How about this:
For a lag of t, the probability that no new samples have been accepted is
(1-P)^|t|.
So the autocorrelation should be:
AF(t) = E[x(n)x(n+t)] = (1-P)^|t| * E[x(n)^2] + (1 -
(1-P)^|t|)*E[x(n)*x_new]
The second term covers the case that a new sample has popped up, so x(n)
and x(n+t) are uncorrelated. So, this term vanishes. The first term is
(1/3)*(1-P)^|t|, so I reckon:
AF(t) = (1/3)*(1-P)^|t|
Does that make sense?
-Ethan
On Wed, Nov 4, 2015 at 8:21 AM, robert bristow-johnson <
r...@audioimagination.com> wrote:
>
>
> ---------------------------- Original Message ----------------------------
> Subject: Re: [music-dsp] how to derive spectrum of random sample-and-hold
> noise?
> From: "Ross Bencina" <rossb-li...@audiomulch.com>
> Date: Wed, November 4, 2015 12:22 am
> To: r...@audioimagination.com
> music-dsp@music.columbia.edu
> --------------------------------------------------------------------------
>
>
>
> with mods....
>
>
> > Using ASDF instead of autocorrelation:
> >
> > let n be an arbitrary time index
> > let t be the ASDF lag time of interest
> >
> > ASDF[t] = (x[n] - x[n-t])^2
> >
> > there are two cases:
> >
> > case 1, (holding): x[n-t] == x[n]
>
> this has probability of P^|t|
>
>
>
>
> > case 2, (not holding) x[n-t] == uniform_random(-1, 1)
>
> this has probability of 1 - P^|t|
>
>
> >
> > In case 1, ASDF[t] = 0
> > In case 2, ASDF[t] = (1/3)^2 (i think)
>
>
>
> so maybe it's
>
>
>
> ASDF[t] = 0 * P^|t| + (1/3)^2 * (1 - P^|t|)
>
>
>
> now the autocorrelation function (AF) is related to the ASDF as
>
>
>
> AF[t] = mean{ x[n] * x[n-t] }
>
> AF[t] = mean{ (x[n])^2 } - (1/2)*mean{ (x[n] - x[n-t])^2 }
>
>
> AF[t] = mean{ (x[n])^2 } - (1/2)*ASDF[t]
>
>
>
> AF[t] = (1/3) - (1/2) * (1/3)^2 * (1 - P^|t|)
>
>
>
> this doesn't quite look right to me. somehow i was expecting AF[t] to go
> to zero as t goes to infinity.
>
>
>
>
> > To get the limit of ASDF[t], weight the values of the two cases by the
> > probability of each case case. (Which seems like a textbook waiting-time
> > problem, but will require me to return to my textbook).
> >
> > Then I just need to convert the ASDF to PSD somehow.
>
>
>
> ASDF[t] = 2*AF[0] - 2*AF[t]
>
>
>
> or
>
>
>
> AF[t] = AF[0] - (1/2)*ASDF[t]
>
>
>
>
>
> PSD = Fourier_Transform{ AF[t] }
>
>
> > Does that seem like a reasonable approach?
>
>
> it's the approach i am struggling with. somehow, i don't like the AF i
> get.
>
>
>
> --
>
>
>
>
> r b-j r...@audioimagination.com
>
>
>
>
> "Imagination is more important than knowledge."
>
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