# Proof regarding autocorrelation of a pulse in white noise

Started by June 30, 2003
```Let  x[n] = s[n] + p[n]
where s[n] is complex Gaussian white noise (GWN)  and p[n] is a
complex signal in the for Ae^{j2pi w + phi} where w is radian
frequency and phi is a phase factor.
I want to show that if I do the autocorrelation of x[n]  (ie
x[n]x[n-1]* , where * denotes the complex conjugate) that the argument
will more closely approach the actual frequency of p[n],
so, if I did  the autocorrelation of x[n] and then took the argument
of the result, it would be with in some % of  the true frequency.
You can assume that the mean of the GWN is zero and the variance is
0.1 or whatever is convenient.

I have taken x[n]x[n-1]* and multiplied it out...and you can make a
few assumptions to limit terms, but I am not seeing a good way to show
what is happening w/ the argument.

Thanks
Craig
```
```crrea2@umkc.edu (Craig) writes:

> Let  x[n] = s[n] + p[n]
> where s[n] is complex Gaussian white noise (GWN)  and p[n] is a
> complex signal in the for Ae^{j2pi w + phi} where w is radian
> frequency and phi is a phase factor.

I assume you mean Ae^{j2pi w n + phi} wotherwise w would not be a
frequency term.

> I want to show that if I do the autocorrelation of x[n]  (ie
> x[n]x[n-1]* , where * denotes the complex conjugate)

That's not really the autocorrelation, it's just the signal times
itself conjugate delayed by one sample.

> that the argument will more closely approach the actual frequency of
> p[n], so, if I did the autocorrelation of x[n] and then took the
> argument of the result, it would be with in some % of the true
> frequency.  You can assume that the mean of the GWN is zero and the
> variance is 0.1 or whatever is convenient.
>
>
> I have taken x[n]x[n-1]* and multiplied it out...and you can make a
> few assumptions to limit terms, but I am not seeing a good way to show
> what is happening w/ the argument.

x[n]x[n-1]* = (Ae^{j2pi w n + phi} + p[n])(Ae^-{j2pi w (n-1) + phi} + p[n-1]*)

= A^2e^{j2pi w n + phi - (j2pi w (n-1) + phi)}
+ p[n] Ae^-{j2pi w (n-1) + phi}
+ p[n-1]*Ae^{j2pi w n + phi}
+ p[n]p[n-1]*

Taking the expectation:

E{ x[n]x[n-1]* } = A^2e^{j2pi w} + 0 + 0 + 0

because E{p[k]} = 0 (p[k] is zero mean) and E{p[k]p[k-1]*} = 0 (p[k]
is white, i.e. uncorrelated).

More generally, the frequency estimator:

argument ( sum over all n ( W[n] x[n]x[n-1]* ) )

is a phase-weighted averager, of which Kay's estimator [2] is the most
well-known.  There is a matlab implementation of several of these:

http://www.itee.uq.edu.au/~kootsoop/wlp.m

The differences between all of them just involve how you choose the
weights W[n].  Kay [2] chooses the weights to be parabolic, which is the
best (minimum variance) answer if your noise power goes to zero. Lank,
Reed and Pollon [4] choose the weights to be uniform, which is the best

For more on this (and other) frequency estimators, have a look at [5].

Ciao,

Peter K.

REFERENCES
==========

[1]  V. Clarkson, P. J.  Kootsookos and B. G.  Quinn,
``Variance Analysis of Kay's Weighted Linear Predictor
Frequency Estimator,'' IEEE Transactions on Signal
Processing, Vol. 42, pp. 2370-2379, 1994.

[2] S. M.  Kay, `` A Fast and Accurate Single Frequency
Estimator,'' IEEE Transactions on Acoustics, Speech
and Signal Processing, Vol. 37(12), pp. 1987-1989,
1989.

[3]  B.C. Lovell and R.C. Williamson,
"The Statistical Performance of Some Instantaneous
Frequency Estimators," IEEE Trans. on Acoustics,
Speech and Signal Processing, Vol. 40, pp. 1708-1723,
1992.

[4] G. W. Lank, I. S.  Reed and G. E.  Pollon,
``A Semicoherent Detection Statistic and Doppler
Estimation Statistic,'' IEEE Transactions on
Aerospace and Electronic Systems, Vol. AES-9(2),
pp. 151-165, 1973.

[5] P. J. Kootsookos, ``A Review of the Frequency Estimation and
Tracking Problems,'' CRASys Technical Report, last updated
21/02/1999, URL:
http://www.itee.uq.edu.au/~kootsoop/comparison-t.pdf, last
accessed: 1/07/2003.

--
Peter J. Kootsookos

"Na, na na na na na na, na na na na"
- 'Hey Jude', Lennon/McCartney
```
```p.kootsookos@remove.ieee.org (Peter J. Kootsookos) wrote in message news:<s68isqngl9y.fsf@mango.itee.uq.edu.au>...
