [ryy3,lags] = xcorr(y.^3);ryy3=ryy3/length(ryy3); plot(lags, sqrt([(15*sigma^4*ryy).^(3/2) , ryy3])); xlim([-400 400]) Peter [1] https://projecteuclid.org/euclid.aop/1176991795

Posted by ●June 3, 2018

On Wednesday, May 30, 2018 at 7:43:53 PM UTC+12, Peter Mairhofer wrote:> Hi, > > Suppose I have the autocorrelation function which results from Gaussian > white noise passing through a 1st order LTI system, analytically: > > Ryy(tau) = 1/RC sigma^2/2 exp(-|tau|/RC) > > How does Rzz(tau) look when z=y^3 ? > > Peter > > > PS: In the absense of the LTI system, trivially Rzz=delta(0) * > 15*sigma^6/2 ... from E((x^3)^2)=E(x^6)=15*sigma^6There is a vast literature on nonlinear systems which goes back to the times of Wiener. The spectrum of white noise passing through nonlinear systems will be there. Certainly a square law, almost certainly a cubic. You may need a volterra series.