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^6

There 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.

Reply by Peter Mairhofer●May 30, 20182018-05-30

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

Reply by robert bristow-johnson●May 30, 20182018-05-30

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^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)
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."

Reply by Peter Mairhofer●May 30, 20182018-05-30

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