> crrea2@umkc.edu (Craig) writes:
>
> > Let  x[n] = s[n] + p[n]
> > where s[n] is complex Gaussian white noise (GWN)  and p[n] is a
> > complex signal in the for Ae^{j2pi w + phi} where w is radian
> > frequency and phi is a phase factor.
>
> I assume you mean Ae^{j2pi w n + phi} wotherwise w would not be a
> frequency term.
>
> > I want to show that if I do the autocorrelation of x[n]  (ie
> > x[n]x[n-1]* , where * denotes the complex conjugate)
>
> That's not really the autocorrelation, it's just the signal times
> itself conjugate delayed by one sample.

This is an interesting discussion. Now, the signal model includes
the noise p[n], but it doesn't show up in Peter's estimate for the
autocorrelation, which I think it should. I can't find any flaw in
Peter's reasoning.

What did I miss?

Rune
```
```allnor@tele.ntnu.no (Rune Allnor) writes:

> p.kootsookos@remove.ieee.org (Peter J. Kootsookos) wrote
> > crrea2@umkc.edu (Craig) writes:
> >
> > > Let  x[n] = s[n] + p[n]
> > > where s[n] is complex Gaussian white noise (GWN)  and p[n] is a
> > > complex signal in the for Ae^{j2pi w + phi} where w is radian
> > > frequency and phi is a phase factor.
> >
> > I assume you mean Ae^{j2pi w n + phi} wotherwise w would not be a
> > frequency term.
> >
> > > I want to show that if I do the autocorrelation of x[n]  (ie
> > > x[n]x[n-1]* , where * denotes the complex conjugate)
> >
> > That's not really the autocorrelation, it's just the signal times
> > itself conjugate delayed by one sample.
>
> This is an interesting discussion. Now, the signal model includes
> the noise p[n], but it doesn't show up in Peter's estimate for the
> autocorrelation, which I think it should. I can't find any flaw in
> Peter's reasoning.
>
> What did I miss?

Hi Rune,

I'm not actually finding an estimate of the autocorrelation, just the
autocorrelation at a lag of 1 sample.

To find the autocorrelation, I'd have to find:

E{x[n]x[n-m]*}

where m varies.  In my analysis, I just set m=1.

Ciao,

Peter K.

--
Peter J. Kootsookos

"Na, na na na na na na, na na na na"
- 'Hey Jude', Lennon/McCartney
```
```Hey guys, thanks a lot, that makes perfect sense, I should have seen that!
It is percisely what I was looking for.
Craig

allnor@tele.ntnu.no (Rune Allnor) wrote in message news:<f56893ae.0306302225.10fbd0bc@posting.google.com>...
> p.kootsookos@remove.ieee.org (Peter J. Kootsookos) wrote in message news:<s68isqngl9y.fsf@mango.itee.uq.edu.au>...
> > crrea2@umkc.edu (Craig) writes:
> >
> > > Let  x[n] = s[n] + p[n]
> > > where s[n] is complex Gaussian white noise (GWN)  and p[n] is a
> > > complex signal in the for Ae^{j2pi w + phi} where w is radian
> > > frequency and phi is a phase factor.
> >
> > I assume you mean Ae^{j2pi w n + phi} wotherwise w would not be a
> > frequency term.
> >
> > > I want to show that if I do the autocorrelation of x[n]  (ie
> > > x[n]x[n-1]* , where * denotes the complex conjugate)
> >
> > That's not really the autocorrelation, it's just the signal times
> > itself conjugate delayed by one sample.
>
> This is an interesting discussion. Now, the signal model includes
> the noise p[n], but it doesn't show up in Peter's estimate for the
> autocorrelation, which I think it should. I can't find any flaw in
> Peter's reasoning.
>
> What did I miss?
>
> Rune
```
```p.kootsookos@remove.ieee.org (Peter J. Kootsookos) wrote in message news:<s683chqbulr.fsf@mango.itee.uq.edu.au>...
> allnor@tele.ntnu.no (Rune Allnor) writes:
>
> > p.kootsookos@remove.ieee.org (Peter J. Kootsookos) wrote
> > > crrea2@umkc.edu (Craig) writes:
> > >
> > > > Let  x[n] = s[n] + p[n]
> > > > where s[n] is complex Gaussian white noise (GWN)  and p[n] is a
> > > > complex signal in the for Ae^{j2pi w + phi} where w is radian
> > > > frequency and phi is a phase factor.
> > >
> > > I assume you mean Ae^{j2pi w n + phi} wotherwise w would not be a
> > > frequency term.
> > >
> > > > I want to show that if I do the autocorrelation of x[n]  (ie
> > > > x[n]x[n-1]* , where * denotes the complex conjugate)
> > >
> > > That's not really the autocorrelation, it's just the signal times
> > > itself conjugate delayed by one sample.
> >
> > This is an interesting discussion. Now, the signal model includes
> > the noise p[n], but it doesn't show up in Peter's estimate for the
> > autocorrelation, which I think it should. I can't find any flaw in
> > Peter's reasoning.
> >
> > What did I miss?
>
> Hi Rune,
>
> I'm not actually finding an estimate of the autocorrelation, just the
> autocorrelation at a lag of 1 sample.

