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

# Proof regarding autocorrelation of a pulse in white noise

Started by ●June 30, 2003

Reply by ●June 30, 20032003-06-30

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 (minimum variance) answer if your noise power goes to infinity. 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

Reply by ●July 1, 20032003-07-01

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

Reply by ●July 1, 20032003-07-01

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

Reply by ●July 1, 20032003-07-01

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

Reply by ●July 1, 20032003-07-01

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.

Reply by ●July 1, 20032003-07-01

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

Reply by ●July 1, 20032003-07-01

"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

Reply by ●July 1, 20032003-07-01

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

Reply by ●July 2, 20032003-07-02