Of course. One of these days I think I'll have to start actually reading
the posts...

Rune

> To find the autocorrelation, I'd have to find:
>
> E{x[n]x[n-m]*}
>
> where m varies.  In my analysis, I just set m=1.
>
> Ciao,
>
> Peter K.
```
```Craig,

Ae^{j2pi w n + phi} should be
Ae^{j(2pi w n + phi)}

throughout this discussion.

Dirk

Dirk A. Bell
DSP Consultant

crrea2@umkc.edu (Craig) wrote in message news:<82396605.0307010443.961f275@posting.google.com>...
> Hey guys, thanks a lot, that makes perfect sense, I should have seen that!
> It is percisely what I was looking for.
> Craig
>
>
> allnor@tele.ntnu.no (Rune Allnor) wrote in message news:<f56893ae.0306302225.10fbd0bc@posting.google.com>...
> > p.kootsookos@remove.ieee.org (Peter J. Kootsookos) wrote in message news:<s68isqngl9y.fsf@mango.itee.uq.edu.au>...
> > > crrea2@umkc.edu (Craig) writes:
> > >
> > > > Let  x[n] = s[n] + p[n]
> > > > where s[n] is complex Gaussian white noise (GWN)  and p[n] is a
> > > > complex signal in the for Ae^{j2pi w + phi} where w is radian
> > > > frequency and phi is a phase factor.
> > >
> > > I assume you mean Ae^{j2pi w n + phi} wotherwise w would not be a
> > > frequency term.
> > >
> > > > I want to show that if I do the autocorrelation of x[n]  (ie
> > > > x[n]x[n-1]* , where * denotes the complex conjugate)
> > >
> > > That's not really the autocorrelation, it's just the signal times
> > > itself conjugate delayed by one sample.
> >
> > This is an interesting discussion. Now, the signal model includes
> > the noise p[n], but it doesn't show up in Peter's estimate for the
> > autocorrelation, which I think it should. I can't find any flaw in
> > Peter's reasoning.
> >
> > What did I miss?
> >
> > Rune
```
```"Dirk Bell" <dirkman@erols.com> wrote

> Craig,
>
> Ae^{j2pi w n + phi} should be
> Ae^{j(2pi w n + phi)}
>
> throughout this discussion.
>
> Dirk
>
> Dirk A. Bell
> DSP Consultant
>

D'oh!  Thanks, Dirk.

Ciao,

Peter K.

--
Peter J. Kootsookos

"Na, na na na na na na, na na na na"
- 'Hey Jude', Lennon/McCartney

```
```dirkman@erols.com (Dirk Bell) wrote in message news:<6721a858.0307011007.31849acd@posting.google.com>...
Think Dirk,
I noticed that as well, no biggie, once I saw how he began, I was able
to put it together readly.  I just couldn't find a good starting
point.  Maybe too much coffee!

> Craig,
>
> Ae^{j2pi w n + phi} should be
> Ae^{j(2pi w n + phi)}
>
> throughout this discussion.
>
> Dirk
>
> Dirk A. Bell
> DSP Consultant
>
>
> crrea2@umkc.edu (Craig) wrote in message news:<82396605.0307010443.961f275@posting.google.com>...
> > Hey guys, thanks a lot, that makes perfect sense, I should have seen that!
> > It is percisely what I was looking for.
> > Craig
> >
> >
> > allnor@tele.ntnu.no (Rune Allnor) wrote in message news:<f56893ae.0306302225.10fbd0bc@posting.google.com>...
> > > p.kootsookos@remove.ieee.org (Peter J. Kootsookos) wrote in message news:<s68isqngl9y.fsf@mango.itee.uq.edu.au>...
> > > > crrea2@umkc.edu (Craig) writes:
> > > >
> > > > > Let  x[n] = s[n] + p[n]
> > > > > where s[n] is complex Gaussian white noise (GWN)  and p[n] is a
> > > > > complex signal in the for Ae^{j2pi w + phi} where w is radian
> > > > > frequency and phi is a phase factor.
> > > >
> > > > I assume you mean Ae^{j2pi w n + phi} wotherwise w would not be a
> > > > frequency term.
> > > >
> > > > > I want to show that if I do the autocorrelation of x[n]  (ie
> > > > > x[n]x[n-1]* , where * denotes the complex conjugate)
> > > >
> > > > That's not really the autocorrelation, it's just the signal times
> > > > itself conjugate delayed by one sample.
> > >
> > > This is an interesting discussion. Now, the signal model includes
> > > the noise p[n], but it doesn't show up in Peter's estimate for the
> > > autocorrelation, which I think it should. I can't find any flaw in
> > > Peter's reasoning.
> > >
> > > What did I miss?
> > >
> > > Rune
```
```"Craig" <crrea2@umkc.edu> wrote

> Maybe too much coffee!

There is no such thing as too much coffee.

Ciao,

Peter K.

--
Peter J. Kootsookos

"Na, na na na na na na, na na na na"
- 'Hey Jude', Lennon/McCartney